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

Answer 1

Answer: (D)

Here, the centre of mass of the system remains unchanged when the mass $m$ moved a distance $L\cos \theta$, let the mass $(m+M)$ moves a distance $x$ in the backward direction.
$\therefore (M+m)x-mL\cos \theta =0$
$\therefore x=(mL\cos \theta )/(m+M)$