Question

All the surface shown in figure are smooth. A block of mass $m=50\text{kg}$ is placed at the top of wedge of mass $M=200\text{kg}$ having angle of inclination of $\theta ={37}^{\circ }$ with horizontal. If block reaches the bottom of the wedge with relative velocity $v_r=20\text{m/s}$ with respect to wedge.The velocity of wedge is

• A. $\frac{16}{5}m/s$
• B. $\frac{8}{5}m/s$
• C. $\frac{16}{3}m/s$
• D. $\frac{1}{2}m/s$

Answer 1

The correct option is A $\frac{16}{5}m/s$


As there are no net external force in horizontal direction.
$\therefore$ Applying PCLM,
$p_i=p_f$
Let the velocity of wedge is $v$
Velocity of block in horizontal direction will be $(-v_r\cos \theta +v)$
$0=m(-v_r$ cos $\theta +v)+Mv$
$\Rightarrow v=\frac{m{v}_r\text{cos}\theta }{m+M}$
$=\frac{50\times 20\times \text{cos}{37}^{\circ }}{250}$
$=4\times \frac{4}{5}=\frac{16}{5}$ m/s