A small body of mass m slides without friction down a curved…

A small body of mass $m$ slides without friction down a curved surface AB from height $h$ and at B moves onto a plank of mass $M$ lying on a smooth floor. Friction between the body and the plank eventually brings them to a common speed. How much energy is dissipated by friction?

Explanation of Method

Energy lost when a body slides onto a plank.

Step 1: Speed at B

\[ v_0 = \sqrt{2gh}. \]

Step 2: Common speed

The friction between body and plank is internal and the floor is smooth, so horizontal momentum is conserved: $mv_0 = (m + M)\,v \;\Longrightarrow\; v = \frac{m}{m + M}\,v_0$.

Step 3: Dissipated energy

\[ \Delta E = \tfrac12 mv_0^2 - \tfrac12 (m + M)v^2, \]

\[ \Delta E = mgh\left(1 - \frac{m}{m + M}\right). \]

\[\Delta E = \frac{mM}{m + M}\,gh\]

$h$ $A$ $B$ $v_0 = \sqrt{2gh}$ $M$ $\text{common speed } v = \frac{m v_0}{m + M}$ $\Delta E = \frac{mM}{m + M}\,gh$

The body reaches the plank with speed $\sqrt{2gh}$. Friction between them is internal, so momentum is conserved; the kinetic energy that disappears in reaching the common speed is turned into heat.

Result

\[\boxed{\,\Delta E = \frac{mM}{m + M}\,gh\,}\]

Remark

Check. The original result is correct. For a very heavy plank ($M \gg m$) all of $mgh$ is dissipated; for a very light one almost nothing is.