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QLDPhysics2024

QCE Physics 2024

Walkthrough of the 2024 QCE physics exam: what it assessed, strategy tips, and the common errors flagged in the official marker report.

Marks: 60
Time: 120 min
Authority: QCAA
Updated: 2026-05-29

Open the official PDF ->Official marking notes

What this paper assessed

This QCAA Physics external assessment is one 60-mark paper sat over 120 minutes (plus 5 minutes perusal). It assesses Unit 3 (Gravity and electromagnetism) and Unit 4 (Revolutions in modern physics):

Unit 3 - Gravity and motion: projectile motion, gravitational fields and orbital motion ( $F = \frac{GMm}{r^2}$ , Kepler's laws), and electromagnetism - charged particles and current-carrying conductors in magnetic fields ( $F = qvB$ , $F = BIl$ ), and electromagnetic induction (Faraday's and Lenz's laws).
Unit 4 - Modern physics: special relativity (time dilation, length contraction, mass-energy $E = mc^2$ ), the quantum nature of light and matter (the photoelectric effect, $E = hf$ ), and the standard model / nuclear physics.

Many items require extracting data from a graph, applying a relationship, then commenting on the limitations of the model. The extended response rewards structured reasoning over volume of writing. The 2024 paper leaned harder on application where the context was unfamiliar but the skill was core.

Structure and timing

60 marks in 120 minutes is 2 minutes per mark:

Multiple choice (~20 marks): ~30-35 minutes.
Short and extended response (~40 marks): ~80-85 minutes at ~2 min/mark. A 6-mark graph-analysis or field-calculation item deserves ~12 minutes.
Keep the final ~10 minutes for unit and significant-figure checks.

For each calculation, list the formula, the substituted values, and the answer with units on separate lines; for conceptual items, name the law before describing the effect.

Worked practice questions (exam-style)

Worked example

Question 1 (5 marks): A satellite orbits Earth in a circular orbit of radius $7.0 \times 10^6\ \text{m}$ . Calculate its orbital speed. (Take $G = 6.67 \times 10^{-11}\ \text{N m}^2\,\text{kg}^{-2}$ , $M_{Earth} = 5.97 \times 10^{24}\ \text{kg}$ .)

For a circular orbit the only force acting is gravity, and that force must supply all of the centripetal acceleration. Setting the two expressions equal lets us solve for the orbital speed.

Step 1: Equate gravitational and centripetal forces

The gravitational force on the satellite is $\dfrac{GMm}{r^2}$ and the centripetal force required is $\dfrac{mv^2}{r}$ . Setting them equal and cancelling the satellite mass $m$ :

\frac{GMm}{r^2} = \frac{mv^2}{r} \quad\Rightarrow\quad v = \sqrt{\frac{GM}{r}}.

Step 2: Substitute the known values

Insert $G = 6.67\times10^{-11}\ \text{N m}^2\,\text{kg}^{-2}$ , $M_{Earth} = 5.97\times10^{24}\ \text{kg}$ , and $r = 7.0\times10^6\ \text{m}$ :

v = \sqrt{\frac{(6.67\times10^{-11})(5.97\times10^{24})}{7.0\times10^6}} = \sqrt{\frac{3.982\times10^{14}}{7.0\times10^6}}.

Step 3: Evaluate and state the result

v = \sqrt{5.69\times10^{7}} \approx 7.5 \times 10^3\ \text{m s}^{-1}.

Final answer: the orbital speed is approximately $7.5 \times 10^3\ \text{m s}^{-1}$ (7.5 km s $^{-1}$ ).

Worked example

Question 2 (5 marks): In a photoelectric experiment, light of frequency $7.5 \times 10^{14}\ \text{Hz}$ ejects electrons from a metal with work function $2.0\ \text{eV}$ . Calculate the maximum kinetic energy of the ejected electrons in joules. (Take $h = 6.63 \times 10^{-34}\ \text{J s}$ , $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$ .)

Einstein's photoelectric equation states that the photon energy goes first into freeing the electron (the work function) and any remainder becomes kinetic energy. We calculate each quantity in joules so the units are consistent throughout.

Step 1: Calculate the photon energy in joules

The energy carried by a single photon of frequency $f$ is $E = hf$ :

E = (6.63\times10^{-34})(7.5\times10^{14}) = 4.97\times10^{-19}\ \text{J}.

Step 2: Convert the work function to joules

The work function is given in electron-volts, so convert using $1\ \text{eV} = 1.60\times10^{-19}\ \text{J}$ :

W = 2.0 \times 1.60\times10^{-19} = 3.20\times10^{-19}\ \text{J}.

Step 3: Apply Einstein's photoelectric equation

Since $hf > W$ , electrons are emitted. The maximum kinetic energy is the photon energy minus the work function:

E_{k,\max} = hf - W = 4.97\times10^{-19} - 3.20\times10^{-19} = 1.77\times10^{-19}\ \text{J}.

Final answer: maximum kinetic energy $\approx 1.8 \times 10^{-19}\ \text{J}$ (about 1.1 eV).

Worked example

Question 3 (5 marks): A muon has a proper lifetime of $2.2\ \mu\text{s}$ and travels at $0.99c$ relative to an observer on Earth. Calculate the lifetime measured by the Earth observer and explain why this allows muons to reach the surface.

Time dilation: $t = \dfrac{t_0}{\sqrt{1 - v^2/c^2}}$ , with $t_0 = 2.2\ \mu\text{s}$ and $v = 0.99c$ .

Lorentz factor: $\gamma = \dfrac{1}{\sqrt{1 - 0.99^2}} = \dfrac{1}{\sqrt{1 - 0.9801}} = \dfrac{1}{\sqrt{0.0199}} \approx \dfrac{1}{0.1411} \approx 7.09.$

Dilated lifetime:

t = \gamma t_0 = 7.09 \times 2.2\ \mu\text{s} \approx 15.6\ \mu\text{s}.

Explanation: in the muon's frame it lives only $2.2\ \mu\text{s}$ , which (even at near light speed) would not allow it to reach the ground. But to the Earth observer the muon's "clock" runs slow, so it survives ~15.6 μs - long enough to travel the distance through the atmosphere. (Equivalently, in the muon's frame the atmosphere is length-contracted.)

Common errors students made

The 2024 QCAA marker report flagged that students:

omitted units in answers;
mixed scalar and vector quantities;
treated projectile motion as one-dimensional rather than resolving components;
forgot that induced EMF depends on the rate of change of flux, not flux itself.

Further traps to avoid:

Inconsistent units - failing to convert eV to J, or μs to s, before substituting.
Applying time dilation/length contraction in the wrong reference frame.
Using the wrong significant figures - match the precision of the data.
Stating a graph trend without quoting gradient/intercept values or their physical meaning.

How to use this paper

Sit the multiple choice in ~30 minutes, then the short/extended responses in ~85 minutes, with calculator and data sheet ready.
Self-mark against the QCAA marking guide and marker report at the authority links in the frontmatter, awarding method marks line by line.
For each calculation, write the relationship, substitute, then evaluate with units and sig figs.
Rewrite one extended response per session as principle, application to the named scenario, and evaluation of assumptions/limitations.

Use this paper well

Sit the paper under exam conditions (120 minutes, 60 marks).
Mark yourself against the official QCAA marking notes.
Compare against the Physics hub to find the syllabus dot points this paper tested.

What this paper assessed

Structure and timing

Worked practice questions (exam-style)

Common errors students made

How to use this paper

Use this paper well

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