A block of mass 'm' is initially lying on a wedge of mass 'M' with an angle of inclination θ, as shown in the figure. The displacement of the wedge when the block is released and reaches to the bottom of the wedge is

The correct option is A

As the block moves down the inclined plain, wedge moves towards left

with respect to the wedge, block's horizontal displacement is equal to the base of the wedge . i.e., hcot\theta

Let 'x' be the displacement of the wedge. Then (hcot\theta - x) is the displacement if the block w.r.t ground

$\therefore$ As $△x_{cm}=\frac{m_1△x_1+m_2△x_2}{m_1+m_2}$

0 = $\frac{m(hcot\theta -x)+M(-x)}{m+M}$

x = $\frac{mhcot\theta }{m+M}$