VCE Physics 2023

a. [3 marks]. On each ball: $T$ along the string, pointing up towards the ceiling; $F_g$ straight down; $F_E$ horizontal, pointing away from the other ball. Each arrow needs one clear head and a label.

b. [2 marks]. Each string is 30° from the vertical. The ball is in equilibrium, so the vertical component of the tension balances the weight ($m = 4.0 \times 10^{-3}$ kg):

c. [2 marks]. The horizontal component of the tension balances $F_E$:

(Equivalently $F_E = T\cos 60°$, or use Pythagoras with $T$ and $mg$.)

From the report. In a, arrows with heads at both ends, $F_E$ pointing inwards (attraction) or $F_E$ drawn at right angles to the string all lost marks. In b, the most common error was converting 4.0 g wrongly. Many wrote working that did not actually give the answer and then wrote "$= 4.5 \times 10^{-2}$ N"; in a "show that" question the marks are for correct working. In c, the wrong trigonometric function was the common error.

a. [2 marks]. Since $g = \tfrac{GM}{r^2}$, the field falls off as the inverse square of the distance from the centre. Phobos is $3390 + 6000 = 9390$ km from the centre:

A ratio means no unit conversion is needed. (Finding $M$ first and then $g$ also works.)

b. [3 marks]. Gravity provides the centripetal force, so $g = \tfrac{4\pi^2 r}{T^2}$ with $r = 9.39 \times 10^6$ m:

(Using the rounded 0.48 gives $2.78 \times 10^4$ s.)

c. [2 marks]. Shorter. From $T = \sqrt{\tfrac{4\pi^2 r^3}{GM}}$, $T \propto r^{3/2}$: a smaller orbital radius gives a shorter period. (A closer orbit is also a faster one, and the path around is shorter.)

From the report. In b, students who carried a mass of $6.41 \times 10^{17}$ kg from part a (from not converting km to m) scored nothing, because an independent method existed. In c, "T is directly proportional to r" was a common misreading, and using $T = \sqrt{\tfrac{4\pi^2 r}{g}}$ without noting that $g$ also changes was the most common error. A bare "shorter" with no reasoning was treated as a guess.

a. [3 marks]. Around each wire, draw concentric circles, closer together near the wire. Use the right-hand grip rule for direction: for current out of the page the field is anticlockwise; for current into the page it is clockwise. Because the currents are opposite, the circles around P and Q circulate in opposite senses, and between the wires both fields point the same way, so the field there is strong (lines crowded).

b. [2 marks]. Repel. The field from P at Q, combined with Q's opposite current, gives a force on Q directed away from P (right-hand slap rule), and by Newton's third law P is pushed away from Q. Equivalently, the fields add between the wires and the field lines push the wires apart.

From the report. In a, many drew on the front view instead of the top view. In b, statements like "the fields are in opposite directions" lost marks unless they said where (around the wires, or between them).

a. [1 mark]. $N_p : N_s = V_p : V_s = 240 : 12 = 20 : 1$.

b. [2 marks]. Total power in the secondary $= 6 \times 20 = 120$ W. For an ideal transformer the same power enters the primary:

(Or: secondary current $= \tfrac{120}{12} = 10$ A, and $I_p = \tfrac{10}{20} = 0.50$ A.)

c. [2 marks]. A transformer works by electromagnetic induction: an EMF is induced in the secondary only when the magnetic flux through it changes. AC in the primary produces a continually changing flux in the core. A steady DC current produces a constant flux, so no EMF is induced in the secondary and the globes stay off.

From the report. In b, the common error was using the current for one globe as the whole secondary current (giving 83.5 mA). In c, vague answers such as "only AC produces an EMF" or "only AC creates a flux" did not score: the key idea is a changing flux.

a. [2 marks]. The flux changes from 0 to $BA$:

b. [1 mark]. Clockwise. The out-of-page flux through the loop is increasing, so by Lenz's law the induced current makes a field into the page inside the loop, which by the right-hand grip rule is a clockwise current.

From the report. In a, the common errors were in squaring 0.040 m (it is 0.0016 m²) and writing 0.00096 in scientific notation.

a. [2 marks]. With the magnets on the axis, the field runs along the axle, parallel to the plane of the loop at every angle of rotation. No field lines pass through the loop, so the flux is always zero. Rotating the loop does not change the flux, and with no change in flux there is no induced EMF (Faraday's law).

b. [1 mark]. Rotate the magnet positions through 90°, so the poles face each other across the loop, one on each side, with the field perpendicular to the axle. The flux through the loop then changes as it turns.

From the report. Many described the device as a motor and talked about forces on a current, but there is no current, so no force. Know the difference between motors and generators.

a. [2 marks]. $P_\text{loss} = I^2 R$:

b. [3 marks]. Voltage drop across the lines: $\Delta V = IR = 700 \times 41 = 2.87 \times 10^4$ V $= 28.7$ kV. Voltage at the town: $500 - 28.7 = 471$ kV. This is less than 480 kV, so the town will not get the voltage it needs.

(Check by power: $350 - 20 = 330$ MW arrives, and $\tfrac{330 \times 10^6}{700} = 471$ kV.)

c. [2 marks]. To send the same power at one-tenth of the voltage, the current must be 10 times larger ($P = VI$), about 7000 A. The line loss $I^2R$ then rises by a factor of $10^2 = 100$. That predicted loss (about 2000 MW) is more than the power station supplies, so in practice very little useful power would reach Town B. This is why transmission uses high voltages.

From the report. In b, some multiplied values by $\sqrt{2}$ for no reason: all values given were RMS. In c, the common error was saying the voltage falls but the current stays the same; some also treated the 41 Ω as the resistance of the whole circuit.

a. [3 marks]. Three forces from the centre of mass: weight $mg$ straight down; the normal force perpendicular to the slope, away from it; and friction (total resistive force) up the slope, parallel to it. There is no separate "force down the slope": that is a component of the weight. Because the velocity is constant, the forces must add to zero.

b. [2 marks]. With zero acceleration, friction balances the component of weight down the slope:

c. [2 marks]. Momentum is conserved: it is transferred through the pole to the Earth, whose enormous mass means its change in velocity is too small to notice. Kinetic energy is not conserved as kinetic energy: it is transformed into other forms, mainly heat in her body and the pole, as her muscles do work to stop her.

From the report. In a, the most common error was adding a force down the slope, and many drew on the wrong diagram. In b, some tried $F = ma$ and stalled because $a = 0$, or used the wrong trigonometric function. In c, a common error was sending the momentum only into the pole and not on to the Earth, or treating momentum and energy the same way.

a. [2 marks]. Take the fall from B back to the shoulder (1.8 m from rest):

The rise takes the same time, so the ball is in the air for $2 \times 0.606 = 1.21 \approx 1.2$ s.

b. [2 marks]. From C (behind, at shoulder level) to B (directly above the shoulder) is a quarter circle of radius 1.8 m, and it must be covered in the time the ball takes to rise, 0.606 s:

Constant speed means the speed at C equals this average speed.

c. [3 marks]. The ball is hit at B, $1.8 + 1.8 = 3.6$ m above the ground. Horizontal motion: time to reach the net $t = \tfrac{12}{24} = 0.50$ s. Vertical motion (starting from zero vertical velocity):

Height above the net $= 3.6 - 1.225 - 0.90 \approx 1.5$ m.

From the report. Part b was new in style and answered poorly: only 26% scored full marks, and the common errors were using the whole circumference and reaching for $F = \tfrac{mv^2}{r}$. In c (31% full marks), many found the 1.225 m drop and then treated it as the height above the ground. The report notes that many students struggle with multi-step problems.

a. [3 marks]. Rearrange $\gamma = \tfrac{1}{\sqrt{1 - v^2/c^2}}$:

b. [1 mark]. The beamline is moving relative to the proton, so it is length-contracted:

c. [2 marks]. Use the relativistic kinetic energy:

From the report. In a, the common errors were giving the speed in m s⁻¹ instead of as a multiple of $c$, and rearranging the Lorentz factor incorrectly. In c, the common error was using the classical $\tfrac12 mv^2$.

a. [1 mark]. One loop is half a wavelength, so $\lambda = 2 \times 0.75 = 1.5$ m and

b. [2 marks]. Waves travel along the string and reflect from the fixed ends, so waves of the same frequency travel in opposite directions and superpose (interfere). When the string length is a whole number of half-wavelengths, the interference is stable: fixed nodes form at the ends (and between loops) with antinodes between them, which is the standing wave.

From the report. Answers that described superposition but left out reflection scored one mark. Over a third of students showed no understanding of standing waves.

a. [2 marks]. At the critical angle the refracted ray travels at 90°:

b. [2 marks]. Yes. The ray passes straight through AB (it enters along the normal), so it meets AC at an angle of incidence of 45°. Since $45° > 41°$, the angle of incidence exceeds the critical angle and the ray is totally internally reflected.

From the report. Part b was not answered well: the common error was reversing the logic and writing "41° is less than 45°, so TIR will not occur".

a. [2 marks]. $P_2$ is the second bright band from the centre, so the path difference is two whole wavelengths:

b. [1 mark]. The spacing increases. The spacing is $\Delta x = \tfrac{\lambda L}{d}$, and red light has a longer wavelength.

c. [3 marks]. The screen shows a pattern of many evenly spaced bright and dark bands. This is interference: bright bands where the paths differ by a whole number of wavelengths (constructive), dark bands where they differ by an odd number of half-wavelengths (destructive). Interference is a wave behaviour. A particle model predicts only two bright regions, one behind each slit, and cannot explain the dark bands where light from the two slits cancels.

From the report. In a, the most common error was not converting units and reporting a path difference of "1020 m", an impossible value that students should have questioned. In c, only about a third answered correctly, even though similar questions are common on past papers; nearly 10% did not attempt it.

a. [2 marks]. From $\lambda = \tfrac{h}{mv}$:

b. [2 marks]. Yes. Significant diffraction happens when the wavelength is similar to the gap size. Here $\tfrac{\lambda}{w} = \tfrac{3.02}{3.62} \approx 0.83$, which is close to 1, so an observable diffraction pattern is expected.

c. [2 marks]. The electron. Equal wavelengths mean equal momentum ($p = \tfrac{h}{\lambda}$), so $mv$ is the same for both. The electron's mass is about 1800 times smaller, so its speed must be about 1800 times larger (about $2.4 \times 10^6$ m s⁻¹).

From the report. In a, errors were the wrong value of Planck's constant or poor rearranging. In b, only about a fifth scored full marks: many found the ratio 0.83 but drew the wrong conclusion, and many wrongly claimed diffraction cannot occur if the ratio is less than 1.

a. [1 mark]. From $E_{k\,\text{max}} = hf - \phi$, the extended line meets the energy axis at $-\phi$. So $\phi = 3.0$ eV. A work function is always positive.

b. [2 marks]. The longest wavelength corresponds to the threshold (cut-off) frequency, where the line meets the frequency axis, $f_0 = 7.2 \times 10^{14}$ Hz:

(The report's value is $4.1 \times 10^{-7}$ m; small differences from reading the intercept are allowed.)

c. [2 marks]. A line parallel to the calcium line (same gradient, Planck's constant), shifted to the right. Its extension meets the energy axis at $-4.70$ eV, and it crosses the frequency axis at $f_0 = \tfrac{\phi}{h} = \tfrac{4.70}{4.14 \times 10^{-15}} \approx 11.4 \times 10^{14}$ Hz.

d. [3 marks]. No. Energy of each photon:

This is less than copper's work function (4.70 eV), so no photoelectrons are emitted. (Equivalently, $f = \tfrac{c}{\lambda} = 7.9 \times 10^{14}$ Hz is below copper's threshold of about $11.4 \times 10^{14}$ Hz.)

From the report. In a, common errors were writing $-3$ eV and trying $\phi = hf_0$ with a misread value. In d, the most common error was calculating the energy or frequency correctly but not comparing it with the work function or threshold to answer the question.

a. [2 marks].

b. [1 mark]. Find two levels 2.84 eV apart: $7.73 - 4.89 = 2.84$ eV. Draw a downward arrow from $n = 4$ (7.73 eV) to $n = 2$ (4.89 eV). Emission is always a downward transition.

c. [3 marks]. The 6.7 eV photon lifts the atom exactly to $n = 3$ (6.70 eV). From there it can return directly or in two steps:

• $n = 3 \to n = 1$: 6.70 eV
• $n = 3 \to n = 2$: $6.70 - 4.89 =$ 1.81 eV
• $n = 2 \to n = 1$: 4.89 eV

From the report. In b, over 40% gave no answer or drew an incorrect arrow, including upward arrows and arrows between the wrong levels. In c, nearly a quarter did not attempt it, and some listed every possible transition in the hope of including the right ones.

a. [1 mark]. The work done by the field becomes kinetic energy: $eV_0 = \tfrac12 mv^2$, so

b. [2 marks]. The magnetic force on the moving electron is always perpendicular to its velocity, so it changes the direction of motion but not the speed. With constant speed and a uniform field, the force also has constant magnitude ($F = evB$). A force of constant size always at right angles to the velocity is a centripetal force, so the path is a circle.

c. [1 mark]. $evB = \tfrac{mv^2}{r}$, which gives

d. [2 marks]

Independent: the accelerating voltage $V_0$. Dependent: the radius $r$. Controlled: the magnetic field strength $B$ (2.0 mT).

e. [2 marks]

$0.052^2 = 0.0027$ m², $0.059^2 = 0.0035$ m², $0.072^2 = 0.0052$ m².

f. [7 marks]

A complete graph has correct axes and labels with units, a sensible scale, five correctly plotted points (including (0, 0)), horizontal uncertainty bars of $\pm 0.0002$ m², and a straight line of best fit through the origin that balances the points rather than joining the first and last.

g. [2 marks]. Take two points on your line, well apart, not data points from the table. For the line above, (0.0010, 400) and (0.0050, 2000):

A value close to $4 \times 10^5$ V m⁻² is expected (a least-squares fit through the origin gives about $3.9 \times 10^5$).

h. [3 marks]. Compare $V_0 = \tfrac{eB^2}{2m} r^2$ with $y = kx$: the gradient is $k = \tfrac{eB^2}{2m}$. So

That is close to the accepted value of $1.76 \times 10^{11}$ C kg⁻¹.

From the report. In b, over half could not give either reason (constant-size force, always perpendicular to the velocity), even though this has appeared many times. In f, just over a third scored full marks, and many forced the line through the first and last points, leaving the point (0.0035, 1500) far off it. In g, the most common errors were using table values instead of points on the line, and using (0.0052, 2000) when it was not on the line. In h (27% full marks), over 40% did not attempt it: link the gradient to the equation before calculating.