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 m is : `v=sqrt((2gR)/(1+m/M))`
B. The x component of displacement of the center 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(2gR)`

Answer

A, B