In the arrangement shown in the figure, the rod of mass m held by two smooth walls, remains always perpendicular to the surface of the wedge of mass $M$. Assuming all the surfaces to be frictionless, find the acceleration of the rod ($a_1$) and that of the wedge($a_2$).

 Let acceleration of m be $a_1$ (absolute) and that of M be $a_2$ (absolute).

Writing equations of motion only in the directions of $a_1$ or $a_2$.

For m

$mg\cos \alpha -N=m{a}_1$

For M,

$N\sin \alpha =M{a}_2$

Here, N= normal reaction between m and M

As discussed above, constraint equation can be written as,

$a_1=a_2\sin \alpha$

Solving above three equations, we get

acceleration of rod,

$a_1=\frac{mg\cos \alpha \sin \alpha }{\left(m\sin \alpha +\frac{M}{\sin \alpha }\right)}$

and acceleration of wedge

$a_2=\frac{mg\cos \alpha }{m\sin \alpha +\frac{M}{\sin \alpha }}$