Edexcel IAL Physics revision · AS — Materials
Deformation of solids
Hooke's law grows up: springs become wires, force-extension becomes stress-strain, and the Young modulus experiment becomes one of the most reliably examined practicals at AS.
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What the syllabus demands
- —Distinguish elastic and plastic deformation; identify the elastic limit
- —Define stress, strain and the Young modulus
- —Describe the Young modulus experiment for a wire
- —Interpret force-extension and stress-strain graphs
- —Calculate strain energy from the area under a force-extension graph (E = ½Fx = ½kx²)
Definitions that earn marks
Clear definitions to practise — check your course mark scheme
- Stress (σ)
- The force applied per unit cross-sectional area: σ = F ÷ A. Units: pascals.
- Strain (ε)
- The extension per unit original length: ε = x ÷ L. Dimensionless.
- Young modulus (E)
- The ratio of stress to strain in the region where they are proportional: E = σ ÷ ε.
- Elastic deformation
- Deformation in which the object returns to its original length when the load is removed; plastic deformation is permanent.
The equations
More equations to practise: the Edexcel IAL formula sheet.
Where the marks die
Common mistakes to check
- 01
Diameter for area. The wire's micrometer reading is a diameter: A = πd²/4, and the d² doubles its percentage uncertainty. Both facts are marks.
- 02
Strain energy read as Fx instead of the area ½Fx. The area under the force-extension line is the energy — for the linear region, a triangle.
- 03
Confusing the elastic limit with the limit of proportionality. Proportionality ends first (graph curves); elasticity can persist slightly beyond it (still returns to original length). They are different points.
- 04
In the Young modulus practical, measuring extension without a reference wire or ignoring temperature — the standard evaluation marks ask for exactly these.
One worked example, done properly
Question
A steel wire of length 2.0 m and diameter 0.50 mm extends by 1.2 mm under a load of 30 N. Calculate the Young modulus.
Method
- 1.A = πd²/4 = π × (0.50 × 10⁻³)² ÷ 4 = 1.96 × 10⁻⁷ m².
- 2.Stress = F/A = 30 ÷ 1.96 × 10⁻⁷ = 1.53 × 10⁸ Pa.
- 3.Strain = x/L = 1.2 × 10⁻³ ÷ 2.0 = 6.0 × 10⁻⁴.
- 4.E = stress ÷ strain = 1.53 × 10⁸ ÷ 6.0 × 10⁻⁴.
E ≈ 2.6 × 10¹¹ Pa (260 GPa)