Question

A wedge of mass $M$ and a cube of mass $m$ are shown in figure. the system is released Considering on frictional force between any two surface, the distance moved by the wedge, when the cube just reaches on the ground is:

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

Answer 1

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

let distance moved by wedge be x and distance moved by block be (2-x)

since there is no external force , centre of mass remains constant

$Mx=m(2-x)$

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