Question

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?

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

Answer 1

Answer: (B), (C)

As external force is zero Center of mass remains in rest.
$( M + m ) x + m \times R =0 $
$x =\frac{- mR }{ M + m }$
Linear momentum conservation
$mV _{1}= MV _{2}$
Energy conservation
$\frac{1}{2} mV _{1}^{2}+\frac{1}{2} MV _{2}^{2}= mgR $
$\frac{1}{2} mV _{1}^{2}+\frac{1}{2} M \left(\frac{ mV _{1}}{ M }\right)^{2}= mgR $
$\frac{1}{2} mV _{1}^{2}\left[1+\frac{ m }{ M }\right]= mgR $
$ V _{1}^{2}=\frac{2 gR }{1+\frac{ m }{ M }} $
$ V _{1}=\sqrt{\frac{2 gR }{1+\frac{ m }{ M }}}$