A block of mass `m` is initially lying on a wedge of mass `M` with an angle of inclination `theta` as shown in figure. Calculate the displacement of the wedge when the block is released and reaches to the bottom of the wedge.

Explanation

`l=hcostheta`
Let displacement of wedge be `x` towards right,

`/_\x_(CM)=(m(1+x)+Mx)/(M+m)`
`0=ml+(M+m)x`
`x=(-ml)/(M+m)=(mhcottheta)/(M+m)`