A ball of mass `m` is distributed from the top of a fixed smooth circular tube in a vertical plane and falls impinging on a ball of mass `2 m` at the bottom. The coefficient of restitution is `1/2`. Find the heights to which the balls rise after a second impact.

Explanation

The situation is shown in the figure. If `u` be the velocity of ball `A` just before the impact. Then

`u=sqrt(2g(2a))`
consider the first impact between `A` and `B` if velocities of the balls become `v_(1)` and `v_(2)`. respectively, according to linear momentum conservation.
`mu+0=mv_(1)+2mv_(2)` ..........i
coefficient of restitution is given as
`e=(v_(2)-v_(1))/u=1/2`........ii
on solving eqn i and ii we get
`v_(1)=0` and `v_(2)=u/2`
ball `A` comes to rest, second impact between `A` and `B` occurs, when ball `B` returns to its initial position owht the sam speed `u/2`. if after the second impact the velocities of the two balls becomes `v_(3)` and `v_(4)` again from linear momentum conservation and coefficient of restitution.
`2m(u/2)+0=mv_(3)+2mv_(4)`..........iii
`e=(v_(3)-v_(4))/(u/2)=1/2`.........iv
on solving eqn iii and iv we get
`v_(3)=u/2` and `v_(4)=u/4`
both the velocities are positive, it implies that both masses will move in the same direction after the second impact. if `h_(1)` and `h_(2)` are the heights to which the masses `m` and `2m` will rise after the second impact according to energy conservation we have for ball of mass `m`,
`1/2m(u/2)^(2)=mgh_(1)` or `h_(1)=u^(2)/(8g)=a/2`
for ball of mass `2m`
`1/2(2m)(u/4)^(2)=2mgh_(2)` or `h_(2)=u^(2)/(32g)=a/8`