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QCE Physics External Assessment strategy: The 2026 guide to Papers 1 and 2

Study guide18 min read

Quick answer

The QCE Physics External Assessment is two 60-mark papers worth 50 percent of the subject result combined, examining Units 3 and 4 cumulatively, and the eight-week preparation plan that scales highest is two weeks of Unit 3 consolidation, two weeks of Unit 4 core (special relativity, photoelectric effect, Bohr atom, nuclear physics), two weeks of timed paper practice with self-marking against QCAA schemes, one week of targeted weak-area drills and a week of taper, paired with deep familiarity with the QCAA Senior Physics Formula and Data Book.

Jump to a section
  1. What this guide is for
  2. Paper structure
  3. The eight-week study plan
  4. The formula sheet and data booklet
  5. In-paper time allocation
  6. The four highest-yield Unit 4 topics
  7. Common pitfalls
  8. Check your knowledge

What this guide is for

The QCE Physics External Assessment is the culminating assessment for General Physics. Two papers, each 90 minutes plus 5 minutes of perusal in 2026, sat in the external assessment block. Combined they are 50 percent of the subject result. The EA is cumulative across Units 3 (Gravity and electromagnetism) and 4 (Revolutions in modern physics), so Unit 3 material remains examinable months after IA1 and IA2 are over.

This guide covers paper structure, the eight-week preparation plan, the formula sheet and data booklet, in-paper time allocation, the four highest-yield Unit 4 topics, and the common pitfalls that cost top-band candidates marks.

Paper structure

Each paper is 90 minutes plus 5 minutes of perusal in 2026. QCAA sets the mark allocation, sections and question types, so check the current syllabus and QCAA's assessment pages for the exact format. Earlier papers have combined multiple choice and short response with stimulus material (graphs, tables, apparatus diagrams) and items integrating data analysis with multi-part calculation and explanation. The pacing targets below are a guide and should be adjusted to the format you are given.

Multiple choice items cover content recall, formula application and basic interpretation. Short response items require calculation, derivation or focused explanation. Data-analysis items reward the same data-test skills tested in IA1: extract values, identify the relevant principle, link to a claim. Pacing target: 1.5 minutes per mark plus 30 seconds of planning before writing an extended item.

The eight-week study plan

Weeks one and two: Unit 3 consolidation

Goals. Refresh Unit 3 content five to seven months after IA1 and IA2.

Tasks. Reread one Unit 3 topic per day. Work three problems per topic. By end of week two, every Unit 3 formula should be retrievable without the booklet.

Time. About 60 minutes per day.

Projectile trajectory with two velocity snapshots, 25 m/s at 40 degrees A parabolic trajectory for a projectile launched from the origin at 25 metres per second at 40 degrees above the horizontal, sampled at 51 points from y equals v zero y times t minus one half g t squared with g equals 9.8. Two velocity snapshot arrows are drawn: one rising at t equals 0.6 seconds with components 19.2 horizontal and 10.2 vertical, and one falling at t equals 2.4 seconds with components 19.2 horizontal and minus 7.5 vertical. The constant horizontal component contrasts with the changing vertical component, so the difference between the two arrows is the action of gravity over 1.8 seconds. x y g v(t1) |v| = 21.7 m/s v(t2) |v| = 20.6 m/s v0 v0 = 25 m/s at 40°; vx is identical at both snapshots, vy changes by −g per second.
Two velocity snapshots make decomposition visible: the horizontal arrows are the same length while the vertical components fall by 9.8 m/s every second, which is the entire content of EA projectile questions.
Centripetal force and tangential velocity at two snapshots on a circular orbit A particle moves anticlockwise around a circle of radius 80 pixels centred at (270, 160). Two snapshots are shown at angles 45 degrees and 135 degrees from the positive x axis. At each snapshot, the velocity vector v is drawn tangent to the circle (perpendicular to the radius) and the centripetal force F is drawn radially inward toward the centre. The equation panel beside the figure shows F equals m v squared over r as a stacked fraction. O r v F v F ω F = mv2 r tangent v, inward F: a closed orbit of radius r
Two snapshots 90° apart make centripetal motion visible: the velocity rotates with the position while the inward force keeps the radial distance constant, so the speed |v| stays the same and the trajectory closes.

Weeks three and four: Unit 4 core

Goals. Master the four highest-yield Unit 4 topics: special relativity, photoelectric effect, Bohr atom, nuclear physics.

Tasks. For each topic, write a one-page summary of the named equations and the conceptual reasoning ('why is time dilation a consequence of the constancy of cc'). Work five calculation problems per topic.

Time. About 75 minutes per day.

Weeks five and six: timed paper practice

Goals. Sit two full EA papers under timed conditions. Self-mark against QCAA marking schemes (released with every past paper on the QCAA Senior Physics subject page).

Tasks. Paper 1 in week five, paper 2 in week six. After each, identify the lowest-scoring question type and design a 90-minute intervention before the next paper.

Time. Two timed papers per week (180 minutes plus rework) plus 30 minutes daily of targeted weak-area drills.

Week seven: integration and weak-area drills

Goals. Address the specific weak areas identified in weeks five and six.

Tasks. One more full timed paper. Targeted drills on the weakest topic.

The error log built in IA1 preparation is extended here. Every error gets one line: question type, mistake, fix.

Week eight: taper

Goals. Reduce volume, increase reliability.

Tasks. Three short sessions (30 to 45 minutes) on three different topic areas. Reread the formula sheet daily. Reread the error log nightly.

The day before each paper. Reread the error log. Light review of the formula booklet. Sleep.

The formula sheet and data booklet

QCAA provides a Senior Physics Formula and Data Book with every EA paper. Every formula required for the assessment is in this booklet, alongside fundamental constants and conversion factors.

Knowing the booklet is part of preparation. Print a copy on day one of the eight-week plan and keep it on your desk for every practice question. By week three, you should know which page contains which family of formulas without checking.

The booklet is helpful but it is not a free lookup. A candidate who has to hunt for E=hfE = hf during paper one has already lost a minute. A candidate who knows it is on page two of the booklet, in the photons section, loses ten seconds.

In-paper time allocation

As a guide on an earlier-style layout: multiple choice about 1.5 minutes per question, short response about 1.5 minutes per mark, with time for stimulus reading and about 30 seconds of planning per extended response. Keep five minutes at the end of each paper for review.

If you fall behind. Skip and return. Multi-mark questions where the first part is unclear can have later parts attempted independently. Mark the skip with a clear circle to find on return.

If you finish early. Reread for sig figs, units, missed parts. Almost every paper has at least one missed mark recoverable on review.

The four highest-yield Unit 4 topics

Special relativity. Time dilation, length contraction, relativistic momentum, relativistic energy. The named equations are Δt=γΔt0\Delta t = \gamma \Delta t_0, L=L0/γL = L_0/\gamma, p=γmvp = \gamma m v and E=γmc2E = \gamma m c^2 where γ=1/1−v2/c2\gamma = 1/\sqrt{1 - v^2/c^2}. Conceptual reasoning. Common questions involve muon decay, GPS satellites, particle accelerators.

Photoelectric effect. The named equation is Ek=hf−ϕE_k = hf - \phi where ϕ\phi is the work function. Concepts. Threshold frequency, stopping voltage, why classical electromagnetic theory cannot explain the observations. Common stimulus types are stopping-voltage graphs and photocurrent-frequency graphs.

Bohr atom and atomic spectra. Energy level diagrams. Transitions emit photons of energy ΔE=hf=hc/λ\Delta E = hf = hc/\lambda. The hydrogen spectrum and the Rydberg formula are typical. Common questions involve calculating the wavelength of a specified transition or identifying which transition produces a given line.

Bohr-model hydrogen energy levels with the Hα and Hβ Balmer transitions Hydrogen energy levels drawn at vertical positions computed from E n equals minus 13.6 over n squared electron-volts, with n equals 1 deep at the bottom and n equals 2 through 5 crowding toward the ionisation continuum at the top. Two visible Balmer transitions, n equals 3 to n equals 2 at 656.3 nanometres and n equals 4 to n equals 2 at 486.2 nanometres, are drawn as downward accent arrows landing on the n equals 2 line. The true-to-scale spacing puts ten point two electron-volts between n equals 1 and 2 and only one point nine between n equals 2 and 3. ionisation continuum (n → ∞) n = 1 n = 2 n = 3 n = 4 n = 5 −13.60 eV −3.40 eV −1.51 eV −0.85 eV −0.54 eV 656 nm (Hα) 486 nm (Hβ) Balmer series (n → 2) ∆E = hc · (Rydberg) eV·nm Levels: En = −13.6 eV ÷ n² photon energy ΔE = hf
The n = 1 to n = 2 gap is over five times the n = 2 to n = 3 gap; this real spacing is what makes the Balmer photons visible while Lyman photons are ultraviolet.

Nuclear physics. Mass defect from atomic mass and nucleon count. Binding energy from E=mc2E = mc^2. Binding energy per nucleon as the stability indicator. Decay equations balancing mass number and atomic number for alpha, beta-minus, beta-plus and gamma decay. Common questions ask for nuclear equation completion or stability comparisons.

Common pitfalls

Sig figs. The QCAA convention is three significant figures unless the data warrants otherwise. Two sig figs is acceptable when the data has only two; four is excessive. Every numerical answer carries explicit sig fig consideration.

Units. Every numerical answer carries explicit units. Lost-mark patterns at top band cluster around units (frequency in Hz not just s−1^{-1}, momentum in kg m s−1^{-1}, energy in J or eV consistently throughout a question).

Sign conventions. Vector quantities require positive-direction consistency within a question. Lenz's law sign, gravitational potential energy sign, work-energy theorem sign. Set a convention at the start and stick to it.

Extended response without planning. Six to ten-mark questions launched into without a 30-second plan typically lose two to three marks for missing structure. The 30 seconds spent listing the principles to invoke is recovered five times over in writing.

Formula booklet hunting. As above. Know the booklet.

Calculator-style answers. A numerical answer without interpretation is half a mark. Every numerical answer in a reasoning context needs a sentence: "this means that ..." or "therefore the claim is supported because ...".

Check your knowledge

Eight EA-style multi-part questions across the Unit 3 and Unit 4 subject matter, with mark allocations shown. Attempt under exam conditions (about 1 minute per mark) before checking against the solutions block. Take g=9.80 m s−2g = 9.80 \ \text{m s}^{-2}, c=3.00×108 m s−1c = 3.00 \times 10^{8} \ \text{m s}^{-1}, h=6.63×10−34 J sh = 6.63 \times 10^{-34} \ \text{J s}, e=1.60×10−19 Ce = 1.60 \times 10^{-19} \ \text{C}, me=9.11×10−31 kgm_e = 9.11 \times 10^{-31} \ \text{kg} where required.

  1. A projectile is launched at 25 m s−1^{-1} at 35 degrees above the horizontal from level ground. Air resistance is negligible. (a) Calculate the maximum height. (b) Calculate the horizontal range. (c) Calculate the speed and direction of the projectile 1.50 s after launch. (8 marks)
  2. A 1500 kg car negotiates a horizontal circular curve of radius 80 m at 20 m s−1^{-1}. (a) Calculate the centripetal acceleration. (b) Calculate the magnitude and direction of the net horizontal force on the car. (c) Calculate the minimum coefficient of static friction between the tyres and the road that allows this turn. (5 marks)
  3. A square loop of side 0.20 m, with 200 turns and resistance 4.0 ohms, lies in a uniform magnetic field that is initially 0.60 T directed perpendicular into the page. The field is reduced linearly to zero over 0.50 s. (a) Calculate the magnitude of the induced EMF. (b) Calculate the magnitude of the induced current. (c) State, with reference to Lenz's law, the direction of the induced current as viewed from in front of the page. (6 marks)
  4. (a, 3) Light of wavelength 400 nm illuminates a sodium metal surface (work function 2.28 eV). Calculate the maximum kinetic energy of emitted photoelectrons in eV. (b, 2) State whether photoelectrons are emitted when the same surface is illuminated by 600 nm light, justifying with calculation. (c, 2) Explain in two sentences why this observation supports the photon model over the wave model of light. (7 marks)
  5. A muon created 60 km above the Earth has a proper lifetime of 2.20 micro-seconds and travels at 0.999c relative to the Earth. (a) Calculate the time-dilated lifetime in the Earth's frame. (b) Calculate the distance the muon travels in the Earth's frame in this lifetime. (c) Calculate the length-contracted distance from 60 km altitude in the muon's frame, then state whether the muon reaches the surface and justify. (7 marks)
  6. The first three Bohr energy levels of hydrogen are E1=−13.6 eVE_1 = -13.6 \ \text{eV}, E2=−3.40 eVE_2 = -3.40 \ \text{eV}, E3=−1.51 eVE_3 = -1.51 \ \text{eV}. (a) Calculate the wavelength of the photon emitted when an electron transitions from n=3n = 3 to n=2n = 2. (b) State the region of the electromagnetic spectrum. (c) Sketch and label an energy-level diagram showing this transition and its emitted photon. (5 marks)
  7. 92238U^{238}_{92}U decays through a series of alpha and beta-minus emissions to 82206Pb^{206}_{82}Pb. (a) Determine the total number of alpha and beta-minus particles emitted. (b) The half-life of 238U^{238}U is 4.5×1094.5 \times 10^{9} years. Calculate the activity of a 1.00 mg sample (Mr=238.05M_r = 238.05, NA=6.02×1023N_A = 6.02 \times 10^{23}). (c) Explain in one sentence why 238U^{238}U remains naturally abundant on Earth despite being radioactive. (6 marks)
  8. A 50 km transmission line carries 200 MW of electrical power from the Townsville coal-fired plant to inland mining customers. The line has total resistance 4.0 Ω4.0 \ \Omega. (a) Calculate the line current and the percent power loss at the operating voltage of 275 kV. (b) Calculate the percent power loss if the same line operated at 110 kV. (c) Explain in two sentences why long-distance transmission uses very high voltages despite the cost of step-up and step-down transformers. (6 marks)
Solutions
Q1
Decompose: ux=25cos⁡35=20.5 m s−1u_x = 25 \cos 35 = 20.5 \ \text{m s}^{-1}; uy=25sin⁡35=14.3 m s−1u_y = 25 \sin 35 = 14.3 \ \text{m s}^{-1}. (a) At max height vy=0v_y = 0, so h=uy2/(2g)=(14.3)2/19.6=10.4 mh = u_y^2 / (2g) = (14.3)^2 / 19.6 = 10.4 \ \text{m}. (b) Time of flight T=2uy/g=2×14.3/9.80=2.92 sT = 2 u_y / g = 2 \times 14.3 / 9.80 = 2.92 \ \text{s}. Range R=uxT=20.5×2.92=59.9 mR = u_x T = 20.5 \times 2.92 = 59.9 \ \text{m}. (c) At t=1.50t = 1.50 s: vx=20.5 m s−1v_x = 20.5 \ \text{m s}^{-1}; vy=14.3−9.80×1.50=−0.40 m s−1v_y = 14.3 - 9.80 \times 1.50 = -0.40 \ \text{m s}^{-1} (just past apex, descending). v=20.52+0.402=20.5 m s−1v = \sqrt{20.5^2 + 0.40^2} = 20.5 \ \text{m s}^{-1}. Angle below horizontal =arctan⁡(0.40/20.5)=1.1= \arctan(0.40/20.5) = 1.1 degrees. The projectile rises to 10.4 m, lands 59.9 m downrange, and at 1.50 s is moving 20.5 m s−1^{-1} at 1.1 degrees below the horizontal.
Q2
(a) ac=v2/r=202/80=5.00 m s−2a_c = v^2 / r = 20^2 / 80 = 5.00 \ \text{m s}^{-2}. (b) Fnet=mac=1500×5.00=7500 NF_{net} = m a_c = 1500 \times 5.00 = 7500 \ \text{N} directed horizontally toward the centre of the circle. (c) Friction provides this force: μmg≥Fnet\mu m g \geq F_{net}, so μ≥Fnet/(mg)=7500/(1500×9.80)=0.510\mu \geq F_{net} / (m g) = 7500 / (1500 \times 9.80) = 0.510. The minimum coefficient of static friction is 0.51.
Q3
(a) Area A=0.202=0.040 m2A = 0.20^2 = 0.040 \ \text{m}^2. Flux change per turn ∣ΔΦ∣=0.60×0.040=0.024|\Delta \Phi| = 0.60 \times 0.040 = 0.024 Wb. ∣ε∣=N∣ΔΦ/Δt∣=200×0.024/0.50=9.6 V|\varepsilon| = N |\Delta \Phi / \Delta t| = 200 \times 0.024 / 0.50 = 9.6 \ \text{V}. (b) I=ε/R=9.6/4.0=2.4 AI = \varepsilon / R = 9.6 / 4.0 = 2.4 \ \text{A}. (c) The flux into the page is decreasing; Lenz's law says the induced current opposes this decrease by generating flux into the page, so by the right-hand rule the current flows clockwise as viewed from in front of the page.
Q4
(a) Photon energy Eph=hc/λ=6.63×10−34×3.00×108/(400×10−9)=4.97×10−19E_{ph} = hc/\lambda = 6.63 \times 10^{-34} \times 3.00 \times 10^{8} / (400 \times 10^{-9}) = 4.97 \times 10^{-19} J =4.97×10−19/1.60×10−19=3.11= 4.97 \times 10^{-19} / 1.60 \times 10^{-19} = 3.11 eV. Kmax=Eph−ϕ=3.11−2.28=0.83K_{max} = E_{ph} - \phi = 3.11 - 2.28 = 0.83 eV. (b) Photon energy at 600 nm =1240/600=2.07= 1240/600 = 2.07 eV (using 1240 eV nm1240 \ \text{eV nm}); less than ϕ=2.28\phi = 2.28 eV, so no photoelectrons are emitted regardless of intensity. (c) The wave model predicts that high-intensity 600 nm light should eventually deliver enough energy to eject electrons, but observation shows no emission at any intensity below threshold frequency; energy of light arrives in discrete photon packets, supporting the photon model.
Q5
(a) Lorentz factor γ=1/1−0.9992=1/1−0.998=1/0.001998=22.4\gamma = 1/\sqrt{1 - 0.999^2} = 1/\sqrt{1 - 0.998} = 1/\sqrt{0.001998} = 22.4. Time in Earth frame t=γτ=22.4×2.20×10−6=4.92×10−5t = \gamma \tau = 22.4 \times 2.20 \times 10^{-6} = 4.92 \times 10^{-5} s. (b) Distance in Earth frame d=vt=0.999×3.00×108×4.92×10−5=1.47×104d = v t = 0.999 \times 3.00 \times 10^{8} \times 4.92 \times 10^{-5} = 1.47 \times 10^{4} m =14.7= 14.7 km. (c) Length-contracted distance in muon's frame: L=L0/γ=60 000/22.4=2.68×103L = L_0/\gamma = 60\,000/22.4 = 2.68 \times 10^{3} m =2.68= 2.68 km. In its own frame, the muon traverses only 2.68 km in 2.20 micro-seconds at 0.999c (which equals 0.999 ×\times 3.00 ×\times 108^{8} ×\times 2.20 ×\times 10−6^{-6} = 660 m), so it does not reach the surface in its own frame's pure decay scenario; however, both frames agree on the outcome. In the Earth frame the muon survives long enough (time-dilated) to cover 14.7 km, still under 60 km, so under these idealised parameters the muon does not reach the surface. (In reality, atmospheric muons travel at varying speeds and 0.9999c muons do reach the surface; this question's data give 14.7 km of travel.) The Earth-frame calculation gives 14.7 km of penetration, less than the 60 km source altitude.
Q6
(a) ΔE=E3−E2=−1.51−(−3.40)=1.89\Delta E = E_3 - E_2 = -1.51 - (-3.40) = 1.89 eV. λ=hc/ΔE=1240 eV nm/1.89=656\lambda = hc/\Delta E = 1240 \ \text{eV nm} / 1.89 = 656 nm. (b) Visible (red) region of the electromagnetic spectrum; this is the H-alpha line of the Balmer series. (c) Diagram: horizontal lines at E=−13.6,−3.40,−1.51E = -13.6, -3.40, -1.51 eV labelled n=1,2,3n = 1, 2, 3; downward arrow from n=3n = 3 to n=2n = 2 labelled "1.89 eV, 656 nm photon emitted". The emitted photon is in the visible region at 656 nm.
Q7
(a) Mass-number balance: 238=206+4a238 = 206 + 4a, so a=8a = 8 alphas. Atomic-number balance: 92=82+2a−b92 = 82 + 2a - b, so b=2a−10=6b = 2a - 10 = 6 beta-minus particles. (b) Number of 238U^{238}U atoms: N=(1.00×10−3/238.05)×6.02×1023=2.53×1018N = (1.00 \times 10^{-3} / 238.05) \times 6.02 \times 10^{23} = 2.53 \times 10^{18}. Decay constant λ=ln⁡2/T1/2=0.693/(4.5×109×3.156×107)=4.88×10−18\lambda = \ln 2 / T_{1/2} = 0.693 / (4.5 \times 10^{9} \times 3.156 \times 10^{7}) = 4.88 \times 10^{-18} s−1^{-1}. Activity A=λN=4.88×10−18×2.53×1018=12.3A = \lambda N = 4.88 \times 10^{-18} \times 2.53 \times 10^{18} = 12.3 Bq. (c) 238U^{238}U remains naturally abundant because its half-life (4.5×1094.5 \times 10^{9} years) is comparable to the age of the Earth, so much of the original supply persists.
Q8
(a) I=P/V=2.00×108/2.75×105=727 AI = P/V = 2.00 \times 10^{8} / 2.75 \times 10^{5} = 727 \ \text{A}. Loss PL=I2R=(727)2×4.0=2.12×106P_L = I^2 R = (727)^2 \times 4.0 = 2.12 \times 10^{6} W =2.12= 2.12 MW. Percent loss =2.12/200×100=1.06= 2.12/200 \times 100 = 1.06 percent. (b) At 110 kV: I=2.00×108/1.10×105=1818 AI = 2.00 \times 10^{8} / 1.10 \times 10^{5} = 1818 \ \text{A}. Loss =18182×4.0=1.32×107= 1818^2 \times 4.0 = 1.32 \times 10^{7} W =13.2= 13.2 MW. Percent loss =13.2/200×100=6.6= 13.2/200 \times 100 = 6.6 percent. (c) The I2RI^2 R loss scales as 1/V21/V^2 for a fixed transmitted power, so raising the transmission voltage from 110 kV to 275 kV reduces losses by a factor of about 6.3; the once-off cost of step-up and step-down transformers is recovered many times over by the saved power over the life of the line.

Sources & how we know this

Based on the official QCAA syllabus and published QCAA past papers and marking guidelines. Not yet individually reviewed by a person; always check QCAA and your school's materials for current rules.

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