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 cut as shown and it slides down.
 
When the mass loses contact with block, its position is x and the velocity is v. At that instant, which of the following options is/are correct? 

There is no net external force acting on the system along the x-axis. Hence, linear momentum of the system (block + point mass) should be conserved along  x - direction.
 
$\therefore$  Along the x-axis
 $mv+MV=0 \mathrm{...}\left(\text{i}\right)$
From conservation of mechanical energy
Loss in GPE of particle of mass m= Gain in kinetic energy of both the block and point mass.
 $$\begin{gathered} \Rightarrow mgh=\frac{1}{2}m{v}^2+\frac{1}{2}M{V}^2 \\ \Rightarrow mgh=\frac{1}{2}m{v}^2+\frac{1}{2}M{\left(\frac{mv}{M}\right)}^2 \\ \Rightarrow 2gh=(1+\frac{m}{M})v^2\Rightarrow v=\sqrt{\frac{2gh}{1+\frac{m}{M}}} \mathrm{...}\left(\text{ii}\right) \end{gathered}$$
Hence, option (A) is correct.
From Eqs. (i) and (ii)
  $V=\frac{m}{M}v=\frac{m\sqrt{2gh}}{\sqrt{M(m+M)}}$
Block ‘M’ will move along –ve x-axis.
Hence option (B) is incorrect.
Displacement of point mass  $\Delta x_M=-X$
Displacement of point mass $\Delta x_m=-(X-R)$
Since the location of centre of mass does not change along the x -axis
$$\begin{gathered} m.\Delta x_m+M.\Delta x_M=0 \\ -m(X-R)-MX=0 \\ -(M+m)X+mR=0 \end{gathered}$$

$\Rightarrow X=\frac{mR}{(M+m)}$
Hence displacement of block  $M=\frac{mR}{(M+m)}$
Displacement of the mass ‘m’  $=(X-R)$
 $\Rightarrow \Delta x_m=-(\frac{mR}{M+m}-R)=\frac{MR}{(M+m)}$
Hence final position of ‘m’
 $x_m=-x+\Delta x_m=-R+\frac{MR}{(M+m)}=\frac{mR}{(M+m)}$
Hence option (C) is incorrect.