AQA A-Level Physics Paper 1, June 2025: Question 4
9 marks · Medium difficulty · Short Answer
Explain units of Young modulus and breaking stress, calculate the material cost of a wire using its dimensions and density, identify identical materials from properties, and deduce from a stress-strain graph whether it corresponds to a given wire.
Practise this questionQuestion
Question text
04.1 Explain why the Young modulus and the breaking stress of a wire have the same
SI unit.
[1 mark]
Table 1 contains data about four metal wires W, X, Y and Z.
Table 1
Wire Length / m Diameter / mm Stiffness / N m−1 Density / kg m−3
W 3.10 1.7 4.90 × 104 2.71 × 103
X 2.17 1.4 5.25 × 104 1.93 × 104
Y 2.50 1.2 5.25 × 104 8.91 × 103
Z 2.50 1.2 3.08 × 104 2.71 × 103
04.2 The metal used to make wire X costs £75 per gram.
Calculate, in £, the cost of wire X.
[3 marks]
cost = £
04.3 State and explain which of the wires in Table 1 are made from the same material.
[1 mark]
04.4 One of the wires in Table 1 was used to obtain the stress–strain data shown
in Figure 5.
*08* Figure 5
Draw a line of best fit on Figure 5.
Go on to deduce, using your line of best fit and Table 1, whether wire W could have
been used to produce Figure 5.
[4 marks]
Mark scheme
Show the mark scheme
Question Answers Additional comments/Guidance Mark AO
04.1 �tensile� stress 1 AO1
Idea that YM = and (tensile) strain has no
�tensile� strain
units
04.2 Evidence of determination of volume of one wire 1 When X is used in MP1, condone the use of 3 AO3
diameter for radius of X.
Expect V = 3.3 × 10−6 m3 Condone PoT error
Uses relationship between density of X Expect M = 64 g or 65 g Condone PoT error
OR
the density of the wire used in MP1
and their volume to determine their mass 2
75 × 64 or 75 × 65
Answer between £4800 and £4900 3
04.3 W and Z, as their densities are the same (and density is Allow suggestion that the Young modulus is 1 AO1
the only one that is a property of the material) the same provided there is evidence of its
determination for both.
04.4 Draws best-fit line 1 For 1 look for an equal scatter of points either 4 AO3
side of the line.
Uses pair of values of stress and strain from point on line
more than halfway along the line
OR Allow half square tolerance on graph read-offs.
uses a difference in two pairs that is more than half the Condone PoT error
line 2
Copy of W data from table 1
Evidence of attempt to use equation(s) to determine E, k,
l, A or d
eg cross-sectional area of W = 2.27 × 10−6 m2
from graph OR for W (for use in a comparison). 3
Condone PoT error
Compares E for W with value from Figure 5
E = 6.8 × 1010 (N m−2)
OR
Accept answers that round in the range
compares k, l, A or d for Figure 5 with value for W
6.7 × 1010 to 6.9 × 1010 for E from graph.
With answer yes 4
Total 9
How to answer it
Properties of Materials & Stress–Strain Analysis
This multi-step question evaluates core knowledge from the Mechanics and Materials topic, specifically:
- Deriving and justifying SI units from definitions (stress, strain, Young modulus).
- Volume, density, and unit conversion calculations applied to financial costing ( m³ to kg to g ).
- Distinguishing between intensive material properties (density, Young modulus) and extensive geometric properties (stiffness, length, diameter).
- Accurate graphical skills: drawing a line of best fit, determining Young modulus from a stress–strain gradient, and relating stiffness k to Young modulus E using E = kL / A .
Identical SI Units for Young Modulus and Breaking Stress
1 Mark • Assessment Objective: AO1
✅ Model Answer
Young modulus is defined as tensile stress divided by tensile strain ( E = σ / ε ).
Because tensile strain is a ratio of two lengths ( ΔL / L ), it has no units (dimensionless). Therefore, the Young modulus must have the same SI unit as stress ( N m⁻² or Pa ).
❌ Common Errors
- Stating merely that "they are both measured in Pascals" without explaining why from the formula.
- Forgetting to explicitly mention that strain is dimensionless / has no units.
Calculating the Cost of Wire X
3 Marks • Assessment Objective: AO3
📐 Step-by-Step Calculation
Given for Wire X:
- Length L = 2.17 m
- Diameter d = 1.4 mm = 1.4 × 10⁻³ m → Radius r = 0.70 × 10⁻³ m
- Density ρ = 1.93 × 10⁴ kg m⁻³
- Cost rate = £75 per gram
Step 1: Calculate volume (V)
A = πr² = π × (0.70 × 10⁻³)² = 1.539 × 10⁻⁶ m²
V = A × L = 1.539 × 10⁻⁶ × 2.17 = 3.340 × 10⁻⁶ m³
Step 2: Calculate mass in grams (m)
m = ρ × V = (1.93 × 10⁴ kg m⁻³) × (3.340 × 10⁻⁶ m³) = 0.06447 kg
Convert to grams: m = 0.06447 × 1000 = 64.47 g (approx. 64 g – 65 g)
Step 3: Calculate total cost
Cost = 64.47 g × £75 = £4835.25
Final Answer: £4800 to £4900 (e.g. £4840)
❌ Common Traps
- Diameter vs Radius: Using d instead of r in πr² will overestimate volume and cost by a factor of 4.
- Unit Prefixes: Forgetting to convert mm to m ( ×10⁻³ ) or kg to g ( ×1000 ).
- Using values from the wrong row of the table.
• Mark 1: Correct volume of wire X ( ~3.3 × 10⁻⁶ m³ ).
• Mark 2: Correct application of m = ρV to find mass in grams ( 64 g or 65 g ).
• Mark 3: Final answer in range £4800 to £4900.
Identifying Wires Made from the Same Material
1 Mark • Assessment Objective: AO1
✅ Model Answer
Wires W and Z are made from the same material.
Reason: They have the exact same density ( 2.71 × 10³ kg m⁻³ ), and density is a bulk property of the material itself (intensive property), independent of wire dimensions.
💡 Material vs Object Properties
- Material properties (Intensive): Density, Young Modulus, Resistivity, Breaking Stress. These remain identical regardless of wire length or thickness.
- Object properties (Extensive): Stiffness ( k ), Mass, Resistance. These depend directly on wire length and cross-sectional area.
Graph Analysis and Wire Verification
4 Marks • Assessment Objective: AO3
🧠 Exam Technique: Drawing the Best-Fit Line
- Use a clear ruler long enough to cover all points.
- The line must be a single, straight line passing through the origin (0, 0) .
- Ensure an even scatter of points above and below the line along its entire length.
📐 Step-by-Step Deduction
Step 1: Young modulus (E) from Figure 5
Young modulus is the gradient of the stress–strain graph:
E = Δσ / Δε
Pick a large triangle (more than halfway along the line), e.g., at strain = 39 × 10⁻⁵ , stress ≈ 26.5 MPa = 26.5 × 10⁶ Pa :
E = (26.5 × 10⁶ Pa) / (39 × 10⁻⁵) = 6.79 × 10¹⁰ Pa (Acceptable: 6.7 × 10¹⁰ to 6.9 × 10¹⁰ Pa )
Step 2: Theoretical Young modulus for Wire W
From Table 1 for Wire W:
- L = 3.10 m
- d = 1.7 mm → r = 0.85 × 10⁻³ m
- A = π × (0.85 × 10⁻³)² = 2.27 × 10⁻⁶ m²
- Stiffness k = 4.90 × 10⁴ N m⁻¹
Using E = (F / A) / (ΔL / L) = (F / ΔL) × (L / A) = k × (L / A) :
E_W = (4.90 × 10⁴ × 3.10) / (2.27 × 10⁻⁶) = 6.69 × 10¹⁰ Pa
Step 3: Comparison & Conclusion
E_graph ≈ 6.8 × 10¹⁰ Pa matches E_W ≈ 6.7 × 10¹⁰ Pa within experimental uncertainty.
Conclusion: Yes, wire W could have been used to produce Figure 5.
❌ Common Pitfalls on 04.4
- Axis Multipliers: Missing the M in MPa (× 10⁶) on the y-axis, or missing 10⁻⁵ on the x-axis.
- Small Gradient Triangles: Using points close to the origin. The mark scheme specifies that readings must be taken from a point more than halfway along the line.
- Reading data points instead of the line: Coordinates for calculating gradient must come directly from the drawn line of best fit, not raw plot points.
• Mark 1: Draws a sensible straight line of best fit with balanced scatter.
• Mark 2: Uses points/difference more than halfway along the line to find gradient.
• Mark 3: Correct attempt to calculate property ( E , k , L , or d ) for comparison (e.g. finding Area = 2.27 × 10⁻⁶ m² and applying E = kL/A ).
• Mark 4: Valid numerical comparison between graph value and table value, concluding with Yes.
Topics
Physics · Practical skills · 3.4 Mechanics and materials · Data analysis
Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.