OCR A Level Mathematics A: Newton's laws, weight and equilibrium in one dimension (H240) – Revision Notes

These notes condense OCR A Level Mathematics A (H240) specification section 3.03, items 3.03a to 3.03k (Version 3.1, August 2026, first assessment 2018), with 3.03e marked A Level only. For full explanations use the study guide. Then test yourself with the practice questions. Course hub: OCR A Level Mathematics. Printable checklist. Free 10-minute diagnostics. Connected particles, pulleys, two-dimensional equilibrium and friction are in separate units.

Key facts

Idea	Statement	Ref
Force	Vector, in newtons; 1 N = 1 kg m s⁻²	3.03a
First law	Constant velocity (or rest) means zero resultant force	3.03b
Second law	Resultant F = ma, same direction as a	3.03c
Vector form	F = ma on vectors, component by component	3.03d
Resolved form	Use F cos θ and F sin θ along and across the motion	3.03e (A Level only)
Weight	W = mg, down; g = 9.8 m s⁻² may be assumed	3.03f, g
Third law	Equal and opposite forces on two different bodies	3.03h
Normal reaction	Perpendicular to the surface; R = 0 means contact is lost	3.03i
Smooth	No friction, so only a normal contact force	3.03j
Light, inextensible string	Same tension along it; same speed and acceleration at both ends	3.03j
Equilibrium	Zero resultant, so zero acceleration	3.03k

Method for any straight-line problem

1. Draw a force diagram for one body only. Mark every force on it.
2. Choose a positive direction, usually the direction of the acceleration.
3. Write F = ma along that line, giving each force its sign.
4. Write a separate equation across the line if a normal reaction is needed.
5. Solve, then use constant-acceleration formulas if time, speed or distance is asked.
6. Check the sign and size of the answer: tensions and reactions must not be negative.

Worked reminders

Lift, passenger of mass 25 kg, accelerating up at 0.4 m s⁻²:

R − 25 × 9.8 = 25 × 0.4
R = 245 + 10 = 255 N

The same passenger with the lift accelerating down at 0.4: R = 25(9.8 − 0.4) = 235 N. At constant speed R = 245 N, which is the weight.

Hoist, load of mass 6 kg on a light inextensible cable:

Accelerating up at 0.5:  T − 58.8 = 6 × 0.5, so T = 61.8 N
Constant velocity:       T = 58.8 N
Decelerating at 0.5 (still rising): 58.8 − T = 6 × 0.5, so T = 55.8 N

Speeding up while rising and slowing down while falling both need T above the weight. Slowing while rising and speeding up while falling need T below it. Check which way the acceleration points before you write the equation.

Falling with resistance: a 2 kg stone falls from rest with constant resistance of 3 N.

2 × 9.8 − 3 = 2a, so a = 8.3 m s⁻²
After 4 s: v = 8.3 × 4 = 33.2 m s⁻¹, distance = ½ × 8.3 × 4² = 66.4 m

Link the second law to the constant-acceleration formulas: v = u + at, then v² = u² + 2as, s = ½(u + v)t, or s = ut + ½at², whichever fits the data.

Vector form: a 2 kg particle with acceleration 3i + 1.5j has resultant 6i + 3j N. Components are independent, so you can also solve i and j parts separately.

Resolved (A Level only): a force P at θ above the horizontal on a smooth horizontal surface gives P cos θ = ma and R + P sin θ = mg. Contact is lost when P sin θ = mg.

Smooth slope (A Level only): the acceleration down a smooth plane at angle θ is g sin θ, and R = mg cos θ.

Must-know distinctions

Pair	Difference
Mass and weight	Mass (kg) is fixed. Weight (N) is mg, so it follows g
Third-law pair and balanced pair	A third-law pair acts on two bodies and can never be balanced. Balanced forces such as weight and R act on one body
R = mg and R	R equals mg only when there is no vertical acceleration. In a lift R changes with a
Light and smooth	Light describes a string, with no mass. Smooth describes a surface, with no friction
Inextensible and light	Inextensible gives equal accelerations at both ends. Light gives equal tension throughout
Constant velocity and constant acceleration	Constant velocity means F = 0. Constant acceleration means F is constant and non-zero

About g: its value varies with place, which is why weight changes from one planet to another while mass does not. Use the value in the question, normally 9.8.

When R becomes 0: the surface no longer pushes and the body leaves it. In a lift this happens when the downward acceleration reaches g. For a pulled body on a smooth surface it happens when the upward component of the pull equals the weight.

When a string goes slack: T = 0. If a string is cut, the tension disappears at once, but velocity does not change at that instant.

Modelling words and what they allow

Word in the question	What you may assume
Particle	Size and rotation are ignored, so all forces act at one point
Light	Negligible mass, so equal tension throughout a single string
Inextensible	Length is fixed, so connected ends share speed and acceleration
Smooth	No friction, so the contact force is normal to the surface
Constant force	Constant acceleration, so the constant-acceleration formulas apply
Air resistance negligible	Weight is the only force on a falling body, so a = g

Quick checks before you finish

• Units: forces in N, masses in kg, accelerations in m s⁻². A mass of 450 g is 0.45 kg.
• Direction: say “upwards” or “downwards” when you quote a force or acceleration, or give the sign you chose.
• Sense check: a lift accelerating upwards needs T greater than the weight; in free fall the floor force is zero; a falling body with resistance has a less than 9.8.
• Method: if the question gives time or distance, expect to use a constant-acceleration formula either before or after the second law.
• Vectors: a resultant such as (5i + 20j) + (7i − 4j) is found by adding components, then divide by the mass for the acceleration. The magnitude is √(x² + y²).
• Accuracy: carry unrounded values through the working and give 3 significant figures at the last step, unless the question says otherwise.

Quick self-test

Use g = 9.8 m s⁻².

1. Find the weight of an 8 kg body.
2. A resultant force of 30 N acts on a 6 kg trolley. Find its acceleration.
3. Forces of 20 N and 8 N act in opposite directions on a 5 kg block on a smooth horizontal surface. Find its acceleration.
4. Find the force on a 2 kg particle with acceleration (3i + 1.5j) m s⁻².
5. A 1000 kg lift rises at constant speed. Find the cable tension.
6. The lift floor supports a 50 kg cleaner while the lift accelerates upwards at 2 m s⁻². Find the force the floor exerts on the cleaner.
7. State the third-law partner of the gravitational pull of the Earth on a book.
8. A 0.5 kg ball falls with constant air resistance of 1.1 N. Find its acceleration.
9. A 10 N force acts at 60° above the horizontal on a 2 kg body on a smooth horizontal surface (A Level only). Find the horizontal acceleration.
10. A 3 kg particle hangs at rest on a light inextensible string. Find the tension.
11. State two assumptions in the term “light inextensible string”.
12. What is the reaction on a passenger in a lift that is in free fall?

Answers

1. 78.4 N
2. 5 m s⁻²
3. Resultant 12 N, so a = 12 ÷ 5 = 2.4 m s⁻²
4. 6i + 3j N
5. 9800 N
6. R = 50(9.8 + 2) = 590 N
7. The pull of the book on the Earth, equal in size and opposite in direction
8. (0.5 × 9.8 − 1.1) ÷ 0.5 = 7.6 m s⁻²
9. 10 cos 60° = 2a, so a = 2.5 m s⁻²
10. T = 3 × 9.8 = 29.4 N
11. The string has negligible mass, so tension is the same throughout; its length does not change, so both ends move together
12. R = 0

Where marks are usually lost

• Forgetting the weight in a vertical equation, or including it in a horizontal one.
• Using R = mg for a body that is accelerating vertically.
• Mixing signs: adding a force that opposes the acceleration.
• Treating weight and R as a third-law pair.
• Using F = ma with F as a single force instead of the resultant.
• Answering with kg where newtons are needed, or leaving out units.
• Rounding g to 10 when the question says 9.8, which gives a different final answer.
• Using sin where cos is needed when resolving (A Level only).
• Omitting a short statement of the model when asked, such as light string or smooth surface.
• Leaving a negative tension or reaction without noticing that the situation cannot occur.

Official syllabus

OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), first assessment 2018, published by OCR: section 3.03, Forces and Newton’s Laws.