Discussion Forum



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?

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A) The velocity of the point mass m is $v= \sqrt{\frac{2gR}{1+\frac{m}{M}}}$

B) The x component of displacement of the centre of mass of the block M is $-\frac{mR}{M+m}$

C) The position of the point mass is $x= -\sqrt{2}\frac{mR}{M+m}$

D) The velocity of the block M is $v= -\frac{m}{M}\sqrt{2gR}$

Answer:

Option A,B

Explanation:

$\triangle x_{cm}$ 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

          $x= -\frac{mR}{M+m}$

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

     0=mv-MV

    $mgR=\frac{1}{2}mv^{2}+\frac{1}{2}MV^{2}$

 Solving these two equations, we get

          $v=\sqrt{\frac{2gR}{1+\frac{m}{M}}}$