Question

A particle of mass $m$ is placed on the smooth face of an inclined plane of mass $M$ and slope $\theta$ which is free to slide on a smooth horizontal plane in a direction perpendicular to the edge. If the particle slides with an accelerating $a$, the acceleration of the inclined plane towards right $A $ will be

• A. $\frac{m a \cos \theta}{(M+m)}$
• B. $\frac{M m \cos \theta}{(M+m)}$
• C. $\frac{M a \cos \theta}{(M+m)}$
• D. $\frac{M \cos \theta}{(M+m)}$

Answer 1

Answer: (A)

As the particle moves down the inclined plane, let $a$ be its acceleration. Now the inclined plane moves horizontally towards right with acceleration $A$.

Considering the vertical component of forces acting on the particle.
$m g-R \cos \theta=m a \sin \theta$ ...(i)
For horizontal components the motion of the particle is considered at the accelerated frame. So a horizontal fictious force acts opposite to $A$. This pseudo force $m A$ acts towards the left.

$R \sin \theta+m A=m a \cos \theta$
$R \sin \theta=m(a \cos \theta-A)$ ...(ii)
Now consider the force on inclined plane.
The forces are shown in figure.
$R \sin \theta=M A$ ...(iii)
$R \cos \theta+M g=S$ .....(iv)
From equations (ii) and (iii), we get
$A=\frac{m a \cos \theta}{(M+ m)}$