Part C) is an entire new question and that is why there are two answers here to the problem.

Taking the positive axis downwards, the distance that the small ball travels is

`s = (v-eu)*t +(g*t^2)/2`

Also this distance for the bucket that has a speed `v` after collision is

`s =v*t`

By equating the above two expressions one gets

`(v-eu)*t +(g*t^2)/2 =v*t`

`(g*t^2)/2 -eu*t =0`

`t*(g*t/2 -e*u) =0`

This equation has two solutions

`t=0` initial time when the ball bounces first from the bucket

`t = (2eu)/g` time when the ball bounces second from the bucket.

A)

The figure is below. Initially the bucket and the counterweight are stationary. After the small ball bounces back, it will have the speed -`e*u` relative to the bucked which will move with the speed `v` . Thus the total speed of the ball after first bounce will be `v-eu`.

In all collisions the linear momentum is the same before and after collision. Initially only the ball has a momentum. Finally both the ball (moving with speed `v-eu`) and the system of bucket plus counterweight (which moves with speed `v`) have momentum.

`m*u = m*(v-eu) +2M*v`

`m*u(1+e) =mv +2Mv`

`v = [m*(1+e)*u]/(m+2M)`

B)

The force in the string is (as momentum theorem says):

`F = (Delta(P))/(Delta(t)) =(M*v)/(Delta(t))`

Thus the momentum in string is just

`P = F*Delta(t) =M*v = [Mm*(1+e)]/(m+ 2M) *u`

**Further Reading**