1)

A block of mass M has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surfaced of a fixed table. Initially, the right edge of the block is at x=0, in a coordinate system fixed in the table. A point mass m is released from rest at the topmost point of the path as shown and slide 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 option is/are correct?

17122019162_suee.PNG


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

B) The x component of displacement of the centre of mass of the block M is mRM+m

C) The position of the point mass is x=2mRM+m

D) The velocity of the block M is v=mM2gR

Answer:

Option A,B

Explanation:

xcm of the block and point mass system=0

    m(x+R)+Mx=0

where x is the displacement of the block

      solving the equation we get

          x=mRM+m

   From the conservation of momentum and mechanical energy of the combined system.

     0=mv-MV

    mgR=12mv2+12MV2

 Solving these two equations, we get

          v=2gR1+mM