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VCE Physics worked exam problems by Area of Study: The 2026 guide

Study guide16 min read

Quick answer

VCE Physics Unit 3-4 exam problems test calculation across all areas of study (projectile and circular motion, gravitational and electric fields, magnetic forces, electromagnetic induction, light and matter waves, photoelectric effect, interference); strong responses identify the physics principle, write the formula, substitute with units, calculate to 3 sig fig, and state the final answer with units.

Jump to a section
  1. What this guide is for
  2. Unit 3 AoS 1: Motion in two dimensions
  3. Unit 3 AoS 2: Fields and forces
  4. Unit 3 AoS 3: Electromagnetic induction
  5. Unit 4 AoS 1: Light and matter
  6. Unit 4 AoS 1 (light as a wave)
  7. Multi-mark working pattern
  8. Check your knowledge

What this guide is for

VCE Physics Unit 3-4 exam preparation requires fluency with calculation patterns across all areas of study. This guide presents worked problems for each AoS, with step-by-step solutions showing the marking conventions VCAA expects.

Unit 3 AoS 1: Motion in two dimensions

Projectile motion

Problem. A ball is thrown horizontally at 15 m/s from a height of 20 m. (a) Time to hit the ground. (b) Horizontal range.

Solution.

(a) Vertical motion: s=ut+12at2s = ut + \frac{1}{2}at^2. u=0u = 0 vertical, a=9.8a = 9.8 m/s2^2 down, s=20s = 20 m.

20=0+12(9.8)t220 = 0 + \frac{1}{2}(9.8)t^2
t2=40/9.8=4.08t^2 = 40/9.8 = 4.08
t=2.02t = 2.02 s.

(b) Horizontal motion: x=vxt=15×2.02=30.3x = v_x t = 15 \times 2.02 = 30.3 m.

Horizontal-launch projectile trajectory with three velocity snapshots A horizontally launched ball traces a true parabola from a 20-metre cliff. The trajectory is plotted as a 41-point polyline computed from y equals h minus one half g t squared using vx equals 15 metres per second and g equals 9.8 metres per second squared. Three snapshots at t equals 0, 1.0 and 2.02 seconds are marked. At each snapshot the constant horizontal velocity vx is drawn as an accent arrow and the growing downward velocity vy as a second accent arrow. A side inset at upper right shows the velocity-triangle decomposition for the t equals 1.0 second snapshot. Curved leader lines connect labels h equals 20 metres, t equals 0, t equals 1.0 s, t equals 2.02 s and range equals 30.3 metres to the relevant features in the surrounding whitespace. (a) trajectory (b) velocity t = 0 t = 1.0 s t = 2.02 s h = 20 m range = 30.3 m v at t = 1.0 s vx = 15 vy = gt v |v| = 17.9 m/s angle = 33 deg
Figure 1. Horizontal-launch projectile: panel (a) shows the true parabolic trajectory (41 computed samples) from a 20-metre cliff with three time snapshots; panel (b) is the velocity-decomposition inset at t = 1.0 s where vx stays at 15 m/s and vy = gt = 9.8 m/s.

Circular motion

Problem. A car of mass 1200 kg goes around a curve of radius 50 m at 20 m/s. Find the centripetal force.

Solution. Fc=mv2/r=1200×400/50=9600F_c = mv^2/r = 1200 \times 400 / 50 = 9600 N.

Uniform circular motion: velocity tangent, centripetal force toward centre A car travels anticlockwise around a true circle of radius r equal to 50 metres at 20 metres per second. Two snapshots ninety degrees apart on the circle are shown: one at the three-o-clock position and one at the twelve-o-clock position. At each snapshot the velocity vector v is drawn as a heavy accent arrow tangent to the circle (anticlockwise convention) and the centripetal force Fc is drawn from the car straight toward the centre. A dashed radial line connects the centre to the three-o-clock snapshot and is labelled r. A subtle dashed grid sits behind both panels. A side inset panel at the right gives the equation Fc equals m v squared over r typeset on three separate text lines, with the substitution Fc equals 9600 newtons for m equals 1200 kilograms, v equals 20 metres per second, r equals 50 metres. (a) two snapshots (b) Fc inset v (tangent) Fc (toward centre) r = 50 m v at top of curve m = 1200 kg, v = 20 m/s, r = 50 m centripetal force Fc = m v 2 r = 1200 × 400 / 50 = 9600 N tyre friction supplies the centripetal force toward the curve centre
Figure 2. Two-snapshot view of uniform circular motion: v is always tangent (anticlockwise) and Fc always radially inward. The side inset gives the symbolic formula and the numerical substitution.

Unit 3 AoS 2: Fields and forces

Gravitational

Problem. A satellite orbits Earth in a circular orbit of radius r=7.0×106r = 7.0 \times 10^6 m. Find its orbital period. (GM=3.99×1014GM = 3.99 \times 10^{14} m3^3/s2^2.)

Solution. T2=4π2r3GM=4π2(7×106)33.99×1014T^2 = \frac{4\pi^2 r^3}{GM} = \frac{4\pi^2 (7 \times 10^6)^3}{3.99 \times 10^{14}}.

r3=3.43×1020r^3 = 3.43 \times 10^{20}. T2=4π2×3.43×1020/3.99×1014=3.39×107T^2 = 4\pi^2 \times 3.43 \times 10^{20} / 3.99 \times 10^{14} = 3.39 \times 10^7. T=5821T = 5821 s ≈97\approx 97 min.

Electric field between parallel plates

Problem. A 12 V battery is connected across parallel plates 2.0 mm apart. (a) Field strength? (b) Force on an electron in the field?

Solution.

(a) E=V/d=12/0.002=6000E = V/d = 12 / 0.002 = 6000 V/m.

(b) F=qE=1.6×10−19×6000=9.6×10−16F = qE = 1.6 \times 10^{-19} \times 6000 = 9.6 \times 10^{-16} N.

Unit 3 AoS 3: Electromagnetic induction

Faraday's law

Problem. A coil of 200 turns with area 0.040 m2^2 is in a magnetic field that changes from 0.20 T to 0.50 T in 0.50 s. Find the induced EMF.

Solution. ΔΦ\Delta \Phi per turn = ΔB×A=0.30×0.040=0.012\Delta B \times A = 0.30 \times 0.040 = 0.012 Wb.

ε=NΔΦΔt=200×0.012/0.50=4.8\varepsilon = N \frac{\Delta \Phi}{\Delta t} = 200 \times 0.012 / 0.50 = 4.8 V.

A coil in a changing external field with induced EMF and current direction Two-panel figure. Panel (a) Coil + circuit: a circular coil of N equals 200 turns sits in an external magnetic field B pointing into the page, shown by a three-by-three grid of crosses inside the loop. By Lenz's law the induced current flows anticlockwise, drawn as two heavy accent arcs on the loop with the label I induced. The loop is connected to an external resistor R that displays plus and minus terminals and the value epsilon equals 4.8 volts. Panel (b) B versus t inset: a small line plot showing B increasing linearly with time, with a delta B and delta t bracket and the Faraday substitution epsilon equals N delta Phi over delta t equals 200 times 0.012 over 0.50 equals 4.8 volts typeset on three lines. Subtle dashed grid in the background. (a) coil + circuit (b) B(t) + Faraday × × × × × × × × × + − B into page (increasing) I induced (anticlockwise) external R N = 200 turns, A = 0.040 m2 B vs t B t slope rises with time ε = N ΔΦ Δt = 200 × 0.012 / 0.50 = 4.8 V
Figure 3. Faraday and Lenz at work: an increasing external field into the page induces an anticlockwise current in the loop (so the coil's own field opposes the increase), and the side inset gives the symbolic and numerical induced EMF for the worked example.

Transformer

Problem. A transformer has 500 primary turns and 100 secondary turns. Input 240 V at 0.5 A. Find secondary voltage and current (ideal transformer).

Solution. Vs/Vp=Ns/Np=100/500=1/5V_s / V_p = N_s / N_p = 100/500 = 1/5. Vs=48V_s = 48 V.

Power conservation: VpIp=VsIsV_p I_p = V_s I_s. Is=VpIp/Vs=240×0.5/48=2.5I_s = V_p I_p / V_s = 240 \times 0.5 / 48 = 2.5 A.

Unit 4 AoS 1: Light and matter

Photoelectric effect

Problem. Light of frequency 7.5×10147.5 \times 10^{14} Hz incident on a metal with work function 2.4 eV. (h=4.14×10−15h = 4.14 \times 10^{-15} eV s.) Find Ek,maxE_{k,max} of ejected electrons.

Solution. Photon energy =hf=4.14×10−15×7.5×1014=3.1= hf = 4.14 \times 10^{-15} \times 7.5 \times 10^{14} = 3.1 eV.

Ek,max=hf−ϕ=3.1−2.4=0.7E_{k,max} = hf - \phi = 3.1 - 2.4 = 0.7 eV.

de Broglie

Problem. An electron is accelerated through 200 V. Find its de Broglie wavelength.

Solution. Shortcut: λ=1.226/V\lambda = 1.226 / \sqrt{V} nm =1.226/200=0.087= 1.226/\sqrt{200} = 0.087 nm.

Unit 4 AoS 1 (light as a wave)

Young's double-slit

Problem. Light λ=600\lambda = 600 nm, slits 0.20 mm apart, screen 1.5 m away. Find fringe spacing.

Solution. Δx=λL/d=600×10−9×1.5/0.20×10−3=4.5×10−3\Delta x = \lambda L / d = 600 \times 10^{-9} \times 1.5 / 0.20 \times 10^{-3} = 4.5 \times 10^{-3} m =4.5= 4.5 mm.

Malus's law (extension)

Malus's law is an extension: it is not in the VCE Physics key knowledge for Units 3 and 4, so use this problem as enrichment only.

Problem. Unpolarised light intensity I0I_0 passes through a polariser, then through a second at 30 degrees to the first. Find the final intensity.

Solution. After first polariser: I0/2I_0/2. After second (Malus): (I0/2)cos⁡2(30)=(I0/2)(3/4)=3I0/8(I_0/2) \cos^2(30) = (I_0/2)(3/4) = 3 I_0/8.

Multi-mark working pattern

For each numerical question:

  1. State the principle (Faraday's law, Newton's second law).
  2. Write the formula in symbolic form.
  3. Substitute values with units.
  4. Calculate with appropriate significant figures.
  5. State the answer with units.

VCAA awards method marks for correct identification of the principle and formula, even if the calculation slips.

Common errors
Unit conversion forgotten
mm, MHz, mW, kV all need converting to SI base units.
Significant figures
Typically 3 sig fig (occasionally 2 for very imprecise data).
Vector vs scalar confusion
Force, momentum, velocity are vectors; speed and energy are scalars.
Sign convention
Choose positive direction, apply consistently.
Wrong hh value
Use h=6.626×10−34h = 6.626 \times 10^{-34} J s, or h=4.14×10−15h = 4.14 \times 10^{-15} eV s, but match units throughout.

Check your knowledge

A broad VCAA-style mix across the four Areas of Study, set as it would appear on the November paper. Attempt under exam conditions before checking the solutions block.

  1. State the difference between displacement and distance for an object in 2-D motion, and give an example where they differ by a factor of two. (2 marks)
  2. A projectile is launched at 25 m s−1^{-1} at 50 degrees above horizontal from ground level on a still day. Taking g=9.80 m s−2g = 9.80 \ \text{m s}^{-2}, calculate (a) the time of flight, (b) the maximum height, (c) the range. (5 marks)
  3. (a, 3) A satellite is placed in low Earth orbit (LEO) at altitude 350 km. Calculate the orbital speed and period. (G=6.67×10−11 N m2kg−2G = 6.67 \times 10^{-11} \ \text{N m}^{2} \text{kg}^{-2}, ME=5.98×1024 kgM_E = 5.98 \times 10^{24} \ \text{kg}, RE=6.37×106 mR_E = 6.37 \times 10^{6} \ \text{m}.) (b, 2) State why LEO satellites must be re-boosted periodically and what eventually happens if they are not. (5 marks)
  4. A current of 8.0 A flows in a 25-turn rectangular coil of dimensions 0.080 m by 0.050 m. The coil is in a 0.30 T uniform magnetic field, with its plane parallel to the field. (a) Calculate the torque on the coil. (b) State the angle at which torque is zero and explain why the coil is unstable at that orientation in an undriven motor. (5 marks)
  5. (a, 3) An electron in a hydrogen atom drops from n=4n = 4 to n=1n = 1. Using En=−13.6/n2 eVE_n = -13.6 / n^2 \ \text{eV}, calculate the photon energy and wavelength. (b, 2) State and identify the spectral region (UV, visible, IR) of the emitted photon. (5 marks)
  6. A spaceship travels from Earth to Alpha Centauri (4.0 light-years in Earth's frame) at v=0.60cv = 0.60c. (a) Calculate the Lorentz factor. (b) Calculate the Earth-frame travel time. (c) Calculate the proper time experienced by the crew. (d) Calculate the Alpha Centauri distance measured in the spaceship's frame. (7 marks)
  7. (a, 3) Two parallel plates separated by 5.0 mm are connected to a 2000 V supply. Calculate the electric field between the plates and the force on an electron between them. (b, 3) The electron, initially at rest at the negative plate, traverses the gap. Calculate its kinetic energy on reaching the positive plate, and its speed (treat non-relativistically; me=9.11×10−31 kgm_e = 9.11 \times 10^{-31} \ \text{kg}). (6 marks)
  8. A practical-investigation data set tests Hooke's law for a steel spring. Force (N) versus extension (m): (1.0, 0.0205), (2.0, 0.0398), (3.0, 0.0612), (4.0, 0.0808), (5.0, 0.1005). (a) State Hooke's law and the expected graph. (b) Calculate the spring constant from the first and last points and comment on whether the data are consistent with Hooke's law across the full range. (c) Estimate the uncertainty in the spring constant if each extension is read to ±0.5 mm\pm 0.5 \ \text{mm} and each force to ±0.05 N\pm 0.05 \ \text{N}. (6 marks)
Solutions
Q1
Displacement is the straight-line vector from start to finish position; distance is the total scalar path length covered. Example: a runner who jogs 200 m around a semicircular track returns 127 m closer to start as a straight line (Δr=2×R=2×200/π=127\Delta r = 2 \times R = 2 \times 200/\pi = 127 m), so displacement is 127 m while distance is 200 m, almost exactly a factor of two apart.
Q2
ux=25cos⁡50∘=16.07 m s−1u_x = 25 \cos 50^\circ = 16.07 \ \text{m s}^{-1}; uy=25sin⁡50∘=19.15 m s−1u_y = 25 \sin 50^\circ = 19.15 \ \text{m s}^{-1}. (a) Time of flight =2uy/g=2×19.15/9.80=3.91 s= 2 u_y / g = 2 \times 19.15 / 9.80 = 3.91 \ \text{s}. (b) Max height =uy2/2g=(19.15)2/19.6=18.7 m= u_y^2 / 2g = (19.15)^2 / 19.6 = 18.7 \ \text{m}. (c) Range =ux×T=16.07×3.91=62.8 m= u_x \times T = 16.07 \times 3.91 = 62.8 \ \text{m}.
Q3
(a) r=6.37×106+3.50×105=6.72×106 mr = 6.37 \times 10^{6} + 3.50 \times 10^{5} = 6.72 \times 10^{6} \ \text{m}. v=GM/r=6.67×10−11×5.98×1024/6.72×106=5.94×107=7.71×103 m s−1v = \sqrt{GM/r} = \sqrt{6.67 \times 10^{-11} \times 5.98 \times 10^{24} / 6.72 \times 10^{6}} = \sqrt{5.94 \times 10^{7}} = 7.71 \times 10^{3} \ \text{m s}^{-1}. T=2πr/v=2π×6.72×106/7.71×103=5.48×103 s=91.3 minutesT = 2\pi r/v = 2\pi \times 6.72 \times 10^{6}/7.71 \times 10^{3} = 5.48 \times 10^{3} \ \text{s} = 91.3 \ \text{minutes}. (b) Atmospheric drag at 350 km is non-negligible; without periodic re-boost, the satellite spirals inward, loses altitude, heats catastrophically and burns up on re-entry within a few years.
Q4
(a) For a coil with its plane parallel to the field, the moment arm is at maximum. Torque τ=NBIA=25×0.30×8.0×(0.080×0.050)=25×0.30×8.0×4.0×10−3=0.24 N m\tau = N B I A = 25 \times 0.30 \times 8.0 \times (0.080 \times 0.050) = 25 \times 0.30 \times 8.0 \times 4.0 \times 10^{-3} = 0.24 \ \text{N m}. (b) Torque is zero when the coil's plane is perpendicular to the field (the field threads the coil along the normal). In an undriven motor (no commutator current reversal) the coil overshoots this position and reverses direction, settling at the zero-torque orientation; the equilibrium is stable if the field still threads in the same direction, but a small perturbation drives the coil back. In a motor with a commutator, this is the "dead point" where current must reverse to maintain rotation.
Q5
(a) E4−E1=−13.6/16−(−13.6/1)=−0.85+13.6=12.75 eVE_4 - E_1 = -13.6/16 - (-13.6/1) = -0.85 + 13.6 = 12.75 \ \text{eV}. λ=hc/E=1240/12.75=97.3 nm\lambda = hc/E = 1240/12.75 = 97.3 \ \text{nm}. (b) 97 nm lies in the deep ultraviolet (the Lyman series), invisible to the human eye and absorbed strongly by atmospheric oxygen.
Q6
(a) γ=1/1−0.36=1/0.64=1/0.80=1.25\gamma = 1/\sqrt{1 - 0.36} = 1/\sqrt{0.64} = 1/0.80 = 1.25. (b) Earth-frame time =4.0 ly/0.60c=4.0/0.60=6.67 years= 4.0 \ \text{ly} / 0.60c = 4.0/0.60 = 6.67 \ \text{years}. (c) Proper time τ0=t/γ=6.67/1.25=5.33 years\tau_0 = t/\gamma = 6.67/1.25 = 5.33 \ \text{years}. (d) Contracted distance L=L0/γ=4.0/1.25=3.20 lyL = L_0/\gamma = 4.0/1.25 = 3.20 \ \text{ly}. Crew traverse 3.20 ly at 0.60c0.60c in their frame, taking 3.20/0.60=5.33 years3.20/0.60 = 5.33 \ \text{years}, matching part (c).
Q7
(a) E=V/d=2000/0.005=4.00×105 V m−1E = V/d = 2000/0.005 = 4.00 \times 10^{5} \ \text{V m}^{-1}. Force F=eE=1.60×10−19×4.00×105=6.40×10−14 NF = eE = 1.60 \times 10^{-19} \times 4.00 \times 10^{5} = 6.40 \times 10^{-14} \ \text{N}. (b) KE=eV=1.60×10−19×2000=3.20×10−16 JKE = eV = 1.60 \times 10^{-19} \times 2000 = 3.20 \times 10^{-16} \ \text{J}. v=2KE/me=2×3.20×10−16/9.11×10−31=7.03×1014=2.65×107 m s−1v = \sqrt{2KE/m_e} = \sqrt{2 \times 3.20 \times 10^{-16}/9.11 \times 10^{-31}} = \sqrt{7.03 \times 10^{14}} = 2.65 \times 10^{7} \ \text{m s}^{-1}. This is roughly 0.09c0.09c, so non-relativistic treatment is acceptable (relativistic correction less than 0.5 percent).
Q8
(a) Hooke's law: F=kxF = kx for a spring within its elastic limit; a graph of FF versus xx is a straight line through the origin with gradient equal to the spring constant kk. (b) Gradient =(5.0−1.0)/(0.1005−0.0205)=4.00/0.0800=50.0 N m−1= (5.0 - 1.0)/(0.1005 - 0.0205) = 4.00/0.0800 = 50.0 \ \text{N m}^{-1}. Checking the intermediate points: at F=3.0F = 3.0, predicted x=0.0600x = 0.0600, measured 0.0612 (within 2 percent); the data are consistent with Hooke's law across the entire range tested. (c) Relative uncertainty in extension over a typical reading Δx/x=0.0005/0.05=1.0\Delta x / x = 0.0005/0.05 = 1.0 percent. Relative uncertainty in force ΔF/F=0.05/3.0=1.7\Delta F / F = 0.05/3.0 = 1.7 percent. Combined (quadrature) for k=F/xk = F/x: 1.02+1.72=2.0\sqrt{1.0^2 + 1.7^2} = 2.0 percent. Absolute Δk=0.02×50.0=1.0 N m−1\Delta k = 0.02 \times 50.0 = 1.0 \ \text{N m}^{-1}. So k=(50.0±1.0) N m−1k = (50.0 \pm 1.0) \ \text{N m}^{-1}.

Sources & how we know this

Based on the official VCAA study design and published VCAA past papers and marking guidelines. Not yet individually reviewed by a person; always check VCAA and your school's materials for current rules.

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