Physics 12 Lesson Notes on Momentum. Momentum is a commonly used term in sports. A team that has the momentum is on the move and is going to take some effort to stop. A team that has a lot of momentum is really on the move and is going to be hard to stop. Momentum is a physics term; it refers to the vector quantity of motion that an object has and defined as the product of mass and velocity.
P = m ∙ v
Check Your Understanding 1. Determine the momentum of a. 60-kg halfback moving eastward at 9 m/s. ______b. 1000-kg car moving northward at 20 m/s. ______c. 40-kg freshman moving southward at 2 m/s. ______
2. A car possesses 20 000 units of momentum. What would be the car's new momentum if ... a. its velocity were doubled. ______b. its velocity were tripled. ______c. its mass were doubled (by adding more passengers and a greater load)
d. both its velocity were doubled and its mass were doubled.
3. A halfback (m = 60 kg), a tight end (m = 90 kg), and a lineman (m = 120 kg) are running down the football field. Consider their ticker tape patterns below.
Compare the velocities of these three players. How many times greater are the velocity of the halfback and the velocity of the tight end than the velocity of the lineman?
Which player has the greatest momentum? Explain.
Momentum and Impulse Connection As mentioned in the previous part of this lesson, the term momentum is a physics concept. Any object with momentum is going to be hard to stop. To stop such an object, it is necessary to apply a force against its motion for a given period of time. The more momentum which an object has, the harder that it is to stop. Thus, it would require a greater amount of force or a longer amount of time or both to bring such an object to a halt. As the force acts upon the object for a given amount of time, the object's velocity is changed; and hence, the object's momentum is changed. A force acting for a given amount of time will change an object's momentum. Put another way, an unbalanced force always accelerates an object - either speeding it up or slowing it down. If the force acts opposite the object's motion, it slows the object down. If a force acts in the same direction as the object's motion, then the force speeds the object up. Either way, a force will change the velocity of an object. And if the velocity of the object is changed, then the momentum of the object is changed. These concepts are merely an outgrowth of Newton's second law as discussed in an earlier unit. Newton's second law (Fnet = m • a) stated that the acceleration of an object is directly proportional to the net force acting upon the object and inversely proportional to the mass of the object. When combined with the definition of acceleration (a = change in velocity / time), the following equalities result.
This equation represents one of two primary principles to be used in the analysis of collisions during this unit. To truly understand the equation, it is important to understand its meaning in words. In words, it could be said that the force times the time equals the mass times the change in velocity. In physics, the quantity Force • time is known as impulse. And since the quantity m•v is the momentum, the quantity m•Δv must be the change in momentum. The equation really says that the
Impulse = Change in momentum
This can be expressed this way: In a collision, an object experiences a force for a specific amount of time which results in a change in momentum. The result of the force acting for the given amount of time is that the object's mass either speeds up or slows down (or changes direction). The impulse experienced by the object equals the change in momentum of the object. In equation form,
Example 1) A n average net force of 17.0 N acts east on an object for 2.5 x 10–2 s. What is the impulse? Example 2) A 26.3 kg object is traveling at 21.0 m/s north. What average net force is required to bring the object to a stop in 2.60 s?
Example 3) An average net force of 31.6 N south is used to accelerate a 15.0 kg object uniformly from rest to 10.0 m/s. What is the time of acceleration?
Example 4) A 3.0 kg is dropped from a height of 6.5 m. How far did the object fall when its momentum is 6.0 kg·m/s down?
Example 5) A rocket with the mass of 9.5 x 103 kg is acted on by an average net force of 1.5 x 105 N up for 15 s, What is the final velocity of the rocket?
Example 6) Without finding the acceleration, calculate the time an average net force of 225 N must act on a 1.0 x 103 kg object to change its velocity from 2.0 m/s east to 5.0 m/s east. Momentum in Two Dimensions 1. A hockey player of mass 82 kg is traveling north with a velocity of 4.1 m/s. He collides with a 76 kg player traveling east at 3.4 m/s. If the two players "lock together momentarily, in what direction will they be going immediately after the collision? How fast will they be moving? 2.7 m/s, 37.5° E of N
2. A 2.5 x 103 kg car traveling west at 6.0 m/s is hit by a 6.0 x 103 kg truck going south at 4.0 m/s. The two vehicles lock together on impact. What is the speed and direction of the wreckage immediately after impact? 3.3 m/s, 58° S of W
3. A frustrated physics student blew up her Physics textbook using a small amount of explosive. It broke into three pieces, which miraculously flew off in directions that were all in the same plane. A 0.200 kg piece flew off at 20.0 m/s and a 0.100 kg piece went off at 90° to the first piece, at 30.0 m/s. a) What was the momentum of the third piece? 5.0 kg m/s
b) If the mass of the third piece was 0.150 kg, what was its velocity right after the explosion? 33 m/s, 37° to the direction of the first mass
4. A proton of mass M collides obliquely with another proton. The first proton is moving with a speed of 6.0 x 106 m/s before it hits the second, stationary proton. Assuming the collision is perfectly elastic (no kinetic energy is lost), and using the fact that the first proton is moved 30° from its initial path after the collision, figure out the speed and direction of each proton after the collision. Note that after the collision the two protons travel at 90˚ to each other. Struck proton 3.0 x 106 m/s, 60° from the incident proton’s original direction; incident ball 5.2 x 106 m/s
5. A 0.40 kg model airplane is traveling 20 km/h toward the south. A 0.50 kg model airplane traveling 25 km/h in a direction 20° east of south collides with the first model airplane. The two planes stick together on impact. What is the direction and magnitude of the velocity of the combined wreckage immediately after the collision? 22 km/h, 12° E of S
6. What impulse is needed to change the velocity of a 90.0 kg football player from 3.6 m/s toward the north and make it 1.2 m/s toward the northeast? In what direction must the force be exerted? ∆p = 259 kg m/s, the force is exerted at 17° E of S. 7) A 30.0 kg object moving to the right at a velocity of 1.00 m/s collides with a 20.0 kg object moving to the left at a velocity of 5.00 m/s. If the 20.0 kg object continues to move left at a velocity of 1.25 m/s, what is the velocity of the 30.0 kg object?
8) A 225 g ball moves with a velocity of 30.0 cm/s to the right. This ball collides with a 125 g ball moving in the same direction at a velocity of 10.0 cm/s. After the collision the velocity of the 125 g ball is 24.0 cm/s to the right. What is the velocity of the 225 g ball after the collision?
9) A 10.0 g object is moving with a velocity of 20.0 cm/s to the right when it collides with a stationary 20.0 g object. After collision the 10.0g object is moving left at a velocity of 12.0 cm/s. What is the velocity of the 20.0 g object after the collision?
Before collision: total momentum = (10 g)(20 cm/s) = 200 g cm/s [right]
After collision : total momentum = 200 g cm/s [right]
+ 200 = (10 ) (–12 ) + (20 )(v) ---- 200 = 20 v – 120 ; 20 v = 320 ; v = 16 cm/s right 10) A 1.4 x 103 kg car is westbound at a velocity of 37.0 km/h when it collides with a 2.0 x 103 kg truck northbound at a velocity of 35.0 km/h. If these two vehicles lock together upon collision. What is the initial velocity of the vehicles after the collision? Before collision: total momentum =
11) A 1.20 kg red ball moving to the right at 17.1m/s strikes a stationary 2.31 kg blue ball. If the final velocity of the red ball is 13.5m/s at 23.0above the horizontal, determine the final velocity of the blue ball.
12) A 1.20 kg red ball moving at 10.0 m/s strikes a 2.31 kg blue ball moving at 15.0 m/s. If the final velocity of the red ball is 13.5 m/s, determine the final velocity of the blue ball.