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 ma

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 [2017]

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? [2017]

Smart Achievers · Accepted Answer

(2, 3)

Let the block of mass M moves by distance $x$ towards left.

$M x = m (R - x)$

$\Rightarrow x = \frac{m R}{M + m} towards left$ $∴$ $x = - \frac{m R}{M + m}$

If $v$ is the velocity of mass $' m'$ as it leaves the block and $V$ is the velocity of block at that instant then according to conservation of linear momentum

$m v = M V$

By energy conservation

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

Solving we get, $v = \sqrt{\frac{2 g R}{1 + \frac{m}{M}}} and V = \frac{m}{M} \sqrt{\frac{2 g R}{1 + \frac{m}{M}}}$

Solution

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

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

3 The $x$ -component of displacement of the center of mass of the block M is: $- \frac{m R}{M + m}$

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

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Solution

1 The position of the point mass m is: x=-2mRM+m

2 The velocity of the point mass m is: v=2gR1+mM

3 The x-component of displacement of the center of mass of the block M is: -mRM+m

4 The velocity of the block M is: V=-mM2gR

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

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

3 The $x$ -component of displacement of the center of mass of the block M is: $- \frac{m R}{M + m}$

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