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Notes/Physics/Paper 1/Deformation of Solids
CAIEAS Level9702§6.1–6.2

Deformation of Solids

Tensile and compressive forces, Hooke's law and the spring constant, stress, strain and the Young modulus with the wire experiment, elastic and plastic behaviour, and the elastic potential energy stored in a deformed material.

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In the AS Work, Energy and Power note, solids stayed rigid: forces did work on them, but nothing changed shape. Real solids do change shape. Pull a wire and it gets longer; squeeze a spring and it gets shorter. This note measures by how much.

You start with the words for stretching and squashing, then Hooke's law and the spring constant of a spring. Next come stress, strain and the Young modulus, which describe the material rather than the object, and the experiment that measures the Young modulus of a wire. After that you learn to tell elastic from plastic behaviour on a graph, and to find the energy stored in a stretched object from the area under its force–extension graph.

Before you start you should be able to

  • Weight W=mgW = mg with g=9.81 N kg−1g = 9.81\ \text{N kg}^{-1}, and resultant force and equilibrium (AS Dynamics)

  • Moments about a pivot, used in two of the worked examples (AS Forces, Density and Pressure)

  • Work done W=FsW = Fs, kinetic and gravitational potential energy, and conservation of energy (AS Work, Energy and Power)

  • Graph skills: reading a gradient with a large triangle, and finding the area under a straight-line graph

  • Unit discipline before any substitution: mm and cm → m, kN → N, and prefix powers of ten handled explicitly

By the end of this page you can

  • Describe deformation as the effect of paired tensile or compressive forces along one dimension, and use the terms load, extension and compression correctly

  • Identify the limit of proportionality on a force–extension graph

  • State Hooke's law in words and recall and use F=kxF = kx and k=F/xk = F/x, including reading kk as the gradient of the straight section of a force–extension graph and combining identical springs in series and parallel

  • Define and use stress, strain and the Young modulus, converting diameters in mm or cm to a cross-sectional area in m² as a separate step

  • Find the Young modulus as the gradient of the straight part of a stress–strain graph, reading axis powers of ten and percentage strains correctly

  • Derive and use k=EA/Lk = EA/L, explain how kk changes with a wire's length and cross-sectional area while EE depends on the material alone, and use x=FL/EAx = FL/EA in ratio questions

  • Describe an experiment to determine the Young modulus of a metal in the form of a wire: apparatus, instrument for each measurement, method, graph plotted and precautions taken

  • Distinguish elastic from plastic deformation and the elastic limit from the limit of proportionality, interpret loading–unloading graphs including permanent extension, and decide whether a new stress stays within the elastic limit

  • Explain why the area under a force–extension graph is the work done, derive and use EP=12Fx=12kx2E_P = \tfrac{1}{2}Fx = \tfrac{1}{2}kx^2 within the limit of proportionality, and estimate the area under a curved graph

  • Use conservation of energy with elastic, kinetic and gravitational potential energy, including changes in stored energy 12k(x22−x12)\tfrac{1}{2}k(x_2^2 - x_1^2)

01

Tension and compression — the language of deformation

Syllabus requirement · §6.1

“

understand that deformation is caused by tensile or compressive forces (forces and deformations will be assumed to be in one dimension only); understand and use the terms load, extension, compression and limit of proportionality

”

Solids give way

Hang a coat on a wire hanger and the wire bows; pump up a bicycle tyre and the walls bulge; sit in a chair and its legs compress invisibly by fractions of a millimetre. No solid is truly rigid: every one changes shape when forces act on it, and this topic is about measuring by how much. The syllabus keeps it simple: one dimension only, a wire, spring or rod pulled or pushed along its own length. Tensile forces stretch an object; compressive forces squash it.

One point is easy to miss: a single force cannot deform anything. Hang a weight from a spring and the spring stretches because two forces act on it — the weight pulling down at one end and the support pulling up at the other. Deformation is always the work of a paired set of equal and opposite forces acting along the axis.

rodFFextension — rod longer than originaltension — tensile forces pullrodFFcompression — rod shorter than originalcompression — forces push inboth forces act along the same line — deformation is in one dimension only

A rod loaded along one axis: paired outward forces put it in tension and it extends; paired inward forces put it in compression and it shortens. The dashed outline is the undeformed rod.

term

meaning

watch out for

load

the force applied to an object to deform it — a force, in newtons

"a load of 200 g" means a weight of mg=2.0 Nmg = 2.0\ \text{N}, not a mass

extension (xx)

stretched length − original (natural) length

a very common lost mark: using the total stretched length instead

compression

the amount by which an object's length is reduced by squeezing

treated as a positive quantity in its own right, not "negative extension"

limit of proportionality

the point beyond which extension stops being proportional to the load

marked as a point P on a force–extension graph — see “Hooke's law and the spring constant”

The four syllabus nouns. Every calculation in this topic begins by converting a length into an extension.

A clean demonstration: naming the deformation

A spring of natural length 52 mm52\ \text{mm} hangs from a support with a 3.0 N3.0\ \text{N} load attached. Its stretched length is 97 mm97\ \text{mm}, so its extension is

x=97−52=45 mm=0.045 mx = 97 - 52 = 45\ \text{mm} = 0.045\ \text{m}

The load is the weight pulling down; the support supplies the equal upward pull at the fixed end — the pair act along the spring's axis, so the deformation is tensile. Reverse the roles: press the same spring between your hands with paired 3.0 N3.0\ \text{N} forces and it shortens to 44 mm44\ \text{mm} — a compression of 8 mm8\ \text{mm}, reported as a positive number in its own right.

Common mistakes

  • "The weight stretches the spring, so one force is enough."

    Two forces act along the axis: the weight at one end, the support's equal upward pull at the other. Remove either and nothing stretches.

    The spring is in equilibrium, so the two forces are equal and opposite. When Hooke's law (next section) uses 'the force F', it means the size of either one of them.

  • Extension of the loaded spring: "x=97 mmx = 97\ \text{mm}".

    x=97−52=45 mmx = 97 - 52 = 45\ \text{mm} — extension is the change in length, not the loaded length.

    Mark schemes can give a separate mark for this subtraction (e.g. x=0.80−0.59x = 0.80 - 0.59 in 9702/21 M/J 2025 Q2(c)(i)).

  • "The load is 200 g, so F = 200."

    A load quoted in grams is a mass; the load force is its weight, F=mg=0.200×9.81=1.96 NF = mg = 0.200\times9.81 = 1.96\ \text{N}.

    Forces in Hooke's law are in newtons. Convert grams to kilograms, then multiply by g.

Your turn

Naming the deformation and computing the extension or compression — the bookkeeping every later section stands on.

  1. 1

    A cable of original length 2.50 m2.50\ \text{m} is used to tow a crate. The towing vehicle pulls one end with a force of 850 N850\ \text{N} and the crate resists with an equal and opposite force at the other end. The cable's length becomes 2.53 m2.53\ \text{m}.

    (i) State whether the cable is in tension or compression, and identify the pair of forces causing the deformation.
    (ii) Calculate the extension of the cable.

    Stuck? Show hint

    Both ends are pulled outwards — which way does the cable deform?

    Show solution
    1. 1

      (i) The forces act outwards at the two ends, so the cable stretches: it is in tension, caused by the paired 850 N850\ \text{N} pulls at its ends.

      Name the pair and their directions — deformation vocabulary questions want the two forces identified, not just a one-word label.

    2. 2

      (ii) x=2.53−2.50=0.03 mx = 2.53 - 2.50 = 0.03\ \text{m}

      Stretched length minus original length. The extension (3 cm), not either length, is the quantity every later formula in this note uses.

    Answer

    (i) Tension — paired 850 N outward pulls at the two ends. (ii) x=0.03 mx = 0.03\ \text{m}.

  2. 2

    A rubber block of thickness 40 mm40\ \text{mm} is placed under a machine foot. The foot and the floor press on the block with a paired compressive force of 250 N250\ \text{N}, and the block thins to 37 mm37\ \text{mm}.

    (i) State the compression of the block.
    (ii) Explain why the block deforms even though the forces on it are balanced.

    Stuck? Show hint

    Balanced forces give zero resultant — but deformation is not about acceleration.

    Show solution
    1. 1

      (i) compression=40−37=3 mm\text{compression} = 40 - 37 = 3\ \text{mm}

      Same subtraction as an extension, reported as a positive shrinkage.

    2. 2

      (ii) Zero resultant force means the block does not accelerate — but the paired surface forces still squeeze it internally, pushing its atoms closer together along the axis. Deformation depends on the forces being applied, not on their resultant.

      The key idea of this section: a body in equilibrium does not accelerate, but it can still be deformed.

    Answer

    (i) 3 mm3\ \text{mm}. (ii) Balanced paired forces squeeze the block internally along the axis; equilibrium prevents acceleration, not deformation.

Paper 1 · MCQReal past-paper questions · Hooke's law and the spring constantPaper 2 · TheoryReal past-paper questions · Hooke's law and the spring constant

The rest of this note

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Can you do all of these?

  • Pair the forces: a load stretches an object only because two opposite forces act along the same axis

  • Compute extension as stretched length minus natural length — never use the total length

  • State Hooke's law as 'extension is proportional to applied force' and use F = kx in both directions

  • Read k as the gradient of the STRAIGHT section of an F–x graph, using two far-apart points on that section

  • Combine springs: parallel adds the k values; series adds the reciprocals; an identical pair gives 2k₀ and k₀/2

  • Convert diameter → radius → A = πd²/4 in SI as its own written step in every stress calculation

  • Calculate stress = F/A in Pa, strain = x/L (no unit), Young modulus E = stress/strain

  • Check a metal's Young modulus is near 10¹¹ Pa before moving on

  • Take E as the gradient of the straight part of a stress–strain graph; watch for '/ 10⁷ Pa' and '%' on the axes

  • In ratio questions use x = FL/EA and change one factor at a time (same volume, double area ⇒ half length)

  • Describe the wire experiment: clamp, long thin wire, marker and rule, micrometer for d, pulley and mass hanger; plot F against x, take the gradient, multiply by L/A

  • Justify the long, thin wire: larger extension and length mean smaller percentage uncertainties

  • Distinguish limit of proportionality (end of the straight line) from elastic limit (end of elastic behaviour)

  • Decide elastic vs plastic from the unloading line: back to the origin = elastic; zero force at positive extension = plastic, with that intercept the permanent extension

  • For 'will it behave elastically?', compare the new stress with the stress at the elastic limit, then conclude

  • Explain work done ≠ energy recovered: some deformation is plastic, and the difference is dissipated as thermal (internal) energy

  • Quote E_P = ½Fx = ½kx² only within the limit of proportionality; for a curved graph, estimate the area (shapes or squares)

  • Price CHANGES in stored energy as ½k(x₂² − x₁²), never ½k(Δx)²

  • Convert mm, cm, kN, N cm⁻¹ and GPa to SI before the first substitution — many wrong MCQ options are unconverted answers

Paper 1 · MCQ
Real past-paper questions, sorted by difficulty, with mark schemes
Paper 2 · Theory
Real past-paper questions, sorted by difficulty, with mark schemes
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