What is the difference between elastic and plastic behaviour in the context of deformation of solids?

In the context of deformation of solids, elastic behaviour refers to the ability of a material to return to its original shape after the removal of an applied force. This occurs when the material is stressed within its elastic limit. Plastic behaviour, on the other hand, occurs when a material is deformed beyond its elastic limit, resulting in permanent deformation. This means the material will not return to its original shape even after the force is removed.

How is the elastic limit related to elastic and plastic behaviour in materials?

The elastic limit is a critical threshold in materials that separates elastic behaviour from plastic behaviour. When a material is stressed within its elastic limit, it exhibits elastic behaviour, meaning it can return to its original shape after the stress is removed. If the stress exceeds the elastic limit, the material undergoes plastic deformation, leading to permanent changes in shape or size.

Why is understanding elastic and plastic behaviour important in Physics, especially for Cambridge International A Levels?

Understanding elastic and plastic behaviour is crucial in Physics because it helps explain how materials respond to forces, which is essential for designing structures and products. For Cambridge International A Levels, this knowledge is fundamental in topics like material science and engineering, where predicting material behaviour under stress is necessary for ensuring safety and functionality.

What role does Hooke's Law play in describing elastic behaviour?

Hooke's Law is a principle that describes the elastic behaviour of materials. It states that the force needed to extend or compress a spring by some distance is proportional to that distance, as long as the elastic limit is not exceeded. This law is fundamental in understanding how materials deform elastically and is often represented by the equation F = kx, where F is the force applied, k is the spring constant, and x is the displacement.

Can you provide examples of materials that exhibit both elastic and plastic behaviour?

Yes, many materials exhibit both elastic and plastic behaviour depending on the amount of stress applied. For example, metals like steel and aluminum show elastic behaviour when subjected to small forces but will undergo plastic deformation if the force exceeds their elastic limit. Polymers and certain alloys also demonstrate similar characteristics, making them useful in various applications where both flexibility and durability are required.

Why is "Confusing limit of proportionality with elastic limit" a common mistake in Elastic and plastic behaviour ?

Why it happens: Both are on stress-strain curve. How to avoid it: Limit of proportionality < elastic limit (typically). Between them: still elastic (returns to original) but not linear. Source: 9702 Examiner Reports 2022-2024

Why is "Applying Hooke's law to rubber" a common mistake in Elastic and plastic behaviour ?

Why it happens: Habit. How to avoid it: Rubber is NON-Hookean. F = kx doesn't apply. Use stress-strain graph or integration. Source: 9702 Examiner Reports 2022-2024

Deformation of SolidsCambridge International A Levels Physics (9702)

Elastic and plastic behaviour

Q: Why is "Applying Hooke's law to rubber" a common mistake in Elastic and plastic behaviour ?

Why it happens: Habit. How to avoid it: Rubber is NON-Hookean. F = kx doesn't apply. Use stress-strain graph or integration. Source: 9702 Examiner Reports 2022-2024

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Elastic and Plastic Behaviour Study Notes — Cambridge International A Level Physics 9702 (2025-2027 syllabus)

Stress-strain curves. Elastic vs plastic deformation. Ductile, brittle, polymeric materials.

What you’ll learn

Mapped to the Cambridge IGCSE 9702 syllabus (2025-2027).

6.2.1 — Distinguish elastic and plastic deformation.
6.2.2 — Read stress-strain curves.
6.2.3 — Classify materials by behaviour.

Stress-strain curve features

Limit, yield, plastic, fracture.

Typical ductile-metal curve.

Linear region (Hookean). Up to limit of proportionality. Slope = $E$ .
Non-linear elastic. Beyond proportionality but body still returns to original.
Yield point. Plastic deformation begins.
Plastic region. Permanent strain. May 'flow' easily.
Maximum stress (ultimate tensile stress).
Necking and fracture.

Brittle (glass). Linear elastic only; sudden fracture. No plastic region.

Polymeric (rubber). Non-linear from very start. Large strain. Hysteresis loop on unload — energy lost as heat.

Cambridge tip. Memorise typical shapes and label features.

Ductile: long plastic region.
Brittle: no plastic region.
Polymeric: hysteresis.

See the full worked example for elastic and plastic behaviour →

Material classifications

Ductile, brittle, polymeric.

Ductile. Plastic deformation occurs before fracture. Can be drawn into wires.

Examples: copper, steel, aluminium.

Brittle. Little or no plastic deformation; fractures suddenly.

Examples: glass, ceramics, cast iron.

Polymeric. Long-chain molecules. Highly elastic at low stress; large strain possible.

Examples: rubber, plastics.
Hysteresis: stress-strain different on loading vs unloading; area enclosed = energy lost as heat.

Cambridge tip. Mark scheme rewards examples + behaviour description.

Ductile: copper/steel.
Brittle: glass.
Polymeric: rubber.

See the full worked example for elastic and plastic behaviour →

Energy stored

Area under $\sigma$ - $\varepsilon$ × volume.

Energy per unit volume. Area under stress-strain graph.

For Hookean (linear) region: $u = \frac{1}{2}\sigma\varepsilon$ .

Total elastic PE. $E_p = u \times V$ where $V$ is volume of the body.

Equivalent form. $E_p = \frac{1}{2}kx^2$ in terms of force-extension.

Example. Wire $1 \times 10^{-6}$ m² × 1 m, stress $4 \times 10^8$ Pa at strain $0.002$ .

$u = \frac{1}{2}(4 \times 10^8)(0.002) = 4 \times 10^5$ J/m³.
$E_p = 4 \times 10^5 \times 10^{-6} = 0.4$ J.

Hysteresis. For rubber, loading and unloading curves DIFFER. Area between curves = heat lost per cycle.

Cambridge tip. Use $\frac{1}{2}\sigma\varepsilon$ only for Hookean materials.

Area under $\sigma$ - $\varepsilon$ × volume.
$\frac{1}{2}\sigma\varepsilon$ for Hookean.
Hysteresis = heat lost.

See the full worked example for elastic and plastic behaviour →

How it’s examined

Solids appear every Paper 1 or 2. Most-tested: stress-strain graph reading (7-8 marks), classification (5 marks).

Sources: Cambridge International A Level Physics 9702 syllabus (2025-2027); 9702 Examiner Reports 2022-2024; 9702/22 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-11.

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Step-by-step worked examples — Elastic and plastic behaviour

Step-by-step solutions to past-paper-style questions on elastic and plastic behaviour , written exactly the way a tutor would explain them at the board.

1Reading a stress-strain graph (8 marks)
Extended• Adapted from 9702/22 May/Jun 2024• stress-strain
Question
A stress-strain graph shows linear behaviour up to (0.001, $2 \times 10^8$ Pa), then curves and yields. Find (a) Young modulus, (b) yield stress. (8 marks)
Step-by-step solution
1. Step 1
  Young modulus = slope of linear region.
  $E = \dfrac{2 \times 10^8}{0.001} = 2 \times 10^{11} \text{ Pa}$
2. Step 2
  Yield stress = stress at start of plastic region. From the graph, $2 \times 10^8$ Pa.
Answer
(a) $E = 2 \times 10^{11}$ Pa. (b) Yield stress $\approx 2 \times 10^8$ Pa.
2Material classification (5 marks)
Extended• material types
Question
Classify (a) steel wire, (b) glass rod, (c) rubber band, and describe their stress-strain behaviour qualitatively. (5 marks)
Step-by-step solution
1. Step 1
  (a) Steel: DUCTILE. Linear up to yield, then plastic flow before fracture. Can be drawn into wires.
2. Step 2
  (b) Glass: BRITTLE. Linear elastic up to fracture; no plastic region. Shatters.
3. Step 3
  (c) Rubber: POLYMERIC. Non-linear from start. Large strains possible. Shows hysteresis on unload.
Answer
Steel ductile (linear + yield + plastic). Glass brittle (linear + sudden break). Rubber polymeric (non-linear + hysteresis).
3Energy stored (6 marks)
Extended• energy
Question
A stress-strain curve for a wire shows linear behaviour: stress $= 4 \times 10^8$ Pa at strain $= 0.002$ . Cross-section is $1 \times 10^{-6}$ m², length 1.0 m. Find the elastic PE stored. (6 marks)
Step-by-step solution
1. Step 1
  Energy per unit volume = $\frac{1}{2}\sigma\varepsilon$ .
  $u = \tfrac{1}{2}(4 \times 10^8)(0.002) = 4 \times 10^5 \text{ J/m}^3$
2. Step 2
  Total energy = $u \times V$ .
  $E_p = 4 \times 10^5 \times (1 \times 10^{-6}) \times 1.0 = 0.4 \text{ J}$
Answer
$E_p = 0.4$ J.

Key Formulae — Elastic and plastic behaviour

The formulae you need to memorise for elastic and plastic behaviour on the Cambridge International A Level 9702 paper, with every variable defined in plain English and a note on when to use it.

Energy stored per unit volume
$u = \tfrac{1}{2}\sigma\varepsilon$
When to use
Area under stress-strain graph for Hookean material. Times volume gives total elastic PE.

Key Definitions and Keywords — Elastic and plastic behaviour

Definitions to memorise and the exact keywords mark schemes credit for elastic and plastic behaviour answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9702 sitting.

Elastic deformation
Examiner keyword
Reversible deformation. Body returns to original shape when load removed.
Plastic deformation
Examiner keyword
Irreversible deformation. Body retains some change when load removed.
Elastic limit
Examiner keyword
Maximum stress beyond which body undergoes plastic deformation.
Limit of proportionality
Examiner keyword
Stress beyond which Hooke's law no longer holds.
Yield stress / yield point
Examiner keyword
Stress at which material begins plastic deformation.
Ductile
Material that undergoes substantial plastic deformation before fracture. e.g. copper, steel.
Brittle
Material that fractures with very little plastic deformation. e.g. glass, ceramics.

Common Mistakes and Misconceptions — Elastic and plastic behaviour

The traps other students keep falling into on elastic and plastic behaviour questions — taken from recent Cambridge International A Level 9702 examiner reports and mark schemes — and how to avoid them.

✕Confusing limit of proportionality with elastic limit
9702 Examiner Reports 2022-2024
Why it happens
Both are on stress-strain curve.
How to avoid it
Limit of proportionality < elastic limit (typically). Between them: still elastic (returns to original) but not linear.
✕Applying Hooke's law to rubber
9702 Examiner Reports 2022-2024
Why it happens
Habit.
How to avoid it
Rubber is NON-Hookean. $F = kx$ doesn't apply. Use stress-strain graph or integration.

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Elastic and plastic behaviour — frequently asked questions

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Elastic and plastic behaviour

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Reading a stress-strain graph (8 marks)

2Material classification (5 marks)

3Energy stored (6 marks)

Energy stored per unit volume

Elastic deformation

Plastic deformation

Elastic limit

Limit of proportionality

Yield stress / yield point

Ductile

Brittle

✕Confusing limit of proportionality with elastic limit

✕Applying Hooke's law to rubber

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Elastic and plastic behaviour — frequently asked questions