Name______KEY______Date______4/25/11______
Physics 12 – Unit Exam 2 Chapters 9,10,11,12,13,14,19
For the questions that require calculations, use the space provided for the question to show how you arrived at your answer. You must show your work in order to receive credit for your answer. Use 10 m/s2 as the acceleration due to gravity at the surface of the earth.
1. A physics student’s mass is 60 kg in Santa Monica. The acceleration due to gravity on the planet Jupiter is approximately 25 m/s2, or 2.5 times greater than the acceleration due to gravity on the earth.
a. How much greater is the mass of the student on the planet Jupiter compared to her mass on the earth? (2)
MASS IS THE SAME. IT DOESN’T CHANGE.
b. How much greater is the weight of the student on the planet Jupiter compared to her weight on the earth? (2)
THE WEIGHT ON JUPITER IS 2.5 TIMES GREATER.
2. A physics student weighs 600 Newtons at the surface of the earth.
a. How much does the student weigh when she is a distance of three earth radii above the surface of the earth? (3)
WEIGHT = 600/42 = 37.5 NEWTONS
b. How much would she weigh at the surface of the earth if the mass of the earth were to double, and the radius of the earth were to double? (3)
WEIGHT = 600 x 2/22 = 300 NEWTONS
3. For the four positions shown in, on and around the earth, rank the weight of a person from greatest to least (that is, where the person weighs the most to where the person weighs the least). (3)
A: WEIGHT = 0 B: WEIGHT = W/2 C: WEIGHT = W D: WEIGHT = W/22 = W/4 R/2
RANKING: C,B,D,A D A B C
R R earth 3. When a student drops a ball from a tall tower, she know that after one second the ball will have fallen 5 meters, after two second it will have fallen 20 meters, and after three seconds it will have
fallen 45 meters, and so on. (Ignore any air resistance.) 15 m/s
a. Suppose she throws a ball horizontally with a velocity of 15 m/s off of a tower that is 20 meters high. How far from the tower will the ball hit the 20 m ground? (3)
TIME TO HIT GROUND = 2 SEC
DISTANCE = 15 M/S x 2 SEC = 30 M
? 3 sec
b. Suppose she throws the ball at an angle 2 sec upwards. If gravity was not present, 40 m the ball will travel along the straight- line path shown, and its position after one, two, and three seconds is indicated 1 sec on the path. However, gravity is 30 m present, so the ball does not travel on 20 M 5 M the straight-line path, but follows a 45 M curved path. Indicate on the diagram where the ball should be after one, two, 20 m and three seconds when gravity is not ignored. (3)
10 m
0 m
c. If the student were to throw the ball horizontally off the tower with a speed of roughly 8 km/sec, where does the ball hit the ground? (Careful!) (2)
8 KM/SEC IS THE SPEED NECESSARY FOR THE BALL TO BECOME A SATELLITE AROUND THE EARTH.
D E 4. A planet is shown circling the sun. C a. What is the shape of the orbit in which the planet is traveling? (2)
sun ELLIPSE
A b. What is the significance of the position of the sun to the orbit in which the planet is traveling? B (2)
THE SUN IS AT THE POSITION OF A FOCI OF THE ELLIPSE.
c. At which point on the orbit is the planet traveling the fastest? (2)
A – CLOSEST TO THE SUN
d. Rank the time it takes the planet to travel the path segments AB, BC, CD, and DE from shortest time to longest time, and explain how you arrived at your ranking. (3)
COMPARE THE AREAS SWEPT OUT:
AB, DE, CD, BC
5. The diagram is a simplified model of the helium atom. Draw a line from the name of each part of the atom to where it is located in the atom. (The atom is not drawn to scale.) (4)
Electron
Nucleus
Neutron
Proton
6. We know that atoms are mostly empty space. We are made up of atoms. Therefore, we are mostly empty space. If this is the case, why don’t we simply ooze through a chair when we sit on it? (2)
ELECTRONS SURROUND THE ATOMS AND MOLECULES THAT MAKE UP US AND THE CHAIR. THE ELECTRONS SURROUNDING OUR ATOMS AND MOLECULES REPEL THE ELECTRONS SURROUNDING THE ATOMS AND MOLECULES OF THE CHAIR. THIS REPULSIVE FORCE PREVENTS US FROM “OOZING” THROUGH THE CHAIR. 7. The atomic number of neon is 10. The atomic mass of one of its isotopes is 21.99. (When rounded off, the atomic mass is equal to the number of particles in the nucleus.) How many protons, neutrons, and electrons are there in this neon isotope? (3)
# of protons = 10 # of neutrons = 22 – 10 = 12 # of electrons = 10 8. The density of gold is 19.3 g/cm3 (19,300 kg/m3), and the density of silver is 10.5 g/cm3 (10,500 kg/m3).
a. Which weighs more, a cubic meter of gold or a cubic meter of silver? (2)
GOLD (19300 VS 10500)
b. Which one occupies a greater volume, one kilogram of gold or one kilogram of silver? (2)
SILVER
2 m 9. The block of wood shown has a mass of 1000 kilograms. Its dimensions are 1 m by 1 m by 2 m. 1 m A 1 m 3 C a. What is the density of the block (in kg/m )? (2) B
1000/(1x1x2) = 500 KG/M3
b. Rank from highest pressure to lowest pressure the face of the block that must be placed on the floor. (2)
SMALLER AREA -> HIGHER PRESSURE: C, A = B
c. What is the numerical value of the least pressure the block can exert on the floor (in N/m2)? (Remember that the pressure will equal the weight of the block divided by the area.) (2)
WEIGHT = 1000 x 10 = 10000 N, LARGEST AREA = 1 x 1 x 2 M3 = 5000 N/M2
10. According the Archimede’s Principle, an object that is floating or submerged in a liquid feels a buoyant force acting on it.
a. What is the buoyant force equal to? (definition) (2)
WEIGHT OF THE LIQUID DISPLACED BY THE OBJECT
b. If an object is floating or submerged in water, what is the cause of the buoyant force? (Don’t just say “the water.”) (2)
THE PRESSURE IS GREATER THE DEEPER YOU GO IN THE WATER. THE BOTTOM OF THE OBJECT FEELS A GREATER PRESSURE UPWARD THAN THE TOP SURFACE FEELS A PRESSURE DOWNWARD. THUS A NET PRESSURE UPWARD.
11. A block of wood weighs 31.2 lbs and its volume is 1 cubic foot (a weight density of 31.2 lbs/ft3). When floating in water (weight density = 62.4 lb/ft3), half of its volume is submerged.
a. What is the magnitude of the buoyant force acting on the wood block? (2)
BF = WEIGHT OF ½ FT3 OF WATER = 31.2 LBS
b. The wood block is now pushed down below the surface of the water. Now what is the magnitude of the buoyant force? (2)
BF = WEIGHT OOF 1 FT3 OF WATER = 62.4 LBS
12. A glass is partially filled with a clear liquid. An ice cube is placed in the liquid and the ice cube sinks to the bottom of the glass. We know that ice has a density of 919 kg/m3 and water has a density of 1000 kg/m3. liquid a. What do we know about the density of the liquid in the glass? (2)
LIQUID DENSITY IS LESS THAN THE DENSITY OF THE ICE.
ice cube b. When the ice cube was submerged in the liquid, the liquid level was marked on the side of the glass. After the ice melts will the level of the liquid be above or below the mark, or remain unchanged? Explain. (2)
BELOW. THE VOLUME THE ICE OCCUPIES IS GREATER THAN THE VOLUME THE WATER RESULTING FROM THE MELTED ICE OCCUPIES.
13. We know that air has weight. Devise a method whereby you can determine the weight of air. (2)
WEIGH A CONTAINER WITH AIR AND WITHOUT THE AIR. THE DIFFERENCE IS THE WEIGHT OF AIR.
14. When sucking soda up through a straw, is it more difficult at sea level or at the top of a high mountain? Explain. (2)
THE SODA IS PUSHED UP THE STRAW BY AIR PRESSURE. THE HIGHER THE AIR PRESSURE, THE EASIER IT IS FOR THE SODA TO MOVE UP THE STRAW. THUS, AT SEA LEVEL DRINKING THE SODA WILL BE EASIER. 15. In class I demonstrated a ping pong ball being held up in a column of air. The air hits the bottom of the ball exerting an upward force on it. But the air also prevented the ball from moving sideways out of the column of air. Explain why the ball did not move sideways and move out of the column of air. You can use the diagrams to the right to help explain what is happening. (3)
THE PRESSURE IS LOW IN A HIGH VELOCITY FLUID AND THE PRESSURE IS HIGH IN A LOW VELOCITY FLUID. WHEN THE BALL BEGINS TO MOVE OUT OF THE STREAM OF AIR TO THE RIGHT, THE FLUID IS MOVING FAST ON THE LEFT SIDE OF THE BALL AND THE FLUID IS MOVING SLOWLY ON THE RIGHT SIDE OF THE FLUID. SO, THE LEFT SIDE OF THE BALL FEELS A LOW PRESSURE AND THE RIGHT SIDE OF THE BALL FEELS A HIGH PRESSURE, THUS, A NET PRESSURE PUSHES THE BALL TO THE LEFT – BACK INTO THE FLUID.
16. A mass is oscillating on the end of a spring with a frequency of 0.5 Hertz.
a. How many complete oscillations does the mass go through in each second? (2)
0.5 OSCILLATIONS – HALF OF AN OSCILLATION
b. How much time does it take the mass to travel from the top of its oscillation to the bottom of its oscillation? (2)
THE PERIOD = 1/FREQUENCY = 1/0.5 = 2 SEC BOTTOM TO TOP IS ½ OF AN OSCILLATION, THEREFORE, 1 SEC
c. Increasing the mass that is oscillating on the end of the spring changes the period of oscillation. If the mass is increased, does the period increase or decrease? (Remember that increased mass means increased inertia.) (2)
PERIOD INCREASES – THE INCREASE IN MASS INCREASES THE INERTIA OF THE SYSTEM
17. In what type of wave do the particles of the medium oscillate parallel to direction of travel of the energy of the wave? (2) LONGITUDINAL WAVES 18. The transverse wave shown is traveling to the right with a velocity of 20 m/s. C
10 cm
10 cm T
a. Label the position of a crest with the letter C, and label the position of a trough with the letter T. (2)
b. How long is the wavelength of the wave? (2)
CREST TO CREST = 2 x 10 CM = 20 CM
c. What is the frequency of the wave? (2)
FREQUENCY = SPEED/WAVELENGTH = 20/0.20 (IN TERMS OF METERS AND METERS/SEC) = 2000/20 (IN TERMS OF CM AND CM/SEC) = 100 HZ
d. How large is the amplitude of the wave? (2)
AMPLITUDE = 5 CM
19. You are standing at the end of the Santa Monica pier and notice debris in the water.
a. Describe the motion of the debris. Is the water wave a transverse wave or a longitudinal wave? (3)
MOVES UP AND DOWN AND BACK AND FORTH – CIRCULAR TYPE – BOTH TRANSVERS AND LONGITUDINAL.
b. You also notice that 10 crests pass your position in 30 seconds, and that the speed of the water waves is 6 m/s. What is the wavelength of the waves? (3)
TIME FOR 1 WAVELENGTH TO PASS IS 3 SEC. TRAVELING AT 6 M/S, IN 3 SEC, THE CREST TRAVELS 18 M. THEREFORE, THE WAVELENGTH IS 18 M.