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. $\frac{ma\cos \theta }{(M+m)}$
• B. $\frac{Mm\cos \theta }{(M+m)}$
• C. $\frac{Ma\cos \theta }{(M+m)}$
• 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.
$mg-R\cos \theta =ma\sin \theta$ ...(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 $mA$ acts towards the left.

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