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

For full explanations and worked examples, use the Mechanics and materials study guide.

These notes cover section 3.4 of AQA A-level Physics (7408), following the AQA AS and A-level Physics specification (7407/7408), version 1.4, for AS and A-level exams from June 2016 onwards. They cover sub-sections 3.4.1.1 to 3.4.2.2. All of section 3.4 is AS and A-level content, examined on A-level Paper 1 and assumed for Paper 2, and its required practicals are assessed in the written papers. Sections 3.9 to 3.13 are options: you study one, assessed in Paper 3 Section B.

Course pages: hub, printable checklist, practice questions. Uncertainty rules for the practicals are in the measurements revision notes.

Equations

Quantity	Equation	Unit
velocity	v = Δs/Δt	m s⁻¹
acceleration	a = Δv/Δt	m s⁻²
uniform acceleration	v = u + at; s = (u + v)t/2; s = ut + ½at²; v² = u² + 2as
Newton’s second law	F = ma (constant mass); F = Δ(mv)/Δt	N
momentum	p = mv	kg m s⁻¹ = N s
impulse	FΔt = Δ(mv)	N s
moment	force × perpendicular distance	N m
couple	one force × perpendicular distance between lines of action	N m
work	W = Fs cos θ	J
power	P = ΔW/Δt = Fv	W
efficiency	useful output power ÷ input power	none or %
energy	ΔEp = mgΔh; Ek = ½mv²	J
density	ρ = m/V	kg m⁻³
Hooke’s law	F = kΔL	N
stress, strain	stress = F/A; strain = ΔL/L	Pa; none
elastic strain energy	½FΔL = area under F–ΔL graph	J
Young modulus	E = stress/strain = FL/(AΔL)	Pa

3.4.1.1 Scalars and vectors

• Scalar: magnitude only (mass, speed, distance). Vector: magnitude and direction (force, weight, velocity, displacement, acceleration).
• Two vectors at right angles: size by Pythagoras, direction by tan⁻¹. Other angles: scale drawing.
• Resolve F at θ: F cos θ along, F sin θ perpendicular. On a slope: mg sin θ along, mg cos θ into the surface.
• Equilibrium: resultant force zero, so at rest or constant velocity. Three forces: closed triangle, or components cancel.

3.4.1.2 Moments

• Moment = force × perpendicular distance to the line of action.
• Couple = equal, opposite, coplanar forces; moment = one force × separation of the lines.
• Principle of moments: in equilibrium, Σ clockwise moments = Σ anticlockwise moments about any point.
• Centre of mass: where the weight acts; at the centre of a uniform regular solid.

Method: beam problems

1. Draw every force, with the weight at the centre of mass.
2. Take moments about the point where an unknown force acts.
3. Use upward forces = downward forces for the last unknown.

3.4.1.3 Motion along a straight line

• Average velocity = total displacement ÷ total time. Instantaneous velocity = gradient of tangent to s–t graph.
• s–t gradient = velocity. v–t gradient = acceleration; v–t area = displacement. a–t area = change in velocity.
• suvat only for uniform acceleration. List s, u, v, a, t; pick the equation missing the unknown you don’t need.
• Bouncing ball v–t (up positive): straight lines of gradient −g, a sudden jump in velocity at each bounce.

Required practical 3 (g by free fall). Plot h against t² for a ball released from rest: gradient = g/2, so g = 2 × gradient. Electronic timing cuts random error; a release delay at the electromagnet is systematic.

3.4.1.4 Projectile motion

Method

1. Split the launch velocity: uₓ = u cos θ, u_y = u sin θ.
2. Horizontal: constant velocity, x = uₓt.
3. Vertical: a = −g, use suvat.
4. Time is the link. Recombine components with Pythagoras for the final velocity.

Qualitative points:

• Friction opposes relative motion. Drag opposes motion through a fluid; lift acts at right angles to the flow.
• Air resistance increases with speed. Terminal speed: drag + other resistive forces = driving force or weight.
• Air resistance lowers the range and peak height and makes the descent steeper than the ascent.
• A vehicle’s top speed is reached when driving force = total resistive force.

3.4.1.5 Newton’s laws

• First: no resultant force → no change in velocity.
• Second: F = ma (mass constant). F is the resultant force.
• Third: equal and opposite forces of the same type on different bodies.

3.4.1.6 Momentum

• Momentum is conserved in any closed system: collisions and explosions (one dimension).
• F = Δ(mv)/Δt; impulse FΔt = Δ(mv); area under F–t graph = change in momentum.
• Longer contact time → smaller force for the same momentum change (crumple zones, packaging, kicking a ball).
• Elastic: Ek conserved. Inelastic: Ek not conserved (momentum still is). Explosion: total momentum unchanged (zero if starting at rest).

3.4.1.7–3.4.1.8 Work, power, energy

• W = Fs cos θ; only the component of force along the motion does work.
• Area under force–displacement graph = work done.
• P = Fv at constant velocity; efficiency = useful output ÷ input power.
• Energy conservation: Ep lost = Ek gained + work done against resistive forces.

3.4.2 Materials

• Limit of proportionality: end of the straight-line (Hooke’s law) region.
• Elastic limit: beyond it, permanent extension remains after unloading.
• Breaking stress: stress at which the material breaks.
• Plastic: large permanent extension for little extra force (copper). Brittle: fractures with almost no plastic deformation (glass).
• Elastic strain energy = ½FΔL = ½kΔL² (Hooke region) = area under F–ΔL graph.
• Spring energy can become Ek and Ep: ½kΔL² = ½mv² or mgh if losses are ignored.
• Loading beyond the elastic limit then unloading: unloading line parallel to loading line; area between = energy to deform.
• Young modulus = gradient of the linear part of the stress–strain graph.

Required practical 4 (Young modulus). Long thin wire; micrometer for diameter (several places, mean); metre rule for length; extension against a fixed marker; plot stress against strain. The diameter term dominates the uncertainty because area depends on d².

Must-know distinctions

Mass vs weight	kg, scalar vs N, vector (W = mg)
Distance vs displacement	scalar vs vector
Elastic vs inelastic collision	Ek conserved vs not; momentum conserved in both
Limit of proportionality vs elastic limit	end of straight line vs onset of permanent deformation
Stiffness k vs Young modulus E	property of one object vs property of the material
Moment vs couple	one force about a point vs a pair of equal, opposite forces

Quick self-test

1. Find the resultant of 6.0 N and 8.0 N forces at right angles.
2. State the condition for three coplanar forces acting at a point to be in equilibrium.
3. A stone falls from rest for 1.5 s. How far does it fall? (g = 9.81 m s⁻²)
4. What does the area under an acceleration–time graph give?
5. Calculate the momentum of a 1500 kg car at 20 m s⁻¹.
6. A 200 N force at 60° to the direction of motion moves an object 15 m. Find the work done.
7. Find the power needed to raise 50 kg vertically at a steady 0.40 m s⁻¹.
8. A spring with k = 40 N m⁻¹ is stretched 0.15 m. Find the energy stored.
9. A rod of area 2.0 × 10⁻⁵ m² carries a tension of 5.0 kN. Find the stress.
10. A ball hits a wall and rebounds more slowly. Is the collision elastic?
11. Describe the force–extension graph of a brittle material.
12. Why does a falling object reach terminal speed?

Answers

1. √(6.0² + 8.0²) = 10 N.
2. Resultant force is zero: the force vectors form a closed triangle.
3. s = ½ × 9.81 × 1.5² = 11.0 m.
4. Change in velocity.
5. 1500 × 20 = 3.0 × 10⁴ kg m s⁻¹.
6. 200 × 15 × cos 60° = 1500 J.
7. P = Fv = 50 × 9.81 × 0.40 = 196 W.
8. ½ × 40 × 0.15² = 0.45 J.
9. 5000 ÷ (2.0 × 10⁻⁵) = 2.5 × 10⁸ Pa.
10. No: kinetic energy is lost, so it is inelastic.
11. Straight line (Hooke’s law) up to fracture, with almost no plastic region.
12. Air resistance rises with speed until it equals the weight; resultant force is then zero, so acceleration is zero.

Where marks are usually lost

• Using the distance along a beam instead of the perpendicular distance to the line of action.
• Swapping sin and cos when resolving weight on a slope.
• Using suvat when acceleration is not uniform (for example, with air resistance).
• Forgetting that the vertical velocity is zero at the top of a projectile’s path, not the speed.
• Treating a rebound as positive: momentum change for a bounce is m(v + u), not m(v − u), when speeds are given.
• Calling weight and normal reaction a Newton’s third law pair.
• Leaving out the work done against resistive forces in energy-conservation answers.
• Using diameter in place of radius for cross-sectional area, or mm instead of m.
• Taking the Young modulus from a point beyond the limit of proportionality.
• Quoting efficiency above 1, or as a percentage and a decimal in the same line.

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.