OxfordAQA International GCSE Physics: Momentum, collisions and safety in road transport (9203)

This guide follows sections 3.1.4 (Momentum) and 3.1.5 (Safety in public transport) of the OxfordAQA International GCSE Physics (9203) specification, for exams May/June 2018 onwards. The qualification is untiered. Section 3.1.4 is marked P in the specification, so it is part of the full Physics award. For short recall notes use the revision notes, and for questions with answers use the practice questions.

Related pages: the OxfordAQA International GCSE Physics hub, the printable checklist, the wider forces overview, the resultant forces guide and the free diagnostics.

What this unit covers

Spec	What you must be able to do
3.1.4 a	Know that moving objects have momentum; use p = mv
3.1.4 b	Use conservation of momentum in a closed system, for two-object collisions and explosions
3.1.4 c	Use F = Δp / t; explain qualitatively air bags, seat belts, side impact bars, crumple zones, crash mats, playground surfaces and cycle helmets
3.1.5 a	Explain that at a steady speed in a straight line the resistive forces balance the driving force
3.1.5 b	Link speed, braking force, deceleration and stopping distance; state the risks of large decelerations
3.1.5 c	Split stopping distance into thinking distance and braking distance; list what affects reaction time
3.1.5 d	Explain braking as work done by friction, reducing kinetic energy and heating the brakes
3.1.5 e	Explain how wet or icy roads and poor brakes or tyres change braking distance

Momentum (3.1.4 a)

Every moving object has momentum. Momentum depends on both mass and velocity, and it has a direction, the same as the velocity.

momentum = mass × velocity, so p = mv

The unit of momentum is the kg m/s. Because it has direction, choose one direction as positive and give the opposite direction a negative sign.

Worked example 1. A trolley of mass 0.75 kg moves at 1.6 m/s. A loaded lorry of mass 8500 kg moves at 12 m/s. Find both momenta.

Trolley: p = 0.75 × 1.6 = 1.2 kg m/s. Lorry: p = 8500 × 12 = 102 000 kg m/s.

Conservation of momentum (3.1.4 b)

In a closed system, the momenta of all the objects add up to the same total before and after they interact. A closed system is one where no external forces act, so nothing outside adds or removes momentum. Collisions and explosions are the two examples named in the specification.

Method for any problem:

1. Choose a positive direction.
2. Write total momentum before, with signs.
3. Write total momentum after, with the unknown velocity.
4. Set them equal and solve.

Collisions where objects stick together

Worked example 2. A tanker of mass 2600 kg travelling at 3.0 m/s hits a stationary car of mass 1300 kg. They lock together. Find their common velocity.

Before: 2600 × 3.0 + 1300 × 0 = 7800 kg m/s. After: (2600 + 1300) × v = 3900v. So 3900v = 7800, giving v = 2.0 m/s in the original direction of the tanker.

Collisions where objects bounce apart

Worked example 3. Trolley A (0.60 kg) moves right at 1.8 m/s. Trolley B (0.80 kg) moves left at 0.40 m/s. After the collision A moves left at 0.20 m/s. Find the velocity of B.

Take right as positive. Before: 0.60 × 1.8 + 0.80 × (−0.40) = 1.08 − 0.32 = 0.76 kg m/s. After: 0.60 × (−0.20) + 0.80 × v = −0.12 + 0.80v. So 0.80v = 0.76 + 0.12 = 0.88, giving v = 1.1 m/s to the right.

The positive answer shows B has reversed direction. A negative sign in your answer always means “opposite to the direction you chose as positive”.

Explosions

In an explosion, objects start together (often at rest) and push apart. The total momentum before is zero, so the total momentum after must also be zero. The two parts move in opposite directions with equal and opposite momenta.

Worked example 4. Two trolleys, masses 0.50 kg and 1.5 kg, sit at rest against a compressed spring. They are released. The 1.5 kg trolley moves right at 0.80 m/s. Find the velocity of the 0.50 kg trolley.

Take right as positive. Before: 0. After: 1.5 × 0.80 + 0.50 × v = 1.2 + 0.50v. So 0.50v = −1.2, giving v = −2.4 m/s. The light trolley moves left at 2.4 m/s.

Force and change in momentum (3.1.4 c)

The specification gives this link between force and momentum change:

F = Δp / t

where F is force in newtons, Δp is the change in momentum in kg m/s and t is time in seconds. Since Δp = mΔv, this is Newton’s Second Law written another way.

A given change in momentum can be produced by a large force acting for a short time or a small force acting for a long time. Making the time longer reduces the force.

Worked example 5. A fielder catches a 0.14 kg ball travelling at 30 m/s. Find the force on the hands if the ball stops in (a) 0.015 s with stiff hands and (b) 0.070 s with hands that move back.

Δp = 0.14 × 30 = 4.2 kg m/s in both cases. (a) F = 4.2 / 0.015 = 280 N. (b) F = 4.2 / 0.070 = 60 N.

The time was about 4.7 times longer, so the force is about 4.7 times smaller.

Safety features explained

Every feature in the specification works the same way: it increases the time over which the momentum of a person changes to zero, so the force on the person falls. Some also spread the force over a bigger area.

Worked example 6. A 75 kg crash-test dummy travelling at 14 m/s stops in 0.12 s with a seat belt and air bag, or 0.03 s against a rigid dashboard.

Δp = 75 × 14 = 1050 kg m/s. With belt and bag: F = 1050 / 0.12 = 8750 N. Without: F = 1050 / 0.03 = 35 000 N, four times larger.

Feature	How it reduces harm
Air bag	Inflates and slows the body over a longer time, spreading the force over a larger area of the chest and head
Seat belt	Stretches slightly, so the body stops over a longer time; also holds you in your seat so you do not hit the windscreen
Crumple zone	Front and rear of the car collapse in a controlled way, increasing the time for the car to stop, so the force on the passengers is smaller
Side impact bar	A strong bar in the door that spreads the force of a side hit across the car body and resists the door being pushed into the passenger
Gymnasium crash mat	Compresses, so the gymnast stops over a longer time and the force is smaller
Playground surface	A soft, cushioned surface (rubber, wood chips) increases the stopping time compared with concrete
Cycle helmet	The inner foam crushes, increasing the time the head takes to stop and reducing the force on the skull

Worked example 7. A 25 kg child lands on a playground surface at 4.0 m/s. Compare the force when the stopping time is 0.020 s on concrete and 0.20 s on a rubber surface.

Δp = 25 × 4.0 = 100 kg m/s. Concrete: F = 100 / 0.020 = 5000 N. Rubber: F = 100 / 0.20 = 500 N.

In an explanation, always give the chain: longer time, so smaller rate of change of momentum, so smaller force.

Steady speed and resistive forces (3.1.5 a)

When a vehicle moves at a steady speed in a straight line, the resultant force on it is zero. The driving force from the engine is equal in size to the resistive forces (air resistance and friction in the moving parts and between tyres and road) and acts in the opposite direction. If the driving force is 1800 N at steady speed, the resistive forces total 1800 N.

Braking force, speed and deceleration (3.1.5 b)

A bigger braking force gives a bigger deceleration (from F = ma, a = F / m). For a given braking force, a faster vehicle takes a longer distance to stop. To stop a faster vehicle in the same distance, a bigger braking force is needed.

Very large decelerations can cause two problems. The brakes can overheat, so they work less well. The wheels can also lose grip, so the driver loses control.

Stopping distance (3.1.5 c)

stopping distance = thinking distance + braking distance

Thinking distance is how far the vehicle moves while the driver reacts, before the brakes are applied. At constant speed, thinking distance = speed × reaction time.

Braking distance is how far the vehicle moves once the braking force acts.

Reaction time becomes longer with tiredness, distractions (such as a phone), drugs and alcohol. A longer reaction time increases thinking distance but does not change braking distance. Speed affects both distances.

Brakes and energy (3.1.5 d)

When brakes are applied, friction between the brake pads and the wheel does work. This work removes kinetic energy from the vehicle and transfers it to thermal energy, so the temperature of the brakes rises. Using work done = force × distance, and kinetic energy = ½mv²:

braking force × braking distance = ½mv²

Worked example 8. A car of mass 950 kg travels at 18 m/s. The driver reacts after 0.60 s. The braking force is 4500 N. Find the stopping distance and the deceleration.

Thinking distance = 18 × 0.60 = 10.8 m. Kinetic energy = ½ × 950 × 18² = 153 900 J. Braking distance = 153 900 / 4500 = 34.2 m. Stopping distance = 10.8 + 34.2 = 45.0 m. Deceleration = 4500 / 950 = 4.7 m/s² (2 significant figures).

Effect of tiredness. With a reaction time of 1.1 s, the thinking distance is 18 × 1.1 = 19.8 m. The braking distance is still 34.2 m, so the stopping distance is 54.0 m.

Effect of doubling speed. At 36 m/s the kinetic energy is ½ × 950 × 36² = 615 600 J. With the same force, the braking distance is 615 600 / 4500 = 136.8 m. Double the speed gives four times the kinetic energy and four times the braking distance.

Adverse conditions and vehicle condition (3.1.5 e)

Braking distance depends on the maximum friction force between tyres and road. Anything that lowers it increases braking distance.

Factor	Effect
Wet road	Water reduces friction between tyres and road, so the braking force is smaller
Icy road	Very low friction; braking distance can be many times longer
Worn brakes	Less friction force is produced by the pads
Worn tyres	Less grip, and worn tread clears water badly, so less friction

Worked example 9. The same car at 18 m/s on a wet road has a maximum braking force of 2700 N. Find the braking distance, stopping distance and deceleration.

Braking distance = 153 900 / 2700 = 57.0 m. Stopping distance = 10.8 + 57.0 = 67.8 m. Deceleration = 2700 / 950 = 2.8 m/s² (2 significant figures).

The thinking distance has not changed, because the weather does not alter the driver’s reaction time.

Common errors

• Forgetting the negative sign for a velocity in the opposite direction.
• Saying air bags “reduce the momentum change”. The change is the same; the time is longer.
• Writing that a crumple zone “absorbs the force”. It increases the time, so the force is smaller.
• Saying that doubling speed doubles the braking distance. It quadruples it, for the same braking force.

Next steps

Work through the practice questions and keep the revision notes to hand.

Official syllabus

OxfordAQA International GCSE Physics (9203) specification, for exams May/June 2018 onwards, published by OxfordAQA.