A block of mass $m$slides down an inclined wedge of same mass $m$shown in figure. Friction is absent everywhere. 

Let $a$ be the acceleration of wedge leftwards and $a_r$ the relative acceleration of block down the plane. Then absolute acceleration of block in horizontal direction will be $(a_r \cos \theta – a)$ towards right. Net force on the system in horizontal direction is zero. Therefore, acceleration of $COM$ in horizontal direction will be zero or acceleration of wedge towards left is equal to the acceleration of block towards right.

$$\begin{gathered} \therefore a_r \cos \theta – a = a \\ or 2a = a_r \cos \theta ....\left(1\right) \end{gathered}$$

Now let $N$be the normal reaction between the block and the wedge. Then free body diagram of wedge gives

$N \sin \theta = ma ....\left(2\right)$

Free body diagram of block with respect to wedge is:

Net force on block perpendicular to plane is zero

Hence,

 $$\begin{gathered} N + ma \sin \theta = mg \cos \theta .... \left(3\right) \\ \mathrm{Solving} \mathrm{eqs}. \left(1\right), \left(2\right) \mathrm{and} \left(3\right), \mathrm{we} \mathrm{get}, \\ {\mathrm{a}}_{\mathrm{r}}=\frac{2\mathrm{g} \sin \mathrm{\theta }}{1+{\sin }^2\mathrm{\theta }} \end{gathered}$$

acceleration of block vertically downwards $a_y=a_r \sin \theta$

${\mathrm{a}}_y=\frac{2\mathrm{g} {\sin }^2 \mathrm{\theta }}{1+{\sin }^2\mathrm{\theta }}$

∴ acceleration of $COM$ is

$a_{com}=\frac{a_y}{2}=\frac{\mathrm{g} {\sin }^2 \mathrm{\theta }}{(1+{\sin }^2\mathrm{\theta })}$