Q.A ball of mass $0.5\ \text{kg}$ is dropped from a height of $5\ \text{m}$ and rebounds to a height of $3.2\ \text{m}$ after striking the ground. If the ball is in contact with the ground for $0.02\ \text{s}$, find the magnitude of the impulse and the average force exerted by the ground on the ball, ignoring the (small) impulse due to the ball's weight during this very short contact time. (Take $g = 9.8\ \text{m/s}^2$.)

Given: mass $m = 0.5\ \text{kg}$, fall height $h_1 = 5\ \text{m}$, rebound height $h_2 = 3.2\ \text{m}$, contact time $\Delta t = 0.02\ \text{s}$, $g = 9.8\ \text{m/s}^2$.

Speed just before impact (from falling height $h_1$): $v_{\text{down}} = \sqrt{2gh_1} = \sqrt{2 \times 9.8 \times 5} = \sqrt{98} \approx 9.90\ \text{m/s (downward)}$

Speed just after impact (needed to rebound to height $h_2$): $v_{\text{up}} = \sqrt{2gh_2} = \sqrt{2 \times 9.8 \times 3.2} = \sqrt{62.72} \approx 7.92\ \text{m/s (upward)}$

Impulse. Taking upward as positive, the ball's momentum changes from $-mv_{\text{down}}$ to $+mv_{\text{up}}$, so $\Delta p = m\big(v_{\text{up}} - (-v_{\text{down}})\big) = 0.5 \times (7.92 + 9.90) = 0.5 \times 17.82 \approx 8.91\ \text{kg}\,\text{m/s (upward)}$ This is the (approximate) net impulse delivered by the ground's impulsive normal force, treating the ball's own weight as contributing negligibly over the very short $0.02\ \text{s}$ contact time.

Average force: $F_{\text{avg}} = \frac{\Delta p}{\Delta t} = \frac{8.91}{0.02} \approx 445.5\ \text{N (upward)}$

Use $v = \sqrt{2gh}$ to find the speeds just before and just after impact, add their magnitudes (since the ball reverses direction) to get $\Delta p$, then divide by the contact time.

• Subtracting the two speeds instead of adding them -- since the ball reverses direction, its momentum change is the sum, not the difference, of the two speeds' magnitudes.
• Forgetting the ball loses height (and hence speed) on rebound, and wrongly assuming $v_{\text{up}} = v_{\text{down}}$.