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2026 VCE Physics exam: What to learn from last year's mistakes
Hi all, in preparation for the upcoming exam, here's what to learn from last year's mistakes.
These are the 2025 questions I would redo first.
Section B, Q11b: average 0.3/3; scored zero 85%; full marks 9%.
Section B, Q2c: average 0.5/3; scored zero 71%; full marks 11%.
Section B, Q6b: average 0.6/2; scored zero 68%; full marks 27%.
Section B, Q19b: average 0.7/2; scored zero 61%; full marks 30%.
Section B, Q19a: average 1.0/3; scored zero 38%; full marks 8%.
VCAA's figures are rounded. These are question averages, not study scores.
1. Q11b: Calculate the power reaching the shed
Where students went wrong
Students used the turbine’s 5.00 kW maximum output as though it was the power being delivered to the lamps. The question gave the voltage at both ends of the transmission wires for a reason.
What they should have done
The turbine supplied 230 V RMS; the shed received 210 V RMS. The total wire resistance was 6.0 Ω.
Voltage lost in the wires = 230 − 210 = 20 V.
Line current = 20 ÷ 6.0 = 3.333… A.
Power at the shed = 210 × 3.333… = 700 W.
Number of 70 W lamps = 700 ÷ 70 = 10.
Use the shed voltage when calculating shed power. Using 230 V includes power dissipated in the wires.
2. Q2c: Area under a force–displacement graph is work
Where students went wrong
Some answers used the 60 N s impulse from part a. That belongs to a force–time calculation. This graph had displacement on the horizontal axis.
What they should have done
The 5.0 kg head slowed from 12 m/s to rest, so it lost ½ × 5.0 × 12² = 360 J of kinetic energy.
The graph was a triangle from 0 to 0.20 m, then a rectangle to 0.40 m:
½ × 0.20 × Fmax + 0.20 × Fmax = 360.
0.30Fmax = 360.
Fmax = 1200 N.
That is the magnitude. Check the axis before choosing momentum or energy.
3. Q6b: Same voltage, same energy gained
Where students went wrong
Students assumed that increasing the separation automatically changed the final kinetic energy, or relied on a vague claim about the particle travelling for longer.
What they should have done
The potential difference was unchanged, so the energy transferred was still qV. Increasing the gap reduces the electric field, E = V/d, and therefore the force. That smaller force acts over a greater distance; the work remains qV. Connect the changed field to the unchanged energy.
4. Q19b: Explain why those sides feel no magnetic force
Where students went wrong
Answers talked about motors or said that currents in opposite directions cancelled. That missed the orientation of each wire relative to the field.
What they should have done
PQ and RS ran parallel to the magnetic field inside the solenoid. For each side, F = BIL sin θ, with θ = 0° or 180°, so F = 0. Opposite current directions alone do not establish zero force.
5. Q19a: Name the actual variables
Where students went wrong
Students confused the current being measured in the wire loop with the current in the solenoid, or named equipment instead of a variable.
What they should have done
The independent variable was the suspended mass. The dependent variable was the current in the wire loop. A control could be the solenoid current, which keeps its magnetic field fixed, or the length QR in the field. Name the circuit component so the marker knows which current you mean.
Source: VCAA 2025 Physics examination report: source. Averages and percentages above are VCAA's published, rounded figures; the explanations are my paraphrases.
Use the VCE Physics hub and past papers and examiner reports alongside this.
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