A body of mass M (figure) with a small disc of mass m placed on it rests on a smooth horizontal plane. The disc is set in motion in the horizontal direction with velocity v. To what height (relative to the initial level) will the disc rise after breaking off the body M? The friction is assumed to be absent.

Explanation

When the disc breaks off the body M, its velocity towards right (along x-axis) equals the velocity of the body M, and let the disc's velocity in upward direction (along y-axis) at that moment be `v_y^'`
From conservation of momentum, along x-axis for the system (disc+body)
`mv=(m+M)v_x^'` or `v_x^'=(mv)/(m+M)`
And from energy conservation, for the same system in the field of gravity:
`1/2mv^2=1/2(m+M)v'_x^2+1/2mv'_y^2+mgh^'`,
where `h^'` is the height of break off point from initial level. So,
`1/2mv^2=1/2(m+M)(m^2v^2)/((M+m))+1/2mv'_y^2+mgh^'`, using (1)
or, `v'_y^2=v^2-(mv^2)/((m+M))-2gh^'`
Also, if `h^('')` is the height of the disc, from the break-off point,
then, `v'_y^2=2gh^(' ')`
So, `2g(h^(' ')+h^')=v^2-(mv^2)/((M+m))`
Hence, the total height, raised from the intital level
`=h^'+h^('')=(Mv^2)/(2g(M+m))`