15.5: Worked Examples
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Example 15.1 Elastic One-Dimensional Collision Between Two Objects

Consider the elastic collision of two carts along a track; the incident cart 1 has mass m1 andmoveswithinitialspeed\(v1,i. The target cart has mass m2=2m1 and is initially at rest, v2,i=0, (Figure 15.7). Immediately after the collision, the incident cart has final speed v1,f and the target cart has final speed v2,f. Calculate the final x-component of the velocities of the carts as a function of the initial speed v1,i.
Solution: The momentum flow diagram for the objects before (initial state) and after (final state) the collision are shown in Figure 15.7. We can immediately use our results above with m2=2m1 and v2,i=0. The final x -component of velocity of cart 1 is given by Equation (15.3.14), where we use v1x,i=v1,i.
v1x,f=−13v1,i
Example 15.2 The Dissipation of Kinetic Energy in a Completely Inelastic Collision Between Two Objects

An incident cart of mass m1 and initial speed v1,i collides completely inelastically with a cart of mass m2 that is initially at rest (Figure 15.7b). There are no external forces acting on the objects in the direction of the collision. Find ΔK/Kinitial =(Kfinal −Kinitial )/Kinitial
Solution: In the absence of any net force on the system consisting of the two carts, the momentum after the collision will be the same as before the collision. After the collision, the carts will move in the direction of the initial velocity of the incident cart with a common speed vf found from applying the momentum condition
m1v1,i=(m1+m2)vf⇒vf=m1m1+m2v1,i
in agreement with Equation (15.4.4).
Example 15.3 Bouncing Superballs

Consider two balls that are dropped from a height hi above the ground, one on top of the other (Figure 15.8). Ball 1 is on top and has mass M1, and ball 2 is underneath and has mass M2 with M2>>M1. Assume that there is no loss of kinetic energy during all collisions. Ball 2 first collides with the ground and rebounds. Then, as ball 2 starts to move upward, it collides with the ball 1 which is still moving downwards (figure below left). How high will ball 1 rebound in the air? Hint: consider this collision as seen by an observer moving upward with the same speed as the ball 2 has after it collides with the ground. What speed does ball 1 have in this reference frame after it collides with the ball 2?
Solution
The system consists of two balls and the earth. There are five special states for this motion shown in the figure below.

Initial State: the balls are released from rest at a height hi above the ground.
State A: the balls just reach the ground with speed va=√2ghi. This follows from ΔEmесh=0⇒ΔK=−ΔU. Thus (1/2)mv2a−0=−mgΔh=mghi⇒va=√2ghi
State B: immediately before the collision of the balls. Ball 2 has collided with the ground and reversed direction with the same speed, va but ball 1 is still moving downward with speed va.
State C: immediately after the collision of the balls. Because we are assuming that m2≫m1 ball 2 does not change its speed as a result of the collision so it is still moving 2 upward with speed va. As a result of the collision, ball 1 moves upward with speed vb
Final State: ball 1 reaches a maximum height hf=v2b/2g above the ground. This again follows from ΔK=−ΔU⇒0−(1/2)mv2b=−mgΔh=−mghf⇒hf=v2b/2g
Choice of Reference Frame:
As indicated in the hint above, this collision is best analyzed from the reference frame of an observer moving upward with speed va the speed of ball 2 just after it rebounded with the ground. In this frame immediately, before the collision, ball 1 is moving downward with a speed v′b that is twice the speed seen by an observer at rest on the ground (lab reference frame). v′a=2va
The mass of ball 2 is much larger than the mass of ball 1, m2≫m1 This enables us to consider the collision (between States B and C) to be equivalent to ball 1 bouncing off a hard wall, while ball 2 experiences virtually no recoil. Hence ball 2 remains at rest in the reference frame moving upwards with speed Va with respect to observer at rest on ground. Before the collision, ball 1 has speed v′a=2va Since there is no loss of kinetic energy during the collision, the result of the collision is that ball 1 changes direction but maintains the same speed, v′b=2va