Year 11 Physics Module 2: Dynamics 40 min Lesson 12 of 15

Conservation of Momentum

Before a gun fires, the total momentum of the gun-bullet system is zero. After — the bullet races forward and the gun kicks back. Add them together: still zero. Momentum was conserved.

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Think First

An astronaut floating in space (no gravity, no air resistance) throws a 2 kg wrench forward at 4 m/s. The astronaut has a mass of 80 kg. What happens to the astronaut — and how fast do they move? Write your prediction before reading on.

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Formula Reference — This Lesson

$\Sigma p_{before} = \Sigma p_{after}$

Conservation of Momentum: total momentum is conserved in a closed system (no net external force)

$m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'$

Expanded form — v₁, v₂ before; v₁', v₂' after. For perfectly inelastic: v₁' = v₂' = v_f

Perfectly inelastic (stick): $v_f = (m_1v_1 + m_2v_2)/(m_1+m_2)$ | Explosion from rest: $0 = m_1v_1' + m_2v_2' \Rightarrow v_2' = -m_1v_1'/m_2$

p=p'
Formula Reference — Conservation of Momentum

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$\Sigma p_{before} = \Sigma p_{after}$

Conservation of Momentum (Law)

Σp = total (vector) momentum of all objects in the system

Applies when: the system is closed — no net external force acts. Internal forces (objects acting on each other) do not change the total momentum.

Common trap: Gravity and friction are external forces that can violate conservation. In practice, for short-duration collisions, these external forces act for too little time to significantly change momentum — but for longer interactions, check whether the system is truly closed.

$m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'$

Conservation — Two Object Collision

Primed (') quantities = after collision. Unprimed = before collision.

Use when: two objects collide and separate. Must know 3 of the 4 velocities to find the fourth — or know the collision type (perfectly inelastic → v₁' = v₂').

Common trap: Objects moving in opposite directions have opposite signs. A ball coming toward you at 5 m/s and one moving away at 5 m/s have momenta of opposite sign — not the same momentum.

$(m_1 + m_2)v_f = m_1v_1 + m_2v_2$

Perfectly Inelastic Collision

Objects stick together after collision → single final velocity v_f

Use when: objects combine (clay hitting clay, car crash where cars lock together, bullet embedding in block)

Common trap: Maximum KE loss occurs in perfectly inelastic collisions — but not all KE is lost unless the combined object is stationary. If v_f ≠ 0, the system still has KE.

$m_1v_1' = -m_2v_2'$

Explosion from Rest

Both objects start at rest → total p = 0 → momenta after are equal and opposite

Use when: system starts at rest and separates (gun + bullet, rocket ejecting gas, astronaut pushing off spacecraft)

Common trap: The two momenta are equal in magnitude and opposite in direction — they do NOT mean equal speeds. A small-mass object gets a high speed; a large-mass object gets a low speed.

Know

Law of conservation of momentum
Condition: closed system, no net external force
Elastic, inelastic, perfectly inelastic definitions
Explosion from rest: m₁v₁' = −m₂v₂'

Understand

Why Newton's Third Law leads to conservation
Why KE is not conserved in most collisions
Why a small-mass object gets a higher recoil speed
How to identify collision type from the scenario

Can Do

Solve perfectly inelastic collisions for v_f
Solve explosion-from-rest for recoil velocity
Calculate KE before and after and find KE lost
Classify a collision as elastic, inelastic, or perfectly inelastic

Misconceptions to Fix

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Wrong: Zero acceleration means an object is stationary.

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Right: Zero acceleration means constant velocity — the object could be moving at constant speed in a straight line.

📚 Core Content

Key Terms

total momentumconserved in a closed system (no net external force)

the systemclosed — no net external force acts

Gravity and frictionexternal forces that can violate conservation

but not all KElost unless the combined object is stationary

momenta afterequal and opposite

The two momentaequal in magnitude and opposite in direction — they do NOT mean equal speeds

01The Law of Conservation of Momentum

The Law of Conservation of Momentum

In a closed system with no net external force, the total momentum before any interaction equals the total momentum after — exactly.

This law emerges directly from Newton's Third Law. When object A exerts a force on object B, object B exerts an equal and opposite force on A. Both forces act for the same time Δt. By the impulse-momentum theorem, the impulse on each object equals its change in momentum — so A's momentum change is equal and opposite to B's momentum change. The total change in momentum of the system is zero.

Derivation from Newton's Third Law:
F_AB = −F_BA (Newton 3)
F_AB × Δt = −F_BA × Δt (same contact time)
J_AB = −J_BA (equal and opposite impulses)
Δp_A = −Δp_B (equal and opposite momentum changes)
Therefore: Δp_A + Δp_B = 0 → total momentum is conserved

Conservation of Momentum — Before and After

Scenario type	Example	Total p before	Total p after	Conserved?
Collision — same direction	Car rear-ends another car	m₁v₁ + m₂v₂	m₁v₁' + m₂v₂'	Yes (closed system)
Collision — opposite directions	Head-on crash	m₁v₁ + (−m₂v₂)	Depends on outcome	Yes
Perfectly inelastic	Objects lock together	m₁v₁ + m₂v₂	(m₁ + m₂)v_f	Yes
Explosion from rest	Gun fires bullet	0	m₁v₁' + m₂v₂' = 0	Yes
Elastic	Billiard ball collision	m₁v₁ + m₂v₂	m₁v₁' + m₂v₂'	Yes (and KE too)

02Applying Conservation in One Dimension

Applying Conservation in One Dimension

Conservation of momentum is an equation. Set up the left side (total p before) equal to the right side (total p after) — then solve for the unknown.

Three Scenario Setups

General collisionObjects separate after

Equation: m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

Unknown: One of v₁', v₂' — need the other given

Key condition: Must know 3 of 4 velocities

Perfectly inelasticObjects stick together

Equation: m₁v₁ + m₂v₂ = (m₁ + m₂)v_f

Unknown: v_f — the single final velocity

Key condition: v₁' = v₂' = v_f (they move as one)

Explosion from restSystem starts at rest

Equation: 0 = m₁v₁' + m₂v₂'

Unknown: One of v₁', v₂' — need the other given

Key condition: Total p before = 0; momenta are equal and opposite

Vector Protocol — conservation problems

Step 1 — Define positive direction and assign signs to all velocities before substituting

Step 2 — Write the full conservation equation before plugging in numbers

Step 3 — Identify the scenario type — perfectly inelastic or explosion changes the number of unknowns

Step 4 — State the direction of the answer, not just the magnitude

03Elastic vs Inelastic Collisions

Elastic vs Inelastic Collisions

All collisions conserve momentum. Only elastic collisions also conserve kinetic energy — in real life, most collisions lose some KE to heat, sound, and deformation.

Elastic

Momentum: Conserved

Kinetic Energy: Conserved

Real example: Billiard balls, ideal gas molecules, some atomic/nuclear collisions

KE check: KE_before = KE_after

Inelastic

Momentum: Conserved

Kinetic Energy: Not conserved — some lost to heat, sound, deformation

Real example: Most real collisions — cars, sports balls, thrown objects

KE check: KE_after < KE_before

Perfectly inelastic

Momentum: Conserved

Kinetic Energy: Maximum KE lost — objects stick together

Real example: Clay-on-clay, car crash (locked bumpers), bullet embedding in block

KE check: KE_after minimum for given Δp

To classify a collision Calculate KE before and after. If KE_after = KE_before → elastic. If KE_after < KE_before → inelastic (or perfectly inelastic if objects stuck). KE_after can never exceed KE_before in a collision — that would violate energy conservation. If your calculation gives KE_after > KE_before, you have made an error.

Common Misconceptions

In a perfectly inelastic collision, all kinetic energy is lost.

KE is only fully lost if the combined object ends up stationary (v_f = 0). This only happens when the two objects have equal and opposite momenta before the collision. In all other perfectly inelastic collisions, the combined object still moves — so it still has KE. The KE lost equals the difference, not the total.

Momentum and kinetic energy are both conserved in all collisions.

Momentum is always conserved in a closed system. KE is only conserved in elastic collisions. Most real collisions are inelastic — KE is lost to internal energy (heat, sound, deformation). The two conservation laws are independent.

In a gun recoil, the gun moves at the same speed as the bullet.

Momenta are equal and opposite (m_gun × v_gun = m_bullet × v_bullet). Since the gun is far heavier than the bullet, the gun's recoil speed is far smaller. A 0.01 kg bullet at 400 m/s gives p = 4 N s. A 2 kg gun therefore recoils at 4/2 = 2 m/s — 200× slower than the bullet.

Interactive: Explosion Momentum Simulator

Interactive: Inelastic Collision (3D)

✏️ Worked Examples

Worked Example 1 Type 5 — Perfectly Inelastic Collision

Problem Setup

Problem type: Type 5 — Perfectly inelastic collision. Objects stick together.

Scenario: A 1200 kg car travelling at +15 m/s collides with a stationary 800 kg car. They lock together. Find: (a) their common velocity after the collision, (b) the KE lost.

m₁ = 1200 kg, v₁ = +15 m/s
m₂ = 800 kg, v₂ = 0
Perfectly inelastic: v₁' = v₂' = v_f
Positive direction: direction of initial motion = positive

Solution

Σp_before = m₁v₁ + m₂v₂ = 1200 × 15 + 800 × 0 = 18 000 N s

Total momentum before. Stationary car has zero momentum.

Σp_after = (m₁ + m₂)v_f = 2000 × v_f

Perfectly inelastic — combined mass after collision. Single unknown v_f.

18 000 = 2000 × v_f → v_f = +9 m/s (forward)

Conservation: p_before = p_after. Positive result confirms forward motion.

(b) KE_i = ½ × 1200 × 15² = 135 000 J. KE_f = ½ × 2000 × 9² = 81 000 J

Calculate KE before and after separately using KE = ½mv².

KE lost = 135 000 − 81 000 = 54 000 J (40% of initial KE)

KE lost = initial − final. Converted to heat, sound, deformation during the crash.

What would change if...

The stationary car were also 1200 kg instead of 800 kg. Would the final speed be higher or lower? Would the KE lost as a percentage of initial KE be higher or lower? Calculate to check your intuition.

Worked Example 2 Type 5 — Explosion from Rest

Problem Setup

Problem type: Type 5 — Explosion from rest. Total p = 0 before and after.

Scenario: A 70 kg astronaut at rest pushes off a 1000 kg spacecraft. The astronaut moves away at +3 m/s. Find the recoil velocity of the spacecraft.

m_ast = 70 kg, v_ast' = +3 m/s
m_sc = 1000 kg, v_sc' = ?
Both at rest initially: p_total = 0
Positive direction: direction astronaut moves = positive

Solution

p_total before = 0 (system at rest)

Both objects stationary before the push. Total momentum = zero.

0 = m_ast × v_ast' + m_sc × v_sc' = 70 × 3 + 1000 × v_sc'

Conservation: p_after = 0. Set up equation with both unknown and known values.

1000 × v_sc' = −210 → v_sc' = −0.21 m/s

Negative sign confirms spacecraft moves in the opposite direction to the astronaut (recoil). Speed = 0.21 m/s — much slower because spacecraft is ~14× heavier.

Check: p_ast = 70 × 3 = +210 N s. p_sc = 1000 × (−0.21) = −210 N s. Sum = 0 ✓

Verification — momenta are equal and opposite, sum to zero. Conservation confirmed.

What would change if...

The astronaut pushed harder and reached +6 m/s instead. What would the spacecraft's recoil velocity be? Does doubling the astronaut's speed double the spacecraft's recoil speed — and does doubling the astronaut's speed double their KE?

Visual Break

Copy into your books

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Conservation Law

Σp_before = Σp_after (closed system)
Derived from Newton's Third Law
Always define positive direction first
Momentum is a vector — signs matter

Collision Types

Elastic: momentum AND KE conserved
Inelastic: momentum conserved, KE lost
Perfectly inelastic: objects stick, max KE lost
Check by calculating KE before and after

Perfectly Inelastic Setup

m₁v₁ + m₂v₂ = (m₁ + m₂)v_f
One unknown: v_f (combined velocity)
KE lost = KE_initial − KE_final
v_f may still be non-zero — KE not all lost

Explosion from Rest

Total p = 0 before → 0 = m₁v₁' + m₂v₂'
Therefore m₁v₁' = −m₂v₂' (opposite directions)
Smaller mass → larger speed (same momentum)
Gun/bullet, rocket, astronaut pushing off spacecraft

🏃 Activities

Activity 01 — Pattern A

Classify and Draw Momentum Diagrams

For each scenario: (a) classify the collision type, (b) draw a labelled before/after momentum vector diagram with arrows proportional to momentum magnitude, (c) state whether KE is conserved.

#	Scenario	Type	KE conserved?
1	Pool cue ball hits 8-ball — 8-ball moves, cue ball stops
2	1000 kg car at 20 m/s hits stationary 1000 kg car — they lock bumpers
3	10 g bullet fired from 3 kg rifle at rest — bullet leaves at 300 m/s
4	400 g clay ball hits 600 g stationary clay ball — they merge
5	Astronaut (80 kg) pushes off 500 kg space station from rest
6	0.5 kg ball at +6 m/s hits 0.5 kg ball — first ball stops, second moves at +6 m/s

Type your classification and KE answers here. Draw diagrams in your book.

Complete the table and draw momentum vector diagrams in your book.

Complete the table and draw diagrams in your book

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Activity 02 — Pattern A

Classify and Calculate

For each problem: identify the collision type, write the correct equation, and solve. For the last problem, determine if the proposed outcome is physically possible.

A 2 kg ball at +5 m/s hits a stationary 3 kg ball. They stick together. Find v_f.
A 0.008 kg bullet is fired from a 4 kg rifle initially at rest. The bullet leaves at +380 m/s. Find the rifle's recoil velocity.
A 600 kg car at +12 m/s collides with a 400 kg car at −8 m/s (opposite direction). They stick together. Find v_f and state the direction.
A student claims that after a collision, a 2 kg object has velocity +8 m/s and a 3 kg object has velocity +4 m/s. Before the collision, the 2 kg object was at +10 m/s and the 3 kg object was at rest. Is this outcome physically possible? Show working to justify.

Type your working for all four parts below.

Complete all four parts in your book — show full working and collision classification.

Complete all four parts in your book

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✅ Check Your Understanding

Revisit Your Thinking

Earlier you were asked: An astronaut throws a 2 kg wrench forward at 4 m/s. What happens to the astronaut?

The full answer: the astronaut moves backward. Before the throw, total p = 0 (both at rest). After: p_wrench = 2 × 4 = +8 N s (forward). By conservation: p_astronaut = −8 N s. The astronaut (80 kg) therefore moves at v = −8/80 = −0.1 m/s (backward at 0.1 m/s).

This is how astronauts manoeuvre in the void — by throwing objects (or expelling gas from thrusters). The same principle governs rocket propulsion: the rocket expels gas backward at high speed, and the rocket itself is pushed forward. There is no external surface to push against — only the conservation of momentum.

Look back at your initial prediction. What did you get right? What changed?

Annotate your initial prediction in your book with what you now understand.

Annotate your initial prediction in your book

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Revisit Your Initial Thinking

Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?

Multiple Choice

5 random questions from a replayable lesson bank — feedback shown immediately

UnderstandBand 3

7. State the law of conservation of momentum and the condition required for it to apply. Use Newton's Third Law to explain why momentum is conserved in a collision between two objects. 3 MARKS

Answer in your book

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ApplyBand 5

8. A 1500 kg car travelling at +20 m/s collides with a stationary 1000 kg car. They lock together after the collision. (a) Calculate the final velocity. (b) Calculate the kinetic energy before and after. (c) Calculate the KE lost and express as a percentage of initial KE. 3 MARKS

Answer in your book

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EvaluateBand 6

9. A 0.5 kg ball travelling at +8 m/s collides with a stationary 1.5 kg ball. After the collision the 0.5 kg ball travels at −2 m/s. (a) Apply the Vector Protocol and find the final velocity of the 1.5 kg ball. (b) Calculate the KE before and after the collision. (c) Classify the collision type with full justification. 4 MARKS

Answer in your book — full Vector Protocol required

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Comprehensive Answers

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Activity 01 — Classification

1. Elastic — cue ball stops, 8-ball takes all momentum. Check: both have same mass, same speed → KE conserved. Yes, KE conserved.

2. Perfectly inelastic — lock bumpers. KE not conserved. p_before = 1000 × 20 = 20 000 N s. v_f = 20 000/2000 = +10 m/s. KE_i = 200 000 J, KE_f = 100 000 J. 50% lost.

3. Explosion from rest. KE increases — chemical energy of gunpowder converts to KE. Total p = 0. Rifle: 0 = 0.01 × 300 + 3 × v_rifle → v_rifle = −1 m/s.

4. Perfectly inelastic (merge). KE not conserved. v_f = (0.4 × v₁)/(0.4 + 0.6) — need v₁ of first ball to complete.

5. Explosion from rest. 0 = 80 × v_ast + 500 × v_station → momenta equal and opposite.

6. Elastic — first ball stops, second takes all momentum, same speed. KE conserved (identical masses, speed transfer). Check: KE_before = ½ × 0.5 × 36 = 9 J. KE_after = ½ × 0.5 × 36 = 9 J. ✓

Activity 02 — Calculate

1. Perfectly inelastic: (2+3)v_f = 2x5 + 0 = 10. v_f = +2 m/s

2. Explosion: 0 = 0.008x380 + 4xv_rifle. v_rifle = -3.04/4 = -0.76 m/s (recoil)

3. Perfectly inelastic: (600+400)v_f = 600x12 + 400x(-8) = 7200 - 3200 = 4000. v_f = +4 m/s (forward)

4. p_before = 2x10 + 3x0 = 20 N s. p_after (claimed) = 2x8 + 3x4 = 16+12 = 28 N s. 20 ≠ 28 — NOT physically possible. Violates conservation of momentum.

Multiple Choice

1. C — +2 N s. p_total = 3×4 + 5×(−2) = 12 − 10 = +2 N s. Objects moving in opposite directions have opposite signs.

2. B — +4 m/s. m₁v₁ + m₂v₂ = (m₁+m₂)v_f: 4×6 + 0 = 6×v_f → v_f = 24/6 = +4 m/s.

3. A — −1.67 m/s. Explosion from rest: 0 = 0.01×500 + 3×v_rifle → v_rifle = −5/3 = −1.67 m/s. Negative sign = recoil backward.

4. D. Elastic collision, equal masses: first ball stops completely, second takes all the momentum at +8 m/s. This is a special result of elastic collisions between equal masses. Option A gives the same total momentum but wrong distribution; option C fails both conservation laws.

5. C. KE lost = KE_initial − KE_final. If the combined object is still moving (v_f ≠ 0), it still has KE — so not all KE is lost. Only if v_f = 0 is all KE lost.

6. B. p_before = 2×10 = 20 N s. p_after = 2×4 + 3×4 = 8 + 12 = 20 N s. Momentum IS conserved — this outcome is physically possible. However, KE_before = ½×2×100 = 100 J; KE_after = ½×2×16 + ½×3×16 = 16 + 24 = 40 J. KE decreased — this is an inelastic collision, not elastic. The outcome is possible, but not elastic.

Short Answer — Model Answers

Q7 (3 marks): The law of conservation of momentum states that the total momentum of a closed system remains constant — the total momentum before any interaction equals the total momentum after. The condition required is that no net external force acts on the system (a closed system). By Newton's Third Law, when object A collides with object B, A exerts a force F on B and B simultaneously exerts an equal and opposite force −F on A. Both forces act for the same contact time Δt. By the impulse-momentum theorem, the impulse on A equals its change in momentum: FΔt = Δp_A. The impulse on B is equal and opposite: −FΔt = Δp_B. Therefore Δp_A + Δp_B = 0 — the total change in momentum of the system is zero — momentum is conserved.

Q8 (3 marks):

(a) p_before = 1500x20 + 0 = 30 000 N s. (1500+1000)v_f = 30 000. v_f = 30 000/2500 = +12 m/s

(b) KE_i = 1/2 x 1500 x 400 = 300 000 J. KE_f = 1/2 x 2500 x 144 = 180 000 J

Q9 (4 marks):

Step 1: Positive direction = initial direction of 0.5 kg ball

(a) 0.5x8 + 1.5x0 = 0.5x(-2) + 1.5xv2' -> 4 = -1 + 1.5v2' -> v2' = 5/1.5 = +3.33 m/s

(b) KE_before = 1/2 x 0.5 x 64 = 16 J. KE_after = 1/2 x 0.5 x 4 + 1/2 x 1.5 x 11.11 = 1 + 8.33 = 9.33 J

(c) KE lost = 16 - 9.33 = 6.67 J. KE is not conserved (KE_after < KE_before). Classification: INELASTIC. Momentum is conserved (p_before = 4 N s, p_after = 0.5x(-2) + 1.5x3.33 = -1 + 5 = 4 N s ✓). KE is not conserved — energy was lost to sound, heat, and deformation. Objects did not stick (not perfectly inelastic). Therefore: inelastic collision.

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