AQA A-Level Physics: Mechanics and materials (7408)

This guide teaches section 3.4, Mechanics and materials, of AQA A-level Physics (7408). It follows the AQA AS and A-level Physics specification (7407/7408), version 1.4, for AS and A-level exams from June 2016 onwards, and covers every sub-section from 3.4.1.1 to 3.4.2.2. Section 3.4 is AS and A-level content (no part of it is A-level only). At A-level it is examined on Paper 1, which covers sections 1 to 5 and 6.1, and it is assumed knowledge for Paper 2; Paper 3 Section A tests practical skills and data analysis, and the written papers assess the two required practicals here. Sections 3.9 to 3.13 are options: you study one, assessed in Paper 3 Section B.

Use it with the revision notes and the practice questions. See also the course hub and the printable checklist. For uncertainties in the required practicals, see Measurements and their errors.

What this section covers

Spec	Content
3.4.1.1	Scalars, vectors, resolving, equilibrium
3.4.1.2	Moments, couples, centre of mass
3.4.1.3	Straight-line motion, graphs, suvat; Required practical 3
3.4.1.4	Projectiles, friction, lift, drag, terminal speed
3.4.1.5	Newton’s laws
3.4.1.6	Momentum, impulse, collisions, explosions
3.4.1.7	Work, power, efficiency
3.4.1.8	Conservation of energy
3.4.2.1	Density, Hooke’s law, stress, strain, strain energy
3.4.2.2	Young modulus; Required practical 4

3.4.1.1 Scalars and vectors

Scalars have magnitude only (speed, distance, mass). Vectors also have direction (velocity, displacement, force, weight, acceleration).

Add two vectors at right angles by Pythagoras and tan⁻¹; for other angles use a scale drawing (tip-to-tail).

To resolve a vector F at angle θ to a direction: component along it = F cos θ, component perpendicular = F sin θ. On a slope of angle θ, weight mg has components mg sin θ down the slope and mg cos θ into the slope.

Equilibrium means zero resultant force: the object is at rest or moving at constant velocity. Two forces in equilibrium are equal and opposite. Three coplanar forces in equilibrium form a closed triangle tip-to-tail, or their components cancel in two perpendicular directions.

Worked example. Forces of 12 N east and 5 N north act on a point. Find the resultant.

size  = √(12² + 5²) = 13 N
angle = tan⁻¹(5/12) = 22.6° north of east

A 4.0 kg box rests on a 30° slope (g = 9.81 m s⁻²). Weight = 39.2 N. Down the slope: 39.2 sin 30° = 19.6 N (balanced by friction). Into the slope: 39.2 cos 30° = 34.0 N (balanced by the normal reaction).

3.4.1.2 Moments

Moment = force × perpendicular distance from the point to the line of action of the force (unit N m).

A couple is a pair of equal and opposite coplanar forces. Moment of a couple = one force × perpendicular distance between the lines of action.

Principle of moments: for an object in equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about that point.

The centre of mass is the point where the whole weight can be taken to act. For a uniform regular solid it is at the centre.

Worked example. A uniform beam 3.0 m long, weight 200 N, rests on supports at each end. A 600 N person stands 1.0 m from the left end. Find each support force.

moments about left end:  R_right × 3.0 = 200 × 1.5 + 600 × 1.0 = 900
R_right = 300 N
vertical equilibrium:    R_left = 800 − 300 = 500 N

3.4.1.3 Motion along a straight line

Velocity v = Δs/Δt and acceleration a = Δv/Δt. Average velocity is total displacement ÷ total time; instantaneous velocity is the gradient of the tangent to the displacement–time graph at that moment.

Graph	Gradient gives	Area gives
displacement–time	velocity	–
velocity–time	acceleration	displacement
acceleration–time	–	change in velocity

A curved s–t or v–t graph means non-uniform acceleration. For a bouncing ball (up positive), the v–t graph is a set of parallel lines of gradient −g, jumping from negative to positive velocity at each bounce.

For uniform acceleration:

v = u + at          s = (u + v)t/2
s = ut + ½at²       v² = u² + 2as

Free-fall acceleration near the Earth is g ≈ 9.81 m s⁻².

Worked example. A car moving at 4.0 m s⁻¹ accelerates at 2.5 m s⁻² over 30 m. Find its final speed and the time taken.

v² = 4.0² + 2 × 2.5 × 30 = 166   →  v = 12.9 m s⁻¹
t = (v − u)/a = (12.88 − 4.0)/2.5 = 3.55 s

Required practical 3: determination of g by a free-fall method. Release a steel ball from rest (electromagnet) and time its fall through height h electronically (light gates or trap door). Repeat for several heights. From h = ½gt², a graph of h against t² is a straight line with gradient g/2. A magnet release delay is a systematic error; scatter in timing is random and is reduced by repeats.

3.4.1.4 Projectile motion

Horizontal and vertical motion are independent. Ignoring air resistance, horizontal velocity is constant and vertical acceleration is g. Time links the two directions.

Worked example. A ball is thrown horizontally at 15 m s⁻¹ from a 20 m cliff.

vertical:   20 = ½ × 9.81 × t²   →  t = 2.02 s
horizontal: range = 15 × 2.02 = 30.3 m
impact:     v_y = 9.81 × 2.02 = 19.8 m s⁻¹
            speed = √(15² + 19.8²) = 24.8 m s⁻¹

Friction opposes relative motion between surfaces (static versus dynamic friction is not tested). Drag opposes motion through a fluid; lift acts perpendicular to the fluid flow. Air resistance increases with speed, so a falling object speeds up until drag equals weight: resultant force zero, terminal speed.

With air resistance a projectile’s range and maximum height fall and the descent is steeper than the ascent. A vehicle reaches maximum speed when driving force equals total resistive force, so this depends on power and shape.

3.4.1.5 Newton’s laws of motion

1. An object stays at rest or at constant velocity unless a resultant force acts on it.
2. Resultant force = mass × acceleration, F = ma, when mass is constant.
3. If A exerts a force on B, B exerts an equal and opposite force of the same type on A (they act on different bodies).

Draw a free-body diagram (one object only) before using F = ma.

Worked example. Find the floor’s reaction on a 60 kg person in a lift accelerating upwards at 1.5 m s⁻².

R − mg = ma   →  R = 60 × (9.81 + 1.5) = 679 N

3.4.1.6 Momentum

Momentum = mass × velocity (kg m s⁻¹ or N s), a vector. In a closed system, total momentum is conserved in collisions and explosions (one-dimensional problems only).

Newton’s second law in general form: F = Δ(mv)/Δt. Impulse FΔt = Δ(mv) for a constant force. The area under a force–time graph is the change in momentum, even for a varying force. For a given momentum change, a longer contact time means a smaller force: crumple zones, airbags and packaging use this, which links momentum to ethical transport design.

Elastic collision: total kinetic energy conserved. Inelastic: it is not. Explosion: total momentum is unchanged (zero if starting at rest); kinetic energy comes from a store such as a spring.

Worked example. A 2.0 kg trolley at 3.0 m s⁻¹ hits a stationary 1.0 kg trolley and they stick together.

2.0 × 3.0 = 3.0 × v   →  v = 2.0 m s⁻¹
Ek before = ½ × 2.0 × 3.0² = 9.0 J;  Ek after = ½ × 3.0 × 2.0² = 6.0 J

Kinetic energy falls, so the collision is inelastic.

3.4.1.7 Work, energy and power

Work done W = Fs cos θ (θ between force and displacement). The area under a force–displacement graph is the work done, even for a variable force. Power P = ΔW/Δt = Fv.

efficiency = useful output power ÷ input power, often given as a percentage.

Worked example. A 50 N pull at 30° above the horizontal drags a sledge 40 m: W = 50 × 40 × cos 30° = 1730 J. A motor with input power 5.0 W lifts 0.50 kg through 1.2 m in 2.0 s:

useful power = mgh/t = 0.50 × 9.81 × 1.2 / 2.0 = 2.94 W
efficiency   = 2.94/5.0 = 0.59 = 59%

3.4.1.8 Conservation of energy

Energy is never created or destroyed. ΔEp = mgΔh and Ek = ½mv². Loss of Ep = gain of Ek + work done against resistive forces.

Worked example. A 70 kg skier starts from rest and descends 25 m vertically, reaching 18 m s⁻¹.

Ep lost = 70 × 9.81 × 25 = 17 168 J
Ek gained = ½ × 70 × 18² = 11 340 J
work against resistance = 5830 J (5.8 kJ)

3.4.2.1 Bulk properties of solids

Density ρ = m/V. A 0.540 kg block of volume 2.0 × 10⁻⁴ m³ has ρ = 2700 kg m⁻³.

Hooke’s law: F = kΔL up to the limit of proportionality; k is the stiffness or spring constant (N m⁻¹). Beyond the elastic limit the material does not return to its original length.

• Tensile stress = F/A (Pa). Tensile strain = ΔL/L (no unit).
• Breaking stress is the stress at which the material breaks.
• Elastic strain energy = ½FΔL = area under the force–extension graph (up to the elastic limit).

On a force–extension graph, plastic behaviour is large extension for little extra force, with permanent deformation (copper). Brittle materials (glass) stay near-linear up to fracture with almost no plastic region. Loaded beyond the elastic limit and unloaded, a metal’s unloading line is parallel to the loading line; the area between them is the energy used to deform it.

The initial gradient of a stress–strain curve is the Young modulus. Crumple zones absorb energy by deforming, another ethical-transport link.

Worked example. A spring with k = 250 N m⁻¹ is compressed 0.080 m and launches a 0.050 kg ball vertically. Find the maximum height, ignoring air resistance.

E = ½kΔL² = ½ × 250 × 0.080² = 0.80 J
h = E/(mg) = 0.80/(0.050 × 9.81) = 1.63 m

3.4.2.2 The Young modulus

Young modulus E = tensile stress / tensile strain = FL/(AΔL)

E is the gradient of the linear part of a stress–strain graph. Unit: Pa.

Worked example. A steel wire 2.0 m long, diameter 0.50 mm, extends by 2.0 mm under a 40 N load.

A = π(0.25 × 10⁻³)² = 1.96 × 10⁻⁷ m²
stress = 40/1.96 × 10⁻⁷ = 2.04 × 10⁸ Pa
strain = 2.0 × 10⁻³/2.0 = 1.0 × 10⁻³
E = 2.04 × 10¹¹ Pa

Required practical 4: determination of the Young modulus by a simple method. Measure a long thin wire’s length (metre rule) and diameter at several points (micrometer). Add masses and read extension against a fixed marker. Plot stress against strain; E is the gradient of the straight part. Long and thin gives a measurable extension; the squared diameter carries the largest percentage uncertainty. Wear eye protection.

Common errors

• Using sin and cos the wrong way round on a slope: mg sin θ acts along the slope.
• Taking moments with a distance that is not perpendicular to the line of action.
• Naming weight and normal reaction as a Newton’s third law pair: they act on the same body.
• Adding speeds in a collision without signs; give each velocity a direction.
• Using diameter instead of radius in A = πr², or forgetting to convert mm to m.
• Calling a collision elastic just because the objects bounce apart; check kinetic energy.

Official syllabus

AQA AS and A-level Physics specification (7407/7408), version 1.4, July 2026, for AS and A-level exams June 2016 onwards, published by AQA. Section 3.4 Mechanics and materials.