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JEE Advanced · Physics · 7. COM & Collisions

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?

  1. A The velocity of the point mass m is v=2gR1+mM
  2. B The velocity of the block M is v=-mM2gR
  3. C The position of the point mass is x=-2mRm+M
  4. D The x component of displacement of the centre of the mass of block M is -mRm+M
Verified Solution

Answer & Solution

Correct Answer

(D) The x component of displacement of the centre of the mass of block M is -mRm+M

Step-by-step Solution

Detailed explanation

When the centre of mass of the system is at origin, MSΔx-cm=m1Δx-+m2Δx-2
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