Momentum and conservation of momentum (HT) — AQA GCSE Physics

SPEC 4.5.7 • MOMENTUM AND CONSERVATION

HIGHER TIER ONLY

Momentum & Conservation

A vector quantity — and the minus sign that decides whether you score full marks or zero on every head-on collision question.

Why This Matters HT only

Momentum appears on every Paper 2 Higher Tier paper — often as a 4-to-6-mark calculation. Examiner reports flag the same two failure modes year after year.

~60%

Defined "momentum is conserved" correctly (Nov 2020)

−

The minus sign for opposing motion: most commonly missed mark

p = mv

Recall equation. Memorise — never given on the sheet

The lesson focus

Two ideas. First: momentum is a vector, so direction (and the right sign) is half of every answer. Second: momentum is conserved in a closed system — total before = total after. Master both, and head-on-collision marks become free.

Momentum — Mass × Velocity

Momentum is the property that determines how hard it is to stop a moving object. A truck at 10 m/s carries more "motion" than a bicycle at 10 m/s — that's its momentum.

Definition (Recall Word-for-Word)

Momentum is the product of an object's mass and velocity.

It is a vector quantity — it has both magnitude and direction.

p = m × v

SymbolQuantitySI Unit

pMomentumkilogram metre per second ( kg m/s )

mMasskilogram (kg)

vVelocitymetre per second (m/s)

Equation status

p = mv is on the recall list — it is not printed on the AQA Physics equation sheet. You must remember it. (It is an HT-only equation, so you'll never need it for Foundation.)

Direction Matters — Sign Convention

Because momentum is a vector, two objects with the same mass and the same speed but moving in opposite directions have momenta with opposite signs. Getting this right is the single biggest mark-earner on this topic.

The Most-Missed Mark

If two objects move towards each other, one velocity is positive and one is negative. Add their momenta with signs intact:

p_total = m₁v₁ + m₂v₂ (not |m₁v₁| + |m₂v₂|)

Examiner reports flag this as the single biggest cause of lost marks in conservation questions.

Conservation of Momentum

The principle that solves every collision and explosion question. Provided no external force acts, the total momentum of a system never changes.

Mark-Scheme Definition

In a closed system, the total momentum before an event is equal to the total momentum after the event.

Σ p_before = Σ p_after

What "closed system" means

A closed system is one where no external force acts on the objects involved. The forces during the collision (e.g. one car pushing on the other) are internal — they cancel out (Newton's 3rd law) and don't change the total momentum. External forces would be friction with the ground, air resistance, or gravity along the direction of motion. Mark schemes typically idealise these as zero unless stated otherwise.

Where this applies

Collisions — two objects hit each other and either bounce apart or stick together

Explosions / recoil — one object splits into two (or more) parts that move apart

The Nov 2020 Q05 hockey question tested exactly this — and only ~60% of students got the 1-mark definition right

Mark scheme alternative

The mark scheme also accepts the shorter wording "(total) momentum stays the same" for the 1-mark definition. But the "before = after" form is more useful for calculations, so learn that one.

Worked Example 1 — Simple p = mv

Start by applying the equation to a single moving object.

A car of mass 1400 kg is travelling north at 18 m/s. Calculate the momentum of the car. 2 MARKS

E

Equation: p = m × v

S

Substitute: p = 1400 × 18

U

Unit: kilogram metre per second (kg m/s)

A

Answer: p = 25 200 kg m/s (or 2.52 × 10⁴ kg m/s) north

💡

Note: Because momentum is a vector, the direction "north" is part of a complete answer. Mark schemes don't always demand the direction explicitly — but they never penalise you for including it.

Worked Example 2 — Sticking Collision

The classic AQA question: one object moving, one stationary, they stick together. Use conservation: total momentum before = total momentum after.

A car of mass 1200 kg moving east at 8.0 m/s collides with a stationary car of mass 800 kg. The two cars stick together. Calculate their combined velocity. 4 MARKS

E

Equation: p_before = p_after, with p = mv applied to each object.

S

Before: p = (1200 × 8.0) + (800 × 0) = 9600 kg m/s east.

After: the combined mass is (1200 + 800) = 2000 kg, moving at velocity v. So 2000 × v = 9600.

U

Rearrange: v = 9600 ÷ 2000

A

Answer: v = 4.8 m/s east

💡

Note: The stationary car still counts toward the final combined mass — it gained the moving car's momentum, not the moving car's velocity.

Worked Example 3 — Head-On Collision

The trap question. Two objects moving towards each other. Set one direction as positive — the other direction must be negative.

Two ice skaters skate towards each other and collide, holding on together after the impact. Skater A has mass 60 kg and velocity +6.0 m/s (east). Skater B has mass 80 kg and velocity −3.0 m/s (west). Calculate their combined velocity after the collision. 4 MARKS

E

Equation: p_before = p_after. Keep signs intact.

S

Before: p = (60 × +6.0) + (80 × −3.0) = (+360) + (−240) = +120 kg m/s.

After: total mass = 60 + 80 = 140 kg, combined velocity v. So 140v = +120.

U

Rearrange: v = 120 ÷ 140

A

Answer: v = +0.86 m/s (to 2 s.f.) — they move east after the collision.

⚠️

Mark loss check: Adding magnitudes — (60 × 6) + (80 × 3) = 600 — gives a velocity of 4.3 m/s. Wrong: the signs of opposing velocities must be tracked. The Nov 2020 examiner report flagged "the minus sign for Player B was often missing" as the biggest mark loss.

Explosions and Recoil

A stationary object splits into pieces. Total momentum before = zero, so total momentum after must also equal zero. The pieces fly apart with equal and opposite momenta.

0 = m₁v₁ + m₂v₂ ⇒ m₁v₁ = −m₂v₂

A stationary 3.0 kg rifle fires a 0.012 kg bullet which leaves the barrel at +500 m/s. Calculate the recoil velocity of the rifle. 3 MARKS

E

Equation: Conservation. Total p before = 0, so total p after = 0.

S

After: (0.012 × 500) + (3.0 × v_rifle) = 0 ⇒ 6.0 + 3.0v = 0 ⇒ v = −6.0 ÷ 3.0

A

Answer: v_rifle = −2.0 m/s (2.0 m/s in the opposite direction to the bullet — the recoil)

💡

Why the rifle moves slowly: the rifle has 250× more mass than the bullet, so its recoil velocity is 250× smaller. Same momentum, much less speed. The same physics governs rocket propulsion: hot gases expelled backwards push the rocket forward.

Investigating Collisions in the Lab

The AQA spec requires that you can describe how to investigate collisions between laboratory trolleys. This isn't a numbered required practical, but it's a fair Paper 2 method-writing question.

ElementWhat you do

SetupLinear (low-friction) track; two trolleys with cards mounted on top; two light gates connected to a data logger.

Measure massWeigh each trolley on a balance. Record m₁ and m₂.

Measure velocityLight gate 1 records the time the card breaks the beam; v = card length ÷ time. Repeat with gate 2 after the collision.

ProcessCalculate p_before = m₁v₁ (Trolley B initially at rest). Calculate p_after = (m₁ + m₂)v_combined. Compare.

ControlSame track angle (level), same trolleys, same light-gate positions — so any difference between the two momenta is due to the collision alone.

Expected resultp_before ≈ p_after — small discrepancies attributable to friction (an external force) acting during the collision.

Weak vs Strong — Sign Handling

Side-by-side, the same scenario worked correctly and incorrectly.

Scenario: a 60 kg object at +6 m/s collides head-on with an 80 kg object at −3 m/s. Stick together.

✗ Wrong (≤1 mark)

p_before = (60 × 6) + (80 × 3) = 600 kg m/sAdded the magnitudes — ignored that the second object moves the opposite way. Result: combined velocity ≈ 4.3 m/s (way too fast, wrong direction logic).

✓ Correct (full marks)

p_before = (60 × +6) + (80 × −3) = +120 kg m/sSigned velocities, signed momenta. Combined velocity = 120/140 ≈ +0.86 m/s east — small and slow, matching the partial cancellation.

How to never forget

Pick the positive direction before you start the calculation, and write it down (e.g. "take east as positive"). Then anything moving the other way is automatically negative. This habit prevents 90% of the marks lost on this topic.

Five Errors That Cost Marks

Verified from examiner reports on AQA Higher Tier Paper 2 over multiple years.

Students often writeCorrect version

"p = m + v"p = m × v (multiplication, not addition)

Adds magnitudes when objects move in opposite directionsApply signs: one velocity positive, the other negative

"Momentum is conserved" with no mention of the closed system"In a closed system , the total momentum before = total momentum after"

Ignores the mass of the stationary object in the "after" totalThe stationary object has zero momentum, but its mass is still part of the combined mass after the collision

Omits the unit "kg m/s" or writes "kg/m/s" / "kg·m·s⁻¹" with wrong layoutThe unit is kilogram metre per second (kg m/s) — three components, one division

Examiner Alert HT only

Nov 2020 P2 Q05.1 — defining "momentum is conserved"

Approximately 60% of students answered this 1-mark definition correctly. The mark scheme accepts either "(total) momentum before = (total) momentum after" or "(total) momentum stays the same". The 40% who lost the mark typically confused conservation with the definition of momentum itself (p = mv), or claimed each object's momentum was conserved — when it is the total that stays the same.

Nov 2020 P2 Q05.2 — the missing minus sign

Two ice hockey players moving towards each other (one at +7.5 m/s, the other at −5.5 m/s) collide and move off together. The examiner report explicitly noted: "The minus sign for player B was often missing." Students who calculated both initial momenta correctly but added magnitudes instead of using signs lost the final mark.

Pattern

Conservation of momentum questions reward structured arithmetic. Show the equation, substitute with signs intact, calculate, give the unit. The marks are at each step — and the sign is its own marking point.

Practice Question

A trolley of mass 0.80 kg moves to the right at 1.2 m/s along a straight track. It collides with a stationary trolley of mass 1.2 kg. The two trolleys do not stick together — after the collision, the first trolley continues to move to the right at 0.30 m/s.

Calculate the velocity of the second trolley immediately after the collision.4 MARKS

E

Equation: p_before = p_after. Take "right" as positive.

S

Before: p = (0.80 × 1.2) + (1.2 × 0) = 0.96 kg m/s.

After: p = (0.80 × 0.30) + (1.2 × v) = 0.24 + 1.2v.

Setting equal: 0.24 + 1.2v = 0.96.

U

Rearrange: 1.2v = 0.96 − 0.24 = 0.72 ⇒ v = 0.72 ÷ 1.2

A

Answer: v = +0.60 m/s (the second trolley moves to the right at 0.60 m/s)

⚠️

Mark loss check: A common mistake is to forget to include the final velocity of the first trolley (0.30 m/s) in the "after" total. If the first trolley keeps moving, its momentum after the collision is non-zero and must be subtracted from the total when finding the second trolley's velocity.

Checkpoint — What You Can Now Do

✅ You can now

Define momentum as the product of mass and velocity, a vector quantity with units kg m/s

Recall and apply p = mv (recall list, HT only)

State conservation of momentum in mark-scheme language: in a closed system, total momentum before = total momentum after

Calculate combined velocity in a sticking collision

Handle signs in head-on collisions — choose a positive direction; opposite motion is negative

Apply conservation to explosions and recoil (total momentum from rest = 0)

Describe the lab investigation using trolleys, light gates and a data logger

Next lesson HT + Physics only

5.13 — Changes in Momentum. The link between force and momentum change: F = m Δv / Δt. The physics behind every car safety feature — crumple zones, airbags, seat belts — extending Δt to reduce the average force. Equation sheet equation, frequently 6 marks.