Centre of Mass 1 Question 12

12. 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?

(2017 Adv.)

(a) The velocity of the point mass $m$ is $v=\sqrt{\frac{2 g R}{1+\frac{m}{M}}}$

(b) The $x$ component of displacement of the centre of mass of the block $M$ is $-\frac{m R}{M+m}$

(c) The position of the point mass is $x=-\sqrt{2} \frac{m R}{M+m}$

(d) The velocity of the block $M$ is $V=-\frac{m}{M} \sqrt{2 g R}$

True/False

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Answer:

Correct Answer: 12. (a)

Solution:

  1. $\Delta x _{cm}$ of the block and point mass system $=0$

$\therefore \quad m(x+R)+M x=0$

where, $x$ is displacement of the block.

Solving this equation, we get

$$ x=-\frac{m R}{M+m} $$

From conservation of momentum and mechanical energy of the combined system

$0=m v-M V \Rightarrow m g R=\frac{1}{2} m v^{2}+\frac{1}{2} M V^{2}$

Solving these two equations, we get

$$ \therefore \quad v=\sqrt{\frac{2 g R}{1+\frac{m}{M}}} $$



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