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 coordinate system fixed to the table. A point mass m is released from rest at the topmost…

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 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 velocity is v. At that instant which of the following options is correct?

Aicharya · Accepted Answer

When the centre of mass of the system is at origin,

$M_{S} Δ {\bar{x}}_{c m} = m_{1} Δ \bar{x} + m_{2} Δ {\bar{x}}_{2}$

$0 = m (+ R + \bar{x}) + m \bar{x}$

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

The linear momentum of the system is conserved.

$0 = m {\bar{v}}_{1} + M {\bar{v}}_{2}$

${\bar{v}}_{2} = - \frac{M {\bar{v}}_{1}}{M}$

From energy conservation,

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

$m g R = \frac{1}{2} m v_{1}^{2} + \frac{1}{2} M {(\frac{m v_{1}}{M})}^{2}$

$m g R = \frac{1}{2} m v_{1}^{2} (1 + \frac{m}{M})$

$\sqrt{\frac{2 g R}{(1 + \frac{m}{M})}} = v_{1}$

A block of mass M has a circular cut with a frictionless surface as shown. The block rests on the horizontal…

Solution