A ball of mass m = 1 kg falling vertically with a velocity v0 = 2 ms-1 strikes a wedge of mass M = 2 kg kept on a smooth, horizontal surface as shown in figure. The coefficient of restitution between the ball and the wedge is e=12. The velocity of the wedge and the ball immediately after collision is α3ms−1.Calculate α.

Let v be velocity of the ball after collision along the normal, v' be velocity of the ball after collision along incline and J be impulse on ball, thenJ=v−−2cos⁡30∘=v+3Impulse on wedge isJsin30∘=Mv1=2v1⇒v+32=2v1⇒ v=4v1−3    ........(1)Coefficient of restitution is defined ase=−v2−v1u2−u1⇒ 12=v+v122cos⁡30∘⇒ v=32−v12    .......(2)Solving (1) and (2), we getv1=13ms−1 and v=13ms−1For the ball, velocity along incline remains constant, so  v′=2sin⁡30∘=1ms−1Hence final velocity of ball isv2=12+132=23ms−1⇒α=2