MYP · 06 · extended response
The spring that stops obeying the first slope
Matter, density, and materials · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 2
- Answer
- Complete
Wire loading curve
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Force / N
Horizontal axis (x): Extension / mm
Read the exact data as a table
| Extension / mm | Force / N |
|---|---|
| 0 | 0 |
| 2 | 40 |
| 4 | 80 |
| 6 | 100 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A wire of original length 1.0 m and cross-sectional area 1.0 mm² is loaded gradually. Its force–extension data are supplied below; segments between points are straight. The first 4 mm is reversible and proportional. After the 6 mm reading, unloading leaves 1 mm permanent extension. Supplied model: Use F = kx, stress = F/A, strain = ΔL/L and Young modulus = stress/strain in the proportional region. Supplied data (Extension / mm; Force / N): (0, 0), (2, 40), (4, 80), (6, 100).
- (a)
Find the initial spring constant and Young modulus, converting the units explicitly.
3 marks - (b)
Calculate the work done loading to 6 mm. Explain why ½Fmax xmax is invalid and whether all the work is recoverable.
3 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Find the initial spring constant and Young modulus, converting the units explicitly.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
Initial stiffness is 40/(0.002) = 2.0×10⁴ N m⁻¹. A millimetre is 10⁻³ m, while 1 mm² is 10⁻⁶ m².
- 2
At 40 N, stress is F/A = 4.0×10⁷ Pa and strain is ΔL/L = 0.002.
- 3
Young modulus = stress/strain = 2.0×10¹⁰ Pa. This material property requires the original geometry; k alone describes this particular wire.
(b)
Calculate the work done loading to 6 mm. Explain why ½Fmax xmax is invalid and whether all the work is recoverable.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
From 0 to 4 mm the area is ½(80)(0.004) = 0.16 J.
- 2
From 4 to 6 mm it is ½(80+100)(0.002) = 0.18 J, so total loading work is 0.34 J.
- 3
The full graph is not one straight proportional line, so ½Fmax xmax would give an incorrect 0.30 J. Permanent extension shows plastic deformation; without an unloading curve the recoverable energy cannot be quantified.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- Initial stiffness is 40/(0.002) = 2.0×10⁴ N m⁻¹. A millimetre is 10⁻³ m, while 1 mm² is 10⁻⁶ m².
- At 40 N, stress is F/A = 4.0×10⁷ Pa and strain is ΔL/L = 0.002.
- Young modulus = stress/strain = 2.0×10¹⁰ Pa. This material property requires the original geometry; k alone describes this particular wire.
- From 0 to 4 mm the area is ½(80)(0.004) = 0.16 J.
- From 4 to 6 mm it is ½(80+100)(0.002) = 0.18 J, so total loading work is 0.34 J.
- The full graph is not one straight proportional line, so ½Fmax xmax would give an incorrect 0.30 J. Permanent extension shows plastic deformation; without an unloading curve the recoverable energy cannot be quantified.
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