In the figure shown the wedge of mass `M` has a semicircular groove. `A` particle of mass `m=(M)/(2)` is released frem `A`. It slides on the smooth circular track and starts climbing on the right face. Find the maximum velocity of wedge during process of motion.

Explanation

Maximum velocity of wedge will be when the ball is at the lowest point in the wedge as till this point the horizontal component of normal on the wedge will be speeding the wedge but after this it will be opposite to the direction of motion of wedge, thereby slowing it down.
Applying conservation of linear momentum in horizontal direction at positions 1 & 2.
Initial momentum, `p_(i)=0`
Fianl momentum, `p_(f)=-Mv+m u`
`p_(i)=p_(f)`
`u=(Mv)/(m)=2v`
Applying conservation of mechanical energy
`U_(i)+K_(i)=U_(f)+K_(f)`
`mgR+0=0+(1)/(2)m u^(2)+(1)/(2)Mv^(2)`
`2mgR=m(2v)^(2)+Mv^(2)`
`2xx(M)/(2)xxgR=4(M)/(2)v^(2)+Mv^(2)`
`MgR=3Mv^(2)`
`v=sqrt((gR)/(3))`