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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The x component of displacement of the centre of mass of the block M is: $- \frac{m R}{M + m}$

The velocity of the point mass m is: $v = \sqrt{\frac{2 g R}{1 + \frac{m}{M}}}$

The position of the point mass is $x = - \sqrt{2} \frac{m R}{M + m}$

The velocity of the block M is: $V = - \frac{m}{M} \sqrt{2 g R}$

answer is A, B.

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Detailed Solution

$1) m V = M V^{1} (v and v^{1} are velocities of m and M)$

$2) m g R = \frac{1}{2} m v^{2} + \frac{1}{2} M {(V^{1})}^{2}$

$\begin{array}{l} m g R = \frac{1}{2} m v^{2} + \frac{1}{2} M {(\frac{m}{M} v)}^{2} \\ Option (A) \sqrt{\frac{2 g R}{1 + \frac{m}{M}}} = V \\ Option (B) m (R - x) - M x = 0 \\ x = \frac{m R}{M + m} \end{array}$

$But in negative x direction hence B is correct.$

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