Question

A block of mass M has a circular cut with a frictionless surface. 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 co-ordinate 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 the velocity is v . At that instant which of the options is /are correct

Answer 1

Let the block be displaced by x. If initially the center of mass of the system is at origin then

$O=\frac{M\times x+m(x+R)}{M+m}$

$O=Mx+mx+mR$

$x=-\frac{mR}{m+M}$

Now, $v$ is the velocity of mass $m$ as it leaves the block and V is the velocity of block at that instant then according to conservation of linear momentum, $mv=MV,$

By energy conservation we get,

$mgR=\frac{1}{2}m{v}^2+\frac{1}{2}m{V}^2$

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

And,

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