Question

A small block of a mass m = 1 kg is placed over a wedge of mass M = 4 kg as shown in the figure. Mass ‘m’ is released from rest. All surfaces are smooth. Origin ‘O’ is as shown in the figure.

The magnitude of velocity of the wedge when ‘m’ leaves the wedge is

• A. $\sqrt{3}m/s$
• B. $\sqrt{2}m/s$
• C. $\frac{1}{\sqrt{2}}m/s$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

The correct option is B $\sqrt{2}m/s$

At the instant ‘m’ leaves the surface of ‘M’, it is moving in horizontal direction. The momentum of the system is conserved along horizontal direction.
For system,
$(P_i{)}_x=(P_f{)}_x$
$\Rightarrow 0=-M{v}_2+m{v}_1$
$\Rightarrow v_2=\left(\frac{m}{M}\right)v_1=\frac{v_1}{4}....\left(1\right)$
Applying energy conservation;
Loss in GPE = Gain in K.E.
$\Rightarrow mgh=\frac{1}{2}m{v}_1^2+\frac{1}{2}M{v}_2^2\Rightarrow 1\times 10\times (4-2)=\frac{1}{2}[v_1^2+4v_2^2]......\left(2\right)$
From eqn (1) and (2), we get
$v_2=\sqrt{2}m/s$ and $v_1=4\sqrt{2}m/s$