Cambridge theory and practical preparation

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Original, AI-authored English practice with worked guidance. Selected topics only; independent teacher review is pending. This is not a complete syllabus, official past-paper collection or predicted exam score. Check your own exam-board specification: these concepts are not certified against a particular board or exam year.

91 original questions across 15 selected topic areas. No account or payment required to practise.

Cambridge International AS & A Level Physics (9702), examination year 2027: AI-assisted conceptual alignment to official sources, not teacher or exam-board certification. Syllabus mapping and remaining coverage.

13 questions in this topic

Exact English revisions checked in an AI-assisted academic review; independent teacher approval is still pending.

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Targeted original worked practice for previously unsupported sections. Selected theoretical support, not complete leaf-outcome coverage or a substitute for supervised practical assessment.

Subtopics in this topic (13)

Physical quantities and defensible estimates
Terminal motion and force balance
Translational and rotational equilibrium
Longitudinal and transverse disturbances
Electromagnetic wavelength, frequency and scale
Diffraction and aperture scale
Gravitational fields and electric point-charge forces
Thermometry and temperature scales
Moles and molecular counting
Magnetic field representation
Fields produced by currents
Practical measurement, graphs and uncertainty
Experimental planning and processed-data evaluation

13 of 13 questions shown

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Question 1A runner covers a measured 100 m in approximately 12 s. Estimate their average speed and the time for 1 km if that average speed were maintained. Give sensible precision, identify the measured quantities and explain why the 1 km prediction is not a measurement.4 indicative marks

Calculate, explain and evaluate as requested4 indicative marksNo calculator

A runner covers a measured 100 m in approximately 12 s. Estimate their average speed and the time for 1 km if that average speed were maintained. Give sensible precision, identify the measured quantities and explain why the 1 km prediction is not a measurement.

Cambridge 2027: Partial conceptual match

Physical quantities and defensible estimates: selected explicit written support for syllabus section 1.1. The source inventory is section-level, not an exhaustive leaf-outcome ledger; one exercise does not establish every outcome or practical bullet. These theoretical calculations and explanations do not demonstrate hands-on laboratory competence. Independent teacher approval is pending.

9702: 1.1: Physical quantities and estimates

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Your response to question 1

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Start with a quantity equation and keep metres and seconds consistent: the result has units m/s, not m/s squared.
The longer distance is ten times as large, so under the constant-average-speed assumption the time is ten times 12 s. Reporting many decimal places would overstate the approximate timing.
This is a model-based estimate. A separate timed 1 km trial is needed to measure actual performance; an estimate must state the assumption that makes extrapolation possible.

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Average speed is distance/time = 100/12, approximately 8.3 m/s.
The modelled time is 1000/(100/12) = 120 s, approximately two minutes.
Distance and elapsed time are the input physical quantities; speed is derived.
The extended prediction assumes the same average speed despite possible fatigue and different pacing.

Examiner tip

Use the stated assumptions and distinguish observed evidence from interpretation. Original AI-reviewed written preparation; independent teacher approval pending. Practical technique requires direct supervised observation.

Question 2A small object of mass 0.020 kg falls vertically through a fluid. Take g=10 m/

s^{2}

and a constant upthrust of 0.040 N. Its downward velocity eventually becomes constant. Find the drag at that terminal speed. Explain the acceleration just after release from rest, assuming drag is then zero, and the subsequent approach to terminal speed.4 indicative marks

Calculate, explain and evaluate as requested4 indicative marksNo calculator

A small object of mass 0.020 kg falls vertically through a fluid. Take g=10 m/ $s^{2}$ and a constant upthrust of 0.040 N. Its downward velocity eventually becomes constant. Find the drag at that terminal speed. Explain the acceleration just after release from rest, assuming drag is then zero, and the subsequent approach to terminal speed.

Cambridge 2027: Partial conceptual match

Terminal motion and force balance: selected explicit written support for syllabus section 3.2. The source inventory is section-level, not an exhaustive leaf-outcome ledger; one exercise does not establish every outcome or practical bullet. These theoretical calculations and explanations do not demonstrate hands-on laboratory competence. Independent teacher approval is pending.

9702: 3.2: Drag and terminal motion

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Your response to question 2

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Draw weight downward and upthrust and drag upward. Constant velocity means zero acceleration, not an absence of forces.
At release, apply F=ma to weight minus upthrust. At terminal speed instead set weight equal to the sum of upthrust and drag; do not set drag equal to weight when buoyancy is included.
The increasing resistive force reduces the rate at which downward velocity changes. Once the forces balance, the object continues at its nonzero terminal velocity under the stipulated steady conditions.

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Weight is 0.020 × 10 = 0.200 N downward.
At terminal speed the resultant force is zero, so upward drag is 0.200-0.040=0.160 N.
Just after release the net downward force is 0.160 N and acceleration is 8.0 m/ $s^{2}$ .
As speed rises, upward drag increases; the downward resultant and acceleration decrease to zero, while velocity need not be zero.

Examiner tip

Question 3A uniform horizontal beam 4.0 m long weighs 100 N. A 200 N load hangs 3.0 m from its left end. Vertical supports at both ends keep it stationary. Calculate both upward reactions, and explain why balancing vertical forces alone does not determine them separately.4 indicative marks

Calculate, explain and evaluate as requested4 indicative marksNo calculator

A uniform horizontal beam 4.0 m long weighs 100 N. A 200 N load hangs 3.0 m from its left end. Vertical supports at both ends keep it stationary. Calculate both upward reactions, and explain why balancing vertical forces alone does not determine them separately.

Cambridge 2027: Partial conceptual match

Translational and rotational equilibrium: selected explicit written support for syllabus section 4.2. The source inventory is section-level, not an exhaustive leaf-outcome ledger; one exercise does not establish every outcome or practical bullet. These theoretical calculations and explanations do not demonstrate hands-on laboratory competence. Independent teacher approval is pending.

9702: 4.2: Force and torque equilibrium

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Choose the left support as pivot so its unknown reaction has no moment. Keep all clockwise load moments on one side of the equation.
The total clockwise moment is 800 N m. Dividing by the right support's four-metre lever arm gives 200 N; the total load is 300 N, leaving 100 N at the left.
Check moments about the right end: the left reaction gives 400 N m, matching 100(2)+200(1). This independently checks the reactions and lever arms.

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The beam's weight acts at its midpoint, 2.0 m from the left end.
Moments about the left support give 4Rright = 100(2)+200(3).
The right reaction is 200 N; vertical force balance then gives the left reaction as 100 N.
Equilibrium requires both zero resultant force and zero resultant moment; force balance alone supplies only the sum of two unknown reactions.

Examiner tip

Question 4A sound wave travels east through air, while a transverse pulse travels east along a horizontal string that moves vertically. Describe the local particle motion in each case. Explain what travels between source and receiver, and predict whether an ideal polarising filter can select a vibration direction of an ordinary longitudinal sound wave in air.4 indicative marks

Calculate, explain and evaluate as requested4 indicative marksNo calculator

A sound wave travels east through air, while a transverse pulse travels east along a horizontal string that moves vertically. Describe the local particle motion in each case. Explain what travels between source and receiver, and predict whether an ideal polarising filter can select a vibration direction of an ordinary longitudinal sound wave in air.

Cambridge 2027: Partial conceptual match

Longitudinal and transverse disturbances: selected explicit written support for syllabus section 7.2. The source inventory is section-level, not an exhaustive leaf-outcome ledger; one exercise does not establish every outcome or practical bullet. These theoretical calculations and explanations do not demonstrate hands-on laboratory competence. Independent teacher approval is pending.

9702: 7.2: Longitudinal and transverse waves

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Define longitudinal and transverse using displacement relative to propagation, not using a diagram's up/down orientation alone.
Follow one marked particle: it oscillates locally rather than being carried east with the crest or compression. Neighbouring interactions transfer energy through the medium.
Polarisation distinguishes possible transverse displacement directions. The stipulated longitudinal sound has particle displacement along its propagation axis, so an optical-style transverse-direction selection is not applicable.

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Air particles oscillate parallel to the sound's eastward propagation, producing compressions and rarefactions.
The string's local displacement is vertical, perpendicular to the eastward pulse motion.
Energy and the disturbance propagate; the material particles do not need to travel from source to receiver.
Ordinary longitudinal sound in air cannot be polarised by selecting a transverse vibration direction; transverse waves can have such distinct directions.

Examiner tip

Question 5In vacuum, radiation A has wavelength 600 nm and radiation B has wavelength 3.0 cm. Use c=3.0×

10^{8}

m/s. Calculate both frequencies, identify A and B in the electromagnetic spectrum and compare their photon energies without assuming their intensities are equal.4 indicative marks

Calculate, explain and evaluate as requested4 indicative marksNo calculator

In vacuum, radiation A has wavelength 600 nm and radiation B has wavelength 3.0 cm. Use c=3.0× $10^{8}$ m/s. Calculate both frequencies, identify A and B in the electromagnetic spectrum and compare their photon energies without assuming their intensities are equal.

Cambridge 2027: Partial conceptual match

Electromagnetic wavelength, frequency and scale: selected explicit written support for syllabus section 7.4. The source inventory is section-level, not an exhaustive leaf-outcome ledger; one exercise does not establish every outcome or practical bullet. These theoretical calculations and explanations do not demonstrate hands-on laboratory competence. Independent teacher approval is pending.

9702: 7.4: Electromagnetic wavelengths

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Convert each wavelength to metres before dividing c by wavelength. Nano means ten to the minus nine, not ten to the minus six.
The shorter wavelength has the greater frequency. The spectrum classifications concern the given wavelengths and do not imply different vacuum propagation speeds.
Taking the frequency ratio avoids needing a numerical Planck constant. Photon energy is not the same quantity as intensity: more low-energy photons can still form a powerful beam.

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600 nm = 6.00× $10^{-7}$ m and 3.0 cm = 3.0× $10^{-2}$ m.
fA=5.0×10^14 Hz and fB=1.0×10^10 Hz.
A is visible light and B is microwave radiation; both travel at c in vacuum.
Since E=hf, each A photon has 5.0×10^4 times B's photon energy; total beam power additionally depends on photon arrival rate.

Examiner tip

Question 6Water waves of wavelength 2.0 cm pass through gaps of width 20 cm and 2.0 cm in a barrier, in otherwise identical conditions. Predict which gives more pronounced spreading. Explain what happens if frequency is increased at fixed water-wave speed and fixed gap width, and why diffraction does not mean the transmitted wave has been reflected backwards.4 indicative marks

Calculate, explain and evaluate as requested4 indicative marksNo calculator

Water waves of wavelength 2.0 cm pass through gaps of width 20 cm and 2.0 cm in a barrier, in otherwise identical conditions. Predict which gives more pronounced spreading. Explain what happens if frequency is increased at fixed water-wave speed and fixed gap width, and why diffraction does not mean the transmitted wave has been reflected backwards.

Cambridge 2027: Partial conceptual match

Diffraction and aperture scale: selected explicit written support for syllabus section 8.2. The source inventory is section-level, not an exhaustive leaf-outcome ledger; one exercise does not establish every outcome or practical bullet. These theoretical calculations and explanations do not demonstrate hands-on laboratory competence. Independent teacher approval is pending.

9702: 8.2: Diffraction at apertures

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Your response to question 6

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Worked guidance

Compare the dimensionless ratio wavelength/gap width: it is 1 for the narrow gap but 0.1 for the wider gap. Do not claim that a wide gap produces absolutely no diffraction.
Use v=fλ with the stipulated fixed speed. Higher frequency means a shorter wavelength, which is smaller relative to the unchanged aperture.
A plan-view sketch should show transmitted wavefronts spreading more widely behind the narrow opening, with arrows into the downstream region. Reflected and diffracted waves are distinct possible effects of a barrier.

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Spreading is more pronounced for the 2.0 cm gap, comparable to the wavelength.
For the 20 cm gap the width is much larger than wavelength, so spreading is relatively small.
At fixed speed, increasing frequency reduces wavelength, so the wavelength/gap-width ratio decreases and spreading becomes less pronounced.
Diffraction is the spreading of transmitted wavefronts into the region beyond an aperture, not the backward propagation characteristic of reflection.

Examiner tip

Question 7A 2.0 kg test mass experiences a gravitational force of 12 N towards the centre of an isolated spherical planet. Find the field strength, describe its field lines outside the planet, and explain the effect of doubling the small test mass on force and field strength. Separately, two positive point charges of 1.0 μC and 2.0 μC are 0.30 m apart in vacuum. Use F = k|Qq|/

r^{2}

and k = 9.0 ×

10^{9}

m^{2}

C^{-2}

to calculate the electric force on each charge and state its direction. Explain what happens when their separation doubles, and why using the centre-to-centre distance for extended conductors requires care.5 indicative marks

Calculate, explain and evaluate as requested5 indicative marksCalculator allowed

A 2.0 kg test mass experiences a gravitational force of 12 N towards the centre of an isolated spherical planet. Find the field strength, describe its field lines outside the planet, and explain the effect of doubling the small test mass on force and field strength. Separately, two positive point charges of 1.0 μC and 2.0 μC are 0.30 m apart in vacuum. Use F = k|Qq|/ $r^{2}$ and k = 9.0 × $10^{9}$ N $m^{2}$ $C^{-2}$ to calculate the electric force on each charge and state its direction. Explain what happens when their separation doubles, and why using the centre-to-centre distance for extended conductors requires care.

Cambridge 2027: Partial conceptual match

Gravitational fields and electric point-charge forces: selected explicit written support for syllabus section 13.1, syllabus section 18.3. The source inventory is section-level, not an exhaustive leaf-outcome ledger; one exercise does not establish every outcome or practical bullet. These theoretical calculations and explanations do not demonstrate hands-on laboratory competence. Independent teacher approval is pending.

9702: 13.1: Gravitational field representation
9702: 18.3: Coulomb forces

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Your response to question 7

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Divide the 12 N gravitational force by the 2.0 kg test mass. Distinguish the planet's field from the force on that particular object; N/kg is equivalent to m/s².
Use inward arrowheads in a radial sketch. The small-test-mass assumption keeps the source field fixed; representative line density is a qualitative indication of strength.
Convert microcoulombs to coulombs: F = 9.0 × $10^{9}$ × 1.0 × $10^{-6}$ × 2.0 × $10^{-6}$ / 0.30² = 0.20 N. Like charges repel, and Newton's third law gives equal magnitudes even though the charges differ.
The inverse-square law gives F(2r) = F(r)/4 = 0.050 N. A spherically symmetric charge distribution has an external field equivalent to a central point charge, but nearby conductors can redistribute charge and lose that symmetry.

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Gravitational field strength is force per unit mass: g = 12/2.0 = 6.0 N/kg.
Outside the isolated spherical planet, field lines point radially inward and spread farther apart with distance, indicating decreasing field strength.
Doubling a sufficiently small test mass doubles its force but leaves the planet's field strength at the same point unchanged.
The electric force magnitude is 0.20 N on each charge; the forces are equal and opposite, directed away from the other positive charge.
Doubling separation gives 0.050 N, one quarter of the original force. Point-charge calculations do not automatically apply to nearby extended conductors whose charge distributions are distorted.

Examiner tip

Question 8A sensor has a stipulated linear resistance-temperature relation between 0°C and 100°C. Its resistances at those points are 100 Ω and 138 Ω. Find the indicated Celsius and kelvin temperatures for 119 Ω. Explain a calibration limitation and why a temperature rise of 50°C is not a rise of 323.15 K.4 indicative marks

Calculate, explain and evaluate as requested4 indicative marksNo calculator

A sensor has a stipulated linear resistance-temperature relation between 0°C and 100°C. Its resistances at those points are 100 Ω and 138 Ω. Find the indicated Celsius and kelvin temperatures for 119 Ω. Explain a calibration limitation and why a temperature rise of 50°C is not a rise of 323.15 K.

Cambridge 2027: Partial conceptual match

Thermometry and temperature scales: selected explicit written support for syllabus section 14.2. The source inventory is section-level, not an exhaustive leaf-outcome ledger; one exercise does not establish every outcome or practical bullet. These theoretical calculations and explanations do not demonstrate hands-on laboratory competence. Independent teacher approval is pending.

9702: 14.2: Thermometry and temperature scales

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Your response to question 8

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Use the two fixed calibration points to interpolate, subtracting the zero-point resistance rather than taking 119/138.
Convert an absolute Celsius temperature by adding 273.15. The 273.15 offset cancels when taking a difference between two temperatures.
A two-point calibration only establishes this stipulated linear model. Additional reference temperatures can reveal deviations; a precise resistance measurement alone does not establish accuracy of the temperature scale model.

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The fractional resistance rise is (119-100)/(138-100)=0.5.
The indicated temperature is 50°C, corresponding to 323.15 K.
The linear relation is an assumption within the calibrated range; real nonlinearity or extrapolation can bias readings.
Temperature intervals have the same numerical size in Celsius and kelvin, so a rise of 50°C is a rise of 50 K, not an absolute temperature.

Examiner tip

Question 9A sealed sample contains 0.250 mol of diatomic nitrogen molecules, N₂. Use NA=6.02×

10^{23}

mol^-1. Find the number of molecules and the number of nitrogen atoms, and explain which count corresponds to N when using pV=NkT with the sample modelled as an ideal molecular gas.4 indicative marks

Calculate, explain and evaluate as requested4 indicative marksNo calculator

A sealed sample contains 0.250 mol of diatomic nitrogen molecules, N₂. Use NA=6.02× $10^{23}$ mol^-1. Find the number of molecules and the number of nitrogen atoms, and explain which count corresponds to N when using pV=NkT with the sample modelled as an ideal molecular gas.

Cambridge 2027: Partial conceptual match

Moles and molecular counting: selected explicit written support for syllabus section 15.1. The source inventory is section-level, not an exhaustive leaf-outcome ledger; one exercise does not establish every outcome or practical bullet. These theoretical calculations and explanations do not demonstrate hands-on laboratory competence. Independent teacher approval is pending.

9702: 15.1: Moles and molecular counting

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State first that the supplied amount is in moles of N₂ molecules. Apply Avogadro's constant to that specified entity.
Multiply the molecular count by two only for the question about constituent atoms. A diatomic molecule is not two independent ideal-gas particles in this model.
The equivalent equation pV=nRT uses the same molecular amount n because R=NAk. Counting atoms instead in pV=NkT would wrongly double the predicted pressure for fixed volume and temperature.

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Molecular count is nNA=0.250(6.02× $10^{23}$ )=1.505× $10^{23}$ .
Each molecule contains two atoms, giving 3.01× $10^{23}$ nitrogen atoms.
In pV=NkT for an ideal molecular gas, N counts gas molecules, not the constituent atom count.
The mole is an amount of specified entities; identifying the entities avoids a factor-of-two ambiguity.

Examiner tip

Question 10A small compass is placed outside a bar magnet, far from other magnetic sources. Explain what its north-seeking end indicates. Describe the direction of field lines outside the magnet, why field lines do not cross at an ordinary point, and whether closer line spacing means a larger force on every possible test charge.4 indicative marks

Calculate, explain and evaluate as requested4 indicative marksNo calculator

A small compass is placed outside a bar magnet, far from other magnetic sources. Explain what its north-seeking end indicates. Describe the direction of field lines outside the magnet, why field lines do not cross at an ordinary point, and whether closer line spacing means a larger force on every possible test charge.

Cambridge 2027: Partial conceptual match

Magnetic field representation: selected explicit written support for syllabus section 20.1. The source inventory is section-level, not an exhaustive leaf-outcome ledger; one exercise does not establish every outcome or practical bullet. These theoretical calculations and explanations do not demonstrate hands-on laboratory competence. Independent teacher approval is pending.

9702: 20.1: Magnetic field representation

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Use the compass orientation to define a local direction, not to claim that the entire field is uniform or that the compass moves along a field line.
Draw continuous field lines with outside arrows from N to S. Lines continue through the magnet as closed patterns; the outside direction alone does not imply isolated magnetic charges.
For a moving point charge the magnetic force magnitude is |q|vB sinθ. A field-strength representation cannot by itself determine the force without the other quantities.

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The compass north-seeking end points along the local magnetic field direction.
Outside a bar magnet, representative field lines run from its north pole towards its south pole.
At an ordinary point a nonzero magnetic field has one resultant direction; intersecting field lines would imply two directions.
Closer lines indicate a stronger field, but magnetic force on a charge also depends on charge, speed and angle; a stationary charge has no magnetic force.

Examiner tip

Question 11An ideal long straight wire carries 4.0 A. Find the magnetic flux density 0.020 m from the wire using B=μ₀I/(2πr), μ₀=4π×

10^{-7}

H/m. Describe the field direction for current out of a page, and predict the changes if current doubles or distance doubles. Contrast the line pattern inside an ideal long solenoid.4 indicative marks

Calculate, explain and evaluate as requested4 indicative marksCalculator allowed

An ideal long straight wire carries 4.0 A. Find the magnetic flux density 0.020 m from the wire using B=μ₀I/(2πr), μ₀=4π× $10^{-7}$ H/m. Describe the field direction for current out of a page, and predict the changes if current doubles or distance doubles. Contrast the line pattern inside an ideal long solenoid.

Cambridge 2027: Partial conceptual match

Fields produced by currents: selected explicit written support for syllabus section 20.4. The source inventory is section-level, not an exhaustive leaf-outcome ledger; one exercise does not establish every outcome or practical bullet. These theoretical calculations and explanations do not demonstrate hands-on laboratory competence. Independent teacher approval is pending.

9702: 20.4: Current-generated magnetic fields

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Worked guidance

Substitute SI values and cancel π: (4× $10^{-7}$ ×4)/(2×0.020)=4.0× $10^{-5}$ T. Do not use centimetres directly for r.
Point the right thumb out of the page; curled fingers give the anticlockwise direction. Reversing current reverses that direction.
Use proportionality in the supplied equation rather than guessing an inverse-square law. The straight-wire field model and the near-uniform interior solenoid model have different geometries and assumptions.

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The wire field magnitude is 4.0× $10^{-5}$ T.
For current out of the page, field lines circle anticlockwise by the right-hand grip rule.
Doubling current doubles B; doubling distance halves B for the stated straight-wire model.
Inside a long solenoid the representative lines are approximately parallel and uniformly spaced away from its ends, rather than circles around a single straight wire.

Examiner tip

Question 12Written preparation, not an actual laboratory assessment: a spring has fictional recorded extensions 0.0, 10.0, 20.0, 30.0 mm under forces 0.00, 0.50, 1.00, 1.50 N. Treat force values as exact for this calculation. Each extension has a separate ±0.5 mm bound; assume the endpoint errors can vary independently. Design a results table and graph, calculate the force-extension gradient in SI units and the extreme two-point gradient bounds from the first and last readings. Explain one measurement improvement and a limitation of these idealised data.5 indicative marks

Calculate, explain and evaluate as requested5 indicative marksCalculator allowed

Written preparation, not an actual laboratory assessment: a spring has fictional recorded extensions 0.0, 10.0, 20.0, 30.0 mm under forces 0.00, 0.50, 1.00, 1.50 N. Treat force values as exact for this calculation. Each extension has a separate ±0.5 mm bound; assume the endpoint errors can vary independently. Design a results table and graph, calculate the force-extension gradient in SI units and the extreme two-point gradient bounds from the first and last readings. Explain one measurement improvement and a limitation of these idealised data.

Cambridge 2027: Partial conceptual match

Practical measurement, graphs and uncertainty: selected explicit written support for internal practical inventory row 1, internal practical inventory row 2, internal practical inventory row 3, internal practical inventory row 4. The source inventory is section-level, not an exhaustive leaf-outcome ledger; one exercise does not establish every outcome or practical bullet. Practical row IDs are internal identifiers, not official numbered syllabus codes. Paper 5 assesses planning and analysis in writing, but adequate preparation requires supervised laboratory experience; actual apparatus use, specimen handling and performance are not demonstrated here. Independent teacher approval is pending.

9702: Paper 3 / Manipulation, measurement and observation: Practical apparatus use and data quality
9702: Paper 3 / Presentation of data and observations: Practical tables and graph conventions
9702: Paper 3 / Analysis, conclusions and evaluation: Practical analysis and uncertainty
9702: Paper 3 / Drawing conclusions: Practical inference and improvements

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Choose headings that separate quantity and unit, use consistent precision appropriate to the resolution and draw a best-fit line rather than joining every point as a rule.
The gradient represents spring stiffness only under the linear elastic model. Convert the 30 mm interval to metres before calculating N/m.
Treat the two endpoint uncertainties as independent worst-case bounds on the difference, giving ±1.0 mm. The question stipulates exact force for this bound; real force uncertainty and a full acceptable-line graphical analysis would add further considerations.

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Use force/N and extension/mm or extension/m as table/axis headings; plot force vertically against extension horizontally with suitable scales.
Nominal gradient is 1.50/0.0300=50.0 N/m.
The endpoint extension difference is 30.0±1.0 mm, so the stipulated force-exact two-point gradient bounds are 1.50/0.0310 and 1.50/0.0290, approximately 48.4 and 51.7 N/m.
Read a scale at eye level with a fixed reference or use a fiducial marker; repeat measurements to assess scatter, but repeats alone do not remove a zero error.
These perfectly aligned fictional values do not establish the full elastic range or demonstrate actual apparatus handling.

Examiner tip

Question 13Written planning exercise: investigate

T^{2}

proportional to pendulum length L for small swings, using

T^{2}

=4π²L/g. Specify how to measure length and period, controls, a graph and the inference from its gradient. For L=0.500±0.002 m and 20 oscillations taking 28.4±0.2 s, find T and the approximate worst-case percentage uncertainty of g calculated from these single measurements. Include a sensible risk control and why a graph using several lengths is preferable.5 indicative marks

Calculate, explain and evaluate as requested5 indicative marksCalculator allowed

Written planning exercise: investigate $T^{2}$ proportional to pendulum length L for small swings, using $T^{2}$ =4π²L/g. Specify how to measure length and period, controls, a graph and the inference from its gradient. For L=0.500±0.002 m and 20 oscillations taking 28.4±0.2 s, find T and the approximate worst-case percentage uncertainty of g calculated from these single measurements. Include a sensible risk control and why a graph using several lengths is preferable.

Cambridge 2027: Partial conceptual match

Experimental planning and processed-data evaluation: selected explicit written support for internal practical inventory row 5, internal practical inventory row 6. The source inventory is section-level, not an exhaustive leaf-outcome ledger; one exercise does not establish every outcome or practical bullet. Practical row IDs are internal identifiers, not official numbered syllabus codes. Paper 5 assesses planning and analysis in writing, but adequate preparation requires supervised laboratory experience; actual apparatus use, specimen handling and performance are not demonstrated here. Independent teacher approval is pending.

9702: Paper 5 / Planning: Experimental planning and risk control
9702: Paper 5 / Analysis, conclusions and evaluation: Processed data and uncertainty evaluation

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Timing many cycles reduces the fractional effect of a fixed start/stop timing uncertainty. Dividing both total time and its uncertainty by 20 leaves the same fractional uncertainty in T.
Plot $T^{2}$ on the vertical axis, so the slope has units $s^{2}$ /m and g = 4π²/slope. The intercept can help diagnose a length offset rather than forcing a line through zero automatically.
The calculation gives a length contribution of 0.4% and a doubled time contribution of about 1.408%, totalling 1.808% under a first-order worst-case rule. This is not a statistical confidence interval and assumes the small-angle model is adequate.

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Measure pivot to bob centre, use a small release angle without a push, and time repeated sets of many complete oscillations at several lengths.
T=28.4/20=1.42 s; T² versus L should be linear with gradient 4π²/g.
For g=4π²L/ $T^{2}$ , the approximate worst-case percentage uncertainty is 100(0.002/0.500+2×0.2/28.4), approximately 1.81%.
Keep the support secure and people clear of the swinging bob; keep amplitude small and release method consistent.
Several lengths reveal model departure, support a gradient/uncertainty estimate and reduce reliance on a single possibly biased timing; no actual laboratory competence is demonstrated here.

Examiner tip

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