A small body of mass m is placed over a larger mass M whose surface...

Since there are no external forces acting on the system in the horizontal direction, the horizontal component of momentum is conserved.

Let $V$ be the velocity of the larger mass $M$ when the small body breaks off. At the point of breaking off, the surface of $M$ is vertical, meaning the small body $m$ has the same horizontal velocity as $M$.

By conservation of horizontal momentum:
$mv = m v_x + M V$
Since $v_x = V$, we have:
$mv = (m + M)V \Rightarrow V = \dfrac{mv}{m + M}$

Let $v_f$ be the net speed of the small body when it breaks off at height $h$. By conservation of mechanical energy:
$\dfrac{1}{2}mv^2 = \dfrac{1}{2}mv_f^2 + \dfrac{1}{2}MV^2 + mgh$

Rearranging for $v_f^2$, we get:
$v_f^2 = v^2 - \dfrac{M}{m}V^2 - 2gh$

Substitute $V = \dfrac{mv}{m + M}$ into the equation:
$v_f^2 = v^2 - \dfrac{M}{m}\left(\dfrac{mv}{m + M}\right)^2 - 2gh$

$v_f^2 = v^2 - \dfrac{Mm}{(m + M)^2}v^2 - 2gh$

$v_f^2 = v^2 \left( 1 - \dfrac{Mm}{(m + M)^2} \right) - 2gh$

$v_f^2 = v^2 \left( \dfrac{m^2 + M^2 + 2Mm - Mm}{(m + M)^2} \right) - 2gh$

$v_f^2 = \dfrac{M^2 + Mm + m^2}{(m + M)^2}v^2 - 2gh$

Taking the square root, the speed of the small body is:
$v_f = \left[ \dfrac{(M^2 + Mm + m^2)}{(M + m)^2}v^2 - 2gh \right]^{1/2}$

Answer: \left[\dfrac{(M^2 + Mm + m^2)}{(M + m)^2}v^2 - 2gh\right]^{1/2}