Materials Problem Solving

This lesson brings together all the concepts from the materials topic into multi-step problems, experimental design questions, and material selection challenges. These are the types of questions that appear frequently in Edexcel A-level Physics papers and require you to combine several equations and concepts.

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Key Equations Summary

Before tackling problems, here is a reference of all the equations used in this topic:

Equation	Variables	When to use
ρ = m/V	Density, mass, volume	Density calculations
p = ρgh	Pressure, density, depth	Fluid pressure at depth
F = kx	Force, spring constant, extension	Hooke’s law (within limit of proportionality)
E = ½kx² = ½Fx	Elastic PE, spring constant, extension	Energy stored in a spring
σ = F/A	Stress, force, area	Stress in a material
ε = ΔL/L	Strain, extension, original length	Strain in a material
E = σ/ε = FL/(AΔL)	Young modulus, stress, strain	Stiffness of a material
F_upthrust = ρVg	Upthrust, fluid density, volume	Archimedes’ principle
A = πd²/4	Area, diameter	Cross-sectional area of a wire

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Multi-Step Calculations

Combining Density and Pressure

Problem 1: A cylindrical storage tank is filled with oil of density 850 kg m⁻³ to a depth of 6.0 m. The tank has a flat circular base of diameter 3.0 m. Calculate (a) the pressure at the base due to the oil, and (b) the force exerted on the base by the oil.

(a) Pressure = ρgh = 850 × 9.81 × 6.0 = 50 031 Pa ≈ 50.0 kPa

(b) Area of base = πr² = π × (1.5)² = 7.069 m² Force = pressure × area = 50 031 × 7.069 = 353 700 N ≈ 354 kN

Note: This is the force due to the oil only. The total force on the base also includes the force from atmospheric pressure above the oil.

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Combining Hooke’s Law and Energy

Problem 2: A spring of spring constant 400 N m⁻¹ is compressed by 0.12 m and used to launch a 0.050 kg ball vertically upward. Assuming all the elastic potential energy is converted to gravitational potential energy, calculate the maximum height reached by the ball.

Energy stored = ½kx² = ½ × 400 × (0.12)² = 2.88 J

At maximum height: mgh = 2.88 h = 2.88 / (0.050 × 9.81) = 2.88 / 0.4905 = 5.87 m ≈ 5.9 m

In reality, the ball will not reach this height because some energy is dissipated as heat (air resistance, friction) and sound.

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Combining Stress, Strain, and Young Modulus

Problem 3: A steel wire of length 4.0 m and diameter 1.0 mm supports a lift cage of mass 200 kg. The Young modulus of steel is 200 GPa.

(a) Calculate the stress in the wire. A = π(0.50 × 10⁻³)² = 7.854 × 10⁻⁷ m² F = 200 × 9.81 = 1962 N σ = F/A = 1962 / 7.854 × 10⁻⁷ = 2.50 × 10⁹ Pa = 2.50 GPa

(b) Calculate the strain. ε = σ/E = 2.50 × 10⁹ / 200 × 10⁹ = 0.0125

(c) Calculate the extension. ΔL = εL = 0.0125 × 4.0 = 0.050 m = 50 mm

(d) Calculate the elastic potential energy stored. E_stored = ½Fx = ½ × 1962 × 0.050 = 49.1 J

Reality check: The stress here (2.50 GPa) exceeds the UTS of most steels (typically 400–550 MPa). In practice, the wire would break. Real lift systems use much thicker cables. Always check whether your calculated stress is realistic.

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Density and Upthrust Combined

Problem 4: A hollow steel sphere has an outer radius of 0.15 m and a wall thickness of 5.0 mm. The density of steel is 7800 kg m⁻³. Will it float in water?

Step 1 — Volume of steel: Outer volume = (4/3)π(0.15)³ = 1.414 × 10⁻² m³ Inner radius = 0.15 − 0.005 = 0.145 m Inner volume = (4/3)π(0.145)³ = 1.278 × 10⁻² m³ Volume of steel = 1.414 × 10⁻² − 1.278 × 10⁻² = 1.36 × 10⁻³ m³

Step 2 — Mass of sphere: m = ρV = 7800 × 1.36 × 10⁻³ = 10.6 kg

Step 3 — Average density: ρ_avg = m / V_outer = 10.6 / 1.414 × 10⁻² = 750 kg m⁻³

Since 750 < 1000 (water), the sphere floats. Fraction submerged = 750/1000 = 75%.

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Experimental Design Questions

Designing an Experiment to Compare Young Modulus

Problem 5: Describe an experiment to determine which of two metal wires (copper and steel) has the greater Young modulus.

Answer structure:

1. Setup: Clamp identical lengths (2–3 m) of each wire to the same support. Attach a pulley and mass hanger to the free end of each.
2. Measure original length (L): Use a metre rule. Record to nearest mm.
3. Measure diameter (d): Use a micrometer screw gauge. Take at least 6 readings at different positions and orientations. Check for zero error.
4. Apply loads: Add equal masses (e.g., 100 g increments) to each wire. After each addition, wait for equilibrium and record the extension using a marker and ruler (or travelling microscope).
5. Plot graph: Plot force against extension for each wire. Draw the line of best fit for the linear region.
6. Calculate E: E = (gradient × L) / A for each wire.
7. Compare: The wire with the steeper gradient (for the same dimensions) or the higher calculated E is stiffer.
8. Safety: Wear goggles. Place a cushion under the masses. Do not exceed the elastic limit.

Key points for full marks:

• Identify the independent variable (force/mass), dependent variable (extension), and control variables (wire length, diameter, temperature).
• Explain how to measure each quantity and state the resolution of each instrument.
• Explain how to process the data (graphical method preferred).
• Discuss at least two sources of uncertainty and how to minimise them.

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Uncertainty in Multi-Step Problems

Problem 6: A student measures the following for a copper wire:

• Length: L = 2.50 ± 0.001 m
• Diameter: d = 0.48 ± 0.02 mm (mean of 6 readings)
• Extension at 30 N: ΔL = 1.5 ± 0.1 mm

Calculate the Young modulus and its percentage uncertainty.

Young modulus: A = π(0.24 × 10⁻³)² = 1.810 × 10⁻⁷ m² E = FL/(AΔL) = (30 × 2.50) / (1.810 × 10⁻⁷ × 1.5 × 10⁻³) = 75 / (2.715 × 10⁻¹⁰) = 2.76 × 10¹¹ Pa ≈ 276 GPa

Percentage uncertainties:

Quantity	Value	Uncertainty	% uncertainty	Notes
Force F	30 N	±0.01 N	0.03%	Negligible
Length L	2.50 m	±0.001 m	0.04%	Negligible
Diameter d	0.48 mm	±0.02 mm	4.17%	Doubled for area = 8.33%
Extension ΔL	1.5 mm	±0.1 mm	6.67%	Largest single contribution

Total % uncertainty in E = 0.03 + 0.04 + 8.33 + 6.67 = 15.1%

E = 276 ± 42 GPa (the accepted value of 130 GPa is well outside this range, indicating significant systematic error — likely in the diameter measurement).

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Material Selection Questions

Choosing a Material for a Specific Application

Problem 7: An aerospace engineer needs to select a material for a landing gear component that must:

• Withstand very high stresses without fracturing
• Absorb energy from the impact of landing (without fracturing)
• Be as light as possible

Material	UTS (MPa)	E (GPa)	Density (kg m⁻³)	Toughness	Behaviour
Mild steel	500	200	7800	High	Ductile, tough
Al alloy	450	70	2700	Moderate	Ductile
Ti alloy	900	110	4500	Very high	Ductile, very tough
CFRP	600	150	1600	Low	Brittle (fails suddenly)

Analysis:

• CFRP is eliminated despite its excellent strength-to-weight ratio because it is brittle — it shatters on impact rather than absorbing energy.
• Mild steel is very tough and ductile but too heavy for aerospace use.
• Aluminium alloy is light and ductile but has lower UTS and toughness compared to titanium.
• Titanium alloy is the best choice: highest UTS, ductile and very tough, and reasonably light.

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Exam-Style Multi-Part Questions

Problem 8: A bungee cord has an unstretched length of 15.0 m and behaves like a spring with k = 50 N m⁻¹ for extensions up to 20 m. A person of mass 70 kg jumps from a bridge.

(a) How far does the person fall before the cord starts to stretch? The person falls 15.0 m (the natural length of the cord) before it begins to extend.

(b) When the cord has extended by 12.0 m, what is the net force on the person? Weight = mg = 70 × 9.81 = 686.7 N (downward) Tension = kx = 50 × 12.0 = 600 N (upward) Net force = 686.7 − 600 = 86.7 N downward

(c) What is the extension at equilibrium? At equilibrium: kx = mg → x = mg/k = 686.7/50 = 13.7 m Total distance = 15.0 + 13.7 = 28.7 m

(d) Using energy conservation, find the maximum extension (lowest point). At the lowest point: mgh = ½kx² where h = 15.0 + x 70 × 9.81 × (15.0 + x) = ½ × 50 × x² 686.7(15.0 + x) = 25x² 25x² − 686.7x − 10 300.5 = 0

Using the quadratic formula: x = (686.7 + √(686.7² + 4 × 25 × 10 300.5)) / (2 × 25) x = (686.7 + √(471 597 + 1 030 050)) / 50 x = (686.7 + 1225.4) / 50 = 38.2 m

Total distance fallen = 15.0 + 38.2 = 53.2 m

Exam insight: Part (d) is the hardest because students must recognise that the gravitational PE lost equals the elastic PE gained over the full distance (15.0 + x), not just the extension x. Setting up the energy equation correctly is worth most of the marks.

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Problem-Solving Strategy

graph TD
    A["Read the question\nIdentify known quantities"] --> B["Convert ALL units to SI\nmm to m, g to kg, etc."]
    B --> C["Identify which equation(s)\nare needed"]
    C --> D{"Multiple steps?"}
    D -- Yes --> E["Solve step by step\nLabel intermediate answers"]
    D -- No --> F["Substitute and solve"]
    E --> G["Check: Are units correct?\nIs the magnitude sensible?"]
    F --> G
    G --> H["Round to appropriate s.f.\n(usually 2 or 3)"]

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Summary Checklist for Materials Problems

Before submitting your answer, check:

• Have you converted all units to SI (mm to m, diameter to radius, GPa to Pa)?
• Have you used the correct formula (ρ = m/V, p = ρgh, F = kx, σ = F/A, ε = ΔL/L, E = σ/ε)?
• Have you calculated area correctly using π(d/2)², not πd²?
• Have you given your answer to an appropriate number of significant figures (usually 2 or 3)?
• Have you included the correct unit?
• For graph-based questions, have you read values from the correct axis and used the gradient or area as appropriate?
• Have you done a reality check — does your answer make physical sense?

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Common Exam Mistakes in Problem Solving

1. Not converting units: This is the single most common source of lost marks. Always convert mm to m and g to kg before substituting.
2. Using πd² instead of πd²/4 for area: Remember that area uses the radius, not the diameter.
3. Calculating work as ½k(Δx)² instead of ½k(x₂² − x₁²): When extending from x₁ to x₂, the work done is NOT ½k(Δx)².
4. Forgetting to check whether the elastic limit has been exceeded: If your calculated stress exceeds the yield stress or UTS, state that the material would have failed.
5. Rounding intermediate values: Keep full precision in intermediate calculations and only round the final answer.

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A-Level Deep Dive: Materials Problem Solving

Spec mapping

Edexcel 9PH0 Topic 4 — Materials, synoptic problem-solving is the integrative end-point of the topic: students combine density, fluid pressure and upthrust, Hooke's law, stress, strain, Young modulus, elastic and plastic deformation, and material-property selection within a single multi-step problem (refer to the official Pearson Edexcel specification document for exact wording). Topic 4 is examined principally on Paper 1, but materials-style synoptic items reappear on Paper 3, which is designed around AO3 problem-solving across topics. The Edexcel formula booklet provides $E_{el} = \tfrac{1}{2}F\Delta L$Eel​=21​FΔL and $p = h\rho g$p=hρg, but does not state $\rho = m/V$ρ=m/V, $\sigma = F/A$σ=F/A, $\varepsilon = \Delta L/L$ε=ΔL/L or $E = \sigma/\varepsilon$E=σ/ε — these must be memorised. Graders reward candidates who write each definition explicitly before substituting numbers.

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Worked example with full mark scheme

Question (12 marks):

A new pedestrian footbridge is to be supported by two parallel steel cables of length $L = 24.0$L=24.0 m. Each cable must safely carry a maximum static load of $F = 8.5 \times 10^4$F=8.5×104 N without exceeding 60% of the steel's yield stress. The steel has yield stress $\sigma_y = 2.50 \times 10^8$σy​=2.50×108 Pa, density $\rho_s = 7850$ρs​=7850 kg m⁻³, and Young modulus $E = 2.10 \times 10^{11}$E=2.10×1011 Pa.

(a) Determine the minimum cable diameter consistent with the safety criterion. (4)

(b) Calculate the cable's extension under the maximum load, using the diameter from (a). (3)

(c) Calculate the elastic strain energy stored in one fully loaded cable. (2)

(d) The cable hangs vertically below the bridge deck during installation. Show whether its self-weight contributes significantly (>5%) to the working stress at the upper attachment point. (3)

Solution with mark scheme:

(a) Step 1 — set the working stress.

Working stress $\sigma_w = 0.60 \times \sigma_y = 0.60 \times 2.50 \times 10^8 = 1.50 \times 10^8$σw​=0.60×σy​=0.60×2.50×108=1.50×108 Pa.

M1 — explicit application of the 60% safety factor; candidates who skip this step and design to the full yield stress lose this mark and propagate the error.

Step 2 — relate stress to cross-sectional area.

$\sigma = F/A \implies A = F/\sigma_w = 8.5 \times 10^4 / 1.50 \times 10^8 = 5.67 \times 10^{-4}$σ=F/A⟹A=F/σw​=8.5×104/1.50×108=5.67×10−4 m².

M1 — correct rearrangement and substitution.

Step 3 — convert area to diameter.

$A = \pi d^2/4 \implies d = \sqrt{4A/\pi} = \sqrt{4 \times 5.67 \times 10^{-4}/\pi} = 0.0269$A=πd2/4⟹d=4A/π​=4×5.67×10−4/π​=0.0269 m $= 26.9$=26.9 mm.

M1 — correct use of $A = \pi d^2/4$A=πd2/4, not $\pi d^2$πd2 or $\pi r^2$πr2 with $r = d$r=d.

A1 — final $d \geq 26.9$d≥26.9 mm (or 27 mm rounded up — accept the next standard manufactured size where stated).

(b) Step 1 — strain.

$\varepsilon = \sigma_w/E = 1.50 \times 10^8 / 2.10 \times 10^{11} = 7.14 \times 10^{-4}$ε=σw​/E=1.50×108/2.10×1011=7.14×10−4.

M1 — explicit use of $\varepsilon = \sigma/E$ε=σ/E; candidates who try $\Delta L = FL/(AE)$ΔL=FL/(AE) in one line and slip on a power of ten get partial credit only.

Step 2 — extension.

$\Delta L = \varepsilon L = 7.14 \times 10^{-4} \times 24.0 = 0.0171$ΔL=εL=7.14×10−4×24.0=0.0171 m $= 17.1$=17.1 mm.

M1 A1 — correct substitution and final value to 3 sf.

(c) Elastic strain energy from the formula booklet expression:

$E_{el} = \tfrac{1}{2}F\Delta L = 0.5 \times 8.5 \times 10^4 \times 0.0171 = 727$Eel​=21​FΔL=0.5×8.5×104×0.0171=727 J.

M1 A1 — direct substitution; both marks awarded for value in range 720–730 J.

(d) Step 1 — cable mass and weight.

Volume $V = AL = 5.67 \times 10^{-4} \times 24.0 = 1.36 \times 10^{-2}$V=AL=5.67×10−4×24.0=1.36×10−2 m³.

Mass $m = \rho_s V = 7850 \times 1.36 \times 10^{-2} = 107$m=ρs​V=7850×1.36×10−2=107 kg, weight $W = mg = 107 \times 9.81 = 1.05 \times 10^3$W=mg=107×9.81=1.05×103 N.

M1 — correct application of $m = \rho V$m=ρV and $W = mg$W=mg with all SI units.

Step 2 — extra stress at the top of the cable.

$\sigma_{self} = W/A = 1.05 \times 10^3 / 5.67 \times 10^{-4} = 1.85 \times 10^6$σself​=W/A=1.05×103/5.67×10−4=1.85×106 Pa.

M1 — recognising that the upper attachment carries the entire self-weight in addition to the load.

Step 3 — compare.

Fractional contribution $= 1.85 \times 10^6 / 1.50 \times 10^8 = 1.23\%$=1.85×106/1.50×108=1.23%, which is well below the 5% threshold.

A1 — explicit numerical comparison and conclusion (statement that self-weight is not significant).

Total: 12 marks (M8 A4, split as shown).

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Specimen question modelled on the Edexcel 9PH0 Paper 3 format