A-Level Maths mechanics revision: forces, motion, moments
A-Level mechanics covers kinematics (the SUVAT equations, motion graphs and calculus-based motion), projectiles, forces and Newton's laws including friction and resolving, connected particles, and moments with rigid-body equilibrium. The marks go to students who draw the diagram first, resolve in consistent directions, and remember that 'modelled as a particle' and 'state an assumption' lines are free marks that appear on nearly every paper.
Mechanics frightens students out of proportion to its difficulty. The maths inside it is mostly Year 1 algebra and resolving; the hard part is translating a worded situation into a correct force diagram. That is a learnable skill with a small set of moves — and once the diagram is right, the equations write themselves.
The specification, in plain English
| Area | What's in it | The trap |
|---|---|---|
| Kinematics | SUVAT, displacement–time and velocity–time graphs, calculus motion | SUVAT needs constant acceleration only |
| Projectiles | Horizontal + vertical components, time of flight, range | Horizontal velocity is constant; only vertical accelerates |
| Forces & Newton's laws | F = ma, resolving, weight vs mass, normal reaction | Forgetting a force, or double-counting weight's components |
| Friction | F ≤ μR, limiting equilibrium F = μR | R ≠ mg on a slope — resolve perpendicular first |
| Connected particles | Pulleys, slopes, towed objects | Each object needs its own F = ma equation |
| Moments | Moment = force × perpendicular distance, equilibrium, tilting | Perpendicular distance; taking moments about the wrong point |
The assumption lines that carry free marks
- "Modelled as a particle" — mass acts at a point, so no rotation and forces are concurrent.
- "Light string" — its mass is ignored; tension is the same throughout.
- "Inextensible string" — both particles have the same acceleration magnitude.
- "Smooth pulley / surface" — no friction at that contact.
- "Ignoring air resistance" — the only acceleration is g.
- Each of these phrases is worth a mark in 'state an assumption' questions — keep the list ready.
Worked questions
A ball is thrown vertically upwards at 14 m/s. Find the time to reach maximum height and the height reached.
- Choose up as positive: u = 14, a = −9.8, v = 0 at the top.
- v = u + at gives 0 = 14 − 9.8t, so t = 14/9.8 ≈ 1.43 s.
- s = ut + ½at² gives s = 14(1.43) − 4.9(1.43²) ≈ 20 − 10 = 10 m.
- Sanity check: v² = u² + 2as gives 0 = 196 − 19.6s, s = 10 m. Two routes, same answer.
t ≈ 1.43 s, s = 10 m. At the top v = 0 — that fact is the whole question.
A 4 kg block on a rough slope at 30° is connected over a smooth pulley to a 6 kg hanging mass. μ = 0.3. Find the acceleration.
- On the slope, resolve perpendicular: R = 4g cos 30° = 4g × 0.866 ≈ 33.9 N.
- Friction: F = μR = 0.3 × 33.9 ≈ 10.2 N, acting down the slope against the motion.
- Along the slope for the 4 kg block: T − 4g sin 30° − F = 4a → T − 19.6 − 10.2 = 4a.
- For the hanging mass: 6g − T = 6a → 58.8 − T = 6a.
- Add the equations: 58.8 − 29.8 = 10a, so a = 29/10 = 2.9 m/s².
a ≈ 2.9 m/s². The two F = ma equations, added to eliminate T, are the standard skeleton — it barely changes between papers.
A uniform 5 m beam of weight 40 N rests on supports at its ends. A 60 N load sits 2 m from the left end. Find the reaction at each support.
- Let the reactions be R₁ (left) and R₂ (right). Vertically: R₁ + R₂ = 100 N.
- Take moments about the left support to eliminate R₁. The beam's weight acts at its centre, 2.5 m out.
- Clockwise = anticlockwise: R₂ × 5 = 40 × 2.5 + 60 × 2 = 100 + 120 = 220.
- R₂ = 44 N, so R₁ = 56 N.
- Check with moments about the right end — R₁ × 5 = 40 × 2.5 + 60 × 3 = 280, giving R₁ = 56. Always verify with the second support.
R₁ = 56 N, R₂ = 44 N. Take moments about a support to kill its unknown — that's the move.
Where marks get dropped
- No diagram, or a diagram missing a force — the normal reaction and friction are the ones left out.
- On a slope, using R = mg instead of mg cos θ — the reaction is always the perpendicular component.
- In SUVAT, mixing signs — pick a positive direction and hold it for every quantity.
- In projectiles, accelerating the horizontal motion — only vertical motion has g.
- Forgetting that both objects in a connected-particles question need their own equation.
- In moments, using distance instead of perpendicular distance when the force is angled.
Your mechanics checklist
- I draw a labelled force diagram before writing a single equation.
- I know all five SUVAT equations and when they do not apply.
- I can resolve on a slope and get R = mg cos θ every time.
- I can set up and solve connected-particles equations on flat and inclined surfaces.
- I can split a projectile into components and find time of flight and range.
- I can take moments about a chosen point and handle tilting problems.
- I have the standard modelling assumptions memorised as one-liners.
Mechanics clicks when the diagram becomes automatic — and it stops clicking the moment a force is missing and nobody tells you which. That is the feedback Lumi gives: it draws the diagram with you, asks which force you think acts next, and shows where the mark scheme got each line.
What to take from this
- Draw the diagram before writing an equation — half of mechanics errors are missing or wrongly-directed forces.
- SUVAT applies only to constant acceleration; calculus handles the rest.
- On a slope, resolve before applying F = μR — the normal reaction is not mg.
- Connected particles: each object gets its own F = ma equation, then solve simultaneously.
- The 'state an assumption' or 'limitation of the model' line is worth a mark on most mechanics questions.
Questions people also ask
Students split about evenly. Mechanics is harder to start — the diagram skill is genuinely new — but its question types repeat more faithfully than any other part of the course, so it rewards drilling unusually well. Statistics is easier to start but fiddlier to finish.
They are not all in the formula booklet, so yes — v = u + at, s = ut + ½at², s = vt − ½at², v² = u² + 2as, s = ½(u + v)t. More useful is knowing which one omits which variable: pick the equation that avoids the unknown you don't have.
It is the contact force perpendicular to the surface. On flat ground with no other vertical forces it equals mg, but on a slope it equals mg cos θ, and with extra vertical forces it changes again. Always resolve perpendicular to the surface to find it.
Most students name moments with tilting — deciding which support loses contact and taking moments about the right point. Connected particles on rough slopes runs it close. Both become routine after five or six worked attempts.