VCE Physics 2021

Every question from the 2021 VCE Physics examination. Multiple-choice answers come with a short reason; Section B solutions sit behind a Show worked solution toggle. This paper was set on the previous study design, so each question also says whether it still matches the course you are studying now.

How to use this page

• Questions are from the 2021 VCE Physics examination, copyright Victorian Curriculum and Assessment Authority (VCAA). Each is summarised briefly here; open the official examination PDF for the full wording, diagrams and the 2021 formula sheet (bound in at the back).
• Answers are original ExamExplained working. Every multiple-choice answer matches the key in the 2021 Physics external assessment report (Word document), and every Section B calculation was redone and compared with the report. Both files are listed on the VCAA Physics examinations page.
• Percentages after multiple-choice answers are the share of students who chose the correct option. In Section B, "From the report" notes summarise what the examiners said students got wrong.
• One disagreement with the report: in Question 9a the report's working gives 17.3 m s⁻¹, which comes from using $g = 10$ m s⁻². The paper says to use $g = 9.8$ m s⁻², which gives 17.1 m s⁻¹. Both are shown below.

Structure and timing

The paper is 130 marks in 150 minutes (plus 15 minutes reading time), with a scientific calculator, the formula sheet and one sheet of pre-written notes (one folded A3 sheet or two A4 sheets). That is about 1.15 minutes per mark. (Papers from 2024 on are 120 marks.)

• Section A (20 marks): 20 multiple-choice questions. Aim for about 20 minutes.
• Section B (110 marks): 20 questions, including a 17-mark practical investigation (Question 20). Show working, and give answers in the unit printed in the answer box. Take $g = 9.8$ m s⁻² unless told otherwise.

Study design check for 2026

The 2021 exam was written for the 2017 to 2023 study design. The current Units 3 and 4 study design (from 2024) keeps motion, fields, electricity generation and transmission, light and matter, and special relativity, but the old "waves and light" topics have gone: Snell's law, total internal reflection, dispersion, the electromagnetic spectrum and wave basics moved to Unit 1, and the Doppler effect and polarisation are now only in Unit 2 options. When you practise with this paper:

• Skip (no longer in Units 3 and 4): Section A Q14 (uses of the electromagnetic spectrum) and Q15 (dispersion in a prism); Section B Q11 (polarisation), Q12 (Snell's law and critical angle in an optical fibre) and Q14 (Doppler effect).
• Adapt: Section A Q13 (reading amplitude and wavelength off a wave graph is now Unit 1 content, though $v = f\lambda$ is still used in Unit 4) and Q19 (the electron-standing-wave picture of the atom is no longer named; quantisation of atomic energy still is).
• Still fully examinable: everything else, including the motion, fields, generator, transmission, photoelectric, de Broglie, relativity and practical (Q20) questions.

Section A: Multiple choice

Q1

Which darts player's throws are precise but inaccurate? Answer: D - Player D's darts are clustered tightly together (precise) but away from the bullseye (inaccurate). (87%)

Q2

Field lines between four equal charges P, Q, R and S. Answer: B - field lines point away from P and R (positive) and into Q and S (negative). (92%)

Q3

Electric field strength against position between charged parallel plates. Answer: A - the field between parallel plates is uniform, so $E$ is the same at X, Y and Z: a horizontal line. (56%)

Q4

Phobetor has 4 times Earth's mass and surface gravity 18 m s⁻². Its radius? Answer: B - $g = \tfrac{GM}{R^2}$, so $R \propto \sqrt{\tfrac{M}{g}}$ and $\tfrac{R_P}{R_E} = \sqrt{\tfrac{4}{18/9.8}} = 1.48 \approx 1.5$. (44%; 39% chose C.)

Q5

A DC motor coil is released from rest when the switch closes. Answer: C - the current runs from J to K in a field pointing from N to S, so the force on side JK is down (and on LM up): the coil turns in direction B. (51%)

Q6

Six turns, flux change 0.05 Wb, average EMF 1.2 V. Time interval? Answer: C - $\Delta t = \tfrac{N\Delta\Phi}{\varepsilon} = \tfrac{6 \times 0.05}{1.2} = 0.25$ s. (73%)

Q7

An ideal 240 V to 5.0 V charger supplies 3.0 A. Current from the mains? Answer: D - power in equals power out: $I_p = \tfrac{5.0 \times 3.0}{240} = 0.0625 \approx 0.06$ A. (69%)

Q8

Function of the split-ring commutator in a generator. Answer: A - it reverses the connections every half turn, so the external circuit (the globe) receives DC. (51%; 34% chose C.)

Q9

Lucy runs off a platform at 6 m s⁻¹. Her horizontal distance $d$ after time $t$? Answer: B - horizontal velocity is constant, so $d = 6t$. (77%)

Q10

Time to fall 8.0 m. Answer: C - $8.0 = \tfrac12 (9.8) t^2$ gives $t = 1.28 \approx 1.3$ s. (69%)

Q11

Spring constant from the force-compression graph (4.0 kN at 0.04 m). Answer: D - the gradient: $k = \tfrac{4.0 \times 10^3}{0.04} = 1.0 \times 10^5$ N m⁻¹. (63%; 27% chose B.)

Q12

Energy stored at 0.02 m compression. Answer: C - area under the graph: $\tfrac12 \times 0.02 \times 2.0 \times 10^3 = 20$ J (or $\tfrac12 k x^2 = \tfrac12 (1.0 \times 10^5)(0.02)^2$). (68%)

Q13

Amplitude and frequency of a wave (displacement peaks at 8 cm, wavelength 6 cm, speed 18 m s⁻¹). Answer: C - amplitude 8 cm; $f = \tfrac{v}{\lambda} = \tfrac{18}{0.06} = 300$ Hz. (55%; 35% chose A.) Study design: adapt (wave basics are now Unit 1 content).

Q14

Match regions of the electromagnetic spectrum with uses. Answer: D - infrared for thermal imaging, visible for optical microscopy, ultraviolet for water sterilisation, X-rays for medical imaging. (70%) Study design: outside Units 3 and 4 (now Unit 1).

Q15

Dispersion of white light by a prism. Answer: A - violet (shorter wavelength) is refracted more than red, and the colours spread at both faces of the prism. (67%) Study design: outside Units 3 and 4 (now Unit 1).

Q16

What is essential to measuring the maximum kinetic energy of photoelectrons? Answer: C - the stopping voltage: the voltmeter reading when the photocurrent falls to its minimum (zero). (35%; 38% chose B.)

Q17

de Broglie wavelength of a 663 kg car at 10 m s⁻¹. Answer: A - $\lambda = \tfrac{h}{mv} = \tfrac{6.63 \times 10^{-34}}{663 \times 10} = 10^{-37}$ m. (68%)

Q18

Power of a 550 nm source emitting 2.8 × 10¹⁶ photons per second. Answer: A - each photon has $E = \tfrac{hc}{\lambda} = 3.62 \times 10^{-19}$ J, so $P = 3.62 \times 10^{-19} \times 2.8 \times 10^{16} = 1.0 \times 10^{-2}$ W. (36%; 31% chose C.)

Q19

Number of whole wavelengths in a de Broglie standing wave around a hydrogen atom. Answer: B - the diagram shows three wavelengths around the orbit, so $n = 3$. (64%) Study design: adapt (the standing-wave model of the atom is no longer named).

Q20

A property of an inertial frame of reference. Answer: D - it is not accelerating, so no observer inside can detect any acceleration of the frame. "Constant speed" (A) is not enough, since moving in a circle at constant speed is accelerated motion. (49%; 37% chose A.)

Section B

Question 1 (3 marks)

Two identical bar magnets are at right angles and the same distance from point P: one lies horizontally to the left of P with its N pole nearest P; the other lies vertically below P with its S pole nearest P.

a. Draw the direction of the combined magnetic field at P. (1 mark)

b. Each magnet's field at P is 10.0 mT. Calculate the combined field strength. (2 marks)

Study design: in the current course.

Show worked solution

a. [1 mark]. Field lines point away from a N pole and towards a S pole. The left magnet's field at P points right (away from its N); the lower magnet's field at P points down (towards its S). The combined field points down and to the right, at 45°.

b. [2 marks]. The two fields are at right angles, so add them as vectors:

$$B = \sqrt{10.0^2 + 10.0^2} = 14.1 \text{ mT}.$$

From the report. In a (39% correct), the most common error was drawing a field line from N to S instead of a single arrow at P. In b (32% full marks), many simply added the two fields to get 20 mT.

Question 2 (4 marks)

A loudspeaker has a coil between a central S pole and a surrounding ring-shaped N pole. Viewed from the front (Figure 3), the current in the coil is clockwise.

a

Draw four magnetic field lines, with arrows. (1 mark)

b

Give the direction of the force on the coil: left, right, up the page, down the page, into or out of the page. (1 mark)

c

The coil has radius 5.0 cm, 20 turns and carries 2.0 A in a 200 mT field. Calculate the force. (2 marks)

Study design: in the current course (force on a current-carrying conductor).

Show worked solution

a. [1 mark]

Four lines running radially inwards, from the outer N ring pole to the central S pole, each with an arrow pointing towards the centre.

b. [1 mark]

E: into the page. At the top of the coil, for example, the current runs to the right and the field points down (inwards); the right-hand rule gives a force into the page. The same is true all the way round the coil.

c. [2 marks]

The field is perpendicular to every part of the wire, and the length of one turn is its circumference:

$$F = nIlB = 20 \times 2.0 \times (2\pi \times 0.050) \times 0.200 = 2.5 \text{ N}.$$

From the report. Part c was answered well by only 23%; the most common error was using the radius instead of the length of wire (the circumference).

Question 3 (3 marks)

The planet Phobetor orbits pulsar PSR B1257+12 at a radius of 6.9 × 10¹⁰ m with a period of 8.47 × 10⁶ s. Assuming a circular orbit, calculate the pulsar's mass.

Study design: in the current course.

Show worked solution

[3 marks]. Gravity provides the centripetal force:

$$\frac{GMm}{r^2} = \frac{4\pi^2 m r}{T^2} \quad\Rightarrow\quad M = \frac{4\pi^2 r^3}{GT^2} = \frac{4\pi^2 (6.9 \times 10^{10})^3}{6.67 \times 10^{-11} \times (8.47 \times 10^6)^2} = 2.71 \times 10^{30} \text{ kg}.$$

From the report. 55% scored full marks. Some lost marks for not showing enough working: the equating of gravitational and centripetal expressions and the substitution both needed to be seen.

Question 4 (2 marks)

Liesel sits on the floor. The action force on her is gravity, 500 N down. Identify and explain the reaction force.

Study design: in the current course (Newton's third law).

Show worked solution

[2 marks]. The reaction force is the gravitational force of Liesel on the Earth: 500 N upwards, acting on the Earth. By Newton's third law, action and reaction forces are equal in size, opposite in direction, of the same type, and act on different bodies. The floor's upward normal force on Liesel is not the reaction force: it acts on Liesel too, and it balances her weight (Newton's first law), which is a different idea.

From the report. This was the worst-answered question in Section B (90% scored zero). Most students named the normal force, confusing Newton's third law with balanced forces.

Question 5 (9 marks)

A stationary electron sits in a uniform magnetic field (out of the page) between parallel plates 6.0 × 10⁻³ m apart, connected through an open switch to a 200 V supply. Ignore gravity.

a

Explain why the magnetic field exerts no force on the electron, using a formula. (2 marks)

b

The switch is closed. Find the magnitude and direction of the electric force on the electron. (3 marks)

c

Ravi says the magnetic force will then have constant magnitude but continually changing direction. Mia says it will steadily increase but always act in the same direction. Evaluate both statements. (4 marks)

Study design: in the current course.

Show worked solution

a. [2 marks]. The magnetic force on a moving charge is $F = qvB$. The electron is stationary, so $v = 0$ and $F = 0$: a magnetic field only exerts a force on a moving charge.

b. [3 marks].

$$E = \frac{V}{d} = \frac{200}{6.0 \times 10^{-3}} = 3.33 \times 10^4 \text{ V m}^{-1}, \qquad F = qE = 1.6 \times 10^{-19} \times 3.33 \times 10^4 = 5.3 \times 10^{-15} \text{ N}.$$

The force on the electron is towards the positive plate: downwards (the lower plate is positive).

c. [4 marks]. Each student is half right.

• Once the switch closes, the electron accelerates from rest, so its speed increases and the magnetic force, $F = qvB$, increases. Ravi is wrong that the magnitude is constant; Mia is right that it increases.
• The magnetic force is always perpendicular to the velocity, so as the path curves the direction of the force keeps changing. Ravi is right that the direction continually changes; Mia is wrong that it stays the same.

From the report. In a, some talked about the electric field or read $v$ in $F = qvB$ as a voltage. In b, algebra and power-of-ten slips were common. Part c was weak (7% full marks): many backed the incorrect parts of the statements or contradicted themselves.

Question 6 (8 marks)

A loop in a simple AC generator (Figure 6) rotates at 50 revolutions per second and generates 4.25 V RMS. At the start, the loop is perpendicular to the uniform field.

a

Sketch the output EMF between P and Q against time for at least two revolutions, with the maximum voltage and a time scale. (4 marks)

b

Describe the function of the slip rings. (1 mark)

c. i

How could it be changed into a DC generator? (1 mark) ii. Sketch the DC output for at least two revolutions, with a time scale. (2 marks)

Study design: in the current course.

Show worked solution

a. [4 marks]. $V_{\text{peak}} = \sqrt{2} \times 4.25 = 6.0$ V, and the period is $T = \tfrac{1}{50} = 0.02$ s. The loop starts perpendicular to the field (maximum flux, zero rate of change), so the EMF starts at zero: a sine curve of amplitude 6.0 V with two full cycles in 0.04 s.

b. [1 mark]

The slip rings keep a continuous connection between the rotating loop and the stationary external circuit without reversing it, so the output is AC.

c. i. [1 mark]

Replace the slip rings with a split-ring commutator.

ii. [2 marks]

The commutator reverses the connections every half turn, so every half-cycle is positive: a rectified sine wave, with a hump every 0.01 s and two revolutions in 0.04 s.

From the report. In a (32% full marks), the common errors were drawing a cosine curve or the wrong period. In b, describing a split-ring commutator was wrong, and in c.i the exact term "split-ring commutator" was required. In c.ii many doubled the period.

Question 7 (7 marks)

A generator delivers 500 MW at 25 kV. The voltage is stepped up to 500 kV for transmission through lines of total resistance 30.0 Ω, then stepped down to 240 V. All transformers are ideal.

a

Explain why the voltage is stepped up for transmission. (2 marks)

b

Calculate the current in the transmission lines. (2 marks)

c

Determine the maximum power available for domestic use. (3 marks)

Study design: in the current course.

Show worked solution

a. [2 marks]. For the same power delivered, $P = VI$, a higher voltage means a smaller current in the lines. The power lost as heat in the lines is $P_{\text{loss}} = I^2R$, so reducing the current greatly reduces the losses.

b. [2 marks].

$$I = \frac{P}{V} = \frac{500 \times 10^6}{500 \times 10^3} = 1.0 \times 10^3 \text{ A} = 1.0 \text{ kA}.$$

c. [3 marks].

$$P_{\text{loss}} = I^2R = (1.0 \times 10^3)^2 \times 30.0 = 30 \text{ MW}, \qquad P_{\text{available}} = 500 - 30 = 470 \text{ MW}.$$

(The ideal step-down transformer passes all 470 MW on at 240 V.)

From the report. In a (12% full marks), many said the current is reduced without saying the power delivered is maintained. In b, a common error was $I = \tfrac{V}{R}$ with 500 kV across 30 Ω. In c, working with 240 V, or stopping at the power loss, did not earn full marks.

Question 8 (11 marks)

NASA's Atlas rocket (531 tonnes) launched the Perseverance capsule to Mars.

a

At lift-off the acceleration was 7.20 m s⁻² with $g = 9.80$ N kg⁻¹. Find the magnitude and direction of the thrust, in MN. (3 marks)

b

The 1000 kg capsule entered Mars's atmosphere at 20 000 km h⁻¹, 300 km above the surface. Calculate its kinetic energy. (2 marks)

c

Figure 8 shows Mars's $g$ falling linearly from 3.7 N kg⁻¹ at the surface to 2.7 N kg⁻¹ at 600 km. Calculate the gravitational potential energy at 300 km relative to the surface. (3 marks)

d

By 10 km altitude, about 99% of the capsule's combined kinetic and gravitational potential energy had gone. Describe how it was transformed and dissipated. (3 marks)

Study design: in the current course (including energy from the area under a field-distance graph).

Show worked solution

a. [3 marks]. Taking up as positive, $F_{\text{thrust}} - mg = ma$:

$$F_{\text{thrust}} = m(g + a) = 531 \times 10^3 \times (9.80 + 7.20) = 9.03 \times 10^6 \text{ N} = 9.03 \text{ MN, upwards}.$$

(The report shows 9.02 MN because it rounds $mg$ and $ma$ separately before adding.)

b. [2 marks]. $v = \tfrac{20\,000}{3.6} = 5.56 \times 10^3$ m s⁻¹.

$$E_k = \tfrac12 mv^2 = \tfrac12 \times 1000 \times (5.56 \times 10^3)^2 = 1.54 \times 10^{10} \text{ J}.$$

c. [3 marks]. The energy per kilogram is the area under the $g$-$h$ graph from 0 to 300 km, a trapezium with $g$ = 3.7 and 3.2 N kg⁻¹:

$$\text{area} = \frac{3.7 + 3.2}{2} \times 300 \times 10^3 = 1.04 \times 10^6 \text{ J kg}^{-1}, \qquad E_g = 1000 \times 1.04 \times 10^6 = 1.04 \times 10^9 \text{ J}.$$

d. [3 marks]. As the capsule descends, gravitational potential energy is converted into kinetic energy. At the same time, friction (drag) from the atmosphere does work against the motion, converting kinetic energy into heat (in the heat shield and surrounding air), light and sound. That heat and sound spread into the atmosphere, so the energy is dissipated.

From the report. In b, the common errors were not converting km h⁻¹ to m s⁻¹ or not squaring $v$. Part c was weak (17% full marks); counting grid squares usually went wrong, although most knew to multiply by the mass. In d (6% full marks), many described only what happened to the kinetic energy and not the gravitational potential energy.

Question 9 (10 marks)

A roller-coaster car starts from rest at A, 15 m above the bottom of a loop, B. C is the top of the loop, of radius $r$. Ignore friction.

a

Find the speed at B. (2 marks)

b

By considering the forces at C, show that the car just stays on the track when $\tfrac{v^2}{r} = g$. (2 marks)

c

What is the maximum height of the loop, $X$, for the car to stay in contact at C? (3 marks)

d

With friction, should the loop radius be increased or decreased? Explain. (3 marks)

Study design: in the current course (circular motion in a vertical plane, at the highest and lowest points).

Show worked solution

a. [2 marks]. Conservation of energy from A to B (the track is curved, so the acceleration is not constant and the constant-acceleration equations do not strictly apply):

$$mgh = \tfrac12 mv^2 \quad\Rightarrow\quad v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 15} = \sqrt{294} = 17.1 \text{ m s}^{-1}.$$

Note: the report's worked answer shows $v^2 = 300$ and $v = 17.3$ m s⁻¹, which corresponds to $g = 10$ m s⁻². With the paper's $g = 9.8$ m s⁻², the answer is 17.1 m s⁻¹.

b. [2 marks]. At the top, both the weight and the normal force point down, towards the centre, and together provide the centripetal force:

$$\frac{mv^2}{r} = mg + F_N.$$

The car only just stays in contact when $F_N = 0$, so $\dfrac{mv^2}{r} = mg$, giving $\dfrac{v^2}{r} = g$.

c. [3 marks]. Energy from A (height 15 m, at rest) to C (height $2r$, with $v^2 = rg$):

$$mg(15) = mg(2r) + \tfrac12 m(rg) \quad\Rightarrow\quad 15 = 2r + 0.5r = 2.5r \quad\Rightarrow\quad r = 6.0 \text{ m}.$$

So the maximum loop height is $X = 2r = 12$ m. (The answer does not depend on $g$.)

d. [3 marks]. The radius must decrease. Friction removes some energy, so the car is slower at C than predicted. The minimum condition $v^2 = rg$, or $r = \tfrac{v^2}{g}$, shows that a smaller speed at the top needs a smaller radius for the car to stay on the track.

From the report. In a, many used $v^2 = u^2 + 2as$, which gives the same number but ignores the varying acceleration. In b, starting from $v = \sqrt{rg}$ and rearranging did not count as considering the forces. Part c was weak (80% scored zero); some substituted the speed from part a. In d, only just over a quarter linked radius to speed through $r = \tfrac{v^2}{g}$.

Question 10 (4 marks)

A spaceship passes a technician at 0.7$c$. The technician measures a different length from the laboratory value, but the same width.

a. Explain why the length changes but not the width. (2 marks)

b. The technician measures its length as 135 m. What was its length at rest on Earth? (2 marks)

Study design: in the current course.

Show worked solution

a. [2 marks]. Length contraction occurs only along the direction of relative motion. The ship moves along its length, so the technician measures a shorter length; its width is perpendicular to the motion and is unchanged.

b. [2 marks].

$$\gamma = \frac{1}{\sqrt{1 - 0.7^2}} = 1.40, \qquad L = \frac{L_0}{\gamma} \;\Rightarrow\; L_0 = 1.40 \times 135 = 189 \text{ m}.$$

The rest (proper) length is the longer one, 189 m.

From the report. In a, some wrote "length dilation". In b, errors were in calculating $\gamma$ or swapping proper and contracted lengths; $\gamma$ can never be less than 1.

Question 11 (2 marks)

Unpolarised light passes through polarising filter F1 (axis vertical) and then F2 (axis horizontal). Will any light reach point P? Explain.

Study design: outside Units 3 and 4 (polarisation is now only in a Unit 2 option).

Show worked solution

[2 marks]. No (or almost no) light reaches P. F1 transmits only light polarised vertically. F2 transmits only horizontally polarised light, at 90° to it, so it blocks all the light that got through F1.

From the report. Answers lost marks for generic descriptions of polarisation, or for saying F1 blocks half the light and F2 the rest without explaining that the axes are perpendicular.

Question 12 (5 marks)

A light ray enters an optical fibre (core $n = 1.46$, cladding $n = 1.42$) from air; inside the core the refracted ray makes 32° with the normal to the end face.

a. Find the angle of incidence $\theta$ at the air-core boundary. (2 marks)

b. Will any of the ray pass into the cladding? Explain, with working. (3 marks)

Study design: outside Units 3 and 4 (Snell's law and total internal reflection are now Unit 1 content).

Show worked solution

a. [2 marks].

$$1.00 \sin\theta = 1.46 \sin 32° \quad\Rightarrow\quad \theta = 51°.$$

b. [3 marks]. The critical angle at the core-cladding boundary is

$$\theta_c = \sin^{-1}\frac{1.42}{1.46} = 76.6°.$$

The ray meets the core-cladding boundary at $90° - 32° = 58°$ to the normal. Since $58° < 76.6°$, total internal reflection does not occur, so some light is transmitted into the cladding (refracted at about 61°), with the rest reflected.

From the report. In b (16% full marks), the most common error was comparing the critical angle with 32° instead of the 58° angle of incidence at the core-cladding boundary.

Question 13 (4 marks)

In Young's double-slit experiment the slits are 2.0 mm apart and 1.0 m from the screen.

a. With 620 nm laser light, find the distance between successive dark bands. (2 marks)

b. Explain how the experiment supports the wave model of light. (2 marks)

Study design: in the current course.

Show worked solution

a. [2 marks].

$$\Delta x = \frac{\lambda L}{d} = \frac{620 \times 10^{-9} \times 1.0}{2.0 \times 10^{-3}} = 3.1 \times 10^{-4} \text{ m} = 0.31 \text{ mm}.$$

b. [2 marks]. The pattern of bright and dark bands is produced by interference: light from the two slits adds constructively where the path difference is $n\lambda$ and cancels where it is $(n + \tfrac12)\lambda$. Interference is a property of waves, which the particle model cannot explain, so the result supports the wave model.

From the report. In a, a common error was simply dividing 620 by 2. In b, the effect on the screen is interference, not diffraction, and generic statements copied from notes did not score.

Question 14 (3 marks)

A fire truck with a 500 Hz siren drives at 20 m s⁻¹ towards Chris, then away from him.

a. Sketch the frequency Chris hears as the truck approaches and then moves away. (2 marks)

b. Name the principle involved. (1 mark)

Study design: outside Units 3 and 4 (the Doppler effect is now only in a Unit 2 option).

Show worked solution

a. [2 marks]. While the truck approaches, the frequency is constant and higher than 500 Hz (a horizontal line above the 500 Hz line). After it passes, the frequency is constant and lower than 500 Hz (a horizontal line below it). The truck's speed is constant, so the heard frequency does not keep changing on either side.

b. [1 mark]. The Doppler effect.

From the report. Part a was poorly done (87% scored zero); many drew waveforms rather than two constant frequency levels.

Question 15 (3 marks)

Students measure a threshold frequency of 6.5 × 10¹⁴ Hz and a work function of 3.2 × 10⁻¹⁹ J. What value of Planck's constant do they get?

Study design: in the current course.

Show worked solution

[3 marks]. At the threshold frequency the photon energy just equals the work function ($E_{k\max} = 0$), so $\phi = hf_0$:

$$h = \frac{\phi}{f_0} = \frac{3.2 \times 10^{-19}}{6.5 \times 10^{14}} = 4.9 \times 10^{-34} \text{ J s}.$$

From the report. Most did well (62% full marks); the common error was dividing the frequency by the work function.

Question 16 (2 marks)

The rapid emission of photoelectrons at very low light intensity supports one model of light. Identify it and give a reason.

Study design: in the current course.

Show worked solution

[2 marks]. The particle (photon) model. Each photon gives all its energy to a single electron in one interaction, so an electron is emitted as soon as the first photon of high enough frequency arrives. The wave model predicts a time delay at low intensity while each electron gradually absorbs enough energy, and no delay is observed.

From the report. Many identified the particle model but then copied general notes on the photoelectric effect without mentioning the absence of a time delay (14% full marks).

Question 17 (4 marks)

Photons of frequency 7.0 × 10¹⁵ Hz strike a space sail at 90° and reflect elastically.

a. Show that each photon's momentum is 1.55 × 10⁻²⁶ kg m s⁻¹. (1 mark)

b. 2.0 × 10¹⁸ photons strike the sail each second. Calculate the force on the sail, to two significant figures. (3 marks)

Study design: in the current course.

Show worked solution

a. [1 mark].

$$p = \frac{h}{\lambda} = \frac{hf}{c} = \frac{6.63 \times 10^{-34} \times 7.0 \times 10^{15}}{3.0 \times 10^8} = 1.55 \times 10^{-26} \text{ kg m s}^{-1}.$$

b. [3 marks]. Each photon reverses direction, so its change in momentum is twice its momentum:

$$\Delta p = 2 \times 1.55 \times 10^{-26} = 3.1 \times 10^{-26} \text{ kg m s}^{-1}.$$

The force is the momentum change per second:

$$F = \frac{\Delta p_{\text{total}}}{\Delta t} = 2.0 \times 10^{18} \times 3.1 \times 10^{-26} = 6.2 \times 10^{-8} \text{ N}.$$

From the report. Part b was weak (5% full marks, 76% scored zero): the most common error was leaving out the factor of 2 for the reflection.

Question 18 (5 marks)

X-ray photons of energy 100 eV pass through a circular aperture of diameter 1.24 µm. Electrons move at 5.0 × 10⁵ m s⁻¹.

a. Show that the electrons' de Broglie wavelength is 1.46 × 10⁻⁹ m. (1 mark)

b. Find the aperture diameter that gives an electron diffraction pattern with the same spacing as the X-ray pattern. (4 marks)

Study design: in the current course (diffraction and the $\tfrac{\lambda}{w}$ ratio).

Show worked solution

a. [1 mark].

$$\lambda_e = \frac{h}{mv} = \frac{6.63 \times 10^{-34}}{9.1 \times 10^{-31} \times 5.0 \times 10^5} = 1.46 \times 10^{-9} \text{ m}.$$

b. [4 marks]. The same pattern spacing needs the same ratio of wavelength to aperture width, $\tfrac{\lambda}{w}$.

• X-ray wavelength: $\lambda_X = \dfrac{hc}{E} = \dfrac{4.14 \times 10^{-15} \times 3.0 \times 10^8}{100} = 1.24 \times 10^{-8}$ m.
• X-ray ratio: $\dfrac{\lambda_X}{w_X} = \dfrac{1.24 \times 10^{-8}}{1.24 \times 10^{-6}} = 1.00 \times 10^{-2}$.
• Electron aperture: $w_e = \dfrac{\lambda_e}{1.00 \times 10^{-2}} = \dfrac{1.46 \times 10^{-9}}{1.00 \times 10^{-2}} = 1.46 \times 10^{-7}$ m.

From the report. 74% scored zero and more than half did not attempt it. Multi-step problems like this reward signposting each step, as above.

Question 19 (4 marks)

An atom has energy levels at 0, 4.9, 6.7, 8.9 and 9.8 eV.

a. Identify the transition that emits a 565 nm photon. (2 marks)

b. Atoms excited to 9.8 eV decay by all possible transitions. How many spectral lines are seen? Explain. (2 marks)

Study design: in the current course.

Show worked solution

a. [2 marks].

$$E = \frac{hc}{\lambda} = \frac{4.14 \times 10^{-15} \times 3.0 \times 10^8}{565 \times 10^{-9}} = 2.2 \text{ eV},$$

which matches a downward transition from 8.9 eV to 6.7 eV.

b. [2 marks]. From the 9.8 eV level there are 4 possible downward transitions, from 8.9 eV there are 3, from 6.7 eV there are 2 and from 4.9 eV there is 1: 10 transitions. But 9.8 → 4.9 eV and 4.9 → 0 eV both release 4.9 eV, giving the same line, so there are 9 lines.

From the report. In a, the common error was drawing the 2.2 eV transition upwards (that would be absorption). In b (27% full marks), most miscounted the transitions.

Question 20 (17 marks)

Jerome and Priya whirl a rubber stopper (mass $m$) in a horizontal circle of radius 0.75 m on a cord that passes through a glass tube to a hanging set of washers (30 g each). They time 20 rotations three times for each trial, adding two washers per trial, starting with ten.

a

Why take repeated time measurements? (1 mark)

b

What causes the tension in the cord? (1 mark)

c

Develop an equation linking $Mg$ with $\pi$, $m$, $R$ and $T$. (3 marks)

d

Complete the table ($g = 10$ N kg⁻¹): $M$ = 0.30, 0.36, 0.42, 0.48, 0.54 kg; time for 20 rotations = 14.0, 12.8, 11.8, 11.0, 10.4 s. (4 marks)

e

Plot $Mg$ against $\tfrac{1}{T^2}$ with a line of best fit and ±0.1 s⁻² uncertainty bars. (4 marks)

f

Calculate the gradient. (2 marks)

g

Use the gradient to show that $m$ is about 50 g. (2 marks)

Study design: in the current course (practical skills, horizontal circular motion and linearising data).

Show worked solution

a. [1 mark]

To reduce the effect of random errors (for example, in starting and stopping the stopwatch), which makes the average time more reliable and less uncertain.

b. [1 mark]

The gravitational force (weight) of the hanging washers.

c. [3 marks]

The washers hang still, so the tension equals their weight, $Mg$. The tension provides the stopper's centripetal force:

$$Mg = \frac{4\pi^2 m R}{T^2}.$$

d. [4 marks]. $Mg = M \times 10$; $T = \tfrac{\text{time}}{20}$.

$M$ (kg)	$Mg$ (N)	Time for 20 (s)	$T$ (s)	$\tfrac{1}{T^2}$ (s⁻²)
0.30	3.0	14.0	0.70	2.04
0.36	3.6	12.8	0.64	2.44
0.42	4.2	11.8	0.59	2.87
0.48	4.8	11.0	0.55	3.31
0.54	5.4	10.4	0.52	3.70

(The report lists 3.30 for line 4; $\tfrac{1}{0.55^2} = 3.306$, which rounds to 3.31. The difference does not change the graph.)

e. [4 marks]. Plot the five points with horizontal bars 0.1 s⁻² each side, and draw a straight line of best fit through them:

f. [2 marks]. Use two points on the line (not data points), far apart, for example (2.3, 3.4) and (3.4, 5.0):

$$\text{gradient} = \frac{5.0 - 3.4}{3.4 - 2.3} = \frac{1.6}{1.1} = 1.5 \text{ N s}^2.$$

(A least-squares fit to the table gives 1.44 N s², so anything close to 1.4 to 1.5 is reasonable.)

g. [2 marks]. Writing part c as $Mg = (4\pi^2 mR) \times \tfrac{1}{T^2}$, the gradient is $4\pi^2 mR$:

$$m = \frac{\text{gradient}}{4\pi^2 R} = \frac{1.5}{4\pi^2 \times 0.75} = 0.051 \text{ kg} \approx 50 \text{ g}.$$

From the report. In b (17% correct), few named the weight of the washers, suggesting few had done the experiment. In c, students mixed up $m$ and $M$. In d, common errors were multiplying by 100 instead of 10 and taking $T = \tfrac{20}{\text{time}}$. In f, many used table values not on their line or assumed the line passed through the origin. Part g was not attempted by many (85% scored zero).

General advice from the 2021 report

• Show working, even when the question says "calculate": the exam rewards demonstrated understanding. Signpost multi-step answers with short statements (Q7c, Q9b, Q17b, Q18b).
• Give final answers as decimals, not fractions or surds, and do not round during a calculation; round only at the end.
• Do not copy from your notes sheet. Answers copied from the A3 sheet that did not address the question scored poorly (Q13b, Q16).
• Plan longer answers, such as evaluations (Q5c), so they do not contradict themselves.