A body of mass `M` with a small block `m` placed on it rests on a smooth horizontal surface. The block is set in motion in the horizontal direction with a velocity `v`. To what height relative to the initial level will the block rise after breaking off from the body `M`. Friction can be assumed to be absent.

Answer

`(M v^(2))/(2(M+m)g)`

Explanation

Just at the time of breaking off, let `m` move with `v_(x)` in horizontal and `v_(y)` in vertical direction. At the instant `M` will also be moving with `v = v_(x)` in horizontal direction.

Applying `COM` along x-axis
`m v = (m+M)v_(x) rArrv_(x) = ((m)/(m+M))v` ..........(i)
Apply `COE`
`(1)/(2)m v^(2) = (1)/(2)m(v_(x)^(2) + v_(y)^(2))+(1)/(2)Mv_(y)^(2) + mgh_(1)` .....(ii)
After breaking off `M`, moves in projectile
`v_(y)^(2) = 2 gh_(2) rArr mgh_(2) - (1)/(2)mv_(y)^(2) = 0` ...(iii)
From (ii) & (iii)
`(1)/(2)m v^(2) = (1)/(2)m v_(x)^(2)+(1)/(2)Mv_(x)^(2)+mg(h_(1)+h_(2))`
`rArr 2ms(h_(1)+h_(2)) = [m v^(2)-mv_(x)^(2)-Mv_(x)^(2)]`
`rArr h_(1)+h_(2) = (Mv^(2))/(2(m+M)g)`