Changes in momentum (HT, physics only) — AQA GCSE Physics

SPEC 4.5.7.3 • CHANGES IN MOMENTUM

HIGHER TIER

PHYSICS ONLY

Changes in Momentum

The link between force and momentum change — and why crumple zones, airbags, seat belts and helmets save lives. Only 10% of students scored full marks on the Jun 2023 crumple-zones question. Here's the chain that wins them.

Why This Matters HT Physics only

This is the single most-failed safety-features question on Paper 2 Higher Tier. The mark scheme is short — three points, three marks — but the answers students write almost never include all three.

10%

Scored all 3 marks on crumple zones (Jun 2023 P2 Q05.4)

F = mΔv/Δt

On the AQA equation sheet — but only if you know to use it

Δt ↑ ⇒ F ↓

Every safety feature works by this single principle

The lesson focus

One equation, one chain. The equation: F = mΔv/Δt. The chain: extending Δt → decreasing the rate of change of momentum → reducing the force. Every safety feature on the spec — crumple zones, airbags, seat belts, helmets — operates on this single principle.

From F = ma to F = mΔv/Δt

The spec explicitly mentions how this equation is derived. Two equations you already know combine to give the new one.

Equation 1 — Newton's 2nd law

F = m × a

Force = mass × acceleration. On the AQA equation sheet.

Equation 2 — definition of a

a = Δv / t (or (v − u)/t)

Acceleration = change in velocity ÷ time. Also on the equation sheet.

Combining them

Substitute a = Δv/Δt into F = ma:

F = m × (Δv / Δt) = mΔv / Δt

Notice that mΔv = the change in momentum (since p = mv, so Δp = mΔv). So:

F = Δp / Δt

Force = the rate of change of momentum. This phrase is the heart of every mark-scheme answer on this topic.

Equation sheet alert

F = mΔv/Δt is on the AQA Physics equation sheet — you don't have to recall it. But you must know how to apply it, and which variables are which.

The Equation — Variables and Units

F = m Δv / Δt

SymbolQuantitySI Unit

FAverage force during the changenewton (N)

mMasskilogram (kg)

ΔvChange in velocity = v_final − v_initialmetre per second (m/s)

ΔtTime over which the change occurssecond (s)

Two equivalent forms

You'll see this equation written in either of two ways — both mean the same thing:

F = mΔv / Δt — solve for force given velocity change and time

F × Δt = m × Δv — the product of force and time equals the change in momentum (Δp)

Δv = final − initial

This is the most common arithmetic error. For a car coming to rest from 15 m/s: Δv = 0 − 15 = −15 m/s. The negative sign means the velocity decreased. The force comes out negative too — meaning it acted against the motion. Most exam answers ask for the magnitude of the force, so the sign is dropped at the end.

Worked Example 1 — Stopping Force

The textbook application: a moving object brought to rest in a known time. Find the average force.

A car of mass 800 kg travels at 15 m/s. It collides with a wall and comes to rest in 0.20 s. Calculate the average force exerted on the car during the collision. 3 MARKS

E

Equation: F = m Δv / Δt (on the equation sheet)

S

Substitute: Δv = 0 − 15 = −15 m/s. F = (800 × −15) / 0.20

U

Calculate: F = −12 000 / 0.20 = −60 000 N

A

Answer: The magnitude of the average force is 60 000 N = 6.0 × 10⁴ N. The negative sign in the working shows the force acts opposite to the car's motion (a decelerating force).

💡

Note: The change in momentum is mΔv = 800 × 15 = 12 000 kg m/s, applied over 0.20 s, gives 60 000 N. The numbers come out cleanly because the time is a round value.

The Key Insight — F ∝ 1/Δt

For the same change in momentum (same mass, same Δv), the average force depends entirely on how long the impact lasts. Double the time, half the force. Quadruple the time, quarter the force.

Why area = momentum change

Force × time has units of N·s — the same units as momentum (since 1 N·s = 1 kg m/s). The area under any force-time graph equals the impulse FΔt = Δp = mΔv. The two collisions above shed the same momentum (12 000 kg m/s); the crumple zone spreads it over 4× more time, so the average force is 4× smaller.

Worked Example 2 — Comparing Collision Times

The same car, the same collision — but two different impact durations. The force changes dramatically.

A car of mass 1200 kg travelling at 12 m/s hits a wall. Case (a): the front of the car is rigid and stops in 0.040 s. Case (b): a crumple zone deforms and the car stops in 0.16 s. Calculate the average force on the car in each case. 4 MARKS

E

Equation: F = m Δv / Δt (equation sheet). Same Δv = 12 m/s for both cases.

S

Case (a): F = (1200 × 12) / 0.040 = 14 400 / 0.040 = 360 000 N = 3.6 × 10⁵ N

S

Case (b): F = (1200 × 12) / 0.16 = 14 400 / 0.16 = 90 000 N = 9.0 × 10⁴ N

A

Ratio: The crumple-zone collision lasts 4× longer, so the average force is 4× smaller (360 kN ÷ 90 kN = 4). This 75% reduction in force is what reduces injury.

Safety Features — All Extend Δt

The spec lists four safety features. Every one of them works on the same principle: extend the time of impact to reduce the average force. Different mechanisms, same physics.

FeatureHow it extends Δt

Crumple zonesThe front (and rear) of the car is designed to crush progressively during impact, increasing the time over which the car decelerates.

AirbagsInflate within milliseconds of a collision and compress as the passenger's head and chest hit them — slowing the body over a longer time than a hard surface would.

Seat beltsStretch slightly during a sudden deceleration, extending the time over which the passenger's body slows down.

Cycle helmetsThe foam interior crushes on impact, slowing the head over a longer time than the bare skull would experience hitting a hard surface.

Beyond cars — also on the spec

The spec also lists two non-car examples that work by the same principle. Be ready to apply F = mΔv/Δt to either of them in an exam.

Gymnasium crash mats — the foam compresses when a gymnast lands, extending the time over which their body decelerates and reducing the force on their joints.

Cushioned surfaces for playgrounds (e.g. rubber matting under climbing frames) — extend the time during which a falling child decelerates, reducing the force of impact. Already tested in Jun 2019 P2 with a child jumping from a swing.

PEE Worked Example — Crumple Zones

The 3-mark question that costs 90% of students at least one mark. The mark scheme is one chain of three links — miss any one and you can't score full marks.

Explain how crumple zones reduce the force exerted on the passengers of a car during a collision. 3 MARKS

P

Point (MP1): Crumple zones increase the time taken for the collision to occur.

E

Evidence (MP2): The change in momentum (Δp = mΔv) of the car is the same, but it happens over a longer time. Therefore the rate of change of momentum decreases.

E

Explain (MP3): Since F = Δp / Δt, a smaller rate of change of momentum means a smaller average force on the passengers.

⚠️

Mark loss check: The 90% who lost marks typically stopped at "the force is smaller" without explaining why. The middle link — the rate of change of momentum — is the marking point most often missed.

Weak vs Strong — Mechanism Language

Side by side, the language that earns marks vs the language that doesn't.

✗ 0 marks

"Crumple zones absorb the force of the collision."Wrong mechanism. Crumple zones don't "absorb" force — they reduce it by extending the impact time. The verb is wrong, the physics is wrong.

✓ 3 marks

"Crumple zones increase the time of the collision, so the rate of change of momentum decreases, reducing the force on the passengers."All three mark scheme points present, in the correct logical order.

✗ Insufficient

"Airbags reduce the impact.""Reduces impact" is the verbatim mark-scheme-rejected phrase. Without a time mechanism and a force consequence, it scores zero — the answer doesn't say how or what physical quantity is reduced.

✓ 3 marks

"Airbags extend the time over which the head decelerates. For the same change in momentum, this reduces the rate of change of momentum, so the force on the head is smaller."Time → rate of change of momentum → force. The full chain.

Five Errors That Cost Marks

Each of these is flagged explicitly in the AQA Jun 2023 mark scheme or examiner report — they will not be credited.

Students often writeWhy it fails / Correct version

"Crumple zones absorb the force"Wrong mechanism. Crumple zones extend Δt ; this reduces the force. The car body doesn't soak up newtons.

"Crumple zones slow down time "Mark scheme verbatim: do not accept . Time doesn't slow; the duration of the impact increases.

"Airbags reduce the impact "Mark scheme verbatim: insufficient . Need to name time as the mechanism and force as the consequence.

"Airbags reduce the momentum "Wrong quantity. Momentum change (Δp = mΔv) is the same in both cases — what changes is the time over which it happens.

Δv = v_initial + v_final (added)Δv = v_final − v_initial (subtracted). A car coming to rest has Δv = 0 − u = −u, not +u.

Examiner Alert HT Physics only

Jun 2023 P2 Q05.4 — crumple zones (3 marks)

Only 10% of students scored all 3 marks. Mark-scheme chain required: (1) increases the time of the collision, (2) so the rate of change of momentum decreases, (3) reducing the force on the passengers. Most students reached point 1 and skipped to "force is reduced" without naming the rate-of-change-of-momentum link in between.

Verbatim rejections from the mark scheme

"slows down time" — DO NOT accept

"reduces impact" — insufficient on its own (must link to force)

"absorbs the force" — wrong mechanism, scores zero

The exact phrase to memorise

"Increases the time of the collision → the rate of change of momentum decreases → the force is reduced." Three links, three marks. Practise writing them as a single sentence with the equation F = Δp/Δt as your justification.

Practice Question

A cyclist of mass 70 kg is travelling at 8.0 m/s. They are thrown from the bike and brought to rest by hitting the ground.

(a) Without a helmet, the cyclist's head stops in 0.020 s. Calculate the average force on the head during this stop. 2 MARKS

(b) With a helmet, the helmet's foam crushes and the head stops in 0.080 s. Calculate the average force on the head during this stop. 2 MARKS

(c) Using the equation F = m Δv / Δt, explain why a cycle helmet reduces the risk of head injury. 2 MARKS

(a)

Without helmet: F = (70 × 8.0) / 0.020 = 560 / 0.020 = 28 000 N = 2.8 × 10⁴ N

(b)

With helmet: F = (70 × 8.0) / 0.080 = 560 / 0.080 = 7 000 N = 7.0 × 10³ N

(c) MP1

The helmet's foam crushes and increases the time over which the head decelerates (from 0.020 s to 0.080 s — 4× longer).

(c) MP2

For the same change in momentum (560 kg m/s), increasing Δt reduces the rate of change of momentum, so the force on the head is smaller (4× smaller, from 28 kN to 7 kN). Less force means less risk of injury.

⚠️

Mark loss check: Part (c) is only 2 marks but tests the same chain as the 3-mark crumple-zone question. Don't stop at "the force is smaller" — say why (time-and-momentum link).

Module 5 Wrap-Up — Forces

This is the last lesson of Module 5. You've now covered the full Forces topic — from vectors and scalars at the start to the link between force and momentum here at the end.

The skills bank you've built

Vectors vs scalars; resolving forces (HT in 2D); free-body diagrams

Weight (W = mg); centre of mass; resultant forces and equilibrium

Work done (W = Fs); elastic potential energy (Eₑ = ½ke²); RP6

Moments, levers, gears; pressure in fluids and atmosphere

Motion graphs: distance-time, velocity-time, and the HT uniform-acceleration equation v² − u² = 2as

Newton's laws; F = ma; RP7; terminal velocity

Stopping distance: thinking + braking; v² dependence; energy chain

Momentum (HT): p = mv; conservation; explosions and recoil

This lesson: F = mΔv/Δt; safety features and the time-force-momentum chain

Cross-topic synthesis (where examiners love to test)

Module 5 connects most strongly to Module 1 (Energy) — KE, work done, and energy transfers underpin every braking and crumple-zone question. The synoptic links from F = ma, energy conservation, and momentum will recur in Paper 2 questions throughout the rest of the course.

Checkpoint — What You Can Now Do

✅ You can now

Derive F = mΔv/Δt from F = ma and a = Δv/Δt

Identify that the equation is on the AQA Physics equation sheet

Interpret F = Δp/Δt as "force = rate of change of momentum"

Calculate the average force in a collision given mass, velocity change and time

Compare forces in two collisions with the same Δp but different Δt

Explain how crumple zones, airbags, seat belts and cycle helmets all work by extending Δt

Write the 3-link mark-scheme chain: time increases → rate of change of momentum decreases → force decreases

Avoid the rejected phrases: "absorbs the force", "slows down time", "reduces the impact" alone

Next module — Waves

Module 6 starts with 6.01 — Transverse and Longitudinal Waves. Wave vocabulary (wavelength, amplitude, frequency, period), the difference between transverse and longitudinal, and the wave equation. EM waves, sound, water waves, and the start of RP8 (measuring wave speed).