Question

A small particle of mass m = 2 kg moving with constant horizontal velocity u = 10 m/s strikes a wedge shaped block of mass M = 4 kg on its inclined surface as shown in figure. After collision particle starts moving up the inclined plane. Calculate the velocity of wedge immediately after collision. ___

Answer 1

2Let $V_{rel}$ be the final velocity of the ball w.r.t. wedge and V be the final velocity of the wedge w.r.t. ground. Now, velocity of ball w.r.t. ground



Horizontal component = $V_n=V_{rel}.cos\alpha +V$
Vertical component =$V_{\gamma }=V_{rel}.sin\alpha$
COM in horizontal direction gives
$mu=m(V_{rel}cos\alpha +V)+MV$ ....(i)
Since velocity of ball along wedge remains constant
$\therefore ucos\alpha =V_{rel}+Vcos\alpha$ ...(ii)
Solving (i) $\&$ (ii) we get
$V=\frac{musi{n}^2\alpha }{M+msi{n}^2\alpha }=2m/s$