Centre Of Mass Ques 9
- 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: 9.(a)
Solution:
Formula:
Motion of Centre of Mass and Conservation of Momentum:
- $\Delta x _{cm}$ 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
$$ \therefore \quad v=\sqrt{\frac{2 g R}{1+\frac{m}{M}}} $$
BETA
