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 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.
and the velocity is \( v \). At that instant, which of the following options is/are correct?
(1) The velocity of the point mass \( m \) is \( v=\sqrt{\frac{2 g R}{1+\frac{m}{M}}} \)
(2) The velocity of the block \( M \) is \( V=-\frac{m}{M} \sqrt{2 g R} \)
(3) The position of the point mass is \( x=-\sqrt{2} \frac{m R}{M+m} \)
(4) The \( x \) component of displacement of the centre of mass of the block \( \mathrm{M} \) is \( -\frac{m R}{M+m} \)
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