A smooth wedge of mass `M` rests on a smooth horizontal surface. A block of mass `m` is projected from its lowermost point with velocity `v_(0)`. What is the maximum height reached by the block?

Explanation

At the instant the block breaks contact with the wedge, they have common `x`-component of velocity. In addition, the block has a vertical component of velocity.

Due to this vertical component, the block rises upwards till the verical component of velocity vanishes.
From momentum conservation along `x`-axis.
`mv_(0)=(m+M)V`...........i
`V=(mv_(0))/((m+M))`............ii
From energy conservation between initial and final positions of the block,
`1/2mv_(0)^(2)=1/2(m+M)V^(2)+mgh`............iii
`1/2mv_0^(2)=1/2(m^(2)/(m+M))v_(0)^(2)+mgh`
`h=v_(0)^(2)/(2g)[M/(m+M)]`