A ball of mass m = 20 g falls from a height h = 2 m onto a…

A ball of mass $m = 20\ \mathrm{g}$ falls from a height $h = 2\ \mathrm{m}$ onto a plane inclined at $\alpha = 45^\circ$. It bounces off horizontally, without any change in the magnitude of its velocity. What is the change in the ball's momentum?

Explanation of Method

Momentum change of a ball bouncing off an inclined plane.

Step 1: Speed at impact

\[ v = \sqrt{2gh} = \sqrt{2 \cdot 9.81 \cdot 2} \approx 6.26\ \mathrm{m/s}. \]

Step 2: Momentum before and after

With $x$ horizontal (the bounce direction) and $y$ up: $\vec p_i = (0,\ -mv)$.

\[ \vec p_f = (mv,\ 0), \]

\[ mv = 0.02 \cdot 6.26 = 0.125\ \mathrm{kg\,m/s}. \]

Step 3: Change of momentum

\[ \Delta\vec p = \vec p_f - \vec p_i = (mv,\ mv), \]

\[ \Delta p = \sqrt2\,mv = 1.414 \cdot 0.125. \]

\[\begin{gathered} \Delta p \approx 0.177\ \mathrm{kg\,m/s} \\ \text{perpendicular to the inclined plane} \end{gathered}\]

$\vec v_i,\ 6.26\ \mathrm{m/s}$ $\vec v_f$ $\Delta\vec p \perp \text{plane}$ $45^\circ$ $m\vec v_i$ $m\vec v_f$ $\Delta\vec p$ $\Delta\vec p = m\vec v_f - m\vec v_i$

The ball arrives vertically and leaves horizontally with the same speed. The momentum change (right) is the difference of the two vectors: $\sqrt2\,mv$, perpendicular to the plane, which is the direction of the plane's push.

Result

\[\boxed{\,\begin{gathered} \Delta p \approx 0.177\ \mathrm{kg\,m/s} \\ \text{perpendicular to the inclined plane} \end{gathered}\,}\]

Remark

Check. The original result is correct. The same answer follows from the components normal to the plane: the normal component $v\cos 45^\circ$ reverses, the component along the plane stays, so $\Delta p = 2mv\cos 45^\circ = \sqrt2\,mv$. The change points along the normal, as it must, since a smooth plane can only push perpendicular to itself.