Specification: AQA A-level Physics (7408) 3.4.2.2 | Paper: 1 (and 2 as assumed knowledge) | Data and formulae booklet: given in the exam
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📌 KEY DEFINITIONS – learn them word for word
Tensile stress: the tensile force applied per unit cross-sectional area, σ = F ÷ A. Unit: Pa (N m⁻²).
Tensile strain: the extension divided by the original length, ε = ΔL ÷ L. It is a ratio of two lengths, so it has no unit.
Young modulus: the tensile stress divided by the tensile strain, for a material below its limit of proportionality. It is a property of the material, not the size of the sample. Unit: Pa.
The equation
tensile stress = tensile force ÷ cross-sectional area σ = F ÷ A
tensile strain = extension ÷ original length ε = ΔL ÷ L
Young modulus = tensile stress ÷ tensile strain E = σ ÷ ε = FL ÷ AΔL
| Quantity | Symbol | Unit |
|---|---|---|
| tensile force | F | newtons (N) |
| cross-sectional area | A | square metres (m²) |
| extension | ΔL | metres (m) |
| original length | L | metres (m) |
| tensile stress | σ | pascals (Pa), 1 Pa = 1 N m⁻² |
| tensile strain | ε | no unit |
| Young modulus | E | pascals (Pa) |
Rearranged: ΔL = FL ÷ AE | F = EAΔL ÷ L | For a wire, A = πd² ÷ 4: halve the diameter to get the radius before squaring, and remember 1 mm² = 10−6 m².
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) A wire of cross-sectional area 2.0 × 10−6 m² carries a tension of 240 N. Calculate the tensile stress in the wire.
Show solution
F: σ = F ÷ A
I: σ = 240 ÷ (2.0 × 10−6)
F: σ is already the subject
A: 1.2 × 108 Pa (120 MPa)
Q2 (Core) A wire of original length 1.80 m extends by 0.0027 m when loaded. Calculate the tensile strain.
Show solution
F: ε = ΔL ÷ L
I: ε = 0.0027 ÷ 1.80
F: ε is already the subject
A: 1.5 × 10−3 (no unit)
Q3 (Core) A copper sample has a tensile stress of 84 MPa when its tensile strain is 7.0 × 10−4. Calculate the Young modulus of copper in Pa and in GPa.
Show solution
Convert first: 84 MPa = 84 × 106 Pa
F: E = σ ÷ ε
I: E = (84 × 106) ÷ (7.0 × 10−4)
F: E is already the subject
A: 1.2 × 1011 Pa = 120 GPa
Q4 (Core) A steel wire with a cross-sectional area of 0.40 mm² supports a tension of 60 N. Calculate the tensile stress in MPa.
Show solution
Convert first: 0.40 mm² = 0.40 × 10−6 m² = 4.0 × 10−7 m²
F: σ = F ÷ A
I: σ = 60 ÷ (4.0 × 10−7)
F: σ is already the subject
A: 1.5 × 108 Pa = 150 MPa
Watch out: 1 mm² = (10−3 m)² = 10−6 m², not 10−3 m².
Q5 (Exam) A copper wire of diameter 0.56 mm carries a tension of 25 N. Calculate the tensile stress in the wire, to 2 s.f.
Show solution
Step 1 – area: r = 0.56 ÷ 2 = 0.28 mm = 0.28 × 10−3 m, so A = πr² = π × (0.28 × 10−3)² = 2.463… × 10−7 m²
Step 2 – stress
F: σ = F ÷ A
I: σ = 25 ÷ (2.463… × 10−7)
F: σ is already the subject
A: 1.015… × 108 Pa = 1.0 × 108 Pa (2 s.f.)
Watch out: using the diameter as the radius gives an area four times too big (9.85 × 10−7 m²) and a stress four times too small.
Q6 (Exam) In the required practical, a student uses a long steel wire with a fixed marker (or Searle’s apparatus). The original length from the clamp to the marker is 2.00 m and a micrometer gives the diameter as 0.38 mm. Adding a 40.0 N load increases the extension by 3.5 mm. Calculate the Young modulus of the steel, to 2 s.f.
Show solution
Convert first: d = 0.38 mm, so r = 0.19 × 10−3 m; ΔL = 3.5 mm = 3.5 × 10−3 m
Step 1 – area: A = π × (0.19 × 10−3)² = 1.134… × 10−7 m²
Step 2 – Young modulus
F: E = FL ÷ AΔL
I: E = (40.0 × 2.00) ÷ (1.134… × 10−7 × 3.5 × 10−3)
F: E is already the subject
A: 2.015… × 1011 Pa = 2.0 × 1011 Pa (2 s.f.), a sensible value for steel
Watch out: L is the original length from the clamp to the marker, not the whole length of wire. Measure the diameter at several points and orientations and take a mean.
Q7 (Exam) A copper wire of length 3.0 m and cross-sectional area 1.5 × 10−7 m² hangs vertically. A 1.0 kg mass is attached to its lower end. The Young modulus of copper is 1.2 × 1011 Pa. Calculate the extension of the wire in mm, to 2 s.f.
Show solution
Step 1 – force: F = mg = 1.0 × 9.81 = 9.81 N
Step 2 – extension
F: E = FL ÷ AΔL
I: 1.2 × 1011 = (9.81 × 3.0) ÷ (1.5 × 10−7 × ΔL)
F: ΔL = FL ÷ AE = 29.43 ÷ (1.5 × 10−7 × 1.2 × 1011) = 29.43 ÷ 18 000
A: 1.635 × 10−3 m = 1.6 mm (2 s.f.)
Q8 (Exam) A stress–strain graph for an aluminium alloy is a straight line from the origin to the point (strain 1.5 × 10−3, stress 105 MPa). Calculate the Young modulus in GPa. A 4.0 m wire of this alloy is under a tensile stress of 56 MPa. Calculate its extension in mm.
Show solution
Step 1 – Young modulus from the gradient
F: E = σ ÷ ε
I: E = (105 × 106) ÷ (1.5 × 10−3)
F: E is already the subject
A: 7.0 × 1010 Pa = 70 GPa
Step 2 – strain: ε = σ ÷ E = (56 × 106) ÷ (7.0 × 1010) = 8.0 × 10−4
Step 3 – extension: ΔL = εL = 8.0 × 10−4 × 4.0 = 3.2 × 10−3 m = 3.2 mm
Watch out: the gradient of a stress–strain graph is E only on the straight-line part, below the limit of proportionality.
Q9 (Stretch) A lift of total mass 1200 kg hangs from a single steel cable and accelerates upwards at 1.5 m s⁻². The breaking stress of the steel (the stress at which it breaks) is 1.6 × 109 Pa, and the engineers require the stress never to exceed one fifth of this. Calculate the minimum diameter of the cable, in mm to 2 s.f.
Show solution
Step 1 – tension (Newton’s second law): T − mg = ma, so T = m(g + a) = 1200 × (9.81 + 1.5) = 13 572 N
Step 2 – maximum working stress: 1.6 × 109 ÷ 5 = 3.2 × 108 Pa
Step 3 – area
F: σ = F ÷ A
I: 3.2 × 108 = 13 572 ÷ A
F: A = 13 572 ÷ (3.2 × 108)
A: A = 4.241… × 10−5 m²
Step 4 – diameter: A = πd² ÷ 4, so d = √(4A ÷ π) = 7.348… × 10−3 m = 7.3 mm (2 s.f.)
Watch out: while the lift accelerates upwards the tension is greater than the weight. Using mg alone underestimates the stress.
Q10 (Stretch) A steel wire and a copper wire, each 1.50 m long and 0.50 mm in diameter, are joined end to end and hung vertically. A 20 N load hangs from the bottom. Calculate the total extension, in mm to 2 s.f. (Young modulus: steel = 2.0 × 1011 Pa, copper = 1.2 × 1011 Pa; ignore the weight of the wires.)
Show solution
Step 1 – area: r = 0.25 × 10−3 m, so A = π × (0.25 × 10−3)² = 1.963… × 10−7 m²
Step 2 – steel extension (both wires carry the full 20 N tension)
F: E = FL ÷ AΔL
I: 2.0 × 1011 = (20 × 1.50) ÷ (1.963… × 10−7 × ΔL)
F: ΔL = FL ÷ AE = 30 ÷ (1.963… × 10−7 × 2.0 × 1011)
A: ΔLsteel = 7.639… × 10−4 m
Step 3 – copper extension: ΔLcopper = 30 ÷ (1.963… × 10−7 × 1.2 × 1011) = 1.273… × 10−3 m
Step 4 – total: 7.639… × 10−4 + 1.273… × 10−3 = 2.037… × 10−3 m = 2.0 mm (2 s.f.)
Watch out: the copper stretches more because its Young modulus is smaller. A stiffer material has a larger E.
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