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

Answer 1

Answer: (B), (D)

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

Using conservation of momentum

$mu=MV$ ....(I)

Using conservation of energy

$mgR=\frac{1}{2}m{u}^2+\frac{1}{2}M{V}^2$ .....(II)

Solving eqn. (I) and (II) we get

$u=\sqrt{\frac{2gR}{1+\frac{m}{M}}}$ and

$V=\frac{m}{M}\sqrt{\frac{2gR}{1+\frac{m}{M}}}$

Change in the center of mass of the system along x axis is zero, since there is no external force acting on the system.
Let the mass $M$ displaces by $-x$, then mass $m$ displaces by $(R-x)$

$\therefore 0=m(R-x)-Mx$

$\Rightarrow x=\frac{mR}{m+M}$
Therefore $x$ component of displacement for the centre of mass of block $M$ is $-\frac{mR}{m+M}$