Deformation of solids
A revision chapter for Cambridge International AS & A Level Physics 9702, topic 6, Deformation of solids, written to the 2028-2030 syllabus, which Cambridge states is unchanged in teaching content from the 2025-2027 syllabus. It is AS Level content, examined in Paper 1 (multiple choice), Paper 2 (AS structured questions) and, as practical context, Paper 3, and assumed knowledge for Papers 4 and 5. The chapter covers all ten learning outcomes in two subtopics. Subtopic 6.1, Stress and strain: deformation caused by tensile or compressive forces acting in one dimension; the terms load, extension, compression and limit of proportionality, with extension always measured from the original length; Hooke's law stated with its condition and written as F = kx; the spring constant k = F/x in N m^-1 as the gradient of a force-extension graph and a property of one specimen; the definitions of stress (force per unit cross-sectional area, pascal), strain (extension per unit original length, no unit) and the Young modulus (stress divided by strain within the limit of proportionality, pascal), with A = pi d^2/4 and careful millimetre-to-metre conversion; why the Young modulus belongs to the material and k = EA/L to the specimen; and the experiment to determine the Young modulus of a metal wire, set out in plan form with a long thin wire over a pulley or Searle's apparatus with a reference wire, a micrometer at several places and orientations, loading and unloading, a graph of load against extension and E = gradient x L/A, the largest uncertainties and safety. Subtopic 6.2, Elastic and plastic behaviour: elastic deformation, plastic deformation and the elastic limit as a point distinct from the limit of proportionality; the unloading line parallel to the loading line and the permanent extension; the area under a force-extension graph as the work done, by the strip argument, for any graph shape; and elastic potential energy within the limit of proportionality from the triangular area, E_P = 1/2 Fx = 1/2 kx^2. Every equation is marked as recall; g = 9.81 m s^-2 comes from the Data and formulas sheet. The chapter includes eight computed figures, a units drill, a force-extension graph studio, six worked examples, a fictional Paper 3 Young modulus dataset giving E = (2.0 +/- 0.2) x 10^11 Pa, a Paper 5-style spring-launcher plan, a mistake clinic, retrieval practice, a mixed exam-style challenge with marking points, a mastery checklist and a spaced-review plan. Shear, bulk modulus, hysteresis, creep and fatigue are excluded as outside the syllabus.Show moreShow less
Revision notes
Interactive notes with exam tips and worked examples.
Study path
Chapter overview
A summary of this Physics chapter — open a section to read it. The full notes, worked examples and practice questions are in the study modules above.
What is Deformation of solids about?
Forces do not only accelerate objects; a pair of equal and opposite forces changes an object's shape. Tensile forces stretch a sample and compressive forces squash it, always along one line. Up to the limit of proportionality the extension is proportional to the load (Hooke's law, \(F = kx\)), and the spring constant \(k = F/x\) describes one particular spring or wire. The Young modulus describes the material instead: \(E = \text{stress}/\text{strain} = FL/(Ax)\), because stress (force per unit area) and strain (extension per unit original length) divide out the sample's size. You must be able to describe the experiment that measures \(E\) for a metal wire. Beyond the elastic limit, which is a different point from the limit of proportionality, deformation is plastic and permanent. The area under a force–extension graph is the work done, for any shape of graph; within the limit of proportionality that area is a triangle, and the elastic potential energy stored is \(E_P = \tfrac{1}{2}Fx = \tfrac{1}{2}kx^2\).
Key ideas to remember
- k belongs to the specimen, E belongs to the material; the limit of proportionality and the elastic limit are two different points; energy is the area, and it is ½Fx only while the line is straight.
- k belongs to the specimen, E to the material; the limit of proportionality and the elastic limit are different points; the area under F–x is the work done, and it is ½Fx only while the line is straight.
What you need to be able to do
- 6.1.1 I can explain that deformation is caused by tensile or compressive forces (forces and deformations will be assumed to be in one dimension only)
- 6.1.2 I can use the terms load, extension, compression and limit of proportionality
- 6.1.3 I can recall and use Hooke's law
- 6.1.4 I can recall and use the formula for the spring constant k = F/x
- 6.1.5 I can define and use the terms stress, strain and the Young modulus
- 6.1.6 I can describe an experiment to determine the Young modulus of a metal in the form of a wire
- 6.2.1 I can use the terms elastic deformation, plastic deformation and elastic limit
- 6.2.2 I can explain that the area under the force–extension graph represents the work done
- 6.2.3 I can determine the elastic potential energy of a material deformed within its limit of proportionality from the area under the force–extension graph
- 6.2.4 I can recall and use EP = ½Fx = ½kx2 for a material deformed within its limit of proportionality
Why Deformation of solids matters
Stored energy becomes kinetic energy. A catapult or spring launcher stores \(\tfrac{1}{2}kx^2\). If all of it is transferred to a projectile of mass \(m\), then \(\tfrac{1}{2}kx^2 = \tfrac{1}{2}mv^2\) (Topic 5). This gives the greatest possible launch speed; energy dissipated in the spring and by air resistance makes the real speed smaller. Worked example 4 uses this, and the Paper 5-style planning item in Practical skills turns it into an experiment.
Common mistakes to avoid
- “The limit of proportionality and the elastic limit are the same point.” Correct They are different points. The limit of proportionality is where the force–extension graph stops being a straight line. The elastic limit is where permanent deformation begins: beyond it the sample does not return to its original length when the load is removed. For a metal wire the elastic limit is at, or a little beyond, the limit of proportionality, so between them the wire is elastic but no longer obeys Hooke's law.
- “The extension is the length of the stretched wire.” Correct Extension = stretched length − original length. Strain divides the extension by the original length too, never by the stretched length.
- “A thicker wire of the same steel has a bigger Young modulus.” Correct A thicker wire has a bigger spring constant \(k\), because \(k = EA/L\). The Young modulus \(E\) is a property of the material: stress and strain have already divided out the area and the length.
- “\(A = \pi d^2\), and 0.50 mm can go straight into the formula.” Correct \(A = \pi r^2 = \pi d^2/4\), with \(d\) in metres: \(\pi(0.50 \times 10^{-3})^2/4 = 1.96 \times 10^{-7}\,\mathrm{m^2}\). And 1 mm2 = 10−6 m2, not 10−3 m2.
- “Strain is measured in metres”, or “the Young modulus is in N m−1.” Correct Strain is a length divided by a length, so it has no unit. The Young modulus is stress divided by strain, so it has the unit of stress, the pascal (N m−2). N m−1 is the unit of \(k\).
- “Energy stored = force × extension.” Correct Within the limit of proportionality the force rises from 0 to \(F\), so the energy stored is the triangle \(\tfrac{1}{2}Fx = \tfrac{1}{2}kx^2\). Beyond the limit of proportionality neither formula applies: use the area under the actual curve.
- “After it has been stretched too far, the wire unloads back down the same curve.” Correct After plastic deformation the unloading line is parallel to the original straight line and meets the extension axis at a permanent extension. Only the area under the unloading line is recovered.
- “Hooke's law, the Young modulus and \(\tfrac{1}{2}kx^2\) are on the Data sheet.” Correct None of them is. Every equation in this topic is recall; the sheet supplies only \(g = 9.81\,\mathrm{m\,s^{-2}}\).
- “The extension is the stretched length of the spring.” Repair Extension = stretched length − original length. A spring that goes from 80 mm to 104 mm has an extension of 24 mm.
- “Hooke's law: the extension is proportional to the load.” Repair Add the condition: "…provided the limit of proportionality is not exceeded." Without it the statement is false for any sample loaded far enough.
- “The limit of proportionality and the elastic limit are the same point.” Repair The first is where the graph stops being straight; the second is where permanent deformation begins. They are different, and a metal can be elastic but not proportional between them.
- “Strain = 1.2 × 10−3 m.” Repair Strain = extension ÷ original length, a ratio of two lengths, so it has no unit: \(\varepsilon = 1.2 \times 10^{-3}\).
- “d = 0.38 mm, so A = π × 0.382 = 0.454.” Repair Two errors: the formula and the unit. \(A = \pi d^2/4\) (or \(\pi r^2\) with \(r = d/2\)), and \(d\) goes in metres: \(\pi(0.38 \times 10^{-3})^2/4 = 1.13 \times 10^{-7}\,\mathrm{m^2}\).
- “A thicker wire has a bigger Young modulus.” Repair The Young modulus is a property of the material. A thicker wire of the same metal has a bigger spring constant, \(k = EA/L\), not a bigger \(E\).
- “Young modulus = force ÷ extension.” Repair Force ÷ extension is \(k\), for one specimen, in N m−1. The Young modulus is stress ÷ strain, \(E = FL/(Ax)\), in Pa.
- “The percentage uncertainty in A is the same as in d.” Repair \(A \propto d^2\), so the percentage uncertainty in \(A\) is twice that in \(d\): 2.6% in \(d\) becomes 5.3% in \(A\).
- “Elastic potential energy = Fx = 20 × 0.080 = 1.6 J.” Repair Within the limit of proportionality the force rises from 0 to \(F\), so the area is a triangle: \(E_P = \tfrac{1}{2}Fx = 0.80\,\mathrm{J}\).
- “Work done stretching the wire to 3.0 mm = ½kx².” (The wire passed its limit of proportionality at 1.5 mm.) Repair Beyond the limit of proportionality the graph is curved, so \(\tfrac{1}{2}kx^2\) does not apply. Find the area under the actual curve, by trapezia or by counting squares.
- “After it is stretched past the elastic limit, the wire unloads back along the same curve.” Repair It unloads along a line parallel to the original straight line, reaching zero force at a permanent extension. Only the area under that unloading line is recovered.
- “The main source of uncertainty was human error in reading the scale.” Repair "Human error" names no quantity and no physical cause, so it tells the reader nothing to improve. Name the quantity and the reason: the extension is only about 3 mm and is the difference of two readings each ±0.1 mm. Then give an improvement that addresses it: a longer wire, or a travelling microscope reading to 0.01 mm.
Examiner tips
- Read the command word before you decide how much to write. This syllabus has fifteen of them: calculate, comment, compare, define, describe, determine, explain, give, identify, justify, predict, show (that), sketch, state and suggest. Define wants a precise meaning — for a physical quantity, usually an equation in words with every quantity named. State and give want a fact and nothing more. Describe wants the points or the features. Explain wants the reasons and the relationships — a describe-level answer to an explain question is incomplete however well written it is. Show (that) gives you the result and asks for the structured evidence that leads to it, so every step must appear — and a final value worked to one more significant figure than the one printed makes it plain that you calculated it rather than copied it. Sketch wants a freehand graph with its key features — intercepts, asymptotes, the shape — correct, but no plotted scale.
- Where does the second force come from? For a wire hanging from a ceiling with a mass on the end, the load (the weight of the mass) pulls down on the bottom end and the support pulls up on the top end with an equal force. Both act on the wire, so the wire is in tension with tension \(F\) everywhere along it. Drawing both forces on the wire makes it clear that the wire is in equilibrium and stretched.
- Sketching this graph. A complete sketch shows: a straight section through the origin; the limit of proportionality at the end of it; the elastic limit marked as a separate point just beyond; a curve that flattens; and an unloading line drawn visibly parallel to the straight section, meeting the extension axis to the right of the origin. Label the permanent extension. The graph studio that follows reads numbers off this graph.
- Interleave with the chapters that use this one. Topic 17 (Oscillations) sets a mass oscillating on a spring because the spring exerts \(F = kx\): re-answer "state Hooke's law" there, and re-derive \(E_P = \tfrac{1}{2}kx^2\) as the area under the \(F\)–\(x\) line. Topic 9 (Electricity) defines resistivity with the same \(L/A\) that turns \(k\) into \(E\) here: compare \(R = \rho L/A\) with \(k = EA/L\). Recalling a topic inside a new context is worth more than another pass over this chapter on its own; at A Level, Paper 4 assumes the whole of the AS content, so nothing here is ever finished with.
How Deformation of solids is examined
- Cambridge International AS & A Level Physics 9702 has five components. Topic 6 is AS Level content, so it is examined in Papers 1, 2 and 3. AS Level content: examined in Paper 1 (multiple choice), Paper 2 (AS structured) and, as practical context, Paper 3. Assumed knowledge for Papers 4 and 5. AS Level candidates take Papers 1, 2 and 3; A Level candidates take all five, either staged over two years (Papers 1–3 in year one, Papers 4 and 5 in year two) or together in one series. Examinations are available in the June and November series, and in March in India.
- Across both the AS Level and the A Level the assessment objectives are weighted AO1 40% (knowledge and understanding), AO2 40% (handling, applying and evaluating information) and AO3 20% (experimental skills and investigations). AS candidates are graded a–e; A Level candidates A*–E. The Data and formulas sheet is printed as page 2 of Papers 1 and 2 and as pages 2 and 3 of Paper 4: it gives the constants and a short list of formulas. Every other equation in this chapter is one the syllabus says you must recall, and this chapter says which is which.
- In a multiple-choice item on this topic, a single slip is enough to land on a wrong option: \(\pi d^2\) instead of \(\pi d^2/4\) for the area, a millimetre not converted, \(Fx\) instead of \(\tfrac{1}{2}Fx\), or the spring constant \(k\) of one wire confused with the Young modulus \(E\) of its material. A Paper 2 question may ask you to define stress, strain or the Young modulus, to state Hooke's law with its condition, to explain the difference between elastic and plastic deformation, or to sketch a force–extension graph with the limit of proportionality, the elastic limit and an unloading line.
- The calculations are \(k = F/x\), stress, strain and \(E = FL/(Ax)\) with careful unit conversion, and energy from a force–extension graph: a triangle \(\tfrac{1}{2}Fx\) within the limit of proportionality, trapezia or counted squares beyond it. None of Hooke's law, \(k = F/x\), stress, strain, the Young modulus or \(E_P = \tfrac{1}{2}Fx = \tfrac{1}{2}kx^2\) is on the Data and formulas sheet; all are recall. The sheet gives \(g = 9.81\,\mathrm{m\,s^{-2}}\) for a load \(F = mg\).
- Topic 6 names its own experiment (6.1.6): load a long, thin wire in equal steps, read the extension against a fixed scale, measure the diameter with a micrometer, plot \(F\) against \(x\) and use \(E = \text{gradient} \times L/A\). The largest uncertainties are in the extension (small, and the difference of two readings) and in the diameter (doubled in \(A\)). Loaded springs and rubber bands are standard Paper 3 contexts, and a spring launcher makes a Paper 5 planning question.
- Read the command word before you decide how much to write. This syllabus has fifteen of them: calculate, comment, compare, define, describe, determine, explain, give, identify, justify, predict, show (that), sketch, state and suggest. Define wants a precise meaning — for a physical quantity, usually an equation in words with every quantity named. State and give want a fact and nothing more. Describe wants the points or the features. Explain wants the reasons and the relationships — a describe-level answer to an explain question is incomplete however well written it is. Show (that) gives you the result and asks for the structured evidence that leads to it, so every step must appear — and a final value worked to one more significant figure than the one printed makes it plain that you calculated it rather than copied it. Sketch wants a freehand graph with its key features — intercepts, asymptotes, the shape — correct, but no plotted scale.
Syllabus reference and sources
Written against: Cambridge International AS & A Level Physics (9702). Syllabus for 2028, 2029 and 2030 (version 1, September 2025); content unchanged from the 2025-2027 syllabus examined now. Topic 6: Deformation of solids.
Written by: Academiq Edu Instructor Panel
Source documents
- Cambridge International AS & A Level Physics 9702
- Section 5 of the same syllabus, “Practical assessment”
- Section 6 of the same syllabus, “Additional information”
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