A Level Physics 9702 — all 25 topics, free.

A complete study guide for Cambridge International AS & A Level Physics 9702, covering all 25 topics and all 76 syllabus sub-topics for exams in 2028, 2029 and 2030.

Sitting exams in 2026 or 2027? You are on the 2026–2027 version. Cambridge states there are no significant changes which affect teaching between the two versions, so this guide covers both — but confirm your exam year with your school.

Topics 1–11 are AS Level (Papers 1, 2 and 3). Topics 12–25 are A2 (Papers 4 and 5) — and Paper 4 also assumes everything from the AS topics. Two of your five papers are practical, so the practical section below is not optional revision.

📄 25 plain-English chapter handouts →✎ Practice & self-test →

CAIE 9702 · exams 2028–203025 topics · 76 sub-topicsAS: Papers 1, 2, 3A Level: adds Papers 4, 535% of the A Level is practical + data-handlingFree & shareable

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The papers

Paper	Length & marks	What it is	Weight
Paper 1 Multiple Choice	1 h 15 min · 40 marks	40 four-option multiple-choice questions on the AS content (topics 1–11).	31% of AS 15.5% of A Level
Paper 2 AS Structured	1 h 15 min · 60 marks	Structured questions on the AS content (topics 1–11).	46% of AS 23% of A Level
Paper 3 Advanced Practical Skills	2 h · 40 marks	A real laboratory paper. Two questions, one hour each, 20 marks each. The context may be outside the syllabus content — you are being tested on skill, not recall.	23% of AS 11.5% of A Level
Paper 4 A Level Structured	2 h · 100 marks	Structured questions on the A2 content (topics 12–25). AS knowledge is still required.	38.5% of A Level
Paper 5 Planning, Analysis and Evaluation	1 h 15 min · 30 marks	A written paper — no equipment. Two questions of 15 marks: plan an experiment, then analyse and evaluate data. Context may be outside the syllabus.	11.5% of A Level

Three routes. AS Level only = Papers 1, 2, 3. A Level staged over two years = Papers 1, 2, 3 in year 1, then Papers 4 and 5 in year 2 (carrying your AS marks forward). A Level in one series = all five papers together.

Add it up: Papers 3 and 5 together are 23% of the full A Level, and they test AO3 — experimental skills — almost exclusively. Students who treat practical work as "the fun lessons" and revise only theory are writing off nearly a quarter of the qualification.

AS Level · Paper 1, 2, 3 · 4 sub-topics

1 · Physical quantities and units

The most under-revised topic in the whole syllabus, and the one that leaks marks into every other. Uncertainty work here is directly examined again in Papers 3 and 5.

1.1Physical quantities

Every physical quantity has a numerical magnitude and a unit. When you estimate a quantity, quote both — "the mass of an adult is about 70 kg", not "about 70".

You are expected to make sensible order-of-magnitude estimates: a person's mass ≈ 70 kg, walking speed ≈ 1.5 m s⁻¹, a car ≈ 1500 kg, room temperature ≈ 293 K, the height of a door ≈ 2 m. Multiple-choice questions use these to eliminate absurd options fast.

1.2SI units

base units: kg, m, s, A, K, mol

All other units are derived from these six. You must be able to express any derived unit in base units and use that to check the homogeneity of an equation — both sides must have the same base units.

Prefixes you must knowpico (10⁻¹²), nano (10⁻⁹), micro (10⁻⁶), milli (10⁻³), centi (10⁻²), deci (10⁻¹), kilo (10³), mega (10⁶), giga (10⁹), tera (10¹²).

Worked example — base units of the pascal

Pressure = force ÷ area, so Pa = N m⁻².
N = kg m s⁻² (from F = ma).
So Pa = kg m s⁻² × m⁻² = kg m⁻¹ s⁻².

Homogeneity is necessary, not sufficientAn equation with matching base units on both sides may still be wrong — a missing factor of ½ or 2π is dimensionless and invisible to the check. Say "the equation is homogeneous", never "the equation is therefore correct".

1.3Errors and uncertainties

adding/subtracting → add absolute uncertainties · multiplying/dividing → add percentage uncertainties · power n → multiply the percentage by n

Systematic error shifts every reading the same way (a zero error, a mis-calibrated scale, a parallax bias). It affects accuracy and is not reduced by repeating. Random error scatters readings either side of the true value. It affects precision and is reduced by averaging repeats.

Accurate vs preciseAccurate = close to the true value. Precise = readings close to each other. A stopwatch started late every time gives precise but inaccurate results.

Worked example — combining uncertainties

A wire has diameter d = 0.40 ± 0.01 mm and length L = 1.500 ± 0.002 m. Find the percentage uncertainty in the cross-sectional area A = πd²/4, and in L/A.

% uncertainty in d = 0.01/0.40 × 100 = 2.5%.
A ∝ d², so % uncertainty in A = 2 × 2.5 = 5.0%.
% uncertainty in L = 0.002/1.500 × 100 = 0.13%.
L/A is a quotient, so add: 0.13 + 5.0 = 5.1% (to 2 s.f.).

Note how the length contributes almost nothing. Improving the ruler would be a waste of effort; the diameter measurement is what limits the experiment. Paper 5 asks exactly this kind of judgement.

The uncertainty of a single ruler reading is two divisions, not oneYou judge both ends of a length against the scale, so a millimetre ruler gives ±1 mm on a length, not ±0.5 mm — unless one end is a fixed zero.

1.4Scalars and vectors

Scalars have magnitude only: distance, speed, mass, time, energy, work, power, pressure, temperature, potential, charge. Vectors have magnitude and direction: displacement, velocity, acceleration, force, weight, momentum, field strength.

resultant of perpendicular vectors: R = √(x² + y²), at angle tan⁻¹(y/x) · components: F_x = F cos θ, F_y = F sin θ

Worked example

A boat is rowed at 3.0 m s⁻¹ due north across a river flowing east at 4.0 m s⁻¹. Find the resultant velocity.

Magnitude: √(3.0² + 4.0²) = √25 = 5.0 m s⁻¹.
Direction: tan⁻¹(4.0/3.0) = 53° east of north.

A vector answer without a direction is incomplete. State it as a bearing, as an angle to a named reference ("53° to the vertical"), or with a sign convention you have defined.

AS Level · 1 sub-topic

2 · Kinematics

2.1Equations of motion

v = u + at · s = ut + ½at² · v² = u² + 2as · s = ½(u + v)t

These apply only to uniform acceleration. Two of them appear on the data sheet; learn all four. Define displacement, speed, velocity and acceleration precisely: acceleration is the rate of change of velocity, which means a change in direction alone is still an acceleration.

Reading graphsOn a displacement–time graph the gradient is velocity. On a velocity–time graph the gradient is acceleration and the area under the line is displacement. Area below the axis counts as negative displacement.

Worked example — projectile motion

A ball is thrown horizontally at 12 m s⁻¹ from a cliff 45 m high. Air resistance is negligible; g = 9.81 m s⁻².

Vertical and horizontal motions are independent. Vertically: u = 0, s = 45 m.
45 = ½ × 9.81 × t² → t² = 9.174 → t = 3.03 s.
Horizontal range = 12 × 3.03 = 36 m (2 s.f.).
Vertical velocity on landing = 9.81 × 3.03 = 29.7 m s⁻¹; resultant speed = √(12² + 29.7²) = 32 m s⁻¹.

Air resistance changes the shape, not just the numbersWith drag, the trajectory is no longer a symmetric parabola: the descent is steeper than the ascent, the range and maximum height are reduced, and the horizontal velocity decreases throughout instead of staying constant.

AS Level · 3 sub-topics

3 · Dynamics

3.1Momentum and Newton's laws of motion

p = mv · F = ma (constant mass) · W = mg

Newton's first law: a body remains at rest or in uniform motion in a straight line unless acted on by a resultant force. Second law: the resultant force is proportional to, and in the direction of, the rate of change of momentum. Third law: when body A exerts a force on body B, B exerts an equal and opposite force of the same type on A.

Newton's second law properly statedF = Δp/Δt. Only when mass is constant does this simplify to F = ma. For a rocket or a falling raindrop that gains mass, use the momentum form.

Third-law pairs act on different bodiesThe weight of a book on a table and the normal contact force from the table are not a third-law pair — they act on the same body and are of different types. The pair to the book's weight is the book's gravitational pull on the Earth.

3.2Non-uniform motion

Air resistance increases with speed. A falling object accelerates, drag grows, the resultant force falls, and the acceleration decreases towards zero: the object reaches terminal velocity when drag equals weight. The velocity–time graph is a curve of decreasing gradient flattening to a horizontal line.

Explain terminal velocity in terms of forces, then resultant force, then acceleration — in that order. "It stops accelerating because of air resistance" is not enough for the marks.

3.3Linear momentum and its conservation

Σp_before = Σp_after (no external resultant force) · elastic: relative speed of approach = relative speed of separation

Momentum is conserved in all collisions. Kinetic energy is conserved only in a perfectly elastic collision; in an inelastic collision some kinetic energy becomes internal energy, and total energy is still conserved.

Worked example

A 0.60 kg trolley moving at 4.0 m s⁻¹ collides with a stationary 0.40 kg trolley and they stick together. Find the common velocity and the kinetic energy lost.

Momentum before = 0.60 × 4.0 = 2.4 kg m s⁻¹.
2.4 = (0.60 + 0.40)v → v = 2.4 m s⁻¹.
KE before = ½ × 0.60 × 4.0² = 4.8 J. KE after = ½ × 1.00 × 2.4² = 2.88 J.
Energy "lost" = 1.9 J, transferred to internal energy and sound.

Momentum is a vector — signs are compulsoryChoose a positive direction, write it down, and give leftward velocities a minus sign. Most collision marks are lost to a dropped sign, not to a wrong method.

AS Level · 3 sub-topics

4 · Forces, density and pressure

4.1Turning effects of forces

moment = F × perpendicular distance from the pivot · torque of a couple = F × perpendicular separation

A couple is a pair of equal, opposite, non-collinear forces. It produces a turning effect with no resultant force, so the body rotates without accelerating linearly.

"Perpendicular distance" means perpendicularIf the force is at angle θ to the arm, the moment is Fd sin θ, not Fd. Sketch the perpendicular from the pivot to the line of action of the force before you substitute.

4.2Equilibrium of forces

equilibrium ⇔ resultant force = 0 and resultant moment about any point = 0

Three coplanar forces in equilibrium must be concurrent (their lines of action meet at a point) and form a closed triangle when drawn head to tail. Use either the closed triangle or resolution into components — the triangle is often faster.

Principle of momentsFor a body in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that point. Choosing the pivot at an unknown force eliminates it from the equation.

Worked example

A uniform beam of weight 120 N and length 4.0 m rests on supports at each end. A 200 N load sits 1.0 m from the left support. Find each support force.

Take moments about the left support: 4.0R = 200 × 1.0 + 120 × 2.0 = 440.
R (right support) = 110 N.
Vertically: L + 110 = 200 + 120 → L = 210 N.

4.3Density and pressure

ρ = m/V · p = F/A · Δp = ρgΔh · upthrust F = ρgV

The hydrostatic pressure difference Δp = ρgΔh follows from the weight of the column of fluid above. Archimedes' principle: the upthrust equals the weight of fluid displaced, and arises because pressure on the bottom face of a submerged body exceeds that on the top face.

Worked example

A block of volume 2.0 × 10⁻³ m³ and mass 1.2 kg is fully submerged in water (ρ = 1000 kg m⁻³). Find the upthrust and the tension in the supporting string.

Upthrust = 1000 × 9.81 × 2.0 × 10⁻³ = 19.6 N.
Weight = 1.2 × 9.81 = 11.8 N.
Upthrust exceeds weight, so the block floats up — a downward string tension of 19.6 − 11.8 = 7.8 N is needed to hold it under.

Δh is vertical depth, not distance along a tubeIn a sloping or U-shaped tube, only the vertical height difference between the two surfaces counts.

AS Level · 2 sub-topics

5 · Work, energy and power

5.1Energy conservation

W = Fs cos θ · P = W/t = Fv · efficiency = useful output ÷ total input

Work is done when the point of application of a force moves in the direction of the force. If the force is perpendicular to the motion — as for the tension in a string in circular motion — no work is done.

Worked example

A car of mass 1200 kg travels at a constant 25 m s⁻¹ against a total resistive force of 600 N. Find the engine's useful output power. If the engine burns fuel at 40 kW, find the efficiency.

Constant speed → driving force = resistive force = 600 N.
P = Fv = 600 × 25 = 15 000 W = 15 kW.
Efficiency = 15/40 = 0.375 = 38%.

5.2Gravitational potential energy and kinetic energy

E_k = ½mv² · ΔE_p = mgΔh (uniform field only)

You should be able to derive both: E_k = ½mv² from W = Fs with v² = u² + 2as, and ΔE_p = mgΔh from work done against weight.

mgΔh only works near the surfaceIt assumes g is constant over the height change. For satellites and escape problems you need topic 13's E_p = −GMm/r instead.

AS Level · 2 sub-topics

6 · Deformation of solids

6.1Stress and strain

σ = F/A · ε = x/L · E = σ/ε · Hooke's law F = kx

A tensile force stretches, a compressive force squashes. Hooke's law holds up to the limit of proportionality. The Young modulus E is the gradient of the straight-line region of a stress–strain graph and is a property of the material, not of the particular specimen — unlike the spring constant k.

Worked example

A steel wire of length 2.00 m and diameter 0.80 mm stretches 1.5 mm under a load of 60 N. Find the Young modulus.

A = π(0.40 × 10⁻³)² = 5.027 × 10⁻⁷ m².
σ = 60 / 5.027 × 10⁻⁷ = 1.194 × 10⁸ Pa.
ε = 1.5 × 10⁻³ / 2.00 = 7.5 × 10⁻⁴.
E = 1.194 × 10⁸ / 7.5 × 10⁻⁴ = 1.6 × 10¹¹ Pa.

The Young modulus experiment is a classic Paper 3 / Paper 5 context. The dominant uncertainty is always the diameter (squared in the area, measured with a micrometer at several points and orientations), and the extension is small — use a vernier scale or a travelling microscope, and a long wire to make the extension measurable.

6.2Elastic and plastic behaviour

elastic strain energy = area under the force–extension graph = ½Fx = ½kx² (Hooke's law region only)

Elastic deformation returns to the original length when the load is removed; plastic deformation does not. The elastic limit is the point beyond which permanent deformation occurs. For a wire loaded past it, the unloading line is parallel to the loading line but offset, and the area between them is the energy dissipated as internal energy.

"Use the graph to estimate the work done" means find the area under the curve — by counting squares if it is not a straight line. Say which method you used.

AS Level · 5 sub-topics

7 · Waves

7.1Progressive waves

v = fλ · f = 1/T · intensity ∝ (amplitude)² · I ∝ 1/r² for a point source

A progressive wave transfers energy without transferring matter. Learn displacement, amplitude, phase difference, period, frequency, wavelength and speed as precise definitions. Intensity is power per unit area, and is proportional to the square of the amplitude — doubling the amplitude quadruples the intensity.

Worked example

A point source emits 60 W uniformly. Find the intensity 3.0 m away, and the ratio of amplitudes at 3.0 m and 6.0 m.

I = P/4πr² = 60 / (4π × 9.0) = 0.53 W m⁻².
Doubling r quarters I; since I ∝ a², the amplitude halves — ratio 2 : 1.

7.2Transverse and longitudinal waves

In a transverse wave the oscillations are perpendicular to the direction of energy transfer (all electromagnetic waves, waves on a string). In a longitudinal wave they are parallel to it (sound), producing compressions and rarefactions.

Sound speed and wave patterns are measured with a cathode-ray oscilloscope: the time-base setting gives the period from the horizontal separation of peaks, and the y-gain gives the amplitude. Read the settings off the question — a common Paper 2 task.

7.3Doppler effect for sound waves

f_o = f_sv / (v ± v_s) — minus when the source approaches

The observed frequency rises as the source approaches (waves bunch up, shorter wavelength) and falls as it recedes. This formula on the data sheet is for a moving source only.

Worked example

An ambulance siren emits 900 Hz and travels at 30 m s⁻¹ towards a stationary observer. Speed of sound = 330 m s⁻¹.

Approaching → denominator v − v_s = 330 − 30 = 300.
f_o = 900 × 330 / 300 = 990 Hz.
Receding: 900 × 330 / 360 = 825 Hz. The pitch drops by 165 Hz as it passes.

The sign is the whole questionApproaching = smaller denominator = higher frequency. Check your answer against that sanity rule before moving on.

7.4Electromagnetic spectrum

all e.m. waves travel at c = 3.00 × 10⁸ m s⁻¹ in free space

In order of increasing wavelength: gamma → X-rays → ultraviolet → visible (400–700 nm) → infrared → microwaves → radio. Gamma rays and X-rays overlap in wavelength and are distinguished by origin — gamma from the nucleus, X-rays from electron transitions or deceleration.

Learn the visible range in both units: 400 nm to 700 nm, i.e. 4 × 10⁻⁷ m to 7 × 10⁻⁷ m. Violet is the short end, red the long end.

7.5Polarisation

Malus's law: I = I₀ cos²θ

Only transverse waves can be polarised — which is why polarisation is evidence that light is transverse and that sound is not. Unpolarised light passing one polariser drops to half intensity; a second polariser at angle θ to the first then transmits I₀ cos²θ.

Worked example

Unpolarised light of intensity 80 W m⁻² passes through a polariser, then an analyser at 60° to it.

After the first polariser: 80 / 2 = 40 W m⁻².
After the analyser: 40 × cos²60° = 40 × 0.25 = 10 W m⁻².

cos²θ, not cos θAnd θ is measured between the transmission axes, not between the light and an axis. Crossed polarisers (θ = 90°) transmit nothing.

AS Level · 4 sub-topics

8 · Superposition

8.1Stationary waves

distance between adjacent nodes = λ/2 · stretched string: f = v/2L for the fundamental

A stationary wave forms when two progressive waves of the same frequency and similar amplitude travel in opposite directions and superpose. Nodes have permanently zero amplitude; antinodes oscillate with maximum amplitude. No energy is transferred along a stationary wave.

Stationary vs progressive — the table examiners wantStationary: amplitude varies with position, energy is not transferred, all particles between adjacent nodes are in phase, wavelength = 2 × node separation. Progressive: constant amplitude, energy transferred, phase varies continuously with position.

Worked example — closed pipe

A tube closed at one end resonates at its fundamental with a 512 Hz tuning fork when the air column is 16.2 cm. Find the speed of sound.

Closed pipe fundamental: L = λ/4, so λ = 4 × 0.162 = 0.648 m.
v = fλ = 512 × 0.648 = 332 m s⁻¹.

A closed pipe has a node at the closed end and an antinode at the open end, so only odd harmonics occur. An open pipe supports all harmonics.

8.2Diffraction

Diffraction is the spreading of a wave as it passes through a gap or around an obstacle. The effect is greatest when the gap is comparable to the wavelength. Demonstrate with a ripple tank.

"Why can you hear round a corner but not see round it?" — sound has a wavelength of order metres, comparable to a doorway, so it diffracts strongly; visible light's wavelength (~5 × 10⁻⁷ m) is far smaller than the gap, so it barely spreads.

8.3Interference

λ = ax/D · constructive: path difference = nλ · destructive: (n + ½)λ

Observable interference requires coherent sources — a constant phase difference, hence the same frequency. In Young's double-slit experiment, a is the slit separation, x the fringe spacing and D the slit-to-screen distance.

Worked example

Double slits 0.50 mm apart are 2.4 m from a screen. Ten fringe spacings measure 30.0 mm. Find the wavelength.

x = 30.0/10 = 3.00 mm = 3.00 × 10⁻³ m.
λ = ax/D = (0.50 × 10⁻³ × 3.00 × 10⁻³) / 2.4
= 6.25 × 10⁻⁷ m = 625 nm (red).

Measure many fringes, not oneMeasuring across ten spacings and dividing reduces the percentage uncertainty tenfold. Note that it is ten spacings, which is eleven fringes — an off-by-one error here is a standard Paper 5 trap.

8.4The diffraction grating

d sin θ = nλ, where d = 1 / (lines per metre)

A grating gives sharper, brighter, more widely separated maxima than double slits, so it measures wavelength far more precisely. White light gives a white zero order and a spectrum for each higher order, with red deviated most (opposite to a prism).

Worked example

A grating has 600 lines per mm. Light of 589 nm is used. How many orders are visible, and at what angle is the second order?

d = 1 / (600 × 10³) = 1.667 × 10⁻⁶ m.
Maximum n: sin θ ≤ 1 → n ≤ d/λ = 1.667 × 10⁻⁶ / 5.89 × 10⁻⁷ = 2.83, so orders 0, 1 and 2 on each side.
sin θ₂ = 2 × 5.89 × 10⁻⁷ / 1.667 × 10⁻⁶ = 0.7068 → θ₂ = 45.0°.

Convert lines-per-mm to d in metres first600 lines per mm is 600 000 lines per metre. Forgetting the factor of 1000 gives an answer a thousand times wrong and is the single most common grating error.

AS Level · 3 sub-topics

9 · Electricity

9.1Electric current

Q = It · I = Anvq (data sheet)

Current is the rate of flow of charge. Charge is quantised in multiples of e = 1.60 × 10⁻¹⁹ C. In I = Anvq, n is the number density of charge carriers and v the mean drift velocity.

Worked example

A copper wire of cross-section 1.0 mm² carries 2.0 A. If n = 8.5 × 10²⁸ m⁻³, find the drift velocity.

v = I/(Anq) = 2.0 / (1.0 × 10⁻⁶ × 8.5 × 10²⁸ × 1.60 × 10⁻¹⁹)
= 2.0 / (1.36 × 10⁴) = 1.5 × 10⁻⁴ m s⁻¹ — about 0.15 mm per second.

The lamp still lights instantly because the electric field propagates near the speed of light, setting all the electrons in the circuit moving at once. Drift speed and signal speed are different things.

9.2Potential difference and power

V = W/Q · P = VI = I²R = V²/R

The p.d. between two points is the energy transferred from electrical to other forms per unit charge. E.m.f. is the energy transferred to electrical form per unit charge. None of the power equations are on the data sheet.

9.3Resistance and resistivity

R = V/I · R = ρL/A

Ohm's law: the current through a metallic conductor is proportional to the p.d. across it, provided the temperature is constant. Know the I–V characteristics of a metallic conductor (straight line through the origin), a filament lamp (curve flattening as temperature and resistance rise), a semiconductor diode (almost no current when reverse biased, sharp rise above ~0.6 V forward) and a thermistor (resistance falls as temperature rises).

Worked example

A nichrome wire of length 1.50 m and diameter 0.30 mm has resistivity 1.1 × 10⁻⁶ Ω m. Find its resistance.

A = π(0.15 × 10⁻³)² = 7.069 × 10⁻⁸ m².
R = 1.1 × 10⁻⁶ × 1.50 / 7.069 × 10⁻⁸ = 23 Ω.

Resistivity is a material property; resistance is notCutting a wire in half halves its resistance but leaves the resistivity unchanged. Stretching a wire to double its length quarters its area (volume is constant), so the resistance goes up by a factor of four, not two.

AS Level · 3 sub-topics

10 · D.C. circuits

10.1Practical circuits

terminal p.d. V = E − Ir · "lost volts" = Ir

A real source has internal resistance r. Maximum power is delivered to the external circuit when the load resistance equals r. A graph of terminal p.d. against current is a straight line of gradient −r and intercept E.

Worked example

A cell of e.m.f. 1.50 V and internal resistance 0.60 Ω drives a 2.40 Ω lamp. Find the current and the terminal p.d.

I = E/(R + r) = 1.50 / 3.00 = 0.50 A.
V = IR = 0.50 × 2.40 = 1.20 V. Lost volts = 0.30 V.

10.2Kirchhoff's laws

ΣI_in = ΣI_out (charge conservation) · ΣE = ΣIR round any loop (energy conservation)

Derive the series and parallel combination rules from these two laws — a standard Paper 2 request. Series: R = R₁ + R₂. Parallel: 1/R = 1/R₁ + 1/R₂ (both on the data sheet).

The parallel formula gives 1/RStudents routinely stop after summing the reciprocals and quote that as the resistance. Invert at the end — and sanity-check: the combined parallel resistance is always smaller than the smallest branch.

10.3Potential dividers

V_out = V_in × R₂/(R₁ + R₂)

A potential divider supplies a variable fraction of a supply p.d. Replacing one resistor with a thermistor gives a temperature-sensitive output; with an LDR, a light-sensitive output. A potentiometer is a continuously variable divider.

Worked example

A 12 V supply drives a divider of a 4.0 kΩ fixed resistor in series with a thermistor. The thermistor is 8.0 kΩ when cold and 2.0 kΩ when hot. Find V across the fixed resistor in each case.

Cold: 12 × 4.0/(4.0 + 8.0) = 4.0 V.
Hot: 12 × 4.0/(4.0 + 2.0) = 8.0 V.
So heating the thermistor raises the output — the basis of a temperature-triggered switch.

Connecting a voltmeter changes the circuitA real voltmeter has finite resistance in parallel with R₂, lowering the combined resistance and therefore lowering the measured output. This is a systematic error, and an excellent Paper 5 evaluation point.

AS Level · 2 sub-topics · last AS topic

11 · Particle physics

11.1Atoms, nuclei and radiation

nuclide notation: ^A_ZX · α: ⁴₂He · β⁻: ⁰₋₁e · β⁺: ⁰₊₁e

The α-particle scattering experiment showed the atom is mostly empty space with a tiny, dense, positively charged nucleus: most α-particles passed straight through, a few deflected, a very few came back.

	Nature	Charge	Stopped by	Ionising
α	helium nucleus	+2e	paper / few cm air	very strong
β	fast electron (or positron)	∓e	few mm aluminium	moderate
γ	e.m. photon	0	cm of lead (never fully)	weak

Both nucleon number and proton number are conserved in every decay equation. β⁻ decay is a neutron becoming a proton, emitting an electron and an antineutrino; β⁺ decay is a proton becoming a neutron, emitting a positron and a neutrino. The existence of the (anti)neutrino was inferred from the continuous energy spectrum of β-particles — energy appeared to be missing.

11.2Fundamental particles

proton = uud · neutron = udd · quark charges: u, c, t = +⅔e · d, s, b = −⅓e

Hadrons are made of quarks and feel the strong force: baryons (three quarks — proton, neutron) and mesons (a quark and an antiquark). Leptons — the electron and the neutrino — are fundamental and do not feel the strong force. Antiquarks have the opposite sign of charge.

Worked example — check the quark composition

Proton uud: +⅔ + ⅔ − ⅓ = +1 ✓
Neutron udd: +⅔ − ⅓ − ⅓ = 0 ✓
β⁻ decay is d → u + e⁻ + ν̄. Charge: −⅓ → +⅔ + (−1) = −⅓ ✓

You will be asked to balance quark equations by charge and to identify a decay as involving the weak interaction — any process that changes quark flavour (β decay) is weak, not strong.

A Level · Paper 4 · 2 sub-topics

12 · Motion in a circle

Topics 12–25 are the A2 content, examined in Paper 4. Paper 4 still assumes everything from topics 1–11.

12.1Kinematics of uniform circular motion

θ in radians = arc/radius · 2π rad = 360° · v = rω · ω = 2π/T = 2πf

In uniform circular motion the speed is constant but the velocity is not, because the direction changes continuously. A changing velocity means an acceleration, which means a resultant force.

12.2Centripetal acceleration

a = rω² = v²/r · F = mrω² = mv²/r, directed towards the centre

The centripetal force is not a new force — it is the name for whatever resultant force points to the centre: tension for a whirled mass, friction for a car on a bend, gravity for a satellite, the normal force component for a banked track.

Worked example

A 0.20 kg ball on a 0.80 m string is whirled in a vertical circle at 4.0 revolutions per second. Find the tension at the top of the circle.

ω = 2π × 4.0 = 25.13 rad s⁻¹.
Required centripetal force = mrω² = 0.20 × 0.80 × 25.13² = 101.1 N.
At the top, weight (0.20 × 9.81 = 1.96 N) points towards the centre and helps: T + 1.96 = 101.1.
T = 99 N. At the bottom, weight opposes: T = 101.1 + 1.96 = 103 N.

There is no "centrifugal force" in a 9702 answerWrite the resultant of the real forces and set it equal to mv²/r. Adding an outward force to your free-body diagram will cost you the mark.

A Level · 4 sub-topics

13 · Gravitational fields

13.1Gravitational field

g = F/m — force per unit mass

A gravitational field is a region in which a mass experiences a force. Near the Earth's surface the field is effectively uniform, with parallel equally-spaced field lines; on a planetary scale it is radial, with lines converging on the centre.

13.2Gravitational force between point masses

F = Gm₁m₂/r² · G = 6.67 × 10⁻¹¹ N m² kg⁻²

Newton's law of gravitation: the attractive force between two point masses is proportional to the product of their masses and inversely proportional to the square of their separation. A uniform sphere behaves as a point mass at its centre.

13.3Gravitational field of a point mass

g = GM/r² · geostationary: T = 24 h, equatorial, west→east

Combining GMm/r² = mrω² gives the orbital relationships. Notice the orbiting mass cancels: every satellite at a given radius has the same period and speed, regardless of mass.

Worked example — geostationary orbit radius

M_Earth = 6.0 × 10²⁴ kg, T = 24 h = 86 400 s.

GM/r² = rω² → r³ = GM/ω² = GMT²/4π².
ω = 2π/86400 = 7.272 × 10⁻⁵ rad s⁻¹, ω² = 5.288 × 10⁻⁹.
r³ = (6.67 × 10⁻¹¹ × 6.0 × 10²⁴)/5.288 × 10⁻⁹ = 7.568 × 10²².
r = 4.2 × 10⁷ m — about 36 000 km above the surface.

r is measured from the centre of the planetNot from the surface. Add the planetary radius to any "height above the surface" before substituting.

13.4Gravitational potential

φ = −GM/r · E_p = −GMm/r (both on the Paper 4 data sheet)

Gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point. It is always negative because gravity is attractive and infinity is defined as the zero — work is done by the field as the mass approaches.

Worked example — escape speed

Show that the escape speed from a planet of mass M and radius R is √(2GM/R), and evaluate it for Earth (M = 6.0 × 10²⁴ kg, R = 6.4 × 10⁶ m).

To just escape, total energy = 0: ½mv² − GMm/R = 0.
v = √(2GM/R) = √(2 × 6.67 × 10⁻¹¹ × 6.0 × 10²⁴ / 6.4 × 10⁶)
= √(1.251 × 10⁸) = 1.1 × 10⁴ m s⁻¹ ≈ 11 km s⁻¹.

The escaping mass cancels — escape speed is independent of the mass of the object.

The minus sign is worth marks. "Gravitational potential is negative because the zero is at infinity and gravitational forces are attractive" is the sentence examiners want.

A Level · 3 sub-topics

14 · Temperature

14.1Thermal equilibrium

Two bodies are in thermal equilibrium when there is no net flow of energy between them — which happens when they are at the same temperature. Temperature is what determines the direction of energy flow, and is a measure of the mean kinetic energy of the molecules, not of the total energy.

Temperature is not "how much heat something has"A bathtub of warm water contains far more internal energy than a red-hot needle, but the needle is at the higher temperature. Energy flows from high to low temperature, not from more to less energy.

14.2Temperature scales

T/K = θ/°C + 273.15 · absolute zero = 0 K = −273.15 °C

The thermodynamic (Kelvin) scale does not depend on the property of any particular substance. Absolute zero is the temperature at which a substance has minimum internal energy. Any physical property that varies with temperature — resistance of a thermistor, e.m.f. of a thermocouple, volume of a liquid — can serve as a thermometric property.

Every gas-law and thermodynamics calculation uses kelvin. Converting is a single mark, but forgetting to convert corrupts everything downstream.

14.3Specific heat capacity and specific latent heat

Q = mcΔθ · Q = mL

Specific heat capacity is the energy needed to raise the temperature of unit mass by one kelvin. Specific latent heat is the energy needed to change the state of unit mass with no change of temperature — the energy goes into breaking intermolecular bonds (increasing potential energy), not into kinetic energy.

Worked example

How much energy converts 0.50 kg of ice at 0 °C into water at 40 °C? (L_f = 3.34 × 10⁵ J kg⁻¹, c_water = 4200 J kg⁻¹ K⁻¹)

Melting: Q₁ = 0.50 × 3.34 × 10⁵ = 1.67 × 10⁵ J.
Heating: Q₂ = 0.50 × 4200 × 40 = 8.4 × 10⁴ J.
Total = 2.5 × 10⁵ J. Note that melting the ice takes twice as much energy as heating the resulting water by 40 K.

In the electrical method for c, the systematic error is energy lost to the surroundings, which makes the measured c too large. Reduce it with lagging, a lid, and by starting below and ending above room temperature so gains and losses roughly cancel.

A Level · 3 sub-topics

15 · Ideal gases

15.1The mole

N = nN_A · N_A = 6.02 × 10²³ mol⁻¹ · 1 u = 1.66 × 10⁻²⁷ kg

One mole contains as many particles as there are atoms in 12 g of carbon-12. Amount of substance is measured in moles.

15.2Equation of state

pV = nRT = NkT · R = 8.31 J K⁻¹ mol⁻¹ · k = 1.38 × 10⁻²³ J K⁻¹

An ideal gas obeys pV ∝ T at all pressures. Real gases approximate this at low pressure and high temperature, where the molecules are far apart and intermolecular forces are negligible.

Worked example

A cylinder holds 0.25 m³ of gas at 1.8 × 10⁵ Pa and 300 K. Find the number of moles, then the pressure if it is heated to 450 K at constant volume.

n = pV/RT = (1.8 × 10⁵ × 0.25)/(8.31 × 300) = 18 mol.
Constant V: p₂ = p₁T₂/T₁ = 1.8 × 10⁵ × 450/300 = 2.7 × 10⁵ Pa.

15.3Kinetic theory of gases

p = ⅓ (Nm/V)⟨c²⟩ · ½m⟨c²⟩ = (3/2)kT

The assumptions: molecules are point particles of negligible volume, in random motion, colliding elastically with each other and the walls, with negligible intermolecular forces except during collisions, and the time of a collision is negligible compared with the time between collisions.

The key result — mean translational kinetic energy is directly proportional to thermodynamic temperature — comes from comparing pV = ⅓Nm⟨c²⟩ with pV = NkT.

Worked example

Find the r.m.s. speed of nitrogen molecules (m = 4.65 × 10⁻²⁶ kg) at 300 K.

½m⟨c²⟩ = 1.5 × 1.38 × 10⁻²³ × 300 = 6.21 × 10⁻²¹ J.
⟨c²⟩ = 2 × 6.21 × 10⁻²¹ / 4.65 × 10⁻²⁶ = 2.671 × 10⁵ m² s⁻².
c_r.m.s. = 517 m s⁻¹.

⟨c²⟩ is the mean of the squaresThe root-mean-square speed is √⟨c²⟩, which is not the mean speed. Take the square root only at the very end.

A Level · 2 sub-topics

16 · Thermodynamics

16.1Internal energy

Internal energy is determined by the state of the system, and is the sum of a random distribution of the kinetic and potential energies of the molecules. A rise in temperature means a rise in internal energy, because the molecular kinetic energies increase.

Why does an ideal gas have no molecular potential energy?Because the model assumes negligible intermolecular forces. So for an ideal gas, internal energy is entirely kinetic and depends only on temperature — which is why an isothermal change of an ideal gas has ΔU = 0.

16.2The first law of thermodynamics

ΔU = q + W · W = pΔV (constant pressure)

ΔU is the increase in internal energy, q the energy transferred to the system by heating, W the work done on the system. Sign convention matters: a gas that expands does work on its surroundings, so W is negative.

Worked example

A gas at a constant pressure of 1.0 × 10⁵ Pa expands from 2.0 × 10⁻³ m³ to 5.0 × 10⁻³ m³ while 800 J is supplied by heating. Find the change in internal energy.

Work done by the gas = pΔV = 1.0 × 10⁵ × 3.0 × 10⁻³ = 300 J.
So work done on the gas, W = −300 J.
ΔU = q + W = 800 − 300 = +500 J.

Get the direction of W right, every timeCompression → work done on the gas → W positive. Expansion → W negative. In an isothermal ideal-gas change ΔU = 0, so q = −W. In an adiabatic change q = 0, so ΔU = W — compressing a gas adiabatically raises its temperature.

A Level · 3 sub-topics

17 · Oscillations

17.1Simple harmonic oscillations

a = −ω²x · x = x₀ sin ωt · v = v₀ cos ωt · v = ±ω√(x₀² − x²) · v₀ = ωx₀

S.h.m. occurs when the acceleration is proportional to the displacement from a fixed point and directed towards that point. That definition — including "in the opposite direction" — is examined verbatim.

Graph relationshipsDisplacement, velocity and acceleration are each quarter of a cycle (90°) out of phase with the next. Acceleration is 180° out of phase with displacement — it is a negative multiple of it. Velocity is maximum at the centre (x = 0) and zero at the extremes; acceleration is the reverse.

Worked example

A mass oscillates with amplitude 5.0 cm and period 0.40 s. Find the maximum speed and the speed at x = 3.0 cm.

ω = 2π/0.40 = 15.71 rad s⁻¹.
v₀ = ωx₀ = 15.71 × 0.050 = 0.79 m s⁻¹.
v = ω√(0.050² − 0.030²) = 15.71 × 0.040 = 0.63 m s⁻¹.
Maximum acceleration = ω²x₀ = 246.7 × 0.050 = 12 m s⁻².

17.2Energy in simple harmonic motion

E_total = ½mω²x₀² · E_k = ½mω²(x₀² − x²) · E_p = ½mω²x²

Total energy is constant (undamped) and proportional to the square of the amplitude. Both E_k and E_p vary with x² — so plotted against displacement they are parabolas, and plotted against time they oscillate at twice the frequency of the motion.

17.3Damped and forced oscillations, resonance

Light damping: amplitude decays exponentially over many oscillations. Critical damping: returns to equilibrium in the shortest time without oscillating. Heavy (over)damping: returns slowly without oscillating.

Resonance occurs when the driving frequency equals the natural frequency of the system; the amplitude is then a maximum and energy transfer from driver to system is greatest. Increasing damping lowers the peak, broadens it, and shifts it slightly to a lower frequency.

Useful resonance: microwave ovens, MRI, radio tuning, musical instruments. Destructive resonance: bridges under marching feet or wind, buildings in earthquakes. Have one example of each ready.

A Level · 5 sub-topics

18 · Electric fields

18.1Electric fields and field lines

E = F/Q — force per unit positive charge, in N C⁻¹ or V m⁻¹

Field lines run from positive to negative, never cross, and their density indicates field strength. Know the patterns for an isolated point charge (radial), two like charges (a neutral point between them), two unlike charges, and a charged sphere.

18.2Uniform electric fields

E = ΔV/d between parallel plates

Between parallel plates the field is uniform. A charged particle entering perpendicular to the field follows a parabolic path — exactly like projectile motion, with the constant electric force playing the role of weight.

Worked example

Plates 4.0 cm apart have a p.d. of 2000 V. An electron enters midway, parallel to the plates, at 6.0 × 10⁶ m s⁻¹. Find its vertical deflection after travelling 5.0 cm horizontally.

E = 2000/0.040 = 5.0 × 10⁴ V m⁻¹.
a = Ee/m = (5.0 × 10⁴ × 1.60 × 10⁻¹⁹)/9.11 × 10⁻³¹ = 8.78 × 10¹⁵ m s⁻².
t = 0.050/6.0 × 10⁶ = 8.333 × 10⁻⁹ s.
y = ½at² = 0.5 × 8.78 × 10¹⁵ × (8.333 × 10⁻⁹)² = 0.305 m.
But the electron started midway, only 2.0 cm from a plate — so this answer is impossible. The electron hits the plate first. Setting y = 0.020 m gives t = √(2 × 0.020/8.78 × 10¹⁵) = 2.13 × 10⁻⁹ s, so it travels only 1.3 cm horizontally before striking the plate.

Always sanity-check a deflection against the plate separation. Examiners set this trap deliberately.

18.3Electric force between point charges

F = Q₁Q₂ / 4πε₀r² · 1/4πε₀ = 8.99 × 10⁹ m F⁻¹

Coulomb's law. Unlike gravitation, the force can be attractive or repulsive, and it is vastly stronger: the electrostatic repulsion between two protons is about 10³⁶ times their gravitational attraction.

18.4Electric field of a point charge

E = Q / 4πε₀r²

Radial and inverse-square, exactly parallel to g = GM/r². Learn the two field families side by side — the algebra is identical and the analogy is frequently examined.

18.5Electric potential

V = Q / 4πε₀r · E_p = Qq / 4πε₀r · E = −dV/dr

Electric potential at a point is the work done per unit positive charge in bringing a small test charge from infinity to that point. Unlike gravitational potential it can be positive or negative — positive near a positive charge, negative near a negative one.

Worked example — closest approach

An α-particle (charge +2e, kinetic energy 5.0 MeV) is fired at a gold nucleus (Z = 79). Find the distance of closest approach.

E_k = 5.0 × 10⁶ × 1.60 × 10⁻¹⁹ = 8.0 × 10⁻¹³ J.
At closest approach all kinetic energy has become electrical potential energy: E_k = Qq/4πε₀r.
Qq = 2 × 79 × (1.60 × 10⁻¹⁹)² = 4.045 × 10⁻³⁶ C².
r = 8.99 × 10⁹ × 4.045 × 10⁻³⁶ / 8.0 × 10⁻¹³ = 4.5 × 10⁻¹⁴ m.

That is about 45 fm — comfortably larger than a nuclear radius, which is why the α-particle is turned back by the Coulomb force without ever touching the nucleus.

Potential goes as 1/r; field goes as 1/r²Mixing the two powers is the most common error in topics 13 and 18 combined. Field is the gradient of potential, which is why it carries the extra power of r.

A Level · 3 sub-topics

19 · Capacitance

19.1Capacitors and capacitance

C = Q/V · parallel: C = C₁ + C₂ · series: 1/C = 1/C₁ + 1/C₂

Capacitance is the charge stored per unit potential difference, in farads (C V⁻¹). Note the combination rules are the opposite way round from resistors: capacitors in parallel add, capacitors in series combine reciprocally.

"Charge stored" means separated, not accumulatedA charged capacitor holds +Q on one plate and −Q on the other; the net charge is zero. Say "charge of magnitude Q on each plate".

19.2Energy stored in a capacitor

W = ½QV = ½CV² = ½Q²/C

The energy is the area under a charge–p.d. graph, which is a triangle because V rises linearly as charge is added. The factor of ½ is there because the early charge is moved across a smaller p.d. than the last.

Worked example

A 470 µF capacitor is charged to 12 V. Find the charge and stored energy. It is then connected to an identical uncharged capacitor — find the new p.d. and the energy now stored.

Q = CV = 470 × 10⁻⁶ × 12 = 5.64 × 10⁻³ C.
W = ½ × 470 × 10⁻⁶ × 144 = 3.4 × 10⁻² J.
Charge is conserved and shared: total C = 940 µF, so V = 5.64 × 10⁻³ / 940 × 10⁻⁶ = 6.0 V.
New energy = ½ × 940 × 10⁻⁶ × 36 = 1.7 × 10⁻² J — half the energy has been dissipated in the connecting wires.

19.3Discharging a capacitor

x = x₀e^−t/RC — works for Q, V and I · time constant τ = RC

After one time constant the quantity has fallen to 1/e ≈ 37% of its initial value. The half-life of the decay is t_½ = RC ln 2 ≈ 0.693RC.

Worked example

A 100 µF capacitor charged to 9.0 V discharges through 47 kΩ. Find the time constant and the p.d. after 10 s.

τ = RC = 47 × 10³ × 100 × 10⁻⁶ = 4.7 s.
V = 9.0 × e^−10/4.7 = 9.0 × e^−2.128 = 9.0 × 0.1191 = 1.1 V.
Time to fall to half: 0.693 × 4.7 = 3.3 s.

To find RC experimentally, plot ln V against t: the gradient is −1/RC. Linearising an exponential by taking logarithms is a Paper 5 staple — expect to be asked to state what to plot and how to get the constant from the gradient and intercept.

A Level · 5 sub-topics

20 · Magnetic fields

20.1Concept of a magnetic field

A magnetic field is a region in which a force acts on a moving charge, a current-carrying conductor, or a magnetic pole. It is produced by moving charges (currents) and by permanent magnets.

20.2Force on a current-carrying conductor

F = BIL sin θ · 1 T = 1 N A⁻¹ m⁻¹

Magnetic flux density B is defined by the force per unit current per unit length on a conductor at right angles to the field. Use Fleming's left-hand rule: First finger = Field, seCond = Current, thuMb = Motion (force).

θ is the angle between the current and the fieldA conductor parallel to the field experiences no force (sin 0 = 0). Maximum force is at 90°.

20.3Force on a moving charge

F = BQv sin θ · Hall voltage V_H = BI/(ntq) · velocity selector: v = E/B

The magnetic force is always perpendicular to the velocity, so it does no work and the speed is unchanged — the charge moves in a circle of radius r = mv/BQ. In a velocity selector, crossed electric and magnetic fields let through only particles for which EQ = BQv, i.e. v = E/B — independent of charge and mass.

Worked example

A proton (m = 1.67 × 10⁻²⁷ kg) moves at 2.0 × 10⁶ m s⁻¹ perpendicular to a 0.35 T field. Find the radius of its path.

BQv = mv²/r → r = mv/BQ.
r = (1.67 × 10⁻²⁷ × 2.0 × 10⁶)/(0.35 × 1.60 × 10⁻¹⁹) = 0.060 m.

20.4Magnetic fields due to currents

Field patterns: a long straight wire gives concentric circles (right-hand grip rule); a flat circular coil gives a field like a short bar magnet; a long solenoid gives a uniform field inside, like a bar magnet outside. A ferrous core greatly increases the solenoid's field.

Two parallel wires carrying current in the same direction attract; in opposite directions they repel. Explain it by taking the field of one wire at the position of the other and applying the left-hand rule.

20.5Electromagnetic induction

Φ = BA · flux linkage = NΦ · E = −N dΦ/dt

Magnetic flux is the product of magnetic flux density and the cross-sectional area perpendicular to the field. Faraday's law: the magnitude of the induced e.m.f. is proportional to the rate of change of magnetic flux linkage. Lenz's law: the induced e.m.f. acts in such a direction as to oppose the change producing it — a direct consequence of conservation of energy.

Worked example

A 250-turn coil of area 4.0 × 10⁻³ m² sits in a field that falls uniformly from 0.80 T to zero in 0.20 s. Find the induced e.m.f.

Initial flux linkage = 250 × 0.80 × 4.0 × 10⁻³ = 0.80 Wb-turns.
E = Δ(NΦ)/Δt = 0.80/0.20 = 4.0 V.

The classic experiments: dropping a magnet through a coil connected to a galvanometer (faster drop → larger deflection), and reversing the deflection by reversing the magnet's direction. Always mention rate of change of flux linkage and opposing the change — those two phrases carry the marks.

A Level · 2 sub-topics

21 · Alternating currents

21.1Characteristics of alternating currents

x = x₀ sin ωt · I_r.m.s. = I₀/√2 · V_r.m.s. = V₀/√2 · mean power = ½ peak power

The root-mean-square value of an alternating current is the value of the direct current that dissipates power in a resistor at the same mean rate. The √2 relationship holds only for a sinusoidal waveform.

Worked example

Mains supply is quoted as 230 V. Find the peak voltage and the mean power dissipated in a 60 Ω heater.

Quoted mains voltage is r.m.s.: V₀ = 230√2 = 325 V.
Mean power = V_r.m.s.²/R = 230²/60 = 882 W.
Peak power = V₀²/R = 325²/60 = 1764 W — exactly twice the mean.

21.2Rectification and smoothing

Half-wave rectification uses a single diode: the negative half-cycles are removed. Full-wave rectification uses a bridge of four diodes: the negative half-cycles are inverted, so the output has twice as many pulses.

A capacitor in parallel with the load smooths the output: it charges at the peak and discharges through the load between peaks, leaving a small ripple. A larger capacitance or a larger load resistance (i.e. a larger RC compared with the period) gives less ripple.

Be ready to sketch input and output waveforms on the same time axis for half-wave, full-wave, and smoothed full-wave — and to explain the smoothing in terms of the capacitor discharging with time constant RC.

A Level · 4 sub-topics

22 · Quantum physics

22.1Energy and momentum of a photon

E = hf = hc/λ · p = h/λ · 1 eV = 1.60 × 10⁻¹⁹ J

Electromagnetic radiation is emitted and absorbed in discrete quanta called photons. The electronvolt is the energy gained by an electron accelerated through 1 V.

Worked example

Find the energy of a 500 nm photon in joules and in electronvolts.

E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/5.00 × 10⁻⁷ = 3.98 × 10⁻¹⁹ J.
In eV: 3.98 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2.49 eV.

22.2Photoelectric effect

hf = Φ + ½mv²_max · threshold: hf₀ = Φ

Four observations that wave theory cannot explain, and which the photon model does:

There is a threshold frequency below which no electrons are emitted, however intense the light.
Emission is instantaneous — no time lag while energy accumulates.
The maximum kinetic energy depends on frequency, not intensity.
Intensity affects the rate of emission (the number of photoelectrons per second), not their energy.

The work function Φ is the minimum energy needed to remove an electron from the surface of the metal.

Worked example

A metal has work function 2.30 eV. Light of wavelength 400 nm falls on it. Find the maximum kinetic energy of the photoelectrons and the threshold wavelength.

Photon energy = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/4.00 × 10⁻⁷ = 4.973 × 10⁻¹⁹ J = 3.11 eV.
E_k,max = 3.11 − 2.30 = 0.81 eV = 1.29 × 10⁻¹⁹ J.
Φ = 2.30 × 1.60 × 10⁻¹⁹ = 3.68 × 10⁻¹⁹ J; λ₀ = hc/Φ = 1.989 × 10⁻²⁵/3.68 × 10⁻¹⁹ = 541 nm.

One photon, one electronAn electron cannot accumulate energy from several photons. That is exactly why a threshold frequency exists and why brighter light of too low a frequency still emits nothing.

22.3Wave–particle duality

de Broglie: λ = h/p = h/mv

Electron diffraction is the evidence that particles behave as waves: a beam of electrons through a thin graphite film produces diffraction rings, and the ring spacing changes with accelerating voltage exactly as λ = h/mv predicts.

Worked example

Find the de Broglie wavelength of an electron accelerated through 100 V.

E_k = eV = 1.60 × 10⁻¹⁹ × 100 = 1.60 × 10⁻¹⁷ J.
v = √(2E_k/m) = √(3.20 × 10⁻¹⁷/9.11 × 10⁻³¹) = 5.93 × 10⁶ m s⁻¹.
λ = h/mv = 6.63 × 10⁻³⁴/(9.11 × 10⁻³¹ × 5.93 × 10⁶) = 1.2 × 10⁻¹⁰ m.

That is comparable to atomic spacings, which is why a crystal acts as a diffraction grating for electrons.

22.4Energy levels in atoms and line spectra

hf = E₁ − E₂

Electrons in an atom occupy discrete energy levels, conventionally negative with the ground state lowest. A photon is emitted when an electron falls between levels, with energy exactly equal to the level difference — which is why spectra are lines, not continuous bands. An emission spectrum is bright lines on a dark background; an absorption spectrum is dark lines at the same wavelengths on a continuous background.

Worked example

An electron falls from −3.40 eV to −13.6 eV. Find the wavelength of the emitted photon.

ΔE = 13.6 − 3.40 = 10.2 eV = 10.2 × 1.60 × 10⁻¹⁹ = 1.632 × 10⁻¹⁸ J.
λ = hc/ΔE = 1.989 × 10⁻²⁵/1.632 × 10⁻¹⁸ = 1.22 × 10⁻⁷ m = 122 nm (ultraviolet).

The existence of discrete lines is the evidence for discrete energy levels. Say that explicitly — it is the mark, not the arithmetic.

A Level · 2 sub-topics

23 · Nuclear physics

23.1Mass defect and nuclear binding energy

E = mc² · 1 u = 931.5 MeV

Mass defect is the difference between the mass of the separate nucleons and the mass of the nucleus. Binding energy is the energy required to separate a nucleus into its constituent nucleons — the energy equivalent of the mass defect.

The binding energy per nucleon curve rises steeply to a maximum near iron-56 (about 8.8 MeV per nucleon) and falls slowly after. Energy is released by fusion of light nuclei and by fission of heavy nuclei — in both cases because the products lie higher on the curve.

Worked example — binding energy of helium-4

Masses: proton 1.00728 u, neutron 1.00867 u, ⁴He nucleus 4.00151 u.

Separate nucleons: 2(1.00728) + 2(1.00867) = 4.03190 u.
Mass defect = 4.03190 − 4.00151 = 0.03039 u.
Binding energy = 0.03039 × 931.5 = 28.3 MeV.
Per nucleon = 28.3/4 = 7.08 MeV.

Binding energy per nucleon, not total, decides stabilityUranium-235 has a much larger total binding energy than iron-56, yet it is far less stable. Always divide by A before comparing.

23.2Radioactive decay

A = λN · x = x₀e^−λt · λ = 0.693/t_½

Radioactive decay is random (you cannot predict which nucleus decays next) and spontaneous (unaffected by temperature, pressure or chemical state). The decay constant λ is the probability of decay per unit time for a nucleus. Activity is measured in becquerels (1 Bq = 1 decay per second).

Worked example

A sample of iodine-131 (t_½ = 8.0 days) has an initial activity of 4.0 × 10⁵ Bq. Find the decay constant, the number of nuclei present, and the activity after 30 days.

λ = 0.693/(8.0 × 24 × 3600) = 1.002 × 10⁻⁶ s⁻¹.
N = A/λ = 4.0 × 10⁵/1.002 × 10⁻⁶ = 3.99 × 10¹¹ nuclei.
After 30 days = 3.75 half-lives: A = 4.0 × 10⁵ × e^{−0.693 × 3.75} = 4.0 × 10⁵ × 0.0743 = 3.0 × 10⁴ Bq.

Keep λ and t in the same unit of timeIf λ is in s⁻¹, t must be in seconds. Working in half-lives (activity halves each one) avoids the conversion entirely when the time is a whole number of half-lives.

A Level · 3 sub-topics

24 · Medical physics

24.1Production and use of ultrasound

Z = ρc (acoustic impedance) · I_R/I₀ = (Z₁ − Z₂)²/(Z₁ + Z₂)² · I = I₀e^−µx

Ultrasound is generated and detected by a piezo-electric transducer: a p.d. across a quartz crystal deforms it, and an alternating p.d. at the crystal's resonant frequency makes it oscillate, emitting ultrasound. The same crystal converts returning pulses back into a p.d.

A coupling gel is essential: the air–skin impedance mismatch is so large that almost all the ultrasound would be reflected at the surface. The gel has an impedance close to skin, so the pulse enters the body.

Worked example

Z(soft tissue) = 1.63 × 10⁶ kg m⁻² s⁻¹, Z(bone) = 6.40 × 10⁶. Find the fraction reflected at a tissue–bone boundary.

Difference = 4.77 × 10⁶; sum = 8.03 × 10⁶.
Ratio = (4.77/8.03)² = 0.594² = 0.35 — about 35% reflected.

Compare with air–tissue, where the ratio is over 0.999: essentially total reflection. That single number explains why the gel exists.

24.2Production and use of X-rays

I = I₀e^−µx · half-value thickness x_½ = ln2/µ

In an X-ray tube, electrons are accelerated through a large p.d. onto a metal target. Most energy becomes internal energy (hence the rotating, cooled anode); a small fraction becomes X-rays — a continuous bremsstrahlung spectrum with sharp characteristic lines from electron transitions in the target atoms.

Contrast is improved by contrast media (barium, iodine) with high atomic number and hence high attenuation. Sharpness is improved by a small focal spot. A CT scan builds a three-dimensional image from many one-dimensional attenuation profiles taken at different angles around the patient, reconstructed by computer.

Worked example

µ = 0.25 cm⁻¹ for a tissue. Find the half-value thickness and the fraction transmitted through 8.0 cm.

x_½ = 0.693/0.25 = 2.8 cm.
I/I₀ = e^{−0.25 × 8.0} = e^−2.0 = 0.135, about 14%.

24.3PET scanning

annihilation: each photon has E = m_ec² = 0.511 MeV

A positron-emitting tracer (bonded to a metabolically active molecule such as glucose) is injected. Emitted positrons annihilate with nearby electrons, producing two gamma photons travelling in opposite directions — required by conservation of momentum. Detectors around the patient record the arrival times; the small difference locates the annihilation event along the detector line, and a computer builds an image of tracer concentration.

Worked example

Find the energy of each annihilation photon. (m_e = 9.11 × 10⁻³¹ kg)

Total rest energy of the pair = 2m_ec² = 2 × 9.11 × 10⁻³¹ × (3.00 × 10⁸)² = 1.640 × 10⁻¹³ J.
Two photons share it equally: 8.20 × 10⁻¹⁴ J each = 0.512 MeV.

"Why two photons, in opposite directions?" — because the electron–positron pair has essentially zero total momentum, so the photons' momenta must cancel. A single photon could not conserve both energy and momentum.

A Level · 3 sub-topics · last topic

25 · Astronomy and cosmology

25.1Standard candles

F = L/(4πd²)

Luminosity is the total power of radiation emitted by a star; radiant flux intensity F is the power received per unit area at the Earth. A standard candle is an object of known luminosity — measure its flux, and the inverse-square law gives its distance, and hence the distance to its galaxy.

Worked example

A type Ia supernova of luminosity 1.0 × 10³⁶ W is observed with flux 2.0 × 10⁻¹² W m⁻². Find its distance.

d = √(L/4πF) = √(1.0 × 10³⁶ / (4π × 2.0 × 10⁻¹²))
= √(3.979 × 10⁴⁶) = 2.0 × 10²³ m — about 21 million light years.

25.2Stellar radii

Wien: λ_max ∝ 1/T · Stefan–Boltzmann: L = 4πσr²T⁴ · σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴

Wien's displacement law gives the surface temperature from the peak of the star's spectrum; the Stefan–Boltzmann law then gives the radius from the luminosity. Together they are how the size of a star that is a mere point of light gets measured.

Worked example

A star's spectrum peaks at 500 nm and its luminosity is 4.0 × 10²⁶ W. Wien's constant is 2.90 × 10⁻³ m K. Find its surface temperature and radius.

T = 2.90 × 10⁻³/5.00 × 10⁻⁷ = 5800 K.
T⁴ = 1.132 × 10¹⁵; 4πσT⁴ = 4π × 5.67 × 10⁻⁸ × 1.132 × 10¹⁵ = 8.064 × 10⁸.
r² = 4.0 × 10²⁶/8.064 × 10⁸ = 4.960 × 10¹⁷ → r = 7.0 × 10⁸ m.

Those are the Sun's numbers, near enough — a useful check that your method is right.

25.3Hubble's law and the Big Bang theory

Δλ/λ ≈ Δf/f ≈ v/c · v ≈ H₀d

Spectral lines from distant galaxies are shifted to longer wavelengths — redshift — which means those galaxies are receding. Hubble found the recession speed is proportional to distance: every galaxy is moving away from every other, so the universe is expanding. Running the expansion backwards implies everything was once concentrated in an extremely hot, dense state — the Big Bang.

Worked example

A line of laboratory wavelength 486.0 nm is observed at 495.7 nm. H₀ = 2.3 × 10⁻¹⁸ s⁻¹. Find the recession speed and distance.

Δλ = 9.7 nm; Δλ/λ = 9.7/486.0 = 0.01996.
v = 0.01996 × 3.00 × 10⁸ = 6.0 × 10⁶ m s⁻¹.
d = v/H₀ = 6.0 × 10⁶/2.3 × 10⁻¹⁸ = 2.6 × 10²⁴ m.

Cambridge requires SI units only for Hubble's law in 9702 — H₀ in s⁻¹, distances in metres. If a question quotes km s⁻¹ Mpc⁻¹, convert before you calculate.

Redshift is not the Doppler effect of things flying through spaceAt A Level you may treat it with the Doppler formula, but say "the galaxies are receding" rather than claiming they are moving through a static space. Cosmological redshift arises from the expansion of space itself.

Reference

Data and formulae you are given

The data and the AS formula list appear on page 2 of Papers 1, 2 and 4. Paper 4 gets a second page of A Level formulae. Knowing exactly what is printed tells you exactly what you must memorise.

GivenData (all papers)

Quantity	Symbol and value
acceleration of free fall	g = 9.81 m s⁻²
speed of light in free space	c = 3.00 × 10⁸ m s⁻¹
elementary charge	e = 1.60 × 10⁻¹⁹ C
unified atomic mass unit	1 u = 1.66 × 10⁻²⁷ kg
rest mass of proton	m_p = 1.67 × 10⁻²⁷ kg
rest mass of electron	m_e = 9.11 × 10⁻³¹ kg
Avogadro constant	N_A = 6.02 × 10²³ mol⁻¹
molar gas constant	R = 8.31 J K⁻¹ mol⁻¹
Boltzmann constant	k = 1.38 × 10⁻²³ J K⁻¹
gravitational constant	G = 6.67 × 10⁻¹¹ N m² kg⁻²
permittivity of free space	ε₀ = 8.85 × 10⁻¹² F m⁻¹ (1/4πε₀ = 8.99 × 10⁹ m F⁻¹)
Planck constant	h = 6.63 × 10⁻³⁴ J s
Stefan–Boltzmann constant	σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴

GivenFormulae printed for you

On every paper (1, 2 and 4): s = ut + ½at² · v² = u² + 2as · Δp = ρgΔh · F = ρgV · Doppler f_o = f_sv/(v ± v_s) · I = Anvq · resistors in series and parallel.

Extra page on Paper 4 only: φ = −GM/r · E_p = −GMm/r · p = ⅓(Nm/V)⟨c²⟩ · a = −ω²x · v = v₀cos ωt · v = ±ω√(x₀²−x²) · V = Q/4πε₀r · E_p = Qq/4πε₀r · capacitors in series and parallel · x = x₀e^−t/RC · V_H = BI/ntq · x = x₀sin ωt · x = x₀e^−λt · λ = 0.693/t_½ · I_R/I₀ = (Z₁−Z₂)²/(Z₁+Z₂)² · L = 4πσr²T⁴ · Δλ/λ ≈ Δf/f ≈ v/c.

What is not given — learn theseF = ma · p = mv · W = Fscos θ · P = Fv · E_k = ½mv² · E_p = mgΔh · σ = F/A, ε = x/L, E = σ/ε · v = fλ · I = I₀cos²θ · λ = ax/D · dsin θ = nλ · V = W/Q · P = VI = I²R = V²/R · R = ρL/A · E = I(R+r) · a = v²/r = rω² · F = Gm₁m₂/r² · g = GM/r² · Q = mcΔθ, Q = mL · pV = nRT = NkT · ½m⟨c²⟩ = (3/2)kT · ΔU = q + W · W = pΔV · E = F/Q = ΔV/d · F = Q₁Q₂/4πε₀r² · C = Q/V, W = ½CV² · F = BILsin θ, F = BQvsin θ · Φ = BA, E = −NdΦ/dt · I_r.m.s. = I₀/√2 · E = hf = hc/λ · hf = Φ + ½mv²_max · λ = h/mv · E = mc² · A = λN · Z = ρc · I = I₀e^−µx · F = L/4πd² · λ_max ∝ 1/T · v ≈ H₀d.

Reference

Uncertainties cheat-sheet

This one page earns marks in Papers 1, 2, 3, 4 and 5. It is the highest return per minute of revision in the whole syllabus.

RulesCombining uncertainties

Operation	Rule	Example
y = a + b or a − b	add absolute uncertainties	(5.0 ± 0.1) − (3.0 ± 0.1) = 2.0 ± 0.2
y = ab or a/b	add percentage uncertainties	2% × 3% → 5%
y = aⁿ	multiply the percentage by n	r³ from r at 2% → 6%
y = ka (k exact)	percentage unchanged	2πr has the same % as r

Absolute ↔ percentage: % = (absolute ÷ value) × 100. Quote the final absolute uncertainty to one significant figure, and round the value to match its decimal place: 4.63 ± 0.2 should be written 4.6 ± 0.2.

From a graph: the uncertainty in a gradient is found by drawing the steepest and shallowest lines that still pass through all the error bars, then taking (max gradient − min gradient)/2. Cambridge accepts either that or (best − worst).

KitTypical instrument uncertainties

Instrument	Resolution	Sensible uncertainty
metre rule	1 mm	±1 mm on a length (both ends judged)
vernier calipers	0.1 mm	±0.1 mm
micrometer screw gauge	0.01 mm	±0.01 mm — check the zero error
stopwatch (hand-operated)	0.01 s	±0.1–0.4 s, dominated by reaction time, not resolution
digital meter	last digit	±1 in the last digit
analogue meter	half a division	±½ division, and beware parallax

Timing many oscillations and dividing is the standard fix for reaction time: timing 20 swings divides the percentage uncertainty by 20. Say why — "to reduce the percentage uncertainty due to reaction time" — not just "to be more accurate".

Papers 3 & 5 · 23% of the A Level

Practical skills: Paper 3

Two questions, one hour and 20 marks each, set in different areas of physics. No prior knowledge of the theory is required — you are marked on skill alone.

P3What each question asks, and where the marks are

Question 1 is a full experiment: collect data, plot a graph, draw conclusions. Question 2 collects data and draws conclusions too, but the method you are given is deliberately inaccurate — you are required to evaluate it and suggest improvements. That is where the marks concentrate.

Skill	Q1 minimum	Q2 minimum
Manipulation, measurement and observation	7 marks	5 marks
Presentation of data and observations	6 marks	2 marks
Analysis, conclusions and evaluation	4 marks	10 marks
The remaining 3 marks in each question are distributed across the skills and may vary between papers.

Question 2 is where students lose the paper. Ten of its twenty marks are for estimating uncertainties, identifying limitations and suggesting improvements — a rehearsable, formulaic skill that most candidates never practise because it feels like "just writing".

P3Tables, graphs and conclusions — the mechanical marks

Table of results. Every column heading needs a quantity and a unit, written as t/s or t (s). Raw readings must all be to the same number of decimal places, consistent with the instrument's resolution. Calculated columns should be to a sensible and consistent number of significant figures.

Graph. Choose scales so the plotted points occupy more than half the grid in both directions. Use sensible scale intervals (1, 2, 5 and their powers of ten — never 3 or 7). Label both axes with quantity and unit. Plot to within half a small square. Draw a single thin line of best fit with roughly equal numbers of points either side.

Gradient and intercept. Use a triangle spanning at least half the drawn line, and show it on the graph. Read the intercept off the graph only if the origin is shown; otherwise substitute a point from the line into y = mx + c.

The three cheapest marks in the paperUnits in column headings. A scale that fills the grid. A gradient triangle actually drawn on the graph. None of them need any physics, and thousands of candidates drop all three every session.

P3Evaluation: limitations and improvements

Marks come in pairs: each limitation you identify must be matched to a specific improvement. Both must be specific to this experiment. Generic answers score zero.

Weak answer (0 marks)	Strong answer (2 marks)
"Human error."	"Reaction time when starting and stopping the stopwatch — use a light gate and timer instead."
"Be more accurate."	"Only two readings were taken — repeat for at least six values of L and plot a graph."
"The equipment was faulty."	"The metre rule was not vertical — clamp it and check with a set square against the bench."
"Air resistance."	"The card may not have fallen squarely through the light gate, so the measured time is too long — use a wider gate and a guide rail."
"Take more readings."	"The temperature rose during the experiment, changing the resistance — switch the current off between readings and let the wire cool."

Structure every evaluation sentence as: problem → why it matters → concrete fix. Three clauses, two marks, every time.

P5Paper 5: planning, analysis and evaluation

Two questions of 15 marks each, written, no equipment. Contexts may lie outside the syllabus — if so, you are given the information you need.

Question 1 — planning. A design task, deliberately unstructured, answered with a labelled diagram and extended writing. Work through this checklist and you will collect most of the 15 marks:

Identify the independent variable, the dependent variable, and the variables kept constant. Name them explicitly — this is the first mark and it is free.
How will you vary the independent variable? Name the apparatus and the range.
How will you measure each variable? Name the instrument for every measurement.
How will you keep the control variables constant? Say how, not just that you will.
Draw a clear labelled diagram of the arrangement — it carries marks by itself, and it must be workable.
State the method of analysis: what you will plot against what, and how the constant follows from the gradient and intercept. Show the rearrangement into y = mx + c form.
Additional detail: repeats, ranges, a relevant safety precaution tied to this experiment (not "wear goggles" by reflex), and one further refinement.

Question 2 — analysis and evaluation. You are given an equation and a set of data, and must find a constant and estimate its uncertainty. The routine:

Linearise the equation. Take logs if the relationship is a power or exponential law; otherwise rearrange to y = mx + c.
Build the table of the new quantities, with headings, units and consistent significant figures — and calculate the absolute uncertainty for every processed value.
Plot with error bars, draw the best line and the worst acceptable line.
Find the gradient and intercept from the best line, then from the worst line to get their uncertainties.
Convert to the constant, propagate the uncertainty, and quote it as value ± absolute uncertainty with a unit.

Worked example — linearising for Paper 5

The period of a pendulum is T = 2π√(L/g). Plan the graph that gives g.

Square both sides: T² = 4π²L/g.
Compare with y = mx + c: plot T² on the y-axis against L on the x-axis.
Gradient = 4π²/g, so g = 4π²/gradient, and the line should pass through the origin.
If instead the relationship were T = kLⁿ with n unknown, plot ln T against ln L: the gradient is n and the y-intercept is ln k.

Log axes need unitless argumentsWrite the column heading as ln(T/s), not ln T. You cannot take the logarithm of a quantity with a unit, and examiners check.

Reference

Definitions bank

Learn these word for word. They are the most reliably scoring marks in Papers 2 and 4 — and the easiest to lose by paraphrasing.

LearnThe definitions examiners want verbatim

Term	Definition
Acceleration	The rate of change of velocity.
Newton's second law	The resultant force is proportional to the rate of change of momentum, and acts in the direction of that change.
Momentum	The product of mass and velocity.
Principle of conservation of momentum	For a system of interacting bodies with no resultant external force, the total momentum in any direction is constant.
Moment of a force	The product of the force and the perpendicular distance of its line of action from the pivot.
Work done	The product of the force and the displacement in the direction of the force.
Power	The rate of doing work, or the rate of energy transfer.
Young modulus	The ratio of stress to strain, within the limit of proportionality.
Intensity	The power transmitted per unit area normal to the direction of propagation.
Coherence	Two sources are coherent if they have a constant phase difference (and hence the same frequency).
Electric current	The rate of flow of charge.
Potential difference	The energy transferred from electrical to other forms per unit charge.
Electromotive force	The energy transferred from other forms to electrical per unit charge.
Resistance	The ratio of the potential difference across a component to the current through it.
Ohm's law	The current in a metallic conductor is proportional to the potential difference across it, provided physical conditions such as temperature remain constant.
Kirchhoff's first law	The sum of the currents into a junction equals the sum of the currents out of it (conservation of charge).
Kirchhoff's second law	The sum of the e.m.f.s round a closed loop equals the sum of the p.d.s round that loop (conservation of energy).
Gravitational field strength	The gravitational force per unit mass.
Newton's law of gravitation	The attractive force between two point masses is proportional to the product of the masses and inversely proportional to the square of their separation.
Gravitational potential	The work done per unit mass in bringing a small test mass from infinity to that point.
Specific heat capacity	The energy required per unit mass to raise the temperature by one kelvin.
Specific latent heat	The energy required per unit mass to change the state of a substance without a change of temperature.
Internal energy	The sum of a random distribution of the kinetic and potential energies of the molecules of a system.
First law of thermodynamics	The increase in internal energy equals the energy transferred to the system by heating plus the work done on the system.
Simple harmonic motion	Motion in which the acceleration is proportional to the displacement from a fixed point and always directed towards that point.
Resonance	The condition in which a system is driven at its natural frequency, giving maximum amplitude and maximum energy transfer from the driver.
Electric field strength	The force per unit positive charge.
Electric potential	The work done per unit positive charge in bringing a small test charge from infinity to that point.
Capacitance	The charge stored on one plate per unit potential difference between the plates.
Magnetic flux density	The force per unit current per unit length on a straight conductor placed at right angles to the field.
Magnetic flux	The product of the magnetic flux density and the area normal to the field.
Faraday's law	The magnitude of the induced e.m.f. is proportional to the rate of change of magnetic flux linkage.
Lenz's law	The induced e.m.f. acts in such a direction as to oppose the change producing it.
R.m.s. current	The value of the direct current that dissipates energy in a resistor at the same mean rate as the alternating current.
Work function	The minimum energy required to remove an electron from the surface of a metal.
Mass defect	The difference between the total mass of the separate nucleons and the mass of the nucleus.
Binding energy	The energy required to separate a nucleus into its constituent nucleons.
Decay constant	The probability of decay of a nucleus per unit time.
Activity	The rate of decay, or the number of disintegrations per unit time.
Half-life	The mean time taken for the number of undecayed nuclei (or the activity) to halve.
Luminosity	The total power of radiation emitted by a star.
Radiant flux intensity	The radiant power passing normally through unit area.
Standard candle	An astronomical object of known luminosity.
Hubble's law	The recession speed of a galaxy is proportional to its distance from us.

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Study planner & progress

Every syllabus unit. Tick one when you can answer a past-paper question on it unaided. Your ticks are saved on this device only — nothing is sent anywhere, and there is no account to create.

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Reference

Free past papers & how to revise

Official (free)

Cambridge International — 9702 subject page: syllabus, specimen papers, past papers, mark schemes and examiner reports.
Examiner reports name the exact questions candidates got wrong each series — read them for every paper you attempt.

Free archives

GCE Guide · PastPapers.co — full CAIE past-paper archives.
Physics & Maths Tutor — topic-sorted questions.

How to revise this subject

Learn what is not on the data sheet. You get a generous list (see below), but F = ma, P = VI, v = fλ, E_k = ½mv², the lens and circuit relationships and every definition are yours to memorise.
Define, don't describe. A large slice of Papers 2 and 4 is definitions. "Define electric potential at a point" has an exact answer worth two marks; a vague sentence gets zero. Use the definitions bank below.
Units and significant figures on every line. Quote answers to the same number of significant figures as the least precise data — usually 2 or 3. Bare numbers lose marks even when the physics is right.
Practise drawing graphs by hand, on graph paper, against a clock. Paper 3 gives marks for scale choice, plotting accuracy, the line of best fit and the gradient triangle. You cannot revise this by reading.
Convert prefixes before substituting. mm to m, g to kg, cm³ to m³, MeV to J. More marks are lost to powers of ten in 9702 than to any single concept.
For "explain" questions, name the physics. Examiners look for the key term — "resonance", "Lenz's law", "the work function", "conservation of momentum" — not a retelling of the scenario.