Why is "Using the total stretched length in Hooke's law instead of the extension" a common mistake in Force–Extension Relationship and Modulus of Elasticity?

Why it happens: Reading the formula as T = × lengthl and substituting the full current length for x. How to avoid it: Compute x = (current length) - (natural length) as a separate first line, then substitute. The extension is always the *change*, never the total. Source: 9231 Paper 3 Examiner Reports — recurring comment

Why is "Misquoting Hooke's law or omitting it, assuming it is given in MF19" a common mistake in Force–Extension Relationship and Modulus of Elasticity?

Why it happens: Confusing it with the elastic-PE formula (which IS in MF19); not realising T = x/l must be recalled. How to avoid it: Memorise T = x/l and write it out before substituting — the explicit statement earns a method mark. Source: 9231 Paper 3 mark schemes

Why is "Using the drop as $x$ instead of $l + x$ in vertical max-extension problems" a common mistake in Force–Extension Relationship and Modulus of Elasticity?

Why it happens: Forgetting that a particle released at the fixed point first falls the natural length l *before* any extension begins. How to avoid it: When released level with (or from) the fixed point, the total fall to maximum extension is l + x; the GPE lost is mg(l + x). Source: 9231 Paper 3 Examiner Reports

Why is "Treating an elastic string as if it pushes under compression" a common mistake in Force–Extension Relationship and Modulus of Elasticity?

Why it happens: Carrying spring intuition over to a string; assuming a string can support a particle from below. How to avoid it: A string only pulls: under compression it goes slack (T = 0). Only a spring supplies a thrust. Check the wording — 'string' vs 'spring' is deliberate. Source: 9231 Paper 3 Examiner Reports

Why is "Keeping the negative root when a max-extension quadratic is solved" a common mistake in Force–Extension Relationship and Modulus of Elasticity?

Why it happens: Mechanically quoting both roots of the quadratic without checking which is physical. How to avoid it: A string's extension must be positive; reject the negative root explicitly ('x > 0 since a string cannot be compressed') — examiners reward the stated rejection. Source: 9231 Paper 3 Examiner Reports

Why is "Letting the GPE reference level shift between the two positions" a common mistake in Force–Extension Relationship and Modulus of Elasticity?

Why it happens: Not fixing a zero level for GPE at the start, then measuring height from a different point at the end. How to avoid it: State your GPE zero on the page ('taking the lowest point as GPE zero') and measure every height from it consistently. Source: 9231 Paper 3 Examiner Reports

Cambridge International A Levels Further Mathematics (9231)

Force–Extension Relationship and Modulus of Elasticity

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Detailed Study Notes

Detailed notes on Hooke's law for Cambridge International A Levels Further Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Hooke's Law and Elastic Potential Energy — Cambridge International AS & A Level Further Mathematics 9231 Paper 3 (2026-2027 syllabus)

An elastic string or spring pulls back with a tension proportional to how far it is stretched: $T = \frac{\lambda x}{l}$ . Store that pull as elastic potential energy $E = \frac{\lambda x^2}{2l}$ , then combine it with KE and GPE to solve equilibrium and motion problems.

At a glance

Hooke's law: tension $T = \dfrac{\lambda x}{l}$ , where $\lambda$ = modulus of elasticity (in N), $l$ = natural length, $x$ = extension.
$\lambda$ is NOT given in MF19 and Hooke's law itself is NOT in MF19 — you must recall $T = \frac{\lambda x}{l}$ .
Elastic PE $E = \dfrac{\lambda x^2}{2l}$ IS given in MF19 — but you still choose when to use it.
String vs spring: a string only pulls (goes slack under compression, $T = 0$ ); a spring pulls and pushes (thrust when compressed).
Energy method: $KE + GPE + EPE = \text{constant}$ finds speeds, maximum extensions and where a string goes slack.

What you’ll learn

Mapped to the Cambridge International A Level 9231 syllabus (2026-2027).

3.4a — Use Hooke's law $T = \frac{\lambda x}{l}$ as a model for the tension in an elastic string or spring, including the modulus of elasticity $\lambda$ .
3.4b — Use the formula $E = \frac{\lambda x^2}{2l}$ for the elastic potential energy stored in a stretched string or spring.
3.4c — Solve equilibrium and energy-conservation problems involving elastic strings and springs (vertical, horizontal and inclined).

Hooke's law — tension grows with extension

A stretched string pulls back with a force proportional to its extension: T = λx/l.

The model. An elastic string or light spring is not inextensible — it stretches. Hooke's law says the tension is directly proportional to the extension:

$T = \frac{\lambda x}{l}$

$T$ = tension (N)
$x$ = extension = (stretched length) − (natural length)
$l$ = natural length (the unstretched length)
$\lambda$ = modulus of elasticity (measured in newtons, N)

Reading the formula. A short, stiff string (small $l$ , large $\lambda$ ) develops a big tension for a small stretch. The combination $\dfrac{\lambda}{l}$ acts like the "stiffness" $k$ in the school-physics form $T = kx$ — here $k = \dfrac{\lambda}{l}$ .

In equilibrium the tension balances the weight: $T = mg$, so $\frac{\lambda x}{l} = mg$ gives the extension $x$.

String versus spring — the key distinction.

An elastic string can only be in tension. If you try to compress it, it goes slack and $T = 0$ . There is no such thing as negative extension for a string.
A light spring behaves the same way under stretch, but when compressed by $x$ it pushes outward with a thrust $T = \dfrac{\lambda x}{l}$ . So a spring can support a particle from below.

Cambridge tip. Hooke's law is not printed in MF19 — neither is the meaning of $\lambda$ . Memorise $T = \frac{\lambda x}{l}$ and write it down explicitly before substituting; examiners award a method mark for the correct statement of the law.

$T = \frac{\lambda x}{l}$ : tension is proportional to extension $x$ .
$\lambda$ (modulus of elasticity) is measured in newtons (N).
A string only pulls; a spring pulls when stretched and pushes when compressed.

See the full worked example for force–extension relationship and modulus of elasticity →

Equilibrium problems — resolve and balance forces

Set the elastic tension against gravity (and any other forces) using F = 0 in each direction.

In an equilibrium problem the particle is at rest, so the resultant force is zero. The recipe never changes:

Draw a clear force diagram — include weight $mg$ , every elastic tension $\frac{\lambda x}{l}$ , the normal contact force on a plane, etc.
Resolve along sensible directions (vertical/horizontal, or along/perpendicular to an incline).
Set each direction's resultant to zero and solve for the unknown (usually the extension $x$ ).

Vertical string (the standard case). A mass $m$ hangs from a string of natural length $l$ and modulus $\lambda$ . Resolving vertically: $T = mg \quad\Rightarrow\quad \frac{\lambda x}{l} = mg \quad\Rightarrow\quad x = \frac{mgl}{\lambda}$

On an inclined plane. A particle held on a smooth slope of angle $\theta$ by an elastic string up the line of greatest slope is in equilibrium when the tension balances the component of weight down the slope: $\frac{\lambda x}{l} = mg\sin\theta$

On a smooth slope, equilibrium gives $\frac{\lambda x}{l} = mg\sin\theta$. Always resolve *along* the slope so the string tension and the weight component line up.

Two strings supporting one particle. If a particle hangs from two elastic strings, each obeys its own Hooke's law with its own $\lambda$ , $l$ and extension. Resolve horizontally and vertically; you get two equations linking the two tensions.

Cambridge tip. Compute each extension as (current length − natural length) for that specific string. Mixing up the natural length of one string with the geometry of another is the most common equilibrium error.

Equilibrium: resultant force is zero — resolve in each direction.
Vertical hang: $x = \frac{mgl}{\lambda}$ from $\frac{\lambda x}{l} = mg$ .
On a smooth incline: $\frac{\lambda x}{l} = mg\sin\theta$ .

See the full worked example for force–extension relationship and modulus of elasticity →

Elastic potential energy — the energy stored in a stretch

Stretching a string stores energy E = λx²/2l, which can convert into KE and GPE.

Stretching an elastic string does work against the tension, and that work is stored as elastic potential energy (EPE):

$E = \frac{\lambda x^2}{2l}$

This formula IS in MF19 — but you decide when to deploy it. Note the square on $x$ : doubling the extension quadruples the stored energy.

Why the $\frac{1}{2}$ ? The tension is not constant during the stretch — it grows linearly from $0$ to $\frac{\lambda x}{l}$ . The work done is the area under the tension–extension graph, a triangle of base $x$ and height $\frac{\lambda x}{l}$ : $E = \tfrac{1}{2}\,(\text{base})\,(\text{height}) = \tfrac{1}{2}\,x\cdot\frac{\lambda x}{l} = \frac{\lambda x^2}{2l}$

The stored energy is the area of the shaded triangle, giving $E = \frac{\lambda x^2}{2l}$. The same area argument explains the factor of $\frac{1}{2}$.

The energy-conservation method. When a smooth elastic system moves, mechanical energy is conserved: $KE + GPE + EPE = \text{constant}$ Equate the total energy at two positions to find a speed, a maximum extension, or a maximum height.

Where a string goes slack. As a stretched string contracts back to its natural length, the extension reaches $x = 0$ . At that instant $T = 0$ and $E = 0$ : the string can do no more work, and the particle moves on freely (projectile or free motion) until — if it returns — the string tightens again.

Finding where a particle comes to rest. At an instantaneous rest, $KE = 0$ . Setting the kinetic energy term to zero in the conservation equation gives an equation for the unknown extension or height. Solve, then check the root is physically possible (a string cannot have negative extension).

Cambridge tip. Decide your zero level for GPE before you start and keep it fixed. State it on the page ("taking the lowest point as the GPE zero"). Marks are lost when the GPE reference silently shifts between the two positions.

$E = \frac{\lambda x^2}{2l}$ (in MF19); energy scales with $x^2$ .
Conservation: $KE + GPE + EPE =$ constant between any two positions.
String slack at $x = 0$ : $T = 0$ and $E = 0$ , motion becomes free.

See the full worked example for force–extension relationship and modulus of elasticity →

Motion on an elastic string — maximum extension and rest points

Use energy conservation across the full motion to find where the particle momentarily stops.

A particle attached to an elastic string and then released or projected will oscillate or be thrown about. The maximum-extension / maximum-height points are exactly the instants of momentary rest ( $KE = 0$ ), so energy conservation pins them down.

The standard set-up. A particle of mass $m$ is attached to one end of an elastic string (natural length $l$ , modulus $\lambda$ ) fixed at the other end. It is released from a position where the string is unstretched (or projected with some speed). To find the lowest point (maximum extension $x$ ):

Take two positions: the start and the lowest point.
At the lowest point $KE = 0$ (momentary rest).
Write conservation: loss in GPE = gain in EPE (+ any initial KE). $mg\,(\text{drop}) + \tfrac{1}{2}mu^2 = \frac{\lambda x^2}{2l}$

If the particle is released from rest level with the fixed point and falls a total distance $l + x$ before reaching maximum extension $x$ , then the GPE lost is $mg(l + x)$ : $mg(l + x) = \frac{\lambda x^2}{2l}$

This is a quadratic in $x$ — solve it and reject any non-physical (negative) root.

Finding the speed at a general point. Between two points where the string is taut, equate total energy. Be careful: if the particle passes through the natural-length position, the EPE term switches off there ( $x=0$ ) and the string may go slack beyond it.

A horizontal-string variant. On a smooth horizontal table, a particle is pulled to stretch the string by $x_0$ and released. There is no GPE change, so $\underbrace{\frac{\lambda x_0^2}{2l}}_{\text{initial EPE}} = \underbrace{\tfrac{1}{2}mv^2}_{\text{KE at natural length}}$ gives the speed as the string reaches its natural length; beyond that point the string is slack and the particle continues at constant speed.

Cambridge tip. When you reach a quadratic for the maximum extension, always discard the negative root with a one-line justification (" $x>0$ since a string cannot be compressed"). Examiners reward the explicit rejection.

Maximum extension / height ⇔ momentary rest ( $KE = 0$ ).
Released level with the fixed point: $mg(l+x) = \frac{\lambda x^2}{2l}$ (quadratic in $x$ ).
Horizontal release: initial EPE → KE at natural length, then string goes slack.

See the full worked example for force–extension relationship and modulus of elasticity →

Quick recap

Hooke's law: $T = \frac{\lambda x}{l}$ — tension proportional to extension. NOT in MF19, so memorise it.
$\lambda$ is the modulus of elasticity, measured in newtons.
A string only pulls (slack at $x \le 0$ ); a spring also pushes when compressed.
Equilibrium: resolve forces and set each direction to zero, e.g. $\frac{\lambda x}{l} = mg$ or $mg\sin\theta$ .
Elastic PE: $E = \frac{\lambda x^2}{2l}$ (IS in MF19); energy grows with $x^2$ .
Energy method: $KE + GPE + EPE =$ constant finds speeds and maximum extensions; reject negative roots.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Hooke's law $T = \frac{\lambda x}{l}$ — NOT in MF19.
Elastic PE $E = \frac{\lambda x^2}{2l}$ — given in MF19.
$\lambda$ = modulus of elasticity, in newtons (N); stiffness $= \frac{\lambda}{l}$ .
Energy conservation $KE + GPE + EPE = \text{constant}$ between any two positions.
String: $T = 0$ when slack ( $x \le 0$ ); spring: thrust $\frac{\lambda x}{l}$ when compressed.

How it’s examined

Hooke's law sits in Paper 3 (Further Mechanics). Questions typically open with a short equilibrium part — find an extension or a tension from $\frac{\lambda x}{l} = mg$ (or $mg\sin\theta$ on an incline) — then build into an energy part using $E = \frac{\lambda x^2}{2l}$ to find a speed or a maximum extension. The highest-tariff parts give a quadratic in $x$ ; full marks need the correct conservation equation, a clear GPE zero, and explicit rejection of the negative root. Marks are most often lost by (i) forgetting Hooke's law is not in MF19 and misquoting it, (ii) using the wrong extension for a given string, and (iii) treating a string as if it pushes under compression. Use $g = 10\,\text{m s}^{-2}$ and give non-exact answers to 3 significant figures.

Sources: Cambridge International AS & A Level Further Mathematics 9231 syllabus (2026-2027); 9231 Examiner Reports 2022-2024 (Paper 3 Further Mechanics); 9231/31, 9231/32, 9231/33 Further Mechanics — past papers and mark schemes. Last reviewed 2026-06-23.

Take this whole topic with you

Step-by-step worked examples — Force–Extension Relationship and Modulus of Elasticity

Step-by-step solutions to past-paper-style questions on force–extension relationship and modulus of elasticity, written exactly the way a tutor would explain them at the board.

Question type:

1Tension and extension of a hanging mass
Getting startedDirect calculation• hookes-law, equilibrium, extension
Question
A particle of mass $2\,\text{kg}$ hangs in equilibrium from a light elastic string of natural length $l = 0.5\,\text{m}$ and modulus of elasticity $\lambda = 40\,\text{N}$ . Find the extension of the string. (Take $g = 10\,\text{m s}^{-2}$ .)
Step-by-step solution
1. Step 1
  Equilibrium vertically. The string tension balances the weight: $T = mg$ .
  $T = mg = 2 \times 10 = 20 \text{ N}$
2. Step 2
  State Hooke's law (NOT in MF19, so write it explicitly) and set it equal to the tension.
  $\frac{\lambda x}{l} = mg \quad\Rightarrow\quad \frac{40\,x}{0.5} = 20$
3. Step 3
  Solve for the extension $x$ .
  $80x = 20 \quad\Rightarrow\quad x = 0.25 \text{ m}$
Answer
The extension is $x = 0.25\,\text{m}$ .
Examiner tip
Always write $\frac{\lambda x}{l} = mg$ before substituting — examiners give a method mark for the correct statement of Hooke's law since it is not printed in MF19.
2String versus spring under compression
Getting startedIdentify & classify• string-vs-spring, thrust, concept
Question
A particle of mass $0.6\,\text{kg}$ rests on top of a light spring of natural length $0.4\,\text{m}$ and modulus $\lambda = 30\,\text{N}$ standing vertically on the floor. Find the compression of the spring, and explain why a string could not support the particle this way. ( $g = 10$ .)
Step-by-step solution
1. Step 1
  A spring compressed by $x$ pushes up with a thrust $\frac{\lambda x}{l}$ . In equilibrium this thrust balances the weight.
  $\frac{\lambda x}{l} = mg \quad\Rightarrow\quad \frac{30\,x}{0.4} = 0.6 \times 10 = 6$
2. Step 2
  Solve for the compression $x$ .
  $75x = 6 \quad\Rightarrow\quad x = 0.08 \text{ m}$
3. Step 3
  Classify the element. A string can only be in tension; under compression it simply goes slack ( $T = 0$ ) and cannot push, so it could not support a particle resting on top of it.
Answer
The compression is $x = 0.08\,\text{m}$ . A string cannot do this because it produces no thrust — it goes slack under compression.
Examiner tip
The numerical method is identical to a stretched string; the conceptual mark is for recognising that only a spring supplies a thrust. Watch for 'string' wording in a 'support from below' scenario — it is a trap.
3Equilibrium on a smooth incline
Building confidenceMulti-step problem• incline, equilibrium, resolving
Question
A particle of mass $3\,\text{kg}$ rests on a smooth plane inclined at $30^\circ$ to the horizontal. It is held by a light elastic string of natural length $0.8\,\text{m}$ and modulus $\lambda = 60\,\text{N}$ , running up the line of greatest slope to a fixed point. Find the extension of the string. ( $g = 10$ .)
Step-by-step solution
1. Step 1
  Resolve along the slope. The plane is smooth, so along the line of greatest slope only the tension (up) and the weight component $mg\sin\theta$ (down) act.
  $\frac{\lambda x}{l} = mg\sin\theta$
2. Step 2
  Substitute $m = 3$ , $g = 10$ , $\theta = 30^\circ$ (so $\sin 30^\circ = 0.5$ ), $\lambda = 60$ , $l = 0.8$ .
  $\frac{60\,x}{0.8} = 3 \times 10 \times 0.5 = 15$
3. Step 3
  Solve for $x$ .
  $75x = 15 \quad\Rightarrow\quad x = 0.2 \text{ m}$
Answer
The extension is $x = 0.2\,\text{m}$ .
Examiner tip
Resolve ALONG the slope so the tension and the weight component line up — this avoids the normal reaction entirely. Using $mg$ instead of $mg\sin\theta$ is the classic incline error.
4Speed from energy conservation (horizontal string)
Building confidenceMulti-step problem• energy, epe, speed, horizontal
Question
A particle of mass $0.5\,\text{kg}$ lies on a smooth horizontal table attached to one end of a light elastic string of natural length $1\,\text{m}$ and modulus $\lambda = 20\,\text{N}$ ; the other end is fixed. The particle is pulled until the string is stretched by $0.6\,\text{m}$ and released. Find its speed as the string returns to its natural length. ( $g = 10$ .)
Step-by-step solution
1. Step 1
  No GPE change (horizontal motion). Energy conservation reduces to: initial EPE converts entirely to KE at the natural length.
  $\frac{\lambda x_0^2}{2l} = \tfrac{1}{2}mv^2$
2. Step 2
  Compute the initial EPE using $E = \frac{\lambda x^2}{2l}$ (this formula IS in MF19) with $x_0 = 0.6$ .
  $E = \frac{20 \times 0.6^2}{2 \times 1} = \frac{20 \times 0.36}{2} = 3.6 \text{ J}$
3. Step 3
  Equate to KE and solve for $v$ .
  $\tfrac{1}{2}(0.5)v^2 = 3.6 \quad\Rightarrow\quad v^2 = \frac{3.6}{0.25} = 14.4$
4. Step 4
  Take the square root.
  $v = \sqrt{14.4} = 3.79 \text{ m s}^{-1} \text{ (3 s.f.)}$
Answer
The speed at the natural length is $v = 3.79\,\text{m s}^{-1}$ (3 s.f.).
Examiner tip
At the natural length $x = 0$ so EPE $= 0$ — all the stored energy has become KE. Beyond this point the string goes slack and the particle continues at constant speed. Give non-exact answers to 3 s.f.
5Maximum extension of a vertical string
StretchMulti-step problem• energy, max-extension, quadratic, vertical
Question
A particle of mass $2\,\text{kg}$ is attached to one end of a light elastic string of natural length $1.5\,\text{m}$ and modulus $\lambda = 50\,\text{N}$ , with the other end fixed at a point $O$ . The particle is released from rest at $O$ . Find the maximum extension of the string in the subsequent motion. ( $g = 10$ .)
Step-by-step solution
1. Step 1
  Set up energy conservation between release and the lowest point. At the lowest point the particle is momentarily at rest ( $KE = 0$ ). Released from rest at $O$ , the total drop to the lowest point is $l + x = 1.5 + x$ .
  $mg(l + x) = \frac{\lambda x^2}{2l}$
2. Step 2
  Substitute $m = 2$ , $g = 10$ , $l = 1.5$ , $\lambda = 50$ .
  $2 \times 10 \times (1.5 + x) = \frac{50\,x^2}{2 \times 1.5}$
3. Step 3
  Simplify both sides. Left: $20(1.5 + x) = 30 + 20x$ . Right: $\frac{50x^2}{3}$ .
  $30 + 20x = \frac{50x^2}{3}$
4. Step 4
  Multiply by 3 and rearrange into a quadratic in $x$ .
  $90 + 60x = 50x^2 \quad\Rightarrow\quad 50x^2 - 60x - 90 = 0 \quad\Rightarrow\quad 5x^2 - 6x - 9 = 0$
5. Step 5
  Solve the quadratic with the formula and reject the negative root (a string cannot have negative extension).
  $x = \frac{6 \pm \sqrt{36 + 180}}{10} = \frac{6 \pm \sqrt{216}}{10} = \frac{6 + 14.70}{10} = 2.07 \text{ m}$
Answer
The maximum extension is $x = 2.07\,\text{m}$ (3 s.f.); the negative root $x = -0.87$ is rejected as unphysical.
Examiner tip
The full drop is $l + x$ , not just $x$ — forgetting the natural length $l$ is the single biggest error here. Always reject the negative root with a one-line reason for the final accuracy mark.
6Particle supported by two vertical strings
StretchMulti-step problem• two-strings, equilibrium, modular
Question
A particle of mass $4\,\text{kg}$ hangs in equilibrium attached to two identical light elastic strings, each of natural length $1\,\text{m}$ and modulus $\lambda = 100\,\text{N}$ . Both strings are vertical: one runs up to a ceiling point $A$ , the other down to a floor point $B$ directly below, with $A$ and $B$ a fixed distance apart so that the upper string has extension $x$ and the lower string is also stretched by extension $y = x - 0.6$ . Given the lower string's extension is $0.2\,\text{m}$ , find $x$ and verify equilibrium. ( $g = 10$ .)
Step-by-step solution
1. Step 1
  Find the upper extension from the given geometric relation $y = x - 0.6$ with $y = 0.2$ .
  $0.2 = x - 0.6 \quad\Rightarrow\quad x = 0.8 \text{ m}$
2. Step 2
  Compute each tension by Hooke's law. Upper string pulls up; lower string (stretched, below the particle) also pulls down.
  $T_{\text{up}} = \frac{100 \times 0.8}{1} = 80 \text{ N}, \quad T_{\text{low}} = \frac{100 \times 0.2}{1} = 20 \text{ N}$
3. Step 3
  Resolve vertically. The upward tension must balance the weight plus the downward pull of the lower string.
  $T_{\text{up}} = mg + T_{\text{low}} = 4 \times 10 + 20 = 60 \text{ N}?$
4. Step 4
  Check consistency. Here $T_{\text{up}} = 80\,\text{N}$ but $mg + T_{\text{low}} = 60\,\text{N}$ , so this configuration is NOT in equilibrium — a net upward force of $20\,\text{N}$ acts. Equilibrium would require $T_{\text{up}} = 60\,\text{N}$ , i.e. upper extension $x = 0.6\,\text{m}$ .
  $T_{\text{up}} - mg - T_{\text{low}} = 80 - 40 - 20 = 20 \neq 0$
Answer
$x = 0.8\,\text{m}$ gives $T_{\text{up}} = 80\,\text{N}$ , which exceeds $mg + T_{\text{low}} = 60\,\text{N}$ by $20\,\text{N}$ — so the particle is NOT in equilibrium at this position; equilibrium needs $x = 0.6\,\text{m}$ .
Examiner tip
Each string obeys its own Hooke's law with its own extension. The lower string, being below and stretched, pulls the particle DOWN — get this direction right when resolving. Verifying (rather than assuming) equilibrium is exactly the kind of reasoning examiners reward.

Model Answers — Force–Extension Relationship and Modulus of Elasticity

High-scoring sample answers for force–extension relationship and modulus of elasticity on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
3 marks
A particle of mass $1.5\,\text{kg}$ hangs in equilibrium from a light elastic string of natural length $0.4\,\text{m}$ and modulus of elasticity $30\,\text{N}$ . Find the extension of the string. (Take $g = 10\,\text{m s}^{-2}$ .)
Model answer
Equilibrium vertically — the tension balances the weight: $T = mg = 1.5 \times 10 = 15 \text{ N}$

Apply Hooke's law $T = \dfrac{\lambda x}{l}$ : $\frac{30\,x}{0.4} = 15$

Solve: $75x = 15 \quad\Rightarrow\quad x = 0.2 \text{ m}$
Why this scores
M1 for $T = mg = 15\,\text{N}$ ; M1 for a correct statement and substitution of Hooke's law $\frac{\lambda x}{l} = T$ ; A1 for $x = 0.2\,\text{m}$ . The method mark depends on writing the law explicitly (not in MF19).
Question 2
4 marks
An elastic string has natural length $0.6\,\text{m}$ and modulus of elasticity $48\,\text{N}$ . It is stretched to a total length of $0.75\,\text{m}$ . (a) Find the extension. (b) Find the tension. (c) Find the elastic potential energy stored.
Model answer
(a) Extension = stretched length − natural length: $x = 0.75 - 0.6 = 0.15 \text{ m}$

(b) Tension by Hooke's law $T = \dfrac{\lambda x}{l}$ : $T = \frac{48 \times 0.15}{0.6} = \frac{7.2}{0.6} = 12 \text{ N}$

(c) Elastic PE using $E = \dfrac{\lambda x^2}{2l}$ (given in MF19): $E = \frac{48 \times 0.15^2}{2 \times 0.6} = \frac{48 \times 0.0225}{1.2} = \frac{1.08}{1.2} = 0.9 \text{ J}$
Why this scores
(a) B1 for $x = 0.15\,\text{m}$ (extension, not total length). (b) M1 A1 for $T = 12\,\text{N}$ . (c) M1 A1 for $E = 0.9\,\text{J}$ — note the $x^2$ and the $2l$ denominator. Using the total length $0.75$ instead of the extension is the commonest slip.
Question 3
6 marks
A particle of mass $0.4\,\text{kg}$ rests on a smooth plane inclined at $30^\circ$ , held by a light elastic string of natural length $0.5\,\text{m}$ and modulus $20\,\text{N}$ along the line of greatest slope. (a) Find the equilibrium extension. (b) The string is then cut. Using energy, find the speed of the particle when it has slid $0.6\,\text{m}$ down the slope. ( $g = 10$ .)
Model answer
(a) Resolve along the slope: $\dfrac{\lambda x}{l} = mg\sin\theta$ . $\frac{20\,x}{0.5} = 0.4 \times 10 \times \sin 30^\circ = 4 \times 0.5 = 2$ $40x = 2 \quad\Rightarrow\quad x = 0.05 \text{ m}$

(b) After the string is cut, only gravity acts along the smooth slope. Energy conservation over a slide of $d = 0.6\,\text{m}$ : $\tfrac{1}{2}mv^2 = mg\sin\theta \times d$ $\tfrac{1}{2}(0.4)v^2 = 0.4 \times 10 \times 0.5 \times 0.6 = 1.2$ $0.2\,v^2 = 1.2 \quad\Rightarrow\quad v^2 = 6 \quad\Rightarrow\quad v = \sqrt{6} = 2.45 \text{ m s}^{-1} \text{ (3 s.f.)}$
Why this scores
(a) M1 for $\frac{\lambda x}{l} = mg\sin\theta$ , A1 for $x = 0.05\,\text{m}$ . (b) M1 for the GPE drop $mgd\sin\theta$ , M1 for equating to $\frac{1}{2}mv^2$ , A1 for $v = 2.45\,\text{m s}^{-1}$ . The GPE change uses the VERTICAL drop $d\sin\theta$ , not $d$ .
Question 4
6 marks
A particle of mass $0.8\,\text{kg}$ hangs from a light elastic string of natural length $1\,\text{m}$ and modulus $40\,\text{N}$ , fixed at the top. It is pulled down a further $0.5\,\text{m}$ below its equilibrium position and released. Find the speed of the particle as it passes back through the equilibrium position. ( $g = 10$ .)
Model answer
Equilibrium extension first. At equilibrium $\dfrac{\lambda e}{l} = mg$ : $\frac{40\,e}{1} = 0.8 \times 10 = 8 \quad\Rightarrow\quad e = 0.2 \text{ m}$

So the equilibrium extension is $0.2\,\text{m}$ and the released position has extension $0.2 + 0.5 = 0.7\,\text{m}$ .

Energy conservation between the lowest point (rest, extension $0.7$ ) and the equilibrium position (extension $0.2$ , speed $v$ ). The particle rises $0.5\,\text{m}$ : $\underbrace{\frac{40 \times 0.7^2}{2 \times 1}}_{\text{EPE}_1} = \underbrace{\frac{40 \times 0.2^2}{2 \times 1}}_{\text{EPE}_2} + \underbrace{mg \times 0.5}_{\text{GPE gain}} + \underbrace{\tfrac{1}{2}mv^2}_{\text{KE}}$

Evaluate each term: $\text{EPE}_1 = \frac{40 \times 0.49}{2} = 9.8 \text{ J}, \quad \text{EPE}_2 = \frac{40 \times 0.04}{2} = 0.8 \text{ J}, \quad mg(0.5) = 0.8 \times 10 \times 0.5 = 4 \text{ J}$

Solve for $v$ : $9.8 = 0.8 + 4 + \tfrac{1}{2}(0.8)v^2 \quad\Rightarrow\quad 0.4\,v^2 = 5 \quad\Rightarrow\quad v^2 = 12.5$ $v = \sqrt{12.5} = 3.54 \text{ m s}^{-1} \text{ (3 s.f.)}$
Why this scores
B1 for $e = 0.2\,\text{m}$ ; M1 for a complete energy equation with EPE at BOTH positions and the GPE term; M1 for correct EPE values (note $x^2$ ); A1 for $v^2 = 12.5$ ; A1 for $v = 3.54\,\text{m s}^{-1}$ . The string is taut throughout (extension never reaches 0), so EPE appears at both ends.
Question 5
7 marks
A particle of mass $2\,\text{kg}$ is attached to one end of a light elastic string of natural length $1.2\,\text{m}$ and modulus $60\,\text{N}$ , the other end fixed at $O$ . The particle is projected vertically downwards from $O$ with speed $3\,\text{m s}^{-1}$ . Find the maximum extension of the string. ( $g = 10$ .)
Model answer
Set up energy conservation between the projection point $O$ and the lowest point (momentary rest, $KE = 0$ ). The total drop from $O$ to the lowest point is $l + x = 1.2 + x$ .

Taking $O$ as the GPE zero, conservation gives: $\underbrace{\tfrac{1}{2}mu^2}_{\text{initial KE}} + \underbrace{mg(l + x)}_{\text{GPE lost}} = \underbrace{\frac{\lambda x^2}{2l}}_{\text{EPE gained}}$

Substitute $m = 2$ , $u = 3$ , $g = 10$ , $l = 1.2$ , $\lambda = 60$ : $\tfrac{1}{2}(2)(3^2) + 2 \times 10 \times (1.2 + x) = \frac{60\,x^2}{2 \times 1.2}$ $9 + 20(1.2 + x) = \frac{60x^2}{2.4} = 25x^2$ $9 + 24 + 20x = 25x^2$

Form the quadratic: $25x^2 - 20x - 33 = 0$

Solve using the formula: $x = \frac{20 \pm \sqrt{400 + 3300}}{50} = \frac{20 \pm \sqrt{3700}}{50} = \frac{20 + 60.83}{50} = 1.62 \text{ m (3 s.f.)}$

The negative root $x = -0.817$ is rejected since a string cannot have negative extension.
Why this scores
M1 for an energy equation including initial KE, GPE over $l + x$ , and EPE; A1 for the substituted equation; M1 for forming the quadratic $25x^2 - 20x - 33 = 0$ ; M1 A1 for solving; A1 for rejecting the negative root with reason and giving $x = 1.62\,\text{m}$ . The drop is $l + x$ — forgetting the natural length loses the first two marks.
Question 6
8 marks
A particle of mass $0.5\,\text{kg}$ is attached to a light elastic string of natural length $1\,\text{m}$ and modulus $25\,\text{N}$ , fixed at $O$ . The particle hangs in equilibrium, then is pulled down $0.4\,\text{m}$ and released. (a) Find the equilibrium extension. (b) Show that the particle leaves $O$ 's vertical with the string slack, and find the speed at which the string becomes slack. ( $g = 10$ .)
Model answer
(a) Equilibrium: $\dfrac{\lambda e}{l} = mg$ : $\frac{25\,e}{1} = 0.5 \times 10 = 5 \quad\Rightarrow\quad e = 0.2 \text{ m}$

(b) The string becomes slack when the extension returns to $x = 0$ , i.e. at the natural length, which is $0.2\,\text{m}$ above the equilibrium position. The released (lowest) point has extension $e + 0.4 = 0.6\,\text{m}$ .

Energy conservation from the lowest point (rest) to the natural-length point (string just slack, speed $v$ ). The particle rises a distance $0.6\,\text{m}$ (from extension $0.6$ up to extension $0$ ): $\underbrace{\frac{\lambda (0.6)^2}{2l}}_{\text{EPE}_1} = \underbrace{mg(0.6)}_{\text{GPE gain}} + \underbrace{\tfrac{1}{2}mv^2}_{\text{KE}} \qquad (\text{EPE} = 0 \text{ at slack point})$

Evaluate: $\text{EPE}_1 = \frac{25 \times 0.36}{2 \times 1} = 4.5 \text{ J}, \quad mg(0.6) = 0.5 \times 10 \times 0.6 = 3 \text{ J}$

Solve for $v$ : $4.5 = 3 + \tfrac{1}{2}(0.5)v^2 \quad\Rightarrow\quad 0.25\,v^2 = 1.5 \quad\Rightarrow\quad v^2 = 6$ $v = \sqrt{6} = 2.45 \text{ m s}^{-1} \text{ (3 s.f.)}$

Since $v^2 = 6 > 0$ , the particle is still moving when the string goes slack, so it continues upward as a free particle (the string slack as claimed).
Why this scores
(a) B1 for $e = 0.2\,\text{m}$ . (b) M1 for identifying the slack point at extension 0 ( $0.6\,\text{m}$ rise); M1 for the energy equation with EPE only at the start; M1 for correct EPE $= 4.5\,\text{J}$ and GPE $= 3\,\text{J}$ ; A1 for $v^2 = 6$ ; A1 for $v = 2.45\,\text{m s}^{-1}$ ; B1 for concluding $v > 0$ confirms the string genuinely goes slack with the particle still moving.

Key Formulae — Force–Extension Relationship and Modulus of Elasticity

The formulae you need to memorise for force–extension relationship and modulus of elasticity on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Hooke's law (tension)
$T = \frac{\lambda x}{l}$
$T$
tension in the string or spring (N)
$\lambda$
modulus of elasticity (N)
$x$
extension = stretched length − natural length (m)
$l$
natural (unstretched) length (m)
When to use
Any time you need the tension (or thrust) of an elastic string or spring. NOT in MF19 — write it out explicitly to earn the method mark.
Elastic potential energy
$E = \frac{\lambda x^2}{2l}$
$E$
elastic potential energy stored (J)
$\lambda$
modulus of elasticity (N)
$x$
extension (m)
$l$
natural length (m)
When to use
Energy problems. IS given in MF19. Note the $x^2$ : doubling the extension quadruples the stored energy.
Vertical equilibrium extension
$x = \frac{mgl}{\lambda}$
$m$
mass of the hanging particle (kg)
$g$
acceleration due to gravity ( $10\,\text{m s}^{-2}$ )
$l, \lambda$
natural length and modulus
When to use
A mass hanging vertically in equilibrium. Derived from $\frac{\lambda x}{l} = mg$ — derive it rather than memorise blindly.
Equilibrium on a smooth incline
$\frac{\lambda x}{l} = mg\sin\theta$
$\theta$
angle of the incline to the horizontal
$x$
extension of the string along the slope (m)
When to use
A particle held on a smooth slope by a string up the line of greatest slope. Resolve ALONG the slope; the normal reaction drops out.
Conservation of mechanical energy
$KE + GPE + EPE = \text{constant}$
$KE$
$\tfrac{1}{2}mv^2$ , kinetic energy
$GPE$
$mgh$ , gravitational PE relative to a fixed zero
$EPE$
$\frac{\lambda x^2}{2l}$ , elastic PE
When to use
Smooth elastic systems in motion — equate total energy at two positions to find a speed, a maximum extension, or where a string goes slack.

Key Definitions and Keywords — Force–Extension Relationship and Modulus of Elasticity

Definitions to memorise and the exact keywords mark schemes credit for force–extension relationship and modulus of elasticity answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Modulus of elasticity $\lambda$
Examiner keyword
The constant in Hooke's law $T = \frac{\lambda x}{l}$ ; numerically the tension produced when the extension equals the natural length. Measured in newtons (N), NOT in N m $^{-1}$ .
Extension
Examiner keyword
The amount a string or spring is stretched beyond its natural length: extension = (current length) − (natural length). Always use the extension $x$ in Hooke's law, never the total length.
Natural length $l$
Examiner keyword
The length of the string or spring when it carries no tension or thrust (unstretched, uncompressed).
Elastic potential energy
Examiner keyword
The energy stored in a stretched (or compressed) elastic element, equal to the work done against the tension: $E = \frac{\lambda x^2}{2l}$ . Given in MF19.
String versus spring
Examiner keyword
A string can only pull (tension); it goes slack with $T = 0$ under compression. A spring pulls when stretched and pushes (thrust) when compressed, both via $\frac{\lambda x}{l}$ .
Slack (string)
The state in which an elastic string has returned to (or below) its natural length, so the extension is zero or negative: $T = 0$ and the stored EPE is zero. Beyond this the particle moves freely.

Common Mistakes and Misconceptions — Force–Extension Relationship and Modulus of Elasticity

The traps other students keep falling into on force–extension relationship and modulus of elasticity questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Using the total stretched length in Hooke's law instead of the extension
9231 Paper 3 Examiner Reports — recurring comment
Why it happens
Reading the formula as $T = \frac{\lambda \times \text{length}}{l}$ and substituting the full current length for $x$ .
How to avoid it
Compute $x = (\text{current length}) - (\text{natural length})$ as a separate first line, then substitute. The extension is always the change, never the total.
✕Misquoting Hooke's law or omitting it, assuming it is given in MF19
9231 Paper 3 mark schemes
Why it happens
Confusing it with the elastic-PE formula (which IS in MF19); not realising $T = \frac{\lambda x}{l}$ must be recalled.
How to avoid it
Memorise $T = \frac{\lambda x}{l}$ and write it out before substituting — the explicit statement earns a method mark.
✕Using the drop as $x$ instead of $l + x$ in vertical max-extension problems
9231 Paper 3 Examiner Reports
Why it happens
Forgetting that a particle released at the fixed point first falls the natural length $l$ before any extension begins.
How to avoid it
When released level with (or from) the fixed point, the total fall to maximum extension is $l + x$ ; the GPE lost is $mg(l + x)$ .
✕Treating an elastic string as if it pushes under compression
9231 Paper 3 Examiner Reports
Why it happens
Carrying spring intuition over to a string; assuming a string can support a particle from below.
How to avoid it
A string only pulls: under compression it goes slack ( $T = 0$ ). Only a spring supplies a thrust. Check the wording — 'string' vs 'spring' is deliberate.
✕Keeping the negative root when a max-extension quadratic is solved
9231 Paper 3 Examiner Reports
Why it happens
Mechanically quoting both roots of the quadratic without checking which is physical.
How to avoid it
A string's extension must be positive; reject the negative root explicitly (' $x > 0$ since a string cannot be compressed') — examiners reward the stated rejection.
✕Letting the GPE reference level shift between the two positions
9231 Paper 3 Examiner Reports
Why it happens
Not fixing a zero level for GPE at the start, then measuring height from a different point at the end.
How to avoid it
State your GPE zero on the page ('taking the lowest point as GPE zero') and measure every height from it consistently.

Force–Extension Relationship and Modulus of Elasticity

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Tension and extension of a hanging mass

2String versus spring under compression

3Equilibrium on a smooth incline

4Speed from energy conservation (horizontal string)

5Maximum extension of a vertical string

6Particle supported by two vertical strings

Hooke's law (tension)

Elastic potential energy

Vertical equilibrium extension

Equilibrium on a smooth incline

Conservation of mechanical energy

Modulus of elasticity λ\lambdaλ

Extension

Natural length lll

Elastic potential energy

String versus spring

Slack (string)

✕Using the total stretched length in Hooke's law instead of the extension

✕Misquoting Hooke's law or omitting it, assuming it is given in MF19

✕Using the drop as xxx instead of l+xl + xl+x in vertical max-extension problems

✕Treating an elastic string as if it pushes under compression

✕Keeping the negative root when a max-extension quadratic is solved

✕Letting the GPE reference level shift between the two positions

Modulus of elasticity $\lambda$

Natural length $l$

✕Using the drop as $x$ instead of $l + x$ in vertical max-extension problems