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

This guide covers sections 3.2.9 (Bulk properties of solids) and 3.2.10 (The Young modulus) of the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4. Both sections are shared content: they are part of International AS and also of International A-level. It includes Required practical 2. Condensed recall is in the revision notes and exam-style questions are in the practice set. Course pages: OxfordAQA A-level Physics hub and the printable checklist. Related: mechanics and materials revision notes. To find weak spots first, try the free diagnostics.

Take g = 9.81 N kg⁻¹ wherever a calculation needs it.

What you must be able to do

Spec section	Skill
3.2.9	Use density ρ = m/V
3.2.9	State Hooke’s law and the elastic limit; use F = kΔL (k is the stiffness or spring constant)
3.2.9	Define tensile stress and tensile strain; use breaking stress
3.2.9	Find elastic strain energy: ½FΔL, the area under a force-extension graph
3.2.9	Describe plastic behaviour, fracture and brittle behaviour from force-extension graphs
3.2.9	Apply energy conservation to elastic strain energy, kinetic energy and gravitational potential energy
3.2.9	Interpret simple stress-strain curves
3.2.10	Use E = stress/strain = FL/(AΔL); find E from a stress-strain graph
3.2.10	Carry out Required practical 2: load-extension graph for a wire and the Young modulus

Density

Density is mass per unit volume: ρ = m/V, in kg m⁻³. It describes the material, not the size of the sample. Convert lengths to metres before finding a volume: 1 cm³ = 1 × 10⁻⁶ m³.

Worked example 1. A brass cylinder has diameter 24.0 mm, length 55.0 mm and mass 0.215 kg. Find its density.

V = π r² l = π × (0.0120)² × 0.0550 = 2.49 × 10⁻⁵ m³. ρ = m/V = 0.215 / 2.49 × 10⁻⁵ = 8.64 × 10³ kg m⁻³.

Hooke’s law and the elastic limit

Hooke’s law: the extension of a material is proportional to the force applied, provided the elastic limit is not exceeded. In symbols, F = kΔL, where ΔL is the extension and k is the stiffness (spring constant), in N m⁻¹. A stiffer spring has a larger k.

Beyond the limit of proportionality the graph curves. The elastic limit is the largest extension from which the material still recovers its original length once the force is removed. For many metals the two points are close together, but they are different ideas. For a spring that obeys Hooke’s law, a force-extension graph is a straight line through the origin whose gradient is k.

Worked example 2. A spring stretches by 62 mm when a 0.450 kg mass hangs from it. Find k, then the extension for a 0.80 kg mass, assuming the elastic limit is not passed.

F = mg = 0.450 × 9.81 = 4.41 N. k = F/ΔL = 4.41 / 0.062 = 71.2 N m⁻¹. For 0.80 kg: F = 7.85 N, so ΔL = F/k = 7.85 / 71.2 = 0.110 m.

Tensile stress, strain and breaking stress

Tensile stress is force per unit cross-sectional area: stress = F/A, in pascals (Pa = N m⁻²). Tensile strain is extension per original length: strain = ΔL/L. Strain has no unit. For a wire of diameter d, A = πd²/4, and the diameter must be in metres.

The breaking stress is the stress at which the material fractures. It depends on the material, not on the thickness of the sample: a thicker wire breaks at a larger force but at the same stress.

Worked example 3. A wire of diameter 0.35 mm breaks under a tension of 38 N. A 2.5 m length of it extends by 4.0 mm under a smaller load. Find the breaking stress and the strain in the second case.

A = π × (0.175 × 10⁻³)² = 9.62 × 10⁻⁸ m². Breaking stress = 38 / 9.62 × 10⁻⁸ = 3.9 × 10⁸ Pa. Strain = 4.0 × 10⁻³ / 2.5 = 1.6 × 10⁻³.

Elastic strain energy

A stretching force does work on the material, and that work is the area beneath the line on a graph of force against extension. For a material obeying Hooke’s law the area is a triangle, so

energy stored = ½FΔL = ½kΔL² (using F = kΔL).

Doubling the extension of a spring multiplies the stored energy by four. For a non-linear graph, find the area by splitting it into triangles and trapeziums, or by counting squares.

Worked example 4. Find the energy stored in the spring of example 2 at an extension of 62 mm.

W = ½FΔL = ½ × 4.41 × 0.062 = 0.137 J.

Worked example 5. A metal strip gives a graph that rises in a straight line from the origin to (0.020 m, 12 N), then in a straight line to (0.060 m, 20 N). Find the work done in stretching it to 0.060 m.

Triangle: ½ × 12 × 0.020 = 0.12 J. Trapezium: ½ × (12 + 20) × (0.060 − 0.020) = 0.64 J. Total = 0.76 J.

Plastic, fracture and brittle behaviour

Describe each behaviour from the graph, using the loading and unloading lines.

• Elastic behaviour: the unloading line retraces the loading line and the sample returns to its original length.
• Plastic behaviour: once the elastic limit is passed, the extension is permanent. On unloading, the line falls parallel to the initial straight part and meets the extension axis at a positive value, the permanent extension. The area between the loading and unloading lines is energy that was not recovered; it is transferred to the internal energy of the material.
• Ductile material (for example, copper): shows a large plastic region before it fractures.
• Brittle material (for example, glass): the graph is a straight line up to the point of fracture, with almost no plastic region. It breaks suddenly at a small extension.

Energy conservation with elastic strain energy

Energy stored in a spring can be transferred to kinetic energy and gravitational potential energy. Write the energy equation between two named positions, then solve. If a question says energy is lost, include it as a separate term.

Worked example 6. A spring with k = 275 N m⁻¹ is compressed by 0.050 m and releases a 0.040 kg dart vertically upwards. Find the maximum height above the release position, and the speed of the dart as it leaves the spring at its natural length. Ignore resistive forces.

Stored energy = ½kΔL² = ½ × 275 × 0.050² = 0.344 J. At the top, all of it is gravitational potential energy: mgh = 0.344, so h = 0.344 / (0.040 × 9.81) = 0.876 m. At natural length the dart has risen 0.050 m: ½mv² = 0.344 − 0.040 × 9.81 × 0.050 = 0.324 J. v = √(2 × 0.324 / 0.040) = 4.03 m s⁻¹.

The Young modulus

Within the elastic region, stress is proportional to strain, and their ratio is a property of the material, the Young modulus:

E = tensile stress / tensile strain = (F/A) / (ΔL/L) = FL/(AΔL)

Its unit is the pascal (Pa), the same as stress. A large E means a stiff material. Unlike k, E does not depend on the size of the sample. Note that k = EA/L for a wire.

Worked example 7. An aluminium wire of original length 1.60 m and diameter 0.80 mm carries a load of 85 N and extends by 3.9 mm. Find the Young modulus.

A = π × (0.40 × 10⁻³)² = 5.03 × 10⁻⁷ m². Stress = 85 / 5.03 × 10⁻⁷ = 1.69 × 10⁸ Pa. Strain = 3.9 × 10⁻³ / 1.60 = 2.44 × 10⁻³. E = 1.69 × 10⁸ / 2.44 × 10⁻³ = 6.9 × 10¹⁰ Pa.

Stress-strain curves

Draw stress up the vertical axis and strain along the horizontal. For a given material this graph is independent of the sample’s dimensions.

• The straight section through the origin shows proportionality. Its gradient is the Young modulus.
• The area under the curve is the energy stored per unit volume, in J m⁻³. For the straight part it is ½ × stress × strain.
• The stress at which the material breaks is the breaking stress. A ductile material may show a clear change in gradient as it deforms plastically, and a brittle material shows a straight line ending at fracture.

Worked example 8. The straight part of a stress-strain graph ends at a strain of 0.0020 and a stress of 130 MPa. Find E and the energy stored per unit volume at that point. The wire has length 1.60 m and diameter 0.80 mm: find the energy stored in the whole wire.

E = 130 × 10⁶ / 0.0020 = 6.5 × 10¹⁰ Pa. Energy per unit volume = ½ × 130 × 10⁶ × 0.0020 = 1.3 × 10⁵ J m⁻³. Volume = 5.03 × 10⁻⁷ × 1.60 = 8.04 × 10⁻⁷ m³. Energy = 1.3 × 10⁵ × 8.04 × 10⁻⁷ = 0.105 J.

Required practical 2: load-extension of a wire

Aim: investigate the load-extension graph for a wire and find the Young modulus of its material.

1. Clamp one end of a long, thin wire (about 2.65 m here) firmly to a fixed support, pass it over a pulley, and hang a weight holder from the other end. A long, thin wire gives an extension large enough to measure.
2. Attach a marker to the wire and find the original length L, from the clamp to the marker, with a metre rule.
3. Use a micrometer to measure the diameter at several places along the wire and in two perpendicular directions. Take the mean and find A = πd²/4.
4. Add masses in equal steps, recording the marker position on a fixed scale each time. Extension is the change in position. Remove the masses in steps and check that the wire returns to its original reading.
5. Wear eye protection, and keep clear of the falling masses if the wire snaps; a box of soft material below them is sensible.

Analysis. Plot extension (vertical) against force (horizontal). The gradient is ΔL/F = L/(EA), so E = L/(A × gradient).

Data: L = 2.65 m; diameters 0.20, 0.21, 0.20, 0.19, 0.20 mm, mean 0.200 mm, so A = 3.14 × 10⁻⁸ m².

Mass / kg	Force / N	Extension / mm
0.2	1.96	0.8
0.6	5.89	2.5
1.0	9.81	4.2
1.2	11.77	5.1

(Intermediate readings were also taken.) The line of best fit through the origin has gradient 4.3 × 10⁻⁴ m N⁻¹.

E = 2.65 / (3.14 × 10⁻⁸ × 4.3 × 10⁻⁴) = 1.96 × 10¹¹ Pa.

The diameter dominates the uncertainty: ±0.01 mm on 0.20 mm is 5%, which is 10% in A because A depends on d². Keep the total load small enough to stay below the elastic limit.

Common errors

• Using the diameter as the radius, or leaving it in millimetres, in A = πd²/4.
• Writing strain with a unit, or using the extended length in place of the original length.
• Using the total length of wire rather than the length from the clamp to the marker.
• Using the gradient of a force-extension graph as E; it gives k, and E needs L and A as well.
• Calling any non-linear region “plastic”: the elastic limit and the limit of proportionality are not always the same point.
• Forgetting the ½ in the area under a straight-line force-extension graph.

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.