VCE Physics 2025

a. [1 mark]. With $u = 0$ and $s = 1.80$ m down:

b. [2 marks]

$v^2 = u^2 + 2as = 0 + 2 \times 9.81 \times 1.80 = 35.3$, so $v \approx 5.94$ m s⁻¹.

c. [1 mark]

The bounce is inelastic: some kinetic energy is transformed into heat, sound and deformation of the ball and floor, so less is available to convert back into gravitational potential energy.

d. [2 marks]

The time will be the same. The horizontal velocity does not affect the vertical motion: the ball still starts with zero vertical velocity, accelerates at $g$ and falls 1.80 m.

From the report. In b, some forgot the final square root. In d (54% full marks), the most common error was claiming the longer, curved path takes more time; some also wrongly gave the ball an initial vertical velocity of 0.75 m s⁻¹.

a. [1 mark]

$I = \Delta p = 5.0 \times 12 = 60$ N s.

b. [2 marks]

The head's change in momentum (60 N s) is the same in both collisions. The airbag makes the collision last longer, and since $F_{\text{av}}\Delta t = \Delta p$, a longer time means a smaller average force on the head, so less chance of injury.

c. [3 marks]

The work done by the airbag equals the head's loss of kinetic energy:

The work is the area under the force-displacement graph, a trapezium with parallel sides 0.20 m and 0.40 m and height $F_{\max}$:

From the report. In b, simply quoting $I = F\Delta t$ without comparing the two collisions was not a justification. Part c was very weak (11% full marks, 71% scored zero): many could not find the area, or wrongly set the area equal to the impulse of 60 N s.

a. [3 marks]. Three forces: the weight (vertically down), the normal force (perpendicular to the slope, away from it) and the resistive forces (friction and air resistance, up the slope, opposite to the motion). There is no separate "net force" arrow: the speed is constant, so the forces balance.

b. [1 mark]. At constant speed, the resistive force balances the component of weight down the slope:

c. [2 marks]. On the flat, the resistive force does work equal to the kinetic energy lost:

From the report. In a, the most common error was drawing an extra $F_{\text{net}}$ arrow down the slope. In b, some used tan instead of sin. In c (46% full marks), the most common error was using $v^2 = u^2 + 2as$ with $a = 9.81$ m s⁻².

a. [2 marks]. The sideways (friction) force provides the centripetal force:

b. [3 marks]. On a banked track with no friction, the horizontal component of the normal force supplies the centripetal force, giving $\tan\theta = \dfrac{v^2}{rg}$:

From the report. Both parts were done well (88% and 76% full marks); errors in b came from using sin or cos instead of tan.

a. [1 mark].

b. [3 marks]. Both charges are positive, so each field points away from its charge. A is directly north of D, so its field at D points south; C is directly east of D, so its field points west. The two equal fields are at right angles:

From the report. Part a errors included not converting cm to m and squaring the distance wrongly. In b (27% full marks) many tried to use $\tfrac{kq_1q_2}{r^2}$ (a force) instead of adding the field vectors.

a. [2 marks].

b. [2 marks]. No change. The kinetic energy gained is $qV$, which depends only on the potential difference (still 250 V), not the distance. Doubling the separation halves the field and force, but the force acts over twice the distance, so the work done is the same.

From the report. Part b (27% full marks): many noted that the force or acceleration decreases or the time increases, but did not explain why the final kinetic energy is unchanged.

a. [1 mark]. Down the page. The right-hand rule gives upwards for a positive charge; the electron is negative, so the force is reversed.

b. [2 marks].

From the report. In a, "upwards" was the common error, ignoring the electron's negative charge.

a. [2 marks]. The orbital radius is Earth's radius plus the altitude: $r = 6.37 \times 10^6 + 1.25 \times 10^6 = 7.62 \times 10^6$ m.

b. [2 marks].

c. [2 marks]. The only force on the satellite is Earth's gravitational pull, directed towards Earth. By Newton's second law ($F = ma$), an unbalanced force produces an acceleration in the direction of the force, so the satellite accelerates towards Earth's centre (changing its direction, not its speed).

From the report. About one-third used the altitude alone, or in km, instead of adding it to Earth's radius; a "show that" also needs the full substitution. In c, citing Newton's first or third law was the common error.

[2 marks]. Tom assumed that $g$ stays constant at 9.81 N kg⁻¹. In fact, the gravitational field strength decreases with altitude, so less work is really needed: his value is an overestimate.

From the report. A small number wrongly said the true work would be greater than Tom's value.

a. [2 marks]. As the magnet approaches and passes through the solenoid, the magnetic flux through the solenoid's turns changes. By Faraday's law, $\varepsilon = -N\dfrac{\Delta\Phi}{\Delta t}$, a changing flux induces an EMF, which drives a current through the ammeter.

b. [3 marks].

• Opposite signs. As the magnet enters, the flux through the solenoid increases; as it leaves, the flux decreases. The change in flux is in opposite senses, so the induced EMFs have opposite signs.
• Larger spike at B. The magnet accelerates as it falls, so it leaves faster than it entered. The same change in flux happens in a shorter time, and since $\varepsilon = N\dfrac{\Delta\Phi}{\Delta t}$, a smaller $\Delta t$ gives a larger EMF.

From the report. In a, vague wording such as "the magnet has a changing field", or not mentioning the solenoid, lost marks. In b (23% full marks), many explained the size difference but not the opposite signs, or wrongly said the magnetic field changes direction.

a. [1 mark]. $I = \dfrac{P}{V} = \dfrac{70}{210} = 0.33$ A.

b. [3 marks]. The wires drop $230 - 210 = 20$ V, so the current in the line (which is also the current into the shed) is

The power delivered to the shed is $P = 210 \times 3.33 = 700$ W, and each lamp uses 70 W, so there are 10 lamps. (Equivalently, $\tfrac{3.33}{0.333} = 10$.)

c. i. [1 mark]. T1: step-up. T2: step-down.

ii. [2 marks]. Stepping up the voltage at the turbine reduces the line current for the same power, and since the power lost in the wires is $P_{\text{loss}} = I^2R$, the losses fall greatly, leaving more power for lamps. Stepping down at the shed returns the voltage to the level the lamps are designed for (about 210 to 230 V).

From the report. Part b was the hardest on the paper: 85% scored zero. Answers needed the voltage drop in the lines, then the line current, then the power in the shed; many just used the 5.00 kW figure without justification. In c.ii, "reduce the current to a safe level" was not a correct reason for the step-down transformer.

a. [1 mark]. The angle between the plane of the coil and the magnetic field changes, so the effective area perpendicular to the field (and the flux) changes.

b. [1 mark]. From maximum flux to zero (or vice versa):

c. [3 marks]. One rotation takes $\tfrac{1}{50} = 0.020$ s, so a quarter turn takes 0.0050 s:

d. [2 marks]. Replace the slip rings with a split-ring commutator. It reverses the coil's connections to the external circuit every half turn, so the output voltage always has the same polarity (DC).

From the report. In c, the most common error was using 0.02 s (a full turn) as the time. In d (36% full marks), many did not name the split-ring commutator correctly or explained its role in a motor instead.

[3 marks]. Find the speed from the de Broglie relation, then the voltage from $qV = \tfrac12 mv^2$:

From the report. Nearly half did not make a meaningful start. Many used $E = \tfrac{hc}{\lambda}$, which applies to photons, not electrons.

[3 marks]. The Michelson-Morley experiment compared the speed of light along two perpendicular paths, expecting Earth's motion through the supposed "aether" to make the speeds differ. Whatever the orientation of the apparatus, no difference was detected (a null result). This supports Einstein's postulate that the speed of light is the same for all observers, regardless of their relative motion, and removes the need for an aether.

From the report. Many knew of the experiment but did not describe that it compared light along two perpendicular paths. Stating the postulate only as "the speed of light is constant" missed "for all observers regardless of their relative motion".

a. [2 marks].

b. [2 marks]. Lower frequency means longer wavelength ($c = f\lambda$). Since $\Delta x = \tfrac{\lambda L}{d}$, the bright bands will be further apart.

From the report. The common error in a was converting mm or nm to metres incorrectly. In b, one mark was for the wider spacing and one for linking it to the longer wavelength; some confused interference with diffraction.

a. [2 marks]. Planck's constant is the gradient, using two points on the trend line:

b. [1 mark]. Extending the line to the vertical axis gives an intercept of $-2.2$ eV, so the work function is 2.2 eV. (Check: $4.4 - 4.4 \times 10^{-15} \times 15 \times 10^{14} = -2.2$.)

From the report. Common errors in a were using data points that were not on the line and misreading the axis scale. In b, many used the accepted value $4.14 \times 10^{-15}$ eV s instead of the graph.

[3 marks]. Rachel measures the proper time $t_0 = 1.000$ s, so

Rachel's speed is 0.14c.

From the report. Some calculated $\gamma = 0.990$, which is impossible: $\gamma$ is never less than 1.

a. [2 marks].

b. [1 mark]. $13.1 - 10.2 = 2.9$ eV, so draw a downward arrow from $n = 5$ to $n = 2$ (downward, because a photon is emitted).

From the report. In a the common error was converting to nanometres. In b, upward arrows (absorption) or lines without a direction lost the mark.

a. [3 marks]

Independent: the mass $m$ (and hence the magnetic force needed). Dependent: the current $I$. Controlled: for example the solenoid current (so the field $B$ is constant), or the length of QR.

b. [2 marks]

PQ and RS run along the solenoid's axis, parallel to the magnetic field. A current parallel to the field experiences no magnetic force ($F = BIl\sin\theta$ with $\theta = 0$), so these sides contribute nothing.

c. [2 marks]

With the pivot at the centre, $mg$ and $F_B$ act at equal distances from the pivot, so their turning effects (torques) are equal when the forces are equal. Only then does balancing the beam mean $F_B = mg$.

d. [1 mark]

Repeating measurements and averaging reduces the effect of random errors.

e. [5 marks]

Use a scale that fills the grid, plot each point with a ±0.2 A vertical bar, and rule a line through the origin that passes through every bar:

f. [2 marks]. Use two well-separated points on the trend line (not data points off the line), for example $(100 \times 10^{-6}, 0.8)$ and $(580 \times 10^{-6}, 4.6)$:

g. i. [3 marks]. $F = BIl$, so $\dfrac{F}{I} = Bl$. The gradient is $\dfrac{I}{F} = \dfrac{1}{Bl}$, so

ii. [2 marks]. $F_B$ would decrease: $F_B = BIl$, and with $B$ and $I$ unchanged, a shorter length $l$ gives a smaller force.

From the report. Part a (8% full marks): common errors were listing several variables in a box, listing equipment, or not saying which variable was which. In b (30%), many did not explain that PQ and RS are parallel to the field. Part d was answered correctly by only 37%, although the study design states that repeated measurements reduce random error. In e, points were sometimes spaced evenly regardless of their values, or the scale wasted space. In f, forgetting the "× 10⁻⁶" was the most common error. In g.i, substituting table data instead of the gradient did not earn marks.