AQA AS Level Physics Paper 2, November 2021: Question 2

8 marks · Hard difficulty · Practical Techniques & Data Analysis

Calculate the extension, tension, Young modulus from a graph, and fundamental base units of a constant in a wire experiment under a suspended weight.

Practise this question

Question

A multi-part physics exam question about determining the Young modulus of a metal wire. Figure 6 shows a horizontal wire of length 3.00 m clamped at points A and B, with midpoint C. Figure 7 shows C displaced vertically downwards by distance y when weight W is suspended. Questions 02.1 through 02.4 ask to show extension, calculate tension, determine Young modulus from a graph of W/y against y squared, and deduce base units for constant k.
Question text

02 A student does an experiment to determine the Young modulus of a metal.

Figure 6 shows a wire made from the metal clamped at points A and B so that the

wire is horizontal. The horizontal distance between A and B = 3.00 m.

C is the mid-point on the wire between A and B.

Figure 6

A mass of weight W is suspended at C to extend the wire. Figure 7 shows that C

moves vertically downwards by a distance y.

Figure 7

02.1 When W is 1.0 N, y is 6.34 cm.

Show that the wire extends by approximately 3 mm.

[1 mark]

02.2 Calculate the tension in the wire when W is 1.0 N.

[2 marks]

tension = N

It can be shown that

EAy2

W = + k

y x3

where E = Young modulus of the metal

A = 1.11 × 10−7 m2

x = 1.50 m

k = a constant.

A student measures y for different values of W and plots the graph shown in Figure 8.

Figure 8

02.3 Determine E using Figure 8.

[4 marks]

E = Pa

02.4 Deduce the fundamental base units for k.

[1 mark]

fundamental base units for k =

Mark scheme

Show the mark scheme Mark scheme for the Young modulus wire experiment question. Gives acceptable answers including calculation steps for extension, trigonometric tension formulas, gradient calculations from the graph for Young modulus, and dimensional analysis for k resulting in kg s^-2.

Question Answers Additional Comments/Guidance Mark AO

02.1 correctly deduces extension is 2.6 or 2.7 mm ✓ Should see AC2 = 1.502 + (6.34 10−2)2; 1 AO2-1h

(new) AC = 1.50134;

Extension of AC = (1.50134 − 1.50 =) 0.00134 m

or 1.34 mm; and then doubles this

Final value must be to at least 2 sf

02.2 evidence of correct working: ✓ For 1✓ acceptable diagrams are shown below 2 AO2-1h

6.34 10−2

sin = or = 2.42° seen

their new AC T T

OR W

W = 2T sin seen

OR T

suitable vector diagram with labelled W

T

1.0 Correct final answer of 11.8 N or 12 N earns both

tension correctly calculated from ✓

2×their sin𝜃 marks

– SICS – – JUNE 2021

02.3 ruled best-fit line between first and sixth points; for 1✓ withhold mark if line is thick, faint or 4 2 AO2-

nd discontinuous 1h

line must pass above 2 point

for 2✓ condone read off errors of 1 division 2 AO3-

and 1b

for ✓ note that 1.503 = 3.375 so allow sub of 3.38

must pass below 4th point ✓

for ✓ reject 2 sf 1.2 1011

gradient calculated from (W y) with y2 0.004 ✓

2 (gradient ~ 3850)

y 2

their gradient 1.503

evidence of using E = −7 3✓

1.11 10

E in range 1.10 1011 to 1.24 1011 (Pa) ✓

02.4 kg s−2 ✓ no credit for N m−1 1 AO1-1b

correct answer only

Total 8

How to answer it

Determining the Young Modulus Using a Stretched Wire

What this question tests

This multi-step synoptic problem assesses your mastery of mechanics and material properties. You will apply Pythagoras' theorem to calculate wire extensions under perpendicular loading, resolve forces using trigonometry to find wire tension, interpret linear graph equations to extract fundamental constants (Young modulus, E ), and perform dimensional analysis to determine fundamental base units.

Question 02.1 [1 mark]

Calculating Wire Extension from Vertical Sag

✅ Correct Answer

Extension = 2.6 mm or 2.7 mm (or 0.0026 m / 0.0027 m )

💡 Key Knowledge

  • Original half-length of the wire x = 1.50 m .
  • Vertical displacement y = 6.34 cm = 0.0634 m .
  • Use Pythagoras theorem to find the new stretched half-length: AC = sqrt(1.50² + 0.0634²) .

📐 Step-by-Step Calculation

  1. Find stretched half-length: AC = sqrt(2.25 + 0.0040196) = 1.501339 m
  2. Find extension for one half: 1.501339 - 1.50 = 0.001339 m
  3. Double for total extension across both halves: 2 × 0.001339 m = 0.00268 m ≈ 2.7 mm

❌ Common Errors

Forgetting to double the extension calculated for the single half-span ( AC ), or failing to convert centimetres into metres before applying Pythagoras' theorem.

Mark awarded for correct deduction of extension to at least 2 significant figures.
Question 02.2 [2 marks]

Calculating Tension in the Stretched Wire

✅ Correct Answer

Tension = 11.8 N or 12 N

💡 Key Knowledge

Resolution of forces: The downward weight W is supported by the vertical components of the tension T acting in both sections of the wire attached to point C: W = 2T sin(θ) .

🧠 Exam Technique & Steps

  1. Determine angle θ to the horizontal using inverse tangent or sine: sin(θ) = 0.0634 / 1.50134 giving θ = 2.42° .
  2. Rearrange the equilibrium equation for tension: T = W / (2 sin(θ)) .
  3. Substitute values: T = 1.0 / (2 × sin(2.42°)) = 11.8 N .

❌ Common Errors

Omitting the factor of 2 in 2T sin(θ) , assuming the tension equals the suspended weight W , or using the wrong trigonometric ratio (using cosine instead of sine relative to the horizontal).

Mark breakdown: 1 mark for evidence of correct working/trigonometry ( sin θ or vector triangle), 1 mark for the correct final evaluated tension.
Question 02.3 [4 marks]

Determining the Young Modulus ( E ) from a Graph

✅ Correct Answer

E in the range 1.10 × 10¹¹ Pa to 1.24 × 10¹¹ Pa

💡 Key Knowledge

By comparing the given equation W/y = (E A y²) / x³ + k to the straight-line equation y = mx + c :

  • y-axis variable: W / y
  • x-axis variable: y²
  • Gradient m = (E A) / x³

🧠 Exam Technique & Calculation Steps

  1. Line of Best Fit: Draw a clean, ruled line passing appropriately between the first and sixth points, respecting scatter.
  2. Gradient Calculation: Pick a large triangle on your line. Gradient ≈ 3850 N m⁻³ .
  3. Rearrange for E: E = (Gradient × x³) / A
  4. Substitute: E = (3850 × 1.50³) / (1.11 × 10⁻⁷) = 1.17 × 10¹¹ Pa

❌ Common Errors

Choosing a gradient calculation triangle that is too small, failing to cube the length x properly ( 1.50³ = 3.375 ), or rounding to 2 significant figures (reject 1.2 × 10¹¹ without proper working).

Mark breakdown: 1 mark for correct line of best fit, 1 mark for correct gradient calculation, 1 mark for substituting into the rearranged formula using x³ , 1 mark for final answer within acceptable range with correct units.
Question 02.4 [1 mark]

Deducing Fundamental Base Units for Constant k

✅ Correct Answer

Fundamental base units: kg s⁻² (or kg / s² )

💡 Key Knowledge

Principle of Homogeneity: Every term in an additive physical equation must share the exact same base units.

🧠 Exam Technique & Derivation

  1. The equation terms are W / y = ... + k . Therefore, k must share the exact same units as W / y .
  2. Units of Weight ( W ) = Newtons ( N ) = kg m s⁻² .
  3. Units of distance ( y ) = metres ( m ).
  4. Divide: (kg m s⁻²) / m = kg s⁻² .

❌ Common Errors

Giving derived units like N m⁻¹ instead of breaking them down fully into SI fundamental base units ( kg , m , s ).

Mark awarded for kg s⁻² only (no credit for N m⁻¹ ).

Topics

Physics · Practical skills · Required Practicals · 3.1 Measurements and their errors · 3.4 Mechanics and materials · Data analysis · AS practicals (1–6)

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.