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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 surfaced of a fixed table. Initially the right edge of the block is at x = 0, in a coordinate 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 option is/are correct ?


A

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

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B

The x component of displacement of the centre of mass of the block M is mRM+m.

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C

The position of the point mass is x= 2mRM+m.

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D

The velocity of the block M is V= mM2gR.

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Solution

The correct option is B

The x component of displacement of the centre of mass of the block M is mRM+m.


As net external force is zero, centre of mass remains at rest.
Hence, (m+M)x+mR=0x=mRm+M

Conserving momentum, mV1=MV2
Conserving energy, 12mV21+12MV22=mgR
Solving we get, V1=2gR1+mM


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