sorry philip now edited please see the qn
A ball of mass M_{1}=1kg falling vertically with a velocity v_{0}=2m/s;strikes a wedge of mass M=2kg kept on a smooth,horizontal surface as shown in figure.The coefficient of restitution between the ball and the wedgel is e=1/2; Find the velocity of the wedge and the ball immediately after collision.

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5 Answers
The coefficient of restitution between the ball and the ball is e=1/2
[3]
This is a sort of standard question...
u got at max 3 variables...
v_{x} of wedge and velocity of ball along normal direction...
and also there are 2 eqns..
1. momentum of system is conserved in horizontal direction..
2. coefficient of restitution...
I will post a complete soln...
velocities shown with red are vel after collision and blue earlier the collision...
initial momentum along horizontal direction of the system is zero
therefore...
Mv_{x}m((2sinÎ¸)cosÎ¸))m(vsinÎ¸)=0
now velocity of separation = e(velocity of approach).. of the point of contact..
therefore..
v+v_{x}sinÎ¸=e(2cosÎ¸)
put the values of Î¸,m,M and e... (M is mass of wedge and m is of ball..)
putting the values..
v=v_{x}=1/√3
so velocity of block 1i+1/√3 j where i is down the incline and j is perpendicularly upward from the incline...
there can be mistake in putting values... [3]