§-Past paper

VICPhysics2023

VCE Physics 2023

Walkthrough of the 2023 VCE physics exam: what it assessed, strategy tips, and the common errors flagged in the official marker report.

Marks: 130
Time: 150 min
Authority: VCAA
Updated: 2026-05-29

Open the official PDF ->Official marking notes

What this paper assessed

The 2023 VCE Physics Units 3 & 4 examination drew across the VCAA study design:

Unit 3 - How do fields explain motion and electricity? gravitational, electric and magnetic fields; motion in fields (orbits, projectiles); the motor effect, electromagnetic induction (Faraday's and Lenz's laws), transformers and AC generation.
Unit 4 - How have understandings of the physical world changed? wave and particle models of light, the photoelectric effect, special relativity (time dilation, length contraction, mass-energy equivalence), and an experimental-design/uncertainty component.

Many items required students to extract data from a graph, apply a relationship, then comment on the limitations of a model. The extended-response items rewarded structured reasoning over volume of writing, with correct units and the relevant law named before the effect was described.

Structure and timing

The paper is 130 marks in 150 minutes, plus 15 minutes reading time. That is roughly 1.15 minutes per mark.

A 4-mark calculation deserves ~4.5 minutes; a longer explanation item scales up accordingly.
Use reading time to scan the graph-based and multi-part questions and to plan the longer responses.
List the formula, the substituted values and the unit-checked answer on separate lines, and reserve ~10 minutes at the end for checking.

Worked practice questions (exam-style)

Worked example

Question 1 (4 marks): A satellite orbits a planet of mass $M = 6.0 \times 10^{24}\ \text{kg}$ at radius $r = 8.0 \times 10^{6}\ \text{m}$ . Calculate its orbital period. ( $G = 6.67 \times 10^{-11}\ \text{N m}^2\text{kg}^{-2}$ .)

Setting gravitational force equal to centripetal force and eliminating the orbital speed gives a formula for the period directly in terms of $G$ , $M$ , and $r$ .

Step 1: Write the period formula

From equating gravitational and centripetal expressions (and using $v = 2\pi r / T$ ), we obtain:

T = 2\pi\sqrt{\frac{r^3}{GM}}.

This is equivalent to Kepler's third law.

Step 2: Compute the required products

Cube the radius: $r^3 = (8.0 \times 10^{6})^3 = 5.12 \times 10^{20}\ \text{m}^3$ . Compute $GM = (6.67 \times 10^{-11})(6.0 \times 10^{24}) = 4.00 \times 10^{14}$ .

Step 3: Substitute and evaluate

T = 2\pi\sqrt{\frac{5.12 \times 10^{20}}{4.00 \times 10^{14}}} = 2\pi\sqrt{1.28 \times 10^{6}} = 2\pi \times 1131 \approx 7.1 \times 10^{3}\ \text{s}.

Final answer: the orbital period is approximately $7.1 \times 10^{3}\ \text{s}$ (2 significant figures).

Worked example

Question 2 (5 marks): A photon of light with frequency $f = 8.0 \times 10^{14}\ \text{Hz}$ strikes a metal with work function $\phi = 2.0\ \text{eV}$ . Determine whether photoelectrons are emitted and, if so, their maximum kinetic energy. ( $h = 6.63 \times 10^{-34}\ \text{J s}$ , $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$ .)

We must compare the photon energy with the work function to decide whether emission occurs, then apply Einstein's photoelectric equation. Working in electron-volts throughout avoids repeated unit conversions.

Step 1: Calculate the photon energy in joules and convert to eV

E = hf = (6.63 \times 10^{-34})(8.0 \times 10^{14}) = 5.30 \times 10^{-19}\ \text{J}.

Converting: $E = \dfrac{5.30 \times 10^{-19}}{1.6 \times 10^{-19}} = 3.31\ \text{eV}$ .

Step 2: Compare photon energy with the work function

Since $E = 3.31\ \text{eV} > \phi = 2.0\ \text{eV}$ , the photon has enough energy to free an electron. Photoelectrons are emitted.

Step 3: Apply Einstein's photoelectric equation for maximum kinetic energy

The excess energy after overcoming the work function becomes kinetic energy:

E_{k,\text{max}} = hf - \phi = 3.31 - 2.0 = 1.31\ \text{eV}\ \big(\approx 2.1 \times 10^{-19}\ \text{J}\big).

Final answer: photoelectrons are emitted with a maximum kinetic energy of $1.31\ \text{eV}$ ( $\approx 2.1 \times 10^{-19}\ \text{J}$ ).

Worked example

Question 3 (5 marks): A spaceship of proper length $120\ \text{m}$ travels at $v = 0.60c$ relative to a station. Calculate its length as measured by an observer on the station, and explain the result.

The proper length $L_0 = 120\ \text{m}$ is measured in the ship's rest frame. The station observer measures the contracted length:

L = L_0\sqrt{1 - \frac{v^2}{c^2}}

Compute the factor:

\sqrt{1 - 0.60^2} = \sqrt{1 - 0.36} = \sqrt{0.64} = 0.80

So:

L = 120 \times 0.80 = 96\ \text{m}

Explanation: length contraction occurs only along the direction of relative motion; the station observer measures the ship as shorter (96 m) than its proper length, a direct consequence of special relativity.

Common errors students made

Recurring errors in VCE Physics responses include:

Missing or inconsistent units in working and final answers.
Mixing scalar and vector quantities (speed vs velocity, distance vs displacement).
Treating projectile motion as one-dimensional instead of resolving into independent components.
Forgetting that induced EMF depends on the rate of change of flux, not flux itself.

Add these subject-specific traps:

Confusing proper length/time with the contracted/dilated value in relativity.
Forgetting to convert between joules and electron-volts in photoelectric calculations.
In Lenz's-law items, stating a direction without linking it to opposing the flux change and to energy conservation.
Mishandling measurement uncertainty - failing to quote uncertainties or significant figures consistent with the data.

How to use this paper

Sit the full 130-mark paper in 150 minutes under timed conditions with the VCAA formula sheet and an approved calculator. Mark against the VCAA examination report (linked in the frontmatter), which lists the most common errors and average marks per question. Rebuild every calculation that lost a method mark from a clean start, listing formula, substitution and unit-checked answer on separate lines. Keep an error log mapping each lost mark to a study-design dot point, and re-attempt graph-interpretation and relativity items under time pressure after a 48-hour gap.

Use this paper well

Sit the paper under exam conditions (150 minutes, 130 marks).
Mark yourself against the official VCAA 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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