Topic 4 asks a single question in many ways: how does a material respond to a force? You will meet density and upthrust, viscous drag and Stokes' law, Hooke's law, stress, strain and the Young modulus, and the difference between elastic, plastic, ductile and brittle behaviour.
stress σ = F/A · strain ε = Δx/x · E = σ/εstress in pascals · strain is a pure number · Young modulus in pascals
The distinction that earns marks:stiffness (large Young modulus) is about how hard it is to stretch. Strength (large breaking stress) is about how much stress it survives. Toughness is about how much energy it absorbs before fracture. They are three different properties.
Work through each screen, answer the questions as you go (several are full A-level calculations) and collect ⭐ stars. Press Start when you're ready.
Topic 4 · density & upthrust
Density and upthrust
ρ = m / Vkg m⁻³ · water is 1000 kg m⁻³ · air is about 1.2 kg m⁻³
Archimedes' principle: the upthrust on a body in a fluid equals the weight of fluid displaced.
Upthrust = ρfluid × Vdisplaced × g
An object floats when the upthrust equals its weight — which happens when its mean density is less than the fluid's.
A steel ship floats because its average density (steel + enclosed air) is far less than that of water, even though steel itself is eight times denser.
Unit trap
A block is 2.0 cm × 3.0 cm × 5.0 cm with a mass of 81 g.
V = 30 cm³ = 30 × 10⁻⁶ m³ = 3.0 × 10⁻⁵ m³ · m = 0.081 kg
ρ = 0.081 ÷ 3.0 × 10⁻⁵ = 2700 kg m⁻³ (aluminium)
Calculate
Your turn — density
1A rectangular block measures 2.0 cm × 3.0 cm × 5.0 cm and has a mass of 81 g. Calculate its density in kg m⁻³.
kg m⁻³
Hint: V = 30 cm³ = 3.0 × 10⁻⁵ m³. m = 0.081 kg. ρ = m ÷ V.
Topic 4 · viscosity
Viscous drag, Stokes' law and terminal velocity
F = 6πηrvStokes' law — a SPHERE of radius r, moving at speed v through a fluid of viscosity η, in LAMINAR flow only
Viscosity η (unit: Pa s) measures a fluid's resistance to flow. Treacle has a high η; water a low one.
Viscosity of a liquid falls as temperature rises (warm honey pours). Viscosity of a gas rises with temperature.
Laminar flow: smooth, ordered layers, no mixing. Turbulent flow: chaotic eddies — Stokes' law then fails.
A sphere falling in a fluid has three forces on it: weight down, upthrust up, viscous drag up. Drag grows with speed, so at terminal velocity:
weight = upthrust + viscous dragat terminal velocity the resultant force — and therefore the acceleration — is zero
Practical: dropping ball bearings through glycerol and timing them between marks is the standard way to measure η. Take readings only after terminal velocity is reached — that is why the top mark is set well below the surface.
Calculate
Your turn — Stokes' law
2A sphere of radius 2.0 mm moves at 0.10 m s⁻¹ through a liquid of viscosity 1.0 Pa s in laminar flow. Calculate the viscous drag force. Give your answer in mN to 3 significant figures.
mN
Hint: F = 6πηrv = 6π × 1.0 × 2.0 × 10⁻³ × 0.10 = 6π × 2.0 × 10⁻⁴ N. Convert to mN by multiplying by 1000.
Quick check
At terminal velocity
?A ball bearing falls at terminal velocity through oil. Which statement is correct?
Topic 4 · Hooke
Hooke's law and elastic strain energy
F = kΔxk = force constant / stiffness in N m⁻¹ · Δx = extension, NOT total length
Hooke's law holds only up to the limit of proportionality. Beyond it, the force–extension graph curves.
Limit of proportionality — the point beyond which F is no longer proportional to Δx.
Elastic limit — just beyond it; stretch further and a permanent extension remains.
Elastic strain energy = the area under the force–extension graph. While Hooke's law holds this is a triangle:
Eel = ½FΔx = ½k(Δx)²joules — and note the SQUARE: double the extension and you store four times the energy
Worked example
A force of 5.0 N produces an extension of 0.040 m.
k = F ÷ Δx = 5.0 ÷ 0.040 = 125 N m⁻¹
E = ½ × 5.0 × 0.040 = 0.10 J = 100 mJ
Calculate
Your turn — spring constant
3A spring extends by 0.040 m when a force of 5.0 N is applied, within its elastic limit. Calculate the spring constant. Give your answer in N m⁻¹.
N m⁻¹
Hint: k = F ÷ Δx = 5.0 ÷ 0.040.
Calculate
Your turn — stored energy
4The same spring is held at an extension of 0.040 m by a force of 5.0 N. Calculate the elastic strain energy stored. Give your answer in mJ.
mJ
Hint: E = ½FΔx = ½ × 5.0 × 0.040 = 0.10 J. Convert to mJ.
Topic 4 · Young modulus
Stress, strain and the Young modulus
Hooke's law describes a particular sample. Stress and strain scale it away so that we describe the material.
σ = F/A · ε = Δx/x · E = σ/εstress: Pa (N m⁻²) · strain: no unit · Young modulus: Pa, usually GPa
On a stress–strain graph, the gradient of the straight section = the Young modulus.
The area under a stress–strain graph is the energy stored per unit volume (J m⁻³).
E = 1.99 × 10⁸ ÷ 9.0 × 10⁻⁴ = 2.21 × 10¹¹ Pa = 221 GPa
Why use a long, thin wire? Long → larger extension for a given strain, so the percentage uncertainty in Δx falls. Thin → larger stress for a given load, so it extends measurably. Use a micrometer and take the mean of several diameters at different points and orientations.
Calculate
Your turn — the Young modulus
5A wire of original length 2.00 m and diameter 0.40 mm extends by 1.8 mm under a load of 25 N. Calculate the Young modulus. Give your answer in GPa to 3 significant figures.
GPa
Hint: A = π(0.20 × 10⁻³)² = 1.257 × 10⁻⁷ m². Stress = 25/A = 1.99 × 10⁸ Pa. Strain = 1.8 × 10⁻³ / 2.00 = 9.0 × 10⁻⁴. E = stress ÷ strain, then divide by 10⁹ to get GPa.
Sort it
Elastic, plastic or fracture?
Tap a statement, then tap the region of the stress–strain graph it describes.
🟢 Elastic region
🟠 Plastic region
🔴 Fracture
Topic 4 · material behaviour
Brittle, ductile and polymeric materials
Brittle (glass, cast iron, ceramics): almost no plastic deformation — the stress–strain line stays straight right up to a sudden fracture. Absorbs little energy.
Ductile (copper, mild steel): a long plastic region — it can be drawn into a wire. Absorbs a lot of energy before breaking, so it is tough.
Polymeric (rubber): huge strains, non-linear, and shows hysteresis — the loading and unloading curves differ, and the area between them is energy converted to thermal energy (which is why tyres get hot).
Stiff is not the same as strong. Glass is stiff (high E) but brittle. Rubber has a tiny E — it is easy to stretch — yet it can absorb enormous energy before failing. Steel manages both stiff and strong.
Quick check
Beyond the elastic limit
?A copper wire is loaded past its elastic limit and then the load is removed. What happens?
Quick check
Reading a stress–strain graph
?Two materials A and B are tested. A has a steeper straight-line region; B has a higher breaking stress. Which statement is correct?
Quick check
Units check
?Which pairing of quantity and unit is correct?
Match it
Match the quantity to its definition
Tap an item on the left, then its partner on the right.
Hooke's law: F = kΔx up to the limit of proportionality; E = ½FΔx = ½k(Δx)²
Young modulus: E = σ/ε = (F/A) ÷ (Δx/x); gradient of the linear part of a stress–strain graph
Behaviour: elastic → recovers · plastic → permanent · brittle → snaps with no plastic region · ductile → drawn into wire · rubber → hysteresis loop = energy lost as heat
Stiff ≠ strong ≠ tough — three separate properties
That is Edexcel Topic 4 in full, including the Young modulus core practical. Press Finish to see your score.
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