Physics Elastic Collision?

Question

A brick (mass M) is fropped from height h onto a ball (mass m). The mass of M is much much greater than the mass of m. How high will the ball bounce after the collision in terms of h? The collision is perfectly elastic.


I would assume that if the collison is elastic, once the brick hits the ball, they would both bounce up together but that is only a guess. The answer is not 0. It is supposed to be some fraction of the original height. Thanks a lot for your help.

Thanks a lot! I was wondering if you could explain your steps though because I am new at this. Where did you get the equation: average velocity = V/2? And how about the equation: V*x/L where x is thheight of the particle. Finally, how did you go from V*x/L to 1/4? Thanks a lot!

Answer 1

We know the brick bounces back to height h. If we think of the ball as a linear spring possessing (small) mass, we note that at the moment the brick separates from the spring at rebound with velocity V, the spring has an average velocity = V/2, since each particle of it moves with velocity V*x/L where x is the height of the particle and L is the current length of the spring. (Fuller explanation: While the spring rebounds, the bottom end of the spring stays on the ground with V=0. The spring expands uniformly so the velocity at any point on the spring = velocity of the top end of the spring * height of that point relative to L. The velocity of the CM of the spring, halfway up, is V/2.) So with 1/2 the velocity of the brick, the spring (or ball) has 1/4 * the specific kinetic energy and thus bounces to a height with 1/4 the specific potential energy, h/4.