A block of mass m is placed on a triangular block of mass m, which in turns placed on a horizontal surface as shown in the figure. Assuming frictionless surfaces find the velocity of the triangular block when the smaller block reaches the bottom end.

Explanation

Let us assume that wedge move leftward with velocity v and block move down ward with velocity u with respect to wedge.
`therefore` Net force is horizontal direction is zero so momentum is conserved in x direction.
Now velocity of block with respect to ground is
`vec(v_(m))=vec(v_(mw))+vec(v_(w))`
`vec(V_(m))=vecu+vecv`
`vec(V_(m))=(ucostheta-v)hati-usinthetahatj`
Now from momentum conservation in x direction

`0=-Mv+mV_("mx")`
`rArrMv=m(ucostheta-v)` .....(1)
From energy conservation
`mgh=1/2mv^(2)+1/2mv_(m)^(2)`
`rArrmgh=1/2mv^(2)+1/2m(u^(2)+v^(2)-2uvcostheta)` ....(2)
from eq. (1) & (2)
`v=[(2m^(2)ghcostheta)/((M+m)(M+msin^(2)theta))]^(1//2)`