Year 11 Physics Module 2 ⏱ ~45 min 5 MC · 3 Short Answer Lesson 14 of 15

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.

Today's hook: At the 2000 Sydney Olympics, Australian M4+ rowers applied 2,880 N·s of impulse per stroke to a 300 kg boat. Before calculating: which of the five Phase 3 formulae would you use to find the velocity change per stroke, and which to find the average force if each stroke lasted 0.8 s?

0/5TASKS

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Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Phase 3 consolidation worksheets.

BuildFoundations & guided practice ApplyApplication practice MasterMastery challenge ExamExam-style questions

Before you read, self-assess

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?

Cross-lesson links: L14 consolidates the entire Phase 3 toolkit from L11 (momentum and impulse), L12 (conservation of momentum, elastic vs inelastic), and L13 (two-stage synthesis). Every formula here connects back to Phase 1 ($F_{net} = ma$ from L04 underlies $F = \Delta p/\Delta t$) and Phase 2 (KE from L06 is tested in elastic collision checks). L15 then integrates all three phases in capstone exam problems.

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Phase 3 Formula Master Table

Core Content

Phase 3 Formula Master Table

+5 XP

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.

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Consolidation Worked Examples

+5 XP

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.

Worked Example ARebound Force

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.

Define positive = toward wall. $v_i = +5\text{ m/s}$; $v_f = -4\text{ m/s}$

$\Delta p = 0.2(-4-5) = 0.2 \times (-9) = -1.8\text{ kg m/s}$

$F = \Delta p/\Delta t = -1.8/0.01 = -180\text{ N}$ (away from wall)

Worked Example BElastic Collision Verification

Ball A (1 kg, 6 m/s) hits stationary Ball B (1 kg). After: A stops, B moves at 6 m/s. Elastic?

$\Sigma p_{before} = 1 \times 6 + 0 = 6\text{ kg m/s}$; $\Sigma p_{after} = 0 + 1 \times 6 = 6\text{ kg m/s}$ ✓

$\Sigma KE_{before} = \frac{1}{2}(1)(6^2) = 18\text{ J}$; $\Sigma KE_{after} = 0 + \frac{1}{2}(1)(6^2) = 18\text{ J}$ ✓ → 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:

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Collision Type Classification

Activities

Collision Type Classification

EvaluateBand 6

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.

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Multi-Concept Momentum Problem

EvaluateBand 6

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.

✓

Show what you have learned

Quick recall, Phase 3 Consolidation

+5 XP

Short Answer, 10 marks

+5 XP

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.

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Retrieve, reflect and finish

Check what actually stuck

Boss Battle, Checkpoint Quiz 3

boss

You have completed Phase 3 (Momentum). This checkpoint covers L11-L14. Beat it to unlock the capstone.

⚔ Enter Checkpoint 3

How did your thinking change?

The 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.

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Module overview · Physics · Checkpoint 3