A ball of mass `m` is pushed with a horizontal velocity `v_(0)` from one end of a sledge of mass `M` and length `l`. if the ball stops after is first collision with the sledge, find the speeds of the ball ad sledge after the second collision of the ball with the sledge.

Explanation

a. Using C.O.L.M, `Mv_(2)=mv_(0)` ………..i
Using Newton's restitution equation

`v_(2)-0=ev_(0)`
Here `m/Mv_(0)=ev_(0)`
which gives `e=m/M` ……….ii
Now the 2nd collisiion left wall the sledge and ball C.O.L.M.
`0+Mv_(2)=mv_(1)+MV_(2)^(')`
`M(m/Mv_(0))=mv_(1)+Mv_(2)^(')`
`implies mv_(1)+Mv_(2)^(')=mv_(0)`

Again using Newton's restitution law
`v_(2)^(')-v_(1)=e(0-m/Mv_(0))`
`v_(2)^(')-v_(1)=m/M(-m/Mv_(0))`
`v_(1)-v_(2)^(')=(m/M)v_(0)`..........iv
From eqn iii and iv we get
`v_(1)=(2v_(0))/((M+m))` and `v_(2)^(')=(mv_(0))/M((M-m)/(M+m))`