What is momentum in physics and how is it calculated?

Momentum in physics is a measure of the quantity of motion an object has. It is calculated as the product of an object's mass and its velocity. The formula is given by p = mv, where p represents momentum, m is the mass in kilograms, and v is the velocity in meters per second. Understanding momentum is crucial for analyzing motion and collisions in AQA GCSE Physics.

How does the principle of conservation of momentum apply in collisions?

The principle of conservation of momentum states that in a closed system, the total momentum before a collision is equal to the total momentum after the collision, provided no external forces act on the system. This principle is fundamental in solving problems related to collisions in AQA GCSE Physics, allowing students to predict the outcomes of interactions between objects.

What is the difference between elastic and inelastic collisions in terms of momentum?

In an elastic collision, both momentum and kinetic energy are conserved. This means that the total momentum and total kinetic energy before and after the collision remain the same. In contrast, in an inelastic collision, momentum is conserved, but kinetic energy is not. Some kinetic energy is transformed into other forms of energy, such as heat or sound. Understanding these differences is essential for AQA GCSE Physics students studying forces and motion.

How does impulse relate to momentum in physics?

Impulse is the change in momentum of an object when a force is applied over a period of time. It is calculated as the product of the force and the time duration over which the force acts, given by the formula Impulse = Force × Time. In AQA GCSE Physics, impulse is a key concept for understanding how forces affect the motion of objects, particularly in scenarios involving sudden changes, such as collisions.

Why is momentum considered a vector quantity?

Momentum is considered a vector quantity because it has both magnitude and direction. The direction of momentum is the same as the direction of the object's velocity. This vector nature is important in AQA GCSE Physics, as it affects how momentum is analyzed in different directions, especially in two-dimensional collision problems where the direction of motion plays a crucial role in solving equations.

A two-trolley collision before and after

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At a glance

$p = m v$ — momentum of a moving object.
Unit: kg·m/s.
Momentum is a vector — direction matters.
Stationary object: p = 0.
Larger mass or higher speed → larger momentum.
HT only.
Conservation of momentum — in a closed system: total momentum before = total momentum after.
Applies to collisions (objects come together) and explosions (objects fly apart).
Momentum is a vector — directions and signs matter.
Two main collision types: elastic (KE conserved) and inelastic (KE not conserved). Momentum is conserved in BOTH.
Before an explosion, total p = 0; afterwards, the two pieces fly in opposite directions with equal and opposite momenta.
HT only — Foundation Tier students do NOT sit this.
Newton's 2nd Law in momentum form: $F = \dfrac{\Delta p}{\Delta t}$ (force = rate of change of momentum).
Δp is the change in momentum (kg·m/s).

What you should be able to do

4.5.7.1 (HT) — Define momentum.
4.5.7.1 (HT) — Recall and apply $p = mv$ .
4.5.7.1 (HT) — Identify momentum as a vector.
4.5.7.2 (HT) — State the principle of conservation of momentum.
4.5.7.2 (HT) — Apply $p = m v$ and conservation of momentum to collisions and explosions in one dimension.
4.5.7.2 (HT) — Carry out calculations involving the conservation of momentum.
4.5.7.3 (HT) — State $F = \dfrac{m \Delta v}{\Delta t}$ as a form of Newton's second law.
4.5.7.3 (HT) — Recall and apply $F = \dfrac{\Delta p}{\Delta t}$ .
4.5.7.3 (HT) — Use the equation in calculations involving collisions.
4.5.7.3 (HT) — Explain how increasing the time over which momentum changes reduces the force experienced — and link this to safety devices (crumple zones, seat belts, airbags, cycle helmets, gymnasium mats).

Study notes

Momentum p = mv

Product of mass and velocity — a vector.

Momentum is defined as the product of mass and velocity:

$p = m v$

$p$ in kg·m/s (no special name).
$m$ in kg.
$v$ in m/s (must include direction).

Because velocity is a vector, momentum is also a vector. Sign matters: a car going +20 m/s has momentum +20m kg·m/s; the same car at −20 m/s has momentum −20m.

Worked example. A 1500 kg car at 25 m/s east. $p = 1500 \times 25 = 37{,}500\,\text{kg·m/s east}$

Worked example. A 0.06 kg tennis ball at 50 m/s. $p = 0.06 \times 50 = 3.0\,\text{kg·m/s}$

Notice the car's momentum is 12,500× larger — heavier objects at moderate speeds dominate over lighter objects at high speeds.

p = mv (kg·m/s).
Vector quantity.
Stationary: p = 0.
HT only.

The principle of conservation of momentum

In a closed system the total momentum is the same before and after any interaction.

Conservation of momentum states:

In a closed system (one with no external forces acting on it), the total momentum is conserved — it is the same before and after any interaction (collision, explosion, push apart…).

Mathematically, for two objects A and B before and after:

$m_A u_A + m_B u_B = m_A v_A + m_B v_B$

where $u$ is initial velocity, $v$ is final velocity.

'Closed system' means no external resultant force acts. In practice this means we ignore friction, air resistance and other external pushes during the very short time of the interaction. (For a real-world collision in the lab, this is a very good approximation.)

Momentum is a vector. When working in 1D, choose a positive direction (say, "right is positive"). Anything moving left has a negative velocity, and therefore a negative momentum. Add momenta with their signs.

Why is momentum conserved? Newton's third law — when A pushes B with a force $F$ , B pushes A with $-F$ . These equal and opposite forces act for the same time, so they produce equal and opposite changes in momentum. The total momentum doesn't change.

A simple two-trolley collision. Whatever momentum the system has before the collision, it has the same total momentum afterwards.

Total p before = total p after.
Only valid for a closed system (no external forces).
Momentum is a vector — use + and − for directions in 1D.
Follows from Newton's third law.

Worked example — a 1D collision

Apply m_A u_A + m_B u_B = m_A v_A + m_B v_B with signs.

Worked example. A 1500 kg car moving at 20 m/s east hits a stationary 1000 kg car. The two stick together. Find their common velocity afterwards.

Step 1 — Write down what we know.

$m_A = 1500$ kg, $u_A = +20$ m/s.
$m_B = 1000$ kg, $u_B = 0$ .
Stick together → both have the same final velocity $v$ .

Step 2 — Apply conservation of momentum.

$m_A u_A + m_B u_B = (m_A + m_B) v$

Step 3 — Substitute.

$1500 \times 20 + 1000 \times 0 = 2500 \, v$

$30{,}000 = 2500 \, v$

$v = 12 \text{ m/s east}$

Worked example 2 — a 'before' and 'after' check. Trolley A (2 kg, 4 m/s east) hits trolley B (3 kg, 1 m/s east). After the collision, A moves at 1 m/s east. Find B's new velocity.

$2(4) + 3(1) = 2(1) + 3(v_B)$

$8 + 3 = 2 + 3 v_B$

$v_B = 3 \text{ m/s east}$

Always check: did the momenta add up the same on both sides? $8 + 3 = 11$ ; $2 + 9 = 11$ ✓.

Write m_A u_A + m_B u_B = m_A v_A + m_B v_B.
Include signs for directions.
If objects stick together, use a single final velocity (m_A + m_B)v.
Always check the total momentum balances.

Explosions and recoil

Before an explosion, p = 0. After, the pieces fly apart with equal and opposite momenta.

An explosion is the time-reversed version of a collision: an object at rest spontaneously breaks into two pieces (e.g. a spring released between two trolleys, or a bullet fired from a gun).

Before: total momentum = 0 (everything at rest). After: by conservation, total momentum must STILL be 0.

So the two pieces fly off with equal-and-opposite momenta.

$m_1 v_1 + m_2 v_2 = 0 \quad\Rightarrow\quad m_1 v_1 = -m_2 v_2$

Worked example — a rifle recoils. A 5 kg rifle fires a 0.020 kg bullet at 400 m/s. Find the recoil speed of the rifle.

Before firing: both at rest, total p = 0.

After: $m_b v_b + m_r v_r = 0$

$0.020 \times 400 + 5 \times v_r = 0$

$8 + 5 v_r = 0 \quad\Rightarrow\quad v_r = -1.6 \text{ m/s}$

The rifle recoils at 1.6 m/s in the OPPOSITE direction. The minus sign tells us the direction.

Notice — the rifle is much heavier than the bullet, so the rifle has a much smaller speed, even though its momentum is equal in magnitude.

Explosion: total p before = 0; total p after = 0.
Pieces fly off with equal and opposite momenta.
Heavier piece → smaller speed (same momentum magnitude).
Examples: recoil of guns, rockets, two trolleys released by a spring.

Common pitfall

Forgetting that momentum is a vector. In an explosion, just because two pieces have 'equal momentum magnitudes' does not mean they have equal speeds — heavier objects move slower for the same magnitude of momentum.

Force = rate of change of momentum (4.5.7.3)

Newton's 2nd Law can be written as F = Δp/Δt. This is the form used in collision and safety problems.

Newton's second law can be written in two equivalent ways:

$F = m a \qquad \text{(force–acceleration form, used earlier)}$

$F = \frac{\Delta p}{\Delta t} = \frac{m \, \Delta v}{\Delta t} \qquad \text{(momentum form, this section)}$

Where:

$F$ is the resultant force acting on the object (N).
$\Delta p = m \Delta v$ is the change in momentum (kg·m/s).
$\Delta t$ is the time taken for the change (s).

The two are equivalent because $a = \Delta v / \Delta t$ , so $F = m a = m \Delta v / \Delta t = \Delta p / \Delta t$ .

Why use the momentum form? Because it makes the time over which a collision happens an explicit factor — and that's exactly what safety design changes.

Worked example. A 60 kg cyclist moving at 8 m/s hits a wall and stops in 0.1 s. Find the average force on the cyclist.

$\Delta p = m \Delta v = 60 \times (0 - 8) = -480$ kg·m/s.
$F = \Delta p / \Delta t = -480 / 0.1 = -4800$ N.

The minus sign just means the force opposes the motion. Magnitude = 4800 N — about 8× the cyclist's body weight.

If the cyclist had been wearing a helmet that allowed the head to slow down over 0.5 s instead of 0.1 s:

$F = -480 / 0.5 = -960$ N.

The force is 5× smaller simply because the time is 5× longer.

F = Δp/Δt — momentum form of Newton's 2nd Law.
Δp = m Δv.
Same Δp over a longer Δt → smaller force.
Equivalent to F = ma.

Crumple zones, airbags and other safety features

All car and bike safety devices work by lengthening the time of the collision so the force on the body is reduced.

If a 70 kg adult in a car stops from 14 m/s in just 0.02 s (hitting a rigid steering column), the force on them is:

$F = \frac{m \Delta v}{\Delta t} = \frac{70 \times 14}{0.02} = 49{,}000 \text{ N}$

That's about 70× body weight — almost certainly fatal.

If the same person decelerates over 0.5 s (because the crumple zone, seat belt and airbag all stretch the deceleration time):

$F = \frac{70 \times 14}{0.5} = 1960 \text{ N}$

Now it's just under 3× body weight — survivable.

Safety feature	What it does	Why it reduces force
Crumple zones (front/rear of car)	Bonnet & boot designed to deform on impact	Increases stopping time Δt; F = Δp/Δt drops
Seat belts	Slightly stretch under load, holding the occupant	Increases the time over which the body decelerates
Airbags	Inflate within ~30 ms and deflate as the head presses into them	Spread the deceleration over a longer time AND a wider area
Cycle / motorcycle helmets	Polystyrene foam crushes on impact	Crushes over ~10 ms instead of 1 ms, cutting force ~10×
Gymnasium / crash mats	Compress under the gymnast	The mat increases the time taken to stop
Playground rubber surfaces	Compress on impact	Increase stopping time → less force on the child

The common principle: same Δp (same momentum change must happen — you ARE going to stop), so increasing Δt is the only lever we have to reduce F.

Don't write: "the helmet absorbs the impact". That sounds vague — examiners want to see Δt explicitly: "the helmet increases the time over which the head stops, so by F = Δp/Δt the force on the head is reduced".

Two identical cars stop with the same change of momentum. The crumple zone makes the stop take longer — slashing the force on the occupants.

Same Δp, larger Δt → smaller F.
Crumple zones, seat belts, airbags, helmets, mats — all increase Δt.
Reduced force → reduced injury.
Use the formula in your explanation, not just the word 'absorbs'.

Common pitfall

Saying 'airbags reduce the change in momentum'. They do NOT — the body must still stop (same Δp). They reduce the FORCE by extending the TIME of the collision.

Calculation practice with F = Δp/Δt

Plug in mass, change in velocity and time — and don't forget directions.

Example A — a tennis ball. A 60 g tennis ball travelling at 30 m/s east is struck by a racket and rebounds at 25 m/s west. The contact lasts 5 ms. Find the average force on the ball.

$m = 0.060$ kg.
Choose east = positive. $u = +30$ m/s, $v = -25$ m/s.
$\Delta v = v - u = -25 - 30 = -55$ m/s.
$\Delta p = m \Delta v = 0.060 \times (-55) = -3.3$ kg·m/s.
$F = \Delta p / \Delta t = -3.3 / 0.005 = -660$ N.

Magnitude 660 N — and the minus sign confirms the force on the ball is westward (back towards the racket).

Example B — a car braking. A 1200 kg car going at 18 m/s brakes uniformly to rest in 6 s. Find the average braking force.

$\Delta p = 1200 \times (0 - 18) = -21{,}600$ kg·m/s.
$F = -21{,}600 / 6 = -3600$ N.

The braking force is 3600 N opposite to the motion.

Example C — choose Δt to limit F. Engineers want to ensure the maximum force on a 70 kg crash-test dummy moving at 20 m/s is no more than 5000 N during a collision. What stopping time is needed?

$\Delta p = 70 \times 20 = 1400$ kg·m/s.
$\Delta t \geq \Delta p / F = 1400 / 5000 = 0.28$ s.

So the airbag and crumple zone together must extend the stopping time to at least 0.28 s.

Δv = v − u (take direction into account).
Force is in the direction of Δp.
If the object rebounds, Δv is larger than just the incoming speed — easy to forget!
Solve for Δt to design a safety feature.

Quick recap

p = mv.
Unit: kg·m/s.
Momentum is a vector — direction matters.
Used in conservation of momentum (4.5.7.2) and impulse (4.5.7.3).
HT only.
Conservation of momentum: total p before = total p after (closed system).
Equation: m_A u_A + m_B u_B = m_A v_A + m_B v_B.
Apply to collisions and explosions in 1D.
Use + and − for directions.
Heavier objects move more slowly for the same momentum.
F = Δp/Δt = m Δv/Δt — momentum form of Newton's 2nd Law.
Δp is the change in momentum (kg·m/s); Δt is the time it takes (s).

Exam tips

Include direction when stating momentum.
Convert g → kg, mph → m/s before calculating.
Use signs (+ / −) to represent opposite directions in 1D problems.
Always state 'in a closed system' when quoting the principle.
Choose a positive direction and stick with it for all velocities.
If two objects stick together, treat them as one combined mass.
Check your answer by computing total p before and total p after.
Recoil questions: total momentum starts at zero, so the gun's momentum = −bullet's momentum.
Quote 'force = rate of change of momentum' or the equation F = Δp/Δt verbatim — both score marks.
Explain safety features using Δt explicitly, NOT vague phrases like 'absorbs the impact'.

Frequently asked questions

Sources: AQA GCSE Physics 8463 specification — section 4.5.7.1 (HT); AQA GCSE Physics 8463 specification — section 4.5.7.2 (HT); AQA GCSE Physics 8463 specification — section 4.5.7.3 (HT) · Last reviewed 2026-05-25 · AQA GCSE · Physics 8463 Higher Tier

Momentum – Study Notes & Past Paper Style Questions | AQA GCSE Physics (8463) Higher Tier | Tutopiya

Momentum

At a glance

What you should be able to do

Study notes

Momentum p = mv

The principle of conservation of momentum

Worked example — a 1D collision

Explosions and recoil

Force = rate of change of momentum (4.5.7.3)

Crumple zones, airbags and other safety features

Calculation practice with F = Δp/Δt

Quick recap

Exam tips

Frequently asked questions

Why is momentum a vector?

What if mass changes (e.g. rocket losing fuel)?

Is momentum conserved in a real-world car crash?

Is kinetic energy conserved too?

What's the difference between a collision and an explosion?

Why does a gun recoil?

What's the difference between this and F = ma?

Why do airbags really help if they only act for a tenth of a second?

Do safety features change the change in momentum?

Why does a rebounding ball experience a bigger force than one that just stops?