Skill Align SACE Stage 2 Physics - Free Online Pack 0
An independently authored current-format practice examination with worked answers and question-specific marking guidance.
- Paper
- Question booklets 1 and 2
- Reading
- Included in total time
- Writing
- 130 minutes total
- Assessment
- 120 marks
The current SACE Physics formula sheet is supplied separately. Approved calculators may be used.
Question booklet 1
Questions 1 to 11 (60 marks). Answer all questions. Allow approximately 65 minutes.
Question 1
8 marksA coastal survey drone flies horizontally at 18.0 m s^-1 and releases a sensor from 45.0 m above level sand. Ignore air resistance in parts (a) and (b).
Question 2
7 marksA mapping relay travels in a circular orbit of radius 2.40 × 10^6 m around an asteroid of mass 8.00 × 10^22 kg. Treat the asteroid as spherical.
Question 3
6 marksA positively charged aerosol ion has charge 3.20 × 10^-19 C and mass 6.40 × 10^-26 kg. It is between parallel plates separated by 0.0400 m with a potential difference of 1.20 kV.
Question 4
3 marksQuestion 5
3 marksQuestion 6
6 marksA 600 nm laser illuminates two slits separated by 0.250 mm. A screen is 2.00 m from the slits.
Question 7
4 marksQuestion 8
4 marksA photoelectric door sensor uses a metal with work function 2.00 eV. Light of frequency 7.25 × 10^14 Hz illuminates the metal.
Question 9
7 marksAn argon guide light is modelled with energy levels at -10.0 eV, -5.00 eV and -3.00 eV.
Question 10
5 marksQuestion 11
7 marksA class varies the rotation frequency of a hand generator while keeping coil area, turns and magnetic field constant. The measured peak emfs are 1.20, 2.39, 3.61 and 4.79 V at 5.00, 10.0, 15.0 and 20.0 Hz respectively.
Question booklet 2
Questions 12 to 21 (60 marks). Answer all questions. Allow approximately 65 minutes.
Question 12
7 marksA pion has a proper mean lifetime of 2.60 × 10^-8 s and moves through a detector at 0.980c.
Question 13
9 marksStudents launch a ball at 45.0° from and to the same height. They measure ranges of 2.54, 3.68, 4.98 and 6.55 m for launch speeds of 5.00, 6.00, 7.00 and 8.00 m s^-1. Air resistance is small.
Question 14
5 marksQuestion 15
5 marksAn ideal transformer powers a 36.0 W aquarium lamp from a 240 V rms supply. The primary has 1200 turns and the secondary has 60 turns.
Question 16
8 marksAn airport inspection X-ray tube accelerates electrons through 80.0 kV before they strike a metal target.
Question 17
3 marksQuestion 18
5 marksA proton travels east at 2.00 × 10^6 m s^-1 into a uniform 0.300 T magnetic field directed into the page.
Question 19
4 marksQuestion 20
11 marksElectrons are accelerated from rest through 150 V and directed at a crystal whose effective plane spacing is 2.00 × 10^-10 m. Use non-relativistic momentum for this question.
Question 21
3 marksWorked Solutions And Marking Guide
General marking principles
- Apply each question-specific criterion to the physical evidence shown; accept equivalent scientifically correct reasoning.
- Where a criterion spans linked steps, award its stated marks only when the complete linked reasoning is demonstrated.
- Carry forward an earlier numerical error when later physics is internally consistent, unless a later criterion independently assesses the same step.
- Do not penalise a missing unit more than once in a linked calculation unless the question separately assesses units.
Question 1
(a) Using 45.0 = 0.5(9.80)t^2 gives t = 3.03 s.
(b) The range is 18.0 × 3.03 = 54.5 m. The vertical impact component is 9.80 × 3.03 = 29.7 m s^-1 downward, so the impact speed is 34.7 m s^-1 at 58.8° below horizontal.
(c) Drag opposes the instantaneous velocity. It reduces the horizontal component throughout the flight and limits the vertical component, so the path has a shorter range, is not symmetric, and the impact speed is lower than the ideal value.
Detailed marking criteria
Part a (1 mark)
Uses the vertical displacement relationship with zero initial vertical speed.
Part a (1 mark)
Calculates a fall time of 3.03 s.
Part b (1 mark)
Calculates the horizontal range as 54.5 m.
Part b (1 mark)
Calculates the downward velocity component as 29.7 m s^-1.
Part b (1 mark)
Combines the perpendicular components to obtain 34.7 m s^-1 at 58.8 degrees below horizontal.
Part c (1 mark)
States that drag acts opposite the instantaneous velocity.
Part c (1 mark)
Explains that the horizontal component and hence range are reduced.
Part c (1 mark)
Explains that the downward component is limited and the impact speed is reduced.
Question 2
(a) g = GM/r^2 = 0.926 m s^-2.
(b) v = sqrt(GM/r) = 1.49 × 10^3 m s^-1 and T = 2πr/v = 1.01 × 10^4 s, or 2.81 h.
(c) Kepler's third law gives T2/T1 = (r2/r1)^(3/2) = 1.50^(3/2) = 1.84.
Detailed marking criteria
Part a (1 mark)
Uses g = GM/r^2 with the distance measured from the asteroid centre.
Part a (1 mark)
Calculates 0.926 m s^-2 with an appropriate unit.
Part b (1 mark)
Equates gravitational acceleration to centripetal acceleration.
Part b (1 mark)
Calculates the orbital speed as 1.49 × 10^3 m s^-1.
Part b (1 mark)
Calculates the period as 1.01 × 10^4 s or 2.81 h.
Part c (1 mark)
Applies the period-radius power relationship for the same central mass.
Part c (1 mark)
Obtains a period ratio of 1.84.
Question 3
(a) E = ΔV/d = 3.00 × 10^4 V m^-1.
(b) F = qE = 9.60 × 10^-15 N and a = F/m = 1.50 × 10^11 m s^-2 in the electric-field direction.
(c) W = qΔV = 3.84 × 10^-16 J = 2.40 × 10^3 eV.
Detailed marking criteria
Part a (1 mark)
Uses the uniform-field relationship E = delta V/d.
Part a (1 mark)
Calculates 3.00 × 10^4 V m^-1.
Part b (1 mark)
Calculates the electric force as 9.60 × 10^-15 N.
Part b (1 mark)
Calculates 1.50 × 10^11 m s^-2 in the field direction.
Part c (1 mark)
Calculates the transferred energy as 3.84 × 10^-16 J.
Part c (1 mark)
Converts the result to 2.40 × 10^3 eV.
Question 4
Indicative answer: B = μ0I/(2πr) = (2.00 × 10^-7)(8.00)/0.0500 = 3.20 × 10^-5 T. The right-hand rule gives a direction into the page at a point east of an upward current.
Detailed marking criteria
Integrated response (1 mark)
Selects the long-straight-conductor field relationship.
Integrated response (1 mark)
Calculates 3.20 × 10^-5 T.
Integrated response (1 mark)
Uses the right-hand rule to identify the field as into the page.
Question 5
Indicative answer: The downward magnetic flux through the ring decreases as the north pole is withdrawn. The ring produces a downward field to oppose that decrease, requiring a clockwise induced current when viewed from above.
Detailed marking criteria
Integrated response (1 mark)
Identifies the decreasing downward magnetic flux.
Integrated response (1 mark)
Applies Lenz's law to require an induced downward field.
Integrated response (1 mark)
Uses the right-hand rule to obtain a clockwise current viewed from above.
Question 6
(a) Δy = λL/d = (600 × 10^-9)(2.00)/(0.250 × 10^-3) = 4.80 × 10^-3 m, or 4.80 mm.
(b) Bright fringes occur where path difference gives waves in phase and constructive interference; dark fringes occur where the waves arrive out of phase and destructively interfere. Since Δy is proportional to wavelength, a longer wavelength produces wider fringe spacing.
Detailed marking criteria
Part a (1 mark)
Converts the wavelength and slit separation consistently to SI units.
Part a (1 mark)
Applies the double-slit fringe-spacing relationship.
Part a (1 mark)
Calculates a separation of 4.80 mm.
Part b (1 mark)
Links bright fringes with in-phase arrival and constructive interference.
Part b (1 mark)
Links dark fringes with destructive interference from out-of-phase arrival.
Part b (1 mark)
Predicts and justifies an increase in fringe spacing for longer wavelength.
Question 7
Indicative answer: Reflection reverses the photon momentum, so the momentum transfer is twice that for absorption. F = 2P/c = 2.67 × 10^-5 N and a = F/m = 3.33 × 10^-8 m s^-2.
Detailed marking criteria
Integrated response (2 marks)
Relates reflected photon momentum to F = 2P/c and explains the factor of two.
Integrated response (1 mark)
Calculates the force as 2.67 × 10^-5 N.
Integrated response (1 mark)
Calculates the acceleration as 3.33 × 10^-8 m s^-2.
Question 8
(a) The photon energy is hf = 4.81 × 10^-19 J = 3.00 eV. Therefore Kmax = 1.00 eV = 1.60 × 10^-19 J and the stopping potential is 1.00 V.
(b) More photons arrive per second, so the emitted-electron rate and photocurrent increase, but each photon has unchanged energy and Kmax does not increase.
Detailed marking criteria
Part a (1 mark)
Calculates the photon energy as about 3.00 eV.
Part a (1 mark)
Subtracts the work function to obtain Kmax = 1.00 eV or 1.60 × 10^-19 J.
Part a (1 mark)
Determines a stopping potential of 1.00 V.
Part b (1 mark)
States that photocurrent increases while maximum kinetic energy remains unchanged.
Question 9
(a) ΔE = 7.00 eV = 1.12 × 10^-18 J, so f = ΔE/h = 1.69 × 10^15 Hz.
(b) The smallest energy step is -3.00 eV to -5.00 eV, so ΔE = 2.00 eV and f = 4.83 × 10^14 Hz.
(c) A population inversion makes stimulated emission more likely than absorption. A metastable upper state holds atoms long enough for the inversion to form. An incident matching photon then triggers an identical photon, building coherent monochromatic light.
Detailed marking criteria
Part a (1 mark)
Determines the transition energy as 7.00 eV or 1.12 × 10^-18 J.
Part a (1 mark)
Calculates the required frequency as 1.69 × 10^15 Hz.
Part b (1 mark)
Identifies the -3.00 eV to -5.00 eV transition as the smallest energy difference.
Part b (1 mark)
Calculates the lowest frequency as 4.83 × 10^14 Hz.
Part c (1 mark)
Links population inversion with stimulated emission predominating over absorption.
Part c (1 mark)
Explains that a metastable state provides time to establish the inversion.
Part c (1 mark)
Connects stimulated emission with identical, coherent photons.
Question 10
Indicative answer: Let x be the distance from +2.00 nC. Equating field magnitudes gives 2/x^2 = 8/(0.300 - x)^2, so 0.300 - x = 2x and x = 0.100 m. Between like charges the fields oppose; the point must be closer to the smaller charge to make its weaker source field equal in magnitude.
Detailed marking criteria
Integrated response (1 mark)
Recognises that the two fields oppose only in the region between the like charges.
Integrated response (2 marks)
Equates the inverse-square field magnitudes and solves the distance ratio correctly.
Integrated response (1 mark)
Locates the point 0.100 m from the +2.00 nC charge.
Integrated response (1 mark)
Explains physically why the zero point is closer to the smaller charge.
Question 11
(a) A line through the trend has gradient approximately (4.79 - 1.20)/(20.0 - 5.00) = 0.239 V Hz^-1.
(b) The data are approximately linear and the extrapolated intercept is close to zero, supporting direct proportionality within the scatter. The small departures from a perfect line mean the conclusion should be qualified by measurement uncertainty.
(c) Keep the coil geometry/turns and magnetic-field strength fixed. Repeat each frequency and average the peak emf, using electronic frequency measurement to reduce speed variation.
Detailed marking criteria
Part a (1 mark)
Uses well-separated points on the overall trend rather than adjacent noisy values.
Part a (1 mark)
Obtains a gradient near 0.240 V Hz^-1 with its unit.
Part b (1 mark)
Identifies the approximately linear trend.
Part b (1 mark)
Uses the near-zero intercept as evidence for direct proportionality.
Part b (1 mark)
Qualifies the conclusion using scatter or measurement uncertainty.
Part c (1 mark)
Identifies two relevant controlled factors such as coil geometry, turns or field strength.
Part c (1 mark)
Proposes repeat measurements and averaging or an equivalent improvement tied to reduced uncertainty.
Question 12
(a) γ = 1/sqrt(1 - 0.980^2) = 5.03.
(b) t = γt0 = 1.31 × 10^-7 s and d = vt = 38.5 m.
(c) Without dilation the mean distance would be vt0 = 7.64 m. A substantial population beyond 30 m is consistent with the longer detector-frame lifetime predicted by special relativity and not with the undilated Newtonian estimate.
Detailed marking criteria
Part a (1 mark)
Substitutes v/c = 0.980 into the Lorentz-factor relationship.
Part a (1 mark)
Calculates gamma as 5.03.
Part b (1 mark)
Uses the proper lifetime in t = gamma t0.
Part b (1 mark)
Calculates the dilated lifetime as 1.31 × 10^-7 s.
Part b (1 mark)
Calculates the detector-frame travel distance as 38.5 m.
Part c (1 mark)
Calculates or states the undilated distance as about 7.64 m.
Part c (1 mark)
Links detections beyond 30 m to the quantitatively larger relativistic lifetime.
Question 13
(a) For a 45.0° launch at equal heights, R = v^2/g, so R is linear in v^2. A best-fit gradient is approximately 0.102 s^2 m^-1.
(b) g = 1/gradient ≈ 9.80 m s^-2. The percentage difference is negligible at the precision of the data, so the result agrees with the accepted value.
(c) A manually judged landing point produces random range uncertainty. Use video or carbon paper to locate first contact, repeat and average each speed, and measure launch speed immediately after release with photogates.
Detailed marking criteria
Part a (1 mark)
Uses R = v^2/g for the stated 45-degree equal-height launch.
Part a (1 mark)
Explains why v squared is the appropriate horizontal variable.
Part a (1 mark)
Determines a best-fit gradient near 0.102 s^2 m^-1.
Part b (1 mark)
Recognises that g is the reciprocal of the gradient.
Part b (1 mark)
Calculates g close to 9.80 m s^-2.
Part b (1 mark)
Makes a justified comparison using precision or percentage difference.
Part c (1 mark)
Identifies a specific limitation and explains its effect on the range data.
Part c (2 marks)
Proposes two relevant improvements tied to landing-point, repeat or launch-speed uncertainty.
Question 14
Indicative answer: At nucleon level n → p + e^- + anti-nu_e. At quark level one down quark changes to an up quark: udd → uud + W^-, followed by W^- → e^- + anti-nu_e. Charge is 0 on each side, baryon number remains 1, and electronic lepton number is 0 because +1 for the electron and -1 for the antineutrino sum to zero.
Detailed marking criteria
Integrated response (1 mark)
Writes the complete nucleon-level beta-minus reaction.
Integrated response (1 mark)
Identifies the down-to-up quark conversion and W-minus boson.
Integrated response (1 mark)
Demonstrates charge and baryon-number conservation.
Integrated response (2 marks)
Correctly assigns electron and antineutrino lepton numbers and demonstrates electronic-lepton-number conservation.
Question 15
(a) Vs/Vp = Ns/Np, so Vs = 240(60/1200) = 12.0 V rms.
(b) Is = P/Vs = 36.0/12.0 = 3.00 A.
(c) Alternating current produces a changing magnetic flux in the common core, which links the secondary and induces an emf. A steady direct current produces no continuing flux change after switching, so it does not maintain a secondary emf.
Detailed marking criteria
Part a (1 mark)
Uses the voltage-turns ratio for one primary-secondary pair.
Part a (1 mark)
Calculates 12.0 V rms.
Part b (1 mark)
Calculates the secondary current as 3.00 A.
Part c (1 mark)
Links alternating primary current to changing core flux and induced secondary emf.
Part c (1 mark)
Explains why steady direct current cannot maintain the induced emf.
Question 16
(a) The greatest photon energy is eΔV = 1.28 × 10^-14 J, so fmax = eΔV/h = 1.93 × 10^19 Hz.
(b) λmin = c/fmax = 1.55 × 10^-11 m. The continuous spectrum occurs because electrons lose variable fractions of their kinetic energy during braking interactions, producing photons over a range of energies.
(c) More electrons strike the target each second, so X-ray intensity increases. The maximum frequency and characteristic frequencies remain unchanged because electron energy and target material are unchanged.
Detailed marking criteria
Part a (1 mark)
Equates maximum photon energy to the electron energy e delta V.
Part a (1 mark)
Converts 80.0 kV and calculates 1.28 × 10^-14 J.
Part a (1 mark)
Calculates fmax = 1.93 × 10^19 Hz.
Part b (1 mark)
Uses c = f lambda to calculate 1.55 × 10^-11 m.
Part b (2 marks)
Explains the continuum through variable electron energy loss during deceleration.
Part c (1 mark)
Predicts an increase in X-ray intensity.
Part c (1 mark)
Explains why the maximum and characteristic frequencies do not change.
Question 17
Indicative answer: Initial momentum is zero, so pC = -(pA + pB) = 6.00 kg m s^-1 west plus 8.00 kg m s^-1 south. Its magnitude is 10.0 kg m s^-1 and its direction is 53.1° south of west.
Detailed marking criteria
Integrated response (1 mark)
Uses conservation of momentum to give fragment C westward and southward components.
Integrated response (1 mark)
Calculates the magnitude as 10.0 kg m s^-1.
Integrated response (1 mark)
Calculates the direction as 53.1 degrees south of west or an equivalent bearing.
Question 18
(a) F = qvB = 9.60 × 10^-14 N. For a positive charge, the right-hand rule gives a force north on the page.
(b) r = mv/(qB) = (1.67 × 10^-27)(2.00 × 10^6)/[(1.60 × 10^-19)(0.300)] = 6.96 × 10^-2 m.
(c) The magnetic force is always perpendicular to the instantaneous velocity, so it supplies centripetal acceleration and does no work. Kinetic energy and speed remain constant.
Detailed marking criteria
Part a (1 mark)
Uses F = qvB for perpendicular velocity and field.
Part a (1 mark)
Calculates 9.60 × 10^-14 N directed north.
Part b (1 mark)
Equates magnetic force with the centripetal-force requirement.
Part b (1 mark)
Calculates a radius of 6.96 × 10^-2 m.
Part c (1 mark)
Explains that perpendicular magnetic force does no work and therefore does not change speed.
Question 19
Indicative answer: Independent detectors can verify that a signal is not local instrumental noise, and separated sites provide timing information that improves source direction and supports shared analysis. Common model assumptions or a systematic calibration method could still bias all inferred source properties, so agreement alone does not eliminate every limitation.
Detailed marking criteria
Integrated response (1 mark)
Explains independent verification against local noise or artefacts.
Integrated response (1 mark)
Explains how separated timing or shared expertise improves localisation or analysis.
Integrated response (1 mark)
Identifies a plausible shared systematic or model limitation.
Integrated response (1 mark)
Forms a qualified conclusion about the strength rather than certainty of the evidence.
Question 20
(a) p = sqrt(2meV) = 6.61 × 10^-24 kg m s^-1 and λ = h/p = 1.00 × 10^-10 m.
(b) d sinθ = λ gives sinθ = 0.500 and θ = 30.0°.
(c) Each electron is detected as a localised event, while many events build an interference distribution associated with a wave. Increasing voltage increases momentum, decreases de Broglie wavelength and moves the first-order maximum to a smaller angle.
(d) Use a position-sensitive detector farther from the crystal and calibrate its zero angle. The longer baseline increases spatial separation for a given angular difference, while zero calibration reduces systematic angular offset.
Detailed marking criteria
Part a (1 mark)
Equates electron kinetic energy to e delta V.
Part a (1 mark)
Obtains p = sqrt(2meV).
Part a (1 mark)
Calculates momentum as 6.61 × 10^-24 kg m s^-1.
Part a (1 mark)
Calculates the de Broglie wavelength as 1.00 × 10^-10 m.
Part b (1 mark)
Applies the first-order grating relationship to the supplied spacing.
Part b (1 mark)
Calculates an angle of 30.0 degrees.
Part c (1 mark)
Identifies localised electron detections as particle-like evidence.
Part c (1 mark)
Identifies the accumulated interference distribution as wave-like evidence.
Part d (1 mark)
Proposes a relevant improvement to angular or position measurement.
Part d (2 marks)
Explains quantitatively or causally how the improvement reduces random or systematic uncertainty in wavelength.
Question 21
Indicative answer: A matching incident photon stimulates an excited atom to emit a second photon with the same energy, phase and direction. Repetition in a population inversion amplifies one frequency into coherent, monochromatic light. Use an enclosed beam path or wavelength-rated eye protection and never view the beam directly.
Detailed marking criteria
Integrated response (1 mark)
Explains that stimulated emission produces a photon identical in energy, phase and direction.
Integrated response (1 mark)
Links repeated stimulated emission in a population inversion to coherent monochromatic amplification.
Integrated response (1 mark)
States a relevant control for the concentrated laser beam.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Stage 2 science as a human endeavour | Q19 | 4 | ___ | Review collaboration, model development, application impacts, limitations and public debate. |
| Stage 2 science inquiry skills | Q11, Q13 | 16 | ___ | Review experimental design, graphing, uncertainty, evaluation and evidence-based conclusions. |
| Topic 1.1 - Projectile motion | Q1 | 8 | ___ | Review component motion, constant acceleration, projectile range and drag. |
| Topic 1.2 - Forces and momentum | Q7, Q17 | 7 | ___ | Review impulse, two-dimensional momentum conservation and propulsion. |
| Topic 1.3 - Circular motion and gravitation | Q2 | 7 | ___ | Review centripetal motion, gravitational fields, Kepler's laws and satellite orbits. |
| Topic 1.4 - Relativity | Q12 | 7 | ___ | Review inertial frames, the two postulates, Lorentz factor, dilation, contraction and relativistic momentum. |
| Topic 2.1 - Electric fields | Q10 | 5 | ___ | Review Coulomb's law, superposition, electric fields, conductors and shielding. |
| Topic 2.2 - Motion of charged particles in electric fields | Q3 | 6 | ___ | Review potential difference, electric-field work and accelerated charged-particle motion. |
| Topic 2.3 - Magnetic fields | Q4 | 3 | ___ | Review magnetic-field direction, straight-conductor fields and superposition. |
| Topic 2.4 - Motion of charged particles in magnetic fields | Q18 | 5 | ___ | Review magnetic force, circular charged-particle motion and cyclotrons. |
| Topic 2.5 - Electromagnetic induction | Q5, Q15 | 8 | ___ | Review flux, Faraday and Lenz laws, generators, transformers and induction applications. |
| Topic 3.1 - Wave behaviour of light | Q6 | 6 | ___ | Review electromagnetic waves, polarisation, interference, double slits and diffraction gratings. |
| Topic 3.2 - Wave-particle duality | Q8, Q16, Q20 | 23 | ___ | Review photons, photoelectric evidence, X-rays and de Broglie waves. |
| Topic 3.3 - Structure of the atom | Q9, Q21 | 10 | ___ | Review spectra, discrete energy levels, fluorescence and laser operation. |
| Topic 3.4 - Standard Model | Q14 | 5 | ___ | Review fundamental particles, gauge bosons, quark composition and conservation laws. |