OCR A Level Physics A: Newton's laws, momentum, impulse and collisions (H556) – Revision Notes

These notes condense section 3.5 of the OCR A Level Physics A (H556) specification, Version 3.0 (March 2026): 3.5.1 Newton’s laws of motion and 3.5.2 Collisions. For fuller explanations, read the study guide. To test yourself, try the practice questions. Two-dimensional collision problems are assessed at A level only. See also the OCR A Level Physics hub, the printable checklist and the free 10-minute diagnostics.

Key definitions and statements

• Newton’s first law (3.5.1(a)): with no resultant force, a body’s velocity does not change.
• Newton’s second law (3.5.1(a), (c)): the resultant force equals the rate of change of momentum, in the direction of the change.
• Newton’s third law (3.5.1(a)): forces between two bodies come in pairs, equal in size, opposite in direction and alike in type, each acting on a different body.
• Linear momentum (3.5.1(b)): mass × velocity. It is a vector.
• Impulse (3.5.1(d)): force × time for which it acts, FΔt. It equals the change in momentum.
• Conservation of momentum (3.5.2(a)): total momentum of a system is constant provided no resultant external force acts on it.
• Perfectly elastic collision (3.5.2(c)): total kinetic energy is unchanged.
• Inelastic collision (3.5.2(c)): total kinetic energy decreases; momentum is still conserved in a closed system.

Formulae and units

Quantity	Equation	Unit
Momentum	p = mv	kg m s⁻¹ (= N s)
Force	F = Δp/Δt	N
Constant mass	F = ma	N
Impulse	FΔt = Δp	N s
Impulse from a graph	area under the F-t graph	N s
Kinetic energy	½mv²	J
Momentum before = after	Σp (before) = Σp (after)	kg m s⁻¹

Why F = ma is a special case: for constant mass, Δp = mΔv, so F = mΔv/Δt = ma. If the mass changes, only F = Δp/Δt works.

Third-law pairs: the four-point check

A genuine pair has all of these. The forces are the same type. They are equal in magnitude. They are opposite in direction. They act on different bodies.

The weight of a stationary book and the push of the table are not a pair, because they act on one body and differ in type. The partner of the weight is the equal upward gravitational pull the book exerts on the Earth. Always name both bodies when you state a pair.

Method boxes

Change in momentum in one dimension

1. Choose a positive direction.
2. Write initial momentum with a sign, then final momentum with a sign.
3. Subtract: final minus initial.
4. If the direction reverses, the magnitudes add.

Force or impulse from F-t data

1. Convert times to seconds.
2. For straight-line sections, use triangle, rectangle or trapezium areas.
3. For a curve, use strips (trapezium rule) or count squares. State that the result is an estimate.
4. Impulse = area. Divide by mass for the velocity change.

1D collision

1. Sketch before and after with arrows, and mark a positive direction.
2. Write the total momentum before and after.
3. Set them equal and solve.
4. Compare total kinetic energy before and after to decide elastic or inelastic.

2D collision (A level only)

1. Choose perpendicular axes, usually along the initial motion.
2. Resolve every velocity: horizontal = v cos θ, vertical = v sin θ.
3. Apply conservation separately along each axis.
4. Combine components with Pythagoras and tan θ for the final speed and direction.
5. Compare kinetic energies.

Worked reminders

Coupling. A 12 000 kg rail truck at 1.5 m s⁻¹ couples with a stationary 8000 kg truck. Common speed = (12 000 × 1.5) ÷ 20 000 = 0.90 m s⁻¹ KE before = ½ × 12 000 × 1.5² = 13 500 J; KE after = ½ × 20 000 × 0.90² = 8100 J Loss = 5400 J, so the collision is inelastic.

Equal masses, elastic, head-on, one at rest. The moving body stops and the other leaves with the original velocity. For two 0.25 kg bodies and an initial speed of 1.8 m s⁻¹, the momentum is 0.45 kg m s⁻¹ before and after.

Recoil. A 5.0 kg launcher fires a 0.050 kg projectile at 40 m s⁻¹. Total momentum starts at zero, so the launcher recoils at (0.050 × 40) ÷ 5.0 = 0.40 m s⁻¹ in the opposite direction.

Safety features. Air bags, seat belts and crumple zones lengthen the stopping time. The change in momentum is the same, so the average force is smaller (F = Δp/Δt). A risk to include in an evaluation is an air bag deploying forcefully close to a small passenger.

Must-know distinctions

• Elastic versus inelastic: both conserve momentum in a closed system. Only elastic conserves kinetic energy. Total energy is always conserved.
• Speed versus velocity: momentum uses velocity, so a rebound reverses the sign.
• Force versus impulse: force is Δp/Δt, impulse is the product FΔt. A graph of force against time has force on the axis; impulse is the area.
• Third-law pair versus balanced forces: the members of a pair act on separate bodies, whereas balanced forces share one body.
• Exact versus estimated area: straight-line graphs give an exact area; curves give an estimate.

Quick self-test

1. A 0.25 kg ball moving at 14 m s⁻¹ rebounds along the same line at 6.0 m s⁻¹. Find the magnitude of its change in momentum.
2. The contact in question 1 lasts 0.020 s. Find the average force.
3. A constant force of 50 N acts for 0.30 s. Find the impulse.
4. A force rises uniformly from 0 to 40 N over 0.50 s and then falls uniformly to 0 over a further 0.50 s. Find the impulse.
5. A 1.8 kg cart at 1.5 m s⁻¹ runs into a stationary 1.2 kg cart and they lock together. Find the common velocity.
6. A 5.0 kg launcher fires a 0.050 kg projectile at 40 m s⁻¹. Find the recoil speed of the launcher.
7. A 0.50 kg ball moves at 3.0 m s⁻¹ at 60° to the x-axis. Find its momentum components along x and y.
8. A 1100 kg van accelerates from rest to 24 m s⁻¹ in 6.0 s. Show that F = ma and F = Δp/Δt give the same resultant force.
9. Give two reasons why the weight of a book and the push from the table are not a third-law pair.
10. In an inelastic collision in a closed system, state which quantities are conserved and which is not.

Answers

1. Take the initial direction as positive. Δp = 0.25 × (−6.0) − 0.25 × 14 = −5.0, so 5.0 kg m s⁻¹.
2. F = 5.0 ÷ 0.020 = 250 N.
3. 50 × 0.30 = 15 N s.
4. Triangle: ½ × 1.0 × 40 = 20 N s.
5. 1.8 × 1.5 = 2.7 = 3.0v, so v = 0.90 m s⁻¹ in the original direction.
6. Launcher momentum equals projectile momentum: 5.0v = 0.050 × 40, so v = 0.40 m s⁻¹.
7. x: 0.50 × 3.0 × cos 60° = 0.75 kg m s⁻¹; y: 0.50 × 3.0 × sin 60° = 1.3 kg m s⁻¹ (2 s.f.).
8. F = ma: a = 24 ÷ 6.0 = 4.0 m s⁻², F = 1100 × 4.0 = 4400 N. F = Δp/Δt: (1100 × 24) ÷ 6.0 = 4400 N. Both give 4400 N, because the mass is constant.
9. They act on the same body, and they are different types of force (gravitational and contact).
10. Momentum and total energy are conserved; kinetic energy is not.

Where marks are usually lost

• Using the difference of speeds when a ball rebounds, not the sum of the magnitudes.
• Leaving a time in ms in F = Δp/Δt or in an area calculation.
• Giving a force as a plain number with no direction when the question asks for direction.
• Treating weight and the table’s push as a third-law pair.
• Stating “momentum is conserved” with no condition about external forces.
• Assuming a collision is elastic without checking kinetic energy.
• In 2D, forgetting that the y-components before the collision may be zero, so the final y-components must cancel.
• Presenting a graph-area estimate as exact, or omitting the unit N s.
• Explaining an air bag only as “it is soft” rather than in terms of a longer time and a smaller force for the same change in momentum.

Official syllabus

OCR, A Level Physics A (H556) specification, Version 3.0, March 2026 (first assessment in 2017), sections 3.5.1 and 3.5.2, Module 3: Forces and motion.