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. $\frac{m L \cos \theta}{m+M}$

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.
$\therefore (M+m) x-m L \cos \theta=0$
$\therefore x =( mL \cos \theta) /( m + M )$