Question

A block Q of mass $M$ is placed on a horizontal frictionless 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. $\frac{m}{M}\text{L cos}\theta$
• B. $\frac{m}{M+m}L$
• C. $\frac{M+m}{\text{mL cos}\theta }$
• D. $\frac{\text{mL cos}\theta }{m+M}$

Answer 1

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

Here, the position of centre of mass of the system remains unchanged along horizontal, when the mass $m$ moved a distance $\text{L cos}\theta$ let the mass $(m+M)$ moves a distance $x$ in the backward direction.
$\therefore (M+m)x-\text{mL cos}\theta =0$
$\therefore x=\frac{\left(\text{mL cos}\theta \right)}{(m+M)}$
Hence, the correct answer is option (d).