A smooth ball of mass $1kg$ is projected with velocity $7m/s$ horizontal from a tower of height $3.5m.$ It collides elastically with a wedge of mass $3kg$ and inclination of ${45}^0$ kept on ground (figure). The ball collides with the wedge at a height of $1m$ above the ground, the velocity of the wedge (in m/s) after collision. (neglect friction at any contact.)

 

$v_x=7m/s,$ velocity along y-axis of the ball just before collision

$v_y=\sqrt{2\times 9.8\times 2.5}=7m/s$ As $v_x=v_y,$ so it strikes the plane of incline perpendicularly

Let the ball rebound with velocity V and let $v_1$ be the velocity of the wedge

Applying the principle conservation of momentum in horizontal direction, we get, $1\times 7=-1\times \frac{v}{\sqrt{2}}+3v_1$

$7\sqrt{2}=3\sqrt{2}{v}_1-v\mathrm{............}\left(i\right)$

Applying the equation for coefficient of restitution, we get, $e=1=\frac{v_1/\sqrt{2}+v}{7\sqrt{2}},7\sqrt{2}=\frac{v_1}{\sqrt{2}}+v\mathrm{..............}\left(ii\right)$

Solving Eqs. (i) and (ii), $v_1=4m/s$ and $v=5\sqrt{2}m/s$