Two spherical bodies of mass `m_(1)` and `m_(2)` fall freely through a distance `h`. before the body `m_(2)` collides with the ground. If the coefficient of restitution of all collisions is `e`, find the velocity of `m_(1)` just after it collides with `m_(2)`

Explanation

As the bodies `m_(1)` and `m_(2)` fall freely through a distance `h`, `m_(2)` collides with ground with velocity `v_(0)=sqrt(2gh)`. Hence `m_(2)` bounces up with a velocity `ev_(0)` just after the collision with the ground.
Hence `m_(2)` bounces up with a velocity `ev_(0)` just after the collision with the ground. then it collides with the freely falling body `m_(1)`. Just after the collision, let us assume that `v_(1)` and `v_(2)` are the velocities of `m_(1)` and `m_(2)` respectively.
ignoring the impulses due to the weights of the bodies during the period of collision. Let us conserve the momentum of the system `(m_(1)+m_(2))`.
Conservation of linear momentum
`P=-m_(1)v_(0)+m_(2)ev_(0)=m_(1)v_(1)+m_(2)v_(2)`.......i

Newton's impact formula `-e[u_(1)-u_(2)]=v_1-v_(2)`
where `u_(1)=-v_(0 )` and `u_(2) =ev_(0)`
Then `-e[(-v_(0))-(ev_(0))]=v_(1)-v_(2)`.............ii
Solving eqn i and ii we have
`v_(1)=(-(m_(1)-em_(2)))/(m_(1)+m_(2))v_(0)+((1+e)m_(2))/(m_(1)+m_(2)) ev_(0)`
Putting `m_(1)=m_(2), e=1/2` and `v_(0)=sqrt(2gh)` we have `v_(1)=sqrt((gh)/32)uarr`