# Ball collision (kleppner 4.23)

1. Jan 22, 2015

### geoffrey159

1. The problem statement, all variables and given/known data

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.)

2. Relevant equations

Collisions

3. 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 ?

2. Jan 22, 2015

### Nathanael

Yes, I believe you are correct, well done. (It rises 9h plus the diameter of the superball above the ground.)

3. Jan 22, 2015

### geoffrey159

oh yes you're right, I forgot the diameter, thank you !