hard p[hysics question, ellastic collison and conservation of energy?

Question

A small ball of mass m is aligned above a larger ball of mass M = 0.77 kg (with a slight separation, as with the baseball and basketball), and the two are dropped simultaneously from height h = 2.1 m. (Assume the radius of each ball is negligible relative to h.)



(a) If the larger ball rebounds elastically from the floor and then the small ball rebounds elastically from the larger ball, what value of m results in the larger ball stopping when it collides with the small ball?
kg
(b) What is the maximum height the small ball then reaches?
m

Answer 1

This is a famous physics challenge test problem. Here's what you did not take into account.
Picture the bottom ball rebounding, and ignore the smaller ball for a moment. You can calculate the energy it bounces back with by conservation of energy. Set that equal to kinetic, and calculate the speed. (Of course, that was the speed it had when it hit the ground in the first place.) Now (here's what people don't think of), the speed would be the speed of the ball's centre of mass. While rebounding, the bottom is touching the ground, at speed zero. If the centre is travelling up at speed v at that last instant when the ball is about to leave the ground, then the top surface of the ball is travelling up at 2v. (That's becaue the ball is expanding uniformly back to its shape after being squished on the ground.)
So the little ball, which was going at speed v when it hits, starts back up at (approximately) speed 2v. (To be a little more accurate, you consider that the upper ball has a collision, not with a stationary floor, but with an object coming at it at 2v, which is the speed of the surface it hits. If the upper ball is much less massive than the lower one, the 2v works OK.) So it goes up much higher than the height is was dropped from. (Height goes as v-squared, and the upper ball is starting up faster.)
So you have to put this little extra bit (the double upward speed) of the top ball into the kinetic energy expression to solve for the initial speeds of each ball on the way up. You can then use the speeds to get their individual final heights.

Answer 2

I am not sure i am doing this right.
Correct me if i am wrong
v^2 = u^2 + 2gs
this gives the velocity of the two balls when they reach the bottom as sqrt(2*9.8*2.1)
As the larger ball rebounds and touches the smaller one, apply law of conservation of momentum (momemta are in opposite directions initially,
(0.77 -m)*sqrt(2*9.8*2.1) = m * sqrt(2*9.8*2.1). Here i have assumed that since elastic collision is involved the smaller ball rebounds with same velocity. Not really sure of this step.
This gives m = 0.77/2 = 0.385 kg. And according to my assumption the smaller ball will rebound to the same height.