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NCFE Review Part A: Multiple TRUE/FALSE

1. Which of the following statements about vectors and scalars are TRUE? List all that apply. 1. A vector is a large quantity and a scalar is a small quantity. 2. A scalar quantity has a magnitude and a vector quantity does not. 3. A vector quantity is described with a direction and a scalar is not. 4. Scalar quantities are path dependent quantities and vector quantities are not. 5. A scalar quantity depends only upon the initial and final values of the quantity; this is not the case for vector quantities. 6. The quantity 20 m/s, north is a speed and as such is a scalar quantity. 7. The quantity 9.8 m/s/s is an acceleration value and as such is a vector quantity.

Answer: CD a. FALSE - This would never be the case. Vectors simply are direction-conscious, path-independent quantities which depend solely upon the initial and final state of an object. Vectors are always expressed fully by use of a magnitude and a direction. b. FALSE - Both scalar and vector quantities have a magnitude or value expressed with a given unit; additionally, a vector quantity requires a direction in order to fully express the quantity. c. TRUE - Vectors are fully described by magnitude AND direction; scalars are not described with a direction. d. TRUE - Scalars such as distance would depend upon the path taken from initial to final location. If you run around the track one complete time, your distance will be different than if you take a step forward and a step backwards. The path MATTERS; distance (like all scalars) depends upon it. On the other hand, the displacement (a vector quantity) is the same for both paths. e. FALSE - Vectors are the types of quantities which depend only upon initial and final state of the object. For instance, the vector quantity displacement depends only upon the starting and final location. f. FALSE - This is certainly not a speed quantity; though the unit is appropriate for speed, the statement of the direction is inconsistent with speed as a scalar quantity. g. FALSE (a rather picky FALSE) - If a direction was included, then this would be an acceleration value. The unit is characteristic of acceleration but the lack of direction is inconsistent with acceleration being a vector quantity.

2. Which of the following statements about distance and/or displacement are TRUE? List all that apply. 1. Distance is a vector quantity and displacement is a scalar quantity. 2. A person makes a round-trip journey, finishing where she started. The displacement for the trip is 0 and the distance is some nonzero value. 3. A person starts at position A and finishes at position B. The distance for the trip is the length of the segment measured from A to B. 4. If a person walks in a straight line and never changes direction, then the distance and the displacement will have exactly the same magnitude. 5. The phrase "20 mi, northwest" likely describes the distance for a motion. 6. The phrase "20 m, west" likely describes the displacement for a motion. 7. The diagram below depicts the path of a person walking to and fro from position A to B to C to D. The distance for this motion is 100 yds. 8. For the same diagram below, the displacement is 50 yds.

Answer: BDF a. FALSE - Distance is the scalar and displacement is the vector. Know this one! b. TRUE - Displacement is the change in position of an object. An object which finishes where it started is not displaced; it is at the same place as it started and as such has a zero displacement. On the other hand, the distance is the amount of ground which is covered. And if it was truly a journey, then there is definitely a distance. c. FALSE - This would only be the case if the person walk along a beeline path from A to B. But if the person makes a turn and veers left, then right and then ..., then the person has a distance which is greater than the length of the path from A to B. Distance refers to the amount of ground which is covered. d. TRUE - If a person never changes direction and maintains the same heading away from the initial position, then every step contributes to a change in position in the same original direction. A 1 m step will increase the displacement (read as out of place-ness) by 1 meter and contribute one more meter to the total distance which is walked. e. FALSE - Distance is a scalar and is ignorant of direction. The "northwest" on this quantity would lead one to believe that this is a displacement (a vector quantity) rather than a distance. f. TRUE - The unit is an appropriate displacement unit (length units) and the direction is stated. Since there is both magnitude and direction expressed, one would believe that this is likely a displacement. g. FALSE - The distance from A to B is 35 yds; from B to C is 20 yds; and from C to D is 35 yds. The total distance moved is 90 yds. h. FALSE (a rather picky FALSE) - Technically, this is not a displacement since displacement is a vector and fully described by both magnitude and direction. The real expression of displacement is 50 yds, left (or west or -)

3. Which of the following statements about velocity and/or speed are TRUE? List all that apply. 1. Velocity is a vector quantity and speed is a scalar quantity. 2. Both speed and velocity refer to how fast an object is moving. 3. Person X moves from location A to location B in 5 seconds. Person Y moves between the same two locations in 10 seconds. Person Y is moving with twice the speed as person X. 4. The velocity of an object refers to the rate at which the object's position changes. 5. For any given motion, it is possible that an object could move very fast yet have an abnormally small velocity. 6. The phrase "30 mi/hr, west" likely refers to a scalar quantity. 7. The average velocity of an object on a round-trip journey would be 0. 8. The direction of the velocity vector is dependent upon two factors: the direction the object is moving and whether the object is speeding up or slowing down. 9. The diagram below depicts the path of a person walking to and fro from position A to B to C to D. The entire motion takes 8 minutes. The average speed for this motion is approximately 11.3 yds/min. 10. For the same diagram below, the average velocity for this motion is 0 yds/min. a. TRUE - Yes! Speed is a scalar and velocity is the vector. Know this one! b. FALSE - Speed refers to how fast an object is moving; but velocity refers to the rate at which one's motion puts an object away from its original position. A person can move very fast (and thus have a large speed); but if every other step leads in opposite directions, then that person would not have a large velocity. c. FALSE - Person Y has one-half the speed of Person X. If person Y requires twice the time to do the same distance, then person Y is moving half as fast. d. TRUE - Yes! That is exactly the definition of velocity - the rate at which velocity changes. e. TRUE - An Indy Race car driver is a good example of this. Such a driver is obviously moving very fast but by the end of the race the average velocity is essentially 0 m/s. f. FALSE - The presence of the direction "west" in this expression rules it out as a speed expression. Speed is a scalar quantity and direction is not a part of it. g. TRUE - For a round trip journey, there is no ultimate change in position. As such, the average velocity is 0 m/t seconds. Regardless of the time, the average velocity will be 0 m/s. h. FALSE - The direction of the velocity vector depends only upon the direction that the object is moving. A westward moving object has a westward velocity. i. TRUE - As discussed in #2g, the distance traveled is 90 meters. When divided by time (8 minutes), the average speed is 11.25 yds/min. j. FALSE - The average velocity would be 0 yds/min only if the person returns to the initial starting position. In this case, the average velocity is 50 yds/8 min, west (6.25 yds/min, west).

4. Which of the following statements about acceleration are TRUE? List all that apply. 1. Acceleration is a vector quantity. 2. Accelerating objects MUST be changing their speed. 3. Accelerating objects MUST be changing their velocity. 4. Acceleration units include the following; m/s2, mi/hr/sec, cm/s2, km/hr/m. 5. The direction of the acceleration vector is dependent upon two factors: the direction the object is moving and whether the object is speeding up or slowing down. 6. An object which is slowing down has an acceleration. 7. An object which is moving at constant speed in a circle has an acceleration. 8. Acceleration is the rate at which the velocity changes. 9. An object that is accelerating is moving fast. 10. An object that is accelerating will eventually (if given enough time) be moving fast. 11. An object that is moving rightward has a rightward acceleration. 12. An object that is moving rightward and speeding up has a rightward acceleration. 13. An object that is moving upwards and slowing down has an upwards acceleration.

Answer: ACEFGHL (and maybe J) a. TRUE - Yes it is. Acceleration is direction-conscious. b. FALSE - Accelerating objects could be changing their speed; but it is also possible that an accelerating object is only changing its direction while maintaining a constant speed. The race car drivers at Indy might fit into this category (at least for certain periods of the race). c. TRUE - Accelerating object MUST be changing their velocity -either the magnitude or the direction of the velocity. d. FALSE - The first three sets of units are acceleration units - they include a velocity unit divided by a time unit. The last set of units is a velocity unit divided by a length unit. This is definitely NOT an acceleration. e. TRUE - This is the case and something important to remember. Consider its application in the last three parts of this question. f. TRUE - Accelerating objects are either slowing down, speeding up or changing directions. g. TRUE - To move in a circle is to change one's direction. As such, there is a change in the velocity (not magnitude, but the direction part); this constitutes an acceleration. h. TRUE - This is the very definition of acceleration. Know this one - its the beginning point of all our thoughts about acceleration. i. FALSE - Accelerating objects are not necessarily moving fast; they are merely changing how fast they are moving (or the direction they are moving). j. FALSE - If the accelerating object is slowing down, then it will eventually stop and not reach a fast speed. And if that doesn't convince you, then consider an object that is accelerating by moving in a circle at constant speed forever; it will accelerate the entire time but never being going any faster than at the beginning. k. FALSE - If an object is moving rightward and slowing down, then it would have a leftward acceleration. l. TRUE - If an object is speeding up, then the direction of the acceleration vector is in the direction which the object is moving. m. FALSE - If an object is slowing down, then the acceleration vector is directed opposite the direction of the motion; in this case the acceleration is directed downwards.

5. Which of the following statements about position-time graphs are TRUE? List all that apply. 1. Position-time graphs cannot be used to represent the motion of objects with accelerated motion. 2. The slope on a position-time graph is representative of the acceleration of the object. 3. A straight, diagonal line on a position-time graph is representative of an object with a constant velocity. 4. If an object is at rest, then the position-time graph will be a horizontal line located on the time- axis. 5. Accelerated objects are represented on position-time graphs by curved lines. 6. An object with a positive velocity will be represented on a position-time graph by a line with a positive slope. 7. An object with a negative velocity will be represented on a position-time graph by a line with a negative slope. 8. An object with a positive acceleration will be represented on a position-time graph by a line which curves upwards. 9. An object with a negative acceleration will be represented on a position-time graph by a line which curves downwards.

Answer: CEFG a. FALSE - Position-time graphs represent accelerated motion by curved lines. b. FALSE - The slope of a position-time graph is the velocity of the object. Some things in this unit are critical things to remember and internalize; this is one of them. c. TRUE - A straight diagonal line is a line of constant slope. And if the slope is constant, then so is the velocity. d. FALSE - Not necessarily true. If the object is at rest, then the line on a p-t graph will indeed be horizontal. However, it will not necessarily be located upon the time axis. e. TRUE - Accelerating objects (if the acceleration is attributable to a speed change) are represented by lines with changing slope - i.e., curved lines. f. TRUE - Since slope on a p-t graph represents the velocity, a positive slope will represent a positive velocity. g. TRUE - Since slope on a p-t graph represents the velocity, a negative slope will represent a negative velocity. h. FALSE - (This is confusing wording here since we might not all agree on what "curving up" means.) A line that slopes upward and has a curve (perhaps you call that "curving up" as I do) has a positive velocity (due to its positive slope). If the curve is "concave down" (you might say leveling off to a horizontal as time progresses) then the object is slowing down and the acceleration is negative. i. FALSE - (Once more, there is confusing wording here since we might not all agree on what "curving downwards" means.) A line that slopes downwards and has a curve (perhaps you call that "curving downwards " as I do) has a negative velocity (due to its negative slope). If the curve is "concave up" (you might say leveling off to a horizontal as time progresses) then the object is slowing down and the acceleration is positive.

6. Which of the following statements about velocity-time graphs are TRUE? List all that apply. 1. The slope on a velocity-time graph is representative of the acceleration of the object. 2. The area on a velocity -time graph is representative of the change in position of the object. 3. An accelerated object's motion will be represented by a curved line on a velocity-time graph. 4. Objects with positive acceleration will be represented by upwardly-curved lines on a velocity-time graph. 5. If an object is at rest, then the velocity-time graph will be a line with zero slope. 6. A line with zero slope on a velocity-time graph will be representative of an object which is at rest. 7. A line with a negative slope on a velocity-time graph is representative of an object with negative velocity. 8. If an object changes its direction, then the line on the velocity-time graph will have a changing slope. 9. An object which is slowing down is represented by a line on a velocity-time graph which is moving in the downward direction.

Answer: ABE (and almost D) a. TRUE - Now this is important! It is the beginning point of much of our discussion of velocity-time graphs. The slope equals the acceleration. b. TRUE - This is equally important. The area is the displacement. c. FALSE - An object which has an acceleration will be represented by an line that has a slope. It may or may not curve, but it must have a slope other than zero. d. FALSE - An object with positive acceleration will have an positive or upward slope on a v-t graph. It does not have to be a curved line. A curved line indicates an object that is accelerating at a changing rate of acceleration. e. TRUE - An object that is at rest has a 0 velocity and maintains that zero velocity. The permanence of its velocity (not the fact that it is zero) gives the object a zero acceleration. and as such, the line on a v-t graph would have a slope of 0 (i.e., be horizontal). f. FALSE - A line with zero slope is representative of an object with an acceleration of 0. It could be at rest or it could be moving at a constant velocity. g. FALSE - A negative slope indicates a negative acceleration. The object could be moving in the positive direction and slowing down (a negative acceleration). h. FALSE - An object which changes its direction will be represented by a line on a v-t graph that crosses over the time-axis from the + velocity region into the - velocity region. i. FALSE - An object which is slowing down has a velocity which is approaching 0 m/s. And as such, on a v-t graph, the line must be approaching the v=0 m/s axis.

7. Which of the following statements about free fall and the acceleration of gravity are TRUE? List all that apply. 1. An object that is free-falling is acted upon by the force of gravity alone. 2. A falling skydiver which has reached terminal velocity is considered to be in a state of free fall. 3. A ball is thrown upwards and is rising towards its peak. As it rises upwards, it is NOT considered to be in a state of free fall. 4. An object in free fall experiences an acceleration which is independent of the mass of the object. 5. A ball is thrown upwards, rises to its peak and eventually falls back to the original height. As the ball rises, its acceleration is upwards; as it falls, its acceleration is downwards. 6. A ball is thrown upwards, rises to its peak and eventually falls back to the original height. The speed at which it is launched equals the speed at which it lands. (Assume negligible air resistance.) 7. A very massive object will free fall at the same rate of acceleration as a less massive object. 8. The value of g on Earth is approximately 9.8 m/s2. 9. The symbol g stands for the force of gravity.

Answer: ADFGH a. TRUE - Yes! This is the definition of free fall. b. FALSE - Skydivers which are falling at terminal velocity are acted upon by large amounts of air resistance. They are experiencing more forces than the force of gravity. As such, they are NOT free-falling. c. FALSE - Any object - whether rising, falling or moving horizontally and vertically simultaneously - can be in a state of free fall if the only force acting upon it is the force of gravity. Such objects are known as projectiles and often begin their motion while rising upwards. d. TRUE - The unique feature of free-falling objects is that the mass of the object does not effect the trajectory characteristics. The acceleration, velocity, displacement, etc. is independent of the mass of the object. e. FALSE - The acceleration of all free-falling objects is directed downwards. A rising object slows down due to the downward gravity force. An upward-moving object which is slowing down is said to have a downwards acceleration. f. TRUE - If the object is truly in free-fall, then the speed of the object will be the same at all heights - whether its on the upward portion of its trajectory or the downwards portion of its trajectory. For more information, see the Projectiles page at The Physics Classroom. g. TRUE - The acceleration of free-falling objects (referred to as the acceleration of gravity) is independent of mass. On Earth, the value is 9.8 m/s/s (the direction is down). All objects - very massive and less massive - experience this acceleration value. h. TRUE - Yes! Know this one! i. FALSE - Nope. A careful physics teacher will never call g the force of gravity. g is known as the acceleration of gravity. It might be best to call it the acceleration caused by gravity. When it comes to the force of gravity, we have yet another symbol for that - Fgrav. But that's a topic to be discussed in a later unit.

Part B: Multiple Choice

8. If an object has an acceleration of 0 m/s2, then one can be sure that the object is not ____. a. moving b. changing position c. changing velocity

Answer: C The object could be moving and could be at rest; however, whether moving or not, it must not have a changing velocity.

9. If car A passes car B, then car A must be ____. 1. accelerating. 2. accelerating at a greater rate than car B. 3. moving faster than car B and accelerating more than car B. 4. moving faster than car B, but not necessarily accelerating. Answer: D All that is necessary is that car A has a greater speed (is moving faster). If so, it will eventually catch up and pass car B. Acceleration is not necessary to overcome car B; a car going 60 mi/hr at a constant speed will eventually pass a car going 50 mi/hr at a constant speed. Surely you have witnessed that while driving down a local highway.

10. Which one of the following is NOT consistent with a car which is accelerating? 1. A car is moving with an increasing speed. 2. A car is moving with a decreasing speed. 3. A car is moving with a high speed. 4. A car is changing direction.

Answer: C An accelerating object must be changing its velocity by either slowing down, speeding up or changing direction. Moving fast merely means that the velocity is high; it says nothing about the acceleration.

11. A fullback is running down the football field in a straight line. He starts at the 0-yard line at 0 seconds. At 1 second, he is on the 10-yard line; at 2 seconds, he is on the 20-yard line; at 3 seconds, he is on the 30- yard line; and at 4 seconds, he is on the 40-yard line. This is evidence that 1. he is accelerating 2. he is covering a greater distance in each consecutive second. 3. he is moving with a constant speed (on average).

Answer: C The fullback is moving 10 yards every second. He has a constant speed and thus covers the same distance (10 yd) in each consecutive second. He is not accelerating.

12. A fullback is running down the football field in a straight line. He starts at the 0-yard line at 0 seconds. At 1 second, he is on the 10-yard line; at 2 seconds, he is on the 20-yard line; at 3 seconds, he is on the 30- yard line; and at 4 seconds, he is on the 40-yard line. What is the player's acceleration?

Answer: 0 m/s/s The fullback is moving 10 yards every second. He has a constant speed. He also is running in a straight line, so he is not changing direction. Thus, his acceleration is 0 m/s/s. Only objects with changing velocity have a nonzero acceleration.

13. Olympic gold medalist Michael Johnson runs one time around the track - 400 meters - in 38 seconds. What is his displacement? ______What is his average velocity? ______

Answer: d = 0 m and v = 0 m/s Michael finishes where he started, so he is not "out of place." His displacement is 0 meters. Since average velocity is displacement over time, his average velocity is also 0 m/s.

14. If an object is moving eastward and slowing down, then the direction of its velocity vector is ____. a. eastward b. westward c. neither d. not enough info to tell

Answer: A The direction of the velocity vector is always in the same direction as the direction which the object moves.

15. If an object is moving eastward and slowing down, then the direction of its acceleration vector is ____. a. eastward b. westward c. neither d. not enough info to tell

Answer: B If an object is slowing down, then the direction of the acceleration vector is in the opposite direction as the direction which the object moves. (If the object were speeding up, the acceleration would be eastward.)

17. Which one of the following quantities is NOT a speed? a. 10 mi/hr b. 10 mi/hr/sec c. 35 m/s d. 20 m/s

Answer: B You can often tell a quantity by its units. 10 mi/hr/sec is an acceleration since there are two time units involved. In fact, the units are velocity change units (mi/hr) per time units (seconds). The quantity speed has units of distance/time.

18. Which one of the following statements is NOT true of a free-falling object? An object in a state of free fall ____. 1. falls with a constant speed of -10 m/s. 2. falls with a acceleration of -10 m/s/s. 3. falls under the sole influence of gravity. 4. falls with downward acceleration which has a constant magnitude.

Answer: A A free-falling object is an object upon which the only force is gravity. As it falls, it accelerates at a rate of approx. 10 m/s/s. This acceleration value is constant during the entire trajectory of the motion. Since this is the case, the speed can not be constant.

19. The average speed of an object which moves 10 kilometers (km) in 30 minutes is ____. a. 10 km/hr b. 20 km/hr c. 30 km/hr d. more than 30 km/hr Answer: B The average speed is distance/time. In this case the distance is 10 km and the time is 0.5 hr (30 minutes). Thus average speed = (10 km)/(0.5 hr) = 20 km/hr

20. What is the acceleration of a car that maintains a constant velocity of 55 mi/hr for 10.0 seconds? a. 0 b. 5.5 mi /hr/s c. 5.5 mi /s/s d. 550 mi/hr/s

Answer: A If the velocity is constant, then there is no acceleration. That is, the value of the acceleration is 0.

21. As an object freely falls, its ____. a. speed increases b. acceleration increases c. both of these d. none of these

Answer: A As an object falls, it accelerates; this means that the speed will be changing. While falling, the speed increases by 10 m/s every second. The acceleration is a constant value of 10 m/s/s; thus, choice b should not be chosen.

22. A speedometer is placed upon a free-falling object in order to measure its instantaneous speed during the course of its fall. Its speed reading (neglecting air resistance) would increase each second by ____. a. about 5 m/s b. about 10 m/s c. about 15 m/s d. a variable amount e. depends on its initial speed.

Answer: B The acceleration of gravity is approximately 10 m/s/s. Acceleration represents the rate at which the velocity changes - in this case, the velocity changes by 10 m/s every second. So the speed will increase by the amount of 10 m/s every second.

23. Ten seconds after being dropped from rest, a free-falling object will be moving with a speed of ____. a. about 10 m/s. b. about 50 m/s. c. about 100 m/s. d. more than 100 m/s.

Answer: C Since the speed of a free-falling object increases by 10 m/s every second, the speed after ten of these seconds will be 100 m/s. You could use the kinematic equation

vf = vi + a*t where vi=0 m/s and a = -10 m/s/s and t=10 s

24. A baseball pitcher delivers a fast ball. During the throw, the speed of the ball increases from 0 to 30.0 m/s over a time of 0.100 seconds. The average acceleration of the baseball is ____ m/s2. a. 3.00 b. 30.0 c. 300. d. 3000 e. none of these

Answer: C Acceleration is velocity change over time. In this problem, the velocity change is +30.0 m/s and the time is 0.100 s. Thus, a = (+30.0 m/s)/(0.100 s) = 300. m/s/s.

25. On takeoff, a rocket accelerates from rest at a rate of 50.0 m/s2 for exactly 1 minute. The rocket's speed after this minute of steady acceleration will be ____ m/s. a. 50.0 b. 500. c. 3.00 x 103 d. 3.60 x 103 e. none of these

Answer: C Use the equation

vf = vi + a*t

3 vf = 0 + (50.0 m/s/s)*(60.0 s) = 3.00 x 10 m/s (Note that the unit on time must be the same as the time units for which the acceleration is given.)

26. When a rock is dropped, it will accelerate downward at a rate of 9.8 m/s2. If the same rock is thrown downward (instead of being dropped from rest), it acceleration will be ____. (Ignore air resistance effects.) a. less than 9.8 m/s2 b. 9.8 m/s2 c. more than 9.8 m/s2

Answer: B Whether rising or falling, if the sole force acting upon the object is gravity, then the acceleration is 9.8 m/s/s (often approximated as 10 m/s/s).

28. Renatta Oyle is again found driving her '86 Yugo down Lake Avenue, leaving the following trail of oil drops on the pavement.

If her car is moving from right to left, then ... 1. her velocity has a rightward direction and her acceleration has a rightward direction. 2. her velocity has a rightward direction and her acceleration has a leftward direction. 3. her velocity has a leftward direction and her acceleration has a rightward direction. 4. her velocity has a leftward direction and her acceleration has a leftward direction.

Answer: D The car is heading leftward and the velocity is always in the same direction as the direction which the object moves. Since the car is speeding up, the acceleration is leftward. Whenever an object speeds up, its acceleration is in the same direction which the object moves. Whenever an object slows down, its acceleration is in the opposite direction which the object moves.

29. On the diagrams below, construct a dot diagram representing the motion of an object with a ... . 1. constant rightward velocity 2. rightward velocity and a rightward acceleration 3. rightward velocity and a leftward acceleration 4. rightward velocity, first slow and constant, and then accelerating to a high speed 5. rightward velocity, first decelerating from a high speed to a rest position, then maintaining the rest position, and finally accelerating at a lower rate than the initial deceleration.

Answer: See diagram above; explanations are given below. a. A constant velocity is depicted by dots which are spaced the same distance apart. b. An object which is accelerating in the same direction as the velocity (both are rightward in this case) is speeding up. And so as you trace your eye from left to right across the diagram (rightward), the dots should be spaced further and further apart to indicate that the object is speeding up. c. An object that moves rightward (i.e., rightward velocity) and accelerating leftwards must be slowing down. And so as you trace your eye from left to right across the diagram (rightward), the dots should be spaced closer and closer together to indicate that the object is slowing down. d. The dots begin by being equally spaced (constant speed) and close together (slow); then the spacing between dots become gradually further and further apart to indicated the speeding up nature of the motion. e. The object is moving rightwards so you will need to trace your eye from left to right across the diagram (rightward). As you do you will note that the dots become closer and closer to indicate that the object is slowing down. Then the dots are piled onto the same location to indicate that the object is at rest. Finally, the dots are spread further and further apart; the rate at which the distance between dots is very gradual - consistent with the statement that the rate of acceleration is less than the original rate of deceleration.

30. On a dot diagram, how does the motion of an object moving to the right and slowing down differ from an object moving to the left and speeding up? Explain. Answer: There is no difference! The two diagrams would be indistinguishable from each other as indicated by the following two diagrams.

There are two ways to accelerate leftward; and regardless of how it happens, the oil drop trace would be identical. A leftward acceleration is a leftward acceleration. Part D: Kinematic Graphing

31. On the position-time graph below, sketch a plot representing the motion of an object which is ... . Label each line with the corresponding letter (e.g., "a", "b", "c", etc.) 1. at rest. 2. moving in the positive direction with constant speed 3. moving in the negative direction and speeding up 4. moving in the positive direction and slowing down 5. moving in the positive direction at a constant speed (slow) and then later fast at constant speed 6. moving with a negative velocity and a negative acceleration 7. moving with a negative velocity and a positive acceleration

Answer: See diagram above; explanations are given below. The governing principle in this problem is that the slope of a position-time graph is equal to the velocity of the object. a. An object at rest (v = 0 m/s) is represented by a line with 0 slope (horizontal line). b. An object with a constant, positive velocity is represented by a straight line (constant slope) which slopes upward (positive slope). c. An object moving in the - direction and speeding up is represented by a line which slopes downward (- slope) and increases its steepness (increasing slope). d. An object moving in the + direction and slowing down is represented by a line which slopes upward (+ slope) and increases its steepness (increasing slope). e. An object moving in the positive direction at constant speed would be represented by a straight diagonal line (constant speed) which slopes upward (+ velocity). So in this case, there will be two straight diagonal lines; the second line will slope more than the first line. f. An object moving in the negative direction with negative acceleration is speeding up (since the a vector is in the same direction as the motion). So on a p-t graph, this object will be represented by a line which slopes downwards (- velocity) and increases its slope over time (speeding up). g. An object moving in the negative direction with positive acceleration is slowing down (since the a vector is in the opposite direction as the motion). So on a p-t graph, this object will be represented by a line which slopes downwards (- velocity) and levels off or becomes more horizontal over time (slowing down).

32. On the velocity-time graph below, sketch a plot representing the motion of an object which is ... . Label each line with the corresponding letter (e.g., "a", "b", "c", etc.) 1. at rest 2. moving in the positive direction at constant speed 3. moving in the negative direction from slow to fast 4. moving in the negative direction from fast to slow 5. moving with a positive velocity and a positive acceleration 6. moving with a positive velocity and a negative acceleration 7. moving with a positive velocity at constant speed and then decelerating to a rest position 8. moving in the positive direction while slowing down, changing directions and moving in the negative directions while speeding up

Answer: See diagram above; explanations are given below. The governing principle in this problem is that the slope of a velocity-time graph is equal to the acceleration of the object. Furthermore, a negative velocity would be a line plotted in the negative region of the graph; a positive velocity would be a line plotted in the positive region of the graph. a. An object at rest (v = 0 m/s) is represented by a line located on the time axis (where v = 0 m/s). b. An object moving in the positive direction at constant speed will be represented on a v-t graph by a horizontal line (slope = a = 0 m/s/s) positioned in the + velocity region. c. An object moving in the negative direction and speeding up will be represented on a v-t graph by a sloped line located in the - velocity region. Since such an object has a - acceleration, the line will slope downwards (- acceleration). d. An object moving in the negative direction and slowing down will be represented on a v-t graph by a sloped line located in the - velocity region. Such an object has a positive acceleration (since it is slowing down, the a vector will be in the opposite direction of the motion). The + acceleration would be consistent with a line that slopes upwards. e. An object moving with a + velocity and a + acceleration would be represented on a v-t graph by a sloped line located in the + velocity region. The + acceleration would be consistent with a line that slopes upwards. f. An object moving with a + velocity and a - acceleration would be represented on a v-t graph by a sloped line located in the + velocity region. The - acceleration would be consistent with a line that slopes downwards. g. An object moving with a constant + velocity would be represented on a v-t graph by a horizontal line (slope = a = 0 m/s/s) located in the + velocity region. The slowing down to rest portion of the graph would be represented by a line which slopes downwards (- acceleration) towards the time axis. 33. Consider the position-time plots below. Sketch the shape of the corresponding velocity-time graphs.

Answer: See diagram above; explanations are given below. In the top graph, the object moves in the + direction with an acceleration from fast to slow until it finally stops. Then the object remains at rest. This means there is a positive velocity and a negative acceleration; the final velocity of the object is 0 m/s. So on a v-t graph, the line needs to be in the +velocity region and the slope needs to be - (for a negative acceleration). The line should end on the v=0 m/s axis (corresponding to its final rest position). In the bottom graph, the object is moving in the - direction with a constant speed. It then gradually slows down until it stops; it then remains at rest. So on a v-t graph, there should be a horizontal line (a=0 m/s/s) in the -velocity region of the graph. Then the line should slope upwards (for the + acceleration that is characteristic of objects moving in the - direction and slowing down). The line should end on the v=0 m/s axis (corresponding to its final rest position). 34. Consider the velocity-time plots below. Sketch the shape of the corresponding position-time graphs.

Answer: See diagram above; explanations are given below. In the top graph, the object moves in the + direction with an acceleration from slow to fast. Then the object maintains a constant speed by walking in the same direction (+ direction). On a p-t graph, this would correspond to a line with positive slope (for moving in the + direction) which eventually straightens out into a diagonal line with constant slope (constant speed) in the positive direction (positive velocity). In the bottom graph, the object is moving in the + direction and slowing down from a high speed to a slow speed until it finally changes direction; the object then moves in the - direction and speeds up. On a p-t graph, this would correspond to a line with positive slope (for moving in the + direction) which gradually levels off to a horizontal (for slowing down); then the line begins to slope downwards (for moving in the - direction) and gradually becomes steeper and steeper (for speeding up).

The velocity-time graph below depicts the motion of an automobile as it moves through Glenview during rush hour traffic. Use the graph to answer questions #35 - #39. 35. Determine the displacement of the automobile during the following intervals of time. PSYW Answer: See diagram above and calculations below. Use area calculations to determine the displacement of the object. t = 0 s - 5 s t = 5 s - 15 s t = 15 s - 20 s

Area of the pink triangle: Area of the green rectangle: Area of the blue triangle A = 0.5*b*h A = b*h plus purple rectangle:

A = 0.5*(5 s)*(10 m/s) A = (10 s)*(10 m/s) A = 0.5*b1*h1 + b2*h2 A = 25 m A = 100 m A = 0.5*(5 s)*(5 m/s) + (5 s)*(5 m/s) A = 12.5 m + 25 m A = 37.5 m

36. Determine the velocity of the automobile at the following instant(s) in time. Answer: See calculations below. It is a velocity graph; so merely read the velocity values off the graph. t = 3 s t = 8 s t = 17 s

v = 6.0 m/s v = 10.0 m/s v = 8.0 m/s

37. Determine the acceleration of the automobile during the following intervals of time. Answer: See calculations below. Do slope calculations to determine the acceleration. Slope is rise/run. t = 0 s - 5 s t = 5 s - 15 s t = 15 s - 20 s

a = slope = rise/run a = slope = rise/run a = slope = rise/run a = (10 m/s)/(5 s) a = (0 m/s)/(10 s) a = (-5 m/s)/(5 s) a = 2.0 m/s/s a = 0.0 m/s/s a = -1.0 m/s/s

39. Supposing the automobile has an oil leak, demonstrate your understanding of its motion by drawing an oil drop diagram for the 20.0 seconds of motion. Divide the diagram into three distinct time intervals (0.0 - 5.0 seconds, 5.0 - 15.0 seconds, 15.0 - 20.0 seconds). Answer: See Diagram and explanation below. In the first five seconds, the object is speeding up so the spacing between dots will gradually increase. In the next ten seconds (5-15 s) the object is maintaining a constant velocity so the spacing between dots remains the same. In the last 5 seconds (15-20 s) the object is slowing down so the spacing between dots is gradually decreasing.

40. For the plots below, determine the velocity of the object ... . a. from 0.0 - 5.0 b. from 5.0 - 10.0 c. at 13.0 seconds seconds seconds

Answer: See explanations and calculations below. For a position-time graph, the velocity is determined from a slope calculation. So in each case, two points must be picked and slope calculation must be performed. a. Pick the two points: (0 s, 20 m) and (5 s, 10 m) slope = rise/run = (-10 m)/(5 s) = -2.0 m/s b. Pick the two points: (5 s, 2 m) and (10 s, 6 m) slope = rise/run = (4 m)/(5 s) = +0.80 m/s c. From 0 to 20 s, the slope is a constant value. So determining the slope at 13 seconds is as simple as merely determining the slope of the line choosing any two points. So just pick the two points: (0 s, 10 m) and (20 s, 50 m) slope = rise/run = (40 m)/(20 s) = +2.0 m/s

41. For the plots below, determine the acceleration of the object ... . a. from 0.0 - 5.0 b. from 5.0 - 10.0 seconds c. at 13.0 seconds seconds

Answer: See explanations and calculations below. For a velocity-time graph, the acceleration is determined from a slope calculation. So in each case, two points must be picked and slope calculation must be performed. a. Pick the two points: (0.0 s, 5.0 m/s) and (5.0 s, 30.0 m/s) slope = rise/run = (25.0 m/s)/(5.0 s) = +5.0 m/s2 b. Pick the two points: (5.0 s, 25.0 m/s) and (10.0 s, 10.0 m/s) slope = rise/run = (-15 m/s)/(5.0 s) = -3.0 m/s2 c. From 10.0 to 15.0 s, the slope is a constant value. So determining the slope at 13 seconds is as simple as merely determining the slope of the line choosing any two points. So just pick the two points: (10.0 s, 15.0 m/s) and (15.0 s, 0.0 m/s) slope = rise/run = (-15.0 m/s)/(5.0 s) = -3.0 m/s 42. For the plots below, determine the displacement of the object ... . a. from 0.0 - 5.0 seconds b. from 5.0 - 10.0 seconds c. from 0.0 - 15.0 seconds

Answer: See explanations and calculations below. For velocity-time graphs, the displacement of an object is found by computing the area between the line and the time axis. The shape is typically a rectangle (area = base*height), a triangle (area = 0.5*base*height) or a trapezoid (which can typically be transformed into a rectangle and a triangle or the area can be computed as 0.5*(h1 + h2)*base). a. Area = 0.5*base*height = 0.50*(5.0 s)*(20.0 m/s) = 50. m b. Area = A1 + A2 (see diagram) = (5.0 s)*(5.0 m/s) + 0.50*(5.0 s)*(10.0 m/s) = 25 m + 25 m = 50. m c. Area = A1 + A2 + A3 + A4 (see diagram) Area = 0.5*(5.0 s)*(20.0 m/s) + (5.0 s)*(20.0 m/s) + (5.0 s)*(20.0 m/s) + 0.5*(5.0 s)*(10.0 m/s) Area = 50 m + 100 m + 100 m + 25 m = 275 m

43. Determine the acceleration (in m/s2) of an object which ... . 1. moves in a straight line with a constant speed of 20.0 m/s for 12.0 seconds 2. changes its velocity from 12.1 m/s to 23.5 m/s in 7.81 seconds 3. changes its velocity from 0.0 mi/hr to 60.0 mi/hr in 4.20 seconds 4. accelerates from 33.4 m/s to 18.9 m/s over a distance of 109 m Answer: See answers, explanations and calculations below. a. If the speed and direction of an object is constant, then the acceleration is 0 m/s2. b. The acceleration is the velocity change per time ratio: a = (Velocity Change)/t = (23.5 m/s - 12.1 m/s) / (7.81 s) = 1.46 m/s2. c. The acceleration is the velocity change per time ratio: a = (Velocity Change)/t = (60.0 mi/hr - 0.0 mi/hr) / (4.20 s) = 14.3 mi/hr/s. 14.3 mi/hr/s * (1.0 m/s) / (2.24 mi/hr) = 6.38 m/s2. d. The acceleration value can also be calculated using kinematic equations if three other kinematic quantities are known. In this case, the know information is: vo = 33.4 m/s; vf = 18.9 m/s; and d = 109 m. 2 2 Using the equation vf = vo + 2*a*d, the acceleration can be computed. 2 2 2 2 2 a = (vf - vo ) / (2*d) = [(18.9 m/s) - (33.4 m/s) ] / (2 * 109 m) = -3.48 m/s .

44. Determine the magnitude of the displacement (in meters) of an object which ... . 1. moves from Hither to Yon (with an average speed of 28.0 m/s) and then back to Hither (with an average speed of 28.0 m/s) if both the forward and the return trip take 46 minutes each. 2. moves at a constant speed of 8.30 m/s in a straight line for 15.0 seconds. 3. decelerates at a rate of -4.35 m/s/s from a speed of 38.1 m/s to a speed of 17.6 m/s 4. accelerates from rest at a rate of 3.67 m/s2 for 12.1 seconds 5. is moving at 12.2 m/s and then accelerates at a rate of +1.88 m/s2 for 17.0 seconds Answer: See answers, explanations and calculations below. a. Since this is a round-trip journey, the overall displacement is 0 m. b. Since the velocity is constant, the displacement can be found by multiplying the velocity by the time. d = v*t = (8.30 m/s) * (15.0 s) = 125 m c. A displacement value can also be calculated using kinematic equations if three other kinematic quantities are known. In this case, the know information is: vo = 38.1 m/s; vf = 17.6 m/s; and a = -4.35 m/s/s. Using 2 2 the equation vf = vo + 2*a*d, the displacement can be computed. 2 2 2 2 d = (vf - vo ) / (2*a) = [(17.6 m/s) - (38.1 m/s) ] / (2 * -4.35 m/s/s) = 131 m. d. A displacement value can be calculated using other kinematic equations if a different set of kinematic quantities is known. Here we know that: vo = 0.0 m/s; t = 12.1 s; and a = 3.67 m/s/s. Using the equation d = 2 vo* t + 0.5*a*t , the displacement can be computed. d = (0 m/s)*(12.1 s) + 0.5*(3.67 m/s/s)*(12.1 s)2 = 269 m. e. Here the displacement value is calculated using the same kinematic equation. We know that: vo = 12.2 2 m/s; t = 17.0 s; and a = 1.88 m/s/s. Using the equation d = vo* t + 0.5*a*t , the displacement can be computed. d = (12.2 m/s)*(17.0 s) + 0.5*(1.88 m/s/s)*(17.0 s)2 = 479 m.

45. The hare is sleeping at a location that is 1200 m from the finish line. The tortoise passes him at a steady speed of 5.0 cm/s. If the hare finally wakes up 6.5 hours later, then what minimum acceleration (assumed constant) must he have in order to pass the tortoise before the finish line. Answer: 0.0067 m/s2 Like a lot of physics word problems, there is more than one path to the final answer. In all such problems, the solution involves thought and good problem-solving strategies (draw a picture, list what you know, list pertinent equations, etc.). The tortoise, moving at a constant speed, will cover the 1200 m in a time of:

ttortoise = d/vtortoise = (1200 m) / (0.050 m/s) = 24000 s = 6.666... hrs The hare will sleep for 6.5 hrs (23400 s) before starting, and so will have only 0.1666... hrs (600 s) to accelerate to the finish line. So the acceleration of the hare can be determine using a kinematic equation. The know information about the hare's motion is: t = 600 s; d = 1200 m; vo = 0 m/s. The best equation is d 2 = vo* t + 0.5*a*t . The vo* t term cancels and the equation can be algebraically rearranged and solved for a: a = 2*d / t2 = 2 * (1200 m) / (600 s)2 = 0.0067 m/s2. 46. A Gold Car moving at 12.0 m/s passes a Green Car while the Green Car is at rest at a stoplight. The Green Car immediately accelerates at a rate of +1.80 m/s/s for 11.0 seconds seconds and then maintains a constant speed. After how much time (relative to the initial starting time) must the Green Car drive before catching up with the Gold Car. Answer: 14.0 s (As mentioned in the previous problem ...) Like a lot of physics word problems, there is more than one path to the final answer. In all such problems, the solution involves thought and good problem-solving strategies (draw a picture, list what you know, list pertinent equations, etc.).

Here the gold car travels with a constant speed for a time of t seconds (where t is the total time of travel for 2 both cars). The distance traveled by the gold car is given by the kinematic equation d = vo* t + 0.5*a*t . The second term cancels and the distance can be expressed as 2 d = vo* t + 0.5*a*t = (12.0 m/s)*t, or

dgold = 12.0* t For the green car, there is an accelerated period and then a constant speed period. The distance traveled during the accelerated period (d1green) is found from the same kinematic equation. For the green car, the first term cancels and the distance is 2 2 2 d1green = vo* t + 0.5*a*t = 0.5*(1.80 m/s )*(11.0 s) , or

d1green = 108.9 m Once the green car has accelerated for 11 seconds, it maintains a constant speed for the remaining time, given by the expression t - 11 s. The speed at which the green car is traveling during this time can be computed using the equation: 2 vfgreen = vo + a*t = (1.80 m/s ) *(11.0 s) = 19.8 m/s

The distance traveled by the green car during this constant speed portion of its motion (d2green) can be 2 computed using the kinematic equation. d = vo* t + 0.5*a*t . The second term cancels and the distance can be expressed as 2 d2green = vo* t + 0.5*a*t = (19.8 m/s) * (t - 11 s) = 19.8*t - 217.8, or

d2green = 19.8*t - 217.8 So the total distance traveled by the green car is given by the expression:

dgreen = d1green + d2green = 108.9 + 19.8*t - 217.8

dgreen = 19.8*t - 108.9 When the green car catches up to the gold car their distance traveled will be the same. So the time t can be determined by setting the two expressions for distance equal to each other and solving for t. 12.0* t = 19.8*t - 108.9 108.9 = 7.80*t t = (108.9) / (7.80) t = 13.96 s = 14.0 s

43. Determine the acceleration (in m/s2) of an object which ... . 1. moves in a straight line with a constant speed of 20.0 m/s for 12.0 seconds 2. changes its velocity from 12.1 m/s to 23.5 m/s in 7.81 seconds 3. changes its velocity from 0.0 mi/hr to 60.0 mi/hr in 4.20 seconds 4. accelerates from 33.4 m/s to 18.9 m/s over a distance of 109 m Answer: See answers, explanations and calculations below. a. If the speed and direction of an object is constant, then the acceleration is 0 m/s2. b. The acceleration is the velocity change per time ratio: a = (Velocity Change)/t = (23.5 m/s - 12.1 m/s) / (7.81 s) = 1.46 m/s2. c. The acceleration is the velocity change per time ratio: a = (Velocity Change)/t = (60.0 mi/hr - 0.0 mi/hr) / (4.20 s) = 14.3 mi/hr/s. 14.3 mi/hr/s * (1.0 m/s) / (2.24 mi/hr) = 6.38 m/s2. d. The acceleration value can also be calculated using kinematic equations if three other kinematic quantities are known. In this case, the know information is: vo = 33.4 m/s; vf = 18.9 m/s; and d = 109 m. 2 2 Using the equation vf = vo + 2*a*d, the acceleration can be computed. 2 2 2 2 2 a = (vf - vo ) / (2*d) = [(18.9 m/s) - (33.4 m/s) ] / (2 * 109 m) = -3.48 m/s .

44. Determine the magnitude of the displacement (in meters) of an object which ... . 1. moves from Hither to Yon (with an average speed of 28.0 m/s) and then back to Hither (with an average speed of 28.0 m/s) if both the forward and the return trip take 46 minutes each. 2. moves at a constant speed of 8.30 m/s in a straight line for 15.0 seconds. 3. decelerates at a rate of -4.35 m/s/s from a speed of 38.1 m/s to a speed of 17.6 m/s 4. accelerates from rest at a rate of 3.67 m/s2 for 12.1 seconds 5. is moving at 12.2 m/s and then accelerates at a rate of +1.88 m/s2 for 17.0 seconds Answer: See answers, explanations and calculations below. a. Since this is a round-trip journey, the overall displacement is 0 m. b. Since the velocity is constant, the displacement can be found by multiplying the velocity by the time. d = v*t = (8.30 m/s) * (15.0 s) = 125 m c. A displacement value can also be calculated using kinematic equations if three other kinematic quantities are known. In this case, the know information is: vo = 38.1 m/s; vf = 17.6 m/s; and a = -4.35 m/s/s. Using 2 2 the equation vf = vo + 2*a*d, the displacement can be computed. 2 2 2 2 d = (vf - vo ) / (2*a) = [(17.6 m/s) - (38.1 m/s) ] / (2 * -4.35 m/s/s) = 131 m. d. A displacement value can be calculated using other kinematic equations if a different set of kinematic quantities is known. Here we know that: vo = 0.0 m/s; t = 12.1 s; and a = 3.67 m/s/s. Using the equation d = 2 vo* t + 0.5*a*t , the displacement can be computed. d = (0 m/s)*(12.1 s) + 0.5*(3.67 m/s/s)*(12.1 s)2 = 269 m. e. Here the displacement value is calculated using the same kinematic equation. We know that: vo = 12.2 2 m/s; t = 17.0 s; and a = 1.88 m/s/s. Using the equation d = vo* t + 0.5*a*t , the displacement can be computed. d = (12.2 m/s)*(17.0 s) + 0.5*(1.88 m/s/s)*(17.0 s)2 = 479 m.

46. A Gold Car moving at 12.0 m/s passes a Green Car while the Green Car is at rest at a stoplight. The Green Car immediately accelerates at a rate of +1.80 m/s/s for 11.0 seconds seconds and then maintains a constant speed. After how much time (relative to the initial starting time) must the Green Car drive before catching up with the Gold Car. Answer: 14.0 s (As mentioned in the previous problem ...) Like a lot of physics word problems, there is more than one path to the final answer. In all such problems, the solution involves thought and good problem-solving strategies (draw a picture, list what you know, list pertinent equations, etc.).

Here the gold car travels with a constant speed for a time of t seconds (where t is the total time of travel for 2 both cars). The distance traveled by the gold car is given by the kinematic equation d = vo* t + 0.5*a*t . The second term cancels and the distance can be expressed as 2 d = vo* t + 0.5*a*t = (12.0 m/s)*t, or

dgold = 12.0* t For the green car, there is an accelerated period and then a constant speed period. The distance traveled during the accelerated period (d1green) is found from the same kinematic equation. For the green car, the first term cancels and the distance is 2 2 2 d1green = vo* t + 0.5*a*t = 0.5*(1.80 m/s )*(11.0 s) , or

d1green = 108.9 m Once the green car has accelerated for 11 seconds, it maintains a constant speed for the remaining time, given by the expression t - 11 s. The speed at which the green car is traveling during this time can be computed using the equation: 2 vfgreen = vo + a*t = (1.80 m/s ) *(11.0 s) = 19.8 m/s

The distance traveled by the green car during this constant speed portion of its motion (d2green) can be 2 computed using the kinematic equation. d = vo* t + 0.5*a*t . The second term cancels and the distance can be expressed as 2 d2green = vo* t + 0.5*a*t = (19.8 m/s) * (t - 11 s) = 19.8*t - 217.8, or

d2green = 19.8*t - 217.8 So the total distance traveled by the green car is given by the expression:

dgreen = d1green + d2green = 108.9 + 19.8*t - 217.8

dgreen = 19.8*t - 108.9 When the green car catches up to the gold car their distance traveled will be the same. So the time t can be determined by setting the two expressions for distance equal to each other and solving for t. 12.0* t = 19.8*t - 108.9 108.9 = 7.80*t t = (108.9) / (7.80) t = 13.96 s = 14.0 s

Newtons Laws

Part A: Mass, Inertia, Weight, and Newton's First Law of Motion

1. Which of the following statements are true of inertia? List all that apply. 1. Inertia is a force. 2. Inertia is a force which keeps stationary objects at rest and moving objects in motion at constant velocity. 3. Inertia is a force which brings all objects to a rest position. 4. All objects have inertia. 5. A more massive object has more inertia than a less massive object. 6. Fast-moving objects have more inertia than slow-moving objects. 7. An object would not have any inertia in a gravity-free environment (if there is such a place). 8. Inertia is the tendency of all objects to resist motion and ultimately stop. 9. In a gravity-free environment (should there be one), a person with a lot of inertia would have the same ability to make a turn as a person with a small amount of inertia.

Answer: DE a. False - Inertia is not a force. b. False - Inertia is NOT a force. c. False - Inertia is NOT a force. Inertia is simply the tendency of an objects to resist a change in whatever state of motion that it currently has. Put another way, inertia is the tendency of an object to "keep on doing what it is doing." Mass is a measure of an object's inertia. The more mass which an object has, the more that it sluggish towards change. d. True - Bet money on this one. Any object with mass has inertia. (Any object without mass is not an object, but something else like a wave.) e. True - Mass is a measure of an object's inertia. Objects with greater mass have a greater inertia; objects with less mass have less inertia. f. False - The speed of an object has no impact upon the amount of inertia that it has. Inertia has to do with mass alone. g. False - Inertia (or mass) has nothing to do with gravity or lack of gravity. In a location where g is close to 0 m/s/s, an object loses its weight. Yet it still maintains the same amount of inertia as usual. It still has the same tendency to resist changes in its state of motion. h. False - Inertia is NOT the tendency to resist motion, but rather to resist changes in the state of motion. For instance, its the tendency of a moving object to keep moving at a constant velocity (or a stationary object to resist changes from its state of rest). i. False - Once more (refer to g), inertia is unaffected by alterations in the gravitational environment. An alteration in the g value effects the weight of an object but not the mass or inertia of the object.

2. Which of the following statements are true of the quantity mass? List all that apply. 1. The mass of an object is dependent upon the value of the acceleration of gravity. 2. The standard metric unit of mass is the kilogram. 3. Mass depends on how much stuff is present in an object. 4. The mass of an object is variable and dependent upon its location. 5. An object would have more mass on Mount Everest than the same object in the middle of Lake Michigan. 6. People in Weight Watcher's are really concerned about their mass (they're mass watchers). 7. The mass of an object can be measured in pounds. 8. If all other variables are equal, then an object with a greater mass would have a more difficult time accelerating. 9. If all other variables are equal, then it would require less exerted force to stop a less massive object than to stop a more massive object. 10. The mass of an object is mathematically related to the weight of the object. Answer: BCFHIJ a. False - Mass is independent of the gravitational environment that an object is in and dependent solely upon the number of atoms in the object and the type of atoms (Carbon: ~12 g/mol; Hydrogen: ~1 g/mol ; Oxygen: ~16 g/mol). Because of this, mass is said to be invariable (unless of course, an object loses some of its atoms) - a constant quantity which is independent of the acceleration of gravity and therefore independent of location. (Weight on the other hand depends upon the gravitational environment.) b. True - Know this one. Kilograms is for mass and Newtons is for force. c. True - This is kind of a simple definition of mass but it does do the job (provided stuff means atoms or material). d. False - See explanation to #2d. e. False - An object has the same mass on Mount Everest as it does at sea level (or near sea level); only the weight of the object would be slightly different in these two locations. f. True - Weight Watcher's participants only use a measurement of their weight as a reflection of how many atoms of flesh that they have burned from their bodies. Their real interest is in losing mass for reasons related to health, appearance, etc. g. False - Pounds is a unit of force commonly used in the British system of measurement. It is not a metric unit and it is not a unit of mass. Kilogram is the standard metric unit of mass and slug is the British unit. h. True - Weight and force of gravity are synonymous terms. You should quickly become comfortable with the terms mass, weight and force of gravity; it will save you many headaches as we continue through the course. i. True - A less massive object has less inertia and as such would offer less resistance to changes in their velocity. For this reason, a less massive object requires less force to bring from a state of motion to a state of rest. j. True - The weight of an object is the mass of the object multiplied by the acceleration of gravity of the object. Mass and weight are mathematically related by the equation: Weight (or Fgrav) = m•g

3. Which of the following statements are true of the quantity weight? List all that apply. 1. The weight of an object is dependent upon the value of the acceleration of gravity. 2. Weight refers to a force experienced by an object. 3. The weight of an object would be less on the Moon than on the Earth. 4. A person could reduce their weight significantly by taking an airplane ride to the top of Mount Everest. 5. Two objects of the same mass can weigh differently. 6. To gain weight, one must put on more mass. 7. The weight of an object can be measured in kilograms. 8. The weight of an object is equal to the force of gravity acting upon the object. 9. When a chemistry student places a beaker on a balance and determines it to be 84.3 grams, they have weighed the beaker. Answer: ABCH and possibly EF a. True - The weight of an object is equal to the force of gravity acting upon the object. It is computed by multiplying the object's mass by the acceleration of gravity (g) at the given location of the object. If the location of the object is changed, say from the Earth to the moon, then the acceleration of gravity is changed and so is the weight. It is in this sense that the weight of an object is dependent upon the acceleration of gravity. b. True - This statement is true in the sense that the weight of an object refers to a force - it is the force of gravity. c. True - The weight of an object depends upon the mass of the object and the acceleration of gravity value for the location where it is at. The acceleration of gravity on the moon is 1/6-th the value of g on Earth. As such, the weight of an object on the moon would be 6 times less than that on Earth. d. False - A trip from sea level to the top of Mount Everest would result in only small alterations in the value of g and as such only small alterations in a person's weight. Such a trip might cause a person to lose a pound or two. e. Mostly True - Two objects of the same mass can weigh differently if they are located in different locations. For instance, person A and person B can both have a mass of 60 kg. But if person A is on the Earth, he will weigh ~600 N, whereas person B would weight ~100 N on the moon. f. Kinda True (Mostly False) - Weight is the product of mass and the acceleration of gravity (g). To gain weight, one must either increase their mass or increase the acceleration of gravity for the environment where they are located. So the statement is true if one disregards the word MUST which is found in the statement. g. False - By definition, a free-falling object is an object upon which the only force is gravity. Such an object is accelerating at a rate of 9.8 m/s/s (on Earth) and as such cannot be experiencing a balance of forces. h. True - This statement is the precise definition of weight. Weight is the force of gravity. i. False - This student has determined the mass of the beaker, not the weight. As such, he/she has massed the beaker, not weighed it.

4. Which of the following statements are true of an object that experiences balanced forces (or unbalanced forces)? List all that apply. 1. If a person is moving to the right, then the forces acting upon it are NOT balanced. 2. A balance of forces is demonstrated by an object which is slowing to a stop. 3. It would take an unbalanced force to keep an object in motion. 4. If an object is moving with a constant speed in a circle, then the forces acting upon the object are balanced. 5. If an object is accelerating at a constant rate of acceleration, then the forces acting upon the object are balanced. 6. It is NOT possible for just three forces to be acting upon an object and they still balance each other. 7. A free-falling object experiences a balance of forces. 8. Balanced forces cause stationary objects to remain at rest and moving objects to come to rest. 9. Unbalanced forces cause objects to move. Answer: None of these are true, though one might make a strong argument for I. a. False - An object which is moving to the right could have unbalanced forces, but only if it is accelerating. The presence of unbalanced forces must always be associated with acceleration, not mere motion. In this case, an object moving to the right could have a balance of forces if it is moving with a constant velocity. b. False - An object would never slow to a stop unless the forces acting upon it were unbalanced. In fact, an object which slows down must have a unbalanced force directed in the direction opposite their motion. c. False - An unbalanced force is only required to accelerate an object. A balance of forces is required to keep an object moving at a constant velocity. For instance, a car moving to the right at constant velocity encounters as much rightward force as leftward force. d. False - An object which moves in a circle has a changing direction. As such, there is an acceleration and this acceleration requires that there be an unbalanced force present on the object. e. False - Any object that accelerates has a changing velocity. An object that accelerates at a constant rate has a velocity that changes by the same amount each second. For instance, a free-falling object changes its velocity by -9.8 m/s ever second. It is said to have a constant acceleration of -9.8 m/s2. A free-falling object, or any object with an acceleration (whether constant or non-constant) must be experiencing an unbalanced force. f. False - Consider an object which weighs 1000 N (a 1000 N downward force) which is being pulled on by two people, each exerting 500 N of upward force. Such an object has three forces acting upon it and the three forces together balance each other. g. False - A free-falling object is an object upon which the only force is gravity. As such, there is an unbalanced force acting upon it; this unbalanced force explains its acceleration. h. False - Balanced forces cause stationary objects to stay at rest. However balanced forces would never cause moving objects to stop; an unbalanced force would be required to stop a moving object. i. False - Unbalanced forces do more than cause objects to move; unbalanced forces cause objects to accelerate. Though one could make a strong argument that an object that is accelerating must also be moving (albeit with a changing velocity). In this sense, this statement is true.

5. Consider Newton's first law of motion to determine which of the following statements are true? List all that apply. 1. Newton's first law of motion is applicable to both moving and nonmoving objects. 2. If a football is moving upwards and rightwards towards the peak of its trajectory, then there are both rightwards and upwards forces acting upon it. 3. It would take an unbalanced force to keep an object in motion. 4. If an object is at rest, then there are no forces acting upon the object. 5. It would take an unbalanced force to keep an object in motion at a constant velocity. 6. It is the natural tendency of all objects to eventually come to a rest position. 7. A pendulum bob is set into its usual back-and-forth periodic motion. After some time (perhaps 10 minutes), the pendulum bob comes to a rest position. This is best explained by the idea of inertia - all objects eventually resist motion. 8. If a 3-kg rock is thrown at a speed of 2 m/s in a gravity-free environment (presuming one could be found), then an unbalanced force of 6 N would be required to keep the rock moving at a constant speed. 9. It would take an unbalanced force to cause an object to accelerate from rest. Answer: AI a. True - Absolutely true. Like all true scientific laws, they govern all objects. In the case of Newton's first law of motion: An object that is nonmoving remains at rest (unless acted upon by an unbalanced force); and a moving object will continue in its motion at a constant velocity (unless acted upon by an unbalanced force). b. False - A football which is moving upwards and rightwards towards its peak, then it has both an upward and a rightward velocity; it does not however have an upward and a rightward force. In fact, if acting as a projectile, it has no horizontal force and maintains a constant horizontal velocity; similarly, it would have a downward force of gravity and a slowing down motion as it rises. If the football were not a projectile, then the horizontal force would be leftward (air resistance opposing its motion) and the vertical force would be gravity and air resistance, both directed downward. c. False - An unbalanced force would accelerate an object. If directed against its motion, then it would actually slow it down rather than keep is motion going. A balance of forces is all that is required to keep an object going at a constant velocity. An unbalanced force directed in the direction of motion would be required to keep an object going with an increasing speed. d. False - If an object is at rest, then there are no unbalanced forces acting upon it. There is a force of gravity and at least one other upward force capable of balancing the force of gravity. e. False - This is dead wrong. It would take a balance of forces to keep an object in motion at constant velocity. An unbalanced force would cause some form of acceleration. f. False - If you answered TRUE, then Galileo and Newton just rolled over in their grave. It is the natural tendency of all objects to maintain their velocity and to resist changes in whatever state of motion that they have. This is the law of inertia. g. False - All objects resist changes in their state of motion. In the absence of unbalanced forces, they maintain their velocity (whether zero or nonzero). The pendulum changes its state of motion due to an unbalanced force - the force of air resistance. h. False - For an object to maintain a constant velocity, 0 Newtons of net force (i.e., a balance of forces) is required. i. True - Unbalanced forces cause stationary objects to accelerate from rest. In the absence of an unbalanced force, a stationary object would remain at rest.

Part B: Force, Acceleration and Newton's Second Law of Motion

6. Which of the following statements are true of the concept of force? List all that apply. a. A force is a push or pull exerted upon an object which results from the interaction of that object with its environment. b. Bubba approaches Billie and gives him a swift shove. Timid little Billie keeps his hands in his pocket during this interaction. Subsequently, while Bubba places a force upon Billie, Billie does not place a force upon Bubba. c. A quarterback throws a football down field. Once thrown, the force from the quarterback persists upon the ball to cause it to continue on its upward trajectory towards its peak. d. A sled slides down the hill and reaches the bottom where it gradually slows to a stop. Once on the level ground, the force of the hill persists upon the sled to allow it to continue its forward motion. e. Forces always cause objects to move. f. An object can experience two or more forces and not accelerate. g. A contact force results from the physical contact between two objects. h. A field force results from the action of two objects which are positioned some distance away. i. Spring and tension forces are examples of field forces. j. A force is a vector quantity; there is always a direction associated with it. k. Force can be measured in kilograms or Newtons depending upon the system of measurement (metric or otherwise). Answer: AFGJ and sort of H. a. True - This is a great definition of force. b. False - According to Newton's third law, one cannot push on an object without being pushed back. The force on Billie is the result of an interaction of Bubba's hands with Billie's body. That force on Billie might cause Billie to go flying, but the reaction force offers resistance to the motion of Bubba's hands and slows them down. In general, forces will always (without exception) come in pairs. c. False - The force of the quarterback on the football is a contact force which can only exist during the interaction (i.e., the contact) between the quarterback's hands and the football. Once thrown, the football continues its horizontal motion due to its own inertia and its vertical motion is effected by the force of gravity. d. False - Be careful if you answered true to this one. If you did, perhaps you believe in the fatal misconception that a rightward force is required to sustain a rightward motion. The sleds motion to the right can be described as a leftward accelerated motion. Such a leftward acceleration demands that there is a leftward force (despite its rightward force). This leftward force slows the rightward-moving sled down. The hill cannot push on the sled unless the hill is in contact with the sled. e. False - Forces, if unbalanced, can cause objects to accelerate (one form of moving; the other form is moving at a constant velocity). But by no means can one say that forces always cause objects to move. For instance, as you sit in your chair, the chair pushes up on your body but your body does not move. f. True - Certainly! As you sit in your chair, the chair pushes up on your body but your body does not accelerate. This upward force (known as the normal force) is balanced by the downward force of gravity. Many objects experience a force yet do not accelerate. g. True - There are two broad categories of forces - contact forces and field forces. Contact forces, by definition, are those which result from the physical contact of two forces. h. True (mostly) - A field force is a force which can acts between two objects even when they are separated by a distance. Field forces have magnitudes which are dependent upon the distance of separation between the two interacting objects. For instance, the force of gravity between the Sun and the earth is a field force whose value depends upon the distance of separation between the center of the Earth and the center of the Sun. In this sense, the force of gravity is a force which acts when two objects are separated in space from each other. Yet field forces can also occur when the two objects are touching each other. In this sense, one can be skeptical of the wording of the statement. i. False - Spring and tension are examples of contact forces. The spring or the rope/cable/wire are in contact with the object upon which it exerts its push or pull. The field forces are electric force, magnetic force, and gravity force. j. True - Forces always have a direction associated with them. As such, force is a vector quantity - a quantity which is fully described by both a magnitude (size, value) and a direction. k. False - Force is measured in Newtons in the metric system and in pounds in the British system. Kilograms is a unit of mass.

7. Consider Newton's second law of motion to determine which of the following statements are true? List all that apply. a. If an object is accelerating to the right, the net force on the object must be directed towards the right. b. If an object is moving to the right and slowing down, then the net force on the object is directed towards the left. c. Accelerating objects are either slowing down or speeding up. d. The acceleration of an object is directly dependent upon its mass and inversely dependent upon its net force. e. An object has an acceleration of 8 m/s/s. If the net force acting upon the object is increased by a factor of 2, then the new acceleration would be 10 m/s/s. f. An object has an acceleration of 8 m/s/s. If the net force acting upon the object is increased by a factor of 3, then the new acceleration would be 11 m/s/s. g. An object has an acceleration of 8 m/s/s. If the mass of the object is increased by a factor of 2, then the new acceleration would be 16 m/s/s. h. An object has an acceleration of 8 m/s/s. If the mass of the object is increased by a factor of 4, then the new acceleration would be 2 m/s/s. i. An object has an acceleration of 8 m/s/s. If the net force acting upon the object is increased by a factor of 2 and the mass of the object is decreased by a factor of 2, then the two factors would offset each other and the acceleration would still be 8 m/s/s. j. An object has an acceleration of 8 m/s/s. If the net force acting upon the object is increased by a factor of 2 and the mass of the object is increased by a factor of 4, then the new acceleration would be 4 m/s/s. k. An object has an acceleration of 8 m/s/s. If the net force acting upon the object is decreased by a factor of 2 and the mass of the object is increased by a factor of 4, then the new acceleration would be 1 m/s/s. l. An object has an acceleration of 8 m/s/s. If the net force acting upon the object is increased by a factor of 4 and the mass of the object is increased by a factor of 2, then the new acceleration would be 16 m/s/s. m. A 2-kg object accelerates from rest to a final velocity of 6 m/s in 3 seconds. The magnitude of the net force acting upon the object is 12 N. n. A 10-kg object slows down from 24 m/s to a final velocity of 9 m/s in 3 seconds. The magnitude of the net force acting upon the object is 80 N. Answer: ABHJKL a. True - The acceleration is directly related to the net force and the direction of the acceleration is always the same as the direction of the net force. When it comes to force, objects can be thought of as being in the middle of a tug-of-war between the individual forces. The force that wins the tug-of-war is the force which determines the direction of the acceleration. So if a rightward force wins over a leftward force, the acceleration will be to the right. b. True - An object which is slowing down has an acceleration which is directed opposite the motion of the object. So an object which moves to the right and slows down experiences a leftward acceleration and therefore a leftward net force. c. False - Acceleration involves a change in velocity and velocity is a vector with a magnitude (15 m/s, 22 m/s, etc.) and a direction (east, northeast, etc.). Accelerating objects are either changing the magnitude of the velocity by speeding up or slowing down or changing the direction of the velocity by turning. d. False - Vice Versa. The acceleration of an object is inversely dependent upon the mass and directly dependent upon the net force. e. False - Acceleration is directly dependent upon the net force. Whatever alteration is made in the net force, the same alteration must be made in the acceleration. So if the net force is increased by a factor of 2, then the acceleration is increased by a factor of 2 from 8 m/s/s to 16 m/s/s. f. False - Whatever alteration is made in the net force, the same alteration must be made in the acceleration. So if the net force is increased by a factor of 3, then the acceleration is increased by a factor of 3 from 8 m/s/s to 24 m/s/s. g. False - Acceleration is inversely dependent upon the mass. Whatever alteration is made in the mass, the inverse must be made of the acceleration. So if the mass is increased by a factor of 2, then the acceleration is decreased by a factor of 2 from 8 m/s/s to 4 m/s/s. h. True - Acceleration is inversely dependent upon the mass. Whatever alteration is made in the mass, the inverse must be made of the acceleration. So if the mass is increased by a factor of 4, then the acceleration is decreased by a factor of 4 from 8 m/s/s to 2 m/s/s. i. False - Acceleration is inversely dependent upon the mass and directly dependent upon the net force. If the net force is increased by a factor of 2, then the acceleration is increased by a factor of 2. If the mass is decreased by a factor of 2, then the acceleration is increased by a factor of 2. The overall result of the two changes is to increase acceleration by a factor of 4 from 8 m/s/s to 32 m/s/s. j. True - Acceleration is inversely dependent upon the mass and directly dependent upon the net force. If the net force is increased by a factor of 2, then the acceleration is increased by a factor of 2. If the mass is decreased by a factor of 4, then the acceleration is decreased by a factor of 4. The overall result of the two changes is to decrease acceleration by a factor of 2 from 8 m/s/s to 4 m/s/s. k. True - Acceleration is inversely dependent upon the mass and directly dependent upon the net force. If the net force is decreased by a factor of 2, then the acceleration is decreased by a factor of 2. If the mass is decreased by a factor of 4, then the acceleration is decreased by a factor of 4. The overall result of the two changes is to decrease acceleration by a factor of 8 from 8 m/s/s to 1 m/s/s. l. True - Acceleration is inversely dependent upon the mass and directly dependent upon the net force. If the net force is increased by a factor of 4, then the acceleration is increased by a factor of 4. If the mass is increased by a factor of 2, then the acceleration is decreased by a factor of 2. The overall result of the two changes is to increase acceleration by a factor of 2 from 8 m/s/s to 16 m/s/s. m. False - The net force is the product m•a. Acceleration (a) can be calculated as the velocity change per time. The velocity change is +6 m/s (from 0 m/s to 6 m/s), so the acceleration is (+6 m/s) / (3 s) = +2 m/s/s. Therefore the net force is (2 kg)•(+2 m/s/s) = +4 N. The + indicates information about the direction; the 4 N is the magnitude. n. False - The net force is the product m•a. Acceleration (a) can be calculated as the velocity change per time. The velocity change is -15 m/s (from 24 m/s to 9 m/s), so the acceleration is (-15 m/s) / (3 s) = -5 m/s/s. Therefore the net force is (10 kg)•(-5 m/s/s) = -50 N. The - indicates information about the direction; the 50 N is the magnitude.

Part C: Forced Choice and Short Answer

8a. Big Bubba has a mass of 100 kg on the earth. What is Big Bubba's mass on the moon where the force of gravity is approximately 1/6-th that of Earth's? ______Explain or show your work. Answer: 100 kg Mass is a quantity which is independent of the location of the object. So if Big Bubba has a mass of 100 kg on Earth, then he also has a mass of 100 kg on the moon. Only the weight would change as Big Bubba is moved from the Earth to the moon. He weighs ~1000 N on Earth and 1/6-th this value (~167 N) on the moon.

8b. Little Billie weighs 360 N on Earth. What is Little Billie's mass on the moon where the force of gravity is approximately 1/6-th that of Earth's? ______Explain or show your work. Answer: ~36 kg The mass of an object is related to weight by the equation W = m•g where g = ~10 m/s/s on Earth and one- sixth this value (~1.67) on the moon. So if Billy weighs 360 N on Earth, then his mass is approximately ~36 kg. His mass on the moon will be the same as his mass on Earth. Only his weight changes when on the moon; rather than being 360 N, it is 60 N. His weight on the moon could be found by multiplying his mass by the value of g on the moon: (36 kg) • (9.8/6 m/s/s) = ~60 N

10. The amount of net force required to keep a 5-kg object moving rightward with a constant velocity of 2 m/s is ____. a. 0 N b. 0.4 N c. 2 N d. 2.5 N e. 5 N

Answer: A Net force is always m•a. In this case, the velocity is constant so the acceleration is zero and the net force is zero. Constant velocity motion can always be associated with a zero net force.

12. Which one(s) of the following force diagrams depict an object moving to the right with a constant speed? List all that apply.

Answer: AC If an object is moving at a constant speed in a constant rightward direction, then the acceleration is zero and the net force must be zero. Choice B and D show a rightward net force and therefore a rightward acceleration, inconsistent with the described motion.

13. According to Newton's third law, every force is accompanied by an equal and opposite reaction force. The reason that these forces do not cancel each other is ____. a. the action force acts for a longer time period b. the two forces are not always in the same direction c. one of the two forces is greater than the other d. the two forces act upon different objects; only forces on the same object can balance each other. e. ... nonsense! They do cancel each other. Objects accelerate because of the presence of a third force. Answer: D Action and reaction forces always act upon the interacting objects for the same amount of time with the same magnitude. So if object A pushes on object B, then object B simultaneously pushes on object A with the same amount of force. The force on object B will be one of perhaps many forces which will govern its motion. But the reaction force is on object A and cannot contribute to object B's motion since it is not acting upon object B. Action-reaction forces can NEVER cancel each other.

14. As you sit in your chair and study your physics (presuming that you do), the force of gravity acts downward upon your body. The reaction force to the force of the Earth pulling you downward is ___. a. the force of the chair pushing you upward b. the force of the floor pushing your chair upward c. the force of the Earth pushing you upward d. the force of air molecules pushing you upwards e. the force of your body pulling the Earth upwards f. ... nonsense! Gravity is a field force and there is no such reaction force. Answer: E The most common wrong answer is a - the force of the chair pushing you upward. As you sit in your chair, the chair is indeed pushing you upward but this is not the reaction force to the force of the Earth pulling you downward. The chair pushing you upward is the reaction force to you sitting on it and pushing the chair downward. To determine the action-reaction force pairs if given a statement of the form object A pulls X-ward on object B, simply take the subject and the object in the sentence and switch their places and then change the direction to the opposite direction (so the reaction force is object B pulls object A in the opposite direction of X). So if the Earth pulls you downward, then the reaction force is you pull the Earth upward.

For the next several questions, consider the velocity-time plot below for the motion of an object along a horizontal surface. The motion is divided into several time intervals, each labeled with a letter.

22. During which time interval(s), if any, are there no forces acting upon the object? List all that apply.

23. During which time interval(s), if any, are the forces acting upon the object balanced? List all that apply.

24. During which time interval(s), if any, is there a net force acting upon the object? List all that apply.

25. During which time interval(s), if any, is the net force acting upon the object directed toward the right? List all that apply.

26. During which time interval(s), if any, is the net force acting upon the object directed toward the left? List all that apply. Answers: See answers and explanations below. 22. None - If an object is on a surface, one can be guaranteed of at least two forces - gravity and normal force. 23. BDFH - If the forces are balanced, then an object is moving with a constant velocity. This is represented by a horizontal line on a velocity-time plot. 24. ACEG - If an object has a net force upon it, then it is accelerating. Acceleration is represented by a sloped line on a velocity-time plot. 25. AE - If the net force is directed to the right, then the acceleration is to the right (in the + direction). This is represented by a line with a + slope (i.e., upward slope). 26. CG - If the net force is directed to the left, then the acceleration is to the left (in the - direction). This is represented by a line with a - slope (i.e., downward slope).

For the next several questions, consider the trajectory diagram shown below for a projectile thrown at an angle to the horizontal. The vector arrows represent the horizontal and vertical components of the projectile's velocity. Several points in the trajectory are labeled with a letter. Use the trajectory diagram to answer the next several questions. (Consider air resistance to be negligible.)

32. At which point(s), if any, are there no forces acting upon the object? List all that apply.

33. At which point(s), if any, are the forces acting upon the object balanced? List all that apply.

34. At which point(s), if any, is there a net force acting upon the object? List all that apply.

35. At which point(s), if any, is the net force acting upon the object directed toward the right? List all that apply.

36. At which point(s), if any, is the net force acting upon the object directed upward? List all that apply. Answers: See answers and explanations below 32. None 33. None 34. ABCDEFG 35. None 36. None This object is a projectile as can be seen by its constant horizontal velocity and a changing vertical velocity (besides that, the problem states that this is a projectile). A projectile is an object upon which the only force is gravity. Gravity acts downward to accelerate an object downward. This force is an unbalanced force or net force. It causes a vertical rising object to slow down and a falling object to speed up. The presence of a horizontal velocity does not demand a horizontal force, only a balance of horizontal forces. Having no forces horizontally would cause the projectile to move at a constant horizontal speed once it is launched. Similarly an upward force is not needed on this projectile. When launched, an upward velocity is imparted to it; this velocity is steadily decreased as the object is acted upon by the downward force of gravity. An upward force would only be required for an object which is speeding up as it rises upward.

Part D: Drawing and Interpreting Free-Body Diagrams

37. Construct free-body diagrams for the following physical situations at the instant in time for which they are described. As is always done in free-body diagrams, label the forces according to type and draw the arrows such that their length reflects the magnitude of the force. a. A book is at rest on top of a b. A book is being pushed to the c. A book is being pushed to the table. right across a table surface with a right across a table surface and constant velocity. (Neglect Fair.) accelerating in the direction of the

push. (Neglect Fair.)

d. A student is pushing lightly upon e. A rightward-moving box (which f. An air track glider is gliding to a large box in an attempt to push it was previously set into rightward the right at constant velocity. to the right across the floor, but the motion across the floor) gradually box fails to move. slows to a stop.

(or replace the Fnorm with Fair)

g. A ball is dropped from rest from h. Several seconds after being i. Several seconds after being the top of a building. (Neglect Fair.) thrown, a football is moving thrown, a football reaches the upwards and rightwards towards precise peak of its trajectory. the peak of its trajectory. (Neglect (Neglect Fair.) Fair.)

j. A falling skydiver is speeding up. k. A falling skydiver has reached a l. After reaching a terminal velocity, terminal velocity. a falling skydiver has opened her parachute.

m. A car is moving to the right at a n. A car is skidding to a stop (with o. An elevator (held by a cable) is high speed across a level roadway wheels locked) while traveling to moving upwards at a constant surface; the driver's foot remains the right across a level roadway velocity. (Neglect Fair.) on the gas pedal. surface.

(Fair could be added to Ffrict)

p. An upward moving elevator q. A downward moving elevator r. A picture hangs symmetrically by (held by a cable) slows down. (held by a cable) slows down. two wires oriented at angles to the (Neglect Fair.) (Neglect Fair.) vertical.

Part E: Force-Mass-Acceleration Relationships

Construct free-body diagrams for the following objects; label the forces according to type. Use the approximation that g=10 m/s2 to determine the magnitude of all forces and the net force and acceleration of the object.

38. A 2-kg box is at rest on a table. 39. A 2-kg box is free- 40. A 2-kg box equipped with a parachute falling from the table to the is falling at a terminal velocity after being ground. dropped from a plane.

("At rest" would indicate a balance of forces and an acceleration of 0 ("Free-falling" indicates that m/s/s.) the only force that influences the motion is the (A "terminal velocity" indicates a constant force of gravity.) velocity and a balance of forces.)

41. A 2-kg box is sliding to the right across a table. 42. An 8-N force is applied to a 2-kg box to move it to The force of friction upon the box is 5 N. the right across the table at a constant velocity of 1.5 m/s.

(Friction is directed opposite the motion and causes (A "constant velocity" indicates an acceleration of 0 a leftward acceleration; no rightward force is m/s/s and a balance of forces.) spoken of, only a rightward motion.)

43. An 8-N force is applied to a 2-kg box to 44. A 500-kg freight elevator is descending down accelerate it to the right across a table. The box through the shaft at a constant velocity of 0.50 m/s. encounters a force of friction of 5 N.

(A "constant velocity" indicates an acceleration of 0 m/s/s and a balance of forces.) (The horizontal forces can be summed as vectors; divide by the mass to obtain the acceleration value.)

45. A 500-kg freight elevator is moving upwards 46. A 150-N rightward force is applied to a 20-kg box towards its destination. Near the end of the ascent, to accelerate it to the right across a rough surface at a the upward moving elevator encounters a downwardrate of 2.0 m/s/s. acceleration of 2.0 m/s/s.

(Begin by multiplying m•a to find the net force - (Begin by multiplying m•a to determine the net force - 1000 N, down. The downward gravity force must be 40 N, right. The rightward applied force must be 40 N 1000 N more than the upward tension force.) more than the leftward friction force.)

Part F: Problem Solving For the following problems, draw free-body diagrams and solve for the requested unknown. Use g = 9.8 m/s/s. 47. A 0.250-kg ball is thrown upwards with an initial velocity of 12.0 m/s at an angle of 30.0 degrees. Determine the acceleration of the ball when it has reached the peak of its trajectory. Assume negligible air resistance. PSYW Answer: 9.8 m/s/s, down

There is only one force upon the ball - the force of gravity. (Air resistance is negligible; the ball is not on a surface, so there is no friction or normal force; the applied force which projects it into motion does not act upon the ball during its trajectory; there are no springs, strings, wires, or cables attached so there is neither a tension force nor a spring force.) The force of gravity acts downward with a magnitude of m•g = (0.250 kg) •(9.8 m/s/s) = 2.45 N. The net force is 2.45 N; when divided by mass, the acceleration can be found.

a = Fnet / m = (2.45 N, down) / (0.250 kg) = 9.8 m/s/s, down

48. A 72-kg skydiver is falling from 10000 feet. At an instant during the fall, the skydiver encounters an air resistance force of 540 Newtons. Determine the acceleration of the skydiver at this instant. PSYW Answer: 2.3 m/s/s, down

There are two forces acting upon the skydiver - gravity (down) and air resistance (up). The force of gravity has a magnitude of m•g = (72 kg) •(9.8 m/s/s) = 706 N. The sum of the vertical forces is

∑Fy = 540 N, up + 706 N, down = 166 N, down

The acceleration of the skydiver can be computed using the equation ∑ Fy = m•ay.

ay = (166 N, down) / (72 kg) = 2.3 m/s/s, down

52. Determine the applied force required to accelerate a 3.25-kg object rightward with a constant acceleration of 1.20 m/s/s if the force of friction opposing the motion is 18.2 N. (Neglect air resistance.) PSYW Answer: 22.1 N

There are four forces acting upon the object as shown in the free-body diagram at the right. The two vertical forces must balance since there is no vertical acceleration. Since the mass and the acceleration are known, the net force can be computed: Fnet = m•a = (3.25 kg, right) • (1.20 m/s/s) = 3.90 N, right Since the acceleration is to the right, the net force is directed to the right. Thus, the rightward force (applied force) must exceed the leftward force (friction force) by an amount equal to the 3.9 N. So the applied force can be computed.

Fapp = Ffrict + Fnet = 18.2 N + 3.90 N = 22.1 N

57. A 4.44-kg bucket suspended by a rope is accelerated upwards from an initial rest position. If the tension in the rope is a constant value of 83.1 Newtons, then determine the speed (in m/s) of the bucket after 1.59 seconds. PSYW Answer: 14.2 m/s

There are two forces acting upon the bucket - the force of gravity (up) and the tension force (down). The magnitude of the force of gravity is found from m•g; its value is 43.5 N. These two forces can be summed as vectors to determine the net force.

Fnet = ·Fy = 83.1 N, up + 43.5 N, down = 39.6 N, up The acceleration can be calculated using Newton's second law of motion.

a = Fnet / m = (39.6 N, up) / (4.44 kg) = 8.92 m/s/s, up

The acceleration value can be used with other kinematic information (vi = 0 m/s, t = 1.59 s) to calculate the final speed of the bucket. The kinematic equation, substitution and algebra steps are shown.

vf = vi + a•t

vf = 0 m/s + (8.92 m/s/s)•(1.59 s)

vf = 14.2 m/s

58. A 22.6-N horizontal force is applied to a 0.0710-kg hockey puck to accelerate it across the ice from an initial rest position. Ignore friction and determine the final speed (in m/s) of the puck after being pushed for a time of .0721 seconds. PSYW Answer: 23.0 m/s Upon neglecting air resistance, there are three forces acting upon the object. The up and down force balance each other and the acceleration is caused by the applied force. The net force is 22.6 N, right (equal to the only rightward force - the applied force). So the acceleration of the object can be computed using Newton's second law.

a = Fnet / m = (22.6 N, right) / (0.0710 kg) = 318 m/s/s, right

The acceleration value can be used with other kinematic information (vi = 0 m/s, t = 0.0721 s) to calculate the final speed of the puck. The kinematic equation, substitution and algebra steps are shown.

vf = vi + a•t

vf = 0 m/s + (318 m/s/s)•(0.0721 s)

vf = 23.0 m/s

59. A train has a mass of 6.32x104 kg and is moving with a speed of 94.3 km/hr. The engineer applies the brakes which results in a net backward force of 2.43x105 Newtons on the train. The brakes are held for 3.40 seconds. How far (in meters) does the train travel during this time? PSYW Answer: 66.8 m

There are three (perhaps four) forces acting upon this train. There is the upward force (normal force) and the downward force (gravity); these two forces balance each other since there is no vertical acceleration. The resistive force is likely a combination of friction and air resistance. These forces act leftward upon a rightward skidding train. In the free-body diagram, these two forces are 5 represented by the Ffrict arrow. The value of this resistive force is given as 2.43x10 N. This is the net force since there are no other horizontal forces; it is the force which causes the acceleration of the train. The acceleration value can be determined using Newton's second law of motion. 5 4 a = Fnet / m = (2.43x10 N) / (6.32x10 kg) = 3.84 m/s/s, left This acceleration value can be combined with other kinematic variables (vi = 94.3 km/hr = 26.2 m/s; t = 3.40 s) in order to determine the distance the train travels in 3.4 seconds. 2 d = vi • t + 0.5 •a • t d = (26.2 m/s) • (3.40 s) + 0.5 • (-3.84 m/s/s) • (3.40 s)2 d = 89.1 m - 22.2 m d = 66.8 m

4. Which of the following statements are true of vector addition, vector subtraction, and vector addition diagrams? List all that apply. a. Vectors A, B, and C are added together as A + B + C. If the order in which they are added is changed to C + B + A, then the result would be different. b. Vectors A, B, and C are added together as A + B + C. If the order in which they are added is reversed to C + B + A, then the result would be a vector with the same magnitude but the opposite direction. c. When constructing a vector diagram for A + B + C, it is not absolutely necessary that vectors B and C use the same scale that is used by vector A. d. The resultant in a vector addition diagram always extends from the head of the last vector to the tail of the first vector. e. If vectors A and B are added at right angles to each other, then one can be sure that the resultant will have a magnitude that is greater than the magnitudes of either one of the individual vectors A and B. f. If vectors A and B are added at right angles to each other, then one can be sure that the resultant will have a magnitude that is less than the arithmetic sum of the magnitudes of A and B. g. Vector addition diagrams cannot be used to determine the resultant when there is a vector subtraction operation. Answer: EF a. FALSE - Altering the order in which three vectors are added does not alter the result of the addition process. A + B + C = C + B + A. Each order of operation yields a resultant with the same magnitude and direction. b. FALSE - As mentioned above in a, altering the order in which three vectors are added does not alter the result of the addition process. Reversing the order produces a resultant with the same magnitude and the same direction. c. FALSE - When constructing a vector addition diagram, a scale must be chosen and adhered to. The scale which used to draw vector A must also be used for vectors B and C. One cannot switch horses in the middle of the stream. d. FALSE - The resultant in a vector addition diagram is drawn from the tail of the first vector (the starting point) to the head of the last vector (the finishing point). e. TRUE - Suppose that A = 3 units and B = 4 units and that the two vectors are directed at right angles to each other. The vector sum or resultant of A + B is 5 units, which is clearly greater than either one of the vectors being added. In general, the resultant in such a case will be represented on a vector addition diagram as the hypotenuse of a right triangle. The hypotenuse is always greater than the other two legs of the triangle. So this statement is always true. f. TRUE - Suppose that A = 3 units and B = 4 units and that the two vectors are directed at right angles to each other. The vector sum or resultant of A + B is 5 units whereas the arithmetic sum is 7 units. In this case and in all cases, the vector sum of two right angle vectors will always be less than the arithmetic sum. That is, Sqrt(a2 + b2) will always be less than a + b. g. FALSE - When a vector subtraction operation is performed, it is usually advisable to simply convert it into a vector addition operation. This is accomplished by adding the negative of the vector which is being subtracted. So A - B would be equivalent to A + (-B). By so doing, a vector addition diagram can be used to determine the resultant.

5. Which of the following descriptions of moving objects accurately portray a projectile? List all that apply. a. an object which is moving through the air and not touching any surface b. a falling skydiver with an open parachute c. any object upon which air resistance is negligible d. a free-falling object e. an object upon which the only significant force is the force of gravity f. a falling feather g. a falling feather in a vacuum chamber h. a falling feather in a falling vacuum chamber. Answer: DEGH A projectile is an object upon which the only force is gravity. Air resistance must be negligible or nonexistent. Other forces resulting from people or things pulling or pushing, attached strings or contact with surfaces must not be present. a. NO - A plane moves through the air and is not touching any surface. Yet, a plane is clearly not a projectile. b. NO - A falling skydiver typically experiences considerable air resistance. It is popular to describe such skydivers as being in free fall. This is an erroneous use of the term. c. NO - As you sit in your chair, air resistance is negligible. You are certainly not a projectile (at least, we hope not). d. YES - A projectile is an object in free fall. e. YES - An object upon which the only significant force is gravity fits the definition of a projectile (provided that significant means "having an influence"). f. NO - Falling feathers encounter air resistance which impedes the downward acceleration and causes the feather to fall at nearly a constant velocity. g. YES - When a feather is allowed to fall in a vacuum, air resistance is eliminated and the feather can free fall. h. YES - When a feather is allowed to fall in a vacuum and the vacuum is free-falling as well, air resistance is eliminated and an observer would notice that both the vacuum chamber and the feather are in free fall.

6. Which of the following statements are true of projectiles? List all that apply. a. A projectile is a free-falling object. b. A projectile experiences negligible or no air resistance. c. A projectile must be moving in the downward direction. d. A projectile must be accelerating in the downward direction. e. A projectile does not have to have horizontal motion. f. A projectile could begin its projectile motion with a downward velocity. g. A projectile does not need to be "falling." Answer: ABDEF and possibly G a. TRUE - Free-falling objects, like projectiles, are objects upon which the only significant force is gravity. b. TRUE - The only force on a projectile is gravity; air resistance must not be present or must not have an influence upon the motion of the projectile. c. FALSE - Projectiles can be moving either upward or downward or at an angle to the vertical. They must however be accelerating downward, consistent with gravity's effect on an object. d. TRUE - The force of gravity acts directly downwards upon an object, causing a downward acceleration. Any projectile must be accelerating downwards regardless of other features of its motion. e. TRUE - A projectile could be moving strictly in a vertical direction with no horizontal motion. A ball thrown straight up in the air would be such a case. f. TRUE - There is no rule about which direction a projectile must be moving at the instant it is projected. It could begin its motion with a initial downward velocity. g. TRUE - The word "falling" can mean different things to different people. If "falling" involves moving in the downward direction at all instants in time, then a projectile does not need to be "falling." To many, "falling" means being pulled downward by gravity's force. In this case, a projectile must be "falling."

7. Which of the following statements are true of the horizontal motion of projectiles? List all that apply. a. A projectile does not have a horizontal velocity. b. A projectile with a rightward component of motion will have a rightward component of acceleration. c. The horizontal velocity of a projectile changes by 9.8 m/s each second. d. A projectile with a horizontal component of motion will have a constant horizontal velocity. e. The horizontal velocity of a projectile is 0 m/s at the peak of its trajectory. f. The horizontal velocity of a projectile is unaffected by the vertical velocity; these two components of motion are independent of each other. g. The horizontal displacement of a projectile is dependent upon the time of flight and the initial horizontal velocity. h. The final horizontal velocity of a projectile is always equal to the initial horizontal velocity. i. As a projectile rises towards the peak of its trajectory, the horizontal velocity will decrease; as it falls from the peak of its trajectory, its horizontal velocity will decrease. j. Consider a projectile launched from ground level at a fixed launch speed and a variable angle and landing at ground level. The horizontal displacement (i.e., the range) of the projectile will always increase as the angle of launch is increased from 0 degrees to 90 degrees. k. Consider a projectile launched from ground level at a fixed launch angle and a variable launch speed and landing at ground level. The horizontal displacement (i.e., the range) of the projectile will always increase as the launch speed is increased. Answer: DFGHK a. FALSE - Many projectiles are moving from left to right and from right to left as they simultaneously free fall. Such projectiles have a horizontal motion. While a projectile can have a horizontal motion, it cannot have a horizontal acceleration. Whatever motion which it has in the horizontal dimension, must be motion with a constant velocity. b. FALSE - A projectile with a rightward motion (in addition to a vertical motion) will have a constant velocity in the rightward direction. This is to say that it has no horizontal acceleration. c. FALSE - A projectile has a constant horizontal velocity. The vertical velocity will change by 9.8 m/s each second. d. TRUE - Absolutely true! Projectiles are objects being acted upon by gravity alone. As such, there is a vertical acceleration but no horizontal acceleration. The horizontal velocity of a projectile is either zero or a constant nonzero value. e. FALSE - The vertical velocity of a projectile is 0 m/s at the peak of its trajectory; but the horizontal component of the velocity at the peak is whatever the value was when first launched. f. TRUE - For any two dimensional motion (whether projectile motion or riverboat problems or ...), perpendicular components of the motion are independent of each other. Any alteration in a vertical component will not effect the horizontal components of motion. g. TRUE - The horizontal displacement (x) can be calculated with the formula x = vox • t, where vox is the initial horizontal velocity and t is the time. These are the two variables which effect the horizontal displacement of a projectile. h. TRUE - Since there is no horizontal acceleration for a projectile, the initial horizontal velocity is equal to the final horizontal velocity. i. FALSE - This is a true description for the vertical component of the velocity. The horizontal velocity is unchanging throughout the trajectory of a projectile. j. FALSE - The range (or horizontal displacement) will increase as the angle is increased from 0 degrees to 45 degrees. The maximum range occurs at 45 degrees. As the angle is further increased to values greater than 45 degrees, the horizontal displacement decreases. k. TRUE - As the launch speed is increased, the components of the initial velocity (both the horizontal and the vertical) increase as well. This causes the projectile to stay in the air for a longer period of time and to be moving faster in the horizontal direction. The result is that increased launch speeds always lead to increased horizontal displacements.

8. Which of the following statements are true of the vertical motion of projectiles? List all that apply. a. The vertical component of a projectile's velocity is a constant value of 9.8 m/s. b. The vertical component of a projectile's velocity is constant. c. The vertical component of a projectile's velocity is changing. d. The vertical component of a projectile's velocity is changing at a constant rate. e. A projectile with an upward component of motion will have a upward component of acceleration. f. A projectile with an downward component of motion will have a downward component of acceleration. g. The magnitude of the vertical velocity of a projectile changes by 9.8 m/s each second. h. The vertical velocity of a projectile is 0 m/s at the peak of its trajectory. i. The vertical velocity of a projectile is unaffected by the horizontal velocity; these two components of motion are independent of each other. j. The final vertical velocity of a projectile is always equal to the initial vertical velocity. k. The vertical acceleration of a projectile is 0 m/s/s when it is at the peak of its trajectory. l. As a projectile rises towards the peak of its trajectory, the vertical acceleration will decrease; as it falls from the peak of its trajectory, its vertical acceleration will decrease. m. As a projectile rises towards the peak of its trajectory, the vertical acceleration is directed upward; as it falls from the peak of its trajectory, its vertical acceleration is directed downward. n. The peak height to which a projectile rises above the launch location is dependent upon the initial vertical velocity. o. As a projectile rises towards the peak of its trajectory, the vertical velocity will decrease; as it falls from the peak of its trajectory, its vertical velocity will decrease. p. Consider a projectile launched from ground level at a fixed launch speed and a variable angle and landing at ground level. The vertical displacement of the projectile during the first half of its trajectory (i.e., the peak height) will always increase as the angle of launch is increased from 0 degrees to 90 degrees. q. Consider a projectile launched from ground level at a fixed launch angle and a variable launch speed and landing at ground level. The vertical displacement of the projectile during the first half of its trajectory (i.e., the peak height) will always increase as the launch speed is increased. Answer: CDFGHINPQ a. FALSE - The vertical component of a projectile's velocity is constantly changing. It is the acceleration which has a value of 9.8 m/s/s. b. FALSE - Projectiles are objects being acted upon by gravity alone. As such, there is a vertical acceleration; the vertical velocity is not constant, but changing. c. TRUE - See part b above. d. TRUE - A projectile has a vertical acceleration of 9.8 m/s/s throughout the entire trajectory. This acceleration value is constant. This means that the vertical velocity changes by the same amount - 9.8 m/s - during each second of its motion. There is a change in the vertical velocity by a constant amount. e. FALSE - All projectiles experience a downward acceleration, whether they are moving upward or downward. The upward-moving projectiles have an upward velocity, but the actual velocity values are getting smaller; that is, the projectile is slowing down on the way to its peak. f. TRUE - This is a true statement. It could also be said that a projectile with an upward component of motion also has a downward acceleration. All projectiles accelerate in the downward direction. Period. g. TRUE - This is absolutely true . h. TRUE - At the peak of its trajectory, a projectile is in the process of changing directions. The vertical velocity must change from a positive value (+ for upward) to a negative value (- for downward). This transition means that the value for the vertical velocity must at sometime be in between a + and - number. The in-between number is 0 m/s and this occurs at the peak. i. TRUE - For any two dimensional motion (whether projectile motion or riverboat problems or ...), perpendicular components of the motion are independent of each other. Any alteration in a vertical component will not effect the horizontal components of motion. j. FALSE - A projectile launched at an angle forms a parabolic trajectory. Suppose that one were to trace a projectile's motion forward in time from the peak and backwards in time from the peak. If done, one would find that the vertical velocity value has the same magnitude for equal amounts of times traced forward and backward from the peak. So for the same time before and after the peak, a projectile has the same speed. However, some projectiles are not launched from the same height at which they land. The final height is not the same as the initial height and as such the time to rise to the peak is not equal to the time to fall from the peak. In such instances, the initial vertical velocity is not equal to the final vertical velocity. k. FALSE - No! No! No! The vertical velocity is 0 m/s at the peak and the vertical acceleration is -9.8 m/s/s throughout the entire trajectory. l. FALSE - This would be a true description of the vertical velocity. But the vertical acceleration is a constant value of 9.8 m/s/s throughout the entire trajectory. m. FALSE - Not only is the magnitude of the vertical acceleration a constant value throughout a projectile's trajectory, the direction is constant as well. Projectile's at all times regardless of any other variable will accelerate downwards at 9.8 m/s/s. This is perhaps the most important truth to digest about projectiles. n. TRUE - The initial vertical velocity has an effect on the time taken by a projectile to rise towards its peak. It also effects the average speed of the projectile as it rises towards its peak. As a result, any alteration in the vertical velocity will alter the peak height of the projectile. o. FALSE - Upward-rising projectiles have a downward acceleration; this means they are slowing down as they rise. The magnitude of their velocity is decreasing. Downward-moving projectiles also have a downward acceleration; this means they are speeding up. The magnitude of their velocity is increasing. p. TRUE - An increase in the angle of launch (from 0 to 90 degrees) will always increase the vertical component of the initial velocity (viy). This increase in viy will lead to increased times for the projectile rising towards its peak. And an increased angle causes the projectile to move with a greater average speed during its path towards its peak. Both of these effects lead to the outcome that the peak height of a projectile will increase as the angle of launch increases from 0 to 90 degrees. q. TRUE - As the launch speed is increased, the components of the initial velocity (both the horizontal and the vertical) increase as well. This causes the projectile to stay in the air for a longer period of time and to be moving faster in the vertical direction. The result is that increased launch speeds always lead to increased heights for projectiles.

9. Which of the following statements are true of the time of flight for a projectile? List all that apply. a. The time that a projectile is in the air is dependent upon the horizontal component of the initial velocity. b. The time that a projectile is in the air is dependent upon the vertical component of the initial velocity. c. For a projectile which lands at the same height that it is projected from, the time to rise to the peak is equal to the time to fall from its peak to the original height. d. For the same upward launch angles, projectiles will stay in the air longer if the initial velocity is increased. e. Assume that a kicked ball in football is a projectile. If the ball takes 3 seconds to rise to the peak of its trajectory, then it will take 6 seconds to fall from the peak of its trajectory to the ground. Answer: BCD a. FALSE - The time for a projectile to rise vertically to its peak (and subsequently fall back to the ground) is dependent upon the initial vertical velocity. Alteration in the horizontal velocity will only cause the projectile to have a greater horizontal displacement (x). b. TRUE - Absolutely true. Projectiles with a greater vertical component of initial velocity will be in the air for longer amount of times (assuming that the direction of viy is upward). An alteration in the viy value will alter the time of flight of the projectile, regardless of the direction of viy. c. TRUE - For projectiles launched at upward angles and landing at the original height, the time to the rise to the peak equals the time to fall from the peak. If it takes 3 seconds to rise upward, it will take 3 seconds to fall. d. TRUE - For a constant launch angle, an increase in the initial velocity (vi) will increase the vertical velocity (viy). This results in an increased time for the projectile to decelerate to 0 m/s as it rises towards its peak. So the projectile takes longer to get to the peak, longer to fall from the peak and overall is in the air for a longer time. e. FALSE - Close, but very false. If it takes 3 seconds to rise to the peak, then it takes 3 seconds to fall from the peak; The 6 seconds is the total time of flight of the projectile.

10. If two displacement vectors of 6 meters and 8 meters (with varying directions) are added together, then the resultant could range anywhere between ___ meters and ___ meters.

a. 0, 48 b. 1.33, 48 c. 0, 14 d. 2, 14

e. ... nonsense! No such prediction can be made.

f. ... nonsense! A prediction can be made but none of these choices are correct.

Answer: D The vector sum of 6.0 meters and 8.0 meters will be the greatest if they are added together in the same direction; that would produce a resultant of 14 meters. The vector sum of 6.0 meters and 8.0 meters will be the smallest if they are added together in the opposite direction; that would produce a resultant of 2.0 meters. An animation depicting the various resultants of 6.0 meters added to 8.0 meters at varying directions is shown on a separate page. See animation.

11. Three vectors are added following the rules of vector addition. A fourth vector is drawn from the tail of the first vector to the head of the last vector. This fourth vector is referred to as ____. a. the equilibrant b. the hypotenuse c. the resultant d. a mistake

Answer: C The resultant represents the result of adding two or more vectors. On a scaled vector addition diagram, the resultant is always drawn from the tail of the first vector to the head of the last vector.

12. TRUE or FALSE: The order in which vectors is added will effect the end result. a. True b. False

Answer: B The order in which vectors is added does not effect the magnitude or direction of the resultant. A + B + C gives the same resultant as B + C + A and the same resultant as A + C + B. An animation depicting the the head-to-tail addition of five vectors in three different orders is shown on a separate page. See animation.

13. Vector A is directed northward and vector B is directed eastward. Which of the following vector addition diagrams best represent the addition of vectors A and B and the subsequent resultant? Answer: E If the head-to-tail method (sometimes referred to as the triangle method) is being used, then the tail of B should be drawn starting at the head of A. Both diagrams D and E show this. Then the resultant should be drawn from the tail of A to the head of B (which is not shown in diagram D). There is also a parallelogram method for adding vectors. In this method, the tails of the two vectors are placed together (as in diagrams A and B). Then a parallelogram should be drawn with the two vectors forming the adjacent sides of the parallelogram. The resultant is drawn from the tails of the vectors to the opposite corner of the parallelogram. This is not done correctly on diagrams A and B. Diagrams C and F do not resemble any (accurate) vector addition method known to humankind.

14. When adding vector B to vector A geometrically (or graphically) using the head to tail method, the resultant is drawn from ____ to the ____. a. head of A, tail of B b. tail of A, head of B c. head of B, tail of A d. tail of B, head of A

Answer: B Adding vector B to vector A is equivalent to saying A + B. In such an instance, A is drawn first, then B is drawn with its tail at the head of A. Finally, the resultant is drawn from the tail of the first vector (A) to the head of the last vector (B).

Use the following vector addition diagrams for Questions #15-#20. 15. Which one of the following vector addition equations is shown in Diagram 1? a. A + B = C b. A + C = B c. B + C = A d. B + A = C e. C + B = A f. C + A = B g. None of these

16. Which one of the following vector addition equations is shown in Diagram 2? a. A + B = C b. A + C = B c. B + C = A d. B + A = C e. C + B = A f. C + A = B g. None of these

17. Which one of the following vector addition equations is shown in Diagram 3? a. A + B = C b. A + C = B c. B + C = A d. B + A = C e. C + B = A f. C + A = B g. None of these

18. Which one of the following vector addition equations is shown in Diagram 4? a. A + B = C b. A + C = B c. B + C = A d. B + A = C e. C + B = A f. C + A = B g. None of these

Answers to questions #15-#18: 15. B 16. A 17. E 18. G Vectors are added by a head-to-tail method and the resultant is drawn from the tail of the first vector to the head of the last vector. So if two vectors are added - say B is added to A (as in A + B) - then first A is drawn and the tail of B is placed at the head of A. The resultant is drawn with its tail at the tail of A and its head at the head of B. Thus, when the tails of two vectors are seen connected, one of the vectors is the resultant and the other is the first vector being added. And when the heads of two vectors are seen connected, one of the vectors is the resultant and the other is the second vector being added. These principles can be applied to answer the above questions. Note that in question 18, there is not a single vector which is drawn from the tail of one vector to the head of another vector. Thus, no resultant is drawn. One could say that the diagram shows that A + B + C = 0.

20. Consider the magnitude and direction of vectors B and C as shown in Diagram 2 above. Which one of the following diagrams would represent C - B = R? Answer: C The subtraction of B from C is equivalent to adding the negative of B to C. That is, C - B = C + (-B). The negative of a vector is simply the same vector drawn in the opposite direction. The correct answer should be a diagram which shows C being drawn first. Then at the head of C, the tail of a vector pointing in the opposite direction of B should be drawn. This is shown in both diagrams B and C. The resultant should then be drawn from the tail of C to the head of -B. This is not shown in diagram B.

22. The vector sum (magnitude only) of 32.0 m, north + 41.0 m, west is ___ m. a. 9.00 b. 36.5 c. 38.0 d. 52.0 e. 73.0 f. 128 g. 2.70 x 103 h. None of these

Answer: D Like question #21 above, these two vectors are directed at right angles to each other. The Pythagorean theorem can be used to determine the resultant of their sum. The resultant (R) is equal to the square root of the sum of the squares of the two vectors being added. R = Sqrt (A2 + B2) R = Sqrt [ (32.0 m)2 + (41.0 m)2 ] = Sqrt [ (1024 m2) + (1681 m2) ] = Sqrt (2705 m2) R = 52.010 m = ~52.0 m

Use the diagram below to answer Questions #23-#28. Each square on the diagram represents a 20-meter x 20-meter area. 23. If a person walks from D to H to G to C, then the distance walked is ____ meters. a. 128 b. 180 c. 401 d. 460 e. 480 f. 533 g. 620 h. None of these

Answer: F Distance is a scalar quantity and adding together three scalars is as simple as adding the three numbers arithmetically. However, since the first of the three legs of this trip is not due East, West, North or South, determining the distance for this leg requires that the Pythagorean theorem is used. That is, walking from D to H is equivalent to walking 140 meters east (7 squares) and 160 meters (8 squares) south. The distance from D to H is thus 212.6 meters - Sqrt [ (140 m)2 + (160 m)2 ]. The walk from H to G is a distance of 80 meters (4 squares) and the walk from G to C is 240 meters (12 squares). The total distance walked is the sum of 212.6 m + 80 m + 240 m = 532.6 m = ~533 m.

25. If a person walks from D to H to G to C, then the direction of the displacement is ___ degrees (as measured counterclockwise from East). a. 39 b. 51 c. 53 d. 217 e. 219 f. 231 g. 233 h. None of these

Answer: E

As discussed in the previous problem, the resultant stretches south 80 meters (4 squares) and west 100 meters (5 squares). This is shown in the diagram at the right. These two parts of the resultant vector can be combined with a trigonometric function to determine the angle theta. The angle theta is approximately 38.7 degrees. This is not the direction of the resultant displacement but simply the angle between the displacement vector and due West. The convention used to express the direction of the vector is to measure the counterclockwise angle of rotation from due East. So in this instance, the direction is 180 degrees + 38.7 degrees or ~219 degrees.

26. If a person walks from H to E to C to G, then the distance walked is ____ meters. a. 80. b. 240. c. 253 d. 333 e. 493 f. 560. g. 640. h. None of these Answer: E Distance is a scalar quantity and adding together three scalars is as simple as adding the individual distances of the three legs is as simple as adding the distances. arithmetically. The problem is simplified in that the first two legs of the trip are along the same line segment. The stop at E can be disregarded since its is merely a point on the line from location H to location C. That is HE + EC = HC. Since the HC line segment does not stretch not due East, West, North or South, determining the distance for this leg requires that the Pythagorean theorem is used. That is, walking from H to C is equivalent to walking 240 meters west (12 squares) and 80 meters (4 squares) north. The distance from D to H is thus 253.0 meters - Sqrt [ (240 m)2 + (80 m)2 ]. The walk from H to E to C is a distance of 253 meters and the walk from C to G is 240 meters (12 squares). The total distance walked is the sum of 253 m + 240 m = ~493 m.

Use the following diagram for Questions #29-#33. In the diagram, a riverboat is shown starting at position A on the east bank of a river. The boat heads towards position B (a point directly across the river from A) with a speed of 3.8 m/s. But because of a current with a speed of 1.8 m/s, the boat lands on the west bank of the river at position C, a location downstream from B. The width of the river (dacross) is 86.4 meters.

29. The magnitude of the resultant velocity of the boat is ____ m/s. a. 1.8 b. 2.0 c. 3.4 d. 3.8 e. 4.2 f. 5.6 g.11 h. None of these

Answer: E This is a case of a boat crossing a river by use of its motor. The motor allows the boat to travel 3.8 meters towards the opposite shore in every second. The river is flowing south, carrying the boat a distance of 1.8 meters down the river every second. The resultant velocity is simply the vector sum of these two individual velocities. Since these two velocities are at right angles to each other, the vector sum can be determined using the Pythagorean theorem.

2 2 2 vresultant = (3.8 m/s) + (1.8 m/s)

2 2 vresultant = Sqrt[ (3.8 m/s) + (1.8 m/s) ]

vresultant = 4.2 m/s

30. The direction of the resultant velocity of the boat is ____ m/s. a. 0 b. 18. c. 712 d. 90. e. 108 f. 162 g. 342 h. None of these

Answer: H

The direction of the resultant velocity is in a southeastern direction. That puts the vector in the third quadrant with a direction somewhere between 180 degrees and 270 degrees. The exact angle can be determined if the angle theta is found using trigonometry. Theta is the angle that the resultant velocity (the red vector in the diagram at the right) makes with due West. This angle can be found suing the tangent function. The work is shown at the right. The angle theta is found to be 25.3 degrees. The actual direction as measured as the counterclockwise angle of rotation with due East is 180 degrees plus 25.3 degrees. This would be 205.3 degrees.

31. The time required for the boat to cross the 86.4 m wide river is ___ seconds. a. 4.2 b. 15 c. 21 d. 23 e. 48 f. None of these

Answer: D The motor allows the boat to travel 3.8 meters towards the opposite shore in every second. The boat has 86.4 meters to travel from shore to shore. (The presence of the current does not alter the width or shore-to- shore distance.) The time to cross the river can be calculated from the river width and the boat velocity using the v = d/t equation. Rearranging the equation to solve for t yields t = d/v t = (86.4 m) / (3.8 m/s) = ~23 s (unrounded 22.7 s)

32. Location C is the location where the boat ultimately lands on the opposite shore. What is the distance from location B to location C. a. 37 b. 41 c. 78 d. 86 e. 96 f. 180 g. 2.0 x 102 h. None of these

Answer: B The distance that location C is downstream from B is mathematically related to the speed of the river and the time to cross the river. The distance can be calculated by multiplying the current speed by the time to cross the river.

ddownstream = vriver • t

ddownstream = (1.8 m/s) • (22.7 s)

ddownstream = ~41 m

33. If the current on a particular day was flowing with two times the velocity, then the time to cross the river would be ____. a. two times greater b. one-half as great

c. greater, but not two times greater d. less, but not one-half the original time

e. the same as it was when the current flowed at 1.8 m/s.

Answer: E For this situation of a boat heading straight across the river, the current velocity is directed perpendicular to the boat velocity. These two components of the boats resulting motion are independent of each other. The boat velocity makes the sole contribution to the ability of the boat to cross the river. The river velocity only carries the boat southward down the river. So an alteration of the river velocity would have no effect on the time required for the boat to cross the river.

35. A football is kicked into the air at an angle of 45 degrees with the horizontal. At the very top of the ball's path, its velocity is ______.

a. entirely vertical b. entirely horizontal

c. both vertical and horizontal d. not enough information given to know.

Answer: B As a projectile rises towards its peak, its horizontal velocity remains constant while its vertical velocity decreases. At its peak, its vertical velocity becomes 0 m/s. At this instant in time, the velocity is entirely horizontal; there is no vertical component to the velocity.

36. A football is kicked into the air at an angle of 45 degrees with the horizontal. At the very top of the ball's path, its acceleration is ______. (Neglect the effects of air resistance.)

a. entirely vertical b. entirely horizontal

c. both vertical and horizontal d. not enough information given to know.

Answer: A As a projectile rises towards its peak, its horizontal velocity remains constant while its vertical velocity decreases. This is to say that the acceleration of the object is vertical, not horizontal. At the peak and everywhere throughout the trajectory, there is a vertical (downward) acceleration. In fact, a projectile is an object upon which the only force is gravity. This force causes an acceleration which is in the same direction as the force - downward.

37. A football is kicked into the air at an angle of 45 degrees with the horizontal. At the very top of the ball's path, the net force acting upon it is ______. (Neglect the effects of air resistance.)

a. entirely vertical b. entirely horizontal c. both vertical and horizontal d. not enough information given to know.

Answer: A A projectile is an object upon which the only force is gravity. Since no other forces act upon the object, the net force would be downward.

38. At what point in its path is the horizontal component of the velocity (vx) of a projectile the smallest?

a. The instant it is thrown. b. Halfway to the top. c. At the top.

d. As it nears the top. e. It is the same throughout the path.

Answer: E As a projectile rises towards its peak, its horizontal velocity remains constant while its vertical velocity decreases. This is to say that the acceleration of the object is vertical, not horizontal. Having a constant horizontal velocity, there is no point along the trajectory where the vx value is smaller than at other points.

39. At what point in its path is the vertical component of the velocity (vy) of a projectile the smallest?

a. The instant it is thrown. b. Halfway to the top. c. At the top.

d. As it nears the top. e. It is the same throughout the path.

Answer: C As a projectile rises towards its peak, its horizontal velocity remains constant while its vertical velocity decreases. During the upward portion of its trajectory, the vy continuously decreases until it becomes 0 m/s at the peak. Thus, the vy is as small as it will ever be when it is at the peak of the trajectory.

40. An airplane that flies at 100 km/h in a 100 km/h hurricane crosswind has a speed (relative to the ground) of ____.

a. 0 km/h b. 100 km/h c. 141 km/h d. 200 km/h

Answer: C When an object such as a plane or a boat moves within a medium which is moving relative to the ground, the speed of the plane or boat (as measured by its speedometer) will not be the same as the speed as measured by a person on the ground (ground speed). The ground speed (i.e., resultant velocity produced by the combination of the plane's speed and the wind speed) can be determined by adding the plane speed and wind speed as vectors. In this case, the two vectors are at right angles, so the resultant velocity can be determined using the Pythagorean theorem. R2 = (100 km/hr)2 + (100 km/hr)2 R = SQRT( (100 km/hr)2 + (100 km/hr)2 ) R = 141 km/hr

42. Roll a bowling ball off the edge of a table. As it falls, its horizontal component of velocity ___.

a. decreases b. remains constant c. increases

Answer: B Once the ball leaves the table's edge, it becomes a projectile. As it falls, its horizontal velocity remains constant while its vertical velocity decreases. This is to say that the acceleration of the object is vertical, not horizontal. In fact, a projectile is an object upon which the only force is gravity. This force causes an acceleration which is in the same direction as the force - downward.

43. A bullet is fired horizontally and hits the ground in 0.5 seconds. If it had been fired with twice the speed in the same direction, it would have hit the ground in ____. (Assume no air resistance.)

a. less than 0.5 s. b. more than 0.5 s. c. 0.5 s.

Answer: C Once the bullet leaves the muzzle, it becomes a projectile (assuming no air resistance). As it falls, its horizontal velocity remains constant while its vertical velocity decreases. The force of gravity acts upon the bullet to cause its downward acceleration. The motion of the bullet in the downward direction is independent of the motion in the horizontal direction. That is to say, any alteration in a horizontal aspect of its motion will not effect the motion in the vertical direction. The time to fall vertically to the ground is not effected by the horizontal speed of the projectile. It would still take 0.5 seconds to fall to the ground from this height regardless of the horizontal speed.

46. Express the direction of each of the following vectors in the diagram below.

A: 0 deg B: 90 deg C: 135 deg D: 205 deg E: 270 deg F: 340 deg Answer: See answers above. The convention that the direction of a vector is measured as the counterclockwise angle of rotation of that vector from due east is used. As such, an eastward vector (A) has a direction of 0 degrees; a northward vector (B) has a direction of 90 degrees; and a southward vector (E) has a direction of 270 degrees. A protractor must be used to determine the directions of C, D and F. Vector C is in the second quadrant, so its direction is between 90 degrees and 180 degrees. The angle between East and vector C can be measured. Vector D is in the third quadrant; its direction is between 180 degrees and 270 degrees. A protractor can be used to measure the angle between West (180 degrees) and vector D. This angle measure can be added to 180 degrees to get the answer. A similar procedure is repeated with Vector F.

47. In the following diagrams, two vectors are being added and the resultant is drawn. For each diagram, identify which vector is the resultant and write the equation (e.g., A + B = C). a. b.

The resultant is vector B. The resultant is vector B. The equation is C + A = B. The equation is A + C = B. c. d.

The resultant is vector C. The resultant is vector A. The equation is A + B = C. The equation is C + B = A.

Answer: See answers above. Vectors are added by a head-to-tail method and the resultant is drawn from the tail of the first vector to the head of the last vector. So if two vectors are added - say B is added to A (as in A + B) - then first A is drawn and the tail of B is placed at the head of A. The resultant is drawn with its tail at the tail of A and its head at the head of B. Thus, when the tails of two vectors are seen connected, one of the vectors is the resultant and the other is the first vector being added. And when the heads of two vectors are seen connected, one of the vectors is the resultant and the other is the second vector being added. These principles can be applied to answer the above questions. 48. Consider the five vectors shown below; they are labeled A, B, C, D and E.

Use the stated magnitude and direction of these vectors to sketch a vector addition diagram for the following vector operations and determine the resultant. You only need to construct an approximate sketch; there is no need to do a scaled diagram.

A + B + C A + C + B A + E + D

E + D + C C - D B + E - C

Answer: See diagrams above. Vectors are added using a head-to-tail method. The magnitude and direction of the vectors are maintained when the addition (and subtraction) operations are performed. In your sketched diagrams, you should be sure to:  place the tail of the second vector in the equation at the head of the first vector in the equation  label each vector (A, B, C ...)  draw the resultant from the tail of the first vector to the head of the last vector  label the resultant A subtraction operation is the same as adding the negative of a vector. So if D is to be subtracted from C, then simply add a vector whose magnitude is the same as D but is directed in the opposite direction.

49. A hiker's motion can be described by the following three displacement vectors. 22.0 km, 45 degrees + 16.0 km, 135 degrees + 12.0 km, 270 degrees Add the three displacement vectors using the head-to-tail method of vector addition. Then answer the following two questions. (a) What is the distance walked by the hiker? (b) What is the resulting displacement of the hiker? Answer: Distance = 50.0 km; Displacement = 15.5 km, 74 degrees Distance is a scalar quantity and does not regard the direction of the individual legs of the hike as being important. So the distance is simply the arithmetic sum of the distances of the three individual legs of the hike. Distance = 22.0 km + 16.0 km + 12.0 km = 50.0 km Displacement is a vector quantity which is equal to the vector sum of the individual displacements of the three-legged hike. A scaled vector addition diagram serves as a useful tool in determining this sum. A convenient scale (such as 1 cm = 4 km or 1 cm = 5 km or 1 cm = 8 km) is chosen and the vectors are added using the head-to-tail method. This is shown in the diagram below.

Once the resultant is drawn, the magnitude (in cm) can be measured and the scale can be used to convert to kilometers. The direction can be measured from the diagram as the counterclockwise angle of rotation from due East. The displacement is 15.5 km at 74 degrees.

50. Solve the following problem using a vector addition diagram.

A hiker hikes for 12.0 km, west. She then makes a turn and hikes in a straight line until she reaches her final destination. If the resultant displacement is 21.0 km at 147 degrees, then what was the displacement (magnitude and direction) of the second leg of the hike? Answer: 12.7 km, 116 degrees This is a unique problem in that the resultant is given and one of the two vectors is given. The problem requests the magnitude and direction of the second vector. In the diagram below, the first vector is drawn and the resultant is drawn beginning at the tail of the first vector. The second vector would stretch from the head of the first vector to the head of the resultant. This can be drawn in the diagram and its magnitude and direction can be determined.

51. Consider the trajectory diagram shown below for a horizontally launched projectile. On the diagram, draw vector arrows representing the vx and vy velocity components during the course of the motion. The length of the arrows should represent the magnitude of the velocity components. Label each component. (Note that the velocity components are already shown for the first position.) Answer: See diagram above. A projectile has a downward acceleration. There is no horizontal acceleration; rather, the horizontal velocity remains constant. Thus, the vx vector is of constant length (i.e., magnitude) throughout the trajectory. The downward acceleration means that the vy vector will be changing. As the object falls, it speeds up. Thus, the vy vector increases its length (i.e., magnitude) throughout the trajectory. This is shown in the diagram above.

52. Consider the trajectory diagram shown below for a non-horizontally launched projectile. On the diagram, draw vector arrows representing the vx and vy velocity components during the course of the motion. The length of the arrows should represent the magnitude of the velocity components. Label each component. (Note that the velocity components are already shown for the first position.) Answer: See diagram above. A projectile has a constant horizontal velocity and a changing vertical velocity. The changing vertical component of velocity is consistent with the fact that there is a vertical acceleration. As the projectile rises towards the peak of its trajectory, the vertical velocity decreases until it reaches 0 m/s at the very peak of the trajectory. As the projectile rises towards the peak of its trajectory, the vertical velocity increases. The vertical velocity upon falling is of the same magnitude and the opposite direction as any corresponding location of the same height during the rising motion. The horizontal velocity remains constant throughout the trajectory. These principles are shown in the diagram above.

53. Consider the diagram below. The blue path represents the trajectory of a projectile dropped from rest from the top of the path. (Each ball location represents the location after a 1-second interval of time.) The red path represents the trajectory of the same ball thrown horizontally in the absence of gravity. Construct a third path accurately showing the trajectory of a projectile launched with the same horizontal speed as the red ball. Show the location during the first four seconds of motion. Finally, the (x, y) coordinate position of location 1 is (0 m, -5 m) and the (x, y) coordinate position of location 2 is (15 m, 0 m). Determine the (x, y) coordinate positions of the four locations in your trajectory. (Assume g ~10 m/s/s.) Answer: See diagram above. A projectile such as this has both a horizontal and vertical component of motion. These two components of motion are independent of each other. The strictly vertical motion of the ball (the blue path above) depicts the vertical displacement at 1-second intervals. If a horizontal motion is imparted to the ball, its vertical displacement will not be effected since these two components are independent of each other. The strictly horizontal motion of the ball (the red path above) portrays the motion of the ball in the absence of gravity. The presence of gravity would cause the ball to accelerate downward. This vertical motion does not effect the horizontal motion since these two components are independent of each other. The green path represents the motion of the projectile when gravity acts upon it and a horizontal motion is imparted to it. The horizontal location at each 1-second instant in time is the same position as it was in the red path above it. And the vertical location at each 1-second instant in time is the same position as it was in the blue path to the left of it. The actual coordinate positions can be determined using the kinematic equations and the given time. For the x-coordinate, the displacement is the same each second - that is, there is a constant horizontal velocity. Since location 2 has a x-coordinate of 15 m, each consecutive location will be 15 m further than the one before it. The y-coordinate is determined by using the 0.5•a•t2 expression with times of 1 s, 2 s, 3 s, and 4 s. The simplification that g is ~10 m/s2 is used to simplify the math and highlight the concept.

54. Consider the map of the United States at the right. Given the scale that 1 cm = 500 km, use a protractor and a ruler to determine the displacement of a traveler for the listed trips. A couple trips include two "legs." Indicate both the magnitude and the direction (using the counterclockwise from East convention) for the resultant displacement vector of the traveler. Trip Magnitude Direction a. From Chicago (CHI) to Washington, DC (WSH) b. From Chicago (CHI) to Seattle (SEA) c. From Jackson Hole (JXH) to Houston (HST) d. From Denver (DEN) to Phoenix (PHX) to Washington (WSH) e. From Salt Lake City (SLC) to Reno (REN) to Miami (MIA)

Answer: See table above. a. The distance from CHI to WSH is ~2.6 cm. According to scale, this would be equivalent to ~1300 km. The direction of this displacement vector is measured as the counterclockwise angle of rotation from due East - 341 degrees. b. The distance from CHI to SEA is ~6.7 cm. According to scale, this would be equivalent to ~3400 km. The direction of this displacement vector is measured as the counterclockwise angle of rotation from due East - 165 degrees. c. The distance from JHX to HST is ~4.3 cm. According to scale, this would be equivalent to ~220 km. The direction of this displacement vector is measured as the counterclockwise angle of rotation from due East - 312 degrees. d. This is a two-legged trip. Yet the displacement only depends upon the initial and final locations. When the traveler is done with the trip, he/she is displaced by an amount equal to the distance from DEN to WSH. This is ~6.4 cm on the map. According to scale, this would be equivalent to ~3200 km. The direction of this displacement vector is measured as the counterclockwise angle of rotation from due East - 356 degrees. e. Again, the displacement only depends upon the initial and final locations. When the traveler is done with the trip, he/she is displaced by an amount equal to the distance from SLC to MIA. This is ~7.7 cm on the map. According to scale, this would be equivalent to ~3900 km. The direction of this displacement vector is measured as the counterclockwise angle of rotation from due East - 334 degrees.

55. A cannonball is launched from the top of a 125-meter high cliff with an initial horizontal speed of 20 m/s. The (x, y) coordinate position of the launch location is designated as the (0, 0) position. Determine the (x, y) coordinate positions of the cannonball at 1-second intervals during its path to the ground. Assume g = ~10 m/s/s, down.

Answer: See diagram above. The horizontal speed is 20 m/s throughout the trajectory. So the projectile will move a distance of 20 meters in each second. This horizontal displacement is combined with a falling motion which drops the projectile some vertical distance below the initial height of the cannon. The vertical displacement is given by the 0.5•a•t2 expression where the a value is stated as ~10 m/s2. Times of 1 s, 2 s, 3 s, etc. can be substituted into this expression to determine the vertical displacements of -5 m, -20 m, -45 m, -80 m, -125 m.

56. In the Vector Addition Lab, Anna starts at the classroom door and walks:  2.0 meters, West  12.0 meters, North,  31.0 meters, West,  8.0 meters, South  3.0 meters, East Using either a scaled diagram or a calculator, determine the magnitude and direction of Anna's resulting displacement. Answer: 30.3 meters, 172 degrees To insure the most accurate solution, this problem is best solved using a calculator and trigonometric principles. The first step is to determine the sum of all the horizontal (east-west) displacements and the sum of all the vertical (north-south) displacements. Horizontal: 2.0 meters, West + 31.0 meters, West + 3.0 meters, East = 30.0 meters, West Vertical: 12.0 meters, North + 8.0 meters, South = 4.0 meters, North The series of five displacements is equivalent to two displacements of 30 meters, West and 4 meters, North. The resultant of these two displacements can be found using the Pythagorean theorem (for the magnitude) and the tangent function (for the direction). A non-scaled sketch is useful for visualizing the situation.

Applying the Pythagorean theorem leads to the magnitude of the resultant (R). R2 = (30.0 m)2 + (4.0 m)2 = 916 m2 R = Sqrt(916 m2) R = 30.3 meters The angle theta in the diagram above can be found using the tangent function. tangent(theta) = opposite/adjacent = (4.0 m) / (30.0 m) tangent(theta) = 0.1333 theta = invtan(0.1333) theta = 7.59 degrees This angle theta is the angle between west and the resultant. Directions of vectors are expressed as the counterclockwise angle of rotation relative to east. So the direction is 7.59 degrees short of 180 degrees. That is, the direction is ~172 degrees.

59. A boat heads straight across a river which is 100. meters wide. For the following two combinations of boat velocities and current velocities, determine the resultant velocity, the time required to cross the river, and the distance traveled downstream. a. b. Given: Given: Boat velocity = 10.0 m/s, East Boat velocity = 8.0 m/s, East River velocity = 4.0 m/s, North River velocity = 5.0 m/s, South Calculate: Calculate: Resultant Vel. (mag. & dir'n): 11 m/s, 22 deg. Resultant Vel. (mag. & dir'n): 9.4 m/s, 328 deg. Time to cross river: 10.0 s Time to cross river: 13 s Distance traveled downstream: 40. m Distance traveled downstream: 63 m

Answer: See table above. The two velocity vectors (boat and river) are directed perpendicular to each other. They can be added using the Pythagorean theorem. The direction is found using the tangent function; it is expressed as the counterclockwise angle of rotation from due East. The time to cross the river is dependent upon the width of the river and the boat velocity. And the distance downstream is dependent upon the time that the boat is moving and the speed at which it moves downstream - the river velocity. a. b. Resultant Vel. (mag. & dir'n): Resultant Vel. (mag. & dir'n): R2 = (10.0 m/s)2 + (4.0 m/s)2 R2 = (8.0 m/s)2 + (5.0 m/s)2 R = SQRT ( (10.0 m/s)2 + (4.0 m/s)2 ) R = SQRT ( (8.0 m/s)2 + (5.0 m/s)2 ) R = SQRT(116 m2/s2) = 11 m/s (rounded from 10.8 R = SQRT(89 m2/s2) = 9.4 m/s (rounded from 9.43 m/s) m/s) dir'n = invtan(4.0/10.0) = 22 degrees dir'n = 360 deg. - invtan(5.0/8.0) = 328 degrees Time to cross river: Time to cross river: d = v*t ---> t = d/v d = v*t ---> t = d/v t = (100. m)/(10.0 m/s) = 10.0 s t = (100. m)/(8.0 m/s) = 13 s (rounded from 12.5 s) Distance traveled downstream: Distance traveled downstream: d = v*t = (4.0 m/s)*(10.0 s) = 40. m d = v*t = (5.0 m/s)*(12.5 s) = ~63 m

62. If a projectile is launched horizontally with a speed of 12.0 m/s from the top of a 24.6-meter high building. Determine the horizontal displacement of the projectile. Answer: x = 27.0 m This horizontally-launched projectile problem can be (and should be) solved in the same manner as the solution to #60 above. While #60 is broken down for you into nice steps, this problem is not so user- friendly. It is strongly recommended that you begin by listing known values for each of the variables in the kinematic equations. It is helpful to organize the information into two columns - a column of known horizontal information and a column of known vertical information.

Horizontal Motion Vertical Motion x = ??? y = -24.6 m (- means moving down)

vix = 12.0 m/s viy = 0.0 m/s (its launched horizontally)

ax = 0 m/s/s (true for all projectiles) ay = -9.8 m/s/s (true for all projectiles)

Since three pieces of y-information are now known, a y-equation can be employed to find the time.

2 y = viy*t + 0.5*ay*t Plugging and chugging the above values into this equation yields a time of 2.25 seconds. Now the t value can be combined with the vix and ax value and used in an x-equation

2 x = vix*t + 0.5*ax*t to yield the answer 27.0 m. More examples and discussion of these types of projectile problems are discussed elsewhere.

64. A projectile is launched horizontally from the top of a 45.2-meter high cliff and lands a distance of 17.6 meters from the base of the cliff. Determine the magnitude of the launch velocity. Answer: 5.79 m/s The best means of starting this problem is to list the known values for each of the variables in the kinematic equations. It is helpful to organize the information into two columns - a column of known horizontal information and a column of known vertical information.

Horizontal Motion Vertical Motion x = 17.6 m (the distance horizontally from cliff base) y = -45.2 m (it falls down from the cliff to the ground)

vix = ??? m/s viy = 0 m/s (it is horizontally launched)

ax = 0 m/s/s (true for all projectiles) ay = -9.8 m/s/s (true for all projectiles)

Since three pieces of y-information are now known, a y-equation can be employed to find the time. One useful equation is

2 y = viy*t + 0.5*ay*t in which case there will be two solutions: t = 3.0372 s and t = -3.0372 s. A full parabola which follows the above function would have to locations where the y coordinate is -45.2 m. One would be "forward in time" at 3.0372 seconds; and the other solution is at a location traced "backwards in time" from the launch time. Of course, the positive answer is the one which we need; it can can be used to determine the initial horizontal velocity (vix). Use the equation:

2 x = vix*t + 0.5*ax*t where t is 3.0372 s, ax = 0 m/s/s and x = 17.6 m. Plugging and chugging the above values into this equation yields the answer of 5.7948 m/s.

65. Two physics students stand on the top of their 3.29-meter second-story deck and launch a water balloon from a homemade winger. The balloon is launched upward at a speed of 45.2 m/s and an angle of 39.1 degrees. The balloon lands in a retention pond whose surface is 2.92 meters below grade. Determine the horizontal distance from launch location to landing location. Answer: 211 m This is a non-horizontally-launched projectile problem in which the initial velocity and launch angle are given. Three initial steps are always wisely taken before starting such a problem. First, determine the initial velocity components (vix and viy) using trigonometric functions. Second, construct a diagram of the physical situation. And third, organize known (and unknown) information in an "x-y table." These three steps are taken here. Quickly barging into a solution before giving the problem some pre-analysis often leads to a wasting of much time and ultimately a lot of confusion. The initial velocity and angle can be resolved into initial velocity components using the equations:

vix = vi * cos(theta) viy = vi * sin(theta)

This yields values of vix = 35.077 m/s and viy = 28.507 m/s. A diagram of the physical situation is shown.

Now the known values for each of the variables in the kinematic equations is listed in a table using a column for known horizontal information and a column for known vertical information.

Horizontal Motion Vertical Motion x = ??? y = -5.49 m (from initial + to height to final - height)

vix = 35.077 m/s (from trig. function) viy = 28.507 m/s (from trig. function)

ax = 0 m/s/s (true for all projectiles) ay = -9.8 m/s/s (true for all projectiles)

Since three pieces of y-information are now known, a y-equation can be employed to find the time. One useful equation is

2 y = viy*t + 0.5*ay*t in which case there will be two solutions: t = -0.1867 s and t = 6.004 s. A full parabola which follows the above function would have two locations where the y coordinate is -5.49 m. One location would be "forward in time" at 6.004 seconds; and the other solution is at a location traced "backwards in time" from the launch time. Of course, the positive answer is the one which we need; it can can be used to determine the horizontal displacement (x). Now use the equation:

2 x = vix*t + 0.5*ax*t where t = 6.004 s, ax = 0 m/s/s and vix = 35.077 m/s (as originally calculated using the trigonometric functions). Plugging and chugging the above values into this equation yields the answer of 211 m. (Wow! Those boys had better be careful.)

69. An unfortunate accident occurred on the toll way. A driver accidentally passed through a faulty barricade on a bridge (quite unfortunately). and landed in a pile of hay (quite fortunately). Measurements at the accident scene reveal that the driver plunged a vertical distance of 8.26 meters. The car carried a horizontal distance of 42.1 meters from the location where it left the bridge. If the driver was in a 65 mi/hr speed zone, then determine the amount by which the driver was exceeding the speed limit at the time of the accident. Assume that the contact with the barricade did not slow the car down. (1.00 m/s = 2.24 mi/hr) Answer: 72.6 mi/hr This is an example of a horizontally-launched projectile problem. Like all projectile problems, the best means of starting the problem is to list the known values for each of the variables in the kinematic equations. It is helpful to organize the information into two columns - a column of known horizontal information and a column of known vertical information.

Horizontal Motion Vertical Motion x = 42.1 m (the horizontal distance which is traveled) y = -8.26 m (it falls down from the cliff to the ground)

vix = ??? m/s viy = 0 m/s (it is horizontally launched)

ax = 0 m/s/s (true for all projectiles) ay = -9.8 m/s/s (true for all projectiles)

Since three pieces of y-information are now known, a y-equation can be employed to find the time. One useful equation is

2 y = viy*t + 0.5*ay*t in which case there will be two solutions: t = 1.2983 s and t = -1.2983 s. A full parabola which follows the above function would have two locations where the y coordinate is -8.26 m. One would be "forward in time" at 1.2983 seconds; and the other solution is at a location traced "backwards in time" from the launch time. Of course, the positive answer is the one which we need; it can can be used to determine the initial horizontal velocity (vix). Use the equation:

2 x = vix*t + 0.5*ax*t where t = 1.2983 s, ax = 0 m/s/s and x = 42.1 m. Plugging a chugging the above values into this equation yields the answer of 32.426 m/s. This is the speed at which the car leaves the bridge at the start of its projectile motion. Converting this to mi/hr involves multiplying by the (2.24 mi/hr) / (1 m/s) conversion ratio. The result is 72.6 mi/hr. (In reality, the car was traveling faster than this speed since the collision with the guard rail likely slowed the car down before it exited the bridge and began its projectile motion.)

56. In the Vector Addition Lab, Anna starts at the classroom door and walks:  2.0 meters, West  12.0 meters, North,  31.0 meters, West,  8.0 meters, South  3.0 meters, East Using either a scaled diagram or a calculator, determine the magnitude and direction of Anna's resulting displacement. Answer: 30.3 meters, 172 degrees To insure the most accurate solution, this problem is best solved using a calculator and trigonometric principles. The first step is to determine the sum of all the horizontal (east-west) displacements and the sum of all the vertical (north-south) displacements. Horizontal: 2.0 meters, West + 31.0 meters, West + 3.0 meters, East = 30.0 meters, West Vertical: 12.0 meters, North + 8.0 meters, South = 4.0 meters, North The series of five displacements is equivalent to two displacements of 30 meters, West and 4 meters, North. The resultant of these two displacements can be found using the Pythagorean theorem (for the magnitude) and the tangent function (for the direction). A non-scaled sketch is useful for visualizing the situation.

Applying the Pythagorean theorem leads to the magnitude of the resultant (R). R2 = (30.0 m)2 + (4.0 m)2 = 916 m2 R = Sqrt(916 m2) R = 30.3 meters The angle theta in the diagram above can be found using the tangent function. tangent(theta) = opposite/adjacent = (4.0 m) / (30.0 m) tangent(theta) = 0.1333 theta = invtan(0.1333) theta = 7.59 degrees This angle theta is the angle between west and the resultant. Directions of vectors are expressed as the counterclockwise angle of rotation relative to east. So the direction is 7.59 degrees short of 180 degrees. That is, the direction is ~172 degrees.

57. In a grocery store, a shopper walks 36.7 feet down an aisle. She then turns left and walks 17.0 feet straight ahead. Finally, she turns right and walks 8.2 feet to a final destination. (a) Determine the magnitude of the overall displacement. (b) Determine the direction of the displacement vector relative to the original line of motion. Answer: (a) 48.0 feet; (b) 21 degrees from the original line of motion This problem is best approached using a diagram of the physical situation. The three displacements are shown in the diagram below on the left. Since the three displacements could be done in any order without effecting the resulting displacement, these three legs of the trip are conveniently rearranged in the diagram below on the right. Now it is obvious from the diagram on the right that the three displacement vectors are equivalent to two perpendicular displacement vectors of 44.9 feet and 17 feet. These two vectors can be added together and the resultant can be drawn from the starting location to the final location. A sketch is shown below.

Since these displacement vectors are at right angles to each other, the magnitude of the resultant can be determined using the Pythagorean theorem. The work is shown below. R2 = (44.9 ft)2 + (17.0 ft)2 = 2305 ft2 R = Sqrt(2305 ft2) R = 48.0 feet The angle theta in the diagram above can be found using the tangent function. tangent(theta) = opposite/adjacent = (17.0 ft) / (44.9 ft) tangent(theta) = 0.3786 theta = invtan(0.3786) theta = 20.7 degrees This is the angle which the resultant makes with the original line of motion (the 36.7 ft displacement vector).

Newton's Laws Applications Review

Part A: Drawing Free-Body Diagrams

1. Construct free-body diagrams for the following physical situations. a. A ball is dropped from rest from b. After being thrown, a football is c. After reaching a terminal the top of a building. Assume moving upwards and rightwards velocity, a falling skydiver then negligible air resistance. towards the peak of its trajectory. opens up the parachute. Assume negligible air resistance.

In the absence of Fair, the only force In the absence of Fair, the only force acting upon the ball is gravity. It is acting upon the ball is gravity. It is a projectile. a projectile. Note that an upwards moving object does not need an When the terminal velocity was upwards force. Only an upwards reached, Fair = Fgrav. Then the accelerating object requires an parachute was opened, making Fair upwards force. even greater than before. This is represented by the larger arrow. d. An air track glider is gliding to e. A car is skidding to a stop while f. A downward moving elevator the right at constant velocity. travling to the right. (held by a cable) slows down.

There is no rightwards force. A There is no rightwards force. A rightwards force would only be rightwards force would only be The cable supplies the tension required if there is a rightwards required if there is a rightwards force. Since the elevator is moving acceleration. If the glider is gliding; acceleration. If the car is skidding downwards and slowing down, there is no mention of it being (wheels are locked), friction acts in there must be more upwards force pushed or pulled (Fapp) and if there a direction opposite its motion to than the downwards gravity force. is a constant velocity, there must be slow it down. balanced forces. g. A 25.0-N force is applied at a 30- h. A picture hangs summetrically i. A large crate slowly accelerated degree angle to a crate in order to by two wires oriented at angles to down a steep and rough inclined accelerate it rightward across a the vertical. plane. rough, horizontal surface.

The force exerted by a wire is a There are only three forces present. The applied force is upwards and tension force. With two wires, there The Fpar and Fperp are merely rightwards. Its upward component would be two upwards-pulling components of gravity; they are not tension forces. The down force is separate forces. The normal force is contributes to the upwards Fnorm to balance the force of gravity. (Note gravity. perpendicular to the surface (drawn in blue). the relative size of Fnorm.)

Part B: Conceptual Notions of Force and Motion Express your understanding of concepts of force, acceleration, and motion by answering the following questions. 2. Which of the following are always true of an object that is at equilibrium? Include all that apply. a. All the forces acting upon the object are equal. b. The object is at rest. c. The object is moving and moving with a constant velocity. d. The object has an acceleration of zero. e. There is no change in the object's velocity. f. The sum of all the forces is 0 N. g. All the forces acting upon an object are balanced.

Answer: DEFG These (DEFG) statements are always true. Statements ABC might be true but are not always true.

3. Which of the following are never true of an object that is at equilibrium? Include all that apply. a. The object is accelerating. b. The object is at rest. c. The object is moving in a circle at constant speed. d. All the forces acting upon the object are equal. e. The object is in free-fall. f. The object is falling and has reached its terminal velocity. g. There is a net force acting upon the object. h. The object is moving and moving with a constant velocity.

Answer: ACEG An object that is at equilibrium can never be accelerating; its acceleration MUST be 0 m/s/s. Thus, A is an answer; and because an object in free-fall (E) and an object moving in a circle (C) are also accelerating, they must be counted as answers as well. (NOTE: an object moving in a circle is changing its direction and as such has an acceleration.) If there is a net force (G), then by definition the object is not at equilibrium. Choices B, D, F and H could be true of an object at equilibrium (though none of them are always true).

4. Big Bubba has a mass of 100 kg on the earth. What is Big Bubba's mass on the moon where the force of gravity is approximately 1/6-th that of Earth's? ______Explain or show your work.

Answer: 100 kg Mass is the amount of matter present in the object and is independent of the weight of the object. The weight or force of gravity acting upon an object depends upon the mass and the acceleration of gravity (Fgrav = m•g). Because g is different on different planets and locations in the universe, the force of gravity or weight of an object will not always be the same. If Big Bubba goes to the moon, he will weigh less due to the decreased acceleration of gravity. However, he will still look the same, still have the same amount of matter, and still have a mass of 100 kg. KNOW the distinction between mass and weight.

5. Little Billie weight 360 N on Earth. What is Little Billie's mass on the moon where the force of gravity is approximately 1/6-th that of Earth's? ______Explain or show your work.

Answer: 37 kg (approx.) The acceleration of gravity on the moon is 1/6-th the value of earth's; this explains why the force of gravity is 1/6-th that of earth's. The force of gravity (Fgrav) on the moon is calculated using the equation Fgrav = m•g where g =1.6 m/s/s (on the moon). If Little Billie's weight on the moon is 1/6-th of 360 N, then Little Billie weighs 60 N on the moon. Inserting 60 N into the equation with 1.6 m/s/s as the value of g, Little Billie's mass can be calculated to be approximately 37 N. (Alternatively, simply find Billie's mass on earth from his weight of 360 N and then conclude that this mass value is not going to change when he goes to the moon.)

8. TRUE or FALSE: For an object resting upon a non-accelerating surface, the normal force is equal to the force of gravity. Answer: FALSE In this instance, the normal force could be equal to the force of gravity. But all that we can conclusively know is that the all the vertical forces sum up to 0 N. If the object is upon an incline, then the normal force will not be equal to the force of gravity. Or if there is another force with an upward or downward component, then the normal force is not equal to the force of gravity.

9. An object rests upon an inclined plane. If the angle of incline is increased, then the normal force would ______. a. increase b. decrease c. remain the same

Answer: B

The normal force is equal to the perpendicular component of the weight vector. So Fnorm = Fperp = m•g•cos(angle). If the incline angle is increased, the cos(angle) decreases towards 0; thus, the normal force decreases as well. Of course, the extreme is when the angle is 90 degrees and there is no normal force.

10. Three pictures of equal weight (20 N) are hung by wires in three different orientations. In which orientation are the wires least likely to break? ______Explain why.

Answer: 60-degree angle to the horizontal The tension in the wire will have the greatest impact upon its tendency to break. As the wire becomes most vertically oriented, the horizontal component of the tension force is reduced and the tension becomes less. The moral of the story -- to support the weight of a picture, one only needs to pull upwards, not leftwards and rightwards. Of course, the best arrangement would be two wires pulling completely vertically.

11. A 50-N force is applied at an angle of 30 degrees north of east. This would be the same as applying two forces at a. 43 N, east and 7 N, north b. 35 N, east and 15 N, north c. 25 N, east and 25 N, north d. 43 N, east and 25 N, north

Answer: D A 50-N force at 30 degrees would have two components - 43 N, east and 25 N, north. These two forces would be equal to the 50-N force at 30 degrees N of E. They are calculated as:

Feast = (50 N)*cos(30 deg)

Fnorth = (50 N)*sin(30 deg)

12. Which one(s) of the following force diagrams depict an object moving to the right with constant speed? Write all possible answers (if any) in the blank: ______

Explain your answer(s) to this question.

Answer: AC Only A and C could have a constant speed. Both B and D would have an acceleration that is directed rightwards. Do not confuse acceleration, motion, and force. A rightwards acceleration requires that there be a rightwards net force, On the other hand, a rightwards motion could have no net force, a leftward net force or a rightwards net force; it depends upon whether the object is slowing down, speeding up or maintaining a constant speed. Only A and C show a balance of forces, so only A and C are moving with constant speed.

Part C: Force-Mass-Acceleration Relationships Use the approximation that g= ~10 m/s2 to fill in the blanks in the following diagrams.

13. 14. 15.

Fgrav = m•g = ~800 N Since Fgrav = m•g, m can be calculated to be ~70 kg ∑Fy = may = (80 kg)•(2.0 Fgrav = m•g = ~800 N m/s/s) (m=Fgrav/g).

Since ay = 0 m/s/s, Fnorm must Since there are two forces pulling ∑Fy = 160 N, down equal Fgrav; so Fnorm = 700 N. upwards and since the sign is hanging The Fgrav (down) and the Fair symmetrically, each force must supply an ∑ (up) must add up to 160 N, Fx = m•ax = (70 kg)•(5.0 m/s/s) upwards pull equal to one-half the object's weight. So the vert pull (Fy) in each force down. Thus, Fair must be ∑Fx = 350 N, left smaller than Fgrav by 160 N. is 400 N. The following triangle can be set (Note that the ·Fx direction is up: Fair = 640 N always the same as the ax direction.)

With Fapp being the only horizontal force, its value must be Using trig, we can write: 350 N - the same as ·Fx. sin(30 deg.)=(400 N)/Ftens

Solving for Ftens yields 800 N.

16. 17.

A quick blank is Fgrav: Fgrav = m•g = ~80 N The first step in an inclined plane problem is to Now resolve the 60-N force into components using resolve the weight vector into parallel and trigonometry and the given angle measure: perpendicular components:

Fpar = m•g•sin(angle) = (420 N)•sin(30 deg) = 210 N

Fperp = m•g•cos(angle) = (420 N)•cos(30 deg) = 364 N

Fx = 60 N•cos(30 deg) = 52 N The mass can be found as m = Fgrav/g

Fy = 60 N•sin(30 deg) = 30 N m = Fgrav/g = (420 N)/(10 m/s/s) = ~42 kg

Since the acceleration is horizontal, the sum of the The Fnorm acts opposite of and balances the Fperp. vertical forces must equal 0 N. So Fgrav = Fy + Fnorm. So Fnorm = Fperp = 364 N Therefore Fnorm = Fgrav - Fy = 50 N. Knowing Fnorm and mu, the Ffrict can be determined: Knowing Fnorm and mu, the Ffrict can be determined: Ffrict = mu•Fnorm = 0.2*(364 N) = 73 N Ffrict = mu•Fnorm = 0.5*(50 N) = 25 N Now the forces parallel to the incline can be Now the horizontal forces can be summed: summed: ∑ Fx = Fx + Ffrict = 52 N, right + 25 N, left ∑F|| = F|| + Ffrict = 210, down to left + 73 N, up to right ∑Fx = 27 N, right

∑F|| = 137 N, down to left Using Newton's second law, ∑Fx = m•ax

So ax = (27 N)/(8 kg) = 3.4 m/s/s, right (3375... m/s/s) Using Newton's second law, ∑F|| = m•a||

So a|| = (137 N)/(42 kg) = 3.3 m/s/s

Part D: Problem-Solving 22. A 945-kg car traveling rightward at 22.6 m/s slams on the brakes and skids to a stop (with locked wheels). If the coefficient of friction between tires and road is 0.972, determine the distance required to stop. PSYW Answer: 26.8 m Like most problems, this problem begins with a free-body diagram (as shown at right). Note that there is no rightwards applied force (a common mistake). Note also that the force of friction is the only force responsible for the acceleration (deceleration) of the car. The Ffrict value is the net force. So determining the acceleration involves finding Fgrav (Fgrav=m•g = 945•9.8 = 9261 N), Fnorm (the same as Fgrav= 9261 N), and Ffrict (Ffrict = mu•Fnorm = 0.972 • 9261 N = 9002 N). With the ∑ Fx = 9002 N, the acceleration can be calculated (ax = ∑ Fx/m = 9002 N/945 kg = 9.53 m/s/s.) This is a leftwards acceleration; ax will be assigned the numerical value of -9.53 m/s/s in the next part of this problem. Now in the kinematic part of this problem, the distance must be found using the known information (vf = 0 m/s, vo = 22.6 m/s and a = 9.53 m/s/s). Use the equation: 2 2 vf = vo + 2 a d (0 m/s)2 = (22.6 m/s)2 + 2•(-9.53 m/s/s)•d d = [(22.6 m/s)2] / [2•(9.53 m/s/s)] = 26.8 m

23. A student pulls a 2-kg backpack across the ice (assume frictionless) by pulling at a 30.0 degrees angle to the horizontal. The velocity-time graph for the motion is shown. Perform a careful analysis of the situation and determine the applied force. PSYW

Answer: 0.289 N The acceleration of the object can be found from the slope of the line on the v-t graph. The acceleration is: a = change in velocity/change in time = (2.0 m/s) / 16 s = 0.125 m/s/s.

The net force is found from m•a; so Fnet = (2 kg)•(0.125 m/s/s) = 0.250 N. The only force that can contribute to the net force is the applied force. The horizontal component of the applied force must therefore be equal to 0.250 N. A diagram depicting the relationship of the lengths of the side for this 30-60-90 triangle is shown at the right. The cosine function can be utilized to determine the magnitude of the tension force.

cos(30 deg) = 0.250 N)/Fapp

Solving for Fapp yields 0.289 N.

28. Consider the two-body system at the right. There is a 0.250-kg object accelerating across a rough surface. The sliding object is attached by a string to a 0.100 kg object which is suspended over a pulley. The coefficient of kinetic friction is 0.183. Calculate the acceleration of the block and the tension in the string. PSYW

Answer: a = 1.52 m/s/s ; Ftens = 0.83 N The solution to this problem begins by drawing a free-body diagram for each object. Note that the + x-axis for the 0.25-kg object is drawn in the direction that the object accelerates; and the + y-axis for the 0.10-kg object is drawn in the direction which it accelerates. Note that while friction acts upon the 0.25-kg object, it does not act upon the 0.10-kg object (since it is not being dragged across the surface). Newton's second law of motion (·F = m•a) can be applied to the motion of the 0.25-kg mass:

∑Fx = m•ax

Ftens - Ffrict = m•ax

The expressioin for Ffrict can be substituted into the equation.

Ftens - µ•Fnorm = m•ax

Since the vertical forces on the 0.250-kg object balance each other, it is known that Fgrav = Fnorm; so Fnorm = m•g = (0.250 kg)•(9.8 m/s/s) = 2.45 N. Since µ = 0.183, µ•Fnorm = 0.448 N. Equation 1 can now be re-written as

Ftens - 0.448 N = (0.250 kg)•ax Newton's second law of motion (∑ F = m•a) can be applied to the motion of the 0.10o-kg mass:

∑Fy = m•ay

Fgrav - Ftens = m•ay

The expression for Fgrav (m•g) can be substituted into the equation.

m•g - Ftens = m•ay

(0.100 kg)•(9.8 m/s/s) - Ftens = m•ay

0.98 N - Ftens = m•ay The mass of the 0.100-kg object can be substituted into the equation to yield equation 3.

0.98 N - Ftens = (0.100 kg)•ay

Equations 2 and 3 each include an acceleration term - ax and ay. These acceleration values are the same since the system accelerates together; thus, ax = ay = a. An inspection of these two equations show that there are now two equations and two unknowns - Ftens and a. These unknown values can be solved for in the customary manner.

First, equation 3 is rearranged to create an expression for Ftens:

0.98 N - (0.100 kg)•a = Ftens

This expression for Ftens is now substituted into equation 2.

Ftens - 0.448 N = (0.250 kg)•a

0.980 N - (0.100 kg)•a - 0.448 N = (0.250 kg)•a This equation can now be solved for acceleration (a) as shown in the following steps.

0.980 N - 0.448 N = (0.250 kg)•a + (0.100 kg)•a 0.532 N = (0.305 kg)•a a = 1.52 m/s/s

Now that a is known, its value can be substituted into equation 4 in order to solve for Ftens.

0.980 N - (0.100 kg)•a = Ftens

0.980 N - (0.10 kg)•(1.52 m/s/s) = Ftens

0.980 N - 0. 152 N = Ftens

0.830 N = Ftens

Momentum and Collisions Review

Part A: Multiple-Multiple Choice

1. Which of the following statements are true about momentum? a. Momentum is a vector quantity. b. The standard unit on momentum is the Joule. c. An object with mass will have momentum. d. An object which is moving at a constant speed has momentum. e. An object can be traveling eastward and slowing down; its momentum is westward. f. Momentum is a conserved quantity; the momentum of an object is never changed. g. The momentum of an object varies directly with the speed of the object. h. Two objects of different mass are moving at the same speed; the more massive object will have the greatest momentum. i. A less massive object can never have more momentum than a more massive object. j. Two identical objects are moving in opposite directions at the same speed. The forward moving object will have the greatest momentum. k. An object with a changing speed will have a changing momentum.

Answer: ADGHK a. TRUE - Momentum is a vector quantity. Like all vector quantities, the momentum of an object is not fully described until the direction of the momentum is identified. Momentum, like other vector quantities, is subject to the rules of vector operations. b. FALSE - The Joule is the unit of work and energy. The kg m/s is the standard unit of momentum. c. FALSE - An object has momentum if it is moving. Having mass gives an object inertia. When that inertia is in motion, the object has momentum. d. TRUE - This is true. However, one should be quick to note that the object does not have to have a constant speed in order to have momentum. e. FALSE - The direction of an object's momentum vector is in the direction that the object is moving. If an object is traveling eastward, then it has an eastward momentum. If the object is slowing down, its momentum is still eastward. Only its acceleration would be westward. f. FALSE - To say that momentum is a conserved quantity is to say that if a system of objects can be considered to be isolated from the impact of net external forces, then the total momentum of that system is conserved. In the absence of external forces, the total momentum of a system is not altered by a collision. However, the momentum of an individual object is altered as momentum is transferred between colliding objects. g. TRUE - Momentum is calculated as the product of mass and velocity. As the speed of an object increases, so does its velocity. As a result, an increasing speed leads to an increasing momentum - a direct relationship. h. TRUE - For the same speed (and thus velocity), a more massive object has a greater product of mass and velocity; it therefore has more momentum. i. FALSE - A less massive object would have a greater momentum owing to a velocity which is greater than that of the more massive object. Momentum depends upon two quantities * mass and velocity. Both are equally important. j. FALSE - When comparing the size of two momentum vectors, the direction is insignificant. The direction of any vector would never enter into a size comparison. k. TRUE - Objects with a changing speed also have a changing velocity. As such, an object with a changing speed also has a changing momentum.

2. Which of the following are true about the relationship between momentum end energy? a. Momentum is a form of energy. b. If an object has momentum, then it must also have mechanical energy. c. If an object does not have momentum, then it definitely does not have mechanical energy either. d. Object A has more momentum than object B. Therefore, object A will also have more kinetic energy. e. Two objects of varying mass have the same momentum. The least massive of the two objects will have the greatest kinetic energy.

Answer: BE a. FALSE - No. Momentum is momentum and energy is energy. Momentum is NOT a form of energy; it is simply a quantity which proves to be useful in the analysis of situations involving forces and impulses. b. TRUE - If an object has momentum, then it is moving. If it is moving, then it has kinetic energy. And if an object has kinetic energy, then it definitely has mechanical energy. c. FALSE - If an object does NOT have momentum, then it definitely does NOT have kinetic energy. However, it could have some potential energy and thus have mechanical energy. d. FALSE - Consider Object A with a mass of 10 kg and a velocity of 3 m/s. And consider Object B with a mass of 2 kg and a velocity of 10 m/s. Object A clearly has more momentum. However, Object B has the greatest kinetic energy. The kinetic energy of A is 45 J and the kinetic energy of B is 100 J. e. TRUE - When comparing the momentum of two objects to each other, one must consider both mass and velocity; both are of equal importance when determining the momentum value of an object. When comparing the kinetic energy of two objects, the velocity of an object is of double importance. So if two objects of different mass have the same momentum, then the object with the least mass has a greater velocity. This greater velocity will tip the scales in favor of the least massive object when a kinetic energy comparison is made.

3. Which of the following statements are true about impulse? a. Impulse is a force. b. Impulse is a vector quantity. c. An object which is traveling east would experience a westward directed impulse in a collision. d. Objects involved in collisions encounter impulses. e. The Newton is the unit for impulse. f. The kg•m/s is equivalent to the units on impulse. g. An object which experiences a net impulse will definitely experience a momentum change. h. In a collision, the net impulse experienced by an object is equal to its momentum change. i. A force of 100 N acting for 0.1 seconds would provide an equivalent impulse as a force of 5 N acting for 2.0 seconds.

Answer: BDFGHI a. FALSE - Impulse is NOT a force. Impulse is a quantity which depends upon both force and time to change the momentum of an object. Impulse is a force acting over time. b. TRUE - Impulse is a vector quantity Like momentum, impulse is not fully described unless a direction is associated with it. c. FALSE - An object which is traveling east could encounter a collision from the side, from behind (by a faster-moving object) or from the front. The direction of the impulse is dependent upon the direction of the force exerted upon the object. In each of these scenarios, the direction of the force would be different. d. TRUE - In a collision, there is a collision force which endures for some amount of time. The combination of force and time is what is referred to as an impulse. e. FALSE - The Newton is the unit of force. The standard metric unit of impulse is the N•s. f. TRUE - The N•s is the unit of momentum. The Newton can be written as a kg•m/s^2. When substituted into the N•s expression, the result is the kg m/s. g. TRUE - In a collision, there is a collision force which endures for some amount of time to cause an impulse. This impulse acts upon the object to change its velocity and thus its momentum. h. TRUE - Yes!!! This is the impulse-momentum change theorem. The impulse encountered by an object in a collision causes and is equal to the momentum change experienced by that object. i. TRUE - A force of 100 N for 0.10 s results in an impulse of 10 N•s. This 10 N•s impulse is equivalent to the impulse created by a force of 5 N for 2.0 seconds.

4. Which of the following statements are true about collisions? a. Two colliding objects will exert equal forces upon each other even if their mass is significantly different. b. During a collision, an object always encounters an impulse and a change in momentum. c. During a collision, the impulse which an object experiences is equal to its velocity change. d. The velocity change of two respective objects involved in a collision will always be equal. e. While individual objects may change their velocity during a collision, the overall or total velocity of the colliding objects is conserved. f. In a collision, the two colliding objects could have different acceleration values. g. In a collision between two objects of identical mass, the acceleration values could be different. h. Total momentum is always conserved between any two objects involved in a collision. i. When a moving object collides with a stationary object of identical mass, the stationary object encounters the greater collision force. j. When a moving object collides with a stationary object of identical mass, the stationary object encounters the greater momentum change. k. A moving object collides with a stationary object; the stationary object has significantly less mass. The stationary object encounters the greater collision force. l. A moving object collides with a stationary object; the stationary object has significantly less mass. The stationary object encounters the greater momentum change.

Answer: ABF a. TRUE - In any collision between two objects, the colliding objects exert equal and opposite force upon each other. This is simply Newton's law of action-reaction. b. TRUE - In a collision, there is a collision force which endures for some amount of time to cause an impulse. This impulse acts upon the object to change its momentum. c. FALSE - The impulse encountered by an object is equal to mass multiplied by velocity change - that is, momentum change. d. FALSE - Two colliding objects will only experience the same velocity change if they have the same mass and the collision occurs in an isolated system. However, their momentum changes will be equal if the system is isolated from external forces. e. FALSE - This statement is mistaking the term velocity for momentum. It is momentum which is conserved by an isolated system of two or more objects. f. TRUE - Two colliding objects will exert equal forces upon each other. If the objects have different masses, then these equal forces will produce different accelerations. g. FALSE - It the colliding objects have different masses, the equal force which they exert upon each other will lead to different acceleration values for the two objects. h. FALSE - Total momentum is conserved only if the collision can be considered isolated from the influence of net external forces. i. FALSE - In any collision, the colliding objects exert equal and opposite forces upon each other as the result of the collision interaction. There are no exceptions to this rule. j. FALSE - In any collision, the colliding objects will experience equal (and opposite) momentum changes, provided that the collision occurs in an isolated system. k. FALSE - In any collision, the colliding objects exert equal and opposite forces upon each other as the result of the collision interaction. There are no exceptions to this rule. l. FALSE - In any collision, the colliding objects will experience equal (and opposite) momentum changes, provided that the collision occurs in an isolated system.

5. Which of the following statements are true about elastic and inelastic collisions? a. Perfectly elastic and perfectly inelastic collisions are the two opposite extremes along a continuum; where a particular collision lies along the continuum is dependent upon the amount kinetic energy which is conserved by the two objects. b. Most collisions tend to be partially to completely elastic. c. Momentum is conserved in an elastic collision but not in an inelastic collision. d. The kinetic energy of an object remains constant during an elastic collision. e. Elastic collisions occur when the collision force is a non-contact force. f. Most collisions are not inelastic because the collision forces cause energy of motion to be transformed into sound, light and thermal energy (to name a few). g. A ball is dropped from rest and collides with the ground. The higher that the ball rises upon collision with the ground, the more elastic that the collision is. h. A moving air track glider collides with a second stationary glider of identical mass. The first glider loses all of its kinetic energy during the collision as the second glider is set in motion with the same original speed as the first glider. Since the first glider lost all of its kinetic energy, this is a perfectly inelastic collision. i. The collision between a tennis ball and a tennis racket tends to be more elastic in nature than a collision between a halfback and linebacker in football.

Answer: AEFGI a. TRUE - A perfectly elastic collision is a collision in which the total kinetic energy of the system of colliding objects is conserved. Such collisions are typically characterized by bouncing or repelling from a distance. In a perfectly inelastic collision (as it is sometimes called), the two colliding objects stick together and move as a single unit after the collision. Such collisions are characterized by large losses in the kinetic energy of the system. b. FALSE - Few collisions are completely elastic. A completely elastic collision occurs only when the collision force is a non-contact force. Most collisions are either perfectly inelastic or partially inelastic. c. FALSE - Momentum can be conserved in both elastic and inelastic collisions provided that the system of colliding objects is isolated from the influence of net external forces. It is kinetic energy that is conserved in a perfectly elastic collision. d. FALSE - In a perfectly elastic collision, in an individual object may gain or lose kinetic energy. It is the system of colliding objects which conserves kinetic energy. e. TRUE - Kinetic energy is lost from a system of colliding objects because the collision transforms kinetic energy into other forms of energy - sound, heat and light energy. When the colliding objects don't really collide in the usual sense (that is when the collision force is a non-contact force), the system of colliding objects does not lose its kinetic energy. Sound is only produced when atoms of one object make contact with atoms of another object. And objects only warm up (converting mechanical energy into thermal energy) when their surfaces meet and atoms at those surfaces are set into vibrational motion or some kind of motion. f. TRUE - See above statement. g. TRUE - If large amounts of kinetic energy are conserved when a ball collides with the ground, then the post-collision velocity is high compared to the pre-collision velocity. The ball will thus rise to a height which is nearer to its initial height. h. FALSE - This is a perfectly elastic collision. Before the collision, all the kinetic energy is in the first glider. After the collision, the first glider has no kinetic energy; yet the second glider has the same mass and velocity as the first glider. As such, the second glider has the kinetic energy which the first glider once had. i. TRUE - There is significant bounce in the collision between a tennis racket and tennis ball. There is typically little bounce in the collision between a halfback and a linebacker (though there are certainly exceptions to this one). Thus, the ball-racket collision tends to be more elastic.

Part B: Multiple Choice 6. Which of the following objects have momentum? Include all that apply. a. An electron is orbiting the nucleus of an atom. b. A UPS truck is stopped in front of the school building. c. A Yugo (a compact car) is moving with a constant speed. d. A small flea walking with constant speed across Fido's back. e. The high school building rests in the middle of town.

Answer: A, C, and D Momentum can be thought of as mass in motion. An object has momentum if it has its mass in motion. It matters not whether the object is of large mass or small mass, moving with constant speed or accelerating; if the object is MOVING, then it has momentum!

7. A truck driving along a highway road has a large quantity of momentum. If it moves at the same speed but has twice as much mass, its momentum is ______. a. zero b. quadrupled c. doubled d. unchanged

Answer: C Momentum is directly related to the mass of the object. So for the same speed, a doubling of mass leads to a doubling of momentum.

8. TRUE or FALSE: A ball is dropped from the same height upon various flat surfaces. For the same collision time, impulses are smaller when the most bouncing take place. a. True b. False

Answer: B Since being dropped from the same height, the balls will be moving with the same pre-collision velocity (assuming negligible air resistance). Upon collision with the ground, the velocity will have to be reduced to zero - that is, the ball will cease moving downwards. This decrease in velocity constitutes the first portion of the velocity change. If the ball bounces, then there is an additional velocity change sending the ball back upwards opposite the original direction. Thus, for the same collision time, bouncing involves a greater velocity change, a greater momentum change, and therefore a greater impulse.

9. Consider a karate expert. During a talent show, she executes a swift blow to a cement block and breaks it with her bare hand. During the collision between her hand and the block, the ___. a. time of impact on both the block and the expert's hand is the same b. force on both the block and the expert's hand have the same magnitude c. impulse on both the block and the expert's hand have the same magnitude d. all of the above. e. none of the above.

Answer: D In any collision, there are always four quantities which are the same for both objects involved in the collision. Each object experiences the same force (Newton's third law) for the same amount of time, leading to the same impulse, and subsequently the same momentum change. Only the acceleration and the velocity change can differ for the two colliding objects. The lower mass object always receives the greater velocity change and acceleration.

10. It is NOT possible for a rocket to accelerate in outer space because ____. List all that apply. a. there is no air in space b. there is no friction in space c. there is no gravity in outer space d. ... nonsense! Rockets do accelerate in outer space.

Answer: D Rockets accelerate in outer space by means of Newton's third law of motion. It does not matter that there is no air outside of the rocket. Rockets produce their own gas by burning fuels. The combustion of rocket fuels produces gaseous products. The rocket's thrusters push these gases backwards (or rightwards, or leftwards, or ...) and the gases push the rocket forwards (or leftwards, or rightwards, or ...). Thus, rockets indeed can and do accelerate in outer space.

12. Suppose that Paul D. Trigger fires a bullet from a gun. The speed of the bullet leaving the muzzle will be the same as the speed of the recoiling gun ____. a. because momentum is conserved b. because velocity is conserved c. because both velocity and momentum are conserved d. only if the mass of the bullet equals the mass of the gun e. none of these

Answer: D In any collision or explosion involving two objects, the momentum change for each object is the same. So both the bullet and the gun encounter the same momentum change. The momentum change is simply the mass multiplied by the velocity change. Thus, the velocity change would only be the same if their masses were the same. Otherwise, the smaller-mass object receives a greater velocity change.

13. Suppose that you're driving down the highway and a moth crashes into the windshield of your car. Which undergoes the greater change is momentum? a. the moth b. your car c. both the same

Answer: C In any collision, there are always four quantities which are the same for both objects involved in the collision. Each object experiences the same force (Newton's third law) for the same amount of time, leading to the same impulse, and subsequently the same momentum change. Only the acceleration and the velocity change can differ for the two objects. The object with the least mass always receives the greatest velocity change and acceleration.

15. Suppose that you're driving down the highway and a moth crashes into the windshield of your car. Which undergoes the greater impulse? a. the moth b. your car c. both the same

Answer: C In any collision, there are always four quantities which are the same for both objects involved in the collision. Each object experiences the same force (Newton's third law) for the same amount of time, leading to the same impulse, and subsequently the same momentum change. Only the acceleration and the velocity change can differ for the two objects. The object with the least mass always receives the greatest velocity change and acceleration.

16. Suppose that you're driving down the highway and a moth crashes into the windshield of your car. Which undergoes the greater acceleration? a. the moth b. your car c. both the same

Answer: A In any collision, there are always four quantities which are the same for both objects involved in the collision. Each object experiences the same force (Newton's third law) for the same amount of time, leading to the same impulse, and subsequently the same momentum change. Only the acceleration and the velocity change can differ for the two objects. The object with the least mass always receives the greatest velocity change and acceleration.

17. Three boxes, X, Y, and Z, are at rest on a table as shown in the diagram at the right. The weight of each box is indicated in the diagram. The net or unbalanced force acting on box Y is _____.

a. 4 N down b. 5 N down c. 5 N up d. 10 N up e. zero

Answer: E If an object is at rest, then all the forces acting upon the object must be zero. The net force on any one of the boxes is 0 Newtons. Subsequently, in each case, the support force (which we have called the "normal force throughout this course) acting upwards on any of the boxes must be equal to the force of gravity on that box (i.e., the weight) plus the amount of load exerted from above (which would be equivalent to the weight of the other boxes located above the box). So for box Y, the support force acting upward would be equal to 9 N while the net force is still 0 Newtons. And for box Z, the support force is 19 N, sufficient to balance the 10- N gravitational force plus the 9-N of force resulting from the other two boxes bearing down on it.

18. In a physics experiment, two equal-mass carts roll towards each other on a level, low-friction track. One cart rolls rightward at 2 m/s and the other cart rolls leftward at 1 m/s. After the carts collide, they couple (attach together) and roll together with a speed of ______. Ignore resistive forces. a. 0.5 m/s b. 0.33 m/s c. 0.67 m/s d. 1.0 m/s e. none of these

Answer: A Use 1 kg as the mass of the carts (or any number you wish) and then set the expression for initial total momentum equal to the expression for the final total momentum: (1 kg)*(2) + (1 kg) *(-1) = (1 kg) *v + (1 kg) *v Now solve for v using the proper algebraic steps. (2 kg•m/s) - (1 kg•m/s) = (2 kg) v 1 kg•m/s = (2 kg)v (1 kg•m/s) / (2 kg) = v 0.5 m/s = v

19. A physics cart rolls along a low-friction track with considerable momentum. If it rolls at the same speed but has twice as much mass, its momentum is ____. a. zero b. four times as large c. twice as large d. unchanged

Answer: C The momentum of an object is calculated as the product of mass and velocity. Thus, the momentum is directly proportional to the mass of the object. If the mass of an object is somehow doubled, the momentum is doubled as well.

20. The firing of a bullet by a rifle causes the rifle to recoil backwards. The speed of the rifle's recoil is smaller than the bullet's forward speed because the ___. a. force against the rifle is relatively small b. speed is mainly concentrated in the bullet c. rifle has lots of mass d. momentum of the rifle is unchanged e. none of these

Answer: C Please don't answer A (for it will make Newton roll over in his grave and he's getting quite tired of that). Perhaps you've heard that "for every action, there is an equal and opposite ...". Choice B is invalid; speed is not something that becomes concentrated or squeezed into an object. Choice D is invalid; ask anyone who's fired a rifle if the rifle is set into motion by the firing of the bullet. (Of course, since it is set in motion, its momentum is not unchanged.) Because of the large mass of the rifle, the acceleration and the recoil speed of the rifle is small.

23. A 4 kg object has a momentum of 12 kg•m/s. The object's speed is ___ m/s. a. 3 b. 4 c. 12 d. 48 e. none of these.

Answer: A This is a relatively simple plug-and-chug into the equation p=m*v with m=4 kg and p=12 kg•m/s.

25. A relatively large force acting for a relatively long amount of time on a relatively small mass will produce a relatively ______. List all that apply. a. small velocity change b. large velocity change c. small momentum change d. small acceleration

Answer: B A large force acting upon a small mass will result in a large acceleration (a=F/m) and subsequently a large velocity change (Delta v = a*t). This rules out choices A and D. A large force and for a long time will result in a large impulse and therefore a large momentum change. This rules out choice C.

26. Consider the concepts of work and energy (presuming you have already studied it) and those of impuse and momentum. Force and time is related to momentum change in the same manner as force and displacement pertains to ______. a. impulse b. work c. energy change d. velocity e. none of these.

Answer: C A force multiplied by a time gives an impulse which will cause (and be equal to) a momentum change. In the same manner, a force multiplied by a displacement gives work which will cause (and be equal to) an energy change. Take the time to reread those two sentences because it relates two big concepts.

27. A 5-N force is applied to a 3-kg ball to change its velocity from +9 m/s to +3 m/s. This impulse causes the momentum change of the ball to be ____ kg•m/s. a. -2.5 b. -10 c. -18 d. -45 e. none of these

Answer: C Don't make this harder than it is; the momentum change of an object can be found if the mass and the velocity change are known. In this equation, m=3 kg and the velocity change is -6 m/s. When finding the velocity change, always subtract the initial velocity from the final velocity (vf - vi). There is a second means of determining the momentum change of an object (though it does not need to be used in this problem). The momentum change can also be found if the force and the time are known. Multiplying force*time yields the impulse and the impulse equals the momentum change.

36. Consider the head-on collision between a lady bug and the windshield of a high speed bus. Which of the following statements are true? List all that apply. a. The magnitude of the force encountered by the bug is greater than that of the bus. b. The magnitude of the impulse encountered by the bug is greater than that of the bus. c. The magnitude of the momentum change encountered by the bug is greater than that of the bus. d. The magnitude of the velocity change encountered by the bug is greater than that of the bus. e. The magnitude of the acceleration encountered by the bug is greater than that of the bus.

Answer: D and E In any collision between two objects, the force, impulse, and momentum change are the same for each object. (This makes statements A, B, and C false.) However, the smaller mass object encounters a greater acceleration and velocity change. (This makes statements D and E true).

Part C: Diagramming and Analysis For Questions #37-#40: Consider the before- and after-collision momentum vectors in the diagram below. Determine the magnitude and direction of the system momentum before and after the collision and identify whether or not momentum is conserved. Finally, determine the magnitude and direction of the net external impulse encountered by the system during the collision.

37.

System Momentum Before Collision: 30 kg•m/s, right

System Momentum After Collision: 30 kg•m/s, right Is momentum conserved? Yes

Net External Impulse During Collision: 0 N

Answer: See table above. Before the collision, the two momentum values can be added: 30 units + 0 = 30 units. (An object at rest has zero momentum.) After the collision, the momentum vectors can be added: 30 units + 0 = 30 units. In each case, the total momentum is in the direction of the arrows. Since the total system momentum before the collision is the same as it is after the collision, the total momentum of the system can be considered to be conserved. As such, there is no net external impulse upon the system. It is considered isolated.

38.

System Momentum Before Collision: 15 kg•m/s, right

System Momentum After Collision: 15 kg•m/s, right

Is momentum conserved? Yes

Net External Impulse During Collision: 0 N•s

Answer: See table above. The dark ball has a leftward momentum, so its momentum must be assigned a - sign to distinguish its momentum from the rightward momentum of the light-colored ball. So before the collision, the system has +25 units + -10 units of momentum - a total of +15 units. After the collision, the total momentum of the system is + 15 units (the "+" is the designated sign for a rightward momentum). Since the system momentum is the same before as after the collision, momentum is conserved and the system is considered isolated from net external impulses.

39.

System Momentum Before Collision: 30 kg•m/s, right System Momentum After Collision: 10 kg•m/s, left

Is momentum conserved? No

Net External Impulse During Collision: 40 N•s, left

Answer: See table above. Since the light-colored ball has a leftward momentum, the leftward direction will be designated as the negative direction. Thus, the before- and after-collision calculations are: Before: +30 units + 0 units = +30 units After: -20 units + 10 units = -10 units The total momentum of the system has changed by -40 units. Thus, momentum is not conserved and the system experiences a net external impulse of -40 units or 40 N•s, leftward.

For Questions #41-#44: Repeat the procedure performed in Questions #37-#40. Note that these diagrams give velocity and mass values before and after the collision.

41.

System Momentum Before Collision: 3 kg•m/s, right

System Momentum After Collision: 3 kg•m/s, right

Is momentum conserved? Yes

Net External Impulse During Collision: 0 N•s

Answer: See table above. Before Collision: 1 kg) * (+2 m/s) + (1 kg) * (1 m/s ) = 3 kg m/s After Collision: (1 kg) * (1 m/s) + (1 kg) * (2 m/s ) = 3 kg m/s The total system momentum is the same before and after the collision. Thus, momentum is conserved and there is no net external impulse on the system.

42.

System Momentum Before Collision: 8 kg•m/s, right

System Momentum After Collision: 10 kg•m/s, right

Is momentum conserved? No

Net External Impulse During Collision: 2 N•s, right

Answer: See table above. Before Collision: (2 kg) * (2 m/s) + (4 kg) * (1 m/s ) = 8 kg m/s After Collision: (1 kg) * (2 m/s) + (4 kg) * (2 m/s ) = 10 kg m/s The total system momentum is NOT the same before and after the collision. Thus, momentum is NOT conserved. There is a net external impulse on the system which is equal to the momentum change of the system. The net external impulse is +2 kg•m/s.

For Questions #45-#49, determine the unknown velocity value. Assume that the collisions occur in an isolated system.

General Note about Questions #45-#49: In each of these problems, an expression for the total momentum before the collision is written by summing the product of mass and velocity for each object. The same is done with the total momentum after the collision. One must be careful to assign a negative momentum to any object which is moving leftward. When the velocity of an object is not known, the variable v should be placed in the expression in place of the actual velocity value. The two expressions for total momentum before and after the collision are set equal to each other and algebraic operations are used to determine the unknown velocities. This method is shown below.

45. Answer: 6 m/s Before Collision: (6 kg) * (10 m/s) + (4 kg) * (4 m/s ) = 76 kg m/s After Collision: (6 kg) * v + (4 kg) * (10 m/s ) = (6 kg) * v + 40 kg m/s (In the following steps, the units are dropped so that the algebra can be more easily followed.) 76 = 6v + 40 36 = 6v v = 6 m/s

56. A cart with a mass of M is moving along a low-friction track with a speed of 60 cm/s. A brick is gently dropped from rest upon the cart. After the collision the cart and brick move together. a. ... If the brick has a mass of 2M, then the post-collision speed of the two objects will be 20 cm/s. b. ... If the brick has a mass of 3M, then the post-collision speed of the two objects will be 15 cm/s. c. ... If the brick has a mass of 4M, then the post-collision speed of the two objects will be 12 cm/s. d. ... If the brick has a mass of 5M, then the post-collision speed of the two objects will be 10 cm/s. e. ... If the brick has a mass of 0.5M, then the post-collision speed of the two objects will be 40 cm/s. f. ... If the brick has a mass of 0.25M, then the post-collision speed of the two objects will be 48 cm/s.

Answer: See Answers above. This is a special case of a perfectly inelastic collision. The total system momentum before the collision is possessed solely by the moving cart. After the collision, the total system momentum is the combined momentum of the brick and the cart. Since the brick and cart travel at the same velocity after the collision, the momentum is simply the sum of their masses multiplied by their velocity. In effect, the total mass which is in motion is increased by some factor as a result of the subsequent motion of the brick. For momentum to be conserved, the velocity of the cart (and the brick which is on top of it) must be decreased by that same factor. This principle is used to reason towards the answers to these questions. a. Adding a brick with a mass of 2M will increase the total mass in motion from M (the cart's mass) to 3M. This threefold increase in mass is accompanied by a threefold decrease in velocity. The new velocity is one-third the original value. The cart and brick move forward with a velocity of 20 cm/s. b. Adding a brick with a mass of 3M will increase the total mass in motion from M (the cart's mass) to 4M. This fourfold increase in mass is accompanied by a fourfold decrease in velocity. The new velocity is one- fourth the original value. The cart and brick move forward with a velocity of 15 cm/s. c. Adding a brick with a mass of 4M will increase the total mass in motion from M (the cart's mass) to 5M. This fivefold increase in mass is accompanied by a fivefold decrease in velocity. The new velocity is one- fifth the original value. The cart and brick move forward with a velocity of 12 cm/s. d. Adding a brick with a mass of 5M will increase the total mass in motion from M (the cart's mass) to 6M. This six-fold increase in mass is accompanied by a six-fold decrease in velocity. The new velocity is one- sixth the original value. The cart and brick move forward with a velocity of 10 cm/s. e. Adding a brick with a mass of 0.5M will increase the total mass in motion from M (the cart's mass) to 1.5M. This increases the mass by a factor of 1.5 and is accompanied by a decrease in velocity by a factor of 1.5. The new velocity is two-thirds the original value. The cart and brick move forward with a velocity of 40 cm/s. f. Adding a brick with a mass of 0.25M will increase the total mass in motion from M (the cart's mass) to 1.25M. This increases the mass by a factor of 1.25 and is accompanied by a decrease in velocity by a factor of 1.25. The new velocity is four-fifths the original value. The cart and brick move forward with a velocity of 48 cm/s.

Work and Energy Review

Part A: Forced Choice Questions 1. Which of the following statements are true about work? Include all that apply. a. Work is a form of energy. b. A Watt is the standard metric unit of work. c. Units of work would be equivalent to a Newton times a meter. d. A kg•m2/s2 would be a unit of work. e. Work is a time-based quantity; it is dependent upon how fast a force displaces an object. f. Superman applies a force on a truck to prevent it from moving down a hill. This is an example of work being done. g. An upward force is applied to a bucket as it is carried 20 m across the yard. This is an example of work being done. h. A force is applied by a chain to a roller coaster car to carry it up the hill of the first drop of the Shockwave ride. This is an example of work being done. i. The force of friction acts upon a softball player as she makes a headfirst dive into third base. This is an example of work being done. j. An eraser is tied to a string; a person holds the string and applies a tension force as the eraser is moved in a circle at constant speed. This is an example of work being done. k. A force acts upon an object to push the object along a surface at constant speed. By itself, this force must NOT be doing any work upon the object. l. A force acts upon an object at a 90-degree angle to the direction that it is moving. This force is doing negative work upon the object. m. An individual force does NOT do positive work upon an object if the object is moving at constant speed. n. An object is moving to the right. A force acts leftward upon it. This force is doing negative work. o. A non-conservative force is doing work on an object; it is the only force doing work. Therefore, the object will either gain or lose mechanical energy.

Answer: ACDHIKNO a. TRUE - Work is a form of energy, and in fact it has units of energy. b. FALSE - Watt is the standard metric unit of power; Joule is the standard metric unit of energy. c. TRUE - A N•m is equal to a Joule. d. TRUE - A kg•m2/s2 is a mass unit times a speed squared unit, making it a kinetic energy unit and equivalent to a Joule. e. FALSE - Work is not dependent on how rapidly the force displaces an object; power is time-based and calculated by force multiplied by speed. f. FALSE - Since Superman does not cause a displacement, no work is done; he is merely holding the car to prevent its descent down the hill. g. FALSE - The upward force does not cause the horizontal displacement so this is a NON-example of work. h. TRUE - There is a component of force in the direction of displacement and so this is an example of work. i. TRUE - There is a force and a displacement; the force acts in the opposite direction as the displacement and so this force does negative work. j. FALSE - For uniform circular motion, the force acts perpendicular to the direction of the motion and so the force never does any work upon the object. k. TRUE - This is clearly work - a force is causing an object to be displaced. l. FALSE - If a force acts at a 90-degree angle to the direction of motion, then the force does not do any work at all. Negative work is done when there is a component of force opposite the direction of motion. m. FALSE - There are many instances in which an individual force does positive work and yet the object maintains a constant speed. Consider a force applied to lift an object at constant speed. The force does positive work. Consider a car moving at constant speed along a level surface. The force of the road on the tires does positive work while air resistance does and equal amount of negative work. n. TRUE - A force which acts in a direction opposite the motion of an object will do negative work. o. TRUE - When non-conservative forces do work upon an object, the object will either gain or lose mechanical energy. Mechanical energy is conserved (neither gained nor lost) only when conservative forces do work upon objects.

2. Which of the following statements are true about power? Include all that apply. a. Power is a time-based quantity. b. Power refers to how fast work is done upon an object. c. Powerful people or powerful machines are simply people or machines which always do a lot of work. d. A force is exerted on an object to move it at a constant speed. The power delivered by this force is the magnitude of the force multiplied by the speed of the object. e. The standard metric unit of power is the Watt. f. If person A and person B do the same job but person B does it faster, then person A does more work but person B has more power. g. The Newton•meter is a unit of power. h. A 60-kg boy runs up a 2.0 meter staircase in 1.5 seconds. His power is approximately 80 Watt. i. A 300-Newton force is applied to a skier to drag her up a ski hill at a constant speed of 1.5 m/s. The power delivered by the toe rope is 450 Watts.

Answer: ABDEI a. TRUE - Power is a rate quantity and thus time-based. b. TRUE - This is the definition of power. c. FALSE - This is not always the case. A machine can do a lot of work but if it fails to do it rapidly, then it is not necessarily powerful. In fact two machines can do the same task (and therefore the same work), yet they can have drastically different power ratings. d. TRUE - An equation for computing work in constant speed situations is P=F•v. e. TRUE - Watt is the unit of power? Yes!! f. FALSE - Vice versa. If two people do the same job, then they're doing the same amount of work. The person who does it fastest generates more power. g. FALSE - A N•m is a Joule and that is a unit of work (not power). Think force (N) times distance (m); that's work (J). h. FALSE - The work would be (m•g)•d or approximately 1200 J. The power is work divided by time - 1200 J/1.5 s = 800 W. i. TRUE - Since force and speed are given, use Power = F•v. The calculation yields 450 W.

3. Consider the following physical situations. For each case, determine the angle between the indicated force (in boldface type) and the displacement ("theta" in the work equation). a. 0 degrees b. 180 degrees c. 90 degrees d. 30 degrees e. 60 degrees a. A rightward applied force is used to displace a television set to the right. - Answer: A - 0 degrees b. The force of friction acts upon a rightward-moving car to bring it to a stop. - Answer: B - 180 degrees c. A waiter uses an applied force to balance the weight of a tray of plates as he carries the tray across the room. - Answer: C - 90 degrees d. The force of air resistance acts upon a vertically-falling skydiver. - Answer: B - 180 degrees e. The force of friction acts upon a baseball player as he slides into third base. - Answer: B - 180 degrees f. An applied force is used by a freshman to lift a World Civilization book to the top shelf of his locker. - Answer: A - 0 degrees g. A bucket of water is tied to a string and tension supplies the centripetal force to keep it moving in a circle at constant speed. - Answer: C - 90 degrees h. An applied force acting at 30-degrees to the horizontal is used to displace an object to the right. - Answer: D - 30 degrees i. A group of football players use an applied force to push a sled across the grass. - Answer: A - 0 degrees j. The tension in the elevator cable causes the elevator to rise at a constant speed. - Answer: A - 0 degrees k. In a physics lab, an applied force is exerted parallel to a plane inclined at 30- degrees in order to displace a cart up the incline. - Answer: A - 0 degrees l. An applied force is exerted upwards and rightwards at an angle of 30- degrees to the vertical in order to displace an object to the right. - Answer: E - 60 degrees m. A child rests on the seat of a swing which is supported by the tension in its cables; he swings from the highest position to its lowest position. - Answer: C - 90 degrees

Answer: See questions above; explanations given below. a. The forward motion is do to the forward pushing; if the force and motion are in the same direction, then the angle is 0 degrees. b. Friction opposes motion and as such does negative work; the angle is 180 degrees. c. The force is vertical and the displacement if horizontal; they make a 90 degree angle. d. Air resistance opposes motion and as such does negative work; the angle is 180 degrees. e. Friction opposes motion and as such does negative work; the angle is 180 degrees. f. The frosh applies an upward force to cause an upward displacement; the angle is 0 degrees. g. For uniform circular motion, the force is inwards and the displacement at each instant is tangent to the circle; these two vectors make a 90 degree angle. h. This is a straightforward question; no tricks here. i. The forward motion is do to the forward pushing; if the force and motion are in the same direction, then the angle is 0 degrees. j. The cable pulls up on the elevator and the elevator is displaced upward; if the force and motion are in the same direction, then the angle is 0 degrees. k. The 30-degree angle is the incline angle, not necessarily the angle between F and d. The force is parallel to the incline and the cart is displaced along the direction of the incline; so the two vectors are in the same direction and the angle between them is 0 degrees. l. Compare the wording of this to part h. This one is tricky because the angle between F and d is 60-degrees. If you missed it, reread the question, paying careful attention to the "with the vertical" part. m. As the child swings, she traces out a circular arc and as such the tension (centripetal) is perpendicular to the direction of motion (tangent).

4. Consider the following physical situations. Identify whether the indicated force (in boldface type) does positive work, negative work or no work. a. Positive Work b. Negative Work c. No Work

Description of Physical Situation +, -, or no Work A. a. A cable is attached to a bucket and the force of tension is used to pull the bucket out of a well. Positive Work A. b. Rusty Nales uses a hammer to exert an applied force upon a stubborn nail to drive it into the wall. Positive Work B. c. Near the end of the Shockwave ride, a braking system exerts an applied force upon the coaster car to bring it to a stop. Negative Work B. d. The force of friction acts upon a baseball player as he slides into third base. Negative Work C. No e. A busy spider hangs motionless from a silk thread, supported by the tension in the thread. Work B. f. In baseball, the catcher exerts an abrupt applied force upon the ball to stop it in the catcher's mitt. Negative Work A. g. In a physics lab, an applied force is exerted parallel to a plane inclined at 30-degrees in order to displace a cart up the incline. Positive Work A. h. A pendulum bob swings from its highest position to its lowest position under the influence of the force of gravity. Positive Work

Answer: See table above; explanations provided below. a. The force is upwards and the displacement is upwards. When the force and the displacement act in the same direction, positive work is done. b. The force is horizontal and the displacement is horizontal. When the force and the displacement act in the same direction, positive work is done. (It is true that the wall is doing negative work upon the nail but this statement is about the hammer's force on the nail.) c. The force is backwards and the displacement is forwards. When the force and the displacement act in the opposite direction, negative work is done. d. The force is backwards and the displacement is forwards. When the force and the displacement act in the opposite direction, negative work is done. e. If the force does not cause the object to be displaced (the object hangs motionless), then no work is done. f. The force is backwards and the displacement is forwards. When the force and the displacement act in the opposite direction, negative work is done. g. The force is upwards and parallel to the incline and the displacement is in the same direction parallel to the incline. When the force and the displacement act in the same direction, positive work is done. h. As the bob swings downwards from its highest position, the motion is downwards (and rightwards); the force is also downwards and as such there is a component of force in the direction of motion. When the force and the displacement act in the same direction, positive work is done. (Note that if the bob was swinging upwards from its lowest position to its highest position, then gravity would be doing negative work.)

5. Which of the following statements are true about conservative and non-conservative forces? Include all that apply. a. A force is regarded as a conservative force if it does work but does not remove mechanical energy from a system of objects. b. A force is regarded as a non-conservative force if it does not add mechanical energy to a system of objects. c. The force of gravity and elastic (spring) force are both examples of a conservative forces. d. Applied forces, air resistance, friction forces, and tension are common examples of non-conservative forces. e. Physicists envy biologists' ability to instill order on the world of animal species through their taxonomic system. So physicists have made a habit of identifying forces as conservative and non-conservative forces in order to instill order on the world of forces. f. If a non-conservative force acts upon an object, then the object will either gain or lose mechanical energy. g. If the only forces which do work upon an object are conservative forces, then the object will conserve its mechanical energy. h. If the sum of an object's KE and PE is remaining constant, then non- conservative forces are NOT doing work. i. If work is NOT done on an object by a non-conservative force, then the object will experience a transformation of energy from kinetic to potential energy (or vice versa). j. An object starts from an elevated position with 50 J of potential energy and begins its fall towards the ground. If non-conservative forces can be assumed to NOT do work, then at some point during the fall the object will have 20 J of potential energy and 30 J of kinetic energy.

Answer: A(sort of) CDGH I(sort of) J a. TRUE (sort of) - If a force does work, yet does not remove mechanical energy from an object, then it is definitely a conservative force. The sort of indicates that a force is also considered a conservative force if it does work and does not add mechanical energy to an object. b. FALSE - If a force does not add mechanical energy to a system of objects, then it is likely a conservative force (provided it doesn't remove mechanical energy either). Non-conservative forces are those which either add or remove energy from a system of objects. c. TRUE - You must know this! d. TRUE - These are all non-conservative forces. You can add normal force to the list as well. e. FALSE - Whether there is envy in a physicist's heart is not for us to tell; the evil found within one's heart is often vast and mysterious ... . We can however definitively say that a physicist classifies forces in order to analyze physical situations in accord with the classification. If only conservative-classified forces do work, then KEi + PEi = KEf + PEf. On the other hand if one or more non-conservative-classified forces are doing work, then KEi + PEi + Wnc = KEf + PEf. f. FALSE - Not only must the force act upon the object, it must also be doing work upon the object. As you sit in your chair, there is a non-conservative force (normal force) acting upon your body. But since it does not do work (it's being assumed that you are not sitting in one of those fancy lounge chairs that has more controls than a TV set), your mechanical energy is not changing. g. TRUE - This is a big principle. You must know this one! h. TRUE - Non conservative forces would alter the total mechanical energy; that is, the PE + KE would not be a constant value. i. TRUE (sort of) - This statement is true (sort of); when only conservative forces are doing work, an object has its kinetic energy transformed into potential energy (or vice versa) without the total amount of the two being altered. It would however be possible that work is not done by a non-conservative force and there be no transformation of energy at all; i.e., the object remains at rest. A conservative force must be doing work in order for there to be a transformation of energy. j. TRUE - One would notice that the PE would begin to drop from 50 J to 0 J and that the KE would increase from 0 J to 50 J. And of course there would be a point at which the PE/KE would be distributed with 20 J to PE and 30 J to KE.

6. Which of the following statements are true about kinetic energy? Include all that apply. a. Kinetic energy is the form of mechanical energy which depends upon the position of an object. b. If an object is at rest, then it does not have any kinetic energy. c. If an object is on the ground, then it does not have any kinetic energy. d. The kinetic energy of an object is dependent upon the weight and the speed of an object. e. Faster moving objects always have a greater kinetic energy. f. More massive objects always have a greater kinetic energy. g. Kinetic energy is a scalar quantity. h. An object has a kinetic energy of 40 J. If its mass were twice as much, then its kinetic energy would be 80 J. i. An object has a kinetic energy of 40 J. If its speed were twice as much, then its kinetic energy would be 80 J. j. Object A has a mass of 1 kg and a speed of 2 m/s. Object B has a mass of 2 kg and a speed of 1 m/s. Objects A and B have the same kinetic energy. k. An object can never have a negative kinetic energy. l. A falling object always gains kinetic energy as it falls. m. A 1-kg object is accelerated from rest to a speed of 2.0 m/s. This object gains 4.0 Joules of kinetic energy. n. If work is done on an object by a non-conservative force, then the object will either gain or lose kinetic energy.

Answer: BGHK a. FALSE - Kinetic energy depends upon the speed of the object; potential energy depends upon the position of the object. b. TRUE - Kinetic energy depends upon speed. If there is no speed (the object is at rest), then there is no kinetic energy. c. FALSE - If an object is on the ground, then it does not have potential energy (relative to the ground). d. FALSE (sort of) - Kinetic energy depends upon mass and speed. Two objects of the same mass could have different weights if in a different gravitational field; so it is not appropriate to say that kinetic energy depends upon weight. e. FALSE - Faster moving objects would have more kinetic energy than other objects of the same mass. However, another object could have less speed and make up for this lack of speed in terms of a greater mass. f. FALSE - More massive objects would have more kinetic energy than other objects with the same speed. However, another object could have less mass and make up for this lack of mass in terms of a greater speed. g. TRUE - Kinetic energy does not have a direction associated with it; it is a scalar quantity. h. TRUE - Kinetic energy is directly related to the mass of an object. i. FALSE - Kinetic energy is directly related to the square of the speed of an object. So a doubling of the speed would result in a quadrupling of the kinetic energy - the new KE would be 160 J. j. FALSE - When it comes to kinetic energy, speed is doubly important (recall v2). So in this case, object A would have more kinetic energy. Doing the calculation yields 2 J for object A and 1 J for object B. k. TRUE - Kinetic energy is determined by the equation 0.5•m•v2. the quantity m is always positive. And even if v is negative, v2 will always be positive. Therefore, kinetic energy can never be a negative value. l. FALSE - If an object is falling at a constant velocity (i.e., the air resistance force equals the downward force of gravity), then there is not an increase in kinetic energy. It is true however that free-falling objects always increase their kinetic energy as they fall. m. FALSE - The kinetic energy increases from 0 J to 2 J (0.5•1•22); that's an increase by 2 J. n. FALSE - Such an object will definitely gain or lose mechanical energy but not necessarily kinetic energy.

7. Which of the following statements are true about potential energy? Include all that apply. a. Moving objects cannot have potential energy. b. Potential energy is the energy stored in an object due to its position. c. Both gravitational and elastic potential energy are dependent upon the mass of an object. d. The gravitational potential energy of an object is dependent upon the mass of the object. e. If the mass of an elevated object is doubled, then its gravitational potential energy will be doubled as well. f. Gravitational potential energy is lost as objects free-fall to the ground. g. The higher that an object is, the more potential energy which it will have. h. The unit of measurement for potential energy is the Joule. i. A 1-kg mass at a height of 1 meter has a potential energy of 1 Joule. j. A 1-kg object falls from a height of 10 m to a height of 6 m. The final potential energy of the object is approximately 40 J. k. If work is done on an object by a non-conservative force, then the object will either gain or lose potential energy.

Answer: BDEFGH a. FALSE - Potential energy has nothing to do with speed; an object could be moving at an elevated position. It is this elevation above zero level which gives an object potential energy. b. TRUE - This is the definition of potential energy. c. FALSE - Gravitational potential energy is dependent upon the mass of the object (PEgrav = m•g•h) but elastic potential energy is dependent upon the spring constant and the compression or stretch length of the 2 spring (PEelastic = 0.5•k•x ). d. TRUE - The equation states that PEgrav = m•g•h; PE is dependent upon mass. e. TRUE - The equation states that PEgrav = m•g•h; if the h is doubled, then the PE will be doubled as well. f. TRUE - As objects free-fall, the height (h) decreases; subsequently, the PE decreases. g. TRUE - The equation states that PEgrav = m•g•h; PE is directly related to height. h. TRUE - The Joule (abbrev. J) is the standard metric unit of energy - all forms of energy. i. FALSE - The final potential energy is calculated as PE = m•g•h = (1 kg)•(~10 m/s/s)•(1 m) = ~10 J. j. FALSE - The final potential energy is calculated as PE = m•g•h = (1 kg)•(~10 m/s/s)•(6 m) = ~60 J; the loss in potential energy during this 4-m fall is -40 J. k. FALSE - The object will either gain or lose mechanical energy, but not necessarily potential energy.

8. Which of the following statements are true about mechanical energy? Include all that apply. a. The total amount of mechanical energy of an object is the sum of its potential energy and the kinetic energy. b. Heat is a form of mechanical energy. c. The mechanical energy of an object is always conserved. d. When non-conservative forces do work, energy is transformed from kinetic to potential (or vice versa), but the total mechanical energy is conserved. e. A bowling ball is mounted from a ceiling by way of a strong cable. It is drawn back and released, allowed to swing as a pendulum. As it swings from its highest position to its lowest position, the total mechanical energy is mostly conserved. f. When a friction force does work on an object , the total mechanical energy of that object is changed. g. The total mechanical energy of an object remains constant if the only forces doing work on the object are conservative forces. h. If an object gains mechanical energy, then one can be certain that a non- conservative force is doing work. Answer: AEFGH a. TRUE - This is the definition of mechanical energy. b. FALSE - Heat or thermal energy is a non-mechanical form of energy. Potential and kinetic energy are the only forms of mechanical energy. c. FALSE - The mechanical energy of an object is only conserved if non-conservative forces do not do work upon the object. d. FALSE- If a non-conservative force does work upon an object, then the total mechanical energy of that object is changed. Energy will not be conserved. e. TRUE - Tension does not do work upon the object and so the total mechanical energy is conserved. The presence of air resistance (a non-conservative force) does a little work and so one might notice a very slight change in mechanical energy. f. TRUE - Friction is a non-conservative force and thus alters the total mechanical energy of an object. g. TRUE - This is the conservation of energy principle and one that you need to firmly understand. h. TRUE - If there is any change in the total mechanical energy of an object (whether a gain or a loss), then you know for certain that there is a non-conservative force doing work.

9. Rank these four objects in increasing order of kinetic energy, beginning with the smallest.

Object B Object C Object D Object A m = 5.0 kg m = 10.0 kg m = 1.0 kg m = 5.0 kg v = 4.0 m/s v = 2.0 m/s v = 5.0 m/s v = 2.0 m/s h = 2.0 m h = 3.00 m h = 5.0 m h = 4.0 m

Answer: D < C < B < A This is probably best done by performing a calculation of KE and comparing the results: Object A: KE = 0.5•(5.0 kg)•(4.0 m/s)2 = 40. J Object B: KE = 0.5•(10.0 kg)•(2.0 m/s)2 = 20. J Object C: KE = 0.5•(1.0 kg)•(5.0 m/s)2 = ~13 J (12.5 J) Object D: KE = 0.5•(5.0 kg)•(2.0 m/s)2 = 10. J The order is evident once the calculations are performed.

10. Rank these four objects in increasing order of potential energy, beginning with the smallest.

Object B Object C Object D Object A m = 5.0 kg m = 10.0 kg m = 1.0 kg m = 5.0 kg v = 4.0 m/s v = 2.0 m/s v = 5.0 m/s v = 2.0 m/s h = 2.0 m h = 3.00 m h = 5.0 m h = 4.0 m

Answer: C < A < D < B This is probably best done by performing a calculation of PE and comparing the results. Using the approximation that g = ~10 m/s/s gives much quicker results. Object A: PE = (5.0 kg)•(~10 m/s2)•(2.0 m) = ~100 J Object B: PE = (10.0 kg)•(~10 m/s2)•(3.00 m) = ~300 J Object C: PE = (1.0 kg)•(~10 m/s2)•(5.0 m) = ~50 J Object D: PE = (5.0 kg)•(~10 m/s2)•(4.0 m) = ~200 J The order is evident once the calculations are performed.

NOTE: The next 15 questions presume that the value of g is ~10 m/s/s. 11. A 1200 kg car and a 2400 kg car are lifted to the same height at a constant speed in a auto service station. Lifting the more massive car requires ____ work. a. less b. the same c. twice as much d. four times as much e. more than 4 times as much Answer: C The amount of work done by a force to displace an object is found from the equation W = F*d*cos(Theta) The force required to raise the car at constant speed is equivalent to the weight (m*g) of the car. Since the 2400-kg car weighs 2X as much as the 1200-kg car, it would require twice as much work to lift it the same distance.

12. An arrow is drawn back so that 50 Joules of potential energy is stored in the stretched bow and string. When released, the arrow will have a kinetic energy of ____ Joules. a. 50 b. more than 50 c. less than 50

Answer: A A drawn arrow has 50 J of stored energy due to the stretch of the bow and string. When released, this energy is converted into kinetic energy such that the arrow will have 50 J of kinetic energy upon being fired. Of course, this assumes no energy is lost to air resistance, friction or any other non-conservative forces and that the arrow is shot horizontally.

13. A child lifts a box up from the floor. The child then carries the box with constant speed to the other side of the room and puts the box down. How much work does he do on the box while walking across the floor at constant speed? a. zero J b. more than zero J c. more information needed to determine

Answer: A For any given situation, the work done by a force can be calculated using the equation W = F*d*cos(Theta) where F is the force doing the work, d is the displacement of the object, and Theta is the angle between the force and the displacement. In this specific situation, the child is applying an upward force on the box (he is carrying it) and the displacement of the box is horizontal. The angle between the force (vertical) and the displacement (upward) vectors is 90 degrees. Since the cosine of 90-degrees is 0, the child does not do any work upon the box. A detailed discussion of a similar situation (the waiter and the tray of food) can be found at The Physics Classroom.

15. A platform diver weighs 500 N. She steps off a diving board that is elevated to a height of 10 meters above the water. The diver will possess ___ Joules of kinetic energy when she hits the water. a. 10 b. 500 c. 510 d. 5000 e. more than 5000 .

Answer: D The use of the work-energy theorem and a simple analysis will yield the solution to this problem. Initially, there is only PE; finally, there is only KE. Assuming negligible air resistance, the kinetic energy of the diver upon hitting the water is equal to the potential energy of the diver on top of the board.

PEi = KEf

m*g*hi = KEf Substituting 500 N for m*g (500 N is the weight of the diver, not the mass) and 10 m for h will yield the answer of 5000 J.

16. A ball is projected into the air with 100 J of kinetic energy. The kinetic energy is transformed into gravitational potential energy on the path towards the peak of its trajectory. When the ball returns to its original height, its kinetic energy is ____ Joules. Do consider the effects of air resistance a. less than 100 b. 100 c. more than 100 d. not enough information given

Answer: A During any given motion, if non-conservative forces do work upon the object, then the total mechanical energy will be changed. If non-conservative forces do negative work (i.e., Fnc*d*cos(Theta) is a negative number), then the final TME is less than the initial TME. In this case, air resistance does negative work to remove energy from the system. Thus, when the ball returns to its original height, their is less TME than immediately after it was thrown. At this same starting height, the PE is the same as before. The reduction in TME is made up for by the fact that the kinetic energy has been reduced; the final KE is less than the initial KE.

18. A 10-Newton object moves to the left at 1 m/s. Its kinetic energy is approximately ____ Joules. a. 0.5 b. 1 c. 10 d. more than 10 Answer: A The KE of any object can be computed if the mass (m) and speed (v) are known. Simply use the equation KE=0.5*m*v2

In this case, the 10-N object has a mass of approximately 1 kg (use Fgrav = m*g). The speed is 1 m/s. Now plug and chug to yield KE of approximately 0.5 J.

19. Luke Autbeloe stands on the edge of a roof throws a ball downward. It strikes the ground with 100 J of kinetic energy. Luke now throws another identical ball upward with the same initial speed, and this too falls to the ground. Neglecting air resistance, the second ball hits the ground with a kinetic energy of ____ J. a. less than 100 b. 100 c. 200 d. more than 200 e. none of these

Answer: B Quite surprisingly to many, each ball would hit the ground with the same speed. In each case, the PE+KE of the balls immediately after being thrown is the same (they are thrown with the same speed from the same height). Upon hitting the ground, they must also have the same PE+KE. Since the PE is zero (on the ground) for each ball, it stands to reason that their KE is also the same. That's a little physics and a lot of logic - and try not to avoid the logic part by trying to memorize the answer.

20. An object at rest may have ______. a. speed b. velocity c. acceleration d. energy e. all of these

Answer: D An object at rest absolutely cannot have speed or velocity or acceleration. However, an object at rest could have energy if there is energy stored due to its position; for example, there could be gravitational or elastic potential energy.

21. A 50-kg platform diver hits the water below with a kinetic energy of 5000 Joules. The height (relative to the water) from which the diver dove was approximately ____ meters. a. 5 b. 10 c. 50 d. 100 Answer: B The kinetic energy of the diver upon striking the water must be equal to the original potential energy. Thus,

m*g*hi = KEf (50 kg)*(~10 m/s/s)*h = 5000 J So, h = ~10 m

23. Which requires more work: lifting a 50.0 kg crate a vertical distance of 2.0 meters or lifting a 25.0 kg crate a vertical distance of 4.0 meters? a. lifting the 50 kg crate b. lifting the 25 kg crate c. both require the same amount of work

Answer: C Work involves a force acting upon an object to cause a displacement. The amount of work done is found by multiplying F*d*cos(Theta). The equation can be used for these two motions to find the work.

Lifting a 50 kg crate vertically 2 Lifting a 25 kg crate vertically 4 meters meters W = (~250 N)*(4 m)*cos(0) W = (~500 N)*(2 m)*cos(0) W = ~1000 N W = ~1000 N (Note: The weight of a 25-kg object is approximately (Note: The weight of a 50-kg object is 250 N; it takes 250 N to lift the object up.) approximately 500 N; it takes 500 N to lift the object up.)

24. A 50.0 kg crate is lifted to a height of 2.0 meters in the same time as a 25.0 kg crate is lifted to a height of 4 meters. The rate at which energy is used (i.e., power) in raising the 50.0 kg crate is ____ as the rate at which energy is used to lift the 25.0 kg crate. a. twice as much b. half as much c. the same Answer: C The power is the rate at which work is done (or energy is used). Power is found by dividing work by time. It requires the same amount of work to do these two jobs (see question #23) and the same amount of time. Thus, the power is the same for both tasks.

25. Using 1000. J of work, a small object is lifted from the ground floor to the third floor of a tall building in 20.0 seconds. What power was required in this task? a. 20 W b. 50 W c. 100 W d. 1000 W e. 20000 W

Answer: B This is a relatively simple plug-and-chug into the equation P=W/t with W=1000. J and t=20.0 s.

Part B: Straightforward Computational Problems 26. Approximate the work required lift a 2.5-kg object to a height of 6.0 meters. PSYW

Answer: ~150 J The work done upon an object is found with the equation W = F*d*cos(Theta) In this case, the d=6.0 m; the F=24.5 N (it takes 24.5 N of force to lift a 2.5-kg object; that's the weight of the object), and the angle between F and d (Theta) is 0 degrees. Substituting these values into the above equation yields W = F*d*cos(Theta) = (24.5 N)*(6 m)*cos(0) = ~150 J (147 J)

28. Eddy, whose mass is 65.0-kg, climbs up the 1.60-meter high stairs in 1.20 s. Approximate Eddy's power rating. PSYW

Answer: P = 849 Watts Eddy's power is found by dividing the work which he does by the time in which he does it. The work done in elevating his 65.0-kg mass up the stairs is determined using the equation W = F*d*cos(Theta) where F = m*g = 637 N (the weight of the 65.0 kg object), d =1.60 m and Theta = 0 degrees (the angle between the upward force and the upward displacement). Solving for W yields 1019.2 Joules. Now divide the work by the time to determine the power: P = W/t = (1019.2 J)/(1.20 s) = 849 Watts

31. A 510-kg roller coaster car starts at a height of 32.0 m. Assuming negligible energy losses to friction and air resistance, determine the PE, KE, and speed of the car at the various locations (A, B, C, D, and E) along the track. Height PE* KE* velocity Location (m) (J) (J) (m/s) Start 32.0 160 000 J 0 J 0 A 28.0 140 000 J 20 000 J 8.9 m/s B 11.0 55 000 J 105 000 J 20.5 m/s C 20.0 100 000 J 60 000 J 15.5 m/s D 5.0 25 000 J 135 000 J 23.2 m/s E 15.0 75 000 J 85 000 J 18.4 m/s F 0 0 J 160 000 J 25.3 m/s

*rounded to the second significant digit Answers: See above table (answers in red) The potential energy for every row can be found using the equation PE = m*g*h where m=510 kg and g = 9.8 m/s/s. In the first row, the total mechanical energy (KE + PE) equals 160 000 J (rounded). Since no work is done by non-conservative forces, the total mechanical energy must be 160 000 J in all the other rows. So the KE can be computed by subtracting the PE from 160 000 J. The velocity can be found using the equation: KE = 0.5*m*v2 where m=510 kg. The algebraic rearrangement of this equation results in v = SQRT(2*KE/m).

32. A 65.8-kg skier accelerates down an icy hill from an original height of 521 meters. Use the work- energy theorem to determine the speed at the bottom of the hill if... a. ... no energy is lost or gained due to friction, air resistance and other non-conservative forces. PSYW b. ... 1.40*105 J of energy are lost due to external forces. PSYW

Answers: (a) v = 101 m/s; (b) v = 77.2 m/s a. Use the work energy theorem:

KEi + PEi + Wnc = KEf + PEf

The PEf can be dropped from the equation since the skier finishes on the ground at zero height. The KEi can also be dropped since the skier starts from rest. The Wnc term is dropped since it is said that no work is done by non-conservative (external) forces. The equation simplifies to

PEi = KEf The expressions for KE (0.5*m*v2) and PE (m*g*h) can be substituted into the equation: 2 m*g*h = 0.5*m*vf where m=65.8 kg, h=521 m, g=9.8 m/s/s. Substituting and solving for vf yields 101 m/s. b. This equation can be solved in a similar manner, except that now the Wext term is -140000 J. So the equation becomes 2 m*g*h - 140000 J = 0.5*m*vf

Now substituting and solving for vf yields 77.2 m/s.

33. Use the work-energy theorem to determine the force required to stop a 988-kg car moving at a speed of 21.2 m/s if there is a distance of 45.7 m in which to stop it. PSYW

Answer: F = 4.86*103 N The work energy theorem can be written as

KEi + PEi + Wnc = KEf + PEf

The PEi and PEf can be dropped from the equation since they are both 0 (the height of the car is 0 m). The KEf can also be dropped for the same reason (the car is finally stopped). The equation simplifies to

KEi + Wnc = 0

2 The expressions for KE (0.5*m*v ) and Wnc (F*d*cos[Theta]) can be substituted into the equation: 2 0.5*m*vi + F*d*cos[Theta] = 0 where m=988 kg, vi=21.2 m/s, d=45.7 m, and Theta = 180 degrees. Substituting and solving for F yields 4.86*103 N.

36. Several physical situations are described below. For each situation, simplify the work-energy equation by canceling any zero terms and any energy terms (whether KE or PE) which are unchanging. Explain each term which gets canceled. The first problem is done as an example. Simplification of Work- Description of Physical Situation Energy Equation a. A ball starts from rest on top of a tall pillar and falls to the ground below. Assume the effect PEi = KEf of air resistance is negligible. (Since initially at rest, KEi = 0 and cancels. Since the final height is 0, PEf = 0 and cancels. Since non-conservative forces are not doing work, Wnc = 0)

b. A ball starts from rest at an elevated position along an inclined plane and rolls to the ground below. Assume that the effect of PEi = KEf friction and air resistance is negligible.

(Initially the ball is at rest, KEi =

0 and cancels. Since the final

height is 0, PEf = 0 and cancels. Since non-conservative forces

are not doing work, Wnc c. A ball starts from rest at an elevated position along an inclined plane and rolls to the ground below. Consider the effect of friction PEi + Wnc = KEf and air (Initially the ball is at rest, KEi = 0 and resistance. cancels. Since the final height is 0, PEf = 0 and cancels. Friction is a non- conservative forces and it does work so Wnc does not cancel.)

d. A track is constructed by stretching a grooved and pliable material between two lab poles. A metal ball starts from rest at PEi + Wnc = PEf point A and rolls to point B. Friction and air resistance have an (The ball is at rest in both its initial and effect upon the ball's motion. final states, so both KEi = 0 and KEf = 0 and cancels. Friction and air resistance are non-conservative forces and do work so Wnc does not cancel.)

e. A pendulum bob is mounted on top of a lab pole and drawn back to a string which it tied between two other poles. The pendulum PEi = KEf + PEf bob is released from rest. Upon reaching the vertical, the string hits (Initially the ball is at rest, KEi = 0 and a barrier and a new pivot point is established as the bob continues cancels. At the point shown in the in motion along an upward trajectory. Ignore the effect of air diagram as the final state, the ball has resistance. both height and would still be moving; so neither final energy term would cancel. The only non-conservative force present - tension - does not do work since it is directed at 90-degrees to the direction of motion; so Wnc cancels.) f. A Hot Wheels car starts from rest on top of an inclined plane and rolls down the incline through a loop and along a horizontal PEi + Wnc = KEf surface. Friction and air resistance have a significant effect on the (Initially the car is at rest, KEi = 0 and car. cancels. Since the final height is 0, PEf = 0 and cancels. Friction is a non- conservative forces and it does work so Wnc does not cancel.) g. An unattended hot dog wagon starts from rest and rolls down a hill and up a second hill. Ignore the effect of friction and air PEi = KEf + PEf resistance. (Initially the wagon is at rest, KEi = 0 and cancels. The wagon has height in its initial state, so the PEi term does not cancel. At the point shown in the diagram as the final state, the wagon has both height and would still be moving; so neither final energy term would cancel. There is no friction nor air resistance and the normal force does not do work since it is perpendicular to the displacement; so Wnc cancels.) h. A roller coaster car is already in motion on the top of the first drop and rolls along the track over a couple of hills. Ignore the PEi = KEf + PEf effect of friction and (Initially the car is at rest, KEi = 0 and air resistance. cancels. The car has height in its initial state, so the PEi term does not cancel. At the point shown in the diagram as the final state, the car has both height and would still be moving; so neither final energy term would cancel. There is no friction nor air resistance and the normal force does not do work since it is perpendicular to the displacement; so Wnc cancels.) i. A cross-country skier is in motion on top of a small hill. He skis down the hill into the valley and up a second smaller hill. He uses KEi + PEi + Wnc = KEf + PEf his poles to propel himself during the entire motion. Ignore the (Initially, the skier has height and motion and so neither initial energy term will cancel. The skier is using his poles to propel himself so there is a non-conservative force doing work; the Wnc term does not cancel. In the final state, the skier is moving and has height (presuming that the zero level is the valley) and so neither final energy term effect of friction and air resistance. will cancel.

41. A 221-gram ball is thrown at an angle of 17.9 degrees and a speed of 36.7 m/s from the top of a 39.8-m high cliff. Determine the impact speed of the ball when it strikes the ground. Assume negligible air resistance. PSYW

Answer: 46.1 m/s

Here is a situation in which the only force doing work -gravity - is a conservative force; so the total mechanical energy of the system is conserved. The conservation of energy equation can be used.

KEi + PEi = KEf + PEf

If we assign the ground below the cliff to be the zero-level of potential energy, then the PEf term can be canceled from the equation.

KEi + PEi = KEf Expressions for kinetic and potential energy can be substituted into the above equation to yield. 2 2 0.5•m•vi + m•g•hi = 0.5•m•vf By dividing each term of the equation by m, the mass can be canceled. 2 2 0.5 • vi + g • hi = 0.5 • vf Now known values can be substituted into this equation and the final velocity can be determined. The work is shown here: 2 2 0.5•(36.7 m/s) + (9.8 m/s/s)•(39.8 m) = 0.5 • vf

2 2 2 2 2 673.4 m /s + 390.0 m /s = 0.5 • vf

2 2 2 1063 m /s = 0.5 • vf

2 2 2 2127 m /s = vf

2 2 SQRT(2127 m /s ) = vf

vf = 46.1 m/s

Circular Motion and Gravitation Review

Part A: Multiple Choice 1. Which of the following statements are true of an object moving in a circle at a constant speed? Include all that apply. a. The object experiences a force which has a component directed parallel to the direction of motion. b. Inertia causes objects to move in a circle. c. There can be a force pushing outwards on the object as long as the net force in inwards. d. Because the speed is constant, the acceleration is zero. e. The acceleration and the net force vector are directed perpendicular to each other. f. If the net force acting upon the object is suddenly reduced to zero, then the object would suddenly depart from its circular path and travel tangent to the circle. g. The acceleration of the object is directed tangent to the circle.

Answer: CF A is false; if the motion is in a circle at constant speed, the net force is perpendicular to the direction of motion and there is neither a component parallel nor anti-parallel to the direction of motion.) B is false; it is centripetal force which causes the circular motion. Inertia (which is NOT a force) is merely the tendency of any moving object to continue in its straight-line constant speed path. C is true; an object which moves in a circle must have a net inward force. There are many instances of individual outwards forces which are exceeded by an individual inward force (e.g., see #5 below). D is false; acceleration occurs when there is a change in velocity. Since the direction of the velocity vector is changing, there is an acceleration - an inward acceleration. E is false; the acceleration and net force are always directed in the same direction. In this case, F and a are directed inward; this happens to be perpendicular to the tangential velocity vector. F is true; if the net force is 0 N, then the moving object will maintain its state of motion. At the instant that Fnet becomes 0 N, the object is moving tangent to the circle. G is false; the acceleration is directed inwards; only the velocity is directed tangent to the circle.

For Questions #2-#5, identify the type of force which causes the following bold-faced objects to travel along a circular path. 2. An eraser is tied to a string swung in a horizontal circle. a. gravity b. normal c. tension d. applied e. friction f. spring g. electrical h. magnetic

3. The moon orbits the earth. a. gravity b. normal c. tension d. applied e. friction f. spring g. electrical h. magnetic

4. A car makes a sharp right-hand turn along a level roadway. a. gravity b. normal c. tension d. applied e. friction f. spring g. electrical h. magnetic

5. A roller coaster car passes through a loop. Consider the car at the bottom of the loop. a. gravity b. normal c. tension d. applied e. friction f. spring g. electrical h. magnetic

Answers: 2. C - Tension (A string is attached to the eraser and pulls it towards the center point of the circle.) 3. A - Gravity (All masses attract with a force of gravity. In the case of the moon and the earth, gravity pulls on the moon in a direction which is roughly perpendicular to its path.) 4. E - Friction (Once the wheels are turned, friction can push perpendicular to the wheels' direction, pushing the car towards the center of the circle.) 5. B - Normal (There are two forces at the bottom of the loop; gravity pulls outwards from the center; but the normal force exceeds the magnitude of gravity, pushing inwards with sufficient force to supply the needed centripetal force.)

6. A physics teacher ties an eraser to the end of a string and then whirls it in a counter-clockwise circle. If the teacher lets go of the string, then the eraser hits a student (or several students) in the classroom. If the string is let go when the eraser is at point X on the diagram at the right, then which student(s) in the class will the eraser hit? Write the initials in this space: ______

Answer: JG and AF Once the centripetal force is no longer present, the eraser will follow its straight-line, inertial path. This path would be directed tangent to the circle and would pass through the locations where JG and AF are sitting..

7. Which of the following statements are true about gravitational force? Identify all that apply. a. The gravitational force only acts between very, very massive objects. b. The gravitational force between an object and the earth is inversely related to the distance between the object's and the earth's center. c. The gravitational force can ALWAYS be accurately calculated by multiplying the object mass by the acceleration of gravity (m•g). d. The gravitational force acting upon an object is the same as the weight of the object. e. The gravitational force between two objects is independent of the mass of the smaller of the two objects. f. If object A gravitationally attracts object B with a force of X Newtons, then object B will also gravitationally attract object A with the same force of X Newtons. g. The doubling of the separation distance (measured from the center) between two objects will halve the gravitational force between the objects. h. It an object is placed two earth-radii above the surface of the earth, then the force of gravitational attraction between the object and the earth will be one- fourth the magnitude as on earth's surface. i. Orbiting astronauts do not experience a force of gravity; this explains why they feel weightless.

Answer: BCDF A is false; gravitational forces are universal (Newton's big idea); they act between any two objects which have mass. B is true; if the distance is increased, then the force is decreased. C is true; this is always the case. It is not true however to say that the gravitational force is equal to mass•9.8 m/s/s. The value of g varies with location and so at distances significantly further from the earth's surface, g is reduced and the gravitational force must be computed using a different value of g. D is true; weight and gravitational force are synonymous. E is false; gravitational force is dependent upon the product of the two masses. Both masses are important in the computation. F is true; for every action there is an equal and opposite reaction. This is Newton's third law, an inescapable reality about forces. G is false; doubling the separation distance will make the force one-fourth the size. H is false; placing the object at two earth-radii above earth's surface will put 3-earth radii from earth's center. This is equivalent to increasing the separation distance by a factor of 3; this decreases the force by a factor of 9. I is false; orbiting astronauts must be experiencing the force of gravity. Without this force, there would be no centripetal force and no orbit.

8. Which of the following statements are true about the acceleration of gravity? Identify all that apply. a. The acceleration of gravity experienced by objects located near to (and far from) from the earth depends upon the mass of the object. b. The acceleration of gravity experienced by objects located near to (and far from) from the earth depends upon the mass of the Earth. c. The acceleration of gravity experienced by objects located near to (and far from) the earth is inversely related to the distance between the center of the object and the center of the earth. d. Increasing the mass of an object will increase the acceleration of gravity experienced by the object. e. Doubling the distance between an object and the earth's center will decrease the acceleration of gravity by a factor of four. f. The acceleration of an orbiting satellite is equal to the acceleration of gravity at that particular location. g. If the mass of the Earth were doubled (without an alteration in its radius), then the acceleration of gravity on its surface would be approximately 20 m/s2. h. If the mass of the Earth were doubled and the radius of the earth were doubled, then the two changes would offset each other and the acceleration of gravity on its surface would still be approximately 10 m/s2.

Answer: BCEFG

2 A is false; check out the equation - g = GMcentral/R . The value of g does not depend upon the object's mass. B is true; check out the equation again. The acceleration of gravity created by the earth depends upon the earth's mass. C is true; check out the equation one more time. The separation distance is located in the denominator of the equation, indicating an inverse relationship. D is false; like statement A, this statement makes a claim that the acceleration of gravity depends upon the mass of the object. It does not. In the derivation of g from the Universal Gravitation equation, the mass cancels. :) E is true; g is inversely proportional to the square of the distance; a doubling of the distance means that you must divide the force of gravity value by 4 (22) to obtain the new force of gravity value. F is true; the acceleration of gravity is the acceleration which is caused by gravity when it is the only force. For an orbiting satellite, gravity is the only force. G is true; according to the equation, the g value is directly proportional to the mass of the earth. An increase in M results in a proportional increase in g. H is false; g is approximately 10 m/s/s on earth's surface. Doubling the mass of the earth would increase g to approximately 20 m/s/s. Then doubling the distance from the surface of the earth to its center would decrease g by a factor of 4. The new acceleration of gravity value would be approximately 5 m/s/s. Clearly, these two alterations do not offset each other.

9. Which of the following statements are true about satellites? Identify all that apply. a. Satellites are falling projectiles. b. All satellites follow circular paths. c. The orbital velocity required of a satellite is dependent upon the mass of the satellite; a more massive satellite would require a greater orbital speed. d. The orbital velocity of a satellite does not depend upon the mass of the planet around which it orbits. e. A high-altitude satellite will require a greater orbital speed than a low- altitude satellite. f. By definition, a geosynchronous satellite orbits the earth in a perfect circle, maintaining the same distance above the surface of the earth. g. Satellites travel faster along their orbital path when they are closest to the earth. h. The acceleration of a satellite varies inversely with its distance from the center of the earth. More distant satellites have smaller accelerations.

Answer: AGH A is true (mostly); satellites are projectiles which are falling towards the Earth without falling into the Earth. They are falling in the sense that (like any projectile), they are dropping below their tangential direction of motion. B is false; some satellites travel in elliptical and even very elliptical orbits. For instance, the planets orbit the Sun in mildly elliptical paths.

C is false; the equation for the orbital velocity of a satellite is v = SQRT(G•Mcentral/R). The Mcentral is the mass of the central body - the body being orbited by the satellite. As seen in the equation, the orbital velocity is independent of the mass of the satellite.

D is false; the equation for the orbital velocity of a satellite is v = SQRT(G•Mcentral/R). The Mcentral is the mass of the central body - the body being orbited by the satellite. Clearly the orbital velocity depends upon the mass of the planet being orbited.

E is false the equation for the orbital velocity of a satellite is v = SQRT(G•Mcentral/R). The R in the denominator inside the radical is the radius of the orbit. The higher altitude satellites have a greaer radius of orbit. As the R in the denominator increases, the required orbital velocity will decrease. F is false; by definition, a geosynchronous satellite has synchronized or matched its orbital period to the rotational period of the Earth about its axis - 24 hours. G is true; this is an idea known since Kepler that is explained by Newton's gravitational law. As a satellite orbits in an elliptical path, the force of gravitation is greatest at the times when it is closest to the Earth; this force causes a greater acceleration and speed of the satellite along these nearest locations of its orbit.

2 H is true; the acceleration of a satellite is given by the equation g=G•Mcentral/R where R is the radius of the orbit. The larger the R value, the smaller the acceleration. This is an inverse square relationship.

10. Which of the following statements are true about the motion of planets about the sun? Identify all that apply. a. The force of gravity is the only force which acts upon the planets. b. Their trajectories are highly elliptical. c. The planets which are furthest from the sun have the greatest period. d. For any given planet, the speed is greatest when the planet is closest to the sun. e. The velocity vector is directed tangent to the elliptical path. f. The net force vector is at all times directed perpendicular to the velocity vector. g. To keep the planet from escaping the sun's gravitational field, the net force vector is greatest when the planet is furthest from the sun.

Answer: ACDE A is true; gravity is the only force acting upon a planet. Think about it - what other physical push or pull can you identify as acting upon the earth. (People pushing on it and pulling it don't count.) B is false; their trajectories are indeed elliptical. Yet in most cases, the ellipses are very close to a circle since the eccentricities are very close to 0. C is true; this is in accord with Kepler's third law: T2/R3 = constant. Expressing this equation another way depicts the direct relationship between T and R: T2 = constant • R3. D is true; this is Kepler's second law of planetary motion. In order to sweep out equal areas in equal times, a satellite must be moving faster when closest to the earth. See a clever animation. E is true; at all times during the orbital path, the velocity of the planet is tangent to the path. Since the path has a slightly elliptical nature to it, the force and acceleration vector are not necessarily perpendicular to this velocity vector. The perpendicular orientation of the F and a vectors relative to the v vector is unique to circular motion. An excellent and informative graphic about this can be found online. F is false; see above explanation and the link to the informative graphic. G is false; the net force is a gravitational force. Gravitational forces vary inversely with distance. Thus, a planet will experience less force when furthest from the sun. You might really need to view the informative graphic.

Part B: Short Answer 11. Explain how something can be moving at a constant speed yet be accelerating at the same time.

Answer: The object would have to be traveling in a curved path since it is accelerating. Acceleration is defined as a change in velocity over a change in time. If the speed is staying the same, then the velocity must be changing by altering the direction in which the object is heading. A force is required to do this. See an informative animation and further explanation.

12. How did Newton come up with the idea that the moon is actually "falling" toward the Earth.

Answer: Newton made the connection between objects falling (accelerating) towards the earth and objects in space which are accelerating towards the earth while they are in circular motion around the earth. Both are being pulled by the earth due to the gravitational force. The moon stays in orbit due to it having the appropriate tangential velocity that keeps it from coming closer to the earth's surface. The moon, however, is still accelerating at the rate any object would have at that distance from the earth. In a sense, the moon is falling around the earth rather than into the earth.

13. Distinguish between true- and apparent-weightlessness.

Answer: True weightlessness occurs when there is an absence of gravitational force on an object. This is not a likely occurrence since every mass in the universe is attracted to every other mass in the universe with some force of gravity. All objects would experience some degree of gravitational attraction (though it may be considerably small). Apparent weightlessness is the sensation of not experiencing any external contact forces. This occurs when an object is in free-fall. Even though the object may have a large force on it, it wouldn't register on a scale that was in free-fall with it. Astronauts appear this way on the space shuttle.

14. Describe the apparent weight of a person in an elevator while upward, accelerating downward, and not accelerating.

Answer: When one stands upright on a scale, the scale measures the force of the scale pushing upwards on the body. As such, the scale does not measure one's true weight, but one's apparent weight. An elevator accelerating someone up must not only apply a force to provide the acceleration but also oppose the gravitational force on the person. Therefore, a scale would register a number larger than the gravitational force if the person is accelerating upward. In this case, a person would experience an apparent weight which is greater than usual. Likewise, if the person is accelerating downward the scale reading must be less than the gravitational force since the gravitational force is accelerating them downward. As such, the person's apparent weight is less than what they are accustomed to experiencing. If the elevator is moving at constant velocity or at rest, the scale reading must match the gravitational force in order to have a net force of 0. The apparent weight would be equal to the actual weight of the person. 16. The diagram at the right shows a satellite orbiting the Earth in an elliptical path in a clockwise direction. a. Draw a vector representing the velocity of the object at position A. b. raw a vector representing the force on the object at position D. c. Draw a vector representing the acceleration of the object at position C. d. At which of the four positions is the satellite moving fastest?

Answer: See diagram for answers. The force and acceleration vectors are directed towards the planet being orbited (which is at a foci of the ellipse) and the velocity vector is directed tangent to the path.

17. A 1.20-kg bucket is held by a string and whirled in a vertical circle. The radius of the circle is 1.30 m. The speed of the bucket is 3.70 m/s at the top of the loop and 7.10 m/s at the bottom of the loop. On the diagrams below, construct a free-body diagram (label all forces according to type) for the bucket at both the top and the bottom of the loop. Fill in the blanks and clearly indicate the magnitude of the individual force values on your free-body diagram.

Answers: See diagram above. In each case, the acceleration is found using the equation: a = v2/R. For the top of the loop, a = (3.70)2/(1.30) = 10.53 m/s2. For the bottom of the loop, a = (7.10)2/(1.30) = 38.78 m/s2. The net force is always found by 2 Fnet = m•a. At the bottom of the loop, Fnet = (1.20 kg)•(10.53 m/s ) = 12.6 N. At the bottom of the loop, Fnet = (1.20 kg)•(38.78 m/s2) = 46.5 N. The force of gravity is always found by m•g where g = 9.8 m/s2. At both top and bottom of the loop, this value is 11.8 N. From the free-body diagram, one can reason the following:

Top: Fnet = Fgrav + Ftens which can be rearranged to Ftens = Fnet - Fgrav

Bottom: Fnet = Ftens - Fgrav which can be rearranged to Ftens = Fnet + Fgrav

Using these equations and the calculated values of Fgrav and Fnet, one can determine the tension at the top and bottom of the circle.

18. Construct a free-body diagram showing the direction and types of forces acting upon the car at the top and the bottom of a loop. Be sure to label the forces according to type.

Anna experiences a downward acceleration of 24.0 m/s2 at the top of the loop and an upwards acceleration of 12.0 m/s2 at the bottom of the loop. Use Newton's second law to fill in all the blanks and to ultimately determine the normal force acting upon Anna's 800. kg roller coaster car. PSYW Top Bottom

Fnet = 19200 N, down Fnet = 9600 N, up

Fgrav = 7840 N, down Fgrav = 7840 N, down

Fnorm = 11360 N, down Fnorm = 17440 N, up Answer: See diagram and table above. This problem is very similar to question #17 above. The net force is found by calculating m•a; its direction is in the same direction as the a vector. The gravitational force is found from m•g; the value of 9.8 m/s/s is used for g. Now the free-body diagram can be used to generate the following equations for the top and the bottom of the loop:

Top: Fnet = Fgrav + Fnorm which can be rearranged to Fnorm = Fnet - Fgrav

Bottom: Fnet = Fnorm - Fgrav which can be rearranged to Fnorm = Fnet + Fgrav

The difference in these two equations is attributed to the fact that at the top of the loop, the Fnorm and the Fgrav point in the same direction as the net force vector; at the bottom of the loop, they point in opposite directions. Using the calculated values of Fnet and Fgrav, the normal force can now be calculated.

]

20. Pete Zaria is riding the Cliff Hanger at Great America. Pete enters a cylindrical barrel which makes 18.5 revolutions every minute. The diameter of the barrel is 6.92 m. Pete's mass is 64.7 kg. Construct a free-body diagram showing the forces acting upon Pete. Calculate the acceleration and net force acting upon Pete Zaria. Finally, determine the coefficient of friction between Pete and the walls that would be required to support Pete's weight. PSYW

Answer: 0.755

As usual, Fgrav = m•g = (64.7 kg)•(9.8 m/s/s) = 634 N. This downward force must be balanced by the friction force since there is no vertical acceleration. Therefore, Ffrict = 634 N. The net inward force is supplied by the normal force; so a calculation of the net force will yield the value of the normal force. Acceleration is found from speed and radius. The speed is: v = (18.5 rev/60 sec)• (2 • pi • 3.46 m/rev) = 6.703 m/s. Now the acceleration is: a = v2/R = (6.703 m/s)2/(3.46 m) = 12.986 m/s2 Now the net force (and the normal force) can be calculated:

Fnet = Fnorm = m•a = (64.7 kg)•(12.986 m/s/s) = 840.20 N Now that the normal force and the friction force have been found, the coefficient of friction can be determined:

µ = (Ffrict)/(Fnorm) = (634 N)/(840.20 N) = 0.755

Part D: Short Computations

21. Two objects attract each other with a force of gravity (Fgrav) of 36 N. If the distance separating the objects is doubled, then what is the new force of gravitational attraction? PSYW

22. Two objects attract each other with a force of gravity (Fgrav) of 36 N. If the distance separating the objects is doubled and the masses one of the objects is tripled, then what is the new force of gravitational attraction? PSYW

23. Two objects attract each other with a force of gravity (Fgrav) of 36 N. If the distance separating the objects is increased by a factor of 4 and the masses of both objects are tripled, then what is the new force of gravitational attraction? PSYW

Answers: For Questions #21-#23, Newton's universal gravitation equation must be used as a guide to thinking about how an alteration in one variable would effect another variable. From the equation, it can be deduced that a change in either one of the masses would produce a proportional change in the force of gravity. However, a change in the separation distance will produce the inverse square effect upon the force of gravity. So increasing separation distance by a factor of x, will decrease the force of gravity by a factor of x2. This reasoning leads to the following solutions. 21. New force = 36 N / 4 = 9.0 N 22. New force = 36 N • 3 / 4 = 27.0 N 23. New force = 36 N •3 •3 / 16 = 20.3 N

Part E: Problem-Solving 29. A roller coaster car loaded with passengers, has a mass of 500. kg; the radius of curvature of the track at the bottom point of a dip is 12.0 m. The vehicle has a speed of 18.0 m/s at this point. a. In the space below, draw a free-body diagram for the car (label forces according to type). b. Calculate the acceleration and the net force acting upon the car. PSYW c. Calculate the force exerted on the vehicle by the track? PSYW

Answers:

The free-body diagram (part a) is shown at the right. The acceleration of the car can be computed as follows: a = v2/R = (18.0 m/s)2/(12.0 m) = 27.0 m/s2 (part b) The net force can be found in the usual manner: 2 Fnet = m•a = (500. kg)•(27.0 m/s ) = 13500 N (part b) Since the center of the circle (see diagram) is above the riders, then both the net force and the acceleration vectors have an upward direction. The force of gravity is downwards, so the net force is equal to the upward force minus the downward force:

Fnet = Fnorm - Fgrav where Fgrav = m • g = (500. kg) • (9.8 m/s/s) = 4900 N Thus,

Fnorm = Fnet + Fgrav = 13500 N + 4900 N = 18400 N (part c)

31. What is the tension in a 0.500 meter rope which carries a 2.50 kg bucket of water in a vertical circle with a velocity of 3.00 m/s when the bucket is at the bottom of its swing?

Answer: 69.5 N Begin with a free-body diagram for the bucket at the bottom of its string. Since the bucket is at the bottom of the circle, the net force is upwards (inwards). Notice that the tension force must be greater than the force of gravity in order to have a net upwards (inwards) force. Also note that while Ftens is directed in the direction of the net force, Fgrav is in the opposite direction. Thus, the equation can be written:

Fnet = Ftens - Fgrav

2 This equation can be rearranged to solve for tension; the expressions for Fgrav (m•g) and Fnet (m•v /R) can be substituted into it: 2 2 Ftens = Fnet + Fgrav = m•v /R + m•g = (2.50 kg)•(3.00 m/s) /(0.500 m) + (2.50 kg)•(9.8 m/s2)

Ftens = 69.5 N

Static Electricity Review

Part A: Multiple Choice 1. Which of the following are true of static charges? Choose all that apply. a. Like charges repel. b. Like charges attract. c. Opposite charges repel. d. Opposite charges attract. e. A positively charged object has lost electrons. f. A positively charged object has gained protons. g. A negatively charged object has lost protons. h. A negatively charged object has gained electrons.

Answer: ADEH A and D are true; they are the simple statement of our essential charge interactions, best remembered by the jingle: "opposites attract, likes repel." B and C are false for these same reasons; they violate the basic statement of charge interactions. E and H are true. For an object to become charged, it must either gain or lose electrons. Losing electrons results in more positive charge than negative charge, making the object charged positively. Gaining electrons results in more negative charge than positive charge, making the object charged negatively. Protons are tightly bound in the nucleus of atoms and can never be added nor removed from atoms by ordinary electrostatic methods. The same reasoning leads one to reject choice F and G as possible true statements; the suggest that protons can be added or removed.

2. Identify the following objects as being either

a. positive b. negative c. neutral

If there is no conclusive evidence, then select all that could be true.

Description of Object a, b, or c? i. An object possesses more protons than electrons. A ii. An object possesses more neutrons than electrons. ABC iii. A formerly neutral object that just lost some electrons. A iv. A formerly neutral object that just gained some electrons. B v. An object which attracts a negatively-charged balloon. AC vi. An object which attracts neutral paper bits and attracts a negatively-charged balloon. A vii. An object which attracts neutral attracts paper bits and repels a negatively-charged balloon. B viii. An object which attracts a negatively-charged balloon and attracts a positively- charged balloon. C ix. An object which attracts a charged balloon (balloon A) which is attracted to a negatively-charged balloon (balloon B). ABC x. An object which attracts a balloon (balloon C) which is repelled by a negatively- charged balloon (balloon D). AC xi. An object which repels a balloon (balloon E) which is repelled by a positively- charged balloon (balloon F). A xii. An object around which the electric field vector is directed inwards. B xiii. An object around which the electric field vector is directed outwards. A

Answer: See table above i. Protons are positive; electrons are negative; more protons than electrons would mean an overall positive charge. ii. Neutrons are neutral and will not have an influence on the overall charge. So this object could be +, -, or neutral depending on the relative number of protons and electrons. iii. If a neutral object loses some electrons, then it will possess more protons (positive charge) than electrons (negative charge); the object will have an overall + charge. iv. If a neutral object gains some electrons, then it will possess more electrons (negative charge) than protons (positive charge); the object will have an overall - charge. v. A negatively charged object would never be attracted to a - balloon. However, a + object would be attracted to a - balloon (opposites attract) and a neutral object would be attracted to a - balloon (since neutral objects are attracted to any charged object). vi. An object which attracts neutral paper bits must be charged. If the object also attracts a - balloon, then it must have a + charge (opposites attract). vii. If an object has a repulsive interaction with any other object, then it is definitely charged. Since like charges repel, the charge which it has is the same type of charge as the charge on the object which it repels. This object has a - charge. viii. A neutral object will be attracted to any charged object, whether + or -. In this case, the object attracts both + balloons and - balloons. Only a neutral object could do this. ix. Balloon A is either + (since it attracts a - object) or neutral (since neutral objects will also attract any charged object). If our object attracts balloon A, then it is either + or - (since such objects would attract balloon A if it were neutral) or even neutral (since it would attract balloon A if balloon A were +). x. Balloon C must be - since it repels another - balloon (like charges repel). If our object attracts balloon C, then it is either + (since opposites attracts) or neutral (since neutral objects are attracted to any charged object). xi. Balloon E must be + if it repels balloon F (like charges repel). If our object repels balloon E, then it to must have a + charge. xii. By definition, the electric field vector points in the direction which a + test charge would be accelerated. So by extension, the E vector points towards - charges and away from + charges. This object must have a - charge. xiii. Using the same reasoning as above, this object must have a + charge.

3. Consider the following statements given below and determine if the charge of Object A is:

a. positive b. negative c. neutral

Description of Charging Method a, b, or c? i. Object A is charged by friction using animal fur. Animal fur has a greater electron A affinity than object A. ii. Object A is charged by contact using a negatively charged object. B iii. Object A is charged by induction using a positively charged object. B iv. Object A is used to charge an aluminum plate by the method of induction. The aluminum plate acquires a positive charge. B v. Object A is used to charge an aluminum plate by the method of A induction. The aluminum plate acquires a negative charge. vi. A rubber rod has a greater electron affinity than animal fur. The rubber rod is charged by friction with animal fur. The rubber rod is then used to charge Object A by B the method of contact. vii. A rubber rod has a greater electron affinity than animal fur. The rubber rod is charged by friction with animal fur. The rubber rod is then used to charge Object A by A the method of induction. viii. A rubber rod has a greater electron affinity than animal fur. The rubber rod is charged by friction with animal fur. The rubber rod is then used to charge an aluminum pop can by the process of induction. The pop can is then contacted to object A A.

Answer: See above table. i. When two objects are charged by friction, electrons are transferred between objects. The object made of a material with the greatest electron affinity is the object which receives the electrons; the other object loses electrons. Thus, Object A loses electrons and acquires a + charge. ii. When charging an object by contact, the object receives the same type of charge as the object used to charge it. iii. When charging an object by induction, the object receives the opposite type of charge as the object used to charge it. The answer to question 12 includes a detailed explanation of the induction charging method. iv. Since induction charging gives objects opposite types of charge, Object A must have an opposite charge as the aluminum plate. v. The same reasoning used in iv. above can be used to explain this question. vi. The rubber rod must become charged negatively since it would acquire electrons from the animal fur. It would then charge Object A with the same type of charge (-) since contact charging results in charging an object with the same type of charge. vii. The rubber rod must become charged negatively since it would acquire electrons from the animal fur. It would then charge Object A with the opposite type of charge (+) since induction charging results in charging an object with the opposite type of charge. viii. The rubber rod must become charged negatively since it would acquire electrons from the animal fur. It would then charge the aluminum plate with the opposite type of charge (+) since induction charging results in charging an object with the opposite type of charge. The aluminum plate would then charge Object A positively since contact charging results in charging an object with the same type of charge.

4. A neutral plastic strip is rubbed with cotton and acquires a positive charge. Which of the following statements are true of the positively-charged strip? a. It lost some electrons to the cotton during the charging process. b. It lost all of its electrons to the cotton during the charging process. c. It has the opposite charge as the cotton. d. It would now be repelled by the piece of cotton which was used to charge it. e. It gained protons during the rubbing process. f. As a material, plastic has a greater affinity for electrons than cotton. g. It could exert either a repulsive or attractive influence upon neutral paper bits. h. It has an excess of protons compared to the number of electrons. i. It could be used to charge an electroscope negatively by the process of induction. j. It lost negative electrons and gained positive electrons during the charging process. k. It lost neutrons during the charging process (or at the very least, its neutrons became ineffective).

Answer: ACHI a. During charging by friction, electrons are transferred from one object to the other object. The object which acquires a + charge is the object which loses the electrons. b. While some electrons are lost, not all electrons lost. c. Charging by friction results in two objects with the opposite type of charge. d. If the two rubbed objects are brought near, then they will attract (rather than repel) since they are charged oppositely. e. Protons are never gained or lost during ordinary electrostatic experiments. They are tightly bound in the nuclei of atoms; it would require an atom-smasher to induce protons to move. f. The object which has the greater electron affinity is the one which acquires the - charge during the friction charging process. g. A neutral object like paper could only attract a charged object like the plastic; repulsion is only observed of two charged objects having the same type of charge. h. Since it has lost electrons, the plastic will have an excess of protons. i. If a positively-charged object is used to charge an electroscope by induction, then the electroscope would acquire a - charge. Induction charging results in objects with opposite types of charge. j. There is no such thing as positive electrons (at least not for our purposes). k. As you may have noticed, talk of neutrons in a unit on electrostatics is largely absent. Neutrons have nothing to do with the discussion.

5. A positively-charged glass rod is touched to the plate of a neutral electroscope. Upon contact, the electroscope becomes charged and the needle deflects. Which of the following statements are true of the charged electroscope? a. The electroscope is now charged positively. b. The electroscope and the glass rod now have the same type of charge. c. The electroscope was charged by the method of induction. d. The electroscope gained protons during the charging process. e. The electroscope gained electrons during the charging process. f. The electroscope lost all of its electrons during the charging process. g. During the charging process, some electrons left the electroscope and entered the glass rod. h. The number of electrons present in the electroscope is less than the number of protons. i. The electroscope needle will deflect even more if the glass rod is brought near it again. j. The electroscope needle would slowly approach the neutral position if a negatively-charged balloon is brought near.

Answer: ABGHIJ a. Charging by contact places the same type of charge on the object being charged. So if a positively charged rod is used to charge the electroscope, the electroscope becomes charged positively. b. Same as part a. c. Induction charging would involve bringing the glass rod near to the electroscope without making contact. The method described above is the contact charging method. d. Protons are never gained or lost during ordinary electrostatic experiments. They are tightly bound in the nuclei of atoms; it would require an atom-smasher to induce protons to move. e. The glass rod, being charged positively, would attract electrons which were present upon the electroscope. These electrons would be transferred to the glass rod. The electroscope would lose electrons, not gain them. In fact, that is why it becomes charged positively. f. While the electroscope does lose electrons, it does not lose all its electrons. g. See the explanation in part e above. h. Having lost electrons, there are now more protons in the electroscope than there are electrons. i. The electroscope and the glass rod are both positively-charged. The presence of the glass rod near the positively-charged electroscope would induce the needle to deflect even more rather than cause it to move towards the neutral position. j. Since the electroscope is now charged positively, bringing a negatively charged object nearby would cause the needle to return to the neutral position.

6. A negatively-charged balloon is brought near to (without touching) a neutral electroscope. With the negatively-charged balloon held near, the electroscope is momentarily touched by a hand (ground). The balloon is then removed and the needle deflects, thus showing a charge. Which of the following statements are true of the charged electroscope? a. The electroscope was charged by the method of induction. b. The charge on the electroscope is the same type of the charge on the balloon. c. The electroscope has more protons than electrons. d. The electroscope gained protons during the charging process. e. The electroscope gained electrons during the charging process. f. The electroscope lost all of its electrons during the charging process. g. The electroscope lost some electrons during the charging process. h. During the charging process, protons moved from the electroscope to the balloon. i. During the charging process, electrons moved from the electroscope to the balloon. j. During the charging process, electrons moved from the electroscope to the hand (ground). k. During the charging process, electrons moved from the hand (ground) to the electroscope.

Answer: ACGJ a. This is a perfect description of the induction charging method. b. Induction charging places a charge on an object which is opposite of the type of charge on the object used to charge it. Thus, the electroscope acquires a positive charge. c. Since the electroscope is positively charged, there are more protons than electrons on the electroscope. d. Protons are never gained or lost during ordinary electrostatic experiments. They are tightly bound in the nuclei of atoms; it would require an atom-smasher to induce protons to move. e. When the - balloon was brought near, electrons in the electroscope move to a location far from the balloon (like-charged objects repel). When touched by the hand (ground), the electrons move even further, entering the hand and the person. So in the induction charging process, the electroscope lost electrons. f. While the electroscope loses electrons, it does not lose all of its electrons. g. See explanation to step e above. h. Protons are never gained or lost during ordinary electrostatic experiments. They are tightly bound in the nuclei of atoms; it would require an atom-smasher to induce protons to move. i. Electrons cannot move from the balloon to the electroscope because there is no available conducting pathway. The air separates the two objects and serves as an insulator, thus preventing the movement of electrons between them. j. See explanation to step e above. k. See explanation to step e above.

7. The outer shell electrons in metals are not tightly bound to the nuclei of their atoms. They are free to roam throughout the material, moving from atom to atom. These materials are good ____. a. conductors b. insulators c. for nothing

Answer: A Metals are great conductors. That is, electrons are free to move across the surface of metals, migrating from atoms to atoms. Being conductors, metals can serve as pathways for the movement of electrons from atom to atom, molecule to molecule, and object to object. All of this is due to the fact that the outer electrons (also known in some chemistry circles as valence electrons) are not tightly bound to the nuclei of their atoms.

For Questions #8 and #9, consider the following situation. Connor Duct (Con to his friends) takes a positively charged rubber rod and touches a metal sphere on an insulated stand as shown at the right. Draw the direction of electron flow. 8. The charge on the metal sphere will end up ___. a. neutral b. negative c. positive

Answer: C This is an example of charging by contact. When a neutral object is charged by contact, the object acquires the same type of charge as the object used to charge it. In this case, the sphere acquires the same type of charge as the rubber rod - a positive charge. (As for how this happens, see the next question.)

9. The sphere acquires this charge because ____. a. electrons move from the rubber rod to the sphere b. electrons move from the sphere to the rubber rod c. protons move from the rubber rod to the sphere d. protons move from the sphere to the rubber rod e. the rubber rod creates a charge on the sphere

Answer: B Once touched, the sphere and the rubber rod transfer electrons between them. Since the rubber rod is + and electrons are -, the rubber rod attracts electrons from the sphere towards itself. Some of these electrons move into the rubber rod, thus reducing the amount of + charge upon it. The sphere in turn becomes charged positively since it has lost electrons.

10. If a positively charged plate is brought near the top of a positively-charged electroscope, then the deflected needle will ____. a. not move at all b. be deflected more c. be deflected less

Answer: B A positively charged electroscope has protons and electrons; only there would be more protons compared to the number of electrons. There is an excess of + charge ("protons without partners") uniformly distributed over the electroscope. Bringing another + object nearby, will draw electrons out of the needle of the electroscope up into the plate. This would leave and even greater excess of charge in the needle of the electroscope. As such, the needle would be seen to deflect even more.

12. A negatively charged balloon is brought near a metal can that rests on a wood table. The side of the can opposite the balloon is momentarily touched. The can is then ______. a. positively charged b. negatively charged c. uncharged

Answer: A This is an example of charging by induction. In all such cases, the object being charged acquires a charge that is opposite in type to the object used to charge it. Bringing the - balloon near the can will chase electrons in the can to the far side of the can. When touched on the far side, electrons are chased further, exiting the can and migrating into the finger and the person. Having lost electrons, the can now has an excess of positive charge. In this example, the finger and person serve as the ground - a seemingly infinite reservoir or holding place of charge.

13. Charge carriers in a metal are electrons rather than protons. This is due to the fact that electrons are ______. a. loosely bound b. lighter c. far from a nucleus d. all of the above e. none of the above

Answer: A Electrons are light and they are far from the nucleus, but neither of these features explain why they are the carriers of charge in electrostatic experiments. Electrons, unlike the protons, are not bound up in an inescapable condition within the atoms of metals. As such, they can move from atom to atom throughout the surface of the metal and can even from object to object. On the other hand, protons are tightly bound in the nucleus of an atom and are unable to escape by ordinary electrostatic means.

14. Two like charges ______. a. attract each other b. repel each other c. neutralize each other d. have no effect on each other e. must be neutrons

Answer: B There are three charge interactions which should be known; this is one of them: Like charged objects repel. The other two: i) oppositely-charged objects attract, and ii) a neutral object and any charged object (whether + or -) will attract.

15. If you comb your hair and the comb becomes positively charged, then your hair becomes ______. a. positively charged b. negatively charged c. uncharged Answer: B This is an example of charging by friction. Two neutral objects, when rubbed together, will transfer electrons between them. The object with the greatest electron affinity will acquire electrons from the other object; the two become charged oppositely. In this case, the hair acquires electrons from the comb, leaving the comb with a shortage of electrons and a + charge; the hair now has an excess of electrons and a - charge.

16. Which of the following charging methods can result in charging an object with a negative charge? Choose all that apply. a. charging by friction b. charging by contact c. charging by induction

Answer: ABC If object A is rubbed with another object having a lower electron affinity, then the atoms of object A will steal electrons from the other object and acquire a negative charge. If object A is touched by a negatively-charged object, then it will become charged negatively. If a positive charge object is used to charge object A by the method of induction, then object A will acquire a negative charge as it draws electrons from the ground.

17. Which of the following charging methods work without ever touching the object to be charged to the object used to charge it? Choose all that apply. a. charging by friction b. charging by contact c. charging by induction

Answer: C In charging by induction, a charged object is brought near an object without touching it. The presence of the charge object induces electron movement and a polarization of the object. Then conducting pathway to ground is established and electron movement occurs between the object and the ground. During the process, the charged object is never touched to the object being charged.

18. Which of the following charging methods result in charging an object opposite to the charge on the object used to charge it? Choose all that apply. a. charging by friction b. charging by contact c. charging by induction Answer: AC In charging by friction, the two objects being rubbed acquire opposite types of charge as electrons are transferred between objects. In charging by contact, both objects acquire the same type of charge. In charging by induction (as described in the answer to Question #17), the objects acquire the opposite type of charge.

19. Which of the following statements are true statements about electric force? Choose all that apply. a. Electric force is a contact force. b. Electric forces can only act between charged objects - either like-charged or oppositely-charged. c. Electric forces between two charged objects increases with increasing separation distance. d. Electric forces between two charged objects increases with increasing quantity of charge on the objects. e. If object A attracts object B with an electric force, then the attractive force must be mutual - i.e., object B also attracts object A with the same force. f. A doubling of the quantity of charge on one of the objects results in a doubling of the electric force. g. A tripling of the quantity of charge on both of the objects results in an increase in the electric force by a factor of 6. h. A doubling of the separation distance between two point charges results in a quadrupling of the electric force. i. A tripling of the separation distance between two point charges results in an electric force which is one-sixth of the original value.

Answer: DEF a. Electric force is a non-contact force (or field force); it can act over separation distances even when the objects do not touch. b. An electrical attraction can even occur between a charged object and a neutral object. The neutral object is first polarized and then the attraction can occur. c. As separation distance increases, the electric force decreases; there is an inverse relationship. d. As the charge on any of the two objects increases, the electric force increases; there is a direct relationship. e. Forces - even electric forces - come in pairs. For every action, there is an equal and opposite reaction force. Indeed, to say that "object A attracts object B" is to automatically infer that object B also attracts object A. f. Whatever alteration is in made of the charge, the same alteration is made of the electric force. Doubling the charge will double the force. g. Whatever alteration is in made of the charge, the same alteration is made of the electric force. Tripling the charge will triple the force; tripling the other charge will triple the force again. Tripling the force (x3) two times will have the net effect of increasing the force by a factor of 9 (not 6). h. Force is inversely proportional to the square of the distance. Whatever alteration is made of the distance, the inverse square alteration is made of the force. So increasing distance by a factor of 2 (doubling it) will decrease force by a factor of 4. i. Force is inversely proportional to the square of the distance. Whatever alteration is made of the distance, the inverse square alteration is made of the force. So increasing distance by a factor of 2 (doubling it) will decrease force by a factor of 9 (3^2).

20. Which of the following statements are true statements about electric field? Choose all that apply. a. The electric field strength created by object A is dependent upon the separation distance from object A. b. The electric field strength created by object A is dependent upon the charge on object A. c. The electric field strength created by object A is dependent upon the charge of the test object used to measure the strength of the field. d. The electric field strength about charged Object A is the force per charge experienced by a test charge placed at some location about Object A. e. As the distance from Object A is doubled, the electric field strength created by object A increases by a factor of 4. f. As the charge of Object A is doubled, the electric field strength created by object A increases by a factor of 2. g. As the charge of Object A is doubled and the distance from object A is doubled, the electric field strength created by object A decreases by a factor of 2. h. Object B is used to test the electric field strength about Object A; as the charge of Object B is doubled, the force which it experiences is doubled but the electric field strength remains the same. i. Object B is used to test the electric field strength about Object A; as the separation distance between Object A and Object B is doubled, the force which it experiences decreases by a factor of 4 but the electric field strength remains the same. j. The electric field strength inside of a closed conducting object (e.g., inside the sphere of the Van de Graaff generator) is zero. k. For irregularly shaped objects, the electric field strength is greatest around points of highest curvature. l. The electric field is a vector which points in the direction which a positive test charge would be accelerated. m. Electric fields are directed inwards in regions around negatively-charged objects and outwards in regions around positively-charged objects. n. Units on electric field are Newtons/Coulomb (N/C).

Answer: ABDFGHJKLMN a. TRUE The E vector depends upon two factors - the charge of the object creating the field and the separation distance from that charge. b. TRUE The E vector depends upon two factors - the charge of the object creating the field and the separation distance from that charge. c. FALSE An object can be used to measure the strength of an electric field; such an object is called a test charge. The test charge encounters an electric force. The electric field is the ratio of the force divided by the charge on the test charge. If the quantity of charge on the test charge is increased, the force is increased proportionately but the ratio remains the same. d. TRUE An object can be used to measure the strength of an electric field; such an object is called a test charge. The test charge encounters an electric force. The electric field is the ratio of the force divided by the charge on the test charge. e. FALSE The electric field strength is inversely proportional to the square of the distance of separation from the charge. If the distance is increased by a factor of 2, then the E is decreased by a factor of 4. f. TRUE The electric field strength is directly proportional to the quantity of charge on the object which creates the field. Doubling the charge (Q) will double the electric field strength. g. TRUE The electric field strength is directly proportional to the quantity of charge on the object which creates the field and inversely proportional to the square of the distance of separation from that charge. Doubling the charge will have the effect of increasing the E by a factor of 2; doubling the distance will have the effect of decreasing the E by a factor of 4. The combined effect of these two factors will result in decreasing the E by a factor of 2. h. TRUE The electric field strength (E) created by object A does not depend upon the charge on the object used to test it. The electric field force (F) would be doubled by the doubling of the charge. But the electric field strength (E) is the force per charge - twice the force on twice the charge yields the same electric field strength. i. FALSE The electric field strength created by object A does depend on how far from object A that the test charge is. Twice the separation distance means one-fourth the electric force (the inverse square law) and therefore one-fourth of the electric field. j. TRUE This is a unique property of the electric field inside a closed conductor. k. TRUE This is another unique property of the electric field. This property explains how lightning rods work. Being pointed, the electric field strength is large around the points (parts of highest curvature). l. TRUE This is the customary convention used to define the direction of the electric field around any charged object. m. TRUE Because electric field direction is defined as the direction which a positively-charged object would accelerate, the direction would be toward negatively charged objects (+ test charges are attracted) and away from positively-charged objects (+ test charges are repelled). n. TRUE Since electric field is the force per charge exerted upon a test charge, the units would be force units per charge units - Newtons/Coulomb.

21. Which of the following statements are true statements about lightning rods? Choose all that apply. a. Lightning rods are placed on homes to reduce the risk of lightning damage to a home. b. Any metallic object which is placed on the roof of a home and grounded by an appropriate conducting pathway can serve as a lightning rod. c. Most lightning rods are pointed as a decorative feature. d. To be totally effective, a lightning rod must stretch high into the sky and draw charge from the lowest clouds by the method of contact. e. Lightning rods are capable of reducing the excess charge buildup in clouds, a characteristic of dangerous thunderstorms. f. Charge is incapable of passing from clouds to a lightning rods since the air between the clouds and the lightning rod has an insulating effect. g. The electric field strength about the points of lightning rods are very high.

Answer: AG a. TRUE The role of lightning rods is to divert charge around your home to the ground in the event of a possible lightning strike. b. FALSE If the lightning rod does not have the characteristic protrusion up into the air above the building it protects, then it might fail to do what it does best - divert charge around the building and preventing it from being conducted through the home. c. FALSE Lightning rods are pointed because the electric field around a sharply curved object is high and serves to ionize air around it; this provides a conducting pathway from the cloud to the ground. The pathway allows charge to slowly flow between cloud and ground, thus preventing a sudden lightning discharge. In more recent years, scientists have found that a blunt tip on the lightning round provides and equal if not greater measure of protection. d. FALSE Lightning rods do not need to contact the clouds to discharge them. The strong electric fields around the points serves to turn air from an insulator to a conductor (by ionizing particles in the air). This provides a conducting pathway between the cloud and the lightning rod. e. FALSE This statement is a good description of the classical dissipation theory of lightning rods. It was once believed (beginning with Ben Frnaklin) that lightning rods served to dissipate or remove electrostatic charge buildup in clouds by removing the charge gradually over time. Scientists generally agree that that dissipation theory is not an accurate view of the role of lightning rods. It is believed that lightning rods serve to divert electrostatic charge around a home during a lightning strike. f. FALSE If this were true, then there wouldn't be any lightning strikes. The role of the lightning rod is to actually make the pathway between the cloud and the ground more conductive by ionizing the air around the rod and providing a more conductive path from cloud to ground through the lightning rod. g. TRUE This explains why lightning rods are often pointed. Though as discussed in more detail in c. above, blunt-tipped lighning rods have been found to provide equal or even greater protection as pointed-tip lightning rods.

22. Charged balloons are used to induce a charge upon neutral metal pop cans. Identify the type of excess charge which would be present on cans G, H, I, J, K, L, M, N and P as shown in the diagrams below. a. positive b. negative c. neutral Diagram Type of Charge on...

G: positive (A) I. H: negative (B)

I: negative (B) II. J: positive (A)

K: positive (A) II. L: negative (B) M: positive (A)

IV. N: negative (B)

V. P: neutral (C) Answer: See table above. I. The presence of the negatively-charged balloon repels electrons which are present in can G. Many of these electrons move into can H. This leaves can G charged positively (it lost electrons) and can H charged negatively (it gained electrons). II. The presence of the negatively-charged balloon next to can J repels electrons which are present in can J. Many of these electrons move into can I. Similarly, the presence of the positively-charged balloon next to can I will attract electrons present in both cans. This combined effect leaves can J to become charged positively (it lost electrons) and can I to become charged negatively (it gained electrons). III. The presence of negatively-charged balloons on each end of the collection of cans causes electrons to be repelled from the balloons. Many electrons leave the two cans on the ends and enter the can in the middle in order to distant themselves from the balloons. This leaves Can L charged negatively (it gained electrons) and Cans K and M charged positively (they lost electrons). IV. The presence of the positively-charged balloon will attract electrons towards itself. There is a migration of electrons from the ground (hand) towards the balloon. This leaves can N charged negatively. V. Electrons in can P will be attracted to the positively-charged balloon. This will polarize the can. But unless there is a source of electrons connected to can P, there is no way for the can to develop an overall charge. The can remains neutral.

For Questions #25 - #27, identify the type of charge on objects A-D based on the electric field lines shown for each configuration of charges. a. positive b. negative c. neutral

25. 26.

A is positive B is negative 27.

C is positive and D is negative.

Answer: See diagrams above. The direction of an electric field is the direction which a positively charged test charge would be accelerated. The electric field lines point in this same direction. As such, electric field lines are always directed away from + charges and towards - charges (or infinity). From this principle, it can be reasoned that A is positive, B is negative, C is positive and D is negative.

Part B: Diagramming and Analysis 28. A neutral conducting sphere is charged by induction using a positively-charged balloon. What will be the charge of the following sphere in step e if steps a - d are followed? ______

Explain how the object acquires this charge. Show the type and location of excess charge on the conducting sphere in each step of the diagram (where appropriate).

Answer: See diagram above. In b., the presence of the positively-charged balloon serves to polarize the sphere. Electrons in the sphere are drawn towards the positively-charged balloon. This leaves a separation of charge in the sphere, with the excess negative charge on the left and the excess positive charge on the right. In c., the sphere is touched to ground and electrons from the ground are drawn upwards and into the right side of the sphere. It is the presence of the + charge on the right side of the sphere which attracts these electrons to itself. At this point the sphere is charged negatively. In d., this excess negative charge is shown distributed on the left side of the sphere, still drawn towards the positively-charged balloon. In e,. the balloon is pulled away and there is a movement of electrons about the sphere so that the excess negative charge is uniformly distributed.

29. A negatively-charged balloon is held above (without touching) a neutral electroscope. The presence of the nearby balloon causes the needle of the electroscope to deflect. a. Draw the location and type of excess charges on the "polarized" neutral electroscope. b. Explain how the balloon has induced the temporary charge upon the electroscope (i.e., describe the direction of electron movement).

Answer: See diagram at right. The negative balloon repels electrons from the plate to the needle and base. Since like charges are repelled, electrons in the top of the electroscope (the plate) are repelled and forced downwards to the bottom of the electroscope (the needle and base). This causes the temporary separation of charge or polarization of the electroscope.

30. Construct electric field lines around the following configuration of charges. Include at least six lines per charge. a. b.

c. d.

Answer: See diagrams above Electric field lines begin at either infinity or a + charge and are drawn either to infinity or to - charges. Avoid crossing lines. At the surface of charges, the lines should be directed radially inwards (for - charges) or radially outwards (for + charges).

31. Use an unbroken arrow to show the direction of electron movement in the following situation. The arrow should extend from the source of the electrons to the final destination of the electrons. a. A positively charged balloon is touched to a neutral b. A glass rod is rubbed with a piece of wool. The conducting sphere. wool has the greater electron affinity.

Electrons move from the sphere to the + Electrons move from the glass rod to the balloon. wool. c. A positively-charged balloon is held near a neutral d. A negatively-charged rod is held near a neutral conducting sphere. The sphere is then touched on the conducting sphere. The sphere is touched on the opposite side. opposite side.

Electrons move from the hand to the sphere. Electrons move from the sphere to the hand.

Answer: See diagrams above. In a, c, and d, the movement of electrons is governed by the principles that opposites attract and like-charged objects repel. Electrons (being negative) are attracted to positively charged objects and repelled by negatively charged objects. In b, the more electron-loving substance (highest electron affinity) gains the electrons from the object with which it is rubbed.

Part C: Short Computations 33. The charge of one electron is -1.6 X 10-19 Coulombs. If a neutral object loses 1.5x106 electrons, then what will be its charge?

Answer: +2.4 x 10-13 C The overall quantity and type of charge on an object is dependent upon the relative number of protons and electrons in the object. To determine the quantity of charge, multiply the charge of one electron by the number of excess or deficient electrons. The type of the charge is dependent upon whether there are more protons or more electrons present in the object. If there are more protons, then the charge is +; otherwise, assign it a - value.

34. Express your understanding of Coulomb's law by filling in the following table.

Q1 (C or µC) Q2 (C or µC) d (m) Felect (N) a. 1.6 x 10-13 C 1.6 x 10-13 C 0.50 m 9.2 x 10-16 N

b. 1.6 x 10-7 C 1.6 x 10-7 C 0.12 m 0.017 N

c. 4.65 µC 7.28 µC 0.658 m 0.704 N

d. 3.21 µC 0.875 µC 0.553 m 0.0827 N

e. 3.16 x 10-7 C 3.16 x 10-7 C 1.50 m 4.00 x 10-4 N

Answer: See table above.

The electric force (Felect) is computed using Coulomb's law: 2 Felect = k•Q1•Q2/d where Q1 and Q2 represent the charges on the two objects, d represents the separation distance between the object's centers and k = 9x109 N/m2/C2. This equation can be rearranged to solve for any quantity in the equation. The substitutions and algebra are shown below.

9 2 2 -13 -13 2 -16 a. Felect = (9x10 N/m /C )•(1.6 x 10 C)•(1.6 x 10 C)/(0.5 m) = 9.2 x 10 N

9 2 2 -7 -7 b. d = SQRT[k•Q1•Q2/Felect] = SQRT[(9x10 N/m /C )•(1.6 x 10 C)•(1.6 x 10 C)/(0.017 N)] = 0.12 N

9 2 2 -6 -6 2 c. Felect = (9x10 N/m /C )•(4.65 x 10 C)•(7.28 x 10 C)/(0.658 m) = 0.704 N

2 2 9 2 2 -6 -7 d. Q2 = (d •Felect )/(k•Q1) = (0.553 m) •(0.0827 N)/[(9x10 N/m /C )•(3.21x10 C)] = 8.75 C = 0.875 µC

2 -4 9 2 2 -7 e. Q = SQRT[(d •Felect )/(k)] = SQRT[(1.50 m)•(4.00x10 N)/(9x10 N/m /C )] = 3.16 x 10 C

36. The following questions check your qualitative understanding of the variables effecting electric field and electric force. Use your understanding to fill in the blanks. a. At a separation distance of 0.500-m, two like-charged balloons experience a repulsive force of 0.320 N. If the distance is doubled, then the repulsive force would be 0.0800 N. b. At a separation distance of 0.500-m, two like-charged balloons experience a repulsive force of 0.320 N. If the distance is halved, then the repulsive force would be 1.28 N. c. At a separation distance of 0.500-m, two like-charged balloons experience a repulsive force of 0.320 N. If the distance is is decreased by a factor of 3 and the charge on one of the balloons is doubled, then the repulsive force would be 5.76 N. d. At a separation distance of 0.500-m, two like-charged balloons experience a repulsive force of 0.320 N. If the distance is is increased by a factor of 3 and the charge on both of the balloons is doubled, then the repulsive force would be 0.142 N. e. At a location of 0.200 m from object A, the electric field strength is 0.0500 N/C. If the electric field strength were measured at a location of 0.400 m from object A, then the field strength would be 0.0125 N/C. f. At a location of 0.200 m from object A, the electric field strength is 0.0500 N/C. If the electric field strength were measured at a location of 1.00 m from object A, then the field strength would be 0.00200 N/C. g. At a location of 0.200 m from object A, the electric field strength is 0.0500 N/C. If the electric field strength were measured using a test charge with twice the charge at a location of 0.200 m from object A, then the field strength would be 0.0500 N/C. h. At a location of 0.200 m from object A, the electric field strength is 0.0500 N/C. If the electric field strength were measured using a test charge with twice the charge at a location of 0.100 m from object A, then the field strength would be 0.200 N/C.

Answer: See sentences above. a. If the distance is doubled, then the electric force is one-fourth the original value: 0.320 N/4 = 0.0800 N. b. If the distance is decreased by a factor of 2 (halved), then the electric force is four times the original value: 0.320 N * 4 = 1.28 N. c. Decreasing the distance by a factor of 3 will increase the force by a factor of 9; increasing the charge by a factor of 2 will increase the force by a factor of 2. So take the original force adjust accordingly: 0.320 N * 9 * 2 = 5.76 N d. Increasing the distance by a factor of 3 will decrease the force by a factor of 9; increasing both the charges by a factor of 2 will increase the force by a factor of 2*2. So take the original force adjust accordingly: 0.320 N / 9 * 2 * 2 = 0.142 N e. If the distance is increased by a factor of 2 (from 0.200 m to 0.400 m) then the electric field strength is decreased by a factor of 4. So take the original E value and adjust accordingly: 0.050 N/C / 4 = 0.0125 N/C. f. If the distance is increased by a factor of 5 (from 0.200 m to 1.000 m) then the electric field strength is decreased by a factor of 25. So take the original E value and adjust accordingly: 0.0500 N/C / 25 = 0.00200 N/C. g. Altering the amount of charge on the test charge will not effect the electric field strength at the same location. So the E value remains the same. h. Altering the amount of charge on the test charge will not effect the electric field strength at the same location. But decreasing the distance by a factor of 2 will increase the electric field strength by a factor of 4. So take the original E value and adjust accordingly: 0.0500 N/C * 4 = 0.200 N/C.

40. For the situation described in Question #39, what would be the magnitude of the net electric force upon a +1.3 µC charge if placed at the origin.

Answer: 0.0070 N This is simply a matter of using the definition of electric field - the electric field is the force per coulomb of charge. Expressed mathematically, E = F/q and rearranged, the force is given by the equation F = E•q By substitution, F = (5.4 x103 N/C)•(1.3x10-6 C) = 0.0070 N

Electric Circuits Review

Part A: Multiple-Multiple Choice 1. Which of the following statements are true about an electric circuit? List all that apply. a. Electrons are the mobile charge carriers in an electric circuit. b. The path of charge flow from the + to the - terminal of the circuit can consist of nonconductive material. c. In an electric circuit of an automobile, the 12-Volt car battery is sometimes referred to as the internal circuit because it is located inside of the hood of the car. d. Charge is supplied with energy in the internal circuit and the energy is transformed into other forms in the external circuit. e. Charge is consumed as it moves through the energy-transforming devices of the external circuit. For instance, the amount of charge entering a light bulb in a second is less than that which exits the light bulb in a second. f. A battery is used to power a flashlight circuit. When the battery no longer works, it is because it has run out of charge. g. The location on an electric circuit where the charge possesses the greatest amount of electric potential energy is the + terminal of the battery.

Answer: ADG a. TRUE - Physicists often speak of conventional current as the direction that positive charge moves through a circuit. This is based on the convention that the direction of the electric field is the direction that a + test charge would be accelerated. In reality, the carrier of electric charge in an electric circuit is the electron. b. FALSE - For charge to flow through an external circuit, the pathway must consist of conductive materials. The presence of a nonconductor would inhibit charge flow and low the current to zero. c. FALSE - The internal circuit is the portion of the circuit inside the battery, located between the - terminal and the + terminal. Whether the battery is inside the hood or not has nothing to do with it. If you were to place your battery on top of your hood, it would still be the internal circuit. d. TRUE - The battery is the internal circuit; it is the place where charge acquires its energy. The external circuit consists of wires, light bulbs, resistors, etc. In the external circuit, charge has its electrical energy transformed into other forms as it passes through the elements of the external circuit. e. FALSE - Charge is neither created nor destroyed, and as such, the amount of charge which enters a light bulb in a second is equal to the amount of charge which exits the light bulb in a second. This quantity (charge per second) is the current and it is everywhere the same in a circuit. f. FALSE - Batteries produce energy through a exothermic chemical reaction. When a battery no longer works, the ability of its chemical reaction to produce products and energy has become limited by the depletion of reaction and the build up of products. It has not run out of charge; it has run of out chemical reactants. g. TRUE - Know this one! The + terminal is the location of high energy in an electric circuit.

2. Which of the following statements are true about an electric potential or electric potential difference? List all that apply. a. Electric potential can be expressed in units of volts. b. Electric potential can also be expressed in units of Joules. c. The electric potential of a charge at a given location provides a measure of the rate at which charge flows past that point. d. Work must be done on a + charge to move it against (i.e., in the opposite direction of) an electric field. e. As a + charge moves in the same direction as an electric field, it gains electric potential. f. The electric potential difference between two points is simply the difference in potential energy possessed by charge between those two points. g. If a unit of charge has a high electric potential at any given location, then that charge possesses a large amount of electric potential energy at that location. h. The + terminal of a battery is a location with a higher electric potential than the - terminal of a battery. i. Charge flowing through a battery will experience a gain in electric potential. j. A 6-Volt battery would provide 12 Joules of energy to 2 Coulombs of charge as it is moved from the - to the + terminal.

Answer: ADGHIJ a. TRUE - Electric potential, also known as voltage, is measured in units of volts. b. FALSE - The electric potential at a given location in a circuit is defined the electric potential energy per charge at the location. Its units would be energy units per charge units - a Joule per Coulomb (not simply a Joule). c. FALSE - The rate at which charge flows past a point is the current. Electric potential is the potential energy per charge at a given location. d. TRUE - By convention, an electric field is the direction which a positive test charge would be pushed or pulled. If a positive charge is moved against an electric field, then work must be done upon the charge to overcome the opposing effect of the field. e. FALSE - A positive charge moving in the direction of an electric field is moving without any external help. Such a charge is analogous to a bicycle moving downhill; the bicyclist would not have to pedal. In such instances, there is a loss in electric potential energy, and thus a loss in electric potential. f. FALSE - Close, but not correct. The difference in electric potential between any two points is simply the different in electric potential energy per charge between those two points. Using a gravitational analogy, gravitational potential is the difference in gravitational potential energy per mass between any two points. Gravitational potential energy is dependent upon the amount of mass present at a location (m•g•h); the more mass present at a location , the more gravitational potential energy. In contrast, gravitational potential reduces the importance of mass by defining itself as the potential energy per kilogram of mass (PE / m). By so doing, gravitational potential is now only location dependent - depending only upon the value of g at a given location and the height above some zero reference level. In the same manner, electric potential energy at a given location is charge-dependent. The more charge present at a location, the more repulsive (or attractive) effects and the greater the potential energy. But electric potential makes the amount of charge unimportant since it is the electric potential energy per charge. Now electric potential is location dependent - dependent for instance on the location in a circuit and not upon the amount of charge flowing past that location in a second. g. TRUE - Electric potential is the potential energy per charge at a given location. The more electric potential which a unit of charge has at a given location, the more electric potential energy that the charge has. h. TRUE - In the battery, the positive terminal is the location of high energy and high potential. The negative terminal is the location of low energy and low potential. i. TRUE - In the battery, a charge is moved from a location of low energy to a location of high energy. The charge would gain electric potential energy and as such, it would also gain electric potential. j. TRUE - A 6-Volt battery supplies every Coulomb of charge with 6 Joules of energy. The energy per charge ratio is 6 Joules per Coulomb. Thus, 2 Coulombs of charge would gain 12 Joules of energy.

3. Which of the following statements are true about electric current? List all that apply. a. Electric current is measured in units of Amperes. b. Electric current and drift speed are synonymous terms. c. Electric current is defined as the number of Coulombs of charge which move past a point on a circuit. d. Electric current is equal to the number of Coulombs of charge which move past a point on a circuit per unit of time. e. Electric current provides a measure of how fast charge moves between two points on a circuit. f. The electric current diminishes in value as charge progresses to locations further and further from the + terminal of the battery. g. The electric current in a circuit will increase as the electric potential impressed across a circuit is increased. h. The electric current in a circuit will triple in value as the electric potential impressed across a circuit is increased by a factor of three. i. Suppose a miniature light bulb is connected to a battery in a circuit. A light bulb with a greater resistance will have a greater current. j. Wider conducting wires are capable of carrying larger currents. k. If one starts getting technical about the meaning of electric current, one could argue that electric current is not something that flows past a point on a circuit, but rather something that exists to a given measure at a point on a circuit.

Answer: ADGHJK a. TRUE - Electric current is the rate at which charge flows past a point on a circuit. It is measured in Coulombs per second, also known as an Ampere or an "Amp." b. FALSE - No! Current refers to how many Coulombs of charge pass a cross-sectional area in a wire in a second of time. Drift speed refers to the distance that a mobile charge carrier travels in a unit of time. One quantity - current - has to do with how many? and the other quantity - drift speed - has to do with how far? c. FALSE - Current is a rate quantity much like velocity or power is a rate quantity. There is a per time basis to it. Current is defined as the Coulombs of charge which pass a point on a circuit per second of time. d. TRUE - Yes! This is the definition of current - a definition which ought to be committed to memory. e. FALSE - See the explanation to part b above. f. FALSE - Charge is marching around a circuit at the same rate everywhere within the circuit. The number of Coulombs which exit the battery at the + terminal in a second is equal to the number of Coulombs which enter the battery at the - terminal in a second. Charge is not created; charge is not destroyed; and charge does not accumulate at a location in a circuit. Charge simply marches together around the loop of a circuit. The quantity which is lost and gained is energy. Work is done on the charge as it passes through the battery, thus pumping the charge up to a high energy. And the charge subsequently does work upon the elements of the external circuit, transforming its electric potential energy to other useful forms (light, sound, mechanical, thermal, etc.). g. TRUE - A higher electric potential impressed across the two terminals of a battery will cause the rate of charge flow (i.e., current) to increase. This is consistent with the V = I•R equation. h. TRUE - Consistent with V = I•R, one could say that the current is directly proportional to the voltage. A tripling of the voltage (while R is held constant) would cause a tripling of the current. i. FALSE - A light bulb with a greater resistance will offer more hindrance to the flow of charge through it and as such lead to less current. Since I = V / R, the quantities current and resistance are inversely proportional. j. TRUE - A wider wire has a greater cross-sectional area and thus offers less resistance. As such, wider wires allow a greater current or flow rate. k. TRUE - Technically speaking, current does not flow or move. Current is a quantity which is measured at a location and is based upon the rate at which charge flows. Teachers and textbooks often get sloppy with their language and say that the current flows. In reality, it is the charge that does the moving; the current is simply measured at a location to reveal the rate at which charge flows past that location.

4. Which of the following statements are true about an electric resistance? List all that apply. a. The resistance of an electric circuit is a measure of the overall amount of hindrance to the flow of charge through the circuit. b. A large resistance value indicates that the charge is encountering a relatively large amount of difficulty in moving through the circuit. c. The unit of resistance is the ampere. d. A 5-ohm resistor would allow a current of 0.5 Amperes when 2.5 Volts is impressed across it. e. A 10-ohm resistor would allow a current of 2 Amperes when 5 Volts is impressed across it. f. The resistance of a conducting wire will increase as the length of the wire is increased. g. The resistance of a conducting wire will increase as the cross-sectional area of the wire is increased. h. Increasing the resistance of an electric circuit will cause the current in the circuit to increase. i. A threefold increase in the resistance of an electric circuit will result in a threefold decrease in the electric current. j. A miniature light bulb with a specific resistance is connected to a 1.5-Volt battery to form a circuit. If it were connected to a 6-Volt battery instead, its resistance would increase by a factor of 4.

Answer: ABDFI a. TRUE - Yes! Remember this as the definition of resistance. b. TRUE - Resistance refers to the amount of hindrance (resistance) to the flow of charge. The more hindrance which charge encounters, the more resistance. c. FALSE - The unit of resistance is the ohm. Ampere or Amps is the unit of current. d. TRUE - The resistance of a resistor is the V/I ratio at that location. A 5-ohm resistor would have a 5:1 ratio of voltage to current. This is a true statement since (2.5 Volt) / (0.5 A) is equal to 5.0 Volt/Amp. e. FALSE - The resistance of a resistor is the V/I ratio at that location. A 10-ohm resistor would have a 10:1 ratio of voltage to current. This is a false statement since (5 Volt) / (2 A) is not equal to 10 Volt/Amp. f. TRUE - The resistance of a wire is mathematically related to the resistively of the material, the length of the wire and the cross-sectional area of the wire. Longer wires have a greater resistance. g. FALSE - The resistance of a wire is mathematically related to the resistively of the material, the length of the wire and the cross-sectional area of the wire. Wider wires have a lower resistance. h. FALSE - More resistance would cause less current. i. TRUE - In accordance with the equation I = V / R, current is inversely proportional to the resistance. Tripling the resistance will cause the current to decrease by a factor of three. j. FALSE - The resistance of a light bulb is an intrinsic property which depends upon the material and dimensions (length and diameter) of the light bulb filament. If the same light bulb is connected to a battery of greater potential, only the current at the bulb location will be modified.

5. Which of the following statements are true about electric power and electric potential energy? List all that apply. a. Power is the amount of energy used by an electrical appliance. b. Power is the rate at which energy is used by an electric circuit. c. A kiloWatt•hour is a unit of electric power. d. A Joule/second is a unit of power. e. A 60-Watt light bulb would consume 30 Joules of electrical energy in 2 seconds. f. A 120-Watt light bulb would consume 1200 Joules of electrical energy in 6 seconds. g. A charge moving through an electric circuit would gain electric potential energy in the internal circuit and lose electric potential energy in the external circuit. h. Charge has the greatest amount of electric potential energy at the - terminal of the battery. i. If a 60-Watt light bulb is placed in a 120-Volt circuit, then the current in the light bulb is 2 Amps. j. Consider two light bulbs with different power ratings. A 60-Watt light bulb has a greater resistance than a 100-Watt light bulb.

Answer: BDGJ a. FALSE - Power is a rate quantity; it is the rate at which energy is used by an electrical appliance. b. TRUE - Yes! Know this definition. c. FALSE - A kiloWatt is a unit of power. An hour is a unit of time. Power is energy per time. So when power is multiplied by time, the result is energy. A kiloWatt•hour is a unit of energy. d. TRUE - Power is the rate at which energy is used by an electrical appliance. As such, power is measure in energy per time units - Joules per second. e. FALSE - Power is the energy per time ratio. A 60-Watt bulb would consume 60 Joules of energy in 1 second, 120 Joules of energy in 2 seconds, and 30 Joules of energy in 0.5 seconds. f. FALSE - Power is the energy per time ratio. A 120-Watt bulb would consume 120 Joules of energy in 1 second, 720 Joules of energy in 6 seconds, and 1200 Joules of energy in 10 seconds. g. TRUE - This is how an electric circuit work. The internal circuit supplies electrical energy to the charge and the electrical energy is converted to other forms of energy in the external circuit. h. FALSE - The - terminal is the location of lowest energy in an electric circuit. the location of highest energy is the + terminal. i. FALSE - Power is related to voltage and current by the equation P = I•V. So a 60-Watt bulb would allow a current of 0.5 Amps if wired into a 120-Volt circuit. j. TRUE - A 60-Watt light bulb and a 100-Watt light bulb have different resistance values, which in turn effects the current at the light bulb location. The bulb with the greatest resistance has the lower current. So which bulb - high power or low power - has the greatest resistance? One must consider the expression for power which includes both the current (I) and the resistance (R) since the choice of bulb will effect both variables. The equation P = I2•R is of utmost importance here. In the equation, the I is of double the importance since it is the square term in the equation. So the bulb with the greatest I is the bulb with the greatest power. The 100-Watt bulb has the greatest current; and therefore it must have the lowest resistance. The 60-Watt bulb has the greatest resistance.

6. Which of the following statements are true about a series circuit? List all that apply. a. Series circuits are characterized by the fact that there is a single pathway by which charge can travel from the + terminal to the - terminal. b. During any single loop of charge around a series circuit, the charge will pass through each resistor in the circuit. c. The overall resistance of a circuit increases as more and more resistors are placed in series in the circuit. d. The total current in a circuit increases as more and more resistors are placed in series in the circuit. e. Suppose that three identical resistors with a resistance of 3-ohms are placed in series with a 12-Volt battery. The current in the circuit is 4 Amps. f. Suppose that three identical resistors with a resistance of 3-ohms are placed in series with a 12-Volt battery. The electric potential difference across each resistor is 12 Volts. g. Suppose that two identical resistors are placed in series with a 12-Volt battery. The current in the circuit is 6 amps. The resistance of each resistor is 4 ohms.

Answer: ABC a. TRUE - Series circuits are circuits which allow a single pathway for charge to follow; there is no no branching locations in a series circuit. b. TRUE - Since there are no branching points in a series circuit, charge is never left with a choice of passing through either Resistor A or Resistor B. It must pass through both resistors in consecutive fashion before returning to the - terminal of the battery. c. TRUE - Adding more resistors in series would mark the overall resistance of the circuit increase since each charge making a loop around the circuit will encounter each of these resistors. d. FALSE - If adding more resistors makes the overall resistance of the circuit increase, then it will make the overall current in the circuit decrease. For a constant battery voltage, current and resistance are inversely proportional. e. FALSE - Three resistors wit a resistance of 3 Ohms placed in series is equivalent to a single resistor with a resistance of 9 Ohms. The current through the circuit would be equal to the battery voltage divided by the overall or equivalent resistance. I = (VTot) / (RTot) = (12 Volts) / (9 Ohms) = 1.33 Amps. f. FALSE - In a series circuit, the voltage gained by the charge in passing through the battery - in this case, 12 Volts - is equal to the overall or cumulative voltage drop of the charge as it passes through the resistors. So the sum of the voltage drops in each of the three individual resistors must be 12 Volts. In the case of three resistors of equal resistance, the charge will lose 4 Volts in each resistors, thus accumulating a total voltage drop of 12 Volts. g. FALSE - The total or equivalent resistance of these two series resistors is simply the ratio of the battery voltage to the overall current. That is, RTot = (VTot) / (ITot) = (12 Volts) / (6 Amps) = 2 Ohms. For resistors in series, the overall or total resistance is the sum of the individual resistances. Thus, each resistor must have a resistance of 1 Ohm.

7. Which of the following statements are true about a parallel circuit? List all that apply. a. Parallel circuits are characterized by the fact that there are multiple pathways by which charge can travel from the + terminal to the - terminal. b. During any single loop of charge around a parallel circuit, the charge will pass through each resistor in the circuit. c. The overall resistance of a circuit increases as more and more resistors are placed in parallel in the circuit. d. The total current in a circuit increases as more and more resistors are placed in parallel in the circuit. e. Suppose that three identical resistors with a resistance of 3-ohms are placed in parallel with a 12-Volt battery. The current in the circuit is 1 Amp. f. Suppose that three identical resistors with a resistance of 3-ohms are placed in parallel with a 12-Volt battery. The electric potential difference across each resistor is 12 Volts. g. Suppose that two identical resistors are placed in parallel with a 12-Volt battery. The overall current in the circuit is 6 amps. The resistance of each resistor is 4 ohms.

Answer: ADFG a. TRUE - This is the definition of a parallel circuit. A parallel circuit contains a branching location. When a charge reaches the branching location, it may choose between a pathway which includes any one of at least two resistors. b. FALSE - This is a characteristic of a series circuit. In a parallel circuit, charge makes a either-or choice at the junction or branching locations, choosing to pass through either Resistor A or Resistor B. It does not pass through both of them in consecutive fashion, but only through one or the other resistor. c. FALSE - Increasing the number of resistors in parallel branches will have the effect of decreasing the overall resistance of the circuit. At first this seems rather counterintuitive. But consider that the new resistor actually opens up a new pathway through which the charge can pass. Not all the charge goes through all the resistors. By opening a new branch in the circuit, the fraction of charge which passes through any one resistor is reduced. So less charge actually encounters a single resistor. Now it makes sense to say that more resistors (i.e., more branches) leads to less overall resistance. d. TRUE - If increasing the number of resistors has the effect of decreasing the overall resistance of the circuit, then it must have the effect of increasing the overall current through the circuit. For a constant battery voltage, current and resistance are inversely proportional. e. FALSE - The overall resistance of a parallel circuit is expressed by the equation:

1/RTot = 1/R1 + 1/R2 + 1/R3 So for this circuit, substituting 3 Ohms for each of the resistors leads to a total resistance of 1 Ohm. The total current through the circuit is given by the equation I = (VTot) / (RTot) = (12 Volts) / (1 Ohm) = 12 Amps. f. TRUE - In any circuit - whether series or parallel, the voltage gained by the charge in passing through the battery - in this case, 12 Volts - is equal to the overall voltage drop of the charge as it passes through the external circuit. In the case of a parallel circuit, the charge makes an either-or choice between resistors. It passes through a single resistor instead of all three resistors. Thus, it must drop 12 Volts during the pass through that resistor. The voltage drop across each branch is 12 Volts. g. TRUE - The total or equivalent resistance of these two parallel resistors is simply the ratio of the battery voltage to the overall current. That is, RTot = (VTot) / (ITot) = (12 Volts) / (6 Amps) = 2 Ohms. For resistors in parallel, the overall or total resistance is expressed by the equation

1/RTot = 1/R1 + 1/R2 Substituting 4 Ohms for each of the individual resistances leads to the overall resistance of 2 Ohms. This is definitely a true statement.

Part B: Multiple Choice 8. If an electric circuit was analogous to a water park, then the battery would be analogous to the ____. a. pipes which carry the water through the water circuit b. pump which supplies energy to move the water from the ground to a high elevation c. the people which flow from the top of the water ride to the bottom of the water ride d. the rate at which water is pumped onto the slide e. the change in potential energy of the riders f. the top of the water slide g. the bottom of the water slide h. the long lines which exist at the park i. the speed at which riders move as they slide from the top to the bottom of the ride

Answer: B A water ride at a water park is analogous to an electric circuit. First of all, there is an entity which flows - water flows in a water park and (in conventional terms) + charge flows in an electric circuit. In each case, the fluid flows spontaneously from a high energy location to a low energy location. The flow is through pipes (or slides) in a water park and through wires in an electric circuit. If the pipes or the wires are broken, then there can be no continuous flow of fluid through the circuit. A complete loop is required to establish the circuit. This flow of fluid - whether of water or charge - is possible when a pressure difference is created between two locations in the circuit. In the water park, the pressure difference is the difference in water pressure created by two locations of different heights. Water flows spontaneously from locations of high pressure (high altitude) to locations of low pressure (low altitude). In an electric circuit, the electric potential difference between the two terminals of a battery or energy source provides the electric pressure which presses on charge to move them from a location of high pressure (high electric potential) to a location of low pressure (low electric potential). Energy is required to move the fluid uphill. In a water park, a water pump is used to do work upon the water in order to raise it from the low height back up to the high height. The water pump does not supply the water; the water which is already in the pipes. Rather, the water pump supplies the energy to pump the water from the location of low energy and low pressure to the location of high energy and high pressure. In an electric circuit, the battery is the charge pump which pumps the charge through the battery from the location of low electric potential energy (the - terminal) to the location of high electric potential energy (the + terminal). The battery does not supply the electric charge; the charge is already in the wires. The battery simply supplies the energy to do work on the charge in pumping it uphill.

9. If an electric circuit was analogous to a water park, then the positive terminal of the battery would be analogous to the ____. a. pipes which carry the water through the water circuit b. pump which supplies energy to move the water from the ground to a high elevation c. the people which flow from the top of the water ride to the bottom of the water ride d. the rate at which water is pumped onto the slide e. the change in potential energy of the riders f. the top of the water slide g. the bottom of the water slide h. the long lines which exist at the park i. the speed at which riders move as they slide from the top to the bottom of the ride

Answer: F A water ride at a water park is analogous to an electric circuit. First of all, there is an entity which flows - water flows in a water park and (in conventional terms) + charge flows in an electric circuit. In each case, the fluid flows spontaneously from a high energy location to a low energy location. The flow is through pipes (or slides) in a water park and through wires in an electric circuit. If the pipes or the wires are broken, then there can be no continuous flow of fluid through the circuit. A complete loop is required to establish the circuit. This flow of fluid - whether of water or charge - is possible when a pressure difference is created between two locations in the circuit. In the water park, the pressure difference is the difference in water pressure created by two locations of different heights. Water flows spontaneously from locations of high pressure (high altitude) to locations of low pressure (low altitude). In an electric circuit, the electric potential difference between the two terminals of a battery or energy source provides the electric pressure which presses on charge to move them from a location of high pressure (high electric potential) to a location of low pressure (low electric potential). Energy is required to move the fluid uphill. In a water park, a water pump is used to do work upon the water in order to raise it from the low height back up to the high height. The water pump does not supply the water; the water which is already in the pipes. Rather, the water pump supplies the energy to pump the water from the location of low energy and low pressure to the location of high energy and high pressure. In an electric circuit, the battery is the charge pump which pumps the charge through the battery from the location of low electric potential energy (the - terminal) to the location of high electric potential energy (the + terminal). The battery does not supply the electric charge; the charge is already in the wires. The battery simply supplies the energy to do work on the charge in pumping it uphill.

10. If an electric circuit was analogous to a water park, then the electric current would be analogous to the ____. a. pipes which carry the water through the water circuit b. pump which supplies energy to move the water from the ground to a high elevation c. the people which flow from the top of the water ride to the bottom of the water ride d. the rate at which water is pumped onto the slide e. the change in potential energy of the riders f. the top of the water slide g. the bottom of the water slide h. the long lines which exist at the park i. the speed at which riders move as they slide from the top to the bottom of the ride

Answer: D The flow of water at a water park is analogous to the flow of charge in an electric circuit. The rate at which charge moves past a point on a circuit as measured in Coulombs of charge per second (or some comparable set of units) is known as the current. In our analogy, the fluid which flows is water and the rate at which the fluid passes any given point is the current.

11. The potential energy possessed per unit of charge at any given location is referred to as the electric ___. a. current b. resistance c. potential d. power

Answer: C This is the definition of electric potential - a concept you should internalize.

12. One ampere is the amount of current that exists when ____ flows by a certain point in a conductor in ____. a. one watt; one second b. one joule; one hour c. one electron; one second d. one electron; one hour e. one volt; one second f. one volt; one hour g. one coulomb; one second h. one coulomb; one hour

Answer: G An ampere is a unit of electric current. And electric current is defined as the rate at which charge moves past a point on a circuit as measured in standard units of Coulombs of charge per second.

13. If 6 coulombs of charge flow past point 'A' in a circuit in 4 seconds, then ____ coulombs of charge will flow past point 'A' in 8 seconds. a. 0.67 b. 1.5 c. 2 d. 3 e. 4 f. 6 g. 8 h. 12 i. 24

Answer: H The current (I) is the quantity of charge flowing past a point (Q) in a given amount of time (t). That is, I = Q/t. So in this case, the current at point A is (6 C) / (4 s) or 1.5 amperes. Thus the Q/t ratio is 1.5 regardless of the time. So solve the equation 1.5 C/s = Q / (8 s) for Q to obtain the answer.

14. In which of the following situations will the light bulb light? List all that apply.

Answer: DF For a circuit to be established, there must be a closed conducting loop from the positive terminal to the negative terminal. This would mean that circuits D, E and F would all be circuits. But in order for the light bulb to light, it must be included as part of the electric circuit. So in E, the bulb does not light since the loop does not extend up into and through the light bulb; charge would simple flow out of the + terminal battery and directly back into - terminal the battery.

For Questions #15-#17: A simple circuit containing a battery and a light bulb is shown in the diagram at the right. Use this diagram to answer the next several questions. 15. The current through the battery is ___. a. greater than that through the light bulb b. less than that through the light bulb c. the same as that through the light bulb d. greater than that through each wire e. less than that through each wire

Answer: C Charge is a conserved quantity; it is never gained nor lost. In an electric circuit, the charge present in the wires and conducting elements is what moves through the circuit. This charge is enclosed in the wires and unable to escape (assuming there is no fault in the circuit). As the charge flows, it does not accumulate in a given location. And charge is not used up as though it were a consumable quantity. Nor is charge transformed into another type of entity. Given all this reasoning, one would conclude that the current at one location in an electric circuit is the same as the current at any other location in an electric circuit.

16. Charge flowing through this circuit is most energized at ____. Choose the one best answer. a. the + terminal of the battery b. the - terminal of the battery c. just prior to entering the light bulb d. just after exiting the light bulb e. ... nonsense! The energy of the charge is the same everywhere throughout the circuit.

Answer: A The + terminal of the battery is the high energy terminal of the battery.

17. The role or purpose of the battery in this circuit is to ____. Choose three. a. supply electric charge so that a current can exist b. supply energy to the charge c. move the charge from the - to the + terminal of the battery d. transform energy from electrical energy into light energy e. establish an electric potential difference between the + and - terminals f. replenish the charge which is lost in the light bulb g. offer resistance to the flow of charge so that the light bulb can get hot

Answer: BCE To establish an electric circuit, charge must be moved from low energy to high energy. Once at high energy, the charge spontaneously flows through the conducting wires and other conducting elements of the circuit back down to the low energy terminal. A battery's role is to supply the energy which is required to move the charge from the - terminal to the + terminal of the battery. By placing a large quantity of like charge at one location, an electric pressure or potential difference is established, forcing the like charges to move away from this location to the location of opposite charge (the - terminal).

18. A 12-Volt battery would supply ___. List all that apply. a. 3 Coulombs of charge with 4 Joules of energy b. 4 Coulombs of charge with 3 Joules of energy c. 12 Coulombs of charge with 1 Joule of energy d. 1 Coulomb of charge with 12 Joules of energy e. 0.5 Coulombs of charge with 24 Joules of energy f. 24 Coulombs of charge with 2 Joules of energy

Answer: D Electric potential (or voltage) is defined as the electric potential energy per charge. It is the Joules of energy per coulomb of charge possessed by some quantity of charge at some location in an electric circuit. A 12 Volt battery moves some quantity of charge from the - terminal to the + terminal, giving the charge energy. Each coulomb of charge would acquire 12 Joules of energy. The energy/charge ratio would be 12 J/C.

19. The charges that flow through the wires in your home ____. a. are stored in the outlets at your home b. are created when an appliance is turned on c. originate at the power (energy) company d. originate in the wires between your home and the power company e. already exist in the wires at your home

Answer: E This question targets a common misconception about electric circuits. The misconception presumes that the role of the electric outlet, the battery, or the power company is to provide the charge required to move through the home. But the power company is only the source of the energy required to set the charge in motion by the establishment of an electric potential difference. The charge itself is present in the wires and conducting elements of your home in the form of mobile electrons.

20. Approximately how long would it take an electron to travel from the battery of a car to a head light and back (complete loop)? a. seconds b. hours c. years d. one-millionth of a second e. one-tenth of a second

Answer: B Electric charge drifting through an electric circuit moves at a rather slow pace. Quite surprising to many, the distance traversed per unit of time is on the order of 1 meter per hour.

21. The electric circuit shown at the right consists of a battery and three identical light bulbs. Which of the following statements are true concerning this circuit? List all that apply. a. The current through point X will be greater than that through point Z. b. The current through point Z will be greater than that through point Y. c. The current will be the same through points X, Y and Z. d. The current through point X will be greater than that through point Y. e. The current through point Y will be greater than that through point X.

22. The electric circuit shown at the right consists of a battery and three identical light bulbs. Which of the following statements are true concerning this circuit? List all that apply. a. The electric potential difference between X and Y is more than that between Y and Z. b. The electric potential difference between X and Z is more than that between Y and W. c. The electric potential difference between X and Y is the same as that between Y and Z. d. The electric potential difference between X and Z is the same as that between Y and W. e. The electric potential difference between Y and W is more than that between X and Y.

Answer: DE The electric potential difference across a light bulb (or any resistor) in an electric circuit is simply the product of the current at that bulb multiplied by the resistance of the bulb. Each bulb has the same resistance (since they're identical) and the same current (since the current is everywhere the same). So the electric potential difference across each bulb is the same. And the potential drop across any two consecutive bulbs is the same. And the potential drop across two bulbs would be greater than that across one bulb.

23. The electric circuit shown at the right consists of a battery and three identical light bulbs. Which of the following statements are true concerning this circuit? List all that apply. a. Conventional current is directed through the external circuit from point X to Y to Z to W. b. Conventional current is directed through the external circuit from point W to Z to Y to X. c. Conventional current is directed through the internal circuit from point W to point X. d. Conventional current is directed through the internal circuit from point X to point W. e. The point where charge possesses the least amount of electric potential energy is point W.

Answer: ACE The battery is referred to as the internal circuit. Charge moves in the internal circuit from the - terminal to the + terminal (in the direction from W towards Z). The wires and light bulbs comprise the external circuit; charge moves through the external circuit from the + terminal to the - terminal (in the direction of X to Y to Z to W).

24. Voltage ____ an electrical circuit. a. goes through b. is expressed across c. is constant throughout d. is the rate at which charges move through

Answer: B Voltage or electric potential is not a thing which moves. Thus, choices A and D are not the answers since they imply movement of voltage. And the voltage or electric potential of a charge is not something which is constant throughout the circuit as choice C suggests. Voltage or electric potential is a measure of how energized a quantity of charge is at a given location relative to the - terminal. It is often expressed as a difference across two points. Perhaps you have noted this language of "the potential across ..." in several of the answers in this Review.

25. Two or more of the following words and phrases mean the same thing. Identify them by listing their letters. a. Voltage b. Wattage c. Electric Potential Difference d. Rate at which charge flows e. Electric Pressure f. Energy

Answer: ACE The voltage or electric potential difference are synonymous terms. Voltage is not synonymous with energy. While voltage (or electric potential difference) is a measure of how energized a quantity of charge is at a given location, voltage is expressed as the energy per charge (and not simply as energy). Drawing from the analogy between a water park and an electric circuit, voltage is a measure of the quantity of electric pressure placed upon a charge in inducing it to move from one location to another location. Wattage is synonymous with power. Current is synonymous with the rate at which charge flows.

26. A high voltage battery can ____. a. do a lot of work on each charge it encounters b. do a lot of work over the course of its lifetime c. push a lot of charge through a circuit d. last a long time

Answer: A Voltage refers to the energy/charge. A battery rated with a high voltage can do a lot of work per every Coulomb of charge which it encounters. It may or may not be able to do a lot of work over its lifetime, depending upon the size of the battery.

27. Which one of the following occurs when a rechargeable battery is recharged? a. The battery, which has run out of watts, has its wattage restored. b. The battery, which has run out of amps, has current placed back into it. c. The battery, which has run out of charge, has charge returned to it. d. The battery, which has run out of chemical reactants, has its chemicals reformed.

Answer: D Batteries perform their energy-supplying tasks by using the energy from an exothermic oxidation-reduction reaction to do work upon charge within the electric circuit. When a battery no longer works, its reactants are consumed to the point that the electric potential which the reactants are capable of producing is small compared to the overall resistance of the circuit. At such an instant in time, the ability to induce a current is limited to the point that the external circuit elements are no longer functional. Not all batteries are rechargeable. Those that are rechargeable can have the products transformed back into reactants. The recharger utilizes electrical energy from an outlet to reverse the previously exothermic reaction, turning its products back into reactants.

28. Birds can safely stand on high voltage electric power lines. This is because ____. a. they are at low potential with respect to the ground. b. they offer no resistance to current. c. they always choose power lines that are not in use. d. the potential difference between their feet is low. e. they are perfect insulators. f. they are perfect conductors.

Answer: D In order for charge to flow between two locations, there must be an electric potential difference established between those two locations. If a bird places its left foot on an electric power line and his right foot a few centimeters away on the same electric power line, then there is little to no difference in potential between his two feet. Without an electric potential difference, charge will not flow through the bird and the bird is safe.

29. When the light bulb in your lamp no longer works, it is because the bulb has _____. a. run out of energy and can no longer pump charge b. run out of voltage and must be recharged c. run out of electrons and so there is no more current d. burned all of its watts and can no longer shine e. tripped a circuit breaker and must be fixed at the fuse box f. a broken filament which has resulted in an open circuit g. ... nonsense! The bulb is fine; your family just needs to fully pay their power bill.

Answer: F The most common cause for the inability of a light bulbs to light is a broken filament. A coiled wire of tungsten stretches between two vertical supports. If disturbed while hot or if overworn, the tungsten metal can break and leave a gap between the two vertical supports. This gap represents a break in the circuit; a closed conducting loop is no longer established and charge will not flow.

30. A battery is needed in the circuit of your flashlight so that ____. a. charge is provided to the wires b. the energy of the light is balanced by the battery c. an exothermic, light-creating reaction is possible d. an electric potential difference is maintained across the circuit e. electrons are supplied in order to light the bulb

Answer: D One of the roles of a battery is to simply establish a difference in electric potential between its two terminals. Charge at the high potential will flow through the external circuit to the low potential location.

31. When you turn on the room lights, they light immediately. This is best explained by the fact that ____. a. electrons move very fast from the switch to the light bulb filament b. electrons present everywhere in the circuit move instantly

Answer: B Electrons move very slowly from one location to another location. But once a circuit is closed, they start moving immediately. While electrons move about a meter or in an hour, the actual signal which tells them to start moving can travel at the speed of light. So once the switch is turned on, a signal is circulated throughout the circuit to start the electrons marching. The electrons present in the filament of the circuit.

34. Which of the following statements represent correct unit equivalencies? List all that apply. a. 1 Ampere = 1 Coulomb / second b. 1 Joule = 1 Volt / Coulomb c. 1 Watt = 1 Joule • second d. 1 Watt = 1 Volt • Coulomb / second e. 1 Joule / Ohm = 1 Ampere • Coulomb f. 1 Joule • Ohm = 1 Volt2 • second

Answer: ADEF This question requires a knowledge of both units for electrical quantities and the equations which relate those quantities. In choice a, an Ampere is a unit of current (I) and a Coulomb/second is a unit of charge per unit of time (Q/t). This is consistent with the equation I = Q/t. In choice b, a Joule is a unit of energy (E) and a Volt / Coulomb is a unit of voltage per unit of charge (V/Q). Since voltage is the energy per charge, we would expect that energy would be equivalent to voltage • charge. Thus, it is incorrect to equate units of energy to units of voltage per charge. In choice c, a Watt is a unit of power (P) and a Joule•second is a unit of energy (E) multiplied by a unit of time (t). But power is energy / time and not energy • time so this is not a correct unit equivalency. In choice d, a Watt is a unit of power (P). On the right side, a Volt is a unit of voltage (V) and a Coulomb / second is a unit of current (I). So since P = I • V, this is a correct unit equivalency. In choice e, a Joule / Ohm is a unit of energy per unit of resistance (E / R). An Ampere • Coulomb is a unit of current multiplied by a unit of charge (I • Q). Thus, the equation is suggesting that E / R = I • Q. This can be rearranged algebraically to say that E / Q = I • R. Since voltage is the energy per charge (E / Q), the equation can be rewritten as V = I • R. This is thus a correct unit equivalency. In choice f, a Joule• Ohm is a unit of energy multiplied by a unit of resistance (E • R). A Volt2 / second is a unit of voltage2 multiplied by a unit of time (V2 • t). So this equation is suggesting that E • R = V2 • t. This can be rearranged algebraically to say that E / t = V2 / R. The right side of the equation is equivalent to power, so the equation can be rewritten as P = V2 / R. Since this is a correct way of writing the power equation, the given unit equivalency is correct.

35. Which of the following diagrams represents resistors connected in in series? List all that apply.

Answer: B A and C represent parallel connections as shown by the branching which occurs before and after the resistors. There is no branching in choice B so it is a series connection of resistors.

Questions #36-#39: The diagram at the right shows two identical resistors - R1 and R2 - placed in a circuit with a 12-Volt battery. Use this diagram to answer the next several questions. 36. These two resistors are connected in ____. a. series b. parallel c. neither

Answer: A One can start at the positive terminal of the battery and begin tracing their finger along the wire. If there is ever a point where the wire comes to a junction and branches in two or more directions, then the circuit has a parallel connection. Otherwise, it is a series circuit. In this diagram, there is no branching. Thus, it is a series circuit.

37. The electric potential difference (voltage drop) across each resistor is ___ Volts. a. 6 b. 12 c. 24 d. ... nonsense!. The electric potential difference is dependent upon the actual resistance of the resistors

Answer: A

Charge receives an increase in electric potential of 12 volts when moving through the internal circuit (the battery). So as a charge leaves the battery and traverses the external circuit, there must be a total drop in electric potential of 12 Volts. This drop in voltage occurs in a two-step fashion as the charge passes through each of the resistors. The charge will lose 6 volts in the first resistor and 6 volts in the second resistor, bringing it back to zero volts by the time it returns to the - terminal of the battery. The potential diagram at the right is a visual means of representing this important concept.

38. If a third resistor (R3), identical to the other two, is added in series with the first two, then the overall resistance will ____ and the overall current will ____. a. increase, increase b. decrease, decrease c. increase, decrease d. decrease, increase e. increase, remain the same f. decrease, remain the same g. remain the same, increase h. remain the same, decrease i. remain the same, remain the same

Answer: C Increasing the number of resistors in a series circuit will increase the overall resistance of that circuit and cause the current to decrease. (The opposite is true of a parallel circuit.)

39. If a third resistor (R3), identical to the other two, is added in series with the first two, then the electric potential difference (voltage drop) across each of the three individual resistors will ____. a. increase b. decrease c. remain the same

Answer: B Using the same reasoning as in Question #37, we can say that the charge will acquire 12 Volts in moving through the battery. It will have to lose this 12 volts in three steps as it passes through the external circuit. Since there are now three voltage drops in the external circuit instead of the original two, each drop must be smaller than before. So there will be a 4-Volt drop through each resistor (instead of the original 6-Volt drop).

Questions #40-#43:

The diagram at the right shows two identical resistors - R1 and R2 - placed in a circuit with a 12-Volt battery. Use this diagram to answer the next several questions.

40. These two resistors are connected in ____. a. series b. parallel c. neither

Answer: B One can start at the positive terminal of the battery and begin tracing their finger along the wire. If there is ever a point where the wire comes to a junction and branches in two or more directions, then the circuit has a parallel connection. Otherwise, it is a series circuit. In this diagram, there is some branching. Once the charge reaches the branching point, it will either pass through the resistor in the left branch (R1) or through the resistor in the right branch (R2). Thus, it is a parallel circuit.

41. The electric potential difference (voltage drop) across each resistor is ___ Volts. a. 6 b. 12 c. 24 d. ... nonsense!. The electric potential difference is dependent upon the actual resistance of the resistors

Answer: B Charge receives an increase in electric potential of 12 volts when moving through the internal circuit (the

battery). So as a charge leaves the battery and traverses the external circuit, there must be a total drop in electric potential of 12 Volts. This drop in voltage occurs in a single step since the charge will only pass through a single resistor on its path back to the battery. So since the charge chooses either the left or the right branch (and not both), either branch must provide the 12-Volt drop in voltage. In parallel circuits, the the electric potential difference across the battery is equal to the electric potential difference across either branch. The potential diagram at the right is a visual means of representing this important concept.

42. If a third resistor (R3), identical to the other two, is added in parallel with the first two, then the overall resistance will ____ and the overall current will ____. a. increase, increase b. decrease, decrease c. increase, decrease d. decrease, increase e. increase, remain the same f. decrease, remain the same g. remain the same, increase h. remain the same, decrease i. remain the same, remain the same

Answer: D Adding an identical resistor in a separate branch will provide more pathways by which charge can traverse through the loop of the circuit. This would be the equivalent of adding another booth at a toll station on a tollway in parallel with an existing booth. Opening up another lane for traffic will reduce the overall resistance and cause an increase in the car flow rate. The same occurs with charge in parallel circuits. More branches means less resistance and an increased current.

43. If a third resistor (R3), identical to the other two, is added in parallel with the first two, then the electric potential difference (voltage drop) across each of the three individual resistors will ____. a. increase b. decrease c. remain the same

Answer: C The electric potential difference across any branch is equal to the voltage of the battery. Adding a new branch can alter the overall resistance and the overall current, but it does not alter the electric potential difference across the battery nor across the branches.

44. The resistance of a charge-carrying conducting wire will increase as the ____. Choose all that apply. a. length of the wire is increased b. cross-sectional area of the wire is increased c. temperature of the wire is increased d. voltage impressed across the ends of the wire is increased e. wire is placed closer and closer to the + terminal of the circuit

Answer: AC Resistance of a wire increases with increasing length and (to a smaller extent) with increasing temperature. Increasing wire length increases the number of atom-charge collisions and thus the amount of resistance. Increasing the temperature increases the resistivity of the material and thus increases the overall resistance.

45. When plugged into a 120-Volt outlet, a light bulb consumes 300 joules of energy over a 5 second time period. The power of the light bulb is ____ Watts. a. 0.0167 b. 0.50 c. 2.0 d. 2.50 e. 60 f. 600 g. 1500 h. 7200

Answer: E Power is simply the rate at which energy is supplied to a circuit or transformed by a circuit. In this case, the power is the energy consumed per time. P = (300 J) / (5 seconds) = 60 Watts

46. A certain electrical circuit contains a battery, wires and a light bulb. If potential energy is gained by charges at the battery location, then charges lose potential energy ____. a. in the wires only b. in the bulb only c. equally in the wires and the bulb d. mostly in the wires but a little in the bulb e. mostly in the bulb but a little in the wires f. nowhere

Answer: E Charge will lose energy as they pass through locations of resistance. When in series, locations of greatest resistance will transform electrical energy into other forms at a greater rate. So energy will be lost in the light bulb and in the wires to a much lesser extent.

47. A high resistance light bulb and a low resistance light bulb are connected in series to a 6-Volt pack of batteries. Which of the two light bulbs will shine the brightest? a. They will have the same brightness. b. The low-R bulb will shine more brightly. c. The high-R bulb will shine more brightly. d. There is no way to make such a prediction since bulb brightness is independent of bulb resistance.

Answer: C Since the two light bulbs are in series, the same current (i) is experienced by each. The power will be given by the i2•R product. Since i is the same for each light bulb, the bulb with the greatest resistance will have the greatest power. So the high-R light bulb will transform electrical energy into light energy at the greatest rate and thus shine most brightly.

49. Three identical light bulbs are connected to a battery as shown at the right. Which adjustments could be made to the circuit that would increase the current being measured at X? Include all that apply. a. increase the resistance of one of the bulbs b. increase the resistance of two of the bulbs c. decrease the resistance of two of the bulbs d. increase the voltage of the battery e. decrease the voltage of the battery f. remove one of the bulbs

Answer: CDF The current in a series circuit (both total current and current through individual resistors) is directly dependent upon the battery voltage and inversely dependent upon the total circuit resistance. This current can be increased by increasing the battery voltage. It can also be increased by decreasing the total resistance. Removing a bulb would decrease the total resistance and decreasing the resistance of any individual bulb would decrease the total resistance.

50. Three identical light bulbs (labeled X, Y and Z) are connected to a battery as shown at the right. Which adjustments could be made to the circuit below that would increase the current at point P? List all that apply. a. increase the resistance of one of the bulbs b. increase the resistance of two of the bulbs c. decrease the resistance of two of the bulbs d. increase the voltage of the battery e. decrease the voltage of the battery f. remove one of the bulbs

Answer: CD Point P represents the location where the total current of this parallel circuit can be measured. The total current would vary directly with the total voltage and inversely with the total resistance. Increasing the battery voltage would increase the current at location P. Decreasing the total resistance would increase the current at location P. The total resistance can be decreased by adding another resistor in a separate branch or by decreasing the resistance of any of the branches.

51. Three identical light bulbs (labeled X, Y and Z) are connected to a battery as shown at the right. Which adjustments could be made to the circuit below that would decrease the current in bulb Z? List all that apply. a. increase the resistance of bulb X b. decrease the resistance of bulb X c. increase the resistance of bulb Z d. decrease the resistance of bulb Z e. increase the voltage of the battery f. decrease the voltage of the battery g. remove bulb Y

Answer: CF The current in bulb Z is dependent upon the voltage drop across bulb Z and the resistance of bulb Z. In equation form,

IZ = VZ / RZ

An increase in battery voltage would increase the voltage drop across bulb Z (VZ) and thus provide a greater current through the bulb. A decrease in the resistance of bulb Z would also increase the current through the bulb. However, making a change in bulb X or Y would have no effect on the VZ / RZ ratio.

54. Use proper schematic symbols to construct a diagram of a circuit powered by a 6-Volt battery that consists of two 3-ohm resistors connected in series. Place ammeters in series at a location such that the current through each resistor can be measured and in a location such that the overall current in the circuit can be measured. On the schematic diagram, use an unbroken arrow to indicate the direction of conventional current. Finally, indicate the ammeter readings on the diagram.

Answer: See diagram below. Since the two resistors are in series, the total or equivalent resistance is simply the sum of the individual resistances. The total resistance is 6 ohms. The total current in the circuit can be found by the ratio of battery voltage to total resistance:

ITOT = ΔVTOT / RTOT = (6 V) / (6 ohms) = 1 Amp Since it is a series circuit, the current through the battery is the same as the current through each of the resistors.

56. Consider the diagram at the right of a series circuit. Each light bulb in the circuit has an identical resistance. Use the labeled points on the diagram to answer the following questions. Each question may have one, less than one, or more than one answer. a. The electric potential at point A is the same as the electric potential at point(s) B. Include all that apply, if any apply. b. The electric potential at point C is the same as the electric potential at point(s) D and E. Include all that apply, if any apply. c. The electric potential at point F is the same as the electric potential at point(s) G. Include all that apply, if any apply. d. The electric potential at point I is the same as the electric potential at point(s) H. Include all that apply, if any apply. e. The electric potential difference between points A and B is the same as the electric potential difference between points C and D (and C and E; and D and E; and F and G; and H and I). Include all that apply, if any apply. f. The electric potential difference between points A and C is the same as the electric potential difference between points E and F (and C and F; and D and F; and E and G; and C and G; and D and G; and G and H; and F and H; and G and I; and F and I). Include all that apply, if any apply. g. The electric potential difference between points A and F is the same as the electric potential difference between points E and H (and between any other two sets of points that are on the opposite side of two adjacent bulbs ... such as points C and H or points D and H or points E and I, etc.). Include all that apply, if any apply. h. The electric potential difference between points D and H is the same as the electric potential difference between points A and F (and between any other two sets of points that are on the opposite side of two adjacent bulbs ... such as points C and H or points D and I or points E and I, etc.). Include all that apply, if any apply. i. The current at point A is the same as the current at point(s) C (and at every other point on the circuit). Include all that apply, if any apply. j. The current at point E is the same as the current at point(s) F (and at every other point on the circuit). Include all that apply, if any apply. k. The current at point G is the same as the current at point(s) G (and at every other point on the circuit). Include all that apply, if any apply.

Answer: See answers above. In an electric circuit, the electric potential for a moving charge is gained in the battery and lost in a light bulb (or some resistor found in the external circuit). So the electric potential of a charge is the same for any two points which are not separated by a battery or by a light bulb. (a through d) In this circuit, the light bulbs have the same resistance; thus, each light bulb causes the same drop in potential (electric potential difference). So the electric potential difference will be the same between any two points which are distanced by a light bulb, or by two light bulbs. (e through h) Finally, in a series circuit the current is the same at every point along the circuit. Since charge is conserved and since there is no location in a circuit where the charge is accumulating, there must be the same charge flow rate at all locations. (i through k)

57. Consider the diagram at the right of a parallel circuit. Each light bulb in the circuit has an identical resistance. Use the labeled points on the diagram to answer the following questions. Each question may have one, less than one, or more than one answer. a. The electric potential at point A is the same as the electric potential at point(s) B (and at any other point before the light bulbs ... such as at points C, D, E and F). Include all that apply, if any apply. b. The electric potential at point D is the same as the electric potential at point(s) B (and at any other point before the light bulbs ... such as at points A, C, E and F). Include all that apply, if any apply. c. The electric potential at point J is the same as the electric potential at point(s) K (and at any other point after the light bulbs ... such as at points G, H, I, and L. Include all that apply, if any apply. d. The electric potential difference between points A and J is the same as the electric potential difference between points B and K (and any other combination of two points located on opposite sides of the light bulbs ... such as points A and K, or points A and L, or points F and G, or points D and G, etc., etc.). Include all that apply, if any apply. e. The electric potential difference between points D and G is the same as the electric potential difference between points A and J (and any other combination of two points located on opposite sides of the light bulbs ... such as points A and K, or points A and L, or points F and G, or points E and H, etc., etc.). Include all that apply, if any apply. f. The current at point A is the same as the current at point(s) J. Include all that apply, if any apply. g. The current at point B is the same as the current at point(s) K. Include all that apply, if any apply. h. The current at point C is the same as the current at point(s) L (and points F and I and E and H and D and G). Include all that apply, if any apply. i. The current at point D is the same as the current at point(s) G (and at points F and I and C and E and H). Include all that apply, if any apply. j. If the light bulb located between points D and G were to be replaced by a bulb of greater resistance, then the current at point(s) A, D, G, and J would be decreased. Include all that apply, if any apply. k. If the light bulb located between points D and G were to be replaced by a bulb of greater resistance, then the electric potential difference between points -- and -- would be increased. Include all that apply, if any apply. (None apply.) l. If the light bulb located between points D and G were to go out, then the current would decrease at point(s) A, D, G and J. Include all that apply, if any apply.

Answer: See answers above. In an electric circuit, the electric potential for a moving charge is gained in the battery and lost in a light bulb (or some resistor found in the external circuit). So the electric potential of a charge is the same for any two points which are not separated by a battery or by a light bulb. Even if the circuit is a parallel circuit, any point between the positive terminal of a battery and light bulb will have the same electric potential; and any point located between the - terminal of the battery and a location after passage through the resistor of a branch have the same electric potential. (a through e) In this circuit, the light bulbs have the same resistance. Thus, when charge reaches the branching location an equal amount of charge will chose the middle branch as does the left branch and the right branch. So with the same resistance, the current in every branch is the same. And the current prior to the branching location and after the branches come together is the same. Finally, locations B and K are at locations where current for two of the three branches will be passing; these points will have the same current. (e through i) In j, removing the bulb in the first branch will not affect the current in the other branches. Such a modification will only reduce the overall circuit current. Less branches would result in more overall resistance and less overall current. Yet the current through the second branch is still the voltage drop across the second branch (which is the battery voltage) divided by the resistance of the second branch. Since removing the bulb in the first branch does not alter either quantity, the current in the middle branch is not altered. In k, the electric potential across a branch is simply equal to the voltage of the battery. Removing a light bulb will not alter the voltage of the battery. Thus, there is no effect. In l, if the light bulb in the first branch burns out, the same effect will occur as occurred in part j.

58. Consider the diagram below of a series circuit. For each resistor, use arrows to indicate the two locations where one would have to tap with the leads of a voltmeter in order to measure the voltage drop across the individual resistor. Finally, indicate the ammeter readings and the voltage readings.

Answer: See diagram above. The total resistance (or equivalent resistance) can first be determined using the equation for series circuits.

RTot = R1 + R2 + R3 = 5 Ohms + 10 Ohms + 15 Ohms = 30 Ohms

Once known, the RTot value can be used with the battery voltage (Δ VTot) to determine the total current in the circuit. ITot = (ΔVTot) / (RTot) = (120 V) / (30 Ohms) = 4 Amps

For a series circuit, the current through each resistor is the same as the total circuit current. Thus, I1 = I2 = I3 = 4 Amps. The voltage drop across a resistor can be determined with a voltmeter by tapping with the leads on the metal wires on the opposite sides of the resistor. By so doing, the voltmeter determines the difference in voltage (i.e., voltage drop or electric potential difference) between the two locations where the leads were tapped. In this circuit, the expected voltage drops (Δ V1, Δ V2, and Δ V3 ) can be computed by determining the IR product for each resistor. This is shown below.

ΔV1 = I1 • R1 = (4 Amps) • (5 Ohms) = 20 Volts

ΔV2 = I2 • R2 = (4 Amps) • (10 Ohms) = 40 Volts

ΔV3 = I3 • R3 = (4 Amps) • (15 Ohms) = 60 Volts

59. Consider the diagram below of a parallel circuit. For each resistor, use arrows to indicate the two locations where one would have to tap with the leads of a voltmeter in order to measure the voltage drop across the individual resistor. Finally, indicate the ammeter readings and the voltage readings.

Answer: See diagram above. The total resistance (or equivalent resistance) can first be determined using the equation for series circuits.

1 / RTot = 1 / R1 + 1 / R2 + 1 / R3 = 1 / (5 Ohms + 1 / (10 Ohms) + 1 / (15 Ohms)

RTot = 2.727 Ohms

Once known, the RTot value can be used with the battery voltage (Δ VTot) to determine the total current in the circuit.

ITot = (ΔVTot) / (RTot) = (120 V) / (2.727 Ohms) = 44.0 Amps The voltage drop across a resistor can be determined with a voltmeter by tapping with the leads on the metal wires on the opposite sides of the resistor. By so doing, the voltmeter determines the difference in voltage (i.e., voltage drop or electric potential difference) between the two locations where the leads were tapped. For a series circuit, the expected voltage drop across each resistor (Δ V1, Δ V2, and Δ V3 ) is the same as the total voltage drop. Thus, Δ V1 = Δ V2 = Δ V3 = 120 Volts. In this circuit, the branch currents can be computed by using the Δ V = I•R equation for each resistor. This is shown below.

R1 = ΔV1 / R1 = (120 Volts) / (5 Ohms) = 24 Amps

I2 = Δ V2 / R2 = (120 Volts) / (10 Ohms) = 12 Amps

I3 = Δ V3 / R3 = (120 Volts) / (15 Ohms) = 8 Amps

Part D: Qualitative Relationships Between Variables 60. A resistor with a resistance of R is connected to a battery with a voltage of V to produce a current of I. What would be the new current (in terms of I) if ... a. ... the resistance is doubled and the same voltage is used? b. ... the voltage is doubled and the same resistance is used? c. ... the voltage is tripled and the resistance is doubled? d. ... the voltage is doubled and the resistance is halved? e. ... the voltage is halved and the resistance is doubled? f. ... five times the voltage and one-third the resistance is used? g. ... one-fifth the voltage and one-fourth the resistance is used?

Answer: See answers below. This question tests your understanding of the current-voltage-resistance relationship. The current is directly proportional to the voltage and inversely proportional to the resistance. Any alteration in the voltage will result in the same alteration of the current. So doubling or tripling the voltage will cause the current to be doubled or tripled. On the other hand, any alteration in the resistance will result in the opposite or inverse alteration of the current. So doubling or tripling the resistance will cause the current to be one-half or one- third the original value. a. The new current will be 0.5 • I. b. The new current will be 2 • I. c. The new current will be 1.5 • I. d. The new current will be 4 • I. e. The new current will be 0.25 • I. f. The new current will be 15 • I. g. The new current will be 0.8 • I.

62. An electric appliance with a current of I and a resistance of R converts energy to other forms at a rate of P when connected to a 120-Volt outlet. What would be the new power rating (in terms of P) if ... a. ... the current is doubled (and the same 120-Volt outlet is used)? b. ... the current is halved (and the same 120-Volt outlet is used)? c. ... the resistance is doubled (and the same 120-Volt outlet is used)? d. ... the resistance is halved (and the same 120-Volt outlet is used)? e. ... the current is tripled (and the same 120-Volt outlet is used)? f. ... the resistance is tripled (and the same 120-Volt outlet is used)? g. ... the same appliance is powered by a 12-Volt supply? h. ... the same appliance is powered by a 240-Volt supply?

Answer: See answers below. This question tests your understanding of the mathematical relationship between power, current, voltage and resistance. There are three equations of importance:

P = I • ΔV P = ΔV2 / R P = I2 • R One must be careful in using the last equation since an alteration in current will also alter the resistance whenever the voltage is held constant. Thus, the first two equations are of greater importance since they represent equations with one independent variable and the other variable held constant. a. 2 • P (doubling the current will double the power) b. (1/2) • P (halving the current will double the power) c. (1/2) • P (doubling the resistance will half the power) d. 2 • P (halving the resistance will double the power) e. 3 • P (tripling the current will triple the power) f. (1/3) • P (tripling the resistance will make the power one-third of the original value) g. (1/100) • P (one-tenth the voltage will make the power one-hundredth of the original value; observe the square on voltage) h. 4 • P (two times the voltage will make the power four times the original value; observe the square on voltage)

66. What is the resistance (in ohms) of a typical 40-Watt light bulb plugged into a 120-Volt outlet in your home?

Answer: 360 Ohms The power dissipated in a circuit is given by the equation P = I•Δ V. Substituting in Δ V/R for the current can lead to an equation relating the resistance (R) to the voltage drop (Δ V) and the power (P). P = I•ΔV = (ΔV/R)•ΔV = ΔV2 / R Rearrangement of the equation and substitution of known values of power (40 Watts) and voltage (120 V) leads to the following solution. R = ΔV2 / P = (120 V)2 / (40 Watts) = 360 Ohms

71. Four resistors - 2.0-Ohms, 5.0-Ohms, 12-Ohms and 15-Ohms - are placed in series with a 12-Volt battery. Determine the current at and voltage drop across each resistor.

Answer: See diagram below. The diagram below depicts the series circuit using schematic symbols. Note that there is no branching, consistent with the notion of a series circuit.

For a series circuit, the overall resistance (RTot) is simply the sum of the individual resistances. That is

RTot = R1 + R2 + R3 + R4

RTot = 2 ½ + 5 ½ + 12 ½ + 15 ½ = 34 ½ The series of three resistors supplies an overall, total or equivalent resistance of 34 Ohms. Since there is no branching, the current is the same through each resistor. This current is simply the overall current for the circuit and can be determined by finding the ratio of battery voltage to overall resistance (VTot/RTot).

ITot = ΔVTot/RTot = (12 Volt) / (34 Ohm) = 0.35294 Amps The current through the battery and through each of the resistors is ~0.353 Amps. The voltage drop across each resistor is equal to the I•R product for each resistor. These calculations are shown below.

ΔV1 = I1 • R1 = (0.35294 Amps) • (2 Ohms) = 0.71 V ΔV2 = I2 • R2 = (0.35294 Amps) • (5 Ohms) = 1.76 V

ΔV3 = I3 • R3 = (0.35294 Amps) • (12 Ohms) = 4.24 V

ΔV4 = I4 • R4 = (0.35294 Amps) • (15 Ohms) = 5.29 V

72. Four resistors - 2-Ohms, 5-Ohms, 12-Ohms and 15-Ohms - are placed in parallel with a 12-Volt battery. Determine the current at and voltage drop across each resistor.

Answer: See diagram below. The diagram below depicts the parallel circuit using schematic symbols. Note that there is a branching, consistent with the notion of a parallel circuit.

For a parallel circuit, the reciprocal of overall resistance (1 / RTot) is simply the sum of the reciprocals of individual resistances. That is

1 / RTot = 1 / R1 + 1 / R2 + 1 / R3 + 1 / R4

1 / RTot = 1 / 2 ½ + 1 / 5 ½ + 1 / 12 ½ + 1 / 15 ½ = 0.850 / ½

RTot = 1.17647 ½ The series of three resistors supplies an overall, total or equivalent resistance of ~1.18 Ohms. This total resistance value can be used to determine the total current through the circuit.

ITot = ΔVTot/RTot = (12 Volt) / (1.17647 Ohm) = 10.2 Amps Since there is branching, the total current will be equal to the sum of the currents at each resistor. The current at each resistor is the voltage drop across each resistor divided by the resistance of each resistor. For series circuits, the voltage drop across each resistor is the same as the voltage gained by the charge in the battery (12 Volts in this case). The branch current calculations are shown below.

I1 = ΔV1 / R1 = (12 Volts) / (2 Ohms) = 6.00 Amp

I 2 = ΔV2 / R2 = (12 Volts) / (5 Ohms) = 2.40 Amp

I 3 = ΔV3 / R3 = (12 Volts) / (12 Ohms) = 1.00 Amp

I 4 = ΔV4 / R4 = (12 Volts) / (15 Ohms) = 0.80 Amp

Basics of Waves Review

1. A single disturbance that moves from point to point through a medium is called a ___. a. period b. periodic wave c. wavelength d. pulse

Answer: D A wave is a continuous and repeating disturbance of a medium and a pulse is a single disturbance.

2. If the particles of the medium are vibrating to and fro in the same direction of energy transport, then the wave is a ____ wave. a. longitudinal b. sound c. standing d. transverse

Answer: A In longitudinal waves, particles of the medium vibrate to and from in a direction parallel to the direction of energy transport. If energy is transmitted along a medium from the east end to the west end, then particles of the medium would vibrate eastward and westward A sound wave is a longitudinal wave but not the answer since a wave which exhibits this characteristic is not necessarily a sound wave.

3. When the particles of a medium are vibrating at right angles to the direction of energy transport, then the wave is a ____ wave. a. longitudinal b. sound c. standing d. transverse

Answer: D In transverse waves, particles of the medium vibrate to and from in a direction perpendicular to the direction of energy transport.

4. A transverse wave is traveling through a medium. See diagram below. The particles of the medium are vibrating _____. a. parallel to the line joining AD. b. along the line joining CI. c. perpendicular to the line joining AD. d. at various angles to the line CI. e. along the curve CAEJGBI.

Answer: A In transverse waves, particles of the medium vibrate to and from in a direction perpendicular to the direction of energy transport. In this case, that would be parallel to the line AD.

5. If the energy in a longitudinal wave travels from south to north, the particles of the medium would be vibrating _____. a. from north to south, only b. both north and south c. from east to west, only d. both east and west

Answer: B In longitudinal waves, particles of the medium vibrate to and from in a direction parallel to the direction of energy transport. If the particles only moved north and not back south, then the particles would be permanently displaced from their rest position; this is not wavelike.

6. As a pulse travels though a uniform medium, the speed of the pulse ____. a. decreases b. increases c. remains the same

Answer: C The speed of a wave or a pulse depends upon the properties of the medium. If the medium is uniform or unchanging, then the speed is constant.

7. The main factor which effects the speed of a sound wave is the ____. a. amplitude of the sound wave b. intensity of the sound c. loudness of the sound d. properties of the medium e. pitch of the sound

Answer: D The speed of a wave is dependent upon the properties of the medium and not the properties of the wave.

8. As a wave travels into a medium in which its speed increases, its wavelength would ____. a. decrease b. increase c. remain the same

Answer: B As a wave crosses a boundary into a new medium, its speed and wavelength change while its frequency remains the same. If the speed increases, then the wavelength must increase as well in order to maintain the same frequency.

9. As a wave passes across a boundary into a new medium, which characteristic of the wave would NOT change? a. speed b. frequency c. wavelength

Answer: B As a wave crosses a boundary into a new medium, its speed and wavelength change while its frequency remains the same.

10. What is the amplitude of the wave in the diagram below? a. 0.03 m. b. 0.04 m. c. 0.05 m. d. 0.06 m.

Answer: A The amplitude of a wave is measured from rest to crest or from rest to trough; but not from crest to trough. Thus, take the 0.06 m measurement and "halve it" to get the answer.

11. The wavelength of the wave in the diagram above (Question #10) is ____ m. a. 0.030 b. 0.040 c. 0.060 d. 0.080

Answer: D The wavelength of a wave is measured as the distance between any two corresponding points on adjacent waves, which would mean from a crest to the next adjacent crest. Thus, the distance from point b to point d is the wavelength - 0.08 m

12. A wave X meters long passes through a medium with a speed of Y meters per second. The frequency of the wave could be expressed as a. Y/X cycles/sec. b. X/Y cycles/sec. c. XY cycles/sec. d. (X + Y) cycles/sec.

Answer: A Let d = wavelength. Since d*f = v, f = v/d.

15. The number of cycles of a periodic wave occurring per unit time is defined as a wave's ____. a. wavelength. b. period. c. amplitude. d. frequency.

Answer: D This is a basic definition which you should know and be able to apply.

16. A periodic and repeating disturbance in a lake creates waves which emanate outward from its source to produce circular wave patterns. If the frequency of the source is 2.00 Hz and the wave speed is 5.00m/s then the distance between adjacent wave crests is ___ meter. a. 0.200 b. 0.400 c. 1.25 d. 2.50 e. 10.0

Answer: D Let w=wavelength; then v = w*f. In this problem, it is given that v=5.00 m/s and f = 2.00 Hz. Substitution and algebra yields w = v / f = 2.50 m.

17. What is the frequency of a wave that has a speed of 0.4 m/s and a wavelength of 0.020 meter? a. 10 hertz. b. 20 hertz. c. 0.008 hertz. d. 0.5 hertz.

Answer: B Let w=wavelength; then v = w*f. In this problem, it is given that v=5 m/s and w = .02 m. Substitution and algebra yields f=v/w=20 Hz.

18. Many wave properties are dependent upon other wave properties. Yet, one wave property is independent of all other wave properties. Which one of the following properties of a wave is independent of all the others? a. wavelength b. frequency c. period d. velocity

Answer: D The speed of a wave is dependent upon the properties of the medium through which it moves, not upon the properties of the wave itself.

19. A pendulum makes exactly 40 vibrations in 20.0 s. Its period is ____. (Be cautious of the units.) a. 0.500 Hz. b. 0.500 s. c. 2.00 Hz. d. 2.00 s. e. 8.00 x 102 Hz.

Answer: B The period is the time for one complete cycle. If the pendulum takes 20 seconds for exactly 40 vibrational cycles, then it must take 0.500 second for one cycle.

20. A period of 0.005 seconds would be equivalent to a frequency of ____ Hz. a. 20 b. 50 c. 200 d. 500 e. 2000

Answer: C The period and the frequency are related by a reciprocal relationship; that is, f = 1/T.

21. TRUE or FALSE: The number of waves generated per second by a source is called the frequency of the source. a. True b. False

Answer: A This is a basic definition which you should know and be able to apply.

22. TRUE or FALSE: The SI unit for frequency is hertz. a. True b. False

Answer: A Know this like the back of your hand.

23. TRUE or FALSE: Doubling the frequency of a wave source (without altering the medium) doubles the speed of the waves. a. True b. False Answer: B Don't be fooled. Wave speed may equal frequency*wavelength. Yet doubling the frequency only halves the wavelength; wave speed remains the same. To change the wave speed, the medium would have to be changed.

24. If the frequency of a wave is doubled and if the speed remains constant, its wavelength is ____. a. quartered. b. halved. c. unchanged. d. doubled.

Answer: B Wave speed equals frequency*wavelength. So doubling the frequency must halve the wavelength in order for wave speed to remain the same.

25. Two different ropes with different mass densities are attached to each other. A pulse is introduced into one end of the rop and approaches the boundary as shown at the right. At the boundary, a portion of the energy is transmitted into the new medium and a portion is reflected. Which one of the diagrams below depicts the possible location and orientation of the pulse shortly after the incident pulse reaches the boundary?

Answer: C The speeds in the two media can be deduced by the distance of the pulses from the boundary. In A and E, the speed is shown as fastest on the right, which makes the transmitted medium the less dense. Rule out A and E since a reflected pulse should not invert when moving from more dense to less dens. Rule out B for just the opposite reasons; the wave is moving from less to more dense and should invert upon reflection. Rule out D because the transmitted pulse never inverts. That leaves C as the answer.

26. When a pulse reaches a boundary between two different media, it will be____. a. reflected, only. b. transmitted, only. c. partly reflected and partly transmitted. d. neither reflected nor transmitted.

Answer: C This is basic information about the boundary behavior of waves.

27. Diagram P at the right shows a transverse pulse traveling along a dense rope toward its junction with a less dense rope. Which of the diagrams (A, B, C, D, or E) below depicts the ropes at the instant that the reflected pulse again passes through its original position marked X? Consider such features as amplitude and relative speed (i.e., the relative distance of the transmitted and reflected pulses from boundary).

Answer: E A, B, and C can be quickly ruled out since it shows the amplitude of the reflected and incident pulse to be the same size. An incident pulse would give up some of its energy to the transmitted pulse at the boundary, thus making the amplitude of the reflected pulse less than that of the incident pulse. Rule out D since it shows the reflected pulse moving faster than the transmitted pulse. This would not happen unless moving from less dense to more dense. This leaves E as the answer.

30. A 2.0-meter long rope is hanging vertically from the ceiling and attached to a vibrator. A single pulse is observed to travel to the end of the rope in 0.50 s. What frequency should be used by the vibrator to maintain three whole waves in the rope? a. 0.75 Hz b. 1.3 Hz c. 4.0 Hz d. 6.0 Hz e. 8.0 Hz

Answer: D The given info allows you to determine the speed of the wave: v=d/t=2 m/0.5 s) = 4 m/s. If there are 3 waves in a 2-meter long rope, then each wave is 2/3-meter long. Now find frequency with the equation v=f*w where v=4 m/s and w=0.667 m. Proper algebra yields 6 Hz as the answer.

31. A standing wave experiment is performed to determine the speed of waves in a rope. The standing wave pattern shown below is established in the rope. The rope makes exactly 90 complete vibrational cycles in one minute. The speed of the waves is ____ m/s.

a. 3.0 b. 6.0 c. 180 d. 360 e. 540

Answer: B If there are exactly 90 vibrations in 60.0 seconds, then there is a frequency of 1.5 Hz. The diagram shows 1.5 waves in 6.0-meters of rope; thus, the wavelength is 4.0 meters. Now use the equation v=f*w to calculate the speed of the wave. Proper substitution yields 6.0 m/s.

33. A node is a point located along the medium where there is always ___. a. a double crest b. c. constructive interference d. destructive interference e. a double rarefaction

Answer: D A node is a point along the medium of no displacement. The point is not displaced because destructive interference occurs at this point.

34. TRUE or FALSE: Constructive interference of waves occurs when two crests meet. a. True b. False

Answer: A Yes! Or when a trough meets a trough or whenever two waves displaced in the same direction (such as both up or both down) meet.

35. Which phenomenon is produced when two or more waves passing simultaneously through the same medium meet up with one another? a. refraction b. diffraction c. interference d. reflection

Answer: C Interference is the meeting of two or more waves when passing along the same medium - a basic definition which you should know and be able to apply.

36. Two pulses are traveling in opposite directions along the same medium as shown in the diagram at the right. Which diagram below best depicts the appearance of the medium when each pulse meets in the middle?

Answer: D When a crest is completely overlapped with a trough having the same amplitude, destructive interference occurs. Complete cancellation takes place if they have the same shape and are completely overlapped.

37. TRUE or FALSE: A vibrating object is necessary for the production of sound. a. True b. False

Answer: A Absolutely! If you don't believe it, then think of some sounds - voice, guitar, piano, tuning fork, chalkboard screech, etc. - and consider what the vibrational source is. All sounds have a vibrating object of some kind as their source.

38. Which one of the following CANNOT transmit sound? a. Liquid air b. Gaseous oxygen c. Liquid water d. Solid steel e. Perfect vacuum

Answer: E Sound is a mechanical wave and as such requires a medium in order to move through space. For this reason, sound cannot move through a vacuum.

3. A ray of light in air is incident on an air-to-glass boundary at an angle of 30. degrees with the normal. If the index of refraction of the glass is 1.65, what is the angle of the refracted ray within the glass with respect to the normal? a. 56 degrees b. 46 degrees c. 30. degrees d. 18 degrees

Answer: D Use Snell's law:

ni * sine(Theta i) = nr * sine(Theta r) where

ni =1.00 (in air), Theta i=30. degrees, nr =1.65 Substitute and solve for Theta r. sine(Theta r) = 1.00 * sine(30. degrees) / 1.65 = 0.3030 Theta r = invsin(0.3030) = 17.6 degrees

4. If the critical angle for internal reflection inside a certain transparent material is found to be 48.0 degrees, what is the index of refraction of the material? (Air is outside the material). a. 1.35 b. 1.48 c. 1.49 d. 0.743

Answer: A The critical angle is the angle of incidence (which is always in the more dense material) for which the angle of refraction is 90 degrees. Apply this to Snell's law equation:

nr * sine(48.0 deg) = 1.00 * sine (90.0 deg) Solve for nr.

nr = 1.00/sin(48.0 deg) = 1.35

6. Carbon tetrachloride (n = 1.46) is poured into a container made of crown glass (n = 1.52). If the light ray in glass incident on the glass-to-liquid boundary makes an angle of 30 degrees with the normal, what is the angle of the corresponding refracted ray with respect to the normal? a. 55.5 degrees b. 29.4 degrees c. 31.4 degrees d. 19.2 degrees

Answer: C Use Snell's law:

ni * sine(Theta i) = nr * sine(Theta r) where

ni =1.52 (in glass), Theta i=30 degrees (angle in glass), nr =1.46 (in carbon tetrachloride) Substitute and solve for Theta r 1.52 * sin(30 deg) = 1.46 * sin(Theta r) [1.52 * sin(30 deg)]/1.46 = sin(Theta r) 0.5205 = sin(Theta r) Theta r = invsine(0.5205) = 31.4 degrees

7. A light ray in air is incident on an air to glass boundary at an angle of 45.0 degrees and is refracted in the glass of 30.0 degrees with the normal. What is the index of refraction of the glass? a. 2.13 b. 1.74 c. 1.23 d. 1.41

Answer: D Use Snell's law:

ni * sine(Theta i) = nr * sine(Theta r) where

ni=1.00 (in air), Theta i=45.0 degrees (angle in air), Theta r=30.0 degrees (in glass)

Substitute and solve for nr.

1.00 * sine(45.0 deg) = nr * sine(30.0 deg)

1.00 * sine(45.0 deg) / sine(30.0 deg) = nr

nr = 1.41

8. A beam of light in air is incident at an angle of 35 degrees to the surface of a rectangular block of clear plastic (n = 1.5). The light beam first passes through the block and reemerges from the opposite side into air at what angle to the normal to that surface? a. 42 degrees b. 23 degrees c. 35 degrees d. 59 degrees

Answer: C The light ray bends towards the normal upon entering and away from the normal upon exiting. If the opposite sides are parallel to each other and surrounded by the same material, then the angle at which the light enters is equal to the angle at which the light exits.

9. A light ray in air enters and passes through a block of glass. What can be stated with regard to its speed after it emerges from the block? a. speed is less than when in glass b. speed is less than before it entered glass c. speed is same as that in glass d. speed is same as that before it entered glass

Answer: D The speed of a light wave (like any wave) is dependent upon the medium through which it moves. In the case of a light wave, the speed is least in the most dense medium. Thus, the light moves slower in glass than in air. However, upon exiting the glass and entering the air, the light returns to the original speed. The speed is the same for the same medium.

10. Which of the following describes what will happen to a light ray incident on an air-to-glass boundary? a. total reflection b. total transmission c. partial reflection, partial transmission d. partial reflection, total transmission Answer: C Upon reaching a boundary, a wave undergoes both reflection and transmission. The only exception is for light in the more dense medium and at angles of incidence greater than the critical angle; in such a case, total internal reflection occurs. Since the question does not specify such conditions, one would have to answer c.

11. Which of the following describes what will happen to a light ray incident on an air-to-glass boundary at an angle of incidence less than the critical angle? a. total reflection b. total transmission c. partial reflection, partial transmission d. partial reflection, total transmission

Answer: C At a boundary between any two materials, there will be both reflection and transmission. The only exception is when the light is incident in the more dense material and at an incident angle greater than the critical angle.

12. Which of the following describes what will happen to a light ray incident on an glass-to-air boundary at an angle of incidence greater than the critical angle? a. total reflection b. total transmission c. partial reflection, partial transmission d. partial reflection, total transmission

Answer: A At a boundary between any two materials, there will be both reflection and transmission. The only exception is when the light is incident in the more dense material and at an incident angle greater than the critical angle. Since the light in this problem is in the more dense medium (glass) and at an angle greater than the critical angle, total internal reflection will occur. No transmission will occur at this boundary for such angles.

13. What is the angle of incidence on an air-to-glass boundary if the angle of refraction in the glass (n = 1.52) is 25 degrees? a. 16 degrees b. 25 degrees c. 40 degrees d. 43 degrees Answer: C Use Snell's law:

ni * sine(Theta i) = nr * sine(Theta r) where ni =1.52 (in glass), Theta i=25 degrees (angle in glass), nr =1.00 (in air) Substitute and solve for Theta r. 1.52 * sine(25 degrees) = 1.00 * sine(Theta r) 1.52 * sine(25 degrees) / 1.00 = sine(Theta r) 0.0.6424 = sine(Theta r) Theta r = invsine(0.6424) = 40.0 degrees

14. A ray of white light, incident upon a glass prism, is dispersed into its various color components. Which one of the following colors experiences the greatest amount of refraction? a. orange b. violet c. red d. green

Answer: B The shorter wavelengths of light undergo the most refraction. Thus, violet is refracted the most and red light is refracted the least. The fact that the various component colors of white light refract different amounts leads to the phenomenon of dispersion.

17. When light from air hits a smooth piece of glass with the ray perpendicular to the glass surface, the part of the light passing into the glass _____. a. will not change its speed b. will not change its direction c. will not change its wavelength d. will not change its intensity

Answer: B When the angle of incidence is 0 degrees (as in this case), there is no bending. The ray still slows down and changes its wavelength. A portion of the wave reflects and so there is a change in intensity within the new medium. Yet, there is no refraction or bending; the direction does not change.

Part C: Diagramming and Analysis

35. The diagram below shows four incident rays Angle of Angle of Angle of Ray traveling through water (n = 1.33) and approaching incidence reflection refraction the boundary with air (n=1.00). For each incident ray, draw the corresponding reflected ray; then 1 0.0 deg. 0.0 deg. 0.0 deg. calculate the angle of refraction and draw the 2 30.0 deg. 30.0 deg. 41.7 deg. corresponding refracted ray. Label or color code 3 45.0 deg. 45.0 deg. 70.1 deg. your reflected and refracted rays to distinguish them from each other and to match them to the incident ray.

4 60.0 deg. 60.0 deg. None

Answer: See diagram above The angle of reflection equals the angle of incidence; this is simply the law of reflection. Each angle is measured relative to the normal. The angle of refraction is determined using Snell's law. For the 30.0-degree angle, the work would be:

nwater * sin(Angle in water) = nair * sin(Angle in air) 1.33 * sin(30.0 deg) = 1.00 * sin(Angle in air) 0.665 = sin(Angle in air) invsin(0.665) = Angle in air 41.7 deg = Angle in air The same calculations are done for the other angles to determine the angle of refraction. In the case of the 60.0 degree angle, the mathematics breaks down and the equation cannot be solved for an angle of refraction. This is because refraction does not occur at the water-air boundaries when light is incident in water at angles of incidence greater than the critical angle (approx. 48 degrees).

36. In the diagrams below, construct refracted rays to indicate the direction which the light rays bend upon crossing the boundary. If light travels faster in medium 1 than in medium 2, If the index of refraction of medium 1 is greater than then upon entering medium 2, the light will bend medium 2, then upon entering medium 2, the light will towards from the normal. bend away from the normal.

Answer: See diagram above. On the left, light is moving from a medium in which it travels fast to one in which it travels slow. In such a case, light refracts towards the normal. (Remember fast to slow bends towards as in "FST" or "Freaky Science Teachers".) The refracted ray should be drawn closer to the normal than the incident ray is. On the right, light is moving from a medium which is more dense or slower to a medium which is less dense or faster. In such a case, light refracts away from the normal. (Remember slow to fast bends away - "SFA" or "Suzie Farts Alot".) The refracted ray should be draw farther away from the normal than the incident ray is.

Part D: Problem-Solving 39. The speed of light in a vacuum is 3.00 x 108 m/s. Determine the speed of light through the following materials. a. air (n=1.00) b. ethanol (n=1.36) c. crown glass (n=1.52) d. zircon (n=1.91) e. diamond (n=2.42)

Answers: For each of these questions, find the speed of light in the material by dividing the 3.00 x108 m/s by the index of refraction of the material. This yields the following results: a. (3.00 x108 m/s)/1.00 = 3 x 108 m/s b. (3.00 x108 m/s)/1.36 = 2.21 x 108 m/s c. (3.00 x108 m/s)/1.52 = 1.97 x 108 m/s d. (3.00 x108 m/s)/1.91 = 1.57 x 108 m/s e. (3.00 x108 m/s)/2.42 = 1.24 x 108 m/s

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