AQA A-Level Physics Paper 1, June 2023: Question 5
10 marks · Hard difficulty · Extended Answer
Determine the Young modulus of a metal from a stress-strain graph, explain its brittle behavior, calculate the extension of identical support wires, and deduce the position of a lamp required to keep a non-identical beam level.
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Question text
05 Figure 11 shows the stress–strain graph for a metal in tension up to the point at which
it fractures.
Figure 11
05.1 Determine, using Figure 11, the Young modulus of the metal.
[1 mark]
Young modulus = Pa
05.2 Explain how the graph shows that this metal is brittle.
[1 mark]
Figure 12 shows a uniform rigid lighting beam AB suspended from a fixed horizontal
support by two identical vertical steel wires. A lamp is attached to the midpoint of AB.
Figure 12
The unloaded length of each steel wire was 1.20 m before it was attached to AB.
AB is horizontal.
mass of AB = 4.4 kg
mass of lamp = 16.0 kg
distance between wires = 2.00 m
diameter of each wire = 0.800 mm
Young modulus of steel = 2.10 × 1011 Pa
05.3 Calculate the extension of each wire.
[3 marks]
19 extension = m
05.4 The right-hand steel wire is removed and replaced with an aluminium wire of
diameter 1.60 mm. The unloaded length of the aluminium wire is the same as that of
the original steel wire.
*18* When the lamp is at the midpoint of AB, one of the wires extends more than the other
so that AB is not horizontal. To make AB horizontal the lamp has to be moved to a
distance x from A. Figure 13 shows the new arrangement.
Figure 13
The Young modulus of aluminium is 7.00 × 1010 Pa.
Deduce distance x.
[5 marks]
x = m
Mark scheme
Show the mark scheme
Question Answers Additional comments/Guidelines Mark AO
05.1 Evidence of appropriate use of Figure 11 e.g. Some evidence that Figure 11 is used: 1 AO1
105 × 106 ÷7.5 × 10−4 calculation based on a point on line between
75 MPa and 125 MPa
OR calculation from point on straight line
extended
OR
Use of triangle from more than half of the
linear section.
leading to an answer in the range 1.38 to 1.42 × 1011 Pa
Allow 2 sf answer 1.4 × 1011 (Pa).
05.2 Idea that wire undergoes only (very) small (increase in) Reject idea that there is no increase in strain. 1 AO1 × 1
strain beyond the linear section before fracture Condone ‘extension’ or ‘(plastic) deformation’
for ‘strain’.
Condone ‘shortly after’ for ‘beyond’
Accept: does not show 'necking' before
fracture
Accept: fracture occurs very near the limit of
proportionality (condone ‘elastic limit’).
Accept references to a particular value of
strain e.g. 9 x 10-4 to 12.7 x 10-4
05.3 Evidence of determination of total load or load on one wire Total load = (4.4 + 16.0) × 9.8(1) = 200(.1) N 3 AO1× 1
Allow ‘g’ for 9.8(1) AO2× 2
(halves load)
(their F)× L Expect to see F =100 N and
Use of E = −7 2
A× ∆L A = 5.03 × 10 m . Condone use of d in
calculation of cross-sectional area A in MP2.
Or separate calculations using σ = F÷A,
E = σ÷strain, strain = ΔL÷L
Condone POT error in MP2.
ΔL = 1.1(4) × 10−3 (m)
05.4 Evidence of extension/strain in each wire is the same 1 ΔL = {FL÷AE} steel = {FL÷AE} aluminium 5 AO3× 5
Substitutes data leading to Fa = 1.33 Fs 2 {F÷d2E} steel = {F÷d2E} aluminium
Fs Fa
2 = 2
0.8 × 210 1.6 × 70
Fa = 1.33 Fs OR Fs = 0.752 Fa 2
1.33 Fs + Fs = 200 N
Calculates Fs or Fa 3 F = 86 N F = 114 N
s , a 3
Attempt to take moments about A or B or
Evidence of an attempt at a moment equation 4
other suitable point, expect to see
16.0gx = 228 − 4.4g 4
Note that an answer of 1.14 m comes from
not taking into account the weight of the
beam
Award max 4 for this approach.
ECF for MP2 and MP3 in MP4
Distance = 1.18 m 5
Total 10
How to answer it
A-Level Physics Study Guide: Materials & Young Modulus
What this question tests
This multi-part exam question evaluates your mastery of bulk material properties, specifically the interpretation of stress-strain graphs, the definition and calculation of the Young modulus ( E = (F × L) / (A × ΔL) ), the distinction between ductile and brittle behavior, and the application of equilibrium conditions (moments and forces) to complex structural systems containing wires of different materials.
Determining the Young Modulus from a Stress-Strain Graph
✅ Correct Answer
Young modulus = 1.38 × 10¹¹ Pa to 1.42 × 10¹¹ Pa (Accept 2 sf: 1.4 × 10¹¹ Pa )
💡 Key Knowledge
- The Young modulus is defined as the gradient of the linear region of a stress-strain graph.
- Always choose a large triangle spanning more than half of the straight section to minimize percentage reading errors.
🧠 Exam Technique
- Check axis multipliers carefully: stress is in MPa ( × 10⁶ Pa ) and strain is multiplied by 10⁻⁴ .
- Show working by writing out coordinates selected from the straight line portion.
❌ Common Errors
- Forgetting to factor in the powers of ten specified on the graph axes.
- Attempting to calculate the gradient using points located in the curved (plastic deformation) region.
📐 Step-by-Step Calculation
- Select coordinates: Take a clear point on the linear line, e.g., stress = 105 × 10⁶ Pa at strain = 7.5 × 10⁻⁴ .
- Apply gradient formula: E = Δstress / Δstrain
- Compute value: 105 × 10⁶ / (7.5 × 10⁻⁴) = 1.40 × 10¹¹ Pa .
Explaining Brittle Behavior from Graphs
✅ Correct Answer
The wire undergoes only a very small (or negligible) increase in strain beyond the linear section before it fractures.
💡 Key Knowledge
A brittle material exhibits little or no plastic deformation before breaking. On a stress-strain curve, the region between the limit of proportionality and fracture is extremely short.
🧠 Exam Technique
Be precise with technical vocabulary. Use terms like "strain", "plastic deformation", or "fracture" rather than vague descriptions like "it snaps suddenly".
❌ Common Errors
Claiming there is "no increase in strain whatsoever" or confusing brittleness with stiffness. Examiners reject statements saying there is no increase in strain at all because microscopic plastic deformation always occurs.
Calculating Wire Extension in a Symmetric System
✅ Correct Answer
Extension = 1.14 × 10⁻³ m (or 1.14 mm )
💡 Key Knowledge
- Total downward force is the combined weight of beam AB and the lamp: W_total = (m_beam + m_lamp) × g .
- Since the lamp is at the midpoint, the load is shared equally between the two identical vertical wires.
🧠 Exam Technique
- Calculate total mass: 4.4 kg + 16.0 kg = 20.4 kg .
- Find total weight using g = 9.81 m s⁻² (or 9.8 m s⁻² ).
- Halve the force to find the tension in a single wire.
- Rearrange Young modulus equation for extension ΔL .
❌ Common Errors
- Omitting the weight of the lighting beam AB ( 4.4 kg ), calculating force using only the lamp's mass.
- Failing to halve the total load across the two supporting wires.
- Radius-diameter mix-ups when calculating cross-sectional area A = πd²/4 or πr² .
📐 Step-by-Step Calculation
- Total Mass: m_total = 4.4 + 16.0 = 20.4 kg
- Total Load / Tension per wire: F_total = 20.4 × 9.81 = 200.0 N . Force on one wire F = 200.0 / 2 = 100.0 N .
- Cross-sectional Area: A = π × (0.800 × 10⁻³ / 2)² = 5.03 × 10⁻⁷ m²
- Rearrange & Solve: ΔL = (F × L) / (A × E) = (100.0 × 1.20) / (5.03 × 10⁻⁷ × 2.10 × 10¹¹) = 1.14 × 10⁻³ m
Deducing Lamp Position for Asymmetric Wires
✅ Correct Answer
Distance x = 1.18 m from A
💡 Key Knowledge
- For beam AB to remain horizontal, both supporting wires must undergo the exact same extension ( ΔL_steel = ΔL_aluminium ).
- Cross-sectional area formula involves diameter squared: A = πd²/4 , meaning area scales with d² .
- Rotational equilibrium requires taking moments about a pivot (such as point A) where the sum of clockwise moments equals the sum of counter-clockwise moments.
🧠 Exam Technique
- Equate the expressions for extension: (F_s × L) / (A_s × E_s) = (F_a × L) / (A_a × E_a) . Simplify by canceling out length L .
- Express the relationship between forces F_s and F_a , and substitute into F_s + F_a = 200 N to find individual tensions.
- Set up a complete moment equation about end A, incorporating the weight of both the uniform beam (acting at its midpoint, 1.00 m from A) and the lamp (acting at distance x ).
❌ Common Errors
- Forgetting to include the weight of beam AB in the moments calculation (leading to an incorrect value of 1.14 m , which caps the mark scheme at max 4 out of 5).
- Incorrectly squaring diameters or mixing up steel and aluminium parameters.
📐 Step-by-Step Calculation
- Equate extensions: Length L and π/4 cancel out: F_s / (d_s² × E_s) = F_a / (d_a² × E_a)
- Substitute values: F_s / (0.800² × 210 × 10⁹) = F_a / (1.60² × 70.0 × 10⁹)
- Find force ratio: Simplifying yields F_a = 1.33 F_s (or F_s = 0.752 F_a ).
- Calculate individual tensions: Since F_s + F_a = 200 N , solving gives F_s = 86 N and F_a = 114 N .
- Take moments about A:
Sum of clockwise moments = Sum of counter-clockwise moments about A
(Weight of lamp × x) + (Weight of beam × 1.00 m) = (Force in right wire B × 2.00 m)
(16.0 × 9.81 × x) + (4.4 × 9.81 × 1.00) = (114 × 2.00)
Solving for x gives x = 1.18 m .
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 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.