Cambridge International AS & A Level Mathematics 9709: Mechanics – Study Guide

This guide teaches Mechanics, topic 4 of the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4). It covers syllabus sections 4.1 to 4.5. The topic is examined only on Paper 4: 1 hour 15 minutes, 50 marks, 6 to 8 structured questions. Paper 4 is worth 40% of the AS Level on the Paper 1 and Paper 4 route, and 20% of the A Level on the Papers 1, 3, 4 and 5 route. It cannot be combined with Paper 6. The syllabus assumes the algebra from Paper 1.

Course hub: Cambridge A Level Mathematics. Printable list: 9709 checklist. Both 10-minute diagnostics include a Mechanics question: the AS diagnostic and the A Level diagnostic.

What this topic covers

Section	What you must be able to do
4.1 Forces and equilibrium	Draw force diagrams; resolve forces; use equilibrium; use friction, F = μR or F ≤ μR; use Newton’s third law
4.2 Kinematics of motion in a straight line	Use displacement–time and velocity–time graphs; use calculus in t; use the constant-acceleration formulae
4.3 Momentum	Use momentum = mv in one dimension; conserve momentum in direct impacts, including coalescence
4.4 Newton’s laws of motion	Use F = ma for a particle and for connected particles; use W = mg; solve vertical and inclined-plane problems
4.5 Energy, work and power	Work done W = Fd cos θ; kinetic and potential energy; the work–energy principle; power P = Fv

Paper conventions from the syllabus. Take g = 10 m s⁻². Questions are mainly numerical. Bodies are modelled as particles, so every force acts at one point. Vector notation is not used in the question papers. You need these trig facts: sin(90° − θ) = cos θ, cos(90° − θ) = sin θ, tan θ = sin θ/cos θ and sin²θ + cos²θ = 1. Give non-exact answers to 3 significant figures, or 1 decimal place for angles in degrees.

4.1 Forces and equilibrium

Start every problem with a clear force diagram: weight, normal contact force, friction, tension or thrust, and any applied force. A force P at angle θ to a direction has component P cos θ along that direction and P sin θ perpendicular to it.

A particle is in equilibrium when the resultant force is zero. In practice, resolve in two perpendicular directions and set each sum to zero. You must calculate; scale drawings earn nothing.

Worked example (two strings). A particle of weight 20 N hangs from two light strings. One string makes 30° with the horizontal and has tension A. The other makes 60° with the horizontal and has tension B. Find A and B.

Horizontal:  A cos 30° = B cos 60°   →   B = A√3
Vertical:    A sin 30° + B sin 60° = 20
             ½A + (A√3)(√3/2) = 20   →   2A = 20
A = 10 N,  B = 10√3 = 17.3 N

Friction. A contact force between two surfaces has a normal component R and a frictional component F. On a smooth contact, F = 0; this model ignores all friction, so it is only an approximation. On a rough contact, F ≤ μR. When the particle is in limiting equilibrium (also described as “about to slip”), F = μR. Friction acts against the direction the particle would move.

Worked example (limiting equilibrium). A box of mass 4 kg rests on rough horizontal ground, μ = 0.35. A rope pulls it with tension T at 20° above the horizontal. The box is about to move. Find T.

Vertical:    R + T sin 20° = 40        →   R = 40 − T sin 20°
Horizontal:  T cos 20° = F = 0.35R
T cos 20° = 0.35(40 − T sin 20°)
T (cos 20° + 0.35 sin 20°) = 14      →   T = 13.2 N

The upward component of T reduces R, so the friction limit is less than 0.35 × 40 = 14 N.

Newton’s third law. If body A exerts a force on B, then B exerts an equal and opposite force on A. The ground pushes up on a particle with R, and the particle pushes down on the ground with R.

4.2 Kinematics of motion in a straight line

Distance and speed are scalars. Displacement, velocity and acceleration are vectors; in one dimension their sign shows direction. “Deceleration” can mean the speed is decreasing.

Graphs.

• Gradient of a displacement–time graph = velocity.
• Gradient of a velocity–time graph = acceleration.
• Area under a velocity–time graph = displacement. Area below the axis counts as negative displacement, but it still adds to the distance travelled.

Calculus. v = ds/dt and a = dv/dt. Going the other way, s = ∫v dt and v = ∫a dt, with a constant found from the starting conditions. Only Paper 1 calculus is needed.

Worked example. A particle starts from rest at O. Its acceleration is a = 8 − 4t m s⁻². Find when it is next at rest, and its displacement from O then.

v = ∫(8 − 4t) dt = 8t − 2t² + c;  v = 0 at t = 0, so c = 0
v = 0:  2t(4 − t) = 0   →   t = 4 s
s = ∫(8t − 2t²) dt = 4t² − (2/3)t³   (s = 0 at t = 0)
s(4) = 64 − 128/3 = 64/3 = 21.3 m

Constant acceleration. The formulae v = u + at, s = ½(u + v)t, s = ut + ½at² and v² = u² + 2as are in the formula list (MF19). Use them only when a is constant. Choose one positive direction and keep to it.

Worked example (vertical motion). A ball is thrown vertically upwards at 15 m s⁻¹ from a point 20 m above the ground. Find the time it takes to reach the ground.

Take upwards as positive: s = −20, u = 15, a = −10. Then −20 = 15t − 5t², so t² − 3t − 4 = 0 and (t − 4)(t + 1) = 0. The ball lands after 4 s; reject t = −1.

4.3 Momentum

Momentum = mv. It is a vector, so in one dimension its sign gives its direction. In a direct impact between two bodies, total momentum before = total momentum after. If the bodies coalesce, they move on together with one common velocity. Impulse and the coefficient of restitution are not in this syllabus.

Worked example. A (3 kg) moves at 5 m s⁻¹ towards B (2 kg), which moves at 4 m s⁻¹ towards A. After the impact A moves at 1 m s⁻¹ in its original direction. Find B’s velocity after impact.

Take A’s original direction as positive: 3(5) + 2(−4) = 3(1) + 2v. So 7 = 3 + 2v and v = 2 m s⁻¹. B now moves at 2 m s⁻¹ in A’s original direction: its direction has reversed.

Kinetic energy is not conserved in general. Here it falls from 53.5 J to 5.5 J.

4.4 Newton’s laws of motion

Newton’s second law for a particle of constant mass: resultant force = ma. Weight W = mg. Resistances such as air resistance are included only when the question says so.

Rough inclined plane. The acceleration up the plane differs from the acceleration down it, because friction reverses.

Worked example. A particle of mass 2 kg is projected at 8 m s⁻¹ up a line of greatest slope of a rough plane. The plane is inclined at α with sin α = 0.6 and cos α = 0.8, and μ = 0.25. Find the distance it travels up the plane and its speed when it returns to the start.

R = 2 × 10 × 0.8 = 16 N,   F = 0.25 × 16 = 4 N
Weight component down the plane = 2 × 10 × 0.6 = 12 N
Up:    −(12 + 4) = 2a   →   a = −8 m s⁻²
       0 = 8² − 2 × 8 × s   →   s = 4 m
Down:  12 − 4 = 2a   →   a = 4 m s⁻²
       v² = 0 + 2 × 4 × 4 = 32   →   v = 5.66 m s⁻¹

It does slide back, because 12 N down the plane is more than the maximum friction of 4 N.

Connected particles. Treat each particle separately, or the whole system when the internal force is not needed. A light inextensible string has the same tension throughout and gives both particles the same speed. A rigid tow-bar can be in tension or in thrust.

Worked example (car and trailer). A car of mass 1200 kg tows a trailer of mass 400 kg with a light rigid tow-bar. The driving force is 2400 N. Resistances are 300 N on the car and 100 N on the trailer.

Whole system: 2400 − 300 − 100 = 1600a, so a = 1.25 m s⁻². Trailer: T − 100 = 400 × 1.25, so the tension is T = 600 N.

If the engine is switched off and the car brakes with a force of 1800 N, the system decelerates at (1800 + 300 + 100)/1600 = 1.375 m s⁻². For the trailer, 100 + T = 400 × 1.375, so the tow-bar pushes back on the trailer with a thrust of 450 N.

4.5 Energy, work and power

• Work done by a constant force: W = Fd cos θ, where θ is the angle between the force and the displacement.
• Kinetic energy = ½mv². Gravitational potential energy = mgh.
• Work–energy principle: the change in total mechanical energy equals the work done by forces other than weight. With no friction or other resistance, mechanical energy is conserved.
• Power is the rate of doing work. Average power = work done ÷ time taken. For a force in the direction of motion, P = Fv.

Worked example (curved slide). A child moves down a smooth curved slide from a point 3.2 m above the bottom, starting at 2 m s⁻¹. Energy: ½v² = ½(2²) + 10 × 3.2, so v² = 68 and v = 8.25 m s⁻¹. The shape of the slide does not matter: only the overall height change counts.

Worked example (car on a hill). A car of mass 1000 kg has an engine working at 30 kW. It moves up a hill inclined at θ to the horizontal, where sin θ = 0.05, against a resistance of 600 N. Find its acceleration when its speed is 20 m s⁻¹.

Driving force = P/v = 30 000/20 = 1500 N
Weight component down the hill = 1000 × 10 × 0.05 = 500 N
1500 − 600 − 500 = 1000a   →   a = 0.4 m s⁻²

At maximum speed a = 0. On level ground with the same power and resistance, the maximum speed is 30 000/600 = 50 m s⁻¹.

Common errors

• Using R = mg when a force at an angle also has a vertical component.
• Using F = μR when the particle is not in limiting equilibrium or moving.
• Using constant-acceleration formulae when a depends on t.
• Adding areas below the time axis as positive when displacement is asked for.
• Giving both momenta the same sign when the bodies move towards each other.
• Using the same acceleration for motion up and down a rough plane.
• Using the full force instead of F cos θ for work done.
• Using P = Fv with the resistance instead of the driving force.

Next steps

Condensed notes and a self-test are in the Mechanics revision notes. Then work through the Mechanics practice questions, which use different numbers and situations from the examples here.

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).