A block of mass M has a circular cut with a frictionless surface as...

When the centre of mass of the system is at origin,

${M}_{S}\Delta{\overset{-}{x}}_{cm} = {m}_{1}\Delta\overset{-}{x} + {m}_{2}\Delta{\overset{-}{x}}_{2}$

$0 = m\left(+ R + \overset{-}{x}\right) + m\overset{-}{x}$

$\overset{-}{x} = \dfrac{- mR}{M + m}$

The linear momentum of the system is conserved.

$0 = m{\overset{-}{v}}_{1} + M{\overset{-}{v}}_{2}$

${\overset{-}{v}}_{2} = - \dfrac{M{\overset{-}{v}}_{1}}{M}$

From energy conservation,

$mgR = \dfrac{1}{2}m{v}_{1}^{2} + \dfrac{1}{2}M{v}_{2}^{2}$

$mgR = \dfrac{1}{2}m{v}_{1}^{2} + \dfrac{1}{2}M{\left(\dfrac{m{v}_{1}}{M}\right)}^{2}$

$mgR = \dfrac{1}{2}m{v}_{1}^{2}\left(1 + \dfrac{m}{M}\right)$

$\sqrt{\dfrac{2gR}{\left(1 + \dfrac{m}{M}\right)}} = {v}_{1}$