All the surfaces shown in figure are assumed to be frictionless. The block of mass m slides on the prism which in turn slides backward on the horizontal surface. Find the acceleration of the smaller block with respect to the prism.

Explanation

Let `a_(0) ` be the acceleration of prism in the backward direction . The mass .m. is placed on non - inertial frame , therefore it is acted on by fictitious force `ma_(0) ` in the forward direction . The forces acting on mass m are
(i) Weight mg acting downward,
(ii) Normal reaction R,
(iii) Fictitious force `ma_(0) `.
According free body diagram of .m. is shown in figure .
Suppose block m slides down the prism with acceleration a .
The equation of motion of .m. parallel to incline is
`ma_(0) cos theta +mg sin theta ` ....(1)

The block is in equilibrium perpendicular to incline , so resolving force perpendicular to incline
`R_(1)+ma_(0) sin theta =mg cos theta ` ....(2)
Now we consider the forces acting on the prism . As prism is on ground (inertial frame ) no fectitious force acts on it . The forces on prism are
(i) Weight (Mg) downward.
(ii) Normal force `R_(1)` exerted by block
(iii) Normal reaction `R_(2)` exerted by ground .
For horizontal motion of prism
`R_(1) sin theta = MA_(0) rArr R_(1) =(Ma_(0)) /(sin theta) ` ....(3)
substituting this value in (2) ,
`(Ma_(0) )/(sin theta) +ma_(0) sin theta =mg cos theta`
This gives `a_(0)=(mg sin theta cos theta )/(M+m sin^(2) theta)` ....(4)
`:.` From (1) , `a=(mg sin theta cos^(2) theta)/(M+m sin ^(2) theta )+g sin theta`
`=((M+m)gsin theta)/(M+m sin^(2) theta)`