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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?

The position of the point mass is: x = −√2mRM+m

The velocity of the block M is: V = −mM√2gR

The x component of displacement of the center of mass of the block M is: −mRM+m

The velocity of the point mass m is: v = √2gR1+mM

Solution:

Using conservation of momentum mu = MV (I)
Using conservation of energy MgR = 1/2mu² + 1/2MV² (II)
Solving eqn. (I) and (II) we get
v = √2gR/(1+m/M)
and V = (m/M)√2gR/(1+m/M)
The position of center of mass as center of mass of system is zero
∴ 0 = −m(R−x) + Mx ⇒ x = mR/(m+M) x − displacement of center of mass.