Question

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?
[A] The x component of displacement of the center of mass of the block M is : -mR/ M+m
[B] The position of the point mass is : x = - √2 mR/ M+m
[C] The velocity of the point mass m is : v = ⎷2gR/ 1+ m/M
[D] The velocity of the block M is: V = -m/M √2gR

Answer 1

Answer:

sorry mate but I don't know the answer

Answer 2

Answer:

Sorry mate !!!!!!! I don't know the answer

but thanks for the free 15 points