Materials
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.
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.
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.
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.
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.
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).
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.
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).
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.
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).