Question

A block $Q$ of mass $M$ is placed on a horizontal frctionless surface $AB$ and a body $P$ of mass $m$ is released on its frictionless slope. As $P$ slides by a length $L$ on this slope of inclination $\theta$, the block $Q$ would slide by a distance

• A. $\frac{m}{M}L\cos \theta$
• B. $\frac{m}{M+m}L$
• C. $\frac{M+m}{mL\cos \theta }$
• D. $\frac{mL\cos \theta }{m+M}$

Answer 1

The correct option is D $\frac{mL\cos \theta }{m+M}$

As there is no external force along X direction. Here, the centre of mass of the system remains unchanged in X direction.
When the mass $m$ moves a distance $L\cos \theta$ on $M$, let the mass $(m+M)$ moves a distance $x$ in the backward direction.
$\therefore (M+m)x-mL\cos \theta =0$
$\therefore x=\left(mL\cos \theta \right)/(m+M)$
Hence, the correct answer is option (d).