Young modulus worksheet answers for A Level physics (free PDF download)

Problem 1 Hooke’s law in words

1. The extension of a spring is directly proportional to the applied force, provided the limit of proportionality is not exceeded.
2. The unit is N/m.
3. A straight line through the origin with gradient k.

Problem 2 Finding k

1. k = F/x = 6.0 ÷ 0.030 = 200 N/m.
2. F = kx = 200 × 0.045 = 9 N.
3. x = F/k = 15 ÷ 200 = 0.075 m (7.5 cm), assuming the limit is not exceeded.

Problem 3 Stress in a wire

1. 2.0 mm² = 2.0 × (10⁻³ m)² = 2.0 × 10⁻⁶ m².
2. σ = F/A = 500 ÷ 2.0 × 10⁻⁶ = 250000000.000 Pa.
3. 2.5 × 10⁸ Pa = 250 MPa.

Problem 4 Strain of a cable

1. 1.5 mm = 0.0015 m.
2. ε = x/L = 0.0015 ÷ 3.00 = 0.0005 (no unit).
3. 0.05 % — 0.050 %.

Problem 5 Young modulus from stress and strain

1. E = stress ÷ strain.
2. E = 1.0 × 10⁸ ÷ 5.0 × 10⁻⁴ = 200000000000.000 Pa.
3. 2.0 × 10¹¹ Pa = 200 GPa, which is steel.

Problem 6 Reading a force–extension graph

1. The graph stops being straight at 8.0 N (extension 0.040 m).
2. k = F/x = 8.0 ÷ 0.040 = 200 N/m.
3. x = 6.0 ÷ 200 = 0.030 m. E = ½Fx = ½ × 6.0 × 0.030 = 0.09 J.

Problem 7 Elastic or plastic?

1. The wire stays longer than the original length: a permanent extension remains.
2. In elastic deformation the material returns to its original shape when the load is removed; in plastic deformation some deformation is permanent.
3. Ductile: copper (or mild steel). Brittle: glass (or ceramic, cast iron).

Problem 8 Copper wire stretch

1. A = π(0.25 × 10⁻³)² = 0.000 m².
2. σ = 30 ÷ 0.000 = 152788745.368 Pa. ε = σ/E = 152788745.368 ÷ 1.2 × 10¹¹ = 0.00127.
3. x = εL = 0.00127 × 1.80 = 0.00229 m = 2.29 mm.

Problem 9 Two wires compared

1. x = FL/(AE) and A ∝ d², so x ∝ L/d².
2. B: L doubled (×2), d² ×4, so x = 4.0 × 2 ÷ 4 = 2.0 mm.
3. d halved: A is ÷4, so x = 4.0 × 4 = 16 mm.

Problem 10 Catapult

1. F = kx = 120 × 0.15 = 18 N.
2. E = ½kx² = ½ × 120 × 0.15² = 1.35 J.
3. ½mv² = E, so v = √(2 × 1.35 ÷ 0.20) = 3.67 m/s.

Problem 11 Energy stored in a wire

1. E = ½Fx = ½ × 200 × 1.0 × 10⁻³ = 0.1 J.
2. V = AL = 1.0 × 10⁻⁶ × 2.0 = 2.0 × 10⁻⁶ m³.
3. 0.1 ÷ 2.0 × 10⁻⁶ = 50000.000 J/m³. Check: σ = 200 ÷ 1.0 × 10⁻⁶ = 2.0 × 10⁸ Pa, ε = 1.0 × 10⁻³ ÷ 2.0 = 5.0 × 10⁻⁴, ½σε = ½ × 2.0 × 10⁸ × 5.0 × 10⁻⁴ = 50000.000 J/m³ ✓.

Problem 12 Reading a stress–strain graph

1. Stress at P = 250 MPa; UTS = 420 MPa.
2. E = 250 × 10⁶ ÷ 1.25 × 10⁻³ = 200000000000.000 Pa (200 GPa).
3. Breaking stress = 380 MPa. The wire narrows (necks) before it breaks, so the actual area is smaller; the calculated stress uses the original area, so it falls even though the real local stress is higher.

Problem 13 Rubber band behaviour

1. No: the force–extension graph is curved, not a straight line through the origin.
2. The area under the loading curve is the work done on the band; the area under the unloading curve is the energy returned. The enclosed area is the energy dissipated as thermal energy inside the rubber.
3. The dissipated energy raises the internal energy of the rubber, so its temperature increases.

Problem 14 Same stress, different strain

1. steel: ε = σ/E = 1.0 × 10⁸ ÷ 200 × 10⁹ = 0.0005
aluminium: ε = σ/E = 1.0 × 10⁸ ÷ 70 × 10⁹ = 0.00143
nylon: ε = σ/E = 1.0 × 10⁸ ÷ 3 × 10⁹ = 0.0333.
2. Steel has the largest Young modulus, so it is stiffest.
3. Nylon extends most. Factor = 66.7 — about 67 times that of steel.

Problem 15 Crane cable

1. 4.0 cm² = 4.0 × 10⁻⁴ m².
2. F = σA = 4.0 × 10⁸ × 4.0 × 10⁻⁴ = 160000.000 N (160 kN).
3. Working load = 160000.000 ÷ 4 = 40000.000 N, so m = F/g = 40000.000 ÷ 9.81 = 4077.472 kg.

Problem 16 Steel and copper in parallel

1. Same x, L and A, so F = EAx/L: F is proportional to E.
2. Steel takes 200 ÷ 320 of the load: 500 N. Copper takes 300 N.
3. x = FL/(EA) = 500 × 1.0 ÷ (2.0 × 10¹¹ × 1.0 × 10⁻⁶) = 2.5 mm. Check with copper: 300 ÷ (1.2 × 10¹¹ × 1.0 × 10⁻⁶) = 2.5 mm ✓.

Problem 17 Sudden loading

1. k = EA/L = 2.0 × 10¹¹ × 1.0 × 10⁻⁶ ÷ 1.5 = 133333.3333 N/m.
2. x = mg/k = 98.1 ÷ 133333.3333 = 0.736 mm.
3. Loss of gravitational energy = gain in elastic energy: mgx_max = ½kx_max², so x_max = 2mg/k = 2 × 0.736 = 1.47 mm. σ_max = kx_max/A = 133333.3333 × 0.00147 ÷ 1.0 × 10⁻⁶ = 196200000.000 Pa (196 MPa).

Problem 18 Polymer strip data

1. The line is straight up to 8 N (40 mm). k = 8 ÷ 0.040 = 200 N/m.
2. E = ½ × 8 × 0.040 = 0.16 J.
3. 8–10 N: ½(8 + 10) × (0.055 − 0.040) = 0.135 J. 10–12 N: ½(10 + 12) × (0.078 − 0.055) = 0.253 J. Total = 0.16 + 0.135 + 0.253 = 0.548 J.

Problem 19 Young modulus from a gradient

1. Gradient = F/x = EA/L, so E = gradient × L/A.
2. A = π(0.45 × 10⁻³)² = 0.000 m². E = 1.86 × 10⁴ × 2.4 ÷ 0.000 = 70169646020.960 Pa (70 GPa, aluminium).
3. Diameter: ±1.1 %, doubled for area = 2.2 %. Length: ±0.083 %. Total ≈ 2.3 % (about ±1.6 GPa). The diameter is the largest source because it is squared.

Problem 20 Evaluate the claims

1. The Young modulus is a property of the material, not of the wire’s size. A thicker wire has the same E, but carries a bigger force for the same stress.
2. The units were mixed: 1.2 mm = 1.2 × 10⁻³ m, so ε = 1.2 × 10⁻³ ÷ 2.0 = 0.0006 (6.0 × 10⁻⁴), not 0.6.
3. The force constant k = EA/L, which grows with the cross-sectional area A.