A Level · AS · structured
AS Level foundations · Question 8
AS Level foundations · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 13
- Topics
- 2
- Answer
- Complete
A vertical steel wire of original length 2.50 m and diameter 0.56 mm hangs from a fixed support. A load of 45 N is attached to its lower end. The Young modulus of steel is 2.0 × 10¹¹ Pa and the wire does not exceed its limit of proportionality.
- (a)
Define Define the Young modulus.
2 marks - (b)
Calculate Calculate the extension of the wire produced by the 45 N load.
4 marks - (c)
Calculate Calculate the elastic potential energy stored in the wire under this load.
2 marks - (d)
Determine The diameter is measured with a micrometer as 0.56 ± 0.01 mm. Determine the percentage uncertainty this contributes to the calculated extension.
3 marks - (e)
Explain The load is increased until the wire is stretched beyond its elastic limit and then removed. Explain how the wire's behaviour differs from that described above.
2 marks
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(a)
Define Define the Young modulus.
Identify what the question is testing, organise the response into distinct mark-earning points, and make every conclusion traceable to a physical principle or to the evidence supplied.
- 1
the ratio of tensile stress to tensile strain
- 2
for a material obeying Hooke's law / below the limit of proportionality
(b)
Calculate Calculate the extension of the wire produced by the 45 N load.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
A = π(0.28 × 10⁻³)² = 2.46 × 10⁻⁷ m²
- 2
stress = F/A = 45 / 2.46 × 10⁻⁷ = 1.83 × 10⁸ Pa
- 3
strain = stress / E = 1.83 × 10⁸ / 2.0 × 10¹¹ = 9.14 × 10⁻⁴
- 4
e = strain × L = 9.14 × 10⁻⁴ × 2.50 = 2.3 × 10⁻³ m (2.3 mm)
(c)
Calculate Calculate the elastic potential energy stored in the wire under this load.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
E = ½Fe = ½ × 45 × 2.3 × 10⁻³
- 2
E = 0.051 J
(d)
Determine The diameter is measured with a micrometer as 0.56 ± 0.01 mm. Determine the percentage uncertainty this contributes to the calculated extension.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
percentage uncertainty in d = (0.01 / 0.56) × 100 = 1.8%
- 2
e is inversely proportional to d², so the percentage uncertainty is doubled
- 3
percentage uncertainty in e = 3.6%
(e)
Explain The load is increased until the wire is stretched beyond its elastic limit and then removed. Explain how the wire's behaviour differs from that described above.
State the outcome first, then link cause to effect with the relevant physical principle. Each link in the reasoning should be explicit enough to earn its own marking point.
- 1
the wire no longer returns to its original length when the load is removed
- 2
it retains a permanent (plastic) extension, because layers of atoms have slipped past one another rather than simply being pulled further apart
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