AQA GCSE Physics 4.5

Momentum

p = mv | Conservation of Momentum | Impulse = FΔt = Δp

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Calculate momentum

Use p = mv to find the momentum of moving objects, including correct SI units (kg m/s).

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Conservation of momentum

State and apply the principle of conservation of momentum to collisions and explosions.

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Collisions vs explosions

Distinguish between collisions (objects come together or bounce) and explosions (objects push apart).

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Impulse

Understand and calculate impulse using FΔt = Δp, and relate it to change in momentum.

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Force and time

Explain how increasing contact time reduces the force required to produce the same change in momentum.

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Vector nature

Treat momentum as a vector quantity and use consistent sign conventions in calculations.

What is Momentum?

Momentum is a fundamental property of any moving object. It describes how difficult it is to stop that object — a heavy lorry moving at speed has far more momentum than a tennis ball moving at the same speed, and is much harder to stop.

Momentum (p) is defined as the product of an object's mass and its velocity.

p = m × v

Symbol	Quantity	SI Unit
p	Momentum	kg m/s
m	Mass	kg
v	Velocity	m/s

Momentum is a vector quantity — it has both magnitude and direction. Direction is determined by the direction of velocity. Always define a positive direction before solving problems!

For example, if we define rightwards as positive (+), then an object moving left has a negative velocity and therefore negative momentum. This sign convention is essential when objects collide or move in opposite directions.

The unit of momentum is kg m/s (sometimes written as N s — these are equivalent). A car of mass 1200 kg travelling at 15 m/s has a momentum of 18 000 kg m/s. A bullet of mass 0.01 kg travelling at 400 m/s has a momentum of only 4 kg m/s — much less than the car, despite its high speed.

Conservation of Momentum

One of the most powerful laws in physics is the conservation of momentum. It applies to any closed system where no external resultant force acts.

Principle of Conservation of Momentum: In a closed system, the total momentum before an event is equal to the total momentum after the event.

Total momentum before = Total momentum after
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Where u₁ and u₂ are the velocities before, and v₁ and v₂ are the velocities after the event.

Collisions

There are two main types of collision:

Elastic collision: Objects bounce off each other. Both momentum and kinetic energy are conserved. (Rare in practice — some bouncing balls approximate this.)
Inelastic collision: Objects stick together after collision (perfectly inelastic) or some kinetic energy is lost as heat/sound. Momentum is still conserved but kinetic energy is NOT.

When objects stick together after a collision, they move with a common final velocity v:

m₁u₁ + m₂u₂ = (m₁ + m₂) × v

Explosions

An explosion is the reverse of a perfectly inelastic collision. Two objects that are initially at rest (total momentum = 0) push apart. After the explosion, one moves left and one moves right. Because total momentum must still equal zero:

m₁v₁ + m₂v₂ = 0 → m₁v₁ = −m₂v₂

This means the two objects have equal and opposite momenta. A classic example is a gun recoiling when a bullet is fired — the bullet goes forwards, the gun kicks backwards.

Conservation of momentum works because of Newton's Third Law: every action has an equal and opposite reaction. The forces objects exert on each other during a collision are equal and opposite, so the changes in momentum cancel out.

Impulse

Newton's Second Law tells us that a resultant force causes a change in momentum. When a force acts over a period of time, we call the product impulse.

Impulse is the product of the resultant force and the time for which it acts. It equals the change in momentum of the object.

Impulse = F × Δt = Δp = m × Δv

Symbol	Quantity	SI Unit
F	Resultant force	N
Δt	Time interval	s
Δp	Change in momentum	kg m/s (= N s)

Notice the units: N × s = kg m/s², so impulse measured in N s is the same as momentum measured in kg m/s.

Practical Importance: Reducing Force by Increasing Time

If the change in momentum is fixed (e.g. bringing a moving car to rest), then from F × Δt = Δp, we can see that increasing Δt decreases F. This is why:

🚗 Crumple zones in cars increase the time of a collision, reducing the force on passengers.
🥊 Crash mats and gym mats extend the time of landing, reducing the impact force.
🎾 Catching a cricket ball with a "give" — drawing your hands back increases Δt and reduces the force felt.
🪂 Parachutes slow descent gradually, reducing the impact force on landing.

On a force–time graph, the area under the graph equals the impulse (= change in momentum). This is true even when the force varies with time.

Vector Signs and Sign Convention

Because momentum is a vector, getting the signs right is critical. Always follow these steps:

Step 1: Choose a positive direction (usually rightwards or the direction of the first object's motion).
Step 2: Assign positive (+) values to velocities in that direction and negative (−) values to velocities in the opposite direction.
Step 3: Calculate total momentum before and set equal to total momentum after.
Step 4: Solve for the unknown. Check the sign of your answer to determine the direction of motion.

If v is positive → object moves in the positive direction
If v is negative → object moves in the negative (opposite) direction

Example of sign importance: A ball of mass 0.5 kg moving right at 6 m/s bounces off a wall and returns left at 4 m/s. Define right as positive:

Momentum before: p = 0.5 × 6 = +3 kg m/s
Momentum after: p = 0.5 × (−4) = −2 kg m/s
Change in momentum: Δp = −2 − (+3) = −5 kg m/s

The wall exerted an impulse of 5 N s to the left on the ball. By Newton's Third Law, the ball exerted 5 N s to the right on the wall.

In explosion problems starting from rest, the total momentum is zero before AND after. The two pieces always move in opposite directions with equal and opposite momenta.

Example 1: A car of mass 1400 kg is travelling at 20 m/s. Calculate its momentum.

1Write out the formula for momentum.
p = m × v

2Substitute the known values.
p = 1400 kg × 20 m/s

3Calculate the result.
p = 28 000 kg m/s

Momentum = 28 000 kg m/s (in the direction of travel)

Example 2 — Collision: A trolley of mass 2 kg moving at 5 m/s to the right collides with a stationary trolley of mass 3 kg. They stick together after the collision. Calculate their common velocity after the collision.

1Define positive direction: rightwards = positive (+).
Write the conservation of momentum equation for objects that stick together.
m₁u₁ + m₂u₂ = (m₁ + m₂) × v

2Substitute known values. The stationary trolley has u₂ = 0.
(2 × 5) + (3 × 0) = (2 + 3) × v
10 + 0 = 5 × v

3Solve for v.
v = 10 ÷ 5 = 2 m/s

4Check the sign: v is positive, so the joined trolleys move to the right. ✓

Common velocity after collision = 2 m/s to the right

Example 3 — Explosion: A gun of mass 3 kg fires a bullet of mass 0.02 kg at 400 m/s. Calculate the recoil velocity of the gun.

1Before the shot, everything is at rest. Total momentum before = 0.
Define rightwards (bullet direction) as positive (+).

2Apply conservation of momentum.
0 = m_bullet × v_bullet + m_gun × v_gun
0 = (0.02 × 400) + (3 × v_gun)
0 = 8 + 3 × v_gun

3Solve for v_gun.
3 × v_gun = −8
v_gun = −8 ÷ 3 = −2.67 m/s

4The negative sign means the gun moves in the negative direction (to the left — recoil). ✓

Recoil velocity of gun = 2.67 m/s to the left (opposite to the bullet)

Example 4 — Impulse: A footballer kicks a ball of mass 0.45 kg, changing its velocity from rest to 18 m/s in 0.08 s. Calculate (a) the impulse and (b) the average force exerted on the ball.

1Calculate the change in momentum (= impulse).
Δp = m × Δv = 0.45 × (18 − 0) = 0.45 × 18 = 8.1 kg m/s

2State the result for impulse.
Impulse = Δp = 8.1 N s

3Use Impulse = F × Δt to find the force.
F × Δt = 8.1
F × 0.08 = 8.1

4Solve for F.
F = 8.1 ÷ 0.08 = 101.25 N

(a) Impulse = 8.1 N s | (b) Average force = 101 N (3 s.f.)

Q1. A ball of mass 0.3 kg moves at 10 m/s. What is its momentum?

Q2. A 5 kg object moving at 4 m/s collides with a stationary 5 kg object. They stick together. What is their velocity after the collision?

Q3. Which of the following is always conserved in a collision, even if kinetic energy is not?

Q4. A force of 50 N acts on a 2 kg ball for 0.4 s. Calculate the change in momentum of the ball. Give your answer in kg m/s.

Q5. A rocket in space (initially at rest) ejects gas of mass 8 kg at 600 m/s backwards. The rocket has mass 500 kg. What is the rocket's forward velocity after the ejection? Give your answer in m/s to 2 decimal places.

Challenge 1 — Head-on Collision:
Object A (mass 4 kg) moves right at 6 m/s. Object B (mass 2 kg) moves left at 3 m/s. They collide and stick together.
(a) Calculate the total momentum before the collision.
(b) Find the velocity of the combined object after the collision, including direction.

Challenge 2 — Impulse and Force:
A car of mass 1200 kg travelling at 25 m/s brakes to a halt.
(a) Calculate the change in momentum of the car.
(b) If the braking force is 6000 N, calculate the time taken to stop.
(c) Explain why crumple zones in cars reduce the risk of injury in a crash.

Challenge 3 — Explosion / Recoil:
A stationary radioactive nucleus of mass 210 u decays. It emits an alpha particle of mass 4 u at 1.5 × 10⁷ m/s.
(a) State the total momentum before decay.
(b) Calculate the recoil velocity of the daughter nucleus (mass 206 u).
(c) In what direction does it move relative to the alpha particle?

Challenge 4 — Extended Response:
A 0.06 kg tennis ball travelling at 20 m/s is hit back at 15 m/s in the opposite direction by a racket. The contact time is 0.005 s.
(a) Calculate the change in momentum of the ball.
(b) Calculate the average force exerted by the racket on the ball.
(c) By Newton's Third Law, what can you say about the force on the racket?