A ball of mass m = 20 g hits a wall perpendicularly at v =…

A ball of mass $m = 20\ \mathrm{g}$ hits a wall perpendicularly at $v = 10\ \mathrm{cm/s}$ and bounces straight back with the same speed. The collision lasts $\Delta t = 0.1\ \mathrm{s}$. Find the impulse the wall gives the ball and the magnitude of the average force of the wall on the ball.

Explanation of Method

A ball bouncing off a wall.

Step 1: Impulse

Take the direction toward the wall as positive: $v_i = +0.1\ \mathrm{m/s}$, $v_f = -0.1\ \mathrm{m/s}$.

\[ J = m\,(v_f - v_i) = 0.02 \cdot (-0.1 - 0.1) = -0.004\ \mathrm{N\,s}. \]

\[\begin{gathered} |J| = 4 \cdot 10^{-3}\ \mathrm{N\,s} \\ \text{directed away from the wall} \end{gathered}\]

Step 2: Average force

\[ \bar F = \frac{|J|}{\Delta t} = \frac{0.004}{0.1}. \]

\[\bar F = 0.04\ \mathrm{N}\]

$v_i = 0.1\ \mathrm{m/s}$ $v_f = 0.1\ \mathrm{m/s}$ $\bar F = 0.04\ \mathrm{N}$ $\text{wall}$

The ball comes in at 0.1 m/s and goes back at 0.1 m/s, so its velocity changes by 0.2 m/s. The wall's average push on it points away from the wall.

Result

\[\boxed{\,\begin{gathered} |J| = 4 \cdot 10^{-3}\ \mathrm{N\,s} \\ \text{directed away from the wall} \end{gathered}\,}\]

\[\boxed{\,\bar F = 0.04\ \mathrm{N}\,}\]

Remark

Check. The original results are correct. The velocity changes by $2v$, not $v$: stopping the ball takes $mv$, and sending it back takes another $mv$.