Question

A body of mass $M$ (figure shown above) 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 $v$. The height (relative to the initial level) to which the disc rise after breaking off the body $M$ is given as $h=\frac{M{v}^2}{xg(M+m)}$. The friction is assumed to be absent. Find '$x$'.

Answer 1

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 ${v^{\prime }}_y$
From conservation of momentum, along x-axis for the system (disc $+$ body)
$mv=(m+M){v^{\prime }}_x$ or ${v^{\prime }}_x=\frac{mv}{m+M}$ (1)
And from energy conservation, for the same system in the field of gravity,
$\frac{1}{2}m{v}^2=\frac{1}{2}(m+m){v^{\prime }}_x^2+\frac{1}{2}m{v^{\prime }}_y^2+mg{h}^{\prime }$, using (1)
or, ${v^{\prime }}_y^2=v^2-\frac{m{v}^2}{(m+M)}-2g{h}^{\prime }$
Also, if $h^{\prime \prime }$ is the height of the disc, from the break-off point,
then ${v^{\prime }}_y^2=2g{h}^{\prime \prime }$
So, $2g(h^{\prime \prime }+h^{\prime })=v^2-\frac{m{v}^2}{(M+m)}$
Hence, the total height, raised from the initial level
$=h^{\prime }+h^{\prime \prime }=\frac{M{v}^2}{2g(M+m)}$