Skill Align QCE Physics - Free Online Pack 0 Paper 2
An independently authored QCE Physics practice paper with worked answers and question-specific marking guidance.
- Paper
- Paper 2 Question and Response Book
- Reading
- 5 minutes perusal
- Writing
- 90 minutes
- Assessment
- 50 marks
The current QCAA Physics formula and data book is supplied separately. A QCAA-approved scientific or graphics calculator may be used.
Short-response questions
Questions 1-9 are short response. Answer every part, show relevant working and justify interpretations using the supplied evidence.
Question 1
2 marksA survey boat moves at 4.00 m s^-1 east relative to the water. The river current is 3.00 m s^-1 north.
Question 2
5 marksAn ideal transformer supplies a 36.0 W field instrument. Its 1500-turn primary is connected to 240 V AC and its secondary has 75 turns.
Question 3
5 marksA maglev carriage has proper length 120 m and passes an inertial track observer at 0.750c.
Question 4
8 marksStar A has a black-body peak at 480 nm and Star B has a peak at 620 nm. Both spectra are broad and contain narrow absorption lines. Use b = 2.898 x 10^-3 m K.
Question 5
5 marksA high-energy electron and positron collide and form a muon-antimuon pair: e- + e+ -> mu- + mu+.
Question 6
11 marksA mapping satellite follows a circular orbit 500 km above Mars. Mars has radius 3.39 x 10^6 m and GM_M = 4.283 x 10^13 m^3 s^-2.
Question 7
5 marksA 400-turn coil of area 0.0150 m^2 is perpendicular to a uniform magnetic field. The field increases uniformly from 0.100 T to 0.580 T in 0.240 s.
Question 8
4 marksA probe clock passes two beacons at rest in one inertial laboratory. The laboratory measures 10.0 s between the passage events, while the probe moves at 0.800c.
Question 9
5 marksA hydrogen atom emits a photon when its electron changes from n = 3 to n = 2. Use E_n = -13.6 eV/n^2 and 1 eV = 1.602 x 10^-19 J.
Worked Solutions And Marking Guide
General marking principles
- Award each listed mark independently when the required physical relationship, working step or evidence statement is demonstrated.
- Accept physically correct equivalent wording and logically equivalent calculations with appropriate units.
- Carry forward a candidate's earlier numerical value when later working is physically consistent, unless the resulting answer is impossible.
Question 1
- The perpendicular components give speed sqrt(4.00^2 + 3.00^2) = 5.00 m s^-1.
- The direction is tan^-1(3.00/4.00) = 36.9 degrees north of east.
Detailed marking criteria
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: The perpendicular components give speed sqrt(4.00^2 + 3.00^2) = 5.00 m s^-1.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: The direction is tan^-1(3.00/4.00) = 36.9 degrees north of east.
Question 2
- The turns relationship is V_s/V_p = N_s/N_p = 75/1500.
- V_s = 240 × 75/1500 = 12.0 V.
- The secondary current is I_s = P/V_s = 36.0/12.0 = 3.00 A.
- The primary current is I_p = P/V_p = 36.0/240 = 0.150 A.
- Ideal power conservation requires V_pI_p = V_sI_s, so the lower-voltage winding carries the proportionally larger current.
Detailed marking criteria
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: The turns relationship is V_s/V_p = N_s/N_p = 75/1500.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: V_s = 240 × 75/1500 = 12.0 V.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: The secondary current is I_s = P/V_s = 36.0/12.0 = 3.00 A.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: The primary current is I_p = P/V_p = 36.0/240 = 0.150 A.
Part c (1 mark)
Awards one mark for accurately establishing this question-specific point: Ideal power conservation requires V_pI_p = V_sI_s, so the lower-voltage winding carries the proportionally larger current.
Question 3
- The proper length 120 m is measured in the carriage rest frame.
- At 0.750c, gamma = 1/sqrt(1 - 0.750^2) = 1.512.
- The track-frame length is L = L_0/gamma = 120/1.512 = 79.4 m.
- The track observer records the positions of the front and rear at the same track-frame time.
- Simultaneous endpoint measurements are necessary because the carriage moves between non-simultaneous observations.
Detailed marking criteria
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: The proper length 120 m is measured in the carriage rest frame.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: At 0.750c, gamma = 1/sqrt(1 - 0.750^2) = 1.512.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: The track-frame length is L = L_0/gamma = 120/1.512 = 79.4 m.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: The track observer records the positions of the front and rear at the same track-frame time.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Simultaneous endpoint measurements are necessary because the carriage moves between non-simultaneous observations.
Question 4
- For Star A, T_A = b/lambda_max = 2.898 × 10^-3/(480 × 10^-9) = 6.04 × 10^3 K.
- For Star B, T_B = 2.898 × 10^-3/(620 × 10^-9) = 4.67 × 10^3 K.
- The shorter-wavelength peak therefore identifies Star A as hotter.
- The broad continuum approximates thermal radiation from a dense photosphere.
- The peak position provides a temperature estimate through Wien's displacement law.
- Narrow dark lines arise when cooler outer gas absorbs photons at element-specific transition energies.
- Real stellar atmospheres alter the spectrum through absorption, scattering and wavelength-dependent opacity.
- Consequently the Wien temperature is a model-based effective surface estimate rather than a complete description of every atmospheric layer.
Detailed marking criteria
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: For Star A, T_A = b/lambda_max = 2.898 × 10^-3/(480 × 10^-9) = 6.04 × 10^3 K.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: For Star B, T_B = 2.898 × 10^-3/(620 × 10^-9) = 4.67 × 10^3 K.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: The shorter-wavelength peak therefore identifies Star A as hotter.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: The broad continuum approximates thermal radiation from a dense photosphere.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: The peak position provides a temperature estimate through Wien's displacement law.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Narrow dark lines arise when cooler outer gas absorbs photons at element-specific transition energies.
Part c (1 mark)
Awards one mark for accurately establishing this question-specific point: Real stellar atmospheres alter the spectrum through absorption, scattering and wavelength-dependent opacity.
Part c (1 mark)
Awards one mark for accurately establishing this question-specific point: Consequently the Wien temperature is a model-based effective surface estimate rather than a complete description of every atmospheric layer.
Question 5
- The electron and muon are negatively charged leptons.
- The positron and antimuon are their positively charged antiparticles.
- Total electric charge is zero before and after the interaction.
- Total lepton number is zero before and after because each lepton is paired with an antilepton.
- Baryon number remains zero because no baryons participate.
Detailed marking criteria
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: The electron and muon are negatively charged leptons.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: The positron and antimuon are their positively charged antiparticles.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Total electric charge is zero before and after the interaction.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Total lepton number is zero before and after because each lepton is paired with an antilepton.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Baryon number remains zero because no baryons participate.
Question 6
- The orbital radius is r = 3.39 × 10^6 + 0.500 × 10^6 = 3.89 × 10^6 m.
- The field strength is g = GM_M/r^2.
- Substitution gives g = 4.283 × 10^13/(3.89 × 10^6)^2 = 2.83 m s^-2.
- Equating gravitational and centripetal acceleration gives v = sqrt(GM_M/r).
- The speed is v = sqrt(4.283 × 10^13/3.89 × 10^6) = 3.32 × 10^3 m s^-1.
- The period relationship is T = 2pi r/v.
- T = 2pi(3.89 × 10^6)/(3.32 × 10^3) = 7.37 × 10^3 s, or about 123 min.
- Gravity is the only modelled force and points continuously towards the centre of Mars.
- The gravitational force supplies centripetal acceleration and changes velocity direction rather than speed.
- The satellite's kinetic and gravitational potential energies remain constant in the ideal circular orbit.
- Atmospheric drag, non-spherical gravity and external bodies are neglected, so a real low orbiter may require corrections.
Detailed marking criteria
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: The orbital radius is r = 3.39 × 10^6 + 0.500 × 10^6 = 3.89 × 10^6 m.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: The field strength is g = GM_M/r^2.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Substitution gives g = 4.283 × 10^13/(3.89 × 10^6)^2 = 2.83 m s^-2.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Equating gravitational and centripetal acceleration gives v = sqrt(GM_M/r).
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: The speed is v = sqrt(4.283 × 10^13/3.89 × 10^6) = 3.32 × 10^3 m s^-1.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: The period relationship is T = 2pi r/v.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: T = 2pi(3.89 × 10^6)/(3.32 × 10^3) = 7.37 × 10^3 s, or about 123 min.
Part c (1 mark)
Awards one mark for accurately establishing this question-specific point: Gravity is the only modelled force and points continuously towards the centre of Mars.
Part c (1 mark)
Awards one mark for accurately establishing this question-specific point: The gravitational force supplies centripetal acceleration and changes velocity direction rather than speed.
Part c (1 mark)
Awards one mark for accurately establishing this question-specific point: The satellite's kinetic and gravitational potential energies remain constant in the ideal circular orbit.
Part c (1 mark)
Awards one mark for accurately establishing this question-specific point: Atmospheric drag, non-spherical gravity and external bodies are neglected, so a real low orbiter may require corrections.
Question 7
- The flux change per turn is Delta Phi = A Delta B = (0.0150)(0.480) = 7.20 × 10^-3 Wb.
- Faraday's law gives |emf| = N Delta Phi/Delta t.
- The average emf is 400(7.20 × 10^-3)/0.240 = 12.0 V.
- The induced field opposes the increase in magnetic flux, as required by Lenz's law.
- Mechanical or electrical work done by the apparatus changing the field supplies the energy dissipated in the coil circuit.
Detailed marking criteria
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: The flux change per turn is Delta Phi = A Delta B = (0.0150)(0.480) = 7.20 × 10^-3 Wb.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Faraday's law gives |emf| = N Delta Phi/Delta t.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: The average emf is 400(7.20 × 10^-3)/0.240 = 12.0 V.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: The induced field opposes the increase in magnetic flux, as required by Lenz's law.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Mechanical or electrical work done by the apparatus changing the field supplies the energy dissipated in the coil circuit.
Question 8
- The probe clock is present at both passage events, so its reading is the proper time Delta t_0.
- At 0.800c, gamma = 1.667.
- Delta t_0 = Delta t/gamma = 10.0/1.667 = 6.00 s.
- Special relativity permits different elapsed times between separated events while every inertial frame still measures light in vacuum at c.
Detailed marking criteria
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: The probe clock is present at both passage events, so its reading is the proper time Delta t_0.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: At 0.800c, gamma = 1.667.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Delta t_0 = Delta t/gamma = 10.0/1.667 = 6.00 s.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Special relativity permits different elapsed times between separated events while every inertial frame still measures light in vacuum at c.
Question 9
- E_3 = -13.6/9 = -1.51 eV and E_2 = -13.6/4 = -3.40 eV.
- The emitted photon energy is |E_2 - E_3| = 1.89 eV.
- In joules this is (1.89)(1.602 × 10^-19) = 3.03 × 10^-19 J.
- Lambda = hc/E = (6.626 × 10^-34)(3.00 × 10^8)/(3.03 × 10^-19) = 6.56 × 10^-7 m, about 656 nm.
- The electron moves to a lower energy level, so the 656 nm photon is emitted.
Detailed marking criteria
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: E_3 = -13.6/9 = -1.51 eV and E_2 = -13.6/4 = -3.40 eV.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: The emitted photon energy is |E_2 - E_3| = 1.89 eV.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: In joules this is (1.89)(1.602 × 10^-19) = 3.03 × 10^-19 J.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Lambda = hc/E = (6.626 × 10^-34)(3.00 × 10^8)/(3.03 × 10^-19) = 6.56 × 10^-7 m, about 656 nm.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: The electron moves to a lower energy level, so the 656 nm photon is emitted.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Unit 3 Topic 1 - Gravity and motion | Q1, Q6 | 13 | ___ | Review vectors, projectiles, circular motion, gravitation, satellites and Kepler's laws. |
| Unit 3 Topic 2 - Electromagnetism | Q2, Q7 | 10 | ___ | Review electric and magnetic fields, forces, induction, generators, transformers and electromagnetic radiation. |
| Unit 4 Topic 1 - Special relativity | Q3, Q8 | 9 | ___ | Review frames, simultaneity, time dilation, length contraction, momentum and mass-energy equivalence. |
| Unit 4 Topic 2 - Quantum theory | Q4, Q9 | 13 | ___ | Review interference, black-body radiation, photons, atomic spectra and matter waves. |
| Unit 4 Topic 3 - The Standard Model | Q5 | 5 | ___ | Review particle types, quark structure, gauge bosons, conservation laws and interaction diagrams. |