Phase 3 Consolidation — Momentum Mastery
At the 2000 Sydney Olympics, the Australian men's coxed four (M4+) rowers each applied an impulse of 2,880 N·s per stroke. The 300 kg boat gained $\Delta v = J/m = 2{,}880/300 = 9.6\text{ m/s}$ in velocity per stroke — a direct application of $J = \Delta p$ that the Australian Institute of Sport used to optimise stroke rate and blade angle for gold-medal performance. This lesson consolidates all five Phase 3 formulae through problems of exactly this type.
Practise this lesson
Phase 3 consolidation worksheets.
Rate your confidence with each of the five Phase 3 formulae (1 = need help, 5 = confident).
Which formula is used when an object bounces off a surface and you need the average force of impact?
Core Content
A cricketer is hit on the helmet by a 156 km/h delivery. The ball contacts the helmet for 0.003 s and bounces back at 80 km/h. You need to find: the momentum change, the impulse delivered, the average force, and whether the collision was elastic. Each answer requires a different formula — and the wrong choice at any step gives a nonsense answer for the next.
| # | Formula | When to use | Key trap |
|---|---|---|---|
| 1 | $p = mv$ | Find momentum from mass and velocity | Vector — always state direction |
| 2 | $J = F\Delta t$ | Impulse from force and contact time | Convert ms to s |
| 3 | $J = \Delta p = mv_f - mv_i$ | Impulse from change in momentum | Direction reversal: use signs |
| 4 | $F = \Delta p/\Delta t$ | Average force from impulse and time | $F$ is average net force — not just applied force |
| 5a | $\Sigma p_{before} = \Sigma p_{after}$ | Any closed-system collision | Requires no external force on system |
| 5b | $\Sigma KE_{before} = \Sigma KE_{after}$ | Elastic collisions only | Test by calculating both sides |
Phase 3 formula selection: bouncing → $\Delta p = m(v_f - v_i)$ with signs, then $F = \Delta p/\Delta t$; collision/explosion → $\Sigma p_{before} = \Sigma p_{after}$; safety device → same $\Delta p$ but compare forces at different $\Delta t$ values; elastic/inelastic → calculate and compare $\Sigma KE$ before and after.
Pause — copy the highlighted Phase 3 selection guide into your book before moving on.
True or false: A closed system always has constant total momentum, regardless of whether the collision is elastic or inelastic.
We just saw the Phase 3 formula table — four scenario types, each with a specific approach. That raises a question: how do you move from recognising the scenario type to executing the calculation without errors? This card answers it → worked examples for each scenario type show the decision sequence in action.
A 0.2 kg ball hits a wall at 5 m/s and rebounds at 4 m/s. Contact time = 10 ms. Find average force.
Ball A (1 kg, 6 m/s) hits stationary Ball B (1 kg). After: A stops, B moves at 6 m/s. Elastic?
Ball A (3 kg, 4 m/s) collides with Ball B (3 kg, stationary). After: A moves at 1 m/s, B moves at 3 m/s. Is momentum conserved?
Two trolleys: A (2 kg, 5 m/s east) and B (3 kg, 2 m/s west). They collide and stick. The final velocity is:
Activities
For each scenario: (a) check if momentum is conserved, (b) calculate $\Sigma KE$ before and after, (c) classify as elastic, inelastic, or perfectly inelastic:
- Ball A (4 kg, 3 m/s) hits Ball B (4 kg, stationary). After: A stops, B moves at 3 m/s.
- Ball A (2 kg, 6 m/s) hits Ball B (2 kg, stationary). They stick together.
A 1500 kg car (north, 20 m/s) and a 3000 kg truck (south, 8 m/s) collide and stick. (a) Find the final velocity. (b) Calculate the impulse on the car during the collision. (c) If contact time is 0.2 s, find the average force on the car.
ApplyBand 4(3 marks) 1. A 0.3 kg ball hits a wall at 10 m/s and rebounds at 8 m/s in the opposite direction. Contact time = 8 ms. Calculate the average force of the wall on the ball.
AnalyseBand 5(3 marks) 2. Ball P (5 kg, 4 m/s east) collides with Ball Q (5 kg, stationary). After: P at 1 m/s east, Q at 3 m/s east. (a) Verify momentum is conserved. (b) Is this elastic? Show your working.
EvaluateBand 6(4 marks) 3. A 1200 kg car (north, 15 m/s) collides with a 900 kg car (south, 10 m/s) and they stick together. (a) Find the final velocity. (b) Calculate KE lost in the collision. (c) Classify the collision and explain where the lost KE went.
Show all answers
Short Answer — Model Answers
Q1 (3 marks): Define positive = toward wall. $v_i = +10\text{ m/s}$; $v_f = -8\text{ m/s}$. $\Delta p = 0.3(-8-10) = 0.3 \times (-18) = -5.4\text{ kg m/s}$. $\Delta t = 0.008\text{ s}$. $F = -5.4/0.008 = -675\text{ N}$ (away from wall, i.e. on the ball in the rebound direction). Magnitude = 675 N.
Q2 (3 marks): (a) $\Sigma p_{before} = 5 \times 4 + 0 = 20\text{ kg m/s}$. $\Sigma p_{after} = 5 \times 1 + 5 \times 3 = 5 + 15 = 20\text{ kg m/s}$ ✓ Momentum conserved. (b) $KE_{before} = \frac{1}{2}(5)(4^2) = 40\text{ J}$. $KE_{after} = \frac{1}{2}(5)(1^2) + \frac{1}{2}(5)(3^2) = 2.5 + 22.5 = 25\text{ J}$. Since $KE_{after} < KE_{before}$, this is an inelastic collision (15 J of KE was lost to heat/sound/deformation).
Q3 (4 marks): Define north as positive. (a) $\Sigma p = 1200 \times 15 + 900 \times (-10) = 18\,000 - 9000 = 9000\text{ kg m/s}$ north. $(1200 + 900)v_f = 9000$; $v_f = 9000/2100 \approx 4.29\text{ m/s}$ north. (b) $KE_{before} = \frac{1}{2}(1200)(15^2) + \frac{1}{2}(900)(10^2) = 135\,000 + 45\,000 = 180\,000\text{ J}$. $KE_{after} = \frac{1}{2}(2100)(4.29^2) \approx \frac{1}{2}(2100)(18.4) \approx 19\,320\text{ J}$. $KE_{lost} = 180\,000 - 19\,320 = 160\,680\text{ J} \approx 161\text{ kJ}$. (c) Perfectly inelastic collision — the cars stick together. The 161 kJ of kinetic energy was converted to heat (friction/deformation of metal), sound, and the plastic deformation of the car bodies during the crash. This energy is non-recoverable.
You have completed Phase 3 (Momentum). This checkpoint covers L11-L14. Beat it to unlock the capstone.
⚔ Enter Checkpoint 3The 2000 Sydney Olympics M4+ rowers illustrate all five formulae in one scenario: $p = mv$ gives boat momentum, $J = \Delta p = 2{,}880\text{ N·s}$ gives the impulse per stroke, $\Delta v = J/m = 9.6\text{ m/s}$ gives the speed gain, $F = \Delta p/\Delta t$ gives average blade force, and $\Sigma p_{before} = \Sigma p_{after}$ applies when blades release at stroke end. The key insight: momentum is always conserved in closed systems, but kinetic energy is only conserved in elastic collisions.