Question

All the surfaces shown in the figure are assumed to be frictionless. The block of mass $m$ slides on the prism which in turn slides backwards on the horizontal surface. Find the acceleration of the smaller block with respect to the prism.

Answer 1

Normal reaction between prism and block be= $N$

Let acceleration of the prism w.r.t ground= $A$

Acceleration of block w.r.t prism be= $a_r$

From the figure, $Nsin\theta$ = mA------- (1)

$m{a}_r=mgSin\theta +mAcos\theta$------- (2)

$N=mgcos\theta -mAsin\theta$------- (3)

Solving (1), (2) and (3), we get

$a_r=gsin\theta \left[\frac{M+m}{M+m{sin}^2\theta }\right]$

Hence acceleration of small block w.r.t prism is = $gsin\theta \left[\frac{M+m}{M+m{sin}^2\theta }\right]$