# Materials

> Edexcel International A-Level Physics · IAL Physics Unit 1 (WPH11)
> Source: https://www.owlsprep.com/study/edexcel-ial-physics-u1-materials/

This guide covers all Edexcel IAL Physics Unit 1 (WPH11) content for the Materials topic, including density, upthrust, Stokes' law, Hooke's law, stress/strain, material deformation, and the two required core practicals with exam-focused worked examples.

**Prerequisites:** [Basic forces, weight and vector addition](https://www.owlsprep.com/study/edexcel-ial-physics-u1-forces-motion/)

## Learning objectives

- Calculate density, upthrust, viscous drag, and elastic strain energy using standard formulae
- Interpret force-extension and stress-strain graphs to identify key material deformation points
- Describe and carry out core practicals for viscosity measurement and Young modulus determination
- Distinguish between elastic, plastic, ductile and brittle material behaviour in exam contexts

## Density and Upthrust

**Density** — Mass per unit volume of a substance, an intensive property independent of sample size.

*Notation:* \rho = \frac{m}{V}

*Example:* Density of pure water at room temperature = 1000 kg m⁻³ = 1 g cm⁻³

Archimedes' principle states that the upthrust acting on an object partially or fully submerged in a fluid is equal to the weight of the fluid displaced by the object. If upthrust equals the weight of the object, it will float; if upthrust is less than weight, it will sink.

**Worked example:** A 0.02 m³ block of aluminium (density = 2700 kg m⁻³) is fully submerged in water. Calculate the upthrust acting on the block, and its apparent weight when submerged.

1. Step 1: Calculate mass of displaced water: $m = \rho_{water} \times V = 1000 \times 0.02 = 20$ kg
2. Step 2: Upthrust = weight of displaced water: $F_{upthrust} = mg = 20 \times 9.81 = 196.2$ N
3. Step 3: Calculate weight of aluminium block: $W = \rho_{Al} \times V \times g = 2700 \times 0.02 \times 9.81 = 529.74$ N
4. Step 4: Apparent weight = actual weight - upthrust = $529.74 - 196.2 = 334$ N (3 significant figures)

> **Exam tip:** Always use $g=9.81$ m s⁻² for calculations unless explicitly told otherwise, and round final answers to 3 significant figures to match standard exam data precision.

## Viscous Drag and Stokes' Law

**Stokes' Law** — Viscous drag force acting on a small sphere moving at low speed through a fluid with laminar (smooth, non-turbulent) flow, where $\eta$ = viscosity of the fluid, $r$ = radius of the sphere, $v$ = speed of the sphere.

*Notation:* F = 6\pi\eta r v

*Example:* Used in Core Practical 2 to measure viscosity of a viscous liquid using falling glass spheres.

> **warning**
>
> Only apply Stokes' Law if the question explicitly states laminar flow, a small spherical object, and low speed. It is invalid for non-spherical objects or turbulent flow conditions.

**Worked example:** A tiny glass sphere of radius 1.5 mm falls at terminal speed through glycerine (viscosity = 1.4 Pa s). Calculate the drag force acting on the sphere if its terminal speed is 0.08 m s⁻¹.

1. Step 1: Convert radius to SI units: $1.5$ mm = $1.5 \times 10^{-3}$ m
2. Step 2: Substitute values into Stokes' Law: $F = 6 \times \pi \times 1.4 \times 1.5 \times 10^{-3} \times 0.08$
3. Step 3: Calculate result: $F \approx 3.2 \times 10^{-3}$ N = 3.2 mN (2 significant figures)

Core Practical 2 (falling-ball viscometry) involves measuring the time taken for spheres of known radius to fall between two marked points in a liquid, calculating their terminal speed, and using Stokes' Law to determine the viscosity of the liquid. Viscosity of liquids decreases as temperature increases.

## Hooke's Law and Elastic Strain Energy

**Hooke's Law** — The extension of an elastic object is directly proportional to the applied force, up to the limit of proportionality, where $k$ = stiffness constant of the object (units: N m⁻¹).

*Notation:* \Delta F = k\Delta x

*Example:* A spring with stiffness $k=100$ N m⁻¹ extends by 0.1 m when a 10 N force is applied.

Elastic strain energy is the energy stored in a stretched or compressed elastic object. It is equal to the area under the force-extension graph for the object. For linear regions following Hooke's Law, this area is a triangle, so $\Delta E_{el} = \frac{1}{2}F\Delta x$. For non-linear graphs, estimate the area by counting grid squares under the curve.

**Worked example:** A spring with stiffness constant 120 N m⁻¹ is stretched by 4 cm. Calculate the elastic strain energy stored in the spring.

1. Step 1: Convert extension to SI units: $4$ cm = $0.04$ m
2. Step 2: Calculate applied force: $F = k\Delta x = 120 \times 0.04 = 4.8$ N
3. Step 3: Calculate strain energy: $\Delta E_{el} = \frac{1}{2} \times 4.8 \times 0.04 = 0.096$ J
4. Note: You can also use the rearranged formula $\Delta E_{el} = \frac{1}{2}k(\Delta x)^2$ to get the same result directly.

> **Exam tip:** Never use $E = F\Delta x$ for strain energy calculations in linear regions, as force increases linearly with extension rather than being constant. You will lose half marks for this error in exams.

## Stress, Strain and Young Modulus

**Young Modulus** — A material-specific measure of stiffness, independent of object dimensions, where $\sigma$ (stress) = $\frac{F}{A}$, $\epsilon$ (strain) = $\frac{\Delta x}{x_0}$, $x_0$ = original length of sample, $A$ = cross-sectional area of sample. Units: Pascals (Pa).

*Notation:* E = \frac{\sigma}{\epsilon} = \frac{kx_0}{A}

*Example:* Young modulus of copper ≈ $2.1 \times 10^{11}$ Pa

Core Practical 3 (determination of Young Modulus) involves measuring the extension of a thin test wire for a range of applied masses, plotting a force-extension graph, using the gradient of the linear region to find the stiffness $k$, and substituting into $E = \frac{kx_0}{A}$ to calculate the Young modulus of the wire material.

**Worked example:** A 2 m long copper wire of cross-sectional area $1.2 \times 10^{-7}$ m² is stretched by 1.2 mm when a 15 N force is applied. Calculate the Young modulus of copper.

1. Step 1: Convert extension to SI units: $1.2$ mm = $1.2 \times 10^{-3}$ m
2. Step 2: Calculate stress: $\sigma = \frac{F}{A} = \frac{15}{1.2 \times 10^{-7}} = 1.25 \times 10^8$ Pa
3. Step 3: Calculate strain: $\epsilon = \frac{\Delta x}{x_0} = \frac{1.2 \times 10^{-3}}{2} = 6 \times 10^{-4}$
4. Step 4: Calculate Young modulus: $E = \frac{\sigma}{\epsilon} = \frac{1.25 \times 10^8}{6 \times 10^{-4}} \approx 2.1 \times 10^{11}$ Pa (2 significant figures)

## Material Deformation and Graph Interpretation

Force-extension graphs are specific to an individual object (depend on dimensions), while stress-strain graphs are characteristic of the material itself. Key points on these graphs include: limit of proportionality (end of linear Hooke's Law region), elastic limit (point after which plastic deformation occurs), yield point (sudden increase in extension for small force increase), and breaking stress (maximum stress before fracture).

> **tip**
>
> When comparing ductile and brittle materials, always reference graph features for marks: ductile materials (e.g. copper) have a large plastic region after the elastic limit, while brittle materials (e.g. glass) fracture immediately after the elastic limit with no plastic deformation.

**Worked example:** A force-extension graph for a metal wire has a linear region from 0 N to 12 N (extension 0 to 3 mm), then curves until it breaks at 15 N (extension 8 mm). Estimate the total strain energy stored in the wire just before breaking.

1. Step 1: Calculate area of linear triangular region: $\frac{1}{2} \times 12 \times 0.003 = 0.018$ J
2. Step 2: Estimate area of non-linear region: approximate as a trapezium with average force $\frac{12+15}{2} = 13.5$ N, and extension change $0.008 - 0.003 = 0.005$ m: area ≈ $13.5 \times 0.005 = 0.0675$ J
3. Step 3: Total strain energy = sum of areas ≈ $0.018 + 0.0675 = 0.086$ J (2 significant figures)

## Common pitfalls

- **Wrong:** Using Reynolds number or Bernoulli principle for fluid-related materials questions
  - Why it fails: These concepts are explicitly out of scope for the Edexcel IAL Unit 1 Materials topic, and you will lose marks for including them
  - Correct: Only use Archimedes' principle and Stokes' Law for all fluid-related materials questions
- **Wrong:** Calculating elastic strain energy as $F\Delta x$ instead of $\frac{1}{2}F\Delta x$ for linear Hooke's Law regions
  - Why it fails: Force is not constant during extension, so the work done formula $W = Fd$ cannot be applied directly
  - Correct: Use $\frac{1}{2}F\Delta x$ for linear regions, or count grid squares to estimate area under the force-extension graph for non-linear regions
- **Wrong:** Applying Stokes' Law to non-spherical objects or turbulent flow conditions
  - Why it fails: Stokes' Law only holds for small spheres moving at low speeds in laminar flow, the formula is invalid otherwise
  - Correct: Only use Stokes' Law if the question explicitly states the required conditions, or describes a small falling ball in a viscous fluid
- **Wrong:** Confusing force-extension (object-specific) and stress-strain (material-specific) graphs
  - Why it fails: Stiffness $k$ is calculated from the gradient of a force-extension graph, while Young modulus $E$ is calculated from the gradient of a stress-strain graph, mixing these leads to incorrect results
  - Correct: Check graph axes first: x-axis extension = object-specific graph, x-axis strain = material-specific graph
- **Wrong:** Forgetting to convert units to SI (mm to m, cm² to m²) when calculating stress or Stokes' Law
  - Why it fails: All standard formulae require base SI units to give correct values for force, stress, viscosity and other quantities
  - Correct: Always convert all given quantities to base SI units before substituting into any formula

## Cheatsheet

| Formula/Concept | Symbol/Definition | Units/Notes |
| --- | --- | --- |
| Density | $\rho = \frac{m}{V}$ | kg m⁻³; 1 g cm⁻³ = 1000 kg m⁻³ |
| Upthrust | Weight of displaced fluid | $= \rho_{fluid} \times V_{displaced} \times g$ |
| Stokes' Law | $F = 6\pi\eta r v$ | Only valid for small spheres, laminar flow, low speed |
| Hooke's Law | $\Delta F = k\Delta x$ | $k$ = stiffness constant, units N m⁻¹ |
| Stress | $\sigma = \frac{F}{A}$ | Units Pa = N m⁻² |
| Strain | $\epsilon = \frac{\Delta x}{x_0}$ | Dimensionless, no units |
| Young Modulus | $E = \frac{\sigma}{\epsilon} = \frac{kx_0}{A}$ | Units Pa; constant for a given material |
| Elastic Strain Energy | $\Delta E_{el} = \frac{1}{2}F\Delta x$ = area under F-x graph | Estimate non-linear area by counting squares |
| Core Practical 2 | Falling-ball viscometry | Measure terminal speed to calculate fluid viscosity |
| Core Practical 3 | Young Modulus determination | Plot F-x graph, use gradient to find $E$ |

## What's next

Now you have mastered the core content for the Edexcel IAL Physics Unit 1 Materials topic, you can move on to practising past paper questions focused on this subtopic, as well as reviewing the rest of Unit 1 Mechanics and Materials to build full exam readiness. Make sure you can answer both calculation and descriptive questions, including those related to the two core practicals, which are frequently tested in IAS Unit 1 papers. Regular practice of graph interpretation questions will also help you avoid common mark-losing errors in exams.

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