A metal ball of mass m = 10 g falls freely from a height H =…

A metal ball of mass $m = 10\ \mathrm{g}$ falls freely from a height $H = 30\ \mathrm{m}$ onto a smooth metal plate and bounces back with the same speed. The contact lasts $\Delta t = 1\ \mathrm{ms}$. Find the magnitude of the impulse the plate gives the ball and the average force of the ball on the plate.

Explanation of Method

Average force of a ball dropped on a metal plate.

Step 1: Speed at impact

\[ v = \sqrt{2gH} = \sqrt{2 \cdot 9.81 \cdot 30} = \sqrt{588.6} \approx 24.26\ \mathrm{m/s}. \]

Step 2: Impulse

The velocity reverses, so it changes by $2v$: $J = 2mv = 2 \cdot 0.01 \cdot 24.26$.

\[J \approx 0.485\ \mathrm{N\,s}\]

Step 3: Average force

\[ \bar F = \frac{J}{\Delta t} = \frac{0.485}{10^{-3}}. \]

\[\bar F \approx 485\ \mathrm{N}\]

By Newton's third law the ball presses on the plate with the same average force, downward.

$H = 30\ \mathrm{m}$ $24.3$ $24.3\ \mathrm{m/s}$ $J = 0.485\ \mathrm{N\,s}$ $\bar F \approx 485\ \mathrm{N}$ $\Delta t = 1\ \mathrm{ms}$ $F$ $t$

Left: the ball hits the plate at 24.3 m/s and leaves upward at the same speed (not to scale). Right: the plate's force during the 1 ms contact, replaced by its average; the area is the impulse.

Result

\[\boxed{\,J \approx 0.485\ \mathrm{N\,s}\,}\]

\[\boxed{\,\bar F \approx 485\ \mathrm{N}\,}\]

Remark

Note. The original results are correct. Strictly, $J/\Delta t$ is the average net force; the plate must also hold up the ball's weight, $mg \approx 0.1\ \mathrm{N}$, during the contact, which is negligible next to 485 N. The force is about 5000 times the ball's weight because the contact is so short.