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Free notes · Secondary 3 and 4

O-Level Physics notes

Free O-Level Physics notes on one page, for the 2026 O-Level (6091) and the 2027 SEC G3 Physics (K323). Full notes for 10 of the 20 syllabus chapters, each with the key idea, the facts and formulas to know, a worked exam question, the common mistakes and exam tips. The other chapters have a checklist of what the syllabus asks for, so the page maps the whole syllabus.

Free to read, print or save as a PDF, with no sign-up. The print layout keeps every chapter and drops the menus. After each chapter, practise on the O-Level Physics prelim papers with answers.

TSS's own revision notes, checked against the syllabus named under Sources. Updated 12 November 2026. Not issued by MOE or SEAB.

How to use these notes

  1. Pick one chapter. Read the key idea and the must-know points.
  2. Close the page and write down what you remember, then check and mark what you missed.
  3. Try the worked exam question before reading its steps, and read the common mistakes.
  4. Do questions on that chapter from a school paper and mark them. Redo the wrong ones a few days later.

Syllabus edition

These notes follow SEAB's O-Level Physics syllabus 6091, examined in 2026, and SEC G3 Physics K323, examined from 2027. The two documents list the same 20 topics and the same three papers. Sources: SEAB, O-Level Physics syllabus 6091 (2026), SEAB, SEC G3 Physics syllabus K323 (2027).

Exam format

PaperWhat it isTimeMarksWeighting
Paper 140 compulsory multiple-choice questions1 h4030%
Paper 2Structured and free-response questions. Section A, 70 marks: compulsory structured questions. The last two carry 20 marks between them, and one is a data-based question worth 8 to 12 marks. Section B, 10 marks: answer one of two questions.1 h 45 min8050%
Paper 3Practical. Section A, 20 marks: one or two experiments in 55 minutes. Section B, 20 marks: one experiment in 55 minutes.1 h 50 min4020%
  • No formula list is printed, so every relationship in the syllabus must be recalled.
  • No notes or textbooks are allowed in the practical.

The whole syllabus, chapter by chapter

All 20 chapters of the syllabus in order. 10 have full notes on this page. The rest have a checklist of what the syllabus asks you to do.

ChapterWhat it coversOn this page
1. Physical Quantities, Units and MeasurementSI units and prefixes, choosing instruments, scalars and vectors.Checklist
2. KinematicsSpeed, velocity, acceleration and motion graphs.Full notes
3. DynamicsForces, mass and weight, Newton's laws, free body diagrams and friction.Full notes
4. Turning Effect of ForcesMoments, the principle of moments, centre of gravity and stability.Full notes
5. PressurePressure, hydraulics, density, liquid pressure and manometers.Full notes
6. EnergyEnergy stores and transfers, kinetic and potential energy, work, power, efficiency and energy resources.Full notes
7. Kinetic Particle Model of MatterSolids, liquids and gases as particles, Brownian motion and gas pressure.Full notes
8. Thermal ProcessesConduction, convection and radiation.Full notes
9. Thermal Properties of MatterInternal energy, specific heat capacity, latent heat and cooling curves.Full notes
10. General Properties of WavesWave terms, wave speed, transverse and longitudinal waves, and sound.Full notes, plus checklist
11. Electromagnetic SpectrumProperties, uses and dangers of electromagnetic waves.Checklist
12. LightReflection, refraction, refractive index, total internal reflection and converging lenses.Full notes
13. Static ElectricityCharges, electric fields, charging by rubbing and by induction.Checklist
14. Current of ElectricityCurrent, e.m.f., potential difference, resistance and I-V graphs.Full notes
15. D.C. CircuitsSeries and parallel rules, effective resistance, potential dividers, thermistors and light-dependent resistors.Full notes
16. Practical ElectricityHeating effect, electrical power and cost, and electrical safety.Checklist
17. MagnetismProperties of magnets, induced magnetism and magnetic fields.Checklist
18. ElectromagnetismThe magnetic effect of a current, the force on a conductor and the d.c. motor.Full notes
19. Electromagnetic InductionInduced e.m.f., the a.c. generator and transformers.Full notes
20. RadioactivityThe nucleus, radioactive decay, half-life, and fission and fusion.Checklist

O-Level Physics at a glance

ChapterThe one thing to remember
KinematicsAverage speed = total distance ÷ total time
Dynamics and Newton's LawsResultant force F = ma, with F in N, m in kg and a in m/s²
Turning Effect of ForcesMoment = force × perpendicular distance from the pivot, in N m
PressurePressure p = F/A, unit Pa, where 1 Pa = 1 N/m²
Energy, Work and PowerWork done W = Fs, in joules
Thermal PhysicsQ = mcΔθ for a temperature change, with c in J/(kg °C)
General Wave PropertiesWave speed v = fλ, in m/s
Light: Reflection, Refraction and LensesRefractive index n = sin i / sin r = c/v
Current ElectricityQ = It, V = W/Q and R = V/I
ElectromagnetismRight-hand grip rule: thumb along the current, curled fingers show the field

Notes 1 of 10

Kinematics

Key idea. Kinematics describes motion using distance, displacement, speed, velocity and acceleration. Speed is distance ÷ time, velocity is displacement ÷ time, and acceleration is change in velocity ÷ time, a = (v − u)/t. On a speed-time graph, the gradient gives the acceleration and the area under the line gives the distance travelled. Near the Earth, objects in free fall accelerate at about 10 m/s².

Must know

  • Average speed = total distance ÷ total time
  • Acceleration a = (v − u)/t, measured in m/s²
  • Speed-time graph: gradient = acceleration, area under the line = distance travelled
  • Use g = 10 m/s² for free fall unless the question gives another value

What the syllabus asks you to do

  • State what is meant by speed and velocity, and calculate average speed as distance travelled ÷ time taken
  • State what is meant by uniform acceleration and calculate acceleration as change in velocity ÷ time taken
  • Interpret given examples of non-uniform acceleration
  • Plot and interpret displacement-time and velocity-time graphs for motion in one dimension
  • Deduce from the shape of a displacement-time graph when a body is at rest, moving with uniform velocity or moving with non-uniform velocity
  • Deduce from the shape of a velocity-time graph when a body is at rest, moving with uniform velocity, or moving with uniform or non-uniform acceleration
  • Calculate the area under a velocity-time graph to find the displacement for motion with uniform velocity or uniform acceleration
  • State that the acceleration of free fall for a body near the Earth is constant and approximately 10 m/s²

What each motion graph tells you

Graph featureDistance-time graphSpeed-time graph
GradientSpeedAcceleration
Horizontal lineObject at restConstant speed
Straight sloping lineConstant speedUniform acceleration or deceleration
Curve getting steeperSpeeding upAcceleration increasing
Area under the lineNo useful meaningDistance travelled

A speed-time graph. The line rises straight from 0 to 20 m/s between 0 and 8 s, stays flat at 20 m/s until 20 s, then falls straight to 0 at 24 s. The areas under the three parts are 80 m, 240 m and 40 m, giving a total distance of 360 m.

Speed-time graph for the worked example. The car accelerates for 8 s, travels at 20 m/s for 12 s, then brakes to rest in 4 s.

Worked exam question

A car starts from rest and accelerates uniformly to 20 m/s in 8.0 s. It then travels at a constant 20 m/s for 12 s before braking uniformly to rest in 4.0 s. (a) Calculate the acceleration in the first 8.0 s. (b) Calculate the total distance travelled. (c) Calculate the average speed for the whole journey.

  1. Acceleration in the first stage: a = (v − u)/t = (20 − 0) ÷ 8.0 = 2.5 m/s².
  2. Distance in the first stage is the area of the triangle under the speed-time graph: ½ × 8.0 × 20 = 80 m.
  3. Distance at constant speed is the area of the rectangle: 12 × 20 = 240 m.
  4. Distance while braking is the area of the second triangle: ½ × 4.0 × 20 = 40 m.
  5. Total distance = 80 + 240 + 40 = 360 m.
  6. Total time = 8.0 + 12 + 4.0 = 24 s, so average speed = total distance ÷ total time = 360 ÷ 24 = 15 m/s.

Answer: (a) 2.5 m/s² (b) 360 m (c) 15 m/s

Common mistakes

  • Reading 20 m off the speed axis of a speed-time graph as the distance travelled.Check the axis labels first, then find distance as the area under the line, split into triangles and rectangles.
  • Averaging the starting and finishing speeds, (0 + 20) ÷ 2, for a journey with several stages or stops.Divide total distance by total time for average speed, after listing the distance of each stage.
  • Writing a = (u − v)/t, so a car slowing from 20 m/s to 10 m/s in 4.0 s gets +2.5 m/s².Write final minus initial every time, then check the sign: slowing down must give a negative acceleration.
  • Saying a heavy stone falls faster than a light one when air resistance is ignored.State that without air resistance every object accelerates at g = 10 m/s², and put any difference down to air resistance.
  • Saying no forces act on a skydiver at terminal velocity.Name both forces, weight and air resistance, say they are equal and opposite, and conclude that the resultant force is zero.
  • Using 72 km/h directly with a time in seconds, or reading minutes on the time axis as seconds.Convert to metres and seconds before calculating. Divide a speed in km/h by 3.6 to get m/s.

Exam tips

  • Before using a graph, check whether it is distance-time or speed-time and note the units on each axis. The same shape means different things on the two graphs.
  • When finding distance from a speed-time graph, label each triangle and rectangle, write each area separately, then add them. Method marks are easier to award when each part is visible.
  • For terminal velocity explanations, give the forces in order: weight constant, air resistance increases with speed, resultant force decreases, acceleration decreases, forces become equal, constant velocity.

Full guide to kinematics, with a quiz and more practice questions.

Notes 2 of 10

Dynamics and Newton's Laws

Key idea. Dynamics links forces to motion. Find the resultant force by adding forces in the same direction and subtracting opposing ones. If the resultant force is zero, an object stays at rest or moves at constant velocity. If it is not zero, the object accelerates according to F = ma. Forces come in equal and opposite pairs acting on different bodies, and weight is W = mg.

Must know

  • Resultant force F = ma, with F in N, m in kg and a in m/s²
  • Weight W = mg, with g = 10 N/kg near the Earth's surface
  • Zero resultant force means rest or constant velocity, not necessarily rest
  • Third-law pairs are equal, opposite, the same type of force and act on different bodies

What the syllabus asks you to do

  • Identify and distinguish between contact forces, such as friction, air resistance, tension and normal force, and non-contact forces, such as gravitational, electrostatic and magnetic forces
  • State that mass is a measure of the amount of matter in a body, and that mass is the property that resists change in motion (inertia)
  • Describe a gravitational field, define gravitational field strength g as gravitational force per unit mass, and apply weight = mass × g
  • Distinguish between mass and weight
  • Apply Newton's laws to describe the effect of balanced and unbalanced forces and to identify action-reaction pairs acting on two interacting bodies
  • Identify the forces acting on a body and draw free-body diagrams for forces in at most two dimensions
  • Recall and apply resultant force = mass × acceleration, F = ma
  • Explain the effects of friction on motion, and describe bodies falling with and without air resistance, including terminal velocity

Common forces on a free-body diagram

ForceDirectionNotes
WeightVertically downwardsW = mg, acting at the centre of gravity
Normal reactionPerpendicular to the surface, pushing away from itActs on an object resting on or pressing against a surface
FrictionAlong the surface, opposing relative motionCauses heating and wear, reduced by lubrication
Air resistanceOpposite to the direction of motionIncreases as speed increases
TensionAlong the string or rope, away from the objectPulls on whatever the rope is attached to

A curve rises from the origin with an initial gradient of 10 m/s², the same as free fall, shown by a dashed straight line. The curve becomes less steep over time and levels off at about 20 m/s, the terminal velocity, where air resistance equals the weight and the acceleration is zero.

Speed-time graph for a ball falling with air resistance (illustrative values). The gradient falls as air resistance grows, until the speed levels off at terminal velocity.

Worked exam question

A box of mass 20 kg is pulled along a horizontal floor by a horizontal rope with a tension of 90 N. A friction force of 30 N acts on the box. Take g = 10 N/kg. (a) Calculate the weight of the box. (b) Calculate the acceleration of the box. (c) The tension is then reduced so that the box moves at constant velocity. State the new tension and explain your answer.

  1. Weight: W = mg = 20 × 10 = 200 N, acting vertically downwards and balanced by the normal reaction from the floor.
  2. Horizontally, the tension and friction act in opposite directions, so the resultant force = 90 − 30 = 60 N in the direction of the pull.
  3. Apply F = ma with the resultant force: a = F ÷ m = 60 ÷ 20 = 3.0 m/s².
  4. Constant velocity means zero acceleration, so by Newton's first law the resultant force must be zero.
  5. For a zero resultant, the tension must equal the friction, so the new tension is 30 N.

Answer: (a) 200 N (b) 3.0 m/s² in the direction of the pull (c) 30 N, because at constant velocity the resultant force is zero, so the pull must exactly balance the 30 N friction.

Common mistakes

  • Putting the 90 N pull straight into F = ma when 30 N of friction also acts.Write the resultant force on its own line, 90 − 30 = 60 N, before dividing by the mass.
  • Saying a box sliding at constant speed needs a forward resultant force to keep moving.Link constant velocity to a resultant force of zero, then find any missing force by balancing.
  • Naming the book's weight and the table's upward push as a third-law pair.Check that the two forces act on different bodies and are the same type of force before calling them a pair.
  • Writing that a bag weighs 5 kg, or giving a mass in newtons.Give mass in kg and weight in N, and convert between them with W = mg, using g = 10 N/kg.
  • Drawing the resultant force forwards for a car that is braking but still moving forwards.Point the resultant force along the acceleration. A car slowing down has a backward resultant force.

Exam tips

  • Draw a quick free-body diagram and write the resultant force as a separate line before you use F = ma. It prevents the most common error in this topic.
  • In third-law answers, name both bodies and state that the forces are equal in size, opposite in direction, of the same type and act on different bodies.
  • Whenever a question says constant velocity or at rest, write resultant force = 0 straight away. It usually leads straight to the unknown force.

Full guide to dynamics and Newton's Laws, with a quiz and more practice questions.

Notes 3 of 10

Turning Effect of Forces

Key idea. The moment of a force is its turning effect about a pivot. It equals the force multiplied by the perpendicular distance from the pivot to the line of action of the force, measured in N m. For a body in equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments about any point, and there is no resultant force.

Must know

  • Moment = force × perpendicular distance from the pivot, in N m
  • In equilibrium, sum of clockwise moments = sum of anticlockwise moments
  • A uniform beam's weight acts at its centre
  • An object topples when the line of action of its weight falls outside its base

What the syllabus asks you to do

  • Describe the moment of a force about a pivot in terms of its turning effect and relate it to everyday examples
  • Recall and apply moment of a force = force × perpendicular distance from the pivot
  • State the principle of moments for a body in equilibrium
  • Apply the principle of moments to new situations or to solve related problems
  • Show an understanding that the weight of a body may be taken as acting at a single point known as its centre of gravity
  • Describe qualitatively the effect of the position of the centre of gravity on the stability of objects

Forces and moments in the worked example (pivot at the 30 cm mark)

ForceDistance from pivotMoment about pivot
Rule's weight, 1.5 N at the 50 cm mark0.20 m to the right0.30 N m clockwise
Load W at the 5 cm mark0.25 m to the leftW × 0.25 anticlockwise
Pivot force, 2.7 N upwards0 mZero, it acts through the pivot
  1. Sketch the beam

    Mark the pivot, every force and its distance from the pivot.

  2. Add the beam's weight

    For a uniform beam it acts at the centre.

  3. Choose the pivot point

    Take moments about an unknown force so that it drops out.

  4. Balance the moments

    Sum of clockwise moments equals sum of anticlockwise moments.

  5. Balance the forces

    Total upward force equals total downward force.

Five steps for solving a principle of moments question on a balanced beam.

Worked exam question

A uniform metre rule of weight 1.5 N is pivoted at its 30 cm mark. A load of unknown weight W hangs from the 5 cm mark, and the rule balances horizontally. Calculate W and the upward force exerted by the pivot on the rule.

  1. The rule is uniform, so its weight acts at its centre, the 50 cm mark, which is 50 − 30 = 20 cm = 0.20 m to the right of the pivot.
  2. Clockwise moment from the rule's weight = 1.5 × 0.20 = 0.30 N m.
  3. The load is 30 − 5 = 25 cm = 0.25 m to the left of the pivot, so its anticlockwise moment = W × 0.25.
  4. By the principle of moments, W × 0.25 = 0.30, so W = 0.30 ÷ 0.25 = 1.2 N.
  5. For no resultant force, the upward pivot force equals the total downward force: 1.2 + 1.5 = 2.7 N.

Answer: W = 1.2 N and the pivot pushes up on the rule with 2.7 N.

Common mistakes

  • Using the full 0.50 m length of a spanner when the force acts at an angle to it.Draw the line of action of the force and measure the perpendicular distance from the pivot to it.
  • Leaving out the beam's own weight when balancing moments.Unless the beam is described as light, add its weight at its centre of gravity before taking moments.
  • Mixing units and writing 40 N × 25 cm = 1000 N m instead of 10 N m.Convert every distance to metres before multiplying, and write the unit N m with the answer.
  • Including an unknown force at the pivot with a non-zero distance.Choose the pivot at the unknown force. Its moment is zero, so it drops out of the equation.
  • Saying an object topples because its centre of gravity is outside the base.Say the object topples when the vertical line of action of its weight falls outside its base.

Exam tips

  • When there is an unknown force at a support, take moments about that support. The unknown then has zero moment and drops out of the equation.
  • Write one clear line, sum of clockwise moments = sum of anticlockwise moments, and label each term with its force and distance before putting in numbers.
  • In stability answers, describe where the line of action of the weight falls relative to the base and whether the moment it produces restores the object or tips it over.

Full guide to turning Effect of Forces, with a quiz and more practice questions.

Notes 4 of 10

Pressure

Key idea. Pressure is force per unit area, p = F/A, measured in pascals, where 1 Pa = 1 N/m². In a liquid, pressure increases with depth and density, p = hρg, and acts equally in all directions. Hydraulic machines transmit pressure through a liquid to multiply force, while barometers and manometers use liquid columns to measure atmospheric and gas pressure.

Must know

  • Pressure p = F/A, unit Pa, where 1 Pa = 1 N/m²
  • Liquid pressure p = hρg, with h the depth below the surface
  • 1 m² = 10 000 cm², so divide an area in cm² by 10 000 to get m²
  • Atmospheric pressure at sea level is about 1.0 × 10⁵ Pa, or about 760 mm of mercury

What the syllabus asks you to do

  • Define pressure in terms of force and area
  • Recall and apply pressure = force ÷ area, p = F/A
  • Describe and explain the transmission of pressure in hydraulic systems, with particular reference to the hydraulic press
  • Recall and apply pressure due to a liquid column = height of column × density of liquid × gravitational field strength, p = hρg
  • Describe how the height of a liquid column may be used to measure atmospheric pressure
  • Describe the use of a manometer to measure a pressure difference

Pressure formulas and units

QuantityFormulaUnit
Pressure on a surfacep = F/APa (N/m²)
Pressure due to a liquid columnp = hρgPa
Total pressure at depth hAtmospheric pressure + hρgPa
Force on a hydraulic pistonF = p × AN
Gas pressure from a manometerAtmospheric pressure + hρg when the open side is higherPa

Two straight lines against depth from 0 to 10 m. The dashed line, pressure due to the water, rises from 0 to 100 kPa at 10 m. The solid line, total pressure, starts at 100 kPa for the atmosphere and rises to 200 kPa at 10 m, staying 100 kPa above the dashed line.

Pressure against depth in fresh water of density 1000 kg/m³, with g = 10 N/kg. Both lines are straight because hρg is proportional to depth.

Worked exam question

A U-tube manometer contains water of density 1000 kg/m³ and is connected to a gas supply. The water level in the open arm is 25 cm higher than the level in the arm connected to the gas. Atmospheric pressure is 1.00 × 10⁵ Pa. Taking g = 10 N/kg, calculate the pressure of the gas.

  1. Convert the height difference to metres: h = 25 cm = 0.25 m.
  2. Pressure difference = hρg = 0.25 × 1000 × 10 = 2500 Pa.
  3. The water is higher in the open arm, so the gas is pushing harder than the atmosphere and its pressure is greater than atmospheric.
  4. Gas pressure = atmospheric pressure + hρg = 100 000 + 2500 = 102 500 Pa.

Answer: 1.025 × 10⁵ Pa (102 500 Pa)

Common mistakes

  • Converting 25 cm² to 0.25 m² by dividing by 100.Remember 1 m² = 100 × 100 = 10 000 cm², so 25 cm² = 0.0025 m².
  • Saying a wider tank has a greater water pressure at the bottom.Use p = hρg. Only the depth, the density and g matter, not the width or shape of the tank.
  • Measuring h from the bottom of the container in p = hρg.Measure h as the vertical depth below the liquid surface, down to the point in question.
  • Subtracting hρg from atmospheric pressure when the liquid stands higher in the open arm of a manometer.Sketch the U-tube. If the gas pushes the liquid down on its side, the gas pressure is atmospheric plus hρg.
  • Giving only hρg when asked for the total pressure on a diver.Add atmospheric pressure, about 1.0 × 10⁵ Pa, whenever the question asks for the total pressure at a depth.
  • Saying a barometer reads higher when the tube is tilted because more of the tube fills with mercury.Always use the vertical height from the mercury surface in the dish to the top of the column.

Exam tips

  • Write every area in m² on its own line before substituting into p = F/A. Area conversions are where most marks are lost in this topic.
  • In manometer questions, sketch the U-tube and mark which side is higher before deciding whether to add or subtract hρg from atmospheric pressure.
  • When describing a barometer, say the height is the vertical distance from the mercury surface in the dish to the top of the mercury column, and mention the vacuum above it.

Full guide to pressure, with a quiz and more practice questions.

Notes 5 of 10

Energy, Work and Power

Key idea. Energy is measured in joules and is never created or destroyed, only transferred between stores. Work done is force × distance moved in the direction of the force, W = Fs. Kinetic energy is ½mv² and the change in gravitational potential energy is mgh. Power is the rate of energy transfer, P = E/t, in watts, and efficiency compares useful output with total input.

Must know

  • Work done W = Fs, in joules
  • Kinetic energy Ek = ½mv², gravitational potential energy change Ep = mgh
  • Power P = W/t = E/t, in watts, where 1 W = 1 J/s
  • Efficiency = useful output ÷ total input × 100%

What the syllabus asks you to do

  • Show an understanding of energy stores, such as kinetic, gravitational, chemical and elastic potential, nuclear and internal, and transfers between them mechanically, electrically, by heating and by waves
  • Recall and apply kinetic energy Ek = ½mv² and gravitational potential energy near the Earth's surface Ep = mgh
  • State the principle of the conservation of energy and apply it to new situations or to solve related problems
  • Recall and apply work done = force × distance moved in the direction of the force
  • Recall and apply power = energy transferred ÷ time taken, P = E/t
  • Calculate the efficiency of an energy transfer as useful energy output ÷ total energy input

Energy, work and power formula sheet

QuantityFormulaUnit
Work doneW = FsJ
Kinetic energyEk = ½mv²J
Change in gravitational potential energyEp = mghJ
PowerP = W/t or P = E/tW
EfficiencyUseful output ÷ total input × 100%No unit, a percentage

Two straight lines against height fallen from 0 to 20 m. Gravitational potential energy falls from 100 kJ to 0, while kinetic energy, shown dashed, rises from 0 to 100 kJ. The lines cross at 10 m, where each is 50 kJ, and their total stays 100 kJ throughout.

Energy stores of the 500 kg roller coaster car in the worked example as it descends 20 m, assuming no energy is dissipated.

Worked exam question

A roller coaster car of mass 500 kg starts from rest at the top of a slope 20 m high. Take g = 10 N/kg. (a) Calculate the gravitational potential energy lost as it reaches the bottom. (b) Calculate its speed at the bottom if no energy is dissipated. (c) Its actual speed at the bottom is 18 m/s. How much energy was transferred to internal energy by friction and air resistance?

  1. Potential energy lost: Ep = mgh = 500 × 10 × 20 = 100 000 J.
  2. With no energy dissipated, all of this becomes kinetic energy: ½ × 500 × v² = 100 000.
  3. So 250 × v² = 100 000, giving v² = 400 and v = 20 m/s.
  4. Actual kinetic energy at 18 m/s: Ek = ½ × 500 × 18² = 250 × 324 = 81 000 J.
  5. Energy dissipated = energy lost from the potential store − energy gained in the kinetic store = 100 000 − 81 000 = 19 000 J.

Answer: (a) 100 000 J (1.0 × 10⁵ J) (b) 20 m/s (c) 19 000 J (1.9 × 10⁴ J)

Common mistakes

  • Calculating ½ × 2.0 × 3.0 = 3.0 J for a 2.0 kg ball moving at 3.0 m/s.Square the speed first, 3.0² = 9.0, then multiply by ½m to get 9.0 J.
  • Using 500 g, or the weight of 5.0 N, as the mass in ½mv².Convert grams to kilograms and use the mass, not the weight, in ½mv².
  • Using the 5.0 m length of a slope as h in mgh.Use only the vertical height change in mgh, however long or curved the path is.
  • Writing that energy is used up or lost in a machine.Say energy is transferred or dissipated, usually to the internal energy of the surroundings, and that the total is conserved.
  • Getting an efficiency of 143% by dividing a 500 J input by a 350 J output.Put the useful output on top, then check that the efficiency is below 100%.
  • Counting work done by the upward force on a bag carried horizontally at constant height.Ask whether the object moves in the direction of the force. If it does not, that force does no work on it.

Exam tips

  • For energy conversion problems, write the stores at the start and at the end, then write the energy equation in words before putting in numbers.
  • State any assumption you make, such as no energy dissipated by friction or air resistance. It often earns a mark and makes your method clear.
  • In efficiency questions, make sure useful output and total input are in the same unit, both in joules or both in watts, before dividing.

Full guide to energy, Work and Power, with a quiz and more practice questions.

Notes 6 of 10

Thermal Physics

Key idea. Thermal physics uses the kinetic particle model to explain temperature, gas pressure and heat transfer. Energy moves by conduction, convection and radiation. The energy needed to change a substance's temperature is Q = mcΔθ, and the energy needed to melt or boil it at constant temperature is Q = ml. Temperature stays constant during a change of state because the energy goes into overcoming forces between particles.

Must know

  • Q = mcΔθ for a temperature change, with c in J/(kg °C)
  • Q = ml for a change of state at constant temperature
  • Radiation needs no medium and is faster from hotter, larger, dull black surfaces
  • The specific heat capacity of water is about 4200 J/(kg °C)

What the syllabus asks you to do

  • Use the kinetic particle model to describe solids, liquids and gases in terms of the arrangement, motion and spacing of particles and the forces between them
  • Infer from a Brownian motion experiment that molecules in a liquid or gas move randomly, and relate a rise in temperature to an increase in the average kinetic energy of the particles
  • Explain the pressure of a gas in terms of the motion of its particles
  • Describe conduction in solids, convection in fluids in terms of density changes, and radiation, which needs no material medium
  • Describe how surface colour and texture, surface temperature and surface area affect the rate of energy transfer by radiation
  • Define heat capacity and specific heat capacity and apply Q = mcΔθ
  • Describe melting and boiling as energy transfer without a change in temperature, and explain the difference between boiling and evaporation
  • Define specific latent heat, apply Q = ml, and sketch and interpret a cooling curve

Thermal physics formulas and values for water

QuantityFormula or valueUnit
Energy for a temperature changeQ = mcΔθJ
Energy for a change of stateQ = mlJ
Heat capacity of an objectC = Q/ΔθJ/K
Specific heat capacity from measurementsc = Q/(mΔθ)J/(kg °C)
Specific heat capacity of waterAbout 4200J/(kg °C)
Specific latent heat of fusion of waterAbout 3.4 × 10⁵J/kg
Specific latent heat of vaporisation of waterAbout 2.3 × 10⁶J/kg

Temperature against time. The line rises from −20 °C to 0 °C in 4.2 s, stays flat at 0 °C until 38.2 s while the ice melts, rises to 100 °C at 80.2 s, then stays flat at 100 °C until 310.2 s while the water boils. Values use c of ice 2100, c of water 4200, latent heats 3.4 × 10⁵ and 2.3 × 10⁶ J/kg.

0.10 kg of ice from −20 °C heated at 1000 W, no losses. Values used: c(ice) 2100, c(water) 4200 J/(kg °C), fusion 3.4 × 10⁵, vaporisation 2.3 × 10⁶ J/kg.

Worked exam question

A 2000 W kettle contains 0.50 kg of water at 30 °C. Take the specific heat capacity of water as 4200 J/(kg °C) and its specific latent heat of vaporisation as 2.3 × 10⁶ J/kg. Assume no energy is lost to the surroundings. (a) Calculate the energy needed to bring the water to 100 °C. (b) How long does this take? (c) The kettle keeps boiling for a further 115 s. What mass of water boils away?

  1. Temperature rise: Δθ = 100 − 30 = 70 °C.
  2. Energy to heat the water: Q = mcΔθ = 0.50 × 4200 × 70 = 147 000 J.
  3. Time: t = E ÷ P = 147 000 ÷ 2000 = 73.5 s.
  4. Energy supplied while boiling: E = Pt = 2000 × 115 = 230 000 J.
  5. The temperature stays at 100 °C during boiling, so use Q = ml: m = Q ÷ l = 230 000 ÷ 2 300 000 = 0.10 kg.

Answer: (a) 147 000 J (1.47 × 10⁵ J) (b) 73.5 s (about 74 s) (c) 0.10 kg

Common mistakes

  • Writing that the particles in a solid do not move.Say solid particles vibrate about fixed positions, and vibrate more as the temperature rises.
  • Saying the particles expand when a metal rod is heated.Say the particles stay the same size and the spaces between them increase.
  • Writing heat rises to explain convection.Say the heated fluid expands, becomes less dense and rises, while cooler, denser fluid sinks to replace it.
  • Using Q = mcΔθ on the flat part of a heating curve, where Δθ = 0.Use Q = ml on flat sections, where the state changes, and Q = mcΔθ only on sloping sections.
  • Explaining a flat section of a heating curve by saying no energy is being absorbed.Say energy is still supplied but goes into overcoming attractive forces, raising potential energy rather than kinetic energy.
  • Calling evaporation slow boiling.Contrast them: evaporation happens at any temperature from the surface only, boiling at a fixed temperature throughout the liquid.

Exam tips

  • On a heating or cooling curve, label each flat section as a change of state and use Q = ml there, and use Q = mcΔθ only on the sloping sections.
  • In gas pressure explanations, mention the frequency of collisions per unit area of wall, and when temperature changes, also mention the average speed of the particles.
  • For heat transfer questions, name the process and then explain it with particles or waves. For convection always include the change in density.

Full guide to thermal Physics, with a quiz and more practice questions.

Notes 7 of 10

General Wave Properties

Key idea. A wave transfers energy without transferring matter. In a transverse wave the particles oscillate at right angles to the direction of energy transfer, and in a longitudinal wave they oscillate parallel to it. Waves are described by amplitude, wavelength, frequency and period, and their speed is given by the wave equation v = fλ, where f = 1/T.

Must know

  • Wave speed v = fλ, in m/s
  • Frequency f = 1/T, in hertz (Hz)
  • Amplitude is measured from the rest position to a crest, not from crest to trough
  • Across n crests there are n − 1 wavelengths

What the syllabus asks you to do

  • Describe wave motion as shown by vibrations in ropes and springs and by waves in a ripple tank, using the term wavefront
  • Show an understanding that waves transfer energy without transferring matter
  • Define and use the terms speed, frequency, wavelength, period and amplitude, including on graphs
  • Recall and apply speed of wave = frequency × wavelength, v = fλ
  • Compare transverse and longitudinal waves and give suitable examples of each
  • Describe the longitudinal nature of sound waves in terms of compression and rarefaction

Transverse and longitudinal waves compared

FeatureTransverseLongitudinal
Direction of oscillationPerpendicular to the direction of energy transferParallel to the direction of energy transfer
FeaturesCrests and troughsCompressions and rarefactions
One wavelengthCrest to neighbouring crestCentre of one compression to the next
ExamplesWater ripples, rope waves, lightSound, a slinky pushed and pulled

A smooth wave plotted against distance from 0 to 0.80 m. Displacement swings between plus 3.0 cm at the crests, at 0.10 m and 0.50 m, and minus 3.0 cm at the troughs, at 0.30 m and 0.70 m. One complete cycle spans 0.40 m, the wavelength, so two wavelengths are shown.

Displacement-distance graph of a wave with amplitude 3.0 cm and wavelength 0.40 m, a snapshot of the whole wave at one instant.

Worked exam question

In a ripple tank, a student measures the distance from the first crest to the sixth crest as 10 cm. She also counts 20 waves passing a fixed point in 4.0 s. Calculate (a) the wavelength, (b) the frequency, (c) the period and (d) the speed of the waves.

  1. From the first crest to the sixth crest there are 6 − 1 = 5 complete wavelengths.
  2. Wavelength λ = 10 ÷ 5 = 2.0 cm = 0.020 m.
  3. Frequency f = number of waves ÷ time = 20 ÷ 4.0 = 5.0 Hz.
  4. Period T = 1/f = 1 ÷ 5.0 = 0.20 s.
  5. Wave speed v = fλ = 5.0 × 0.020 = 0.10 m/s.

Answer: (a) 2.0 cm (0.020 m) (b) 5.0 Hz (c) 0.20 s (d) 0.10 m/s (10 cm/s)

Common mistakes

  • Giving 1.20 m as the amplitude when crests are 0.60 m above and troughs 0.60 m below the rest line.Measure amplitude from the rest position to a crest, which is half the crest-to-trough distance.
  • Dividing 10 cm by 6 when six crests span 10 cm.Count the gaps. Across n crests there are n − 1 wavelengths, so six crests span five wavelengths.
  • Reading a wavelength off a displacement-time graph.Check the horizontal axis first. Distance gives the wavelength, time gives the period.
  • Saying a floating cork is carried along by the passing wave.Say the particles oscillate about fixed positions while the energy moves along the wave.
  • Saying that doubling the frequency doubles the wave speed in the same medium.Keep the speed fixed for a given medium, then use v = fλ, so doubling f halves λ.

Exam tips

  • Convert to base units before using v = fλ: kHz to Hz, MHz to Hz, cm to m and ms to s. A wrong power of ten is the most common lost mark here.
  • Check the horizontal axis label on any wave graph first. Distance gives the wavelength, time gives the period.
  • When describing a wave type, use the phrases oscillate about a fixed position and direction of energy transfer, and state whether they are perpendicular or parallel.

Full guide to general Wave Properties, with a quiz and more practice questions.

Notes 8 of 10

Light: Reflection, Refraction and Lenses

Key idea. Light reflects with the angle of incidence equal to the angle of reflection, both measured from the normal. It refracts when it changes speed, and the refractive index is n = sin i / sin r = c/v. Total internal reflection happens when light in a denser medium meets a boundary at more than the critical angle, where sin c = 1/n. A converging lens forms real or virtual images.

Must know

  • Refractive index n = sin i / sin r = c/v
  • Critical angle: sin c = 1/n, about 42° for glass with n = 1.5
  • Plane mirror image: same size, upright, laterally inverted, virtual, as far behind as the object is in front
  • Speed of light in a vacuum c = 3.0 × 10⁸ m/s

What the syllabus asks you to do

  • Use the terms normal, angle of incidence and angle of reflection, and apply angle of incidence = angle of reflection in constructions, measurements and calculations
  • Use the terms normal, angle of incidence and angle of refraction to describe the refraction of light
  • Recall and apply sin i / sin r = constant
  • Define refractive index as the ratio of the speed of light in vacuum to the speed of light in the medium
  • Explain the terms critical angle and total internal reflection, and apply total internal reflection to optical fibres in telecommunication and medicine
  • Describe the action of a thin converging lens on a beam of light and define its focal length
  • Draw ray diagrams to show the formation of real and virtual images by a thin converging lens
  • Describe the characteristics of the images formed by a thin converging lens: real or virtual, magnified or diminished, upright or inverted

Images formed by a thin converging lens

Object positionImageUse
Beyond 2FReal, inverted, diminished, between F and 2F on the other sideCamera, the eye
At 2FReal, inverted, same size, at 2F on the other sidePhotocopier making a same-size copy
Between F and 2FReal, inverted, magnified, beyond 2F on the other sideProjector
Closer than FVirtual, upright, magnified, on the same side as the objectMagnifying glass

A straight line through the origin with sin r on the horizontal axis from 0 to 0.7 and sin i on the vertical axis from 0 to 1.0. It passes through (0.2, 0.3), (0.4, 0.6) and (0.6, 0.9), so its gradient is 1.5, the refractive index of the glass.

Graph of sin i against sin r for light entering glass of refractive index 1.5. The straight line through the origin has a gradient equal to n.

Worked exam question

A ray of light in air strikes the surface of a glass block with an angle of incidence of 50°. The angle of refraction in the glass is 30°. The speed of light in air is 3.0 × 10⁸ m/s. Calculate (a) the refractive index of the glass, (b) the speed of light in the glass and (c) the critical angle of the glass.

  1. Refractive index: n = sin i / sin r = sin 50° ÷ sin 30° = 0.766 ÷ 0.500 = 1.532, which is 1.53 to 3 significant figures.
  2. Speed in glass: v = c/n = 3.0 × 10⁸ ÷ 1.532 = 1.96 × 10⁸ m/s.
  3. Critical angle: sin c = 1/n = 1 ÷ 1.532 = 0.653.
  4. c = sin⁻¹ 0.653 = 40.7°, keeping the unrounded value of n to avoid rounding errors.

Answer: (a) 1.53 (b) 1.96 × 10⁸ m/s (c) 40.7°

Common mistakes

  • Measuring a 30° angle from the mirror surface and calling it the angle of incidence.Draw the normal first, at 90° to the surface, then measure every angle from it.
  • Writing n = sin r / sin i and getting n = 0.65 for glass.Put the angle in air on top, and check that the refractive index comes out greater than 1.
  • Saying total internal reflection can happen as light enters glass from air.Check direction first: light must travel from the denser medium towards the less dense one, beyond the critical angle.
  • Describing a plane mirror image as inverted.Describe it as upright, laterally inverted, virtual, the same size and as far behind the mirror as the object is in front.
  • Drawing ray diagrams freehand, without arrows, with virtual rays as solid lines.Use a ruler and sharp pencil, add arrows to every ray, and draw virtual rays as dashed lines.
  • Treating an object 15 cm from a lens of focal length 10 cm as the object-at-2F case.Mark F and 2F on both sides of the lens before drawing any rays or naming the case.

Exam tips

  • In refraction calculations, check that your refractive index is greater than 1 and that the ray bends towards the normal on entering the denser medium. Both are quick checks on your working.
  • When explaining total internal reflection, state both conditions in full. Answers that give only one condition usually lose a mark.
  • For ray diagrams, use a sharp pencil and ruler, draw at least two standard rays, and finish by stating three properties of the image: real or virtual, upright or inverted, magnified or diminished.

Full guide to light: Reflection, Refraction and Lenses, with a quiz and more practice questions.

Notes 9 of 10

Current Electricity

Key idea. Current is the rate of flow of charge, I = Q/t. Potential difference is the energy converted per unit charge, V = W/Q, and resistance is R = V/I. In a series circuit the current is the same everywhere and resistances add. In a parallel circuit each branch has the same p.d., the branch currents add, and 1/R = 1/R₁ + 1/R₂ gives a smaller combined resistance.

Must know

  • Q = It, V = W/Q and R = V/I
  • Series: R = R₁ + R₂, with the same current everywhere
  • Parallel: 1/R = 1/R₁ + 1/R₂, with the same p.d. across each branch
  • A wire's resistance is proportional to its length and inversely proportional to its cross-sectional area

What the syllabus asks you to do

  • State that current is the rate of flow of charge, measured in amperes, distinguish conventional current from electron flow, and apply Q = It
  • State that e.m.f. and potential difference are work done per unit charge, measured in volts
  • State that resistance = p.d. ÷ current and apply R = V/I
  • Recall and apply the proportionality between the resistance of a wire and its length and cross-sectional area
  • Describe the effect of a temperature increase on the resistance of a metal, and sketch and interpret I-V graphs for an ohmic conductor, a filament lamp and a diode
  • Apply the rules for current and potential difference in series and parallel circuits
  • Recall and apply the formulae for the effective resistance of resistors in series and in parallel
  • Describe the action of NTC thermistors and light-dependent resistors and their use as input transducers in potential dividers

Series and parallel circuits compared

FeatureSeriesParallel
CurrentSame at every pointBranch currents add up to the supply current
Potential differenceP.d.s add up to the e.m.f.Same p.d. across each branch
Combined resistanceR = R₁ + R₂1/R = 1/R₁ + 1/R₂, smaller than either
If one lamp breaksAll the lamps go outLamps in the other branches stay lit

Current against p.d. from 0 to 12 V. The resistor gives a straight line through the origin reaching 2.0 A at 12 V, so its resistance stays 6.0 Ω. The filament lamp starts steep and bends over to about 1.8 A at 12 V, showing its resistance rising as the filament gets hotter.

I-V graphs for a 6.0 Ω resistor and a typical filament lamp (illustrative values). The resistor's line is straight, while the lamp's curve flattens as it heats up.

Worked exam question

A 12 V supply is connected to a 4.0 Ω resistor in series with a parallel combination of a 6.0 Ω resistor and a 12 Ω resistor. Calculate (a) the total resistance of the circuit, (b) the current from the supply, (c) the p.d. across the 4.0 Ω resistor and (d) the current in the 6.0 Ω resistor.

  1. Parallel part: 1/R = 1/6.0 + 1/12 = 2/12 + 1/12 = 3/12, so R = 12 ÷ 3 = 4.0 Ω.
  2. Total resistance in series: 4.0 + 4.0 = 8.0 Ω.
  3. Supply current: I = V/R = 12 ÷ 8.0 = 1.5 A.
  4. P.d. across the 4.0 Ω resistor: V = IR = 1.5 × 4.0 = 6.0 V.
  5. The rest of the 12 V is across the parallel part: 12 − 6.0 = 6.0 V, and both branches share this p.d.
  6. Current in the 6.0 Ω resistor: I = 6.0 ÷ 6.0 = 1.0 A. Check: the 12 Ω branch carries 6.0 ÷ 12 = 0.5 A, and 1.0 + 0.5 = 1.5 A.

Answer: (a) 8.0 Ω (b) 1.5 A (c) 6.0 V (d) 1.0 A

Common mistakes

  • Saying the current after a lamp is smaller than the current before it.Remember the current is the same at every point in a series circuit, before and after each component.
  • Giving 0.50 Ω as the combined resistance of two 4.0 Ω resistors in parallel.Invert 1/R at the end, then check the answer is smaller than the smallest branch resistance.
  • Saying a filament lamp obeys Ohm's law.Explain that the filament heats up and its resistance rises, so its I-V graph curves.
  • Drawing a voltmeter in series or an ammeter in parallel.Put the ammeter in the loop, in series, and connect the voltmeter across the component, in parallel.
  • Thinking that doubling the diameter of a wire halves its resistance.Doubling the diameter makes the area four times larger, so the resistance falls to a quarter.
  • Saying a thermistor's resistance rises as it gets hotter.Remember an NTC thermistor's resistance falls as it warms, and an LDR's resistance falls as the light gets brighter.

Exam tips

  • Redraw complicated circuits neatly, replace each parallel group with its combined resistance first, then work outwards to the supply.
  • Check every answer with the circuit rules: currents into a junction equal currents out, and p.d.s around a series loop add up to the e.m.f.
  • When describing an I-V graph, state what happens to the resistance and give the physical reason, such as the filament getting hotter.

Full guide to current Electricity, with a quiz and more practice questions.

Notes 10 of 10

Electromagnetism

Key idea. Electromagnetism links current and magnetism. A current produces a magnetic field, whose direction is given by the right-hand grip rule. A current-carrying wire in a magnetic field feels a force, given by Fleming's left-hand rule, which drives a d.c. motor. A changing magnetic field induces an e.m.f., which is used in generators and in transformers, where Vs/Vp = Ns/Np.

Must know

  • Right-hand grip rule: thumb along the current, curled fingers show the field
  • Fleming's left-hand rule: First finger Field, seCond finger Current, thuMb Motion
  • Transformer: Vs/Vp = Ns/Np, and VpIp = VsIs if ideal
  • Induced e.m.f. increases with faster motion, a stronger magnet and more turns

What the syllabus asks you to do

  • Draw the magnetic field patterns due to currents in straight wires and solenoids, and state the effect of changing the size or direction of the current
  • Describe the use of the magnetic effect of a current in electromagnets, such as circuit breakers
  • Describe the force on a current-carrying conductor in a magnetic field, including the effect of reversing the current or the field
  • Use Fleming's left-hand rule to deduce the relative directions of force, field and current
  • Explain the turning effect on a current-carrying coil in a motor, and describe the split-ring commutator and the effect of a soft-iron core
  • Deduce that a changing magnetic field induces an e.m.f., that its direction opposes the change producing it, and the factors affecting its size
  • Describe a simple a.c. generator and the use of slip rings, and sketch its output voltage against time
  • Describe a simple iron-cored transformer, apply Vp/Vs = Np/Ns and VpIp = VsIs for an ideal transformer, and explain the advantages of high-voltage transmission

Which rule to use in electromagnetism

SituationRuleWhat it gives
Field around a straight wire or solenoidRight-hand grip ruleDirection of the magnetic field, or the north pole of a solenoid
Force on a current in a magnetic fieldFleming's left-hand ruleDirection of the force and so the motion
Current induced by a changing fieldLenz's lawA current that opposes the change causing it

Two wave-shaped curves of e.m.f. against time from 0 to 20 ms. The solid curve has a peak of 6 V and completes one cycle in 20 ms, a frequency of 50 Hz. The dashed curve, for twice the rotation speed, has a peak of 12 V and completes two cycles in 20 ms, a frequency of 100 Hz.

Output e.m.f. of a simple a.c. generator (illustrative values). Rotating the coil twice as fast doubles both the peak e.m.f. and the frequency.

Worked exam question

An ideal transformer has 2000 turns on its primary coil, which is connected to a 240 V a.c. supply. Its secondary coil lights a lamp rated 12 V, 36 W at normal brightness. Calculate (a) the number of turns on the secondary coil, (b) the current in the lamp and (c) the current in the primary coil.

  1. Use the turns ratio: Ns/Np = Vs/Vp, so Ns = Np × Vs ÷ Vp = 2000 × 12 ÷ 240 = 100 turns.
  2. The lamp works normally, so it has 12 V across it and takes 36 W. Current in the lamp: Is = P ÷ V = 36 ÷ 12 = 3.0 A.
  3. For an ideal transformer, power in = power out, so VpIp = VsIs = 36 W.
  4. Primary current: Ip = 36 ÷ 240 = 0.15 A.

Answer: (a) 100 turns (b) 3.0 A (c) 0.15 A

Common mistakes

  • Using the right hand to find the direction of the force on a current-carrying wire.Use Fleming's left-hand rule for the force. Keep the right-hand grip rule for the field around a current.
  • Pointing the first finger from the S pole of the magnet towards its N pole.Remember that field lines run from the north pole to the south pole outside a magnet.
  • Saying the split-ring commutator reverses the direction of rotation.Say it reverses the coil current every half turn, so the coil keeps turning the same way.
  • Saying a transformer will work from a battery.Explain that only a.c. gives a changing field in the core, so a steady d.c. induces nothing in the secondary.
  • Getting 4800 V from a 240 V supply with 1000 primary turns and 50 secondary turns.Write Vs/Vp = Ns/Np with the secondary on top on both sides, then check that a step-down transformer gives a lower voltage.
  • Saying a current flows while a magnet rests inside a coil.Say an e.m.f. is induced only while the field linking the coil changes, through motion or a changing current.

Exam tips

  • Before using Fleming's left-hand rule, mark the field direction and current direction on the diagram, then orient your hand. Write the resulting force direction in words as well as with an arrow.
  • In Lenz's law answers, name the pole formed at the end of the coil nearest the magnet and say that it opposes the motion of the magnet.
  • In transformer calculations, write the equation in symbols first and finish with a sense check, such as whether the voltage should go up or down.

Full guide to electromagnetism, with a quiz and more practice questions.

Syllabus checklist for the rest

What the syllabus asks you to do in the chapters, and parts of chapters, that do not have full notes yet. Each point is the syllabus learning outcome in plain words. Tick each one you can do without looking.

Chapter 1. Physical Quantities, Units and Measurement

4 points to check
  • Recall the base quantities and units: mass (kg), length (m), time (s), current (A), temperature (K) and amount of substance (mol).
  • Use the prefixes nano, micro, milli, centi, deci, kilo, mega, giga and tera.
  • Choose a measuring instrument for a quantity, thinking about its range and precision.
  • State what scalar and vector quantities are, with examples, and add two vectors by a scale drawing.

Chapter 10. General Properties of Waves: also in the syllabus

2 points to check
  • Relate the loudness of a sound to its amplitude and its pitch to its frequency.
  • Describe how an echo forms and how it is used to measure distance, and how ultrasound is used in sonar and in medical scanning of soft tissue.

Chapter 11. Electromagnetic Spectrum

4 points to check
  • State that all electromagnetic waves are transverse and travel at the same speed in a vacuum.
  • Describe the main regions of the spectrum in order of wavelength and frequency.
  • State typical uses of radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays.
  • Describe how over-exposure can harm living cells and tissue, for example by heating or ionising effects.

Chapter 13. Static Electricity

4 points to check
  • State that there are positive and negative charges, measured in coulombs, and that like charges repel while unlike charges attract.
  • Describe an electric field as a region where a charge feels a force, and draw the field of one point charge and between two point charges.
  • Explain charging by rubbing as a transfer of electrons, and describe experiments that show charging by induction.
  • Describe hazards of static charge and uses such as the electrostatic precipitator.

Chapter 16. Practical Electricity

4 points to check
  • Describe the heating effect of electricity in kettles, ovens and heaters.
  • Recall and apply P = VI and E = VIt, and calculate the cost of electricity in kWh.
  • State the hazards of using electricity, and explain fuses, fuse ratings, circuit breakers, earthing and double insulation.
  • State the meaning of live, neutral and earth, describe the wiring of a mains plug, and explain why switches, fuses and circuit breakers go on the live wire.

Chapter 17. Magnetism

3 points to check
  • State the properties of magnets and describe induced magnetism.
  • Distinguish temporary magnets such as iron from permanent magnets such as steel.
  • Use a compass to find the direction of a magnetic field, and draw the field around a bar magnet and between two bar magnets.

Chapter 20. Radioactivity

5 points to check
  • Describe the atom as a positive nucleus of protons and neutrons with electrons around it, and use proton number, nucleon number, isotope and nuclide notation.
  • Show that decay is random and spontaneous, and describe alpha, beta and gamma radiation, including how ionising and penetrating each is.
  • Write decay equations in nuclide notation, and explain background radiation.
  • Use half-life in simple calculations from tables or decay curves, and discuss uses and hazards of radioactivity.
  • State the meaning of nuclear fission and fusion and link them to the energy released by nuclear fuels.

If your child is struggling

  • Find the exact chapter and point. The checklists and common mistakes on this page show where an answer breaks down.
  • Go back to that chapter's worked question and do it again without the steps.
  • Ask the school teacher which chapters the class has covered and which are tested next.
  • If several chapters are weak, work on one at a time, with short timed questions, rather than rereading everything.
  • A tutor can take your child through the weak chapters in order. See O-Level Physics tuition.

Revise with these next

Questions students ask

What do these O-Level Physics notes cover?

The whole syllabus of 20 chapters. 10 chapters have full notes: Kinematics, Dynamics and Newton's Laws, Turning Effect of Forces, Pressure, Energy, Work and Power, Thermal Physics, General Wave Properties, Light: Reflection, Refraction and Lenses, Current Electricity and Electromagnetism. Chapters without full notes yet have a syllabus checklist on this page instead, and get full notes as their topic guides are written.

What is the O-Level Physics exam format?

Paper 1, 1 h, 40 marks. Paper 2, 1 h 45 min, 80 marks. Paper 3, 1 h 50 min, 40 marks. No formula list is printed, so every relationship in the syllabus must be recalled. The full table is under Exam format on this page.

Which syllabus do the O-Level Physics notes follow?

They are written for the 2026 O-Level (6091) and the 2027 SEC G3 Physics (K323). The syllabus documents are linked under Sources at the end of this page.

Is there a formula list in the O-Level Physics exam?

The O-Level Physics paper prints no formula list, so every formula in these notes must be memorised.

Can I print these notes or save them as a PDF?

Yes. Use the print button at the top of the page, then choose Save as PDF. The print layout drops the menus and keeps each chapter readable on A4.

What is the difference between speed and velocity?

Speed is distance travelled per unit time and has size only, so it is a scalar. Velocity is the rate of change of displacement, so it also has a direction. A runner who completes a full lap of a track has a non-zero average speed but zero average velocity, because she finishes where she started.

What is the unit of moment?

The newton metre, N m. It comes from multiplying a force in newtons by a perpendicular distance in metres. If distances are given in centimetres, convert them to metres first, so a 40 N force at 25 cm gives 40 × 0.25 = 10 N m.

Is work done when you hold a heavy bag still?

No work is done on the bag, because it does not move in the direction of the force you apply. Your muscles still convert chemical potential energy to internal energy, which is why your arms tire, but the bag itself gains no energy.

What is the difference between transverse and longitudinal waves?

In a transverse wave, such as light or a water ripple, the particles oscillate at right angles to the direction the energy travels. In a longitudinal wave, such as sound, they oscillate along the direction of travel, forming compressions and rarefactions.

Sources

These are TSS's own notes. If anything here differs from the syllabus document, the syllabus document is right. Tell us and we will fix it.