Ball collision (kleppner 4.23)

In summary, when a small ball of mass m is placed on top of a "superball" of mass M and both balls are dropped from height h, the small ball rises to a height of approximately 9h plus the diameter of the superball after the collision, assuming elastic collisions and m<<M. This is due to conservation of energy and momentum.
  • #1
geoffrey159
535
72

Homework Statement



A small ball of mass m is placed on top of a “superball” of mass M, and the two balls are dropped to the floor from height h. How high does the small ball rise after the collision? Assume that collisions with the superball are elastic, and that m<<M. To help visualize the problem, assume that the balls are slightly separated when the superball hits the floor. (If you are surprised by the result, try demonstrating the problem with a marble and a superball.)

Homework Equations



Collisions

The Attempt at a Solution



When the superball touches ground, both balls have speed ##v_i = \sqrt{2gh} ## by conservation of energy.
The collision of the superball with the ground is elastic, kinetic energy is conserved, so it rebounds with the same speed and then collides with the small ball, according to the assumption made in the text.
Again, the collision is elastic, so momentum is conserved as well as kinetic energy, so we must find ##v_m## and ##v_M## such that:

##\left\{
\begin{array}{}
(M-m)v_i = Mv_M + mv_m \\
(M+m) v_i^2= M v_M^2 + m v_m^2
\end{array}
\right.
##

I find that the speed of the small ball after collision is ## v_m = \frac{3M-m}{M+m}v_i ##, and the speed of the superball is ##v_M = \frac{M-3m}{M+m}v_i ##.

Now using conservation of mechanical energy, the highest height h' attained by the small ball is :
## \frac{1}{2}mv_m^2 = mg h' \Rightarrow h' = \frac{1}{2g}
(\frac{3M-m}{M+m})^2v_i^2 =(\frac{3M-m}{M+m})^2 h \approx 9h ##

Can it be that big ?
 
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  • #2
Yes, I believe you are correct, well done. (It rises 9h plus the diameter of the superball above the ground.)
 
  • #3
oh yes you're right, I forgot the diameter, thank you !
 

1. What is Ball Collision?

Ball Collision refers to the physical phenomenon in which two or more balls come into contact and exchange energy. This can occur in various contexts, such as in sports, games, or scientific experiments.

2. How is Ball Collision described in Kleppner 4.23?

In Kleppner 4.23, Ball Collision is described as an inelastic collision between two balls of different masses and velocities. The collision results in a transfer of momentum and energy between the balls, causing them to move in new directions and speeds.

3. What is the difference between elastic and inelastic Ball Collision?

Elastic Ball Collision is a type of collision in which the total kinetic energy of the system is conserved. This means that the two balls bounce off each other without any loss of energy. In contrast, inelastic Ball Collision involves a loss of kinetic energy due to deformation or heat generation during the collision.

4. How is the conservation of momentum applied in Ball Collision?

In Ball Collision, the total momentum of the system (the sum of the individual momentums of each ball) is conserved. This means that before and after the collision, the total momentum remains the same. This principle is used to calculate the velocities of the balls after the collision.

5. Can Ball Collision be used to study real-life situations?

Yes, Ball Collision can be applied to real-life situations, such as analyzing the movement of billiard balls on a pool table or understanding the impact of collisions in sports like soccer or basketball. It is also an important concept in physics and engineering, used to study the motion and energy transfer in various systems.

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