Density, Hooke's Law and Elasticity Revision Notes | A Level AQA Physics

1. Density — Packing Mass into Space

Picture two boxes of identical size. One is filled with lead shot, the other with polystyrene beads. Both occupy the same volume, yet one is dramatically heavier. That difference is captured by a single quantity: density.

The equation $\rho = \dfrac{m}{V}$ tells you how tightly matter is packed. A material with a high density has a large mass squeezed into a small volume — think steel at around 7800 kg m⁻³ compared with wood at roughly 600 kg m⁻³.

For a regularly shaped object (a cuboid, cylinder, or sphere), you calculate volume from its dimensions. For an irregularly shaped object, you use Archimedes' displacement method: submerge the object in water and measure the volume of water displaced. That displaced volume equals the object's volume.

Worked example: Calculating Density of a Metal Cylinder

A metal cylinder has a mass of 0.540 kg. Its radius is 0.020 m and its length is 0.215 m. Calculate its density. [3 marks]

1. $\rho = \dfrac{m}{V}$ — [M1] Correct formula stated. Always write the equation first.
2. $V = \pi r^2 l = \pi \times (0.020)^2 \times 0.215$ $= \pi \times 0.020 \times 0.020 \times 0.215$ $= \pi \times 0.000400 \times 0.215$ $= \pi \times 8.60 \times 10^{-5}$ $= 2.70 \times 10^{-4} \text{ m}^3$ — [M1] Correct volume formula for a cylinder with r² expanded as 0.020 × 0.020 = 0.000400. Keep units as m³ throughout.
3. $\rho = \dfrac{0.540}{2.70 \times 10^{-4}}$ — Mass in kg, volume in m³ — units are consistent.
4. $\rho = 2000 \text{ kg m}^{-3}$ — [A1] Final answer with correct unit. This density is consistent with a light alloy such as magnesium.

Answer: ρ = 2000 kg m⁻³

Density is a property of the material itself, not the object's size. Doubling the volume of a steel block doubles its mass, so the ratio m/V stays the same. This is why density is described as an intensive property.

2. Hooke's Law and the Spring Constant

Now that you have a handle on density as a material property, the next question is: how do materials respond when you pull or compress them? The starting point is Hooke's law — one of the most tested relationships in A Level Physics.

Hang a mass from a spring and it stretches. Double the mass and the extension doubles. This proportional relationship holds — but only up to a point. That point is the elastic limit.

The spring constant $k$ is a measure of stiffness. A large $k$ means the spring is stiff — you need a large force to produce a small extension. A small $k$ means the spring is compliant — a small force produces a large extension.

The elastic limit is the point beyond which the material no longer returns to its original shape when the force is removed. Below the elastic limit, deformation is elastic (fully recoverable). Beyond it, deformation becomes plastic (permanent).

On a force–extension graph, the Hooke's law region is the straight-line section through the origin. The gradient of this straight line equals the spring constant $k$. Once the line curves, the elastic limit has been passed and the material no longer obeys Hooke's law.

Hooke's law applies to springs, wires, and many other materials in their elastic region. The key condition is always stated in the specification: provided the elastic limit is not exceeded. Include this condition in any exam definition for full marks.

3. Tensile Stress and Tensile Strain

Hooke's law describes a spring, but engineers need to compare different materials regardless of their dimensions. A thick steel rod and a thin steel wire are both steel — yet they behave very differently under the same force. To make fair comparisons, physicists use tensile stress and tensile strain.

Think of stress as the 'intensity' of the force — how concentrated it is across the material's cross-section. A force of 1000 N spread across a large area causes far less stress than the same force concentrated on a tiny area.

Strain tells you the proportional deformation. An extension of 2 mm in a 1 m wire is a strain of 0.002. The same 2 mm extension in a 10 mm wire is a strain of 0.2 — a very different situation, even though the raw extension is identical.

Worked example: Calculating Tensile Stress and Tensile Strain

A steel wire of original length 2.50 m and circular cross-section of diameter 0.80 mm is stretched by a force of 150 N, producing an extension of 3.0 mm. Calculate (a) the tensile stress and (b) the tensile strain. [4 marks]

1. Diameter = 0.80 mm = 0.80 × 10⁻³ m Radius = 0.40 × 10⁻³ m Extension ΔL = 3.0 mm = 3.0 × 10⁻³ m — Always convert to SI before substituting. Mixing mm and m is one of the most common errors in this topic.
2. $A = \pi r^2 = \pi \times (0.40 \times 10^{-3})^2$ $= \pi \times 0.40 \times 10^{-3} \times 0.40 \times 10^{-3}$ $= \pi \times 1.60 \times 10^{-7}$ $= 5.03 \times 10^{-7} \text{ m}^2$ — Use radius (half the diameter). Expanding (0.40 × 10⁻³)² longhand: 0.40 × 0.40 = 0.16 and 10⁻³ × 10⁻³ = 10⁻⁶, giving 1.60 × 10⁻⁷.
3. $\sigma = \dfrac{F}{A} = \dfrac{150}{5.03 \times 10^{-7}} = 2.98 \times 10^{8} \text{ Pa}$ — [M1, A1] Stress is approximately 300 MPa — a realistic value for steel under moderate load.
4. $\varepsilon = \dfrac{\Delta L}{L_0} = \dfrac{3.0 \times 10^{-3}}{2.50} = 1.2 \times 10^{-3}$ — [M1, A1] Strain is dimensionless — do not attach a unit. This value (0.0012) is typical for steel in its elastic region.

Answer: (a) Tensile stress σ ≈ 2.98 × 10⁸ Pa (b) Tensile strain ε = 1.2 × 10⁻³ (no unit)

4. Elastic Strain Energy and Breaking Stress

When you stretch a spring within its elastic limit, you do work on it. That work is stored as elastic strain energy — and it is fully released when the force is removed. This is the same energy that makes a stretched elastic band snap back, or a compressed spring launch a projectile.

On a force–extension graph, the elastic strain energy stored equals the area under the graph up to the point of interest. For the linear (Hooke's law) region, this area is a triangle, giving the familiar formula $E = \frac{1}{2}F\Delta L$. Substituting $F = k\Delta L$ gives the equivalent form $E = \frac{1}{2}k(\Delta L)^2$.

Beyond the elastic limit, the area under the graph still represents energy — but not all of it is recovered. Some energy is dissipated as heat during plastic deformation. This is why a permanently stretched spring feels warm after being over-extended.

Breaking stress is a critical design parameter. Engineers must ensure that the stress in any structural component remains well below the breaking stress of the material used, with an appropriate safety factor applied.

Worked example: Calculating Elastic Strain Energy Stored in a Spring

A spring with spring constant k = 250 N m⁻¹ is extended by 0.12 m within its elastic limit. Calculate the elastic strain energy stored. [2 marks]

1. $E = \dfrac{1}{2}k(\Delta L)^2$ — [M1] This form is used when k and ΔL are known. It avoids needing to calculate F separately.
2. $(\Delta L)^2 = 0.12 \times 0.12 = 0.0144 \text{ m}^2$ — Always expand the square explicitly. A common error is writing 0.12² = 0.24 (doubling instead of squaring).
3. $E = \dfrac{1}{2} \times 250 \times 0.0144 = 125 \times 0.0144 = 1.80 \text{ J}$ — [A1] 125 × 0.0144: 125 × 0.01 = 1.25 and 125 × 0.0044 = 0.55, giving 1.80 J.

Answer: Elastic strain energy E = 1.80 J

The concepts in this topic are well-defined, but the marks are often lost in the details of method, units, and precise language. The guidance below targets the patterns that appear most frequently in AQA mark schemes.

5. Exam Technique and Common Mistakes

When an exam question asks you to explain why a material obeys or does not obey Hooke's law, structure your answer as: observation + because + scientific principle. For example: 'The extension doubles when the force doubles, because the atoms in the material are displaced proportionally from their equilibrium positions, and the restoring force is proportional to displacement within the elastic region.'

For evaluate questions about material behaviour — for instance, comparing two materials from a stress–strain graph — always make a direct comparison using data from the graph, state which material has the higher breaking stress or greater stiffness, and give a conclusion. Vague qualitative statements without reference to values will not reach the top mark band.

Across this topic, every key quantity — density, spring constant, tensile stress, tensile strain, and elastic strain energy — is defined precisely, has a specific unit (or is dimensionless in the case of strain), and appears in at least one equation. Knowing which equation to reach for, and applying it with consistent SI units, is what separates a confident answer from a dropped mark.