AQA A-level Physics (7408) · 3.4 Mechanics and materials
Mini-Lesson
Mechanics and materials
This mini-lesson covers AQA 3.4 — Mechanics and materials: moments and equilibrium, projectile motion, Newton’s laws, momentum and impulse, work, energy and power, Hooke’s law, stress, strain and the Young modulus.
Section 3.4 is compulsory and appears on Paper 1.
Where this sits: AQA sections 3.1–3.8 are compulsory for every A-level Physics student. (Sections 3.9–3.13 are the five options — you study just one of those.)
Work through each screen, answer the questions as you go (many are full A-level calculations) and collect ⭐ stars. Press Start when you're ready.
3.4.1 · forces in equilibrium
Scalars, vectors, moments and couples
Scalars have magnitude only (mass, energy, speed). Vectors have magnitude and direction (force, velocity, momentum). Vectors are added tip-to-tail, or resolved into perpendicular components:
Fx = F cosθ · Fy = F sinθa body is in equilibrium when the resultant force is zero AND the resultant moment is zero
Moment = force × perpendicular distance from the pivot (N m).
Couple = two equal, antiparallel forces that are not in line. Its torque = one force × the perpendicular separation. A couple produces rotation with no resultant force.
Principle of moments: for equilibrium, total clockwise moment = total anticlockwise moment about any point.
The weight of a uniform beam acts at its centre of mass — the midpoint.
Worked example
A light rod is pivoted at its centre. A 20 N weight hangs 1.2 m to the left. What force F, placed 0.80 m to the right, balances it?
F × 0.80 = 20 × 1.2 = 24 N m → F = 24 ÷ 0.80 = 30 N.
Calculate
Your turn — calculation 1
1A light rod is pivoted at its centre. A 20 N weight hangs 1.2 m to the left of the pivot. Calculate the force needed 0.80 m to the right of the pivot to keep the rod horizontal.
N
Hint: Anticlockwise moment = 20 × 1.2 = 24 N m. So F × 0.80 = 24.
3.4.1 · motion & projectiles
Motion, the suvat equations and projectiles
For uniform acceleration in a straight line:
v = u + at · s = ut + ½at² · v² = u² + 2as · s = ½(u + v)t
Projectiles are the classic vector problem. Treat the two directions completely separately:
Horizontal: no force (ignoring air resistance) → velocity is constant, so s = vxt.
Vertical: acceleration is g = 9.81 m s−2 downwards, so use suvat.
The time of flight is set by the vertical motion and links the two.
Worked example — horizontal launch
A ball is thrown horizontally at 12 m s−1 from a cliff 20 m high.
Vertical: 20 = ½ × 9.81 × t² → t² = 4.08 → t = 2.02 s.
Horizontal: range = 12 × 2.02 = 24.2 m.
Air resistance reduces the range and makes the path asymmetric: the descent is steeper than the ascent, and the peak comes past the halfway point of the trajectory.
Calculate
Your turn — calculation 2
2A ball is thrown horizontally at 12 m s−1 from a cliff 20 m above the sea. Calculate its horizontal distance from the cliff when it lands. (g = 9.81 m s−2, ignore air resistance)
m
Hint: Vertical: 20 = ½ × 9.81 × t² → t = 2.02 s. Horizontal: distance = 12 × t.
Quick check
Think it through
?A skydiver reaches terminal velocity. Which statement is correct?
3.4.1 · Newton & momentum
Newton’s laws, momentum and impulse
Newton’s laws: (1) a body stays at rest or at constant velocity unless acted on by a resultant force; (2) the resultant force equals the rate of change of momentum; (3) forces act in equal and opposite pairs on different bodies.
F = Δp ÷ Δt → FΔt = Δ(mv)impulse = force × time = change in momentum (N s = kg m s−1) = area under a force–time graph
Momentum is always conserved in a collision or explosion (no external resultant force). Momentum is a vector — watch the signs.
Elastic collision: kinetic energy is also conserved. Inelastic: some KE becomes internal energy; momentum is still conserved.
Crumple zones, airbags, catching a ball by pulling back all increase Δt for the same Δp, so the force is reduced.
Worked example
A 0.15 kg ball hits a wall at 20 m s−1 and rebounds at 15 m s−1.
Taking towards the wall as positive: Δp = 0.15 × (−15) − 0.15 × (+20) = −2.25 − 3.00 = −5.25 kg m s−1.
Magnitude of the impulse = 5.25 N s. If the contact lasts 0.020 s, the mean force = 5.25 ÷ 0.020 = 263 N.
Calculate
Your turn — calculation 3
3A 0.15 kg ball hits a wall at 20 m s−1 and rebounds along the same line at 15 m s−1. Calculate the magnitude of the impulse on the ball.
N s
Hint: The ball reverses direction, so the speeds ADD: impulse = 0.15 × (20 + 15).
Sort it
Elastic collision, inelastic collision, or true of both?
Tap a statement, then tap the box it belongs in.
🎱 Elastic only
💥 Inelastic only
🔁 True of both
3.4.1 · work, energy, power
Work, energy and power
W = Fs cosθ · Ek = ½mv² · ΔEp = mgΔhP = ΔW ÷ Δt = Fv · efficiency = useful output ÷ total input
The cosθ matters: a force perpendicular to the motion does no work (which is why the centripetal force in circular motion does none).
Work done = area under a force–displacement graph.
Conservation of energy: energy cannot be created or destroyed. In a fall with no air resistance, mgh converts entirely into ½mv².
Worked example — power from P = Fv
A car travels at a steady 25 m s−1 against a total resistive force of 600 N.
At constant speed the driving force also equals 600 N, so P = Fv = 600 × 25 = 15 kW of useful output power.
Calculate
Your turn — calculation 4
4A crane lifts a 250 kg load through a height of 12 m in 20 s at a steady speed. Calculate the crane’s useful output power, in kW. (g = 9.81 m s−2)
kW
Hint: Work done = mgΔh = 250 × 9.81 × 12 = 29 430 J. Power = W ÷ t = 29 430 ÷ 20 W. Convert to kW.
3.4.2 · materials
Hooke’s law, elastic strain energy and density
F = kΔL · E = ½FΔL = ½k(ΔL)² · ρ = m ÷ Vk = stiffness (spring constant, N m−1) · elastic strain energy = area under a force–extension graph
Hooke’s law holds up to the limit of proportionality. Beyond the elastic limit the material no longer returns to its original length — it deforms plastically.
Springs in series: each carries the full load, so the extensions add → the combination is less stiff. In parallel: they share the load → stiffer (k values add).
In a loading–unloading cycle beyond the elastic limit, the area between the curves is the energy dissipated as internal energy (hysteresis, as in rubber).
Worked example
A spring of stiffness 200 N m−1 is extended 0.15 m elastically.
E = ½k(ΔL)² = ½ × 200 × 0.15² = 0.5 × 200 × 0.0225 = 2.25 J.
Calculate
Your turn — calculation 5
5A spring with spring constant 200 N m−1 is stretched elastically by 0.15 m. Calculate the elastic strain energy stored.
J
Hint: E = ½k(ΔL)² = 0.5 × 200 × 0.15².
3.4.2 · Young modulus
Stress, strain and the Young modulus
stress σ = F ÷ A · strain ε = ΔL ÷ L · E = σ ÷ εstress in Pa (N m−2) · strain has no units · Young modulus E in Pa, and equals the gradient of the linear part of a stress–strain graph
Breaking stress is the stress at which a material fractures — a property of the material, not of the sample size.
Brittle (glass): obeys Hooke’s law right up to fracture, with no plastic region. Ductile (copper): a long plastic region, so it can be drawn into wires.
Required practical: measure the diameter of a wire in several places with a micrometer (it is squared in A = πd²/4, so it dominates the uncertainty), use a long thin wire for a measurable extension, and wear eye protection.
Worked example
Wire: L = 2.0 m, d = 0.50 mm, F = 40 N, ΔL = 2.4 mm.
A = π(0.25×10−3)² = 1.96×10−7 m². σ = 40 ÷ 1.96×10−7 = 2.04×108 Pa.
ε = 2.4×10−3 ÷ 2.0 = 1.2×10−3. E = 2.04×108 ÷ 1.2×10−3 = 1.7×1011 Pa — about right for steel.
Calculate
Your turn — calculation 6
6A steel wire is 2.0 m long with diameter 0.50 mm. A force of 40 N produces an extension of 2.4 mm. Calculate the Young modulus. Give the number in front of ×1011 Pa.
× 10¹¹ Pa
Hint: A = πr² = π(0.25×10−3)² = 1.96×10−7 m². Stress = 40/A; strain = 2.4×10−3/2.0; E = stress ÷ strain.
Match it
Match each definition to its quantity
Tap a card on the left, then its partner on the right.
Statement
Answer
Quick check
Think it through
?The Young modulus of a material is best found from a stress–strain graph as
Quick check
Think it through
?A 2 kg trolley moving at 3 m s−1 collides with a stationary 4 kg trolley and they stick together. What is their common velocity afterwards?
Recap
The big ideas to know
Vectors: resolve with F cosθ and F sinθ · equilibrium = zero resultant force AND zero resultant moment
Moments: moment = F × perpendicular distance · couple = torque with no resultant force
Projectiles: horizontal velocity constant; vertical acceleration = g; time links the two
Newton & momentum: F = Δp/Δt · impulse = FΔt = Δp · momentum conserved in ALL collisions
Energy: W = Fs cosθ · Ek = ½mv² · P = Fv · efficiency = useful ÷ total
Springs: F = kΔL · E = ½FΔL · series = less stiff, parallel = stiffer
Young modulus: E = stress ÷ strain = gradient of the linear part of the stress–strain graph
You have covered the whole of AQA 3.4 — forces, energy, momentum and the mechanical properties of materials. Press Finish to see your score.
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