Question

A wedge of mas M and cube of mass m are shown in figure. The system is released. Considering no friction moved by the wedge, then the C.O.M when the cube just reaches on the ground is :

• A. $\frac{m2\sqrt{2}}{(m+M)}$
• B. $2m$
• C. $2\sqrt{2}m$
• D. $\frac{2m}{m+M}$

Answer 1

The correct option is D $\frac{2m}{m+M}$

Let us assume that the C.O.M. of the wedge from the origin is $x$ distance apart and $x_1$ be C.O.M. with origin

Since no external force is acting on the horizontal direction. Then,

$\frac{\left(M{x}_1+ml\frac{1}{\sqrt{2}}\right)}{m+M}=\frac{\left(M\left(x_1+x\right)+mx\right)}{m+M}$

$\Rightarrow x=\frac{ml\cos \theta }{m+M}$

$\sin {45}^{\circ }=\frac{2}{l}=2\sqrt{2}$

By substituting all the values in above equation we get the

$x=\frac{2m}{m+M}$