Physics • Year 11 • Module 2: Dynamics • Lesson 11
Momentum and Impulse
Apply the impulse-momentum theorem to real data, direction-reversal problems, force-time graphs, and safety device analysis.
1. Calculate from data — sports ball impacts
The table below records data for six sports ball impacts. Complete all missing cells. Define the initial direction of the ball as positive for each row. Show working. 12 marks (2 per row)
| Sport / scenario | m (kg) | vi (m/s) | vf (m/s) | Δp (N s) | Δt (s) | Favg (N) |
|---|---|---|---|---|---|---|
| Cricket bat striking ball | 0.16 | +40 | −35 | 0.002 | ||
| Tennis serve (racquet on ball) | 0.058 | 0 | +62 | 0.004 | ||
| Rugby tackle (player stopped) | 100 | +8 | 0 | 0.3 | ||
| Soccer kick | 0.43 | 0 | +22 | 0.008 | ||
| Baseball pitch caught in glove | 0.145 | +40 | 0 | 0.05 | ||
| Squash ball off wall | 0.024 | +35 | −28 | 0.003 |
1.1 Which scenario produces the largest impulse magnitude? Which produces the smallest? Explain why large impulse does not necessarily mean large average force. 2 marks
2. Interpret a force-time graph — cricket ball impact
The graph below shows how the force on a 0.16 kg cricket ball varies during a bat impact. The x-axis shows time (ms) and the y-axis shows force (N). 8 marks
Figure 2. Force on a 0.16 kg cricket ball during bat contact. Peak force = 5000 N; contact time = 2.0 ms. Illustrative data.
2.1 The graph shows a triangular pulse. Calculate the impulse (area under the F-t graph) using the formula for the area of a triangle. Remember to convert contact time to seconds. 2 marks
2.2 Using J = Δp, and the impulse you calculated above, find the speed of the ball just after the bat impact, if the ball was initially at rest. 2 marks
2.3 The average force during this impact is NOT 5000 N. Explain why, and calculate the correct average force. 2 marks
2.4 A heavier cricket bat increases the contact time to 3.0 ms with the same peak force. Sketch the new triangular pulse on the axes above (label it “heavier bat”) and explain whether the impulse is larger or smaller. 2 marks
3. Compare braking vs collision — same Δp, different Δt
Complete the table for a 1200 kg car initially travelling at 25 m/s (define this as positive). 8 marks
| Feature | Normal braking (Δt = 4 s) | Wall collision (Δt = 0.1 s) |
|---|---|---|
| Initial momentum pi (kg m/s) | ||
| Final momentum pf (kg m/s) | ||
| Δp (kg m/s) | ||
| Average force F (N) | ||
| F as multiple of car’s weight (approx.) |
3.1 Crumple zones extend the wall collision time from 0.1 s to 0.2 s. What is the average force with crumple zones? Express as a fraction of the no-crumple-zone force. 2 marks
3.2 A student claims: “Crumple zones just make the car weaker.” Using the impulse-momentum theorem, explain why this reasoning is incorrect and why crumple zones save lives. 3 marks
4. Predict and justify — catching technique
A fielder catches a cricket ball (m = 0.16 kg, speed = 35 m/s) in two ways: with hands rigid (stopping time = 0.01 s) and with hands “giving” (stopping time = 0.08 s). 6 marks
4.1 Calculate the average force on the ball in each case. Show full working with the Vector Protocol (define positive direction first). 3 marks
4.2 By Newton’s third law, the force on the fielder’s hands equals the force on the ball. By what factor is the force reduced by “giving” hands? Explain the physiological implication for injury prevention. 3 marks
Q1 — Sports ball data table
Cricket bat: Δp = 0.16(−35 − 40) = 0.16 × (−75) = −12 N s; F = −12/0.002 = −6000 N.
Tennis serve: Δp = 0.058(62 − 0) = +3.596 N s; F = 3.596/0.004 = +899 N.
Rugby tackle: Δp = 100(0 − 8) = −800 N s; F = −800/0.3 = −2667 N.
Soccer kick: Δp = 0.43(22 − 0) = +9.46 N s; F = 9.46/0.008 = +1183 N.
Baseball catch: Δp = 0.145(0 − 40) = −5.8 N s; F = −5.8/0.05 = −116 N.
Squash ball: Δp = 0.024(−28 − 35) = 0.024 × (−63) = −1.512 N s; F = −1.512/0.003 = −504 N.
1.1 Largest impulse magnitude: rugby tackle (−800 N s). Smallest: tennis serve (3.596 N s, about 3.6 N s). Large impulse does not mean large force: the rugby tackle has a large impulse because a large Δp is spread over a long time (0.3 s), making the average force only 2667 N. J = FΔt — the same impulse can be achieved with a small F over a long Δt.
Q2 — Force-time graph
2.1 Δt = 2.0 ms = 0.002 s. Area = ½ × 0.002 × 5000 = 5 N s.
2.2 J = Δp = mvf − 0 ⇒ vf = J/m = 5/0.16 = 31.25 m/s.
2.3 The average force is not 5000 N because 5000 N is the peak force at the midpoint. The force varies (rises then falls), so the average force = J/Δt = 5/0.002 = 2500 N (half the peak, as expected for a symmetric triangular pulse).
2.4 Heavier bat: Δt = 3.0 ms = 0.003 s. Area = ½ × 0.003 × 5000 = 7.5 N s. The impulse is larger (7.5 vs 5 N s) because the same peak force acts for a longer time. The triangle is wider on the x-axis but the same height.
Q3 — Braking vs collision comparison
pi = 1200 × 25 = +30 000 kg m/s (same in both). pf = 0 (same in both). Δp = −30 000 N s (same in both).
Braking: F = −30 000/4 = −7500 N ≈ 0.64 × car’s weight.
Wall collision: F = −30 000/0.1 = −300 000 N ≈ 25.4 × car’s weight.
3.1 With crumple zones: F = −30 000/0.2 = −150 000 N. This is half the no-crumple-zone force (150 000/300 000 = 0.5). Crumple zones halve the average force by doubling Δt.
3.2 The student’s reasoning is incorrect. Crumple zones do not change Δp (the car still goes from 25 m/s to 0 in either case). They deliberately extend Δt so that the same Δp is delivered over a longer time. By J = FΔt = Δp, a longer Δt means a smaller F. The force on occupants is what injures or kills them — reducing F from 300 000 N to 150 000 N halves the deceleration force, which can be the difference between survival and fatality. The “weakness” is intentional: it is what saves lives.
Q4 — Catching technique
4.1 Positive direction: initial direction of ball = positive. vi = +35 m/s; vf = 0.
Δp = 0.16(0 − 35) = −5.6 N s (same in both cases).
Rigid hands: F = −5.6/0.01 = −560 N. “Giving” hands: F = −5.6/0.08 = −70 N.
4.2 Factor = 560/70 = 8×. The “giving” technique reduces the force on the hands by a factor of 8. Physiologically, 560 N can cause bruising, fractures, or dislocation in the finger joints. 70 N is well within the pain-free range for most people. Cricketers, outfielders in baseball, and goalkeepers intuitively extend Δt to avoid injury — this is the impulse-momentum theorem applied in real life.