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Materials

Year 1 / ASYear 2 / A-Level AQA, Edexcel, OCR, WJEC/Eduqas, CCEA

A-Level Physics materials revision: Hooke's law, stress and strain, Young modulus with derivation, stress-strain graphs, elastic strain energy and the Young modulus required practical - AQA 7408 3.4.2, OCR A H556 3.4, Edexcel 9PH0 Topic 4.

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🎯 Learning Objectives

📋 Key Definitions and Core Concepts

Hooke's law: The force needed to extend a spring is proportional to the extension, F = kx, up to the limit of proportionality.
Spring constant (k): Force per unit extension, k = F/x, in N/m; the gradient of the linear part of a force-extension graph.
Stress (σ): Force per unit cross-sectional area, σ = F/A, in Pa (N/m^2).
Strain (ε): Fractional extension, ε = x/L, where L is the original length; dimensionless (often quoted as a percentage).
Young modulus (E): The ratio of stress to strain, E = σ/ε = FL/(Ax), in Pa; a property of the material, not the sample's shape.
Limit of proportionality: The point beyond which force is no longer proportional to extension; Hooke's law stops applying.
Elastic limit: The maximum extension (or stress) from which the material still returns to its original length; beyond it deformation is plastic.
Ultimate tensile stress (UTS): The maximum stress a material can withstand before it starts to neck and fail; the peak of the stress-strain graph.
Brittle material: A material that breaks with little or no plastic deformation - a straight stress-strain graph ending suddenly (glass, ceramics).
Elastic strain energy: Energy stored in a stretched material, E = ½kx^2 = ½Fx for a linear force-extension graph; equal to the area under the graph.

📌 Key Points

Hooke's law: F = kx up to the limit of proportionality; springs in series share 1/k_total = Σ1/k; parallel k_total = Σk.
Elastic strain energy: E = ½kx^2 = ½Fx = area under the F-x graph (linear).
Stress σ = F/A (Pa); strain ε = x/L; Young modulus E = σ/ε = FL/(Ax).
Stress-strain graph: linear to the limit of proportionality, then elastic limit, yield point, UTS, breaking point.
Brittle = breaks with no plastic deformation; ductile = deforms plastically before breaking (drawn into wires).
Energy recovered on unloading = area under the unloading curve; the loop area between loading and unloading is energy dissipated (heat).

Density, Hooke's law and elastic strain energy

Density is mass per unit volume, ρ = m/V in kg/m^3 (1 g/cm^3 = 1000 kg/m^3) - a property of the material, useful in identifying substances and in floating arguments (an object floats if its mean density is less than the fluid's).

Hooke's law (AQA 7408 3.4.2.1; OCR A H556 3.4; Edexcel 9PH0 Topic 4): up to the limit of proportionality the extension x of a spring is proportional to the applied force, F = kx, where k is the spring constant in N/m. This is an empirical law - it describes a region of behaviour, not a universal truth, so always state 'up to the limit of proportionality'. Springs in parallel share the load, so k_total = k_1 + k_2; springs in series each carry the full force and their extensions add, so 1/k_total = 1/k_1 + 1/k_2 (the same structure as capacitors in series and resistors in parallel - remember by which quantity adds).

Elastic strain energy is the energy stored in a stretched material. Derivation: the force starts at 0 and rises linearly to F while the extension grows to x, so the average force during stretching is F/2. Work done = average force × distance = (F/2)x, so E = ½Fx = ½kx^2 (substituting F = kx). Geometrically, it is the area under the force-extension graph. When the force is removed, this energy is recovered - provided the material is within its elastic limit.

Worked example: springs in series and energy stored

Two springs with k = 200 N/m and k = 300 N/m are connected in series and support a 10 N load.

Line 1: Series: 1/k_total = 1/200 + 1/300 = 0.005 + 0.00333 = 0.00833 (M1).

Line 2: k_total = 120 N/m (A1).

Line 3: Total extension x = F/k_total = 10/120 = 0.083 m (A1).

Line 4: Energy stored = ½Fx = ½ × 10 × 0.083 = 0.42 J (A1: ½Fx form used directly).

Check: energy is also ½kx^2 per spring summed - ½ × 200 × 0.05^2 + ½ × 300 × 0.033^2 ≈ 0.25 + 0.17 = 0.42 J - the series extensions are x_1 = 10/200 = 0.05 m and x_2 = 10/300 = 0.033 m, agreeing.

Stress, strain and the Young modulus

Force-extension data describes a particular sample; to compare materials, normalise it (AQA 7408 3.4.2.2; OCR A H556 3.4). Stress σ = F/A (force divided by cross-sectional area, in Pa or N/m^2) spreads out the effect of different thicknesses. Strain ε = x/L (extension divided by original length, dimensionless) spreads out the effect of different lengths. The Young modulus is the gradient of the linear (Hookean) region of the stress-strain graph:

E = σ/ε = (F/A)/(x/L) = FL/(Ax)

in pascals, typically GPa for metals (steel ≈ 200 GPa, copper ≈ 120 GPa, rubber ≈ 0.01-0.1 GPa). It is a property of the material only - a thin long wire and a thick short rod of the same steel have the same Young modulus but very different spring constants. Units discipline: convert diameters to radii carefully (A = πr^2 = πd^2/4), watch M (mega) vs m (milli) prefixes, and quote strain as a fraction unless told otherwise.

Worked example: Young modulus of a wire

A wire of length 2.50 m and diameter 0.40 mm extends 2.0 mm under a load of 50 N.

Line 1: A = πd^2/4 = π × (0.40 × 10^-3)^2/4 = 1.26 × 10^-7 m^2 (M1: area from diameter).

Line 2: Stress = F/A = 50/1.26 × 10^-7 = 4.0 × 10^8 Pa (A1).

Line 3: Strain = x/L = 2.0 × 10^-3/2.50 = 8.0 × 10^-4 (A1).

Line 4: E = stress/strain = 4.0 × 10^8/8.0 × 10^-4 = 5.0 × 10^11 Pa (A1). Note: this is higher than steel - realistic wires give about 2 × 10^11 Pa, so always check magnitudes.

Why stress and strain rather than force and extension? Because the comparison of materials is the point - the Young modulus lets an engineer predict the extension of any sample of the material: x = FL/(AE).

Stress-strain graphs and material behaviour

The stress-strain graph is the examinable picture (AQA 7408 3.4.2.2). Described in words: starting at the origin the graph is a straight line through it - Hooke's law region, where the gradient is the Young modulus and the material behaves elastically. The limit of proportionality is where the line begins to curve (hard to pin down precisely); the elastic limit follows - beyond this point the material no longer returns to its original length when unloaded. For a ductile metal the graph then rises through a yield point (strain increases with little extra stress - large plastic deformation), reaching a maximum at the ultimate tensile stress (UTS), after which necking occurs: the sample thins at one point, the force (and engineering stress) falls, and the curve ends at the breaking point/fracture stress.

Material adjectives - know the precise meanings: strong = high breaking stress; stiff = high Young modulus (steep gradient); brittle = breaks with little or no plastic deformation (straight graph ending abruptly - glass, concrete, ceramics); ductile = undergoes large plastic deformation before breaking, so can be drawn into wires (copper); malleable = deforms plastically under compression (hammered into sheets); hard = resists indentation/scratching; tough = absorbs a lot of energy before fracture (large area under the graph) - the opposite of brittle.

Energy interpretation: the area under a stress-strain graph is energy per unit volume (units J/m^3); the area under a force-extension graph is energy (J). On unloading within the elastic limit, the unloading curve retraces the loading curve and all energy is recovered. Beyond the elastic limit the unloading line is parallel to the initial linear section but does not pass through the origin - it intercepts the x-axis at the permanent (plastic) extension. The area of the hysteresis loop between the loading and unloading curves is the energy dissipated as heat per cycle - this is why a repeatedly stretched rubber band warms up.

Worked example: loading and unloading a wire

A wire is stretched beyond its elastic limit to a force of 60 N (extension 4.0 mm), then unloaded. On reaching zero force the permanent extension is 1.0 mm. The initial linear gradient was 2.0 × 10^4 N/m.

Line 1: The unloading line is parallel to the loading line, gradient 2.0 × 10^4 N/m (B1).

Line 2: It passes through (4.0 mm, 60 N) and drops 60 N at gradient 2.0 × 10^4 N/m: extension change = 60/2.0 × 10^4 = 3.0 mm (M1).

Line 3: Final extension = 4.0 - 3.0 = 1.0 mm - the permanent set (A1).

Line 4: Energy dissipated = area of the loop = area under loading minus area under unloading: ½ × 60 × 4.0 × 10^-3 - ½ × 60 × 3.0 × 10^-3 = 0.12 - 0.09 = 0.03 J (A1).

Required practical: determining the Young modulus (AQA RP4, OCR PAG2, Edexcel CP5)

The set-up (AQA required practical 4; OCR PAG2 materials; Edexcel CP5): a long thin wire (typically ~2 m of steel or copper, diameter measured with a micrometer) is clamped at one end over a pulley (or clamped to a bench edge) and loaded with masses. A sticky-tape marker on the wire moves past a fixed millimetre rule as weights are added. Measure the original length L from the clamp point to the marker with a tape measure; the diameter with a micrometer screw gauge at several points and orientations (the wire may not be uniform or circular) and average; add masses in equal increments, recording the marker reading each time (load and unload to check for the elastic limit and zero error).

Analysis: plot a graph of load against extension (or stress against strain); the Young modulus is the gradient of the linear region times the geometry factor - for a force-extension graph, E = gradient × L/A. Using the gradient rather than a single data pair averages random error. Plot force against extension and compute E from the gradient to avoid dividing small differences.

Uncertainty and evaluation - the marks live here: the wire diameter is small (≈0.2-0.4 mm), so its uncertainty dominates. A micrometer resolution of 0.01 mm on a 0.30 mm diameter gives 3.3% uncertainty, and since E depends on A = πd^2/4, the diameter uncertainty doubles to about 7% in E (power rule). Reduce it by measuring the diameter at many points and averaging. Extensions are small (a few mm on a mm rule: ±0.5 mm resolution) - use a longer wire, larger loads (staying elastic), or a vernier scale/Searle's apparatus to improve resolution. Other error sources: kinks in the wire (straighten before measuring L), the marker slipping (use a fine wax pencil line), thermal expansion, and masses below the elastic limit assumption (verify by unloading to check the wire returns to its original length). Quote the final result as E ± ΔE and give the percentage difference from the accepted value for the material.

Worked example: percentage uncertainty in E

L = 2.00 ± 0.01 m (0.5%), d = 0.30 ± 0.01 mm (3.3%), extension at 50 N = 1.8 ± 0.5 mm (28%), F = 50 ± 0.5% (balance resolution).

Line 1: E = FL/(Ax); A ∝ d^2, so percentage uncertainty = ΔF/F + ΔL/L + 2Δd/d + Δx/x (M1: power rule on d).

Line 2: = 0.5 + 0.5 + 6.6 + 28 = 35.6% ≈ 36% (A1).

Line 3: The extension term dominates: increase the load (staying elastic) or use a more sensitive extension measurement - a Searle's apparatus with a vernier - to cut it (A1: evaluation).

💡 Worked Example

Exam-style: A wire of original length 2.50 m and diameter 0.40 mm extends by 2.0 mm when a load of 50 N is applied. Calculate the stress, the strain and the Young modulus of the wire. (AQA 7408-style, 6 marks)
Mark-scheme logic: Cross-sectional area: A = πd^2/4 = π × (0.40 × 10^-3)^2/4 = 1.26 × 10^-7 m^2 (M1 A1: area from diameter, correct unit). Stress σ = F/A = 50/1.26 × 10^-7 = 4.0 × 10^8 Pa (M1 A1: stress with unit). Strain ε = x/L = 2.0 × 10^-3/2.50 = 8.0 × 10^-4 (A1: dimensionless strain). Young modulus E = σ/ε = FL/(Ax) = 4.0 × 10^8/8.0 × 10^-4 = 5.0 × 10^11 Pa (M1 A1: gradient ratio quoted with unit). Sanity check: steel is about 2 × 10^11 Pa, so comment that the value is of the right order for a metal.

❓ Practice Questions & Mark Schemes

Two springs with spring constants 200 N/m and 300 N/m are connected in series. A 10 N load is applied. Calculate the total extension and the energy stored.
Answer: 1/k_total = 1/200 + 1/300 = 0.00833, k_total = 120 N/m (M1 A1). x = F/k = 10/120 = 0.083 m (A1). Energy = ½Fx = ½ × 10 × 0.083 = 0.42 J (A1: ½Fx = area under F-x graph).
A wire of length 1.50 m and cross-sectional area 2.0 × 10^-7 m^2 extends 1.2 mm under a 40 N load. Calculate the Young modulus.
Answer: Stress = F/A = 40/2.0 × 10^-7 = 2.0 × 10^8 Pa (M1 A1). Strain = 1.2 × 10^-3/1.50 = 8.0 × 10^-4 (A1). E = 2.0 × 10^8/8.0 × 10^-4 = 2.5 × 10^11 Pa (A1) - consistent with a metal such as steel.
Sketch (describe) the stress-strain graph for a ductile metal and label the limit of proportionality, elastic limit, yield point and ultimate tensile stress.
Answer: Straight line through the origin (gradient = Young modulus) up to the limit of proportionality (B1); curve begins; elastic limit beyond which unloading leaves permanent extension (B1); yield point where strain grows with little stress increase (B1); curve peaks at the UTS then falls as necking occurs before fracture (B1). Ductile = large plastic strain before breaking (A1).
Explain why the micrometer measurement of wire diameter contributes most of the uncertainty in a Young modulus experiment, and state two ways to reduce it.
Answer: Diameter is small (fraction mm), so resolution uncertainty is a large percentage; E ∝ 1/d^2 so the percentage uncertainty in E is double that in d (M1 A1: power rule). Reduce: measure at several points/orientations and average (A1); use a larger-diameter wire or a higher-resolution instrument (A1).
A wire is loaded beyond its elastic limit and then unloaded. Describe what the unloading curve looks like and what the area of the loading-unloading loop represents.
Answer: The unloading line is parallel to the initial linear (Hookean) section but does not pass through the origin - it intercepts the extension axis at the permanent plastic extension (M1 A1). The loop area between loading and unloading curves is the energy dissipated as heat during the cycle (A1); energy is only fully recovered for loading within the elastic limit (A1).
A material has a Young modulus of 70 GPa and breaks at a stress of 100 MPa. A second material has E = 200 GPa and breaks at 400 MPa. Compare their stiffness, strength and toughness.
Answer: Stiffness: the second material has the higher Young modulus (200 vs 70 GPa), so it is stiffer - less strain for the same stress (A1). Strength: the second breaks at a higher stress (400 vs 100 MPa), so it is stronger (A1). Toughness needs the area under the stress-strain graph - breaking stress and modulus alone do not determine it, since a brittle material can be strong but fail with little energy absorbed (A1: correct reasoning about what the data does not show).

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