Unit 3 · Forces, Energy and Deformation of Solids
The Complete Cambridge International AS & A Level Physics (9702) Guide · Cambridge International AS & A Level Physics (9702)Unit 3 · Forces, Energy and Deformation of Solids
- The Complete Cambridge International AS & A Level Physics (9702) Guide
- Cambridge International AS & A Level Physics (9702)
- 12 sections
Cambridge International does not publish a per-unit weighting for this exam. Exam structure (official): AS Level: Multiple Choice (75 min, 40 questions, 31%) · AS Level Structured Questions (75 min, 46%) · Advanced Practical Skills (120 min, 2 questions, 23%); A Level: Multiple Choice (75 min, 40 questions, 15.5%) · AS Level Structured Questions (75 min, 23%) · Advanced Practical Skills (120 min, 2 questions, 11.5%) · A Level Structured Questions (120 min, 38.5%) · Planning, Analysis And Evaluation (75 min, 2 questions, 11.5%). Be ready to carry the same unit skills and representations into a complete answer on every paper. The 120 N load acts over 1.50 × 10⁻⁶ m², so stress is a force-per-area quantity. Strain instead compares the 1.20 mm extension with the original 2.00 m length and has no unit.
- How Cambridge International AS & A Level Physics (9702) assesses this Cambridge International publishes no per-unit percentage for Cambridge International A Level Physics; Paper 1 accounts for 31% of the AS Level total and 15.5% of the A Level total. · AS Level: Multiple Choice (75 min, 40 questions, 31%) · AS Level Structured Questions (75 min, 46%) · Advanced Practical Skills (120 min, 2 questions, 23%) | A Level: Multiple Choice (75 min, 40 questions, 15.5%) · AS Level Structured Questions (75 min, 23%) · Advanced Practical Skills (120 min, 2 questions, 11.5%) · A Level Structured Questions (120 min, 38.5%) · Planning, Analysis And Evaluation (75 min, 2 questions, 11.5%) · show the model with A graph of force (N) against extension (mm)., A formula card that explains the modulus relation combines force, original length, area and extension in one expression., A table linking key concept, why it matters, visible method.
- Key skills Moments, Couples, Torque and Equilibrium of Forces, Density, Pressure in Fluids and Upthrust, Work, Conservation of Energy, Efficiency and Power, Gravitational Potential Energy and Kinetic Energy, Stress, Strain, Hooke's Law and the Young Modulus, Elastic and Plastic Behavior and Elastic Potential Energy
- How to study for Unit 3 This page turns A graph of force (N) against extension (mm)., A formula card that explains the modulus relation combines force, original length, area and extension in one expression., A table linking key concept, why it matters, visible method. into one route: Divide the 120 N force by cross-sectional area.
- The organizing decision Distinguish the plotted force–extension gradient from Young modulus: the latter also depends on original length and cross-sectional area.
What Cambridge International AS & A Level Physics (9702) Unit 3 covers
Use this map to connect each assessed skill to the relationship or representation that makes it visible.
Moments, Couples, Torque and Equilibrium of Forces
Young modulus; elastic strain energy; moment of a force02Density, Pressure in Fluids and Upthrust
Young modulus; elastic strain energy; moment of a force03Work, Conservation of Energy, Efficiency and Power
Young modulus; elastic strain energy; moment of a force04Gravitational Potential Energy and Kinetic Energy
Young modulus; elastic strain energy; moment of a force05Stress, Strain, Hooke's Law and the Young Modulus
Young modulus; elastic strain energy; moment of a force06Elastic and Plastic Behavior and Elastic Potential Energy
Young modulus; elastic strain energy; moment of a forceCambridge International does not publish a per-unit weighting for this exam.
Official unit name and weighting: Cambridge International syllabus.
Forces, Energy and Deformation of Solids
Connect the published share to the unit model
The 120 N load acts over 1.50 × 10⁻⁶ m², so stress is a force-per-area quantity. Strain instead compares the 1.20 mm extension with the original 2.00 m length and has no unit.
Young modulus connects those two results only after extension has been converted to 0.00120 m. Keeping stress and strain distinct prevents the force–extension gradient from being mistaken for the material modulus.
Report 8.00 × 10⁷ Pa, 6.00 × 10⁻⁴ and 1.33 × 10¹¹ Pa in their proper roles. The dimensionless strain sits between two pressure-valued quantities.
Begin by isolating the system and naming the target: equilibrium, pressure, transferred energy or deformation. Select the matching relation and derive any needed expression before inserting numbers. For material response, distinguish a force-extension graph from stress-strain quantities; the graph gradient and Young modulus are not interchangeable unless the specimen geometry has been accounted for.
The decision that organizes this unit
Define the system and choose the route before calculating
Distinguish the plotted force–extension gradient from Young modulus: the latter also depends on original length and cross-sectional area.
Divide the 120 N force by cross-sectional area
Mechanism route and repair branches
Relationships to preserve
- Stress: Area is 1.50 × 10⁻⁶ m².
- Strain: Extension and original length need one unit.
- Young modulus: Stress divided by strain retains pascals.
Representations to read
- A graph of force (N) against extension (mm).
- A formula card that explains the modulus relation combines force, original length, area and extension in one expression.
- A table linking key concept, why it matters, visible method.
Branches to reject
- Dividing by 1.20 mm without converting it to 0.00120 m
- Swapping the force and extension axes
- Labeling an extension ratio as a force ratio
| Key concept | Why it matters | Visible method |
|---|---|---|
| Stress | Area is 1.50 × 10⁻⁶ m² | Divide the 120 N force by cross-sectional area |
| Strain | Extension and original length need one unit | Use 0.00120 m / 2.00 m |
| Young modulus | Stress divided by strain retains pascals | Connect 8.00 × 10⁷ Pa with 6.00 × 10⁻⁴ |
| System and target | Equilibrium, pressure, energy and deformation use different models. | Name the target before selecting the relation. |
| Geometry in modulus | Material modulus is not just a force-extension gradient. | Retain original length and cross-sectional area. |
How Cambridge International AS & A Level Physics (9702) assesses Forces, Energy and Deformation of Solids
What a complete response must make visible
Match the task to evidence that a reader can audit, then check the most likely reasoning failure before finalizing the response.
| Task | Evidence to show | Hurdle |
|---|---|---|
| Moments, Couples, Torque and Equilibrium of Forces | Young modulus; elastic strain energy; moment of a force | Dividing by 1.20 mm without converting it to 0.00120 m |
| Density, Pressure in Fluids and Upthrust | Young modulus; elastic strain energy; moment of a force | Swapping the force and extension axes |
| Work, Conservation of Energy, Efficiency and Power | Young modulus; elastic strain energy; moment of a force | Labeling an extension ratio as a force ratio |
| Gravitational Potential Energy and Kinetic Energy | Young modulus; elastic strain energy; moment of a force | Dividing by 1.20 mm without converting it to 0.00120 m |
| Stress, Strain, Hooke's Law and the Young Modulus | Young modulus; elastic strain energy; moment of a force | Swapping the force and extension axes |
| Elastic and Plastic Behavior and Elastic Potential Energy | Young modulus; elastic strain energy; moment of a force | Labeling an extension ratio as a force ratio |
Worked example: equilibrium, pressure, energy and elasticity
Read each question as it appears on the exam paper, choose your answer, then tap it to see the feedback.
Practice 1: Stress, strain and Young modulus
A steel wire extends under a tensile load.
A steel wire is 2.00 m long, has cross-sectional area 1.50 × 10⁻⁶ m² and extends 1.20 mm under a 120 N load. Calculate its stress, strain and Young modulus.
[5]Show all necessary working.
A. Stress = F/A, strain = ΔL/L, and Young modulus = stress/strain.
✅ Correct! 😊 Correct. These relations give 8.00 × 10⁷ Pa, 6.00 × 10⁻⁴, and 1.33 × 10¹¹ Pa respectively.
B. Stress = F/L, strain = ΔL/A, and Young modulus = stress × strain.
❌ Not quite 😭 This pairs force and extension with the wrong geometric quantities. The resulting expressions do not have the required dimensions.
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C. Stress = F/ΔL and Young modulus is the force–extension gradient.
❌ Not quite 😭 F/ΔL is stiffness, not stress or Young modulus. Young modulus also depends on original length and cross-sectional area.
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D. Use 1.20 as the extension in meters and divide strain by stress.
❌ Not quite 😭 The extension must be converted from millimetres to 0.00120 m. Reversing the modulus ratio also gives reciprocal units.
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Full explanation
- Step 1Stress = F/A = 120/(1.50 × 10⁻⁶) = 8.00 × 10⁷ Pa.
- Step 2Convert the extension: 1.20 mm = 0.00120 m.
- Step 3Strain = ΔL/L = 0.00120/2.00 = 6.00 × 10⁻⁴.
- Step 4Young modulus = stress/strain = 1.33 × 10¹¹ Pa.
Practice 2: Work done by a varying force
The force on an object increases uniformly with displacement.
A force increases uniformly from 0 N to 40 N as an object moves 0.30 m. Determine the work done by the force.
[3]Show all necessary working.
A. Use the gradient, 40/0.30, to obtain 133 J.
❌ Not quite 😭 The gradient describes how force changes with displacement, not energy transferred. It is measured in N/m rather than J.
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B. Use the rectangular area 40 × 0.30 to obtain 12 J.
❌ Not quite 😭 This assumes 40 N acts throughout the displacement. The force rises from zero, so the area is triangular.
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C. Use the triangular area ½ × 40 × 0.30 to obtain 6.0 J.
✅ Correct! 😊 Correct. Work is the area under the force–displacement graph, which is a triangle here.
D. Use ½ × 40 × 0.30² to obtain 1.8 J.
❌ Not quite 😭 The displacement is not squared when finding graph area. Multiplying force by displacement already gives joules.
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Full explanation
- Step 1Work done equals the area under a force–displacement graph.
- Step 2A uniform rise from zero produces a triangular region.
- Step 3Calculate W = ½ × base × height = ½(0.30 m)(40 N).
- Step 4The work done is 6.0 J.
Key terms for Unit 3: Forces, Energy and Deformation of Solids
Models, uses, and boundaries
- Forces, Energy and Deformation of Solids
- The modulus relation combines force, original length, area and extension in one expression. Use this relationship when the modulus relation combines force, original length, area and extension in one expression..
- Worked example: equilibrium, pressure, energy and elasticity
- Converting 1.20 mm to 0.00120 m keeps strain dimensionless in the modulus calculation. Use this relationship when converting 1.20 mm to 0.00120 m keeps strain dimensionless in the modulus calculation..
- Forces, Energy and Deformation of Solids: change one condition
- The labeled force–extension axes identify a stiffness comparison rather than a modulus by themselves. Use this relationship when the labeled force–extension axes identify a stiffness comparison rather than a modulus by themselves..
- Forces, Energy and Deformation of Solids: test proposed results
- A plausible modulus option must retain pressure units after stress is divided by strain. Use this relationship when a plausible modulus option must retain pressure units after stress is divided by strain..
Cambridge International AS & A Level Physics (9702) Unit 3 FAQ
How much of Cambridge International AS & A Level Physics (9702) does Unit 3 carry?
Cambridge International does not publish a per-unit weighting for this exam.
What is the first move on a Forces, Energy and Deformation of Solids problem?
Divide the 120 N force by cross-sectional area
Which relationships should I preserve?
Stress: Area is 1.50 × 10⁻⁶ m². Strain: Extension and original length need one unit. Young modulus: Stress divided by strain retains pascals.
Which representations should I practice?
Practice moving among A graph of force (N) against extension (mm)., A formula card that explains the modulus relation combines force, original length, area and extension in one expression., A table linking key concept, why it matters, visible method..
What error should I check before submitting an answer?
Check for Dividing by 1.20 mm without converting it to 0.00120 m; Swapping the force and extension axes; Labeling an extension ratio as a force ratio.
Evidence workshop
Continue from the free model into complete practice
The full unit guide continues with the chapter’s worked examples, figures, scoring tables, and answer checks.
- Worked example: equilibrium, pressure, energy and elasticity
- Forces, Energy and Deformation of Solids: change one condition
- Forces, Energy and Deformation of Solids: test proposed results
- Check the worked-example evidence for Forces, Energy and Deformation of Solids
Full unit practice. Open the complete guide for the full evidence workshop and synthesis.
Related Cambridge International AS & A Level Physics (9702) unit guides
Cambridge International AS & A Level Physics (9702) Exam Guide & Review
The whole exam and its official unit sequence.01Physical Quantities and Units
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Cambridge International does not publish a per-unit weighting for this exam.How to study Cambridge International AS & A Level Physics (9702) Unit 3
Start with the organizing decision
Before solving, restate the decision in operational terms: Distinguish the plotted force–extension gradient from Young modulus: the latter also depends on original length and cross-sectional area.. Your first written move should be to Divide the 120 N force by cross-sectional area.
Practice the same idea in several representations
Rotate through A graph of force (N) against extension (mm)., A formula card that explains the modulus relation combines force, original length, area and extension in one expression., A table linking key concept, why it matters, visible method.. Use each representation to practice Moments, Couples, Torque and Equilibrium of Forces, Density, Pressure in Fluids and Upthrust, Work, Conservation of Energy, Efficiency and Power, Gravitational Potential Energy and Kinetic Energy, Stress, Strain, Hooke's Law and the Young Modulus, Elastic and Plastic Behavior and Elastic Potential Energy, and explain what stays invariant when the surface form changes.
Turn each error into a repair check
After every attempt, audit the response for Dividing by 1.20 mm without converting it to 0.00120 m; Swapping the force and extension axes; Labeling an extension ratio as a force ratio. Then redo only the first step that made the reasoning diverge, keeping units, direction, and model conditions visible.
Confirm current course details in the official Cambridge International syllabus for the 2025-2027 administration.
Where to go next
Return to the Cambridge International AS & A Level Physics (9702) to see every unit and plan your revision.