CP3 – Young Modulus | Edexcel A Level Physics

Theory & Background

Stress, strain, the Young Modulus and elastic deformation

Objectives

Measure the extension of a wire under increasing load and determine its Young Modulus E
Plot a force–extension graph and identify the limit of proportionality
Convert to a stress–strain graph and calculate E directly from the gradient
Evaluate sources of uncertainty and compare with accepted values

Stress, Strain and Young Modulus

When a tensile force F is applied to a wire of cross-sectional area A and original length L₀, it stretches by extension ΔL. We define three key quantities:

Tensile stress σ = F / A (Pa or N m⁻²) Tensile strain ε = ΔL / L₀ (dimensionless) Young Modulus E = σ / ε = (F · L₀) / (A · ΔL) (Pa)

E is constant within the elastic region — up to the limit of proportionality.

Hooke's Law and the Force–Extension Graph

Within the elastic region, extension ΔL is proportional to force F. The gradient gives stiffness k:

F = k · ΔL → k = ΔF / ΔΔL (N m⁻¹) E = k · L₀ / A

Area under F–ΔL graph = elastic strain energy stored = ½F·ΔL

Material Properties

Copper: E ≈ 130 GPa · UTS ≈ 210 MPa Steel: E ≈ 200 GPa · UTS ≈ 400 MPa

Steel is stiffer (higher E) and stronger (higher UTS) than copper. Both are ductile — they deform plastically before fracture.

Step-by-Step Procedure

Follow these steps in the Simulation tab

Equipment

Long wire (copper or steel, ~2 m) · Clamp and bench pulley · 100 g hanger + slotted masses · Metre rule · Micrometer · Reference marker · Safety goggles

Choose material and set wire dimensions

Select copper or steel. Adjust diameter d (measured with a micrometer at 3 points in a real experiment) and original length L₀.

Hang the initial load (hanger only)

The hanger alone (100 g) straightens any kinks. This is your zero-extension reference — note the ruler reading through the zoom lens.

Add slotted masses one at a time

Click + Add 100 g to place a disc on the hanger. After each addition, wait for oscillations to stop, read the ruler zoom, then click Record Reading.

F = mg → 5 discs = 500 g = 4.91 N

Read extension using the zoom lens

The zoom panel shows a magnified view of the ruler at the reference marker position. The green marker line indicates your zero; the amber line shows the current wire position. Read ΔL to 0.1 mm precision.

Collect at least 8 readings

Record readings at 8+ different loads. Watch for the red warning beyond the limit of proportionality.

Analyse both graphs

Graph 1 (F vs ΔL) gives k. Graph 2 (σ vs ε) gives E directly from the gradient.

E = gradient of σ–ε graph | E = kL₀/A

Controls

Wire material

E = 130 GPa

Wire diameter d 0.40 mm

0.30 mm0.60 mm

Wire length L₀ 2.00 m

1.00 m3.00 m

Slotted Masses 100 g (hanger)

Click Remove or tap top disc to remove

Force F

0.98 N

Extension ΔL

0.000 mm

Stress σ

— MPa

Strain ε

—

Readings recorded: 0/8 minimum

🔍 Ruler Zoom

— Green line: zero ref
— Amber line: wire marker
— Read ΔL between lines

RULER READING
0.000 mm

Add masses using the panel on the left, then click Record Reading to log each data point.

Results Data Table

Auto-filled from simulation. Stress and strain calculated automatically.

#	Discsadded	m/ g	F = mg/ N	ΔL/ mm	ΔL/ m	σ = F/A/ MPa	ε = ΔL/L₀× 10⁻³	Region
No data yet — go to Simulation tab and record readings.

Graph & Analysis

Graph 1 gives stiffness k. Graph 2 gives Young Modulus E directly from the stress–strain gradient.

Stiffness k

—Record ≥3 readings first

Graph Parameters

Gradient k—

R²—

Elastic pts used—

E from k·L₀/A

Calculated E—

Accepted E—

% Difference—

Young Modulus E

—Record ≥3 readings first

Graph Parameters

Gradient = E—

R²—

Accepted E—

% Difference—

Interpretation

Complete your experiment to see analysis here.

Discussion Questions

Write your answers below and reveal model answers when ready.

Question 1

Explain why a long, thin wire is used rather than a short, thick one.

Extension ΔL = FL₀/AE. A longer wire (larger L₀) gives a proportionally larger extension for the same force, making ΔL easier to measure accurately. A thinner wire (smaller A) also increases ΔL. Both reduce percentage uncertainty in ΔL, which directly reduces percentage uncertainty in E.

Question 2

Describe the difference between the limit of proportionality and the elastic limit.

The limit of proportionality is where stress is no longer proportional to strain — the F–ΔL graph begins to curve. The elastic limit is slightly beyond this — the maximum stress at which the wire still returns to its original length on unloading. Beyond the elastic limit, permanent plastic deformation occurs. The elastic limit cannot be identified from the loading curve alone; you must unload and measure permanent set.

Question 3

Show that the gradient of the stress–strain graph equals E, and that this is consistent with E = kL₀/A from the force–extension graph.

Stress–strain gradient = Δσ/Δε = E by definition. From F–ΔL: gradient = k = ΔF/ΔΔL. Since σ = F/A and ε = ΔL/L₀: E = σ/ε = (F/A)÷(ΔL/L₀) = FL₀/AΔL = (ΔF/ΔΔL)·(L₀/A) = kL₀/A. Both methods are equivalent. The σ–ε method is more fundamental because E is independent of geometry — it is a pure material property.

Question 4

Identify two sources of systematic error and explain how each affects your calculated E.

1. Wire kinks — kinks straighten before elastic stretching occurs. Initial extension readings are too large, making gradient k smaller, and E is underestimated. Applying a small pre-load before zero reference removes kinks. 2. Parallax in ruler reading — if the eye is not level with the reference marker, ΔL readings are systematically over- or under-estimated. This shifts the gradient and gives an incorrect E. A travelling microscope or vernier scale attached to the wire eliminates parallax.

Question 5

Steel has E ≈ 200 GPa while copper is ≈ 130 GPa. What does this tell you about interatomic bonding?

Young Modulus measures the stiffness of interatomic bonds — the gradient of the force–separation curve at equilibrium. Higher E means bonds resist stretching more strongly per unit strain. Steel (iron + carbon) has stronger, stiffer metallic bonds than pure copper. Carbon atoms in steel occupy interstitial lattice positions, pinning dislocations and increasing resistance to deformation. Copper has weaker bonds and a face-centred cubic structure allowing easier slip, giving lower E and greater ductility.