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Q.

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?

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a

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

b

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

c

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

d

The velocity of the block M is: V=−mM2gR

answer is A.

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Detailed Solution

1) mV=MV1v and v1 are velocities of m and M )  2) mgR=12mv2+12MV12mgR=12mv2+12MmMv2 Option (A)2gR1+mM=V Option (B)m(R−x)−Mx=0x=mRM+m But in negative x direction hence B is correct.
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