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Rotational Motion

Section: Physics

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Question : 31 of 59

Marks: +1, -0

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?

[JEE Adv 2017 P1]

The velocity of the point mass m is: $v=\sqrt{\frac{2gR}{1+\frac{m}{M}}}$
The velocity of the block M is : $V=-\frac{m}{2m}\sqrt{2gR}$
The position of the point mass is : $x=-\sqrt{2}\frac{mR}{M+m}$
The x component of displacement of the center of mass of the block M is : $-\frac{mR}{M+m}$

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