Materials

Edexcel International A-Level Physics· 2018 Spec Issue 3, Statements 23-32· 25 min read

1. Density and Upthrust★★☆☆☆⏱ 5 min

📘 Definition

Density

\rho = \frac{m}{V}

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

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: $m = ρ_{w a t er} \times V = 1000 \times 0.02 = 20$ kg
2
Step 2: Upthrust = weight of displaced water: $F_{u pt h r u s t} = m g = 20 \times 9.81 = 196.2$ N
3
Step 3: Calculate weight of aluminium block: $W = ρ_{A l} \times V \times g = 2700 \times 0.02 \times 9.81 = 529.74$ 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.

2. Viscous Drag and Stokes' Law★★★☆☆⏱ 6 min

📘 Definition

Stokes' Law

F = 6\pi\eta r v

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

Example:

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

📐 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 = $1.5 \times 1 0^{- 3}$ m
2
Step 2: Substitute values into Stokes' Law: $F = 6 \times π \times 1.4 \times 1.5 \times 1 0^{- 3} \times 0.08$
3
Step 3: Calculate result: $F \approx 3.2 \times 1 0^{- 3}$ 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.

3. Hooke's Law and Elastic Strain Energy★★☆☆☆⏱ 5 min

📘 Definition

Hooke's Law

\Delta F = k\Delta x

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).

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 $Δ E_{e l} = \frac{1}{2} F Δ x$ . 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: $F = k Δ x = 120 \times 0.04 = 4.8$ N
3
Step 3: Calculate strain energy: $Δ E_{e l} = \frac{1}{2} \times 4.8 \times 0.04 = 0.096$ J
4
Note: You can also use the rearranged formula $Δ E_{e l} = \frac{1}{2} k (Δ x)^{2}$ to get the same result directly.

Exam tip:

Never use $E = F Δ x$ 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.

4. Stress, Strain and Young Modulus★★★★☆⏱ 7 min

📘 Definition

Young Modulus

E = \frac{\sigma}{\epsilon} = \frac{kx_0}{A}

A material-specific measure of stiffness, independent of object dimensions, where $σ$ (stress) = $\frac{F}{A}$ , $ϵ$ (strain) = $\frac{Δ x}{x _{0}}$ , $x_{0}$ = original length of sample, $A$ = cross-sectional area of sample. Units: Pascals (Pa).

Example:

Young modulus of copper ≈ $2.1 \times 1 0^{11}$ 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 $E = \frac{k x _{0}}{A}$ to calculate the Young modulus of the wire material.

📐 Worked Example

A 2 m long copper wire of cross-sectional area $1.2 \times 1 0^{- 7}$ 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 = $1.2 \times 1 0^{- 3}$ m
2
Step 2: Calculate stress: $σ = \frac{F}{A} = \frac{15}{1.2 \times 1 0 ^{- 7}} = 1.25 \times 1 0^{8}$ Pa
3
Step 3: Calculate strain: $ϵ = \frac{Δ x}{x _{0}} = \frac{1.2 \times 1 0 ^{- 3}}{2} = 6 \times 1 0^{- 4}$
4
Step 4: Calculate Young modulus: $E = \frac{σ}{ϵ} = \frac{1.25 \times 1 0 ^{8}}{6 \times 1 0 ^{- 4}} \approx 2.1 \times 1 0^{11}$ Pa (2 significant figures)

5. Material Deformation and Graph Interpretation★★★☆☆⏱ 5 min

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).

📐 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)

6. Common Pitfalls

Wrong move:

Using Reynolds number or Bernoulli principle for fluid-related materials questions

Why:

These concepts are explicitly out of scope for the Edexcel IAL Unit 1 Materials topic, and you will lose marks for including them

Correct move:

Only use Archimedes' principle and Stokes' Law for all fluid-related materials questions

Wrong move:

Calculating elastic strain energy as $F Δ x$ instead of $\frac{1}{2} F Δ x$ for linear Hooke's Law regions

Why:

Force is not constant during extension, so the work done formula $W = F d$ cannot be applied directly

Correct move:

Use $\frac{1}{2} F Δ x$ for linear regions, or count grid squares to estimate area under the force-extension graph for non-linear regions

Wrong move:

Applying Stokes' Law to non-spherical objects or turbulent flow conditions

Why:

Stokes' Law only holds for small spheres moving at low speeds in laminar flow, the formula is invalid otherwise

Correct move:

Only use Stokes' Law if the question explicitly states the required conditions, or describes a small falling ball in a viscous fluid

Wrong move:

Confusing force-extension (object-specific) and stress-strain (material-specific) graphs

Why:

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 move:

Check graph axes first: x-axis extension = object-specific graph, x-axis strain = material-specific graph

Wrong move:

Forgetting to convert units to SI (mm to m, cm² to m²) when calculating stress or Stokes' Law

Why:

All standard formulae require base SI units to give correct values for force, stress, viscosity and other quantities

Correct move:

Always convert all given quantities to base SI units before substituting into any formula

7. Quick Reference Cheatsheet

Formula/Concept	Symbol/Definition	Units/Notes
Density	$ρ = \frac{m}{V}$	kg m⁻³; 1 g cm⁻³ = 1000 kg m⁻³
Upthrust	Weight of displaced fluid	$= ρ_{f l u i d} \times V_{d i s pl a ce d} \times g$
Stokes' Law	$F = 6 π η r v$	Only valid for small spheres, laminar flow, low speed
Hooke's Law	$Δ F = k Δ x$	$k$ = stiffness constant, units N m⁻¹
Stress	$σ = \frac{F}{A}$	Units Pa = N m⁻²
Strain	$ϵ = \frac{Δ x}{x _{0}}$	Dimensionless, no units
Young Modulus	$E = \frac{σ}{ϵ} = \frac{k x _{0}}{A}$	Units Pa; constant for a given material
Elastic Strain Energy	$Δ E_{e l} = \frac{1}{2} F Δ x$ = 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 $E$

Going deeper

official_specEdexcel IAL Physics 2018 Specification Unit 1Refer to statements 23-32 for materials content

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.