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

This guide follows the OCR A Level Physics A (H556) specification, Version 3.0 (March 2026), and covers section 3.5: 3.5.1 Newton’s laws of motion and 3.5.2 Collisions. It is written for the full A level. Two-dimensional collision problems are assessed at A level only, so if you are working towards the AS qualification you can leave out the 2D material. For a quick-recall version, use the revision notes. To test yourself, use the practice questions. See also the OCR A Level Physics hub, the printable checklist and the free 10-minute diagnostics.

What this unit covers

Spec reference	What you must be able to do
3.5.1(a)	State and apply Newton’s three laws of motion
3.5.1(b)	Use linear momentum p = mv and treat it as a vector
3.5.1(c)	Use net force = rate of change of momentum, F = Δp/Δt, and know F = ma is a special case
3.5.1(d)	Use impulse = FΔt
3.5.1(e)	Find impulse as the area under a force-time graph, including estimating the area under a non-linear graph
3.5.2(a)	Apply the principle of conservation of momentum
3.5.2(b)	Analyse collisions and interactions in one dimension and (A level only) in two dimensions
3.5.2(c)	Distinguish perfectly elastic and inelastic collisions

3.5.1(a) Newton’s three laws of motion

First law. With zero resultant force, an object’s velocity stays constant, so its acceleration is zero.

Second law. The resultant force on an object is proportional to the rate of change of its momentum, and acts in the direction of that change. Written as an equation, F = Δp/Δt (see 3.5.1(c)).

Third law. Forces between two bodies come in pairs: body A pushes or pulls on body B, and B pushes or pulls back on A with a force of the same size, opposite direction and the same kind.

A third-law pair has four features. The forces are the same type (both gravitational, both contact, both electrostatic). They have equal magnitude and opposite direction. They act on different bodies. They exist at the same time.

For a book on a table, the Earth’s pull on the book pairs with the book’s pull on the Earth. The table’s push on the book pairs with the book’s push on the table. The weight and the table’s push are equal and opposite, yet they are not a pair, because they share one body and differ in type.

3.5.1(b) Linear momentum

Linear momentum is mass multiplied by velocity:

p = mv

The unit is kg m s⁻¹, which is the same as N s. Momentum is a vector, so it has the direction of the velocity. In one dimension, choose a positive direction and give opposite directions opposite signs.

Worked example 1. A 0.057 kg tennis ball travelling at 32 m s⁻¹ towards a wall rebounds along the same line at 24 m s⁻¹. Find its change in momentum.

Take the initial direction as positive. Initial momentum = 0.057 × 32 = +1.824 kg m s⁻¹ Final momentum = 0.057 × (−24) = −1.368 kg m s⁻¹ Change = −1.368 − 1.824 = −3.192, so the magnitude is 3.2 kg m s⁻¹ (2 s.f.), directed away from the wall.

The velocity change is 56 m s⁻¹, not 8, because the direction reverses.

3.5.1(c) Force as rate of change of momentum

The net force equals the rate of change of momentum:

F = Δp/Δt

For a body of constant mass, Δp = mΔv, so F = mΔv/Δt = ma. Newton’s second law in the form F = ma is therefore a special case, valid when the mass does not change. The form F = Δp/Δt is the more general one, and you need it when mass is gained or lost, or when you only know momentum before and after.

Worked example 2. The tennis ball in example 1 is in contact with the wall for 4.5 ms. Find the average force on the ball.

F = Δp/Δt = 3.192 ÷ 0.0045 = 709 N, so 7.1 × 10² N (2 s.f.), directed away from the wall.

Worked example 3. Water from a hose leaves at 14 m s⁻¹ and 0.45 kg of it hits a wall each second, stopping without rebounding. Find the force on the water, and state the force on the wall.

Momentum lost per second = 0.45 × 14 = 6.3 kg m s⁻¹ per second, so the wall exerts 6.3 N on the water. By the third law, the water exerts 6.3 N on the wall. The water is a stream, not one body of fixed mass, so F = Δp/Δt is the natural form to use.

3.5.1(d) Impulse

Rearranging the second law gives FΔt = Δp. The quantity FΔt is the impulse of the force. Its unit is N s, equal to kg m s⁻¹. Impulse is a vector in the direction of the force, and the impulse of the net force equals the change in momentum.

If Δp is fixed, a longer time Δt means a smaller average force.

Air bags, seat belts and crumple zones do exactly this. A person stopping from a given speed has a fixed change of momentum, and a longer stopping time gives a smaller average force on the body. The benefit is lower injury risk. A risk is an air bag deploying forcefully close to a small passenger. An evaluation should name both a benefit and a risk.

3.5.1(e) Impulse as the area under a force-time graph

On a graph of force against time, the area between the curve and the time axis equals the impulse, because area = F × Δt summed over the interval. For a constant force the area is a rectangle. For a force that rises and falls, split the area into triangles, rectangles and trapeziums, or count squares.

Worked example 4 (straight lines). A 0.16 kg ball at rest is struck. The force rises uniformly from 0 to 800 N in 6.0 ms and then falls uniformly to 0 at 18 ms. Find the speed.

Impulse = ½ × 0.018 × 800 = 7.2 N s v = impulse ÷ m = 7.2 ÷ 0.16 = 45 m s⁻¹

Worked example 5 (non-linear). A force acts on the same ball in a different strike. Readings at 2.0 ms intervals are:

t / ms	0	2	4	6	8	10	12
F / N	0	150	380	520	400	180	0

Estimate the impulse using trapeziums of width 2.0 ms (0.0020 s).

Area = 0.0020 × (0/2 + 150 + 380 + 520 + 400 + 180 + 0/2) = 0.0020 × 1630 = 3.3 N s (2 s.f.) Initial speed 0, so v = 3.26 ÷ 0.16 = 20 m s⁻¹.

This is an estimate, because straight lines approximate the curve. Narrower strips improve it. Convert ms to s first.

3.5.2(a) Conservation of momentum

The principle of conservation of momentum: for a system of interacting bodies, the total momentum in a given direction before an interaction equals the total momentum in that direction after it, provided no resultant external force acts on the system.

It follows from the third law: the two bodies exert equal and opposite forces for the same time, so they receive equal and opposite impulses. One gains the momentum the other loses.

3.5.2(b) Collisions in one dimension

Method: choose a positive direction, write total momentum before and after, set them equal, and solve.

Worked example 6. A 0.42 kg air-track glider moving at 0.90 m s⁻¹ hits a stationary 0.28 kg glider. They stick together. Find their common speed and the kinetic energy lost.

Before: 0.42 × 0.90 = 0.378 kg m s⁻¹ After: (0.42 + 0.28)v = 0.70v v = 0.378 ÷ 0.70 = 0.54 m s⁻¹ KE before = ½ × 0.42 × 0.90² = 0.1701 J KE after = ½ × 0.70 × 0.54² = 0.10206 J Loss = 0.068 J (2 s.f.)

Explosion. If two bodies at rest push apart, total momentum stays zero, so their momenta are equal and opposite. The lighter body gets the higher speed.

3.5.2(c) Elastic and inelastic collisions

• Perfectly elastic: total kinetic energy is the same before and after.
• Inelastic: total kinetic energy is lower after the collision, because some is transferred to other forms such as thermal energy and sound.

Momentum is conserved in both types if the system is closed. Total energy is always conserved, but kinetic energy need not be.

Worked example 7. A 0.60 kg trolley moving at 2.4 m s⁻¹ collides head-on with a stationary 0.20 kg trolley. The collision is perfectly elastic. Find both velocities after.

Momentum: 0.60 × 2.4 = 0.60v₁ + 0.20v₂, so 1.44 = 0.60v₁ + 0.20v₂ Kinetic energy: ½ × 0.60 × 2.4² = ½ × 0.60 × v₁² + ½ × 0.20 × v₂², so 1.728 = 0.30v₁² + 0.10v₂² From the momentum equation, v₂ = 7.2 − 3v₁. Substituting gives 1.728 = 0.30v₁² + 0.10(7.2 − 3v₁)², which reduces to v₁² − 2.4v₁ + 1.44 = 0, so v₁ = 1.2 m s⁻¹. Then v₂ = 7.2 − 3.6 = 3.6 m s⁻¹. Check: momentum after = 0.72 + 0.72 = 1.44 kg m s⁻¹, and kinetic energy after = 0.432 + 1.296 = 1.728 J. Both match.

In a perfectly elastic head-on collision, the speed of approach equals the speed of separation (2.4 m s⁻¹ and 3.6 − 1.2 = 2.4 m s⁻¹). This is a useful check on your answer.

Collisions in two dimensions (A level only)

Momentum is a vector, so it is conserved separately along two perpendicular directions. Choose axes along and perpendicular to the initial motion, resolve each velocity, and apply conservation on each axis.

Worked example 8. A 0.16 kg puck moving along the x-axis at 5.0 m s⁻¹ hits a stationary 0.24 kg puck. The first puck leaves at 3.0 m s⁻¹ at 35° above the x-axis. Find the velocity of the second puck and say whether the collision is elastic.

x-direction: 0.16 × 5.0 = 0.16 × 3.0 cos 35° + 0.24vₓ 0.80 = 0.393 + 0.24vₓ, so vₓ = 1.695 m s⁻¹ y-direction: 0 = 0.16 × 3.0 sin 35° + 0.24v_y v_y = −0.2753 ÷ 0.24 = −1.147 m s⁻¹ (downwards) Speed = √(1.695² + 1.147²) = 2.0 m s⁻¹, at tan⁻¹(1.147 ÷ 1.695) = 34° below the x-axis. KE before = ½ × 0.16 × 5.0² = 2.0 J. KE after = ½ × 0.16 × 3.0² + ½ × 0.24 × 2.047² = 0.72 + 0.503 = 1.22 J. Kinetic energy has fallen, so the collision is inelastic.

Common errors

• Subtracting speeds instead of velocities when direction reverses.
• Calling weight and the normal contact force a third-law pair.
• Using F = ma when mass changes, as with water jets.
• Forgetting to convert ms to s.
• Applying kinetic energy conservation to every collision. It holds only if the question says the collision is perfectly elastic or you have shown it.
• Stating conservation of momentum without the condition of no resultant external force.

Next steps

Next, try the practice questions. If resolving vectors is slow, revisit units, measurement, uncertainty and vectors, and for force diagrams and F = ma problems see dynamics, free-body diagrams, drag and terminal velocity.

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.