Cambridge International AS & A Level Mathematics 9709: Mechanics – Revision Notes

These notes condense Mechanics, topic 4 of the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4), sections 4.1 to 4.5. The content is examined only on Paper 4 (1 hour 15 minutes, 50 marks), which counts towards both the AS Level and the A Level. For full explanations and worked examples, read the Mechanics study guide first.

Practise with the Mechanics practice questions. Course hub: Cambridge A Level Mathematics. Checklist: 9709 printable checklist. Quick checks: AS diagnostic and A Level diagnostic.

Use g = 10 m s⁻² throughout, as the syllabus expects.

Formula table

Quantity	Formula	In MF19?
Constant acceleration	v = u + at, s = ½(u + v)t, s = ut + ½at², v² = u² + 2as	Yes
Weight	W = mg	No
Friction	F ≤ μR; F = μR when limiting or sliding	No
Newton’s second law	resultant force = ma	No
Momentum	mv	No
Work done by a constant force	Fd cos θ	No
Kinetic energy	½mv²	No
Gravitational potential energy	mgh	No
Power	work done ÷ time; P = Fv	No
Calculus in t	v = ds/dt, a = dv/dt; s = ∫v dt, v = ∫a dt	No

4.1 Forces and equilibrium

• Resolve along and perpendicular to a convenient direction (on a slope: along and perpendicular to the plane).
• Weight components on a plane at angle α: mg sin α down the slope, mg cos α into the slope.
• Equilibrium: the sum of components in every direction is zero.
• Smooth contact: no friction. This is a model, not reality.
• “About to slip” means limiting equilibrium, so F = μR.
• Newton’s third law: forces between two bodies are equal in size and opposite in direction.

Method: equilibrium with friction

1. Draw the diagram with every force labelled.
2. Decide which way the particle would move, and draw friction opposite to it.
3. Resolve perpendicular to the surface to find R.
4. Resolve along the surface.
5. Use F = μR only if the particle is limiting or moving; otherwise F ≤ μR.

4.2 Kinematics

• Distance and speed are scalars; displacement, velocity and acceleration are vectors.
• s–t graph: gradient = velocity. v–t graph: gradient = acceleration; area = displacement.
• Constant a → MF19 formulae. Variable a (a function of t) → calculus.
• Maximum or minimum velocity: set a = dv/dt = 0.
• At rest (instantaneously): set v = 0.

Worked reminder: two particles. P starts from rest at O with acceleration 2 m s⁻². At the same moment, Q passes O at a constant 12 m s⁻¹ in the same direction. P catches Q when t² = 12t, so t = 12 s, 144 m from O.

4.3 Momentum

• Total momentum before = total momentum after (direct impact, one dimension).
• Coalesce: the bodies join and move with one common velocity.
• Choose a positive direction first; velocities the other way are negative.
• A negative answer for a velocity means the body moves in the negative direction.
• Kinetic energy is usually lost in an impact, even though momentum is conserved.

Worked reminder: a rebound. A 0.4 kg ball moving at 5 m s⁻¹ hits a stationary 0.6 kg ball and rebounds at 1 m s⁻¹. Taking the ball’s first direction as positive: 0.4 × 5 = 0.4 × (−1) + 0.6v, so 2 = −0.4 + 0.6v and v = 4 m s⁻¹. The rebound velocity is entered as −1, not +1.

4.4 Newton’s laws

Method: connected particles

1. Draw a separate diagram for each particle.
2. Write resultant force = ma for each particle in its own direction of motion.
3. Use the fact that the string (or tow-bar) gives both the same acceleration.
4. Add the equations to eliminate the tension, or use the whole system.
5. Find the tension from one particle’s equation.

Worked reminder: particles over a pulley. Particles of masses 3 kg and 2 kg hang on either side of a smooth pulley. System: 30 − 20 = 5a, so a = 2 m s⁻². For the 2 kg particle: T − 20 = 2 × 2, so T = 24 N. Check with the 3 kg particle: 30 − T = 3 × 2 gives T = 24 N.

• Smooth plane at α: a = g sin α down the plane (5 m s⁻² at 30°).
• Rough plane: acceleration going up is not the same as coming down, because friction reverses.
• A string once taut then slack: after it goes slack, each particle moves under its own forces only.

4.5 Energy, work and power

Method: work–energy

1. Choose the start and end positions.
2. Write the KE and PE at each.
3. Find the work done against resistance (resistance × distance) and by any driving force.
4. Initial energy + work by driving forces = final energy + work against resistances.

Method: power and acceleration

1. Driving force D = P/v, with v the speed at that instant.
2. Resultant force = D − resistance (− mg sin θ on a hill going up).
3. Resultant force = ma. At maximum (steady) speed, a = 0.

Worked reminder: average power. Raising 50 kg through 6 m at constant speed in 4 s needs 50 × 10 × 6 = 3000 J of work, so the average power is 750 W.

Must-know distinctions

• Mass vs weight: mass in kg; weight mg in newtons.
• Distance vs displacement: distance adds all motion; displacement is the net change in position.
• Speed vs velocity: velocity has a sign; speed does not.
• Tension vs thrust: a string can only pull; a rod or tow-bar can pull (tension) or push (thrust).
• Limiting vs not limiting: F = μR only at the point of slipping or while sliding.
• Momentum vs kinetic energy: momentum is conserved in impacts; kinetic energy is generally not.
• Driving force vs resultant force: P = Fv uses the driving force, not the resultant.

Quick self-test

1. Find the magnitude of the resultant of forces 6 N and 8 N at right angles, and its angle with the 6 N force.
2. A 5 kg box is on rough horizontal ground with μ = 0.4. Find the least horizontal force that makes it move.
3. A particle starts from rest with acceleration 3 m s⁻². Find its speed and distance after 4 s.
4. A stone is dropped from rest and falls 45 m. Find the time taken and its final speed.
5. A car accelerates uniformly from rest to 10 m s⁻¹ in 5 s, stays at 10 m s⁻¹ for 10 s, then decelerates uniformly to rest in 5 s. Find the total distance.
6. A particle has velocity v = 3t² − 12t m s⁻¹. When, for t > 0, is it instantaneously at rest?
7. A 2 kg particle moving at 6 m s⁻¹ hits a stationary 4 kg particle and they coalesce. Find their common speed.
8. A 60 kg person stands in a lift accelerating upwards at 1.5 m s⁻². Find the normal contact force from the floor.
9. Find the work done by a 20 N force at 60° to the direction of motion over 5 m.
10. A car moves at a steady 30 m s⁻¹ on a level road with its engine working at 24 kW. Find the resistance.
11. Find the gain in potential energy when a 3 kg mass is raised 4 m.
12. A car of mass 800 kg works at 16 kW against a resistance of 400 N on level ground. Find its acceleration at 10 m s⁻¹.

Answers

1. √(6² + 8²) = 10 N; tan θ = 8/6, θ = 53.1°.
2. R = 50 N, so the force is 0.4 × 50 = 20 N.
3. v = 3 × 4 = 12 m s⁻¹; s = ½ × 3 × 16 = 24 m.
4. 45 = 5t², t = 3 s; v = 10 × 3 = 30 m s⁻¹.
5. Area = 25 + 100 + 25 = 150 m.
6. 3t(t − 4) = 0, so t = 4 s.
7. 2 × 6 = 6v, v = 2 m s⁻¹.
8. R − 600 = 60 × 1.5, R = 690 N.
9. 20 cos 60° × 5 = 50 J.
10. Steady speed so D = resistance: 24 000/30 = 800 N.
11. 3 × 10 × 4 = 120 J.
12. D = 16 000/10 = 1600 N; 1600 − 400 = 800a, a = 1.5 m s⁻².

Where marks are usually lost

• Leaving out a force from the diagram, most often friction or the normal contact force on a slope.
• Taking R = mg on a slope or when a pulling force has a vertical component.
• Drawing friction the wrong way; it opposes the motion that would happen.
• Using constant-acceleration formulae when acceleration varies with time.
• Confusing distance with displacement when a particle changes direction.
• Not choosing a positive direction before a momentum calculation.
• Missing the weight component mg sin θ when a car climbs a hill.
• Using P = Fv with the resultant force instead of the driving force.
• Rounding intermediate values to 2 or 3 significant figures, so the final answer is inaccurate.

Official syllabus

Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027, Version 4, Cambridge Assessment International Education (part of Cambridge University Press & Assessment). Topic 4, Mechanics (for Paper 4).