A ball of mass m = 1 kg falling vertically with a velocity $v_0$ = 2 m/s 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 =0.5 then the ratio of velocity of ball to the velocity of the wedge just after collision is __
 

Given M = 2kg, m = 1kg and $v_0$ = 2m/s

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 $J \sin 30° = M{v}_1 = m{v}_2$
or  $\frac{J}{2}$  = $2V_1 = V_2$                             ..... (1)
$J \cos 30° = m (v_3 + v_0)$
or  $\frac{\sqrt{3}}{2}$$J = (v_3 + 2)$                    .... (2)
Applying, relative speed of separation = e (relative speed of approach) in common normal direction, we get

$(v_1 + v_2) \sin 30° + v_3 \cos 30° = (v_0 \cos 30°) ; or v_1 + v_2 + \sqrt{3}{v}_3=\sqrt{3}$      .... (3)
Solving Eqs. (1), (2) and (3), we get  $v_1=\frac{1}{\sqrt{3}}m/s$  ;       $v_2=\frac{2}{\sqrt{3}}m/s$