Tennis ball was carefully placed on top of solid rubber ball of same size?

Question

and the two balls were dropped from height 10m on concrete floor. During entire fall tennis ball remained on top of rubber ball.

Ignoring air resistance and other losses of energy, to what altitude will the tennis ball bounce?

Answer 1

Assume an infinitesimal separation between the balls when they hit the ground with velocity v0=sqrt(2gh) = 14.01/s. Then the bottom ball (m1) bounces with v1(i) = -v0 and immediately collides with the top ball (m2) whose precollision velocity v2(i) = v0. The relative precollision velocity vrel(i) = v2(i)-v1(i) = 2v0. The combined CM velocity vcm = m1v1+m2v2/(m1+m2) = -m1v0 + m2v0/m1+m2 = v0*(m2-m1)/(m1+m2).
With energy conservation they then bounce apart with relative velocity vrel(f) = -vrel(i) = -2v0. By momentum conservation the velocity of ball2 relative to the CM vrel(f,2) = vrel(f)*m1/(m1+m2). So m2's total velocity
v2(f)= vcm + vrel(2,f) =
v0*(m2-m1)/(m1+m2) - 2*m1/(m1+m2) =
v0*((m2-m1) - 2m1) / (m1+m2).
This means that as m2/m1 approaches infinity, v2(f) approaches 3v0. If you imagine an infinitesimal mass with vi = -1 colliding with a large mass with vi = 1, it rebounds with a relative velocity = 2 thus a total velocity = 3. The ref. contains a calculator to try out example collisions.