Question

A block of mass M has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surfaced 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. When the mass loses contact with the block, its position is x and the velocity is v. At that instant, which of the following option is/are correct ?

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

Answer 1

Answer: (A), (B)

The correct options are
A

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

B

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

As net external force is zero, centre of mass remains at rest.
Hence, $(m+M)x+mR=0\Rightarrow x=\frac{-mR}{m+M}$

Conserving momentum, $m{V}_1=M{V}_2$
Conserving energy, $\frac{1}{2}m{V}_1^2+\frac{1}{2}M{V}_2^2=mgR$
Solving we get, $V_1=\sqrt{\frac{2gR}{1+\frac{m}{M}}}$