JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Conservation of Momentum

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 position of the point mass $m$ is: $x = - 2 \frac{m R}{M + m}$
(B)The velocity of the point mass $m$ is: $v = \frac{2 g R}{1 + \frac{m}{M}}$ ✓ Correct
(C)The $x$ component of displacement of the center of mass of the block $M$ is: $- \frac{m R}{M + m}$ ✓ Correct
(D)The velocity of the block $M$ is: $V = - \frac{m}{M} 2 g R$

Answer: (B), (C)

Solution

Since there are no external horizontal forces, the center of mass of the system remains stationary. When the mass

m

loses contact with the block, it is at the bottom of the circular path. Using conservation of momentum in the horizontal direction:

m v_{m} + M V_{M} = 0

, where

v_{m}

and

V_{M}

are the velocities of

m

and

M

respectively. From the center of mass condition:

m (x + R) + M x = 0

, giving the displacement of block

M

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

, confirming option (C). Using energy conservation:

m g R = \frac{1}{2} m v_{m}^{2} + \frac{1}{2} M V_{M}^{2}

. Combining with momentum conservation yields

v_{m} = \frac{2 g R}{1 + \frac{m}{M}}

, confirming option (B).

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · One or more correct options

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JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Conservation of Momentum

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