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
$m{V}_1=M{V}_2$
Energy conservation
$\frac{1}{2}m{V}_1^2+\frac{1}{2}M{V}_2^2=mgR$
$\frac{1}{2}m{V}_1^2+\frac{1}{2}M{\left(\frac{m{V}_1}{M}\right)}^2=mgR$
$\frac{1}{2}m{V}_1^2\left[1+\frac{m}{M}\right]=mgR$
$V_1^2=\frac{2gR}{1+\frac{m}{M}}$
$V_1=\sqrt{\frac{2gR}{1+\frac{m}{M}}}$