OxfordAQA International AS and A-level Physics: Bulk properties of solids and the Young modulus (9630) – Revision Notes

These notes condense sections 3.2.9 and 3.2.10 of the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4. This is shared content for International AS and International A-level. For full explanations and worked examples, read the study guide, then test yourself with the practice set. Course pages: OxfordAQA A-level Physics hub and the printable checklist. Not sure where to start? Take a free diagnostic. Related: mechanics and materials revision notes.

Use g = 9.81 N kg⁻¹.

Definitions

• Density: mass per unit volume.
• Hooke’s law: extension is proportional to the applied force, up to the elastic limit.
• Elastic limit: the point beyond which the material does not return to its original length when the force is removed.
• Stiffness (spring constant) k: force per unit extension, in N m⁻¹.
• Tensile stress: the force acting on each unit of cross-sectional area.
• Tensile strain: the extension divided by the original length.
• Breaking stress: the stress at which the material fractures.
• Young modulus: the ratio tensile stress : tensile strain, in the region where stress is proportional to strain.

Formulas and units

Quantity	Formula	Unit
Density	ρ = m / V	kg m⁻³
Spring law (Hooke)	F = kΔL	N
Stress	F/A	Pa (N m⁻²)
Strain	ΔL/L	none
Strain energy	½FΔL = ½kΔL²	J
Young modulus	E = (F/A)/(ΔL/L) = FL/(AΔL)	Pa
Wire stiffness	k = EA/L	N m⁻¹
Wire area	A = πd²/4	m²
Strain energy per unit volume	½ × stress × strain	J m⁻³

The ½kΔL² form only works when F = kΔL holds. For a non-linear graph, use the area.

Method boxes

Energy from a force-extension graph

1. Convert extension to metres.
2. Break the shape into triangles, rectangles and trapeziums, or count squares.
3. Add the areas. Units: N × m = J.

Energy conservation with a spring

1. Name the start and end positions.
2. Write the elastic strain energy at the start: ½kΔL².
3. Write the kinetic and gravitational terms at the end. Add any energy transferred to the surroundings as its own term.
4. Solve. Watch the height: does it include the compression distance?

Finding E from a stress-strain graph

1. Choose two points on the straight part, well apart.
2. Gradient = change in stress / change in strain.
3. State the answer in Pa.

Required practical 2 in five lines

1. Long thin wire, clamped firmly, with a marker and a fixed scale.
2. Measure L (clamp to marker) and the diameter at several places and angles with a micrometer.
3. Add equal loads, record the extension, then unload to check the reading returns.
4. Plot extension against force. Gradient = L/(EA).
5. E = L/(A × gradient). The diameter gives the largest percentage uncertainty because A depends on d².

Worked reminders

Reminder 1. Stress 4.5 × 10⁸ Pa with strain 2.5 × 10⁻³ gives E = 4.5 × 10⁸ / 2.5 × 10⁻³ = 1.8 × 10¹¹ Pa.

Reminder 2. A spring stretched by 0.050 m by a force of 8.0 N has k = 8.0 / 0.050 = 160 N m⁻¹ and stores ½ × 8.0 × 0.050 = 0.20 J.

Reminder 3. Replace a wire by one of the same material and length but twice the diameter. The area is four times larger, so for the same force the stress and the extension are four times smaller.

Reminder 4. A rod of cross-sectional area 4.0 × 10⁻⁶ m² breaks at a force of 2.4 kN. Breaking stress = 2.4 × 10³ / 4.0 × 10⁻⁶ = 6.0 × 10⁸ Pa. A rod of the same material with area 1.0 × 10⁻⁵ m² breaks at 6.0 × 10⁸ × 1.0 × 10⁻⁵ = 6.0 kN.

Reminder 5. A 0.20 kg trolley is pushed back against a spring of k = 500 N m⁻¹ by 0.030 m, on a smooth slope rising at 30°. The stored energy is ½ × 500 × 0.030² = 0.225 J. Written as a gain in height, 0.225 = 0.20 × 9.81 × h, so the trolley rises 0.115 m vertically from release.

Must-know distinctions

Compare	Key difference
Limit of proportionality and elastic limit	The first is where force stops being proportional to extension. The second is where the deformation stops being reversible.
Elastic and plastic	Elastic: the sample returns to its original length. Plastic: there is permanent extension.
Ductile and brittle	Ductile: large plastic region before fracture. Brittle: fractures with little or no plastic region.
k and E	k depends on the sample’s size and shape. E is a property of the material.
Force-extension and stress-strain	The first depends on dimensions. The second does not. The gradient of the first is k; the gradient of the second is E.
Breaking stress and breaking force	The force depends on the cross-section of the sample. The stress is a property of the material.

Stress-strain curves in words

For a metal such as copper, describe the curve in order. The line is straight and steep at first, then the gradient decreases at the limit of proportionality. Beyond the elastic limit, strain keeps increasing and is permanent. The curve ends at the breaking stress. For glass, the line is straight and stops at fracture. Always say which axis you are describing, and give the gradient in the first straight region as the Young modulus.

Reading graphs

• Straight line through the origin: Hooke’s law holds. Gradient is k (force on the vertical axis).
• Curve that flattens: beyond the limit of proportionality, the material becomes easier to stretch.
• Unloading line parallel to the loading line, ending at a positive extension: plastic deformation has occurred. The gap on the extension axis is the permanent extension.
• Area between loading and unloading lines: energy transferred to internal energy of the material, not recovered.
• Straight line ending at fracture: brittle behaviour.

Quick self-test

1. A block of volume 3.2 × 10⁻⁴ m³ has mass 0.88 kg. Find its density.
2. A spring has k = 180 N m⁻¹. What force extends it by 35 mm?
3. A tension of 540 N acts in a wire of diameter 1.2 mm. Find the stress.
4. A 3.20 m length of wire extends by 2.4 mm. Find the strain.
5. A force of 14 N extends a spring by 0.035 m. Find k and the stored energy.
6. By what factor does the stored energy change if the extension of a spring (within the limit) is doubled?
7. A straight line on a stress-strain graph passes through a strain of 8.0 × 10⁻⁴ at a stress of 160 MPa. Find E.
8. A wire of length 0.90 m and cross-sectional area 3.0 × 10⁻⁷ m² extends by 1.8 mm under a force of 60 N. Find E.
9. A spring with k = 300 N m⁻¹ compressed by 0.040 m launches a 0.050 kg trolley along a smooth horizontal track. Find its speed.
10. What does the area between the loading and unloading lines of a force-extension graph represent?

Answers

1. ρ = 0.88 / 3.2 × 10⁻⁴ = 2.75 × 10³ kg m⁻³.
2. F = kΔL = 180 × 0.035 = 6.3 N.
3. A = π × (0.60 × 10⁻³)² = 1.13 × 10⁻⁶ m²; stress = 540 / 1.13 × 10⁻⁶ = 4.8 × 10⁸ Pa.
4. Strain = 2.4 × 10⁻³ / 3.20 = 7.5 × 10⁻⁴.
5. k = 14 / 0.035 = 400 N m⁻¹; energy = ½ × 14 × 0.035 = 0.245 J.
6. Four times larger.
7. E = 160 × 10⁶ / 8.0 × 10⁻⁴ = 2.0 × 10¹¹ Pa.
8. E = FL/(AΔL) = 60 × 0.90 / (3.0 × 10⁻⁷ × 1.8 × 10⁻³) = 1.0 × 10¹¹ Pa.
9. ½ × 300 × 0.040² = 0.24 J; v = √(2 × 0.24 / 0.050) = 3.1 m s⁻¹.
10. Energy that is not recovered: it is transferred to internal energy of the material.

Where marks are usually lost

• Using diameter as radius in the area, or leaving the diameter in millimetres.
• Forgetting the ½ in the strain energy of a straight-line graph, or using ½kΔL² when the graph is not straight.
• Leaving extensions in millimetres when the formula needs metres.
• Using the stretched length in the strain, rather than the original length.
• Calling the gradient of a force-extension graph the Young modulus.
• Describing a brittle material as “weak”: brittle means it fractures with little or no plastic deformation, and it can still be strong.
• Writing “it is elastic” for a curved graph without saying where the elastic limit lies.
• Giving a practical uncertainty without saying that the diameter, squared in the area, dominates it.
• Ignoring the compression distance when working out the height reached by a spring-launched object.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams May/June 2020 onwards, published by OxfordAQA. Sections 3.2.9 and 3.2.10.