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

Quick recall for statements P2.2i, P2.2j, P2.2k, P2.2l (Higher Tier only) and P2.2q of OCR GCSE (9-1) Gateway Science Physics A (J249), Version 5.0 (July 2026). For full explanations, start with the study guide. Then test yourself with the practice questions, tick off the printable checklist, see the course hub and take the free 10-minute diagnostics.

Equations and units

Idea	Equation	Units
Newton’s second law (P2.2i)	F = m a	N, kg, m/s²
Acceleration from a speed change	a = (v - u) / t	m/s², m/s, s
Inertial mass (P2.2j)	m = F / a	kg, N, m/s²
Momentum (P2.2k)	p = m v	kg m/s, kg, m/s
Force and momentum (P2.2l, Higher Tier only)	F = (change in p) / t	N, kg m/s, s

Always convert to kg, m/s, N and s before substituting. The force in F = ma is the resultant force.

Newton’s second law

• Resultant force and acceleration have the same direction.
• Double the resultant force on the same mass: double the acceleration.
• Double the mass under the same force: half the acceleration.
• No resultant force means no acceleration: constant velocity (or at rest).

Method for F = ma questions

1. Draw or list the forces. Find the resultant (subtract opposing forces).
2. Write the equation, then rearrange before substituting.
3. Substitute with units. Give the answer with a unit and a sensible number of significant figures.

Reminder example. A 1300 kg car has driving force 900 N and resistive force 640 N.

resultant = 900 - 640 = 260 N
a = 260 / 1300 = 0.20 m/s²

Reminder example. A skier of mass 64 kg accelerates at 1.5 m/s² down a slope under a resultant force of 96 N: F = 64 × 1.5 = 96 N.

Force and acceleration experiment

• The specification suggests light gates, weights and trolleys (PAG P2).
• Keep the total mass of the system the same by moving weights between trolley and hanger.
• Result: acceleration is proportional to resultant force, and F ÷ a gives the mass of the system.
• Scatter in the F ÷ a column shows measurement uncertainty.

Inertia and inertial mass

• Inertia: a measure of how difficult it is to change the velocity of an object.
• Inertial mass: the ratio of force over acceleration, m = F / a.
• Large inertial mass: small acceleration for a given force. Small inertial mass: large acceleration.

Object	Force	Acceleration	Inertial mass
Cart A	24 N	3.0 m/s²	8.0 kg
Cart B	24 N	0.60 m/s²	40 kg

Cart B has more inertia. It is harder to speed up, slow down or turn.

Momentum and collisions

Momentum = mass × velocity. It is a vector, in kg m/s, with the direction of the velocity.

Conservation of momentum: with no external force, total momentum before = total momentum after.

Method for collision questions

1. Choose a positive direction. Movement the other way is negative.
2. Write total momentum before as a sum, with a zero term for any stationary object.
3. Write total momentum after, using the combined mass if the objects join.
4. Set before equal to after and solve. State the direction.

Type	Reminder example	Result
Join together	2.5 kg cart at 4.8 m/s runs into a resting 1.5 kg cart and they lock	12 / 4.0 = 3.0 m/s
Push apart from rest	Total momentum before = 0, so the two momenta are equal and opposite	Lighter object gets the greater speed
Bounce apart	Before: sum of p. After: sum of p. Solve for the unknown velocity	Include signs

The water rocket: the water goes 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).

More worked reminders

Bounce apart. A 0.80 kg trolley moving at 3.0 m/s hits a stationary 0.40 kg trolley. Afterwards the 0.80 kg trolley moves on at 1.0 m/s and the 0.40 kg trolley moves off.

before: 0.80 x 3.0 + 0.40 x 0 = 2.4 kg m/s
after:  0.80 x 1.0 + 0.40 x v = 0.80 + 0.40 v
2.4 = 0.80 + 0.40 v, so v = 4.0 m/s (same direction)

Push apart. A 50 kg skater and a 40 kg skater at rest push apart. The 50 kg skater moves at 1.2 m/s one way. Momentum of the first skater = 50 × 1.2 = 60 kg m/s, so the 40 kg skater has 60 kg m/s the other way: v = 60 / 40 = 1.5 m/s.

Words to use in explanations

• “The resultant force is zero, so there is no acceleration.”
• “The total momentum before the collision equals the total momentum after.”
• “Momentum is a vector, so the direction matters.”
• “The velocity is changing because the direction is changing, so the object is accelerating.”
• (Higher Tier only) “The change in momentum is the same, but it happens over a longer time, so the force is smaller.”

Higher Tier only: force and change in momentum

F = ma and a = (v - u) / t combine to give F = (m v - m u) / t. So force = change in momentum ÷ time.

• Same change in momentum, longer time: smaller force.
• Seat belts, air bags and crumple zones lengthen the stopping time and reduce the force.
• Reminder example: a 0.30 kg ball stopped from 12 m/s in 0.020 s needs F = 0.30 × 12 / 0.020 = 180 N.

Circular motion at constant speed

• Speed is constant. Direction keeps changing.
• Velocity is a vector, so its direction changing means the velocity is changing.
• A changing velocity is an acceleration, so a resultant force towards the centre is needed.
• Rubber bung on a string: if the string breaks, the bung moves in a straight line along its direction at that instant.

Must-know distinctions

Pair	Difference
Speed and velocity	Speed has size only. Velocity has size and direction.
Force and momentum	Force changes momentum. Momentum is mass × velocity.
Momentum before and after	The total is equal. One object’s momentum can change.

Quick self-test

1. A 6.0 kg cart has a resultant force of 15 N. Find its acceleration.
2. A resultant force of 48 N gives an object an acceleration of 6.0 m/s². Find its mass.
3. A force of 7.5 N gives an acceleration of 1.5 m/s². What is the inertial mass?
4. Find the momentum of a 1400 kg car moving at 15 m/s.
5. A 9.0 kg trolley has a momentum of 27 kg m/s. Find its velocity.
6. A 2.5 kg cart at 4.8 m/s runs into a resting 1.5 kg cart and the two lock together. Find their speed.
7. A gun of mass 4.0 kg fires a 0.020 kg bullet at 300 m/s. Find the recoil speed of the gun.
8. (Higher Tier only) A 0.30 kg ball moving at 12 m/s is stopped in 0.020 s. Find the average force.
9. Why does a car going round a roundabout at constant speed have a changing velocity?
10. A van at 0.50 m/s² under a resultant force of 650 N: what is its mass?

Answers

1. a = 15 / 6.0 = 2.5 m/s².
2. m = 48 / 6.0 = 8.0 kg.
3. m = 7.5 / 1.5 = 5.0 kg.
4. p = 1400 × 15 = 21 000 kg m/s.
5. v = 27 / 9.0 = 3.0 m/s.
6. Before: 2.5 × 4.8 = 12 kg m/s. After: 4.0 × v. So v = 3.0 m/s.
7. Bullet: 0.020 × 300 = 6.0 kg m/s forwards. The gun has 6.0 kg m/s backwards, so v = 6.0 / 4.0 = 1.5 m/s.
8. F = 0.30 × 12 / 0.020 = 180 N.
9. The direction of travel keeps changing. Velocity is a vector, so a change in direction is a change in velocity, even though the speed is constant.
10. m = 650 / 0.50 = 1300 kg.

Where marks are usually lost

• Using one force instead of the resultant force.
• Substituting grams or tonnes without converting to kg.
• Forgetting the zero term for a stationary object in a collision.
• Using the mass of one object, not the combined mass, after objects stick.
• Dropping the sign when an object moves in the opposite direction.
• Writing “momentum is lost” instead of “momentum is transferred, and the total is conserved”.
• Saying the speed changes in circular motion, rather than the velocity.
• Leaving out units, or the direction of a vector answer.
• Answers about safety that mention force but not the longer time (Higher Tier only).

Official syllabus

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