A ball of mass m=1kg falling vertically with a velocity `v_(0)=2m//s` strikes a wedge of mass M=2kg kept o a smooth, horizontal surface as shown in figure. The coefficient of resitution between the ball and the wege is `e=1//2`. Find the velocity of the wedge is e=1/2. find the velocity of the wedge and the ball immediately after collison.

Explanation

Given M = 2kg and m = 1kg

Let, J be the impulse between ball and wedge during collision and `v_(1), v_(2)` and `v_(3)` be the components of velocity of the wedge and the ball in horizontal and vertical directions respectively.
Applying impulse = change in momentum
we get `J sin 30^(@) = Mv_(1) = mv_(2)`
or `J/2=2v_(1)=v_(2)`
`Jcos30^(@)=m(v_(3)+v_(0))` ....(i)
or `(sqrt3)/2J=(v_(3)+2)` ....(ii)
Applying, relative speed of separation = e
(relative speed of approach) in common normal direction, we get
`(v_(1)+v_(2))sin30^(@)+v_(3)cos30^(@)=1/2(v_(0)cos30^(@))`
or `v_(1)+v_(2)+sqrt3v_(3)=(sqrt3)/2` .....(iii)
Solving Eqs. (i), (ii) and (iii), we get

`v_(1)=1/(sqrt3)`m/s
`v_(2)=2/(sqrt3)` m/s and `v_(3)=0`
Thus, velocities of wedge
and ball are `v_(1)=1/(sqrt3)` m/s

and `v_(2)=2/(sqrt3)` m/s in horizontal direction as shown in figure.