The shape of the lesson
By the end of the lesson, learners can
- Load it, watch it extend: Add load to the wire and trace the force-extension graph.
- From spring to material: The spring constant describes one sample; the Young modulus describes the material.
- Measuring the Young modulus of a wire: The standard experiment turns four measurements into one material constant.
- Spring: F = kx, so k = F / x.
- Material: σ = F / A, ε = x / L.
- Young modulus: E = σ / ε.
One picture every idea returns to
Below the limit of proportionality, extension is proportional to load: F = kx, with k the spring constant. To describe the material, use tensile stress σ = F / A and tensile strain ε = x / L. Their ratio is the Young modulus, E = stress / strain, a property of the material independent of the sample size.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, How much it stretches.. Pull on a wire and it lengthens. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Load it, watch it extend | Add load to the wire and trace the force-extension graph. Project tab 2 of the lab (Stress, strain and Young modulus) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: From spring to material | The spring constant describes one sample; the Young modulus describes the material. | Pairs |
| 21 to 30 min | Teach and check: Measuring the Young modulus of a wire | The standard experiment turns four measurements into one material constant. | Anticipation / Reaction Guide |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 14 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Measuring the Young modulus of a wire, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Anticipation / Reaction Guide
Before the lesson, students mark five statements true or false. After the teaching they mark them again and justify any change of mind in one sentence each. The statements come from the examiner traps below.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Keep the limits straight: the limit of proportionality is where the graph stops being straight, while the elastic limit is where permanent deformation begins; they are close but not the same. | Stress uses the cross-sectional area (A = πd²/4 for a round wire), not the diameter or length. Strain is a ratio and has no units, so an answer with units of metres is wrong. |
Stress, strain and the Young modulus: worksheet
6 questions, 14 marks. Written in the style of the papers; no past paper question is reproduced.
State Hooke law and the condition under which it holds.
A spring extends by 25 mm under a load of 5.0 N. Find its spring constant.
A wire of diameter 0.50 mm carries a tension of 30 N. Find the tensile stress.
A wire of original length 1.6 m stretches by 0.80 mm under load. Find the strain.
For the wire in questions 3 and 4, the stress is 1.5 × 10⁸ Pa at a strain of 5.0 × 10⁻⁴. Find the Young modulus of the material.
Distinguish between the limit of proportionality and the elastic limit.
Answer key and marking notes
- The extension is proportional to the applied load (F = kx) ✓
- This holds up to the limit of proportionality ✓
- k = F / x = 5.0 / (25 × 10⁻³) ✓
- k = 200 N m⁻¹ ✓
- Area A = πd²/4 = π × (0.50 × 10⁻³)² / 4 = 1.96 × 10⁻⁷ m² ✓
- σ = F / A = 30 / (1.96 × 10⁻⁷) ✓
- σ = 1.5 × 10⁸ Pa ✓
- ε = x / L = (0.80 × 10⁻³) / 1.6 ✓
- ε = 5.0 × 10⁻⁴ ✓
- E = σ / ε = (1.5 × 10⁸) / (5.0 × 10⁻⁴) ✓
- E = 3.0 × 10¹¹ Pa ✓
- The limit of proportionality is where the load-extension graph stops being a straight line (Hooke law fails) ✓
- The elastic limit is the point beyond which the material no longer returns to its original length (permanent deformation begins) ✓
20 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Hooke law holds up to the:
- Alimit of proportionality
- Bbreaking point
- Cyield stress
- DYoung modulus
A spring extends 0.060 m under 9.0 N. Its spring constant is:
- A0.0067 N m⁻¹
- B540 N m⁻¹
- C150 N m⁻¹
- D15 N m⁻¹
Tensile stress is defined as:
- Aforce per unit length
- Bforce per unit cross-sectional area
- Cextension per unit length
- Dforce times area
Tensile strain is:
- Aforce divided by area
- Bstress times area
- Cextension times length
- Dextension divided by original length
The Young modulus is:
- Atensile stress divided by tensile strain
- Bstress times strain
- Cforce times extension
- Dstrain divided by stress
A wire of area 1.0 × 10⁻⁶ m² carries 50 N. The stress is:
- A5.0 × 10⁶ Pa
- B2.0 × 10⁻⁸ Pa
- C5.0 × 10⁷ Pa
- D50 Pa
A 4.0 m wire stretches 2.0 mm. The strain is:
- A2.0 × 10⁻³
- B5.0 × 10⁻⁴
- C8.0 × 10⁻³
- D0.50
A material has stress 2.0 × 10⁸ Pa at strain 1.0 × 10⁻³. Its Young modulus is:
- A2.0 × 10⁵ Pa
- B5.0 × 10⁻¹² Pa
- C2.0 × 10¹² Pa
- D2.0 × 10¹¹ Pa
The unit of the Young modulus is the:
- Apascal
- Bnewton
- Cjoule
- Dnewton per metre
Two wires of the same material but different lengths and areas have the same Young modulus because it is:
- Athe spring constant
- Bproportional to length
- Ca property of the material, not the sample
- Dproportional to area
A spring of constant 50 N m⁻¹ supports a 2.0 N weight. Its extension is:
- A100 m
- B0.040 m
- C25 m
- D0.40 m
The gradient of the straight part of a load-extension graph gives the:
- AYoung modulus
- Bstress
- Cenergy stored
- Dspring constant
Beyond the elastic limit, a material:
- Adoes not return to its original length
- Bobeys Hooke law
- Chas zero stress
- Dhas zero strain
A wire of diameter 0.40 mm has a cross-sectional area of about:
- A4.0 × 10⁻⁴ m²
- B5.0 × 10⁻⁷ m²
- C1.3 × 10⁻⁷ m²
- D1.6 × 10⁻⁷ m²
A 2.0 m wire (E = 2.0 × 10¹¹ Pa, area 1.0 × 10⁻⁷ m²) carries 40 N. Its extension is:
- A2.0 × 10⁻³ m
- B4.0 × 10⁻³ m
- C8.0 × 10⁻⁴ m
- D1.0 × 10⁻³ m
Strain has no units because it is:
- Aa force
- Ban energy
- Ca pressure
- Da ratio of two lengths
A spring constant is 250 N m⁻¹. The load needed for a 0.020 m extension is:
- A5.0 N
- B12 500 N
- C0.50 N
- D50 N
For a given material and load, a thinner wire (smaller area) has a:
- Asmaller stress
- Bgreater Young modulus
- Cgreater stress and so a greater extension
- Dsmaller strain
Compressive deformation occurs when a force:
- Astretches a material
- Bsquashes a material, reducing its length
- Ctwists a material
- Dhas no effect
To find the Young modulus of a wire you plot stress against strain. The Young modulus is the:
- Aarea under the line
- Bintercept on the stress axis
- Creciprocal of the gradient
- Dgradient of the straight-line region
Quiz key
Extension is proportional to load only up to the limit of proportionality.
k = F/x = 9.0/0.060 = 150 N m⁻¹.
Stress = F/A.
Strain = x/L, a dimensionless ratio.
E = stress/strain.
σ = F/A = 50/(1.0 × 10⁻⁶) = 5.0 × 10⁷ Pa.
ε = x/L = (2.0 × 10⁻³)/4.0 = 5.0 × 10⁻⁴.
E = (2.0 × 10⁸)/(1.0 × 10⁻³) = 2.0 × 10¹¹ Pa.
Strain is dimensionless, so E has the units of stress, the pascal.
The Young modulus depends only on the material.
x = F/k = 2.0/50 = 0.040 m.
For F against x, the gradient is k = F/x.
Past the elastic limit the deformation is partly permanent (plastic).
A = πd²/4 = π(0.40 × 10⁻³)²/4 = 1.3 × 10⁻⁷ m².
x = FL/(EA) = (40 × 2.0)/((2.0 × 10¹¹)(1.0 × 10⁻⁷)) = 4.0 × 10⁻³ m.
Strain = extension/length, length over length, so it is dimensionless.
F = kx = 250 × 0.020 = 5.0 N.
Smaller area gives larger stress (F/A), and with the same E, larger strain and extension.
Compression shortens the material; tension lengthens it.
E = stress/strain = gradient of the stress-strain graph.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.