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

This guide follows the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018. It covers section 3.03 (Forces and Newton’s Laws), items 3.03a to 3.03k, with the vector form of the second law (3.03d) and the extension to resolved forces (3.03e, A Level only). The remaining parts of 3.03k (connected particles and smooth pulleys) are taught in a separate unit, as are two-dimensional equilibrium and friction. Next: revision notes and practice questions. Course hub: OCR A Level Mathematics. Printable checklist. The free 10-minute diagnostics show where to start.

What this unit covers

Ref	Skill	Level
3.03a	Treat a force as a vector; draw force diagrams	AS and A Level
3.03b	Use Newton’s first law; complete a diagram for equilibrium	AS and A Level
3.03c	Use F = ma in a straight line	AS and A Level
3.03d	Use F = ma with forces and accelerations as 2-D vectors	AS and A Level
3.03e	Use F = ma when forces must be resolved (two dimensions)	A Level only
3.03f, g	Use W = mg; know g = 9.8 m s⁻² may be assumed and that g depends on location	AS and A Level
3.03h, i	Use Newton’s third law; use the normal reaction; know R = 0 means contact is lost	AS and A Level
3.03j	Use the smooth-contact model; light, inextensible strings; know its limits	AS and A Level
3.03k	Combine equilibrium with motion in a straight line	AS and A Level

SI units and constant-acceleration formulas are in quantities, units and kinematics. Vector arithmetic is in vectors in two and three dimensions.

Forces as vectors (3.03a)

A force has magnitude and direction, measured in newtons (N). One newton is 1 kg m s⁻². A force can change the velocity of an object that has a given mass. On a straight line, choose one direction as positive and write each force with a sign. The resultant is the signed sum.

A force diagram shows only the forces acting on the body you have chosen, each as an arrow from the body in the direction of the force. Keep velocity and acceleration off a force diagram, or draw them as clearly different arrows.

Typical forces at this level: weight (down), tension (along a string, away from the body), normal reaction (perpendicular to a surface, away from it), driving force, resistance, and applied forces such as a push.

Newton’s first law (3.03b)

A particle at rest, or moving with constant velocity, stays that way unless an external force acts on it. The working form is: if the velocity is constant, the resultant force is zero. Constant velocity does not mean zero speed. A body moving at 12 m s⁻¹ steadily is in equilibrium.

Example 1. A parachutist of mass 70 kg falls at a steady speed. Find the air resistance.

Constant speed, so resultant force = 0.
Weight = 70 × 9.8 = 686 N downwards.
Air resistance R acts upwards: R − 686 = 0
R = 686 N

Completing a diagram. A box rests on a table. Two forces act: weight W downwards and the normal reaction R upwards. For equilibrium R = W. If a diagram shows only the weight, the missing force is R.

Newton’s second law in one dimension (3.03c)

For a body of constant mass under constant forces, along a straight line:

resultant force = mass × acceleration, F = ma

where F is measured in newtons, m in kilograms and a in metres per second squared. The resultant must be taken in the direction of the acceleration. Write one equation along the line of motion, giving every force its sign.

Example 2. A van of mass 900 kg has a driving force of 1800 N and resistance of 450 N. It starts from rest. Find the acceleration and the speed after 8 s.

Resultant forwards = 1800 − 450 = 1350 N
1350 = 900a, so a = 1.5 m s⁻²
v = u + at = 0 + 1.5 × 8 = 12 m s⁻¹

The second law gives a. The constant-acceleration formulas then give speeds, times and distances.

Newton’s second law with vectors (3.03d)

In vector form, F = ma applies to the whole vector. Column vectors and i, j notation are both used, so F = (F₁, F₂) or F = F₁i + F₂j. Each component obeys the law separately.

Example 3. Find the force, in vector form, that gives a 3 kg body an acceleration of (2i − 5j) m s⁻². Find its magnitude.

F = 3(2i − 5j) = 6i − 15j N
|F| = √(6² + 15²) = √261 = 16.2 N (3 s.f.)

Example 4. Forces (5i + 20j) N and (7i − 4j) N act on a 4 kg particle. Find the acceleration, its magnitude and its direction.

Resultant = (12i + 16j) N
a = (12i + 16j)/4 = 3i + 4j m s⁻²
|a| = √(3² + 4²) = 5 m s⁻²
Angle with i: tan⁻¹(4/3) = 53.1° (1 d.p.)

Resolved forces (3.03e) — A Level only

If a force acts at angle θ to the direction of motion, only its component in that direction contributes: F cos θ. The perpendicular component F sin θ changes the contact force instead.

Example 5. A 5 kg body is on a smooth horizontal surface. A 30 N force acts at 25° above the horizontal. Find the acceleration and the normal reaction.

Horizontal: 30 cos 25° = 5a, so a = 5.44 m s⁻² (3 s.f.)
Vertical (no vertical acceleration): R + 30 sin 25° = 5 × 9.8
R = 49 − 12.68 = 36.3 N (3 s.f.)

The pull partly lifts the body, so R is less than the weight.

Example 6. A particle slides down a line of greatest slope of a smooth plane inclined at 30°. Find its acceleration.

Weight component down the slope = mg sin 30°
m a = m × 9.8 × sin 30°
a = 4.9 m s⁻²

The mass cancels. The normal reaction (mg cos 30°) does not appear in the equation along the slope.

Weight and the value of g (3.03f, 3.03g)

The weight of a body is the gravitational force on it, W = mg, directed towards the centre of the planet. Mass, in kg, stays the same wherever the body is. Weight is a force, in N, and changes with g.

You may assume g = 9.8 m s⁻². The specification wants you to know that g varies with location in the universe, so a 60 kg person weighs less on the Moon than on Earth while their mass stays 60 kg. Use the value the question gives and write answers to the accuracy that value supports. The inverse square law of gravitation is not required.

Example 7. A 0.4 kg ball falls with constant air resistance of 1.2 N. Find its acceleration.

Take downwards as positive.
0.4 × 9.8 − 1.2 = 0.4a
3.92 − 1.2 = 0.4a, so a = 6.8 m s⁻²

On Earth, with no resistance, any falling body has a = g, whatever its mass.

Newton’s third law and contact forces (3.03h, 3.03i)

Every action has an equal and opposite reaction. If A pushes B with force F, then B pushes A with force F in the opposite direction. The two forces:

• have equal magnitude and opposite directions,
• are of the same type (both contact, or both gravitational),
• act on different bodies.

Because they act on different bodies they never cancel in one equation of motion. A book on a table: the table pushes up on the book (R); the book pushes down on the table (R). The weight of the book and R are not a third-law pair, because both act on the book.

Normal reaction. The force between a body and a surface is perpendicular to the surface. For a body at rest on a horizontal surface, R = W. In general R comes from the equation of motion in the perpendicular direction. R = 0 means contact is lost. R can never be negative, because a surface can push but not pull.

One particle. If parts of a system have no relative motion, you may model them as a single particle with their combined mass.

Example 8. A passenger of mass 75 kg stands in a lift that accelerates downwards at 1.5 m s⁻². Find the force from the floor.

Downwards positive: 75 × 9.8 − R = 75 × 1.5
R = 735 − 112.5 = 622.5 N

The passenger presses on the floor with 622.5 N by the third law. If the lift’s acceleration reached 9.8 m s⁻² downwards, R = 75(9.8 − 9.8) = 0 and contact would be lost.

Example 9. A lift of mass 280 kg carries a 65 kg passenger. The cable tension T accelerates both upwards at 0.7 m s⁻². Find T and the force between passenger and floor.

Whole system as one particle (345 kg):
T − 345 × 9.8 = 345 × 0.7, so T = 3622.5 N
Passenger alone:
R − 65 × 9.8 = 65 × 0.7, so R = 682.5 N

Smooth contact and light strings (3.03j)

• Smooth means no friction. The contact force on a smooth surface is purely normal.
• Light means the mass of the string is negligible, so the tension is the same at every point of a single string with no pulley or peg.
• Inextensible means its length does not change, so two ends moving together have the same speed and acceleration.
• A string can only pull. Tension T ≥ 0; T = 0 means it is slack.

Limits: no real surface is exactly smooth, a real string has mass and stretches, and an object treated as a particle has size and may rotate. The model suits cases where friction is small and the string is light compared with the load.

Equilibrium with motion in a straight line (3.03k)

The same equation of motion handles every case: constant velocity gives zero resultant, so the tension equals the weight. Any change in speed gives a resultant, so the tension differs from the weight.

Example 10. A 2 kg particle on a light inextensible string is raised from rest, rising 3 m in 2 s with constant acceleration. Find the tension.

s = ut + ½at²: 3 = ½ × a × 4, so a = 1.5 m s⁻²
T − 2 × 9.8 = 2 × 1.5, so T = 22.6 N

If the string can take at most 50 N, then 50 − 19.6 = 2a gives a = 15.2 m s⁻² as the largest upward acceleration.

Example 11. A lift of mass 500 kg (with load) accelerates from rest at 1.2 m s⁻² for 3 s, moves at constant speed, then decelerates at 0.8 m s⁻² to rest.

Phase 1: T = 500(9.8 + 1.2) = 5500 N, speed reached = 1.2 × 3 = 3.6 m s⁻¹
Phase 2: constant speed, T = 500 × 9.8 = 4900 N
Phase 3: T − 4900 = 500(−0.8), so T = 4500 N; time = 3.6 ÷ 0.8 = 4.5 s

If the cable is cut, only weight remains, so the acceleration is 9.8 m s⁻² downwards at once, though the body keeps its velocity at that instant.

Common errors

• Leaving out the weight in a vertical problem, or using R = mg when the body is accelerating.
• Writing an equation with forces from different bodies, or from different directions, without signs.
• Calling the weight and the normal reaction a third-law pair.
• Using mass where a force is needed (kg where N is needed).
• Giving a negative tension or normal reaction without comment: it means the model has failed, not that the string pushes.
• Resolving a force with cos where sin is needed: check which angle is measured from the direction of motion.

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.