In the arrangement shown in figure, the rod of mass `m`held by two smooth walls, remains always perpendicular to the surface of the wedge of mass `M`. Assumming all the surface to be frictionless, find the acceleration of the rod and that of the wedge.

Explanation

Let acceleration of `m` be `a_(1)` (absolute) and that of `M` be `a_(2)` (absolute).
Writing equations of motion only in the direction of `a_(1)` or`a_(2)`
For `m mg cos alpha -N =ma_(1)` …(i)
For `M`. `N sin alpha = ma_(1)` …(ii)
Here `N =` normal reation between `m` and `M`
As discussed above, constraint rquation can be written as,
`a_(1) =a_(2) sin alpha` ...(iii)
Solving above three equation, we get
acceleration of rod `a_(1) =(mg cos alpha sin alpha)/((m sin alpha +(M)/(sin alpha)))`
and acceleration of wedge `a_(2) =(mg cos alpha)/(m sin alpha +(M)/(sin alpha)`