Velocity after Elastic Collision!

Both spheres are made of steel, so they have the same density. This implies that if the mass of the smaller sphere is $m$ then the mass of the larger sphere is $8m$ .

Both of the spheres fall the same distance, $h-2a$ acquiring a speed of

$V_i=\sqrt{2g(h-2a)}$ .

The larger sphere collides elastically with the floor so it's momentum is reversed and it is now traveling upwards at a speed of $V_i$ .

(Of course, this can only happen in the limit as the Earth's mass goes to infinity, that way the Earth can conserve the momentum of the larger sphere without gaining any energy from it. But this is a very good approximation.)

So now the smaller sphere is traveling downwards with momentum $mV_i$ and the larger sphere is traveling upwards with momentum $8mV_i$ . Let's call the final velocity (after the collision) of the small and larger spheres $V_{fs}$ and $V_{fL}$ respectively. (Also suppose both end the collision traveling upwards.) We then get the following equation for conservation of momentum:

$8mV_i-mV_i=8mV_{fL}+mV_{fs}$

The problem says the collision is elastic so we can also write an equation that says the kinetic energy will be the same before and after the collision. (We neglect the change in height during the collision by assuming it lasts a negligible amount of time.)

$0.5(8mV_i^2+mV_i^2)=0.5(8mV_{fL}^2+mV_{fs}^2)$

Solving these two equations yields $V_{fs}=\frac{23}{9}V_i$