JEE Advanced2017PhysicsCenter of Mass, Momentum and CollisionActual
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
Options
- AThe velocity of the point mass m is v = 2 g R 1 + m M
- BThe velocity of the block M is v =- m M 2 g R
- CThe position of the point mass is x = - 2 m R m + M
- DThe x component of displacement of the centre of the mass of block M is - m R m + M
Correct answer
A. The velocity of the point mass m is v = 2 g R 1 + m M
Step-by-step solution
When the centre of mass of the system is at origin, M S Δ x - c m = m 1 Δ x - + m 2 Δ x - 2 0 = m + R + x - + m x - x - = - m R M + m The linear momentum of the system is conserved. 0 = m v - 1 + M v - 2 v - 2 = - M v - 1 M From energy conservation, m g R = 1 2 m v 1 2 + 1 2 M v 2 2 m g R = 1 2 m v 1 2 + 1 2 M m v 1 M 2 m g R = 1 2 m v 1 2 1 + m M 2 g R 1 + m M = v 1