A block of mass M has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fixed table. Initially the right edge of the block is at x = 0, in a co-ordinate system fixed to the table. A point mass m is released from rest at the topmost point of the path as shown and it slides down. When the mass loses contact with the block, its position is x and the velocity is v. At that instant, which of the following options is/are correct?

Options

A. The velocity of the point mass mis: `v=sqrt(2gR/(1+m/M)`
B. The x component of displacement of the centre mR of mass of the block M is `(mR)/(M+m)`
C. The position of the point mass is :`x=-sqrt2(mR)/(M+m)`
D. The velocity of the block M is `V=m/Msqrt(gR)`