Skill Align WACE Physics ATAR Year 12 - Free Online Pack 0
An independently authored 2026 WACE Physics ATAR practice examination with worked answers and question-specific marking guidance.
- Paper
- Written examination
- Reading
- 10 minutes
- Writing
- 3 hours
- Assessment
- 178 raw marks, weighted to 100 examination marks
The official 2026 SCSA Physics Formulae and Data booklet is supplied separately. Up to three approved calculators may be used; calculators must not create or store programs or text.
Section One: Short response
Answer all questions. Show relevant working and reasoning. This section is worth 30% of the examination.
Question 1
4 marksQuestion 2
5 marksQuestion 3
4 marksQuestion 4
4 marksQuestion 5
4 marksQuestion 6
4 marksQuestion 7
4 marksQuestion 8
5 marksQuestion 9
6 marksQuestion 10
8 marksQuestion 11
6 marksSection Two: Problem-solving
Answer all questions. Use the supplied stimulus, diagrams and data where provided. This section is worth 50% of the examination.
Question 12
11 marksA 420 kg rescue craft travels in a level circle of radius 38.0 m at 12.0 m s^-1. The horizontal water force provides the centripetal force. The crew then increases the speed to 15.0 m s^-1 without changing the radius.
Question 13
16 marksA 920 kg communications relay is in a circular orbit 1.20 × 10^7 m from the centre of Europa. Use MEuropa = 4.80 × 10^22 kg. A second relay orbits at 1.80 × 10^7 m. Ignore Jupiter's influence for this model.
Question 14
17 marksSingly ionised neon-20 ions of mass 3.32 × 10^-26 kg pass undeflected through crossed fields E = 2.40 × 10^4 V m^-1 and B = 0.180 T. They then enter a second 0.180 T magnetic field perpendicular to their motion. Each ion carries charge +e.
Question 15
14 marksA wind-turbine generator has a 160-turn rectangular coil of area 0.0850 m^2 rotating at 25.0 revolutions per second in a uniform 0.420 T field. The coil's normal is parallel to the field at t = 0. The generator is connected to a resistive load through an ideal transformer.
Question 16
16 marksA diffraction grating with 600 lines per millimetre is used to observe a bright sodium line. The first-order maximum occurs at 20.4° from the central maximum. In a separate discharge, an atom emits a photon of this wavelength after occupying an excited level.
Question 17
10 marksA rectangular 80-turn motor coil is 0.120 m by 0.0800 m and carries 2.40 A in a 0.650 T uniform field. At one instant the coil's normal is 35.0° to the field. The motor draws 72.0 V and has winding resistance 4.00 ohm while rotating.
Section Three: Comprehension and data analysis
Answer both questions. Question 18 uses an unfamiliar-context comprehension stimulus and Question 19 requires analysis of experimental data. This section is worth 20% of the examination.
Question 18
20 marksTwo independent laser-interferometer observatories have a straight-line separation of 3.60 × 10^3 km. At each site, two perpendicular detector arms each have reference length L = 3.00 km. For this simplified detector model, strain is h = ΔL/L, where ΔL is the change in one arm length. A short transient increases one arm length by 1.20 × 10^-18 m while the perpendicular arm shortens. Both sites record matching differential transients after allowing for a 10.0 ms arrival delay. Local vibration monitors show no coincident disturbance. The source note states that a gravitational wave is a propagating distortion of spacetime produced by accelerating, asymmetric masses; it travels at c and alternately stretches and squeezes perpendicular directions.
Question 19
20 marksA student places a 0.0800 m wire segment perpendicular to a nominally uniform magnetic field and measures magnetic force as current changes. The plotted mean values are (0.00 A, 0.000 N), (0.50 A, 0.012 N), (1.00 A, 0.024 N), (1.50 A, 0.036 N), (2.00 A, 0.048 N) and (2.50 A, 0.060 N). Force uncertainty is ±0.002 N and current uncertainty is negligible at the graph scale.
Worked Solutions And Marking Guide
General marking principles
- Apply each question-specific criterion to the physical evidence shown; equivalent scientifically correct reasoning may receive the same credit.
- Where a criterion spans linked steps, award its stated marks only when the complete linked reasoning is demonstrated.
- Carry forward an earlier numerical error when later physics is internally consistent, unless a later criterion independently assesses the same step.
- Do not penalise a missing unit more than once in a linked calculation unless the question separately assesses units.
Section One: Short response Question 1
Indicative answer: Taking moments about the hinge gives T(2.10) = (18.0 × 9.80)(1.20) + 320(1.80), so T = 375 N. Vertical equilibrium then gives R + 375 - 176.4 - 320 = 0, hence the hinge force is 121 N upward.
Detailed marking criteria
Integrated response (2 marks)
Forms the correct moment equation about the hinge using the sign's centre-of-mass distance and the lamp distance.
Integrated response (1 mark)
Obtains a cable tension of 375 N to three significant figures.
Integrated response (1 mark)
Applies vertical force equilibrium to obtain a 121 N upward hinge force.
Section One: Short response Question 2
Indicative answer: The components are ux = 24.0 cos35.0° = 19.7 m s^-1 and uy = 24.0 sin35.0° = 13.8 m s^-1. Solving -3.20 = 13.8t - 4.90t^2 gives the positive root t = 3.03 s. The range is x = 19.7 × 3.03 = 59.6 m.
Detailed marking criteria
Integrated response (2 marks)
Resolves the launch velocity into 19.7 m s^-1 horizontally and 13.8 m s^-1 vertically.
Integrated response (2 marks)
Uses vertical displacement -3.20 m and selects the positive flight time 3.03 s.
Integrated response (1 mark)
Calculates the horizontal range as 59.6 m.
Section One: Short response Question 3
Indicative answer: The required centripetal force is mv^2/r = 208 N. At the top, mg - Ntop = 208, so Ntop = 400 N. At the bottom, Nbottom - mg = 208, so Nbottom = 816 N.
Detailed marking criteria
Integrated response (1 mark)
Calculates the inward centripetal-force requirement as 208 N.
Integrated response (1 mark)
Uses mg - N = mv^2/r at the top of the wheel.
Integrated response (1 mark)
Obtains the top normal force as 400 N.
Integrated response (1 mark)
Uses N - mg = mv^2/r at the bottom to obtain 816 N.
Section One: Short response Question 4
Indicative answer: For 0.980c, gamma = 5.03, so the laboratory lifetime is gamma times the proper lifetime = 11.1 microseconds. The laboratory travel time is 9000/(0.980c) = 30.6 microseconds, which is about 2.76 dilated lifetimes rather than 13.9 proper lifetimes. The survival fraction is therefore much larger than a classical-lifetime prediction.
Detailed marking criteria
Integrated response (1 mark)
Calculates the Lorentz factor as approximately 5.03.
Integrated response (1 mark)
Obtains the dilated mean lifetime 11.1 microseconds in the laboratory frame.
Integrated response (1 mark)
Calculates the laboratory journey time as 30.6 microseconds.
Integrated response (1 mark)
Compares the journey with the dilated and proper lifetimes to explain enhanced muon survival.
Section One: Short response Question 5
Indicative answer: Equilibrium requires qE = mg. Thus E = (1.50 × 10^-12 × 9.80)/(3.20 × 10^-15) = 4.59 × 10^3 N C^-1. Because the charge is positive, the electric field and electric force are upward.
Detailed marking criteria
Integrated response (1 mark)
Sets the upward electric force equal to the particle's weight.
Integrated response (2 marks)
Calculates the field magnitude as 4.59 × 10^3 N C^-1.
Integrated response (1 mark)
States that the field is upward for the positively charged particle.
Section One: Short response Question 6
Indicative answer: B = μ0I/(2πr) = (4π × 10^-7 × 420)/(2π × 0.0650) = 1.29 × 10^-3 T. Taking east, north and upward as a right-handed coordinate set, the right-hand grip rule gives a field directed south at the point vertically above the eastward-current busbar.
Detailed marking criteria
Integrated response (1 mark)
Converts 65.0 mm to 0.0650 m and selects B = μ0I/(2πr).
Integrated response (2 marks)
Calculates the field magnitude as 1.29 × 10^-3 T.
Integrated response (1 mark)
Uses the right-hand grip rule to identify the field as southward at the point vertically above the eastward current.
Section One: Short response Question 7
Indicative answer: The induced emf is Blv = 0.840 × 1.25 × 18.0 = 18.9 V. The current is 18.9/0.350 = 54.0 A, so the magnetic force is BIl = 0.840 × 54.0 × 1.25 = 56.7 N, opposing the motion by Lenz's law.
Detailed marking criteria
Integrated response (1 mark)
Applies the perpendicular motional-emf relationship ε = Blv.
Integrated response (1 mark)
Obtains an induced emf of 18.9 V.
Integrated response (1 mark)
Calculates the induced current as 54.0 A.
Integrated response (1 mark)
Calculates a 56.7 N magnetic force and identifies it as opposing the motion.
Section One: Short response Question 8
Indicative answer: The line current is P/V = 8.00 × 10^6/(132 × 10^3) = 60.6 A. The cable loss is I^2R = 13.2 kW. The fraction lost is 13.2 kW/8.00 MW × 100% = 0.165%.
Detailed marking criteria
Integrated response (2 marks)
Uses P = VI at 132 kV to calculate a line current of 60.6 A.
Integrated response (2 marks)
Uses I^2R to calculate a cable loss of 13.2 kW.
Integrated response (1 mark)
Calculates the transmitted-power loss as 0.165%.
Section One: Short response Question 9
Indicative answer: The photon energy is hc/λ = 3.40 eV, so Kmax = 3.40 - 2.10 = 1.30 eV and the stopping potential is 1.30 V. Doubling intensity doubles the photon arrival rate and therefore the saturation photocurrent, but does not change Kmax or the stopping potential because photon energy is unchanged.
Detailed marking criteria
Integrated response (2 marks)
Calculates the 365 nm photon energy as approximately 3.40 eV.
Integrated response (2 marks)
Obtains a maximum kinetic energy of 1.30 eV and stopping potential of 1.30 V.
Integrated response (2 marks)
Explains that doubled intensity increases photocurrent but not maximum energy or stopping potential.
Section One: Short response Question 10
Indicative answer: Systematic galaxy redshift and the Hubble relationship support present cosmic expansion but do not by themselves establish the early thermal history. The near-uniform 2.7 K cosmic microwave background with a black-body spectrum supports a hot, dense early universe and later cooling. The observed primordial hydrogen, helium and trace lithium abundances broadly match nucleosynthesis predictions. Agreement among these independent lines is stronger than redshift alone, while each remains subject to measurement and model assumptions.
Detailed marking criteria
Integrated response (2 marks)
Distinguishes evidence for current expansion from proof of the complete Big Bang history.
Integrated response (2 marks)
Links the cosmic microwave background temperature and black-body spectrum to a hot, dense early universe.
Integrated response (2 marks)
Links primordial light-element abundances to early-universe nucleosynthesis.
Integrated response (2 marks)
Evaluates convergent independent evidence while acknowledging measurement or model limitations.
Section One: Short response Question 11
Indicative answer: The period is 31.6/20 = 1.580 s. Hence g = 4π^2(0.625)/(1.580)^2 = 9.88 m s^-2. The percentage uncertainty is (0.003/0.625)×100% + 2(0.2/31.6)×100% = 1.75%, giving an absolute uncertainty of about 0.17 m s^-2. The result is (9.88 ± 0.17) m s^-2.
Detailed marking criteria
Integrated response (1 mark)
Calculates the period as 1.580 s from the 20-oscillation timing.
Integrated response (2 marks)
Substitutes consistently to obtain g = 9.88 m s^-2.
Integrated response (2 marks)
Combines the length percentage uncertainty with twice the timing percentage uncertainty to obtain 1.75%.
Integrated response (1 mark)
Reports the result as approximately (9.88 ± 0.17) m s^-2.
Section Two: Problem-solving Question 12
(a) F = mv^2/r = 420 × 12.0^2/38.0 = 1.59 × 10^3 N, horizontally towards the centre of the circular path.
(b) At 15.0 m s^-1, F = 2.49 × 10^3 N. The force factor is (15.0/12.0)^2 = 1.56.
(c) vmax = sqrt(Fr/m) = sqrt(2800 × 38.0/420) = 15.9 m s^-1. In a steady circle, centripetal acceleration changes only direction; a tangential acceleration changes speed as well, so the net acceleration is the vector sum of radial and tangential components.
Detailed marking criteria
Part a (1 mark)
Selects the horizontal centripetal-force relationship mv^2/r.
Part a (2 marks)
Calculates 1.59 × 10^3 N and states that it is directed towards the centre.
Part b (2 marks)
Calculates the new centripetal force as 2.49 × 10^3 N.
Part b (1 mark)
Shows that the force increases by the speed-squared factor 1.56.
Part c (3 marks)
Rearranges the centripetal-force equation and obtains a greatest safe speed of 15.9 m s^-1.
Part c (2 marks)
Distinguishes radial acceleration, which changes velocity direction, from tangential acceleration, which changes speed.
Section Two: Problem-solving Question 13
(a) v = sqrt(GM/r) = 517 m s^-1. The period is 2πr/v = 1.46 × 10^5 s, or 40.5 h.
(b) T2/T1 = (r2/r1)^(3/2) = 1.5^(3/2) = 1.837. Therefore T2 = 2.68 × 10^5 s, or 74.4 h.
(c) The orbiting clock has a different speed and is at a different gravitational potential from the surface clock. Special-relativistic motion slows the orbiting clock relative to the chosen surface frame, while general-relativistic gravitational time dilation changes its rate in the weaker orbital field. A navigation system must model both effects and the chosen reference frame.
Detailed marking criteria
Part a (2 marks)
Equates gravitational force to centripetal force and obtains v = sqrt(GM/r).
Part a (3 marks)
Calculates the inner-relay speed as 517 m s^-1 and period as 1.46 × 10^5 s.
Part b (1 mark)
Sets up the Kepler ratio T2/T1 = (r2/r1)^(3/2).
Part b (2 marks)
Obtains the period ratio 1.837.
Part b (2 marks)
Calculates the outer period as 2.68 × 10^5 s or 74.4 h.
Part c (2 marks)
Identifies special-relativistic clock-rate change due to orbital speed.
Part c (2 marks)
Identifies gravitational clock-rate change due to different gravitational potential.
Part c (2 marks)
Explains that both corrections and a defined reference frame are required for precise comparison.
Section Two: Problem-solving Question 14
(a) For no deflection, qE = qvB, so v = E/B = 1.33 × 10^5 m s^-1. Charge cancels because both field forces are proportional to q; mass does not enter the transverse force balance.
(b) Using qvB = mv^2/r gives r = mv/(qB) = 0.153 m.
(c) At fixed q, v and B, radius is proportional to mass. Thus r22 = r20(22/20) = 0.168 m. The larger-radius trajectory reaches a different detector position, allowing spatial separation.
(d) The speed ratio is v/c = 4.44 × 10^-4, so gamma differs from 1 by only about 9.9 × 10^-8. Classical momentum is therefore adequate at the precision of the supplied data.
Detailed marking criteria
Part a (2 marks)
Balances the electric and magnetic forces for an undeflected ion.
Part a (1 mark)
Calculates the selected speed as 1.33 × 10^5 m s^-1.
Part a (1 mark)
Explains why charge and mass do not appear in the selector speed.
Part b (1 mark)
Equates magnetic force to the centripetal-force requirement.
Part b (1 mark)
Rearranges to r = mv/(qB).
Part b (1 mark)
Calculates the neon-20 path radius as 0.153 m.
Part c (2 marks)
Recognises that analyser radius is proportional to ion mass at fixed charge and speed.
Part c (2 marks)
Calculates the neon-22 radius as 0.168 m.
Part c (1 mark)
Explains separation by distinct impact positions for the two radii.
Part d (2 marks)
Compares the selected speed with c and obtains v/c approximately 4.44 × 10^-4.
Part d (2 marks)
Concludes quantitatively that the Lorentz factor is effectively 1 at the stated precision.
Part d (1 mark)
Justifies use of classical momentum for this analyser.
Section Two: Problem-solving Question 15
(a) The angular speed is 2πf = 157 rad s^-1. The peak emf is NABω = 160 × 0.0850 × 0.420 × 157 = 897 V, and the rms emf is 897/sqrt(2) = 634 V.
(b) At t = 0 the linked flux NAB is maximum and its instantaneous rate of change is zero, so emf is zero. During the quarter-cycle the flux falls as a cosine to zero while the magnitude of its rate of change rises; the emf reaches its peak magnitude at one quarter-period.
(c) Ns/Np = Vs/Vp = 230/634 = 0.363. An ideal transformer trades voltage for current: VpIp = VsIs, so its output power equals input power and energy conservation forbids a power gain.
Detailed marking criteria
Part a (1 mark)
Converts 25.0 revolutions per second to angular speed 157 rad s^-1.
Part a (2 marks)
Uses εpeak = NABω to obtain 897 V.
Part a (2 marks)
Calculates the rms emf as 634 V.
Part b (2 marks)
States that linked flux is maximum and emf is zero at t = 0.
Part b (1 mark)
Describes linked flux falling to zero during the first quarter-cycle.
Part b (2 marks)
Explains that emf reaches peak magnitude when the flux crosses zero most rapidly.
Part c (2 marks)
Uses the rms voltage ratio consistently for an ideal transformer.
Part c (1 mark)
Calculates Ns/Np = 0.363.
Part c (1 mark)
Uses VpIp = VsIs to explain equal ideal input and output power.
Section Two: Problem-solving Question 16
(a) The slit spacing is d = 1/(600 × 10^3) = 1.667 × 10^-6 m. For m = 1, λ = d sin20.4° = 5.81 × 10^-7 m, or 581 nm.
(b) E = hc/λ = 3.42 × 10^-19 J. Its momentum is p = h/λ = 1.14 × 10^-27 kg m s^-1.
(c) Bound electrons occupy discrete allowed energy states. A transition can release only the difference between its initial and final energies, so the photon has Eγ = Ei - Ef. Different permitted transitions produce distinct photon energies and wavelengths; the photon does not carry the electron's entire bound-state energy.
Detailed marking criteria
Part a (1 mark)
Converts 600 lines per millimetre to a spacing of 1.667 × 10^-6 m.
Part a (2 marks)
Applies d sinθ = mλ to first order.
Part a (2 marks)
Calculates the wavelength as 5.81 × 10^-7 m or 581 nm.
Part b (2 marks)
Uses E = hc/λ to obtain 3.42 × 10^-19 J.
Part b (1 mark)
Uses p = h/λ to obtain 1.14 × 10^-27 kg m s^-1.
Part b (1 mark)
Reports both photon quantities with appropriate units.
Part c (2 marks)
States that bound atomic electrons occupy discrete allowed energy levels.
Part c (2 marks)
Relates photon energy to the difference Ei - Ef between two levels.
Part c (2 marks)
Connects distinct permitted transitions with the observed line spectrum.
Part c (1 mark)
Distinguishes transition energy from the electron's complete bound-state energy.
Section Two: Problem-solving Question 17
(a) The area is 0.00960 m^2. The torque is NIAB sin35.0° = 0.687 N m.
(b) The resistive voltage is IR = 24.0 V, so the back emf is 72.0 - 24.0 = 48.0 V. The back emf opposes the supply and represents electrical input being converted to mechanical power rather than winding heat.
(c) The winding loss is I^2R = 144 W. The source input is VI = 432 W, leaving 288 W converted mechanically; equivalently EbackI = 48.0 × 6.00 = 288 W.
Detailed marking criteria
Part a (1 mark)
Calculates the coil area as 0.00960 m^2.
Part a (1 mark)
Uses τ = NIAB sinθ with θ between the normal and field.
Part a (1 mark)
Obtains a torque of 0.687 N m.
Part b (1 mark)
Applies V = Eback + IR to the rotating motor.
Part b (1 mark)
Calculates the back emf as 48.0 V.
Part b (1 mark)
Explains that back emf opposes the supply and is associated with mechanical energy conversion.
Part c (1 mark)
Calculates the winding loss as 144 W.
Part c (1 mark)
Calculates the source input power as 432 W.
Part c (2 marks)
Obtains 288 W mechanical conversion by energy balance or EbackI.
Section Three: Comprehension and data analysis Question 18
(a) The strain is ΔL/L = 1.20 × 10^-18/3000 = 4.00 × 10^-22. The supplied observations show opposite changes in perpendicular arms and matching transients at two independent sites with no coincident local vibration. Motion of one mirror does not account for that full spatial and temporal pattern; these observations support a gravitational-wave interpretation without proving that all possible noise sources have been excluded.
(b) The path-length projection difference is cΔt = 3.00 × 10^6 m, or 3000 km. The largest possible delay is separation/c = (3.60 × 10^6)/(3.00 × 10^8) = 0.0120 s = 12.0 ms, so 10.0 ms is possible. As the projected distance is smaller than the baseline, a single delay permits a range of directions; another independent baseline or directional observation is needed.
(c) A consistent light-speed, correlated strain signal supports the prediction that changing mass distributions can radiate gravitational disturbances. The matching differential signals and lack of a coincident vibration-monitor disturbance make a single-site mechanical explanation inadequate. The observations do not directly image the bodies or establish a unique source mass, distance or orientation; those claims require further observations and a tested source model.
Detailed marking criteria
Part a (2 marks)
Calculates the strain as 4.00 × 10^-22.
Part a (3 marks)
Distinguishes the correlated differential spacetime signature from a local single-mirror vibration.
Part b (2 marks)
Uses cΔt to obtain a projected path difference of 3.00 × 10^6 m.
Part b (2 marks)
Explains that one delay supplies a locus rather than a unique sky position.
Part b (2 marks)
Identifies a need for another detector or independent directional information.
Part b (1 mark)
Calculates the maximum light-speed delay across the 3600 km baseline as 12.0 ms.
Part b (1 mark)
Checks that the measured 10.0 ms delay is below this maximum and hence physically possible.
Part c (2 marks)
Links the correlated strain to the supplied prediction of propagating spacetime distortion.
Part c (2 marks)
Uses the supplied two-site correlation and vibration-monitor evidence to evaluate a local mechanical explanation.
Part c (2 marks)
Identifies at least one source property that is not uniquely established by strain alone.
Part c (1 mark)
Forms a qualified conclusion that distinguishes observational support from complete proof.
Section Three: Comprehension and data analysis Question 19
(a) The gradient is ΔF/ΔI = 0.060/2.50 = 0.0240 N A^-1. Since F = BIl for a perpendicular wire, gradient = Bl and B = 0.0240/0.0800 = 0.300 T.
(b) The values lie on a straight line through the origin within ±0.002 N, and equal 0.50 A current steps produce equal 0.012 N force steps. The data therefore support F proportional to I over the measured range, although they do not test behaviour outside it.
(c) For B = gradient/l, percentage uncertainty = (0.0008/0.0240 + 0.0005/0.0800)×100% = 3.96%. The absolute uncertainty is 0.0396 × 0.300 = 0.0119 T, so B = (0.300 ± 0.012) T.
(d) For example, take half the difference between signed force readings at equal and opposite currents to cancel a constant force-sensor zero offset. Independently measure and constrain the active wire length within the uniform field to improve the length used in F = BIl. These address force offset and geometry respectively. Repeating readings and averaging to reduce random scatter is another valid improvement. Any two suitable improvements addressing different uncertainty sources can receive full credit; all three examples are not required.
Detailed marking criteria
Part a (2 marks)
Determines the best-fit gradient as 0.0240 N A^-1 using well-separated points.
Part a (2 marks)
Identifies the graph gradient as Bl for a perpendicular wire.
Part a (1 mark)
Calculates the magnetic-field magnitude as 0.300 T.
Part b (2 marks)
Uses the straight-line trend and near-zero intercept as evidence for direct proportionality.
Part b (2 marks)
Relates equal current increments to equal force increments.
Part b (1 mark)
Qualifies the conclusion to the measured current range and stated uncertainty.
Part c (2 marks)
Combines the independent percentage uncertainties for gradient divided by length.
Part c (2 marks)
Calculates the percentage uncertainty as approximately 3.96%.
Part c (1 mark)
Reports B as approximately (0.300 ± 0.012) T.
Part d (2 marks)
Proposes a first suitable specific improvement (1 mark) and explains how it improves the evidence (1 mark). Accept current reversal, zero checks, active-length control, repetition or another valid method.
Part d (2 marks)
Proposes a second suitable specific improvement (1 mark) and explains how it improves the evidence (1 mark). Accept any independent valid method; no particular pair of examples is required.
Part d (1 mark)
Explains why the two improvements address different uncertainty sources, such as force offset, active-length uncertainty or random scatter.
Diagnostic Checklist
| Review focus | Section | Reflection and next action |
|---|---|---|
| Unit 3 - Equilibrium and motion | Q1, Q2, Q3, Q12 | Review torque, equilibrium, force resolution, projectiles and circular-motion energy. |
| Unit 3 - Gravitation and relativity | Q4, Q13, Q18 | Review gravitational fields, satellite motion, Kepler's law and relativistic frames. |
| Unit 4 - Electric and magnetic fields | Q5, Q6, Q17 | Review electric fields, potential difference, magnetic fields, forces and motor torque. |
| Unit 4 - Induction and field applications | Q7, Q8, Q14, Q15 | Review flux, Faraday and Lenz laws, generators, transformers, accelerators and relativistic energy. |
| Unit 4 - Quantum physics and cosmology | Q9, Q10, Q16 | Review waves, photons, atomic spectra, matter waves, redshift and Big Bang evidence. |
| Units 3 and 4 - Science inquiry skills | Q11, Q19 | Review uncertainty, graph analysis, experimental validity, evidence and justified improvements. |