Assuming all the surfaces to be smooth, find the acceleration of the triangular block of mass `M` when the block of mass `m` slides on it.

Explanation

When the block of mass `m` slides in downward direction, the triangular block will move towards left.

Let the acceleration of triangular block w.r.t. the ground be `a_(0)`
Acceleration of the block w.r.t. to the triangular block `=a` (as shown)

`larrNsintheta=Ma_(0)` (i)

`rarrNsintheta=m(acostheta-a_(0))` (ii)
`darr:mg-Ncostheta=masintheta` (iii)
Solving `a_(0)=(mgsinthetacostheta)/(M+msin^(2)theta)`
OR
If we consider the motion of `m` along the plane, for this break `a_(0)` into component along and `bot^(ar)` to the plane.


Along the plane`:` `mgsintheta=m(a-a_(0)costheta)` (i)
`bot^(ar)` to the plane`:` `mgcostheta-N=ma_(0)sintheta` (ii)
Solving (i) and (ii), we get
`a_(0)=(mgsinthetacostheta)/(M+msin^(2)theta)`
OR
For smaller block, if we work from the reference frame attached to the triangular block (a non`-`inertial frame, because the triangular block is acceleration with acceleration `a_(0)`) we have to apply a psuedo force `ma_(0)` on the small block towards right.

`bot^(ar)` to the plane`:` `N+ma_(0)sintheta-mgcostheta=ma'=0`
`N=mgcostheta-ma_(0)sintheta`
For `M:Nsintheta=ma_(0)`
`(mgcostheta-ma_(0)sintheta)sintheta-Ma_(0)`
`mgsinthetacostheta=M(M+msin^(2)theta)a_(0)`
`a_(0)=(mgsinthetacostheta)/(M+msin^(2)theta)`