OCR GCSE Physics A: Newton's second law, inertia, momentum and circular motion (J249)

This guide covers statements P2.2i, P2.2j, P2.2k, P2.2l and P2.2q of the OCR GCSE (9-1) Gateway Science Physics A specification (J249), Version 5.0 (July 2026), first assessment 2018. They sit in the Newton’s laws part of Topic P2 Forces. Statement P2.2l is Higher Tier only. Everything else is for both tiers.

Related pages: the OCR GCSE Physics hub, the printable checklist, the revision notes and the practice questions. The wider topic is in OCR GCSE Physics: Forces. To find your weak spots first, try the free 10-minute diagnostics.

What this guide covers

Statement	What you must be able to do	Tier
P2.2i	Apply Newton’s second law in calculations relating force, mass and acceleration	Both
P2.2j	Explain that inertia measures how hard it is to change an object’s velocity; inertial mass is force divided by acceleration	Both
P2.2k	Define momentum and describe examples of momentum in collisions (the idea of conservation of momentum)	Both
P2.2l	Apply formulae relating force, mass, velocity and acceleration to explain how the changes involved are inter-related	Higher Tier only
P2.2q	Explain why an object moving in a circle at constant speed has a changing velocity (qualitative only)	Both

Work done, power and Newton’s third law are separate statements and are not taught here.

Newton’s second law (P2.2i)

When a resultant force acts on an object, the object accelerates. The acceleration is bigger for a bigger resultant force and smaller for a bigger mass. The acceleration is in the same direction as the resultant force.

force = mass x acceleration        F = m a

Quantity	Symbol	Unit
Resultant force	F	newton, N
Mass	m	kilogram, kg
Acceleration	a	metre per second squared, m/s²

One newton is the resultant force that gives a 1 kg mass an acceleration of 1 m/s². Rearranged: a = F ÷ m and m = F ÷ a.

The force in F = ma is the resultant (net) force. If several forces act, find the resultant first by adding forces in the same direction and subtracting forces in opposite directions.

Worked example 1: finding acceleration. A cyclist and bike have a combined mass of 85 kg. The driving force is 120 N and the resistive forces total 45 N.

resultant force = 120 - 45 = 75 N
a = F / m = 75 / 85 = 0.88 m/s²  (2 s.f.)

Worked example 2: finding mass. A resultant force of 3600 N gives a lorry an acceleration of 1.5 m/s².

m = F / a = 3600 / 1.5 = 2400 kg

Worked example 3: acceleration from a speed change. A 1100 kg car brakes from 18 m/s to a stop over 6.0 s. What resultant force acts on it?

a = (v - u) / t = (0 - 18) / 6.0 = -3.0 m/s²
F = m a = 1100 x (-3.0) = -3300 N

The negative sign shows that the force acts opposite to the motion. The size of the force is 3300 N. You can quote it as 3300 N backwards.

Investigating force and acceleration

The specification suggests using light gates, weights and trolleys to investigate the link between force and acceleration (PAG P2). Light gates measure the speeds of the trolley, from which acceleration is found. The weights provide the pulling force.

A fair test keeps the total mass of the system fixed. If you add weights to the hanger, take the same weights off the trolley, so the mass being accelerated does not change.

Worked example 4: interpreting data. A trolley system has the results below.

Force F (N)	Acceleration a (m/s²)	F ÷ a (kg)
0.2	0.50	0.40
0.4	1.02	0.39
0.6	1.48	0.41
0.8	2.01	0.40
1.0	2.49	0.40

Acceleration is proportional to force, because doubling the force roughly doubles the acceleration (0.2 N gives 0.50 m/s² and 0.4 N gives 1.02 m/s²). F ÷ a is almost constant at about 0.40 kg, which is the mass of the system. Small differences between the F ÷ a values are measurement uncertainty, not a failure of the law.

Inertia and inertial mass (P2.2j)

Inertia is a measure of how difficult it is to change the velocity of an object. A stationary object resists being set moving. A moving object resists being slowed, sped up or turned.

Inertial mass is defined as the ratio of force over acceleration:

inertial mass = force / acceleration        m = F / a

For the same force, an object with a large inertial mass has a small acceleration. An object with a small inertial mass has a large acceleration.

Worked example 5. The same 20 N resultant force acts separately on two trolleys. Trolley A accelerates at 4.0 m/s². Trolley B accelerates at 0.80 m/s².

mass A = 20 / 4.0 = 5.0 kg
mass B = 20 / 0.80 = 25 kg

Trolley B has five times the inertial mass, so it has more inertia and is harder to accelerate.

Everyday examples to explain: a loaded shopping trolley is harder to start and harder to stop than an empty one; a loaded lorry is harder to stop than an empty one when the same braking force is applied; passengers lurch forward when a bus brakes because their bodies tend to keep moving at the old velocity.

Momentum and collisions (P2.2k)

Momentum is the product of an object’s mass and its velocity.

momentum = mass x velocity        p = m v

Momentum is measured in kilogram metres per second (kg m/s). It is a vector, so it has a direction, the same as the velocity. Choose one direction as positive and give movement the other way a negative sign.

Worked example 6. A 1800 kg van travels at 12 m/s. Its momentum is 1800 × 12 = 21 600 kg m/s. A 72 kg runner at 6.5 m/s has 468 kg m/s. A heavy, slow object can have the same momentum as a light, fast one.

Conservation of momentum

In a collision or explosion, if no external force acts on the objects, the total momentum before equals the total momentum after. Momentum is conserved. Momentum is not lost: some may move from one object to another.

Worked example 7: objects that join. A 4000 kg truck moving at 3.0 m/s hits a stationary 6000 kg truck. They couple together.

momentum before = 4000 x 3.0 + 6000 x 0 = 12 000 kg m/s
mass after = 4000 + 6000 = 10 000 kg
v = 12 000 / 10 000 = 1.2 m/s

Worked example 8: objects that bounce apart. A 2.0 kg trolley moves at 1.5 m/s towards a stationary 1.0 kg trolley. After the collision the 1.0 kg trolley moves at 2.0 m/s in the same direction.

before: 2.0 x 1.5 + 1.0 x 0 = 3.0 kg m/s
after:  2.0 x v + 1.0 x 2.0 = 2.0 v + 2.0
3.0 = 2.0 v + 2.0, so v = 0.50 m/s (same direction)

Worked example 9: an explosion or push-apart. Two skaters, 60 kg and 45 kg, stand at rest. They push apart and the 60 kg skater moves east at 1.5 m/s.

total momentum before = 0
60 x 1.5 + 45 x v = 0, so v = -90 / 45 = -2.0 m/s

The 45 kg skater moves west at 2.0 m/s. The two momenta are equal in size and opposite in direction, so they add to zero. The lighter skater has the greater speed.

The specification gives the water rocket as a demonstration: the explosion propels the water down with the same momentum as the rocket gains going up. Light gates, weights and trolleys can measure momentum before and after collisions (PAG P3).

Force, mass, velocity and acceleration (P2.2l) – Higher Tier only

The specification asks Higher Tier candidates to apply formulae relating force, mass, velocity and acceleration, and explain how the changes involved are inter-related. You can link F = ma, p = mv and a = (v - u) / t.

Start with F = ma and replace a with (v - u) / t:

F = m (v - u) / t = (m v - m u) / t

So force = change in momentum ÷ time taken, or F = Δp ÷ t. A resultant force changes an object’s momentum. For a given change in momentum, a longer time means a smaller force.

Worked example 10. A 0.45 kg football at rest is kicked and leaves the boot at 22 m/s. The contact time is 0.015 s.

change in momentum = 0.45 x 22 - 0 = 9.9 kg m/s
F = 9.9 / 0.015 = 660 N

Worked example 11: why longer stopping times are safer. A 70 kg driver travelling at 14 m/s is brought to rest.

change in momentum = 70 x 14 = 980 kg m/s
stopping in 0.10 s: F = 980 / 0.10 = 9800 N
stopping in 0.40 s: F = 980 / 0.40 = 2450 N

The change in momentum is the same, but a stopping time four times longer gives a force four times smaller. Seat belts, air bags and crumple zones increase the stopping time and so reduce the force on the body.

If the mass stays the same, this is just F = ma again: the change in velocity per second is the acceleration, and multiplying by the mass gives the force.

Circular motion (P2.2q)

Speed is a scalar: it has size only. Velocity is a vector: it has size and direction. An object moving round a circle at constant speed has a velocity that keeps changing, because its direction keeps changing.

Position of a car on a roundabout	Speed	Direction of travel
Start	8 m/s	North
Quarter turn	8 m/s	East
Half turn	8 m/s	South
Three-quarter turn	8 m/s	West

The speed is always 8 m/s but the four velocities are all different. At the start and at the half turn the speeds are equal, but the velocities are opposite.

A change in velocity is an acceleration. So an object in circular motion at constant speed is accelerating. By F = ma, a resultant force is needed, and it acts towards the centre of the circle. The specification suggests spinning a rubber bung on a string: the string provides the inward force. If the string breaks, there is no resultant force, so the bung travels in a straight line at constant velocity, along the direction it was moving at that instant.

Model answer: “The speed is constant but the direction of motion is continually changing. Velocity is a vector, so a change in direction means a change in velocity.”

Common errors

• Using a single force in F = ma when several forces act. Use the resultant.
• Writing mass in grams or force in kN without converting to kg and N.
• Treating momentum as always conserved by one object. It is the total that is conserved.
• Forgetting directions in a collision. A trolley moving the other way has negative momentum.
• Saying “speed changes” in circular motion. At constant speed, the velocity changes because of direction.
• In Higher Tier answers, comparing forces without saying the change in momentum is the same.

Official syllabus

OCR GCSE (9-1) Gateway Science Physics A (J249), Version 5.0 (July 2026), first assessment 2018, published by OCR.