Processing math: 100%
Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Physics LibreTexts

15.5: Worked Examples

( \newcommand{\kernel}{\mathrm{null}\,}\)

Example 15.1 Elastic One-Dimensional Collision Between Two Objects

clipboard_effa9af20de1f4daa3d634497913a457f.png
Figure 15.7 Elastic collision between two non-identical carts

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

The final x-component of velocity of cart 2 is given by Equation (15.4.17) v2x,f=23v1,i

Example 15.2 The Dissipation of Kinetic Energy in a Completely Inelastic Collision Between Two Objects

clipboard_ea5cfebd74d5d365992d52c0409ee5448.png
Figure 15.7b Inelastic collision between two non-identical carts

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)vfvf=m1m1+m2v1,i

The initial relative speed is vreli=v1,i. The final relative velocity is zero because the carts stick together so using Equation (15.3.26), the change in kinetic energy is ΔK=12μ(vreli)2=12m1m2m1+m2v21,i
The ratio of the change in kinetic energy to the initial kinetic energy is then ΔK/Kinitial=m2m1+m2
As a check, we can calculate the change in kinetic energy via ΔK=(KfKi)=12(m1+m2)v2f12v21,i=12(m1+m2)(m1m1+m2)2v21,i12v21,i=(m1m1+m21)(12m1v21,i)=12m1m2m1+m2v21,i

in agreement with Equation (15.4.4).

Example 15.3 Bouncing Superballs

clipboard_e49700c848071b8adf577d1cad329673a.png
Figure 15.8b Two superballs dropping

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.

clipboard_ef4a3fb93e1c1b5e02e0d0515af39882f.png
part a)

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=0ΔK=ΔU. Thus (1/2)mv2a0=mgΔh=mghiva=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 m2m1 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=ΔU0(1/2)mv2b=mgΔh=mghfhf=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 vb that is twice the speed seen by an observer at rest on the ground (lab reference frame). va=2va

The mass of ball 2 is much larger than the mass of ball 1, m2m1 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 va=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, vb=2va

However, according to an observer at rest on the ground, after the collision ball 1 is moving upwards with speed vb=2va+va=3va
While rebounding, the mechanical energy of the smaller superball is constant (we consider the smaller superball and the Earth as a system) hence between State C and the Final State, ΔK+ΔU=0
The change in kinetic energy is ΔK=12m1(3va)2
The change in potential energy is ΔU=m1ghf
So the condition that mechanical energy is constant (Equation (15.5.10)) is now 12m1(3v1a)2+m1ghf=0
We can rewrite Equation (15.5.13) as m1ghf=912m1(va)2
Recall that we can also use the fact that the mechanical energy doesn’t change between the Initial State and State A yielding an equation similar to Equation (15.5.14), m1ghi=12m1(va)2
Now substitute the expression for the kinetic energy in Equation (15.5.15) into Equation (15.5.14) yielding m1ghf=9m1ghi
Thus ball 1 reaches a maximum height hf=9hi


This page titled 15.5: Worked Examples is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Peter Dourmashkin (MIT OpenCourseWare) via source content that was edited to the style and standards of the LibreTexts platform.

Support Center

How can we help?