A small disc of mass $m$ slides down a smooth hill of height $h$ without initial velocity and gets onto a plank of mass $M$ lying on the horizontal plane at the base of the hill shown in figure. Due to friction between the disc and the plank, the disc slows down and, beginning with a certain moment, moves in one piece with the plank. Find the total work performed by the friction forces in this process.

Velocity of the disc just before comes in contact with the plank can be obtained as

$mgh=\frac{1}{2}m{\nu }^2=\nu =\sqrt{\left(2gh\right)}$

Let ${\nu }_1$ be the combined velocity of both the disc and the plank when they move together. In this process momentum of the system remain constant, therefore

$$\begin{gathered} m\nu +0=\left(M+m\right){\nu }_1 \\ {\nu }_1=m\nu /\left(M+m\right) \end{gathered}$$

Initial K.E. of the system $=\frac{1}{2}m{\nu }^2$

Final K.E. of the system $=\frac{1}{2}\left(M+m\right){\nu }_1^2$

Work performed by the friction in this process $W$ $=$ change in K.E. of the system

$=\frac{1}{2}\left(M+m\right){\nu }_1^2-\frac{1}{2}m{\nu }^2$

Substituting the value of $\nu$ and ${\nu }_1$ in the above expression, we get

$W=-\left[\frac{ghmM}{m+M}\right]$