Skill Align TASC Physics Level 4 - Free Online Pack 0
An independently authored four-section PHY415115 practice examination themed around Tasmanian coastal monitoring, hydro-electric infrastructure and medical technology.
- Paper
- Written practice examination
- Reading
- 15 minutes preparation
- Writing
- 3 hours
- Assessment
- 180 marks
The current TASC PHY415115 Physics Information Sheet is approved for use throughout the examination and is linked as a separate official resource.
Section A - Newtonian Physics
5 compulsory questions, 45 marks. Suggested working time: 45 minutes. This section assesses Criterion 5.
Question 1
6 marksA coastal survey crew releases equipment from a drone above a level beach while maintaining constant horizontal velocity. The sensor package is released at 14.0 m s^-1 horizontally, 28.0 m above its landing level. Ignore air resistance.
Question 2
8 marksDuring a Hobart training session, a wicketkeeper redirects a moving practice ball through a measured contact interval. A 0.180 kg practice ball initially moves east at 6.00 m s^-1. For 0.120 s it experiences constant force components -3.00 N east and 8.00 N north.
Question 3
9 marksAn electric maintenance cart travels along a straight service tunnel while its traction system works against rolling resistance. A 420 kg maintenance cart travels at 4.00 m s^-1 along a level path. A constant drive force of 1200 N acts for 18.0 m while total resistance is 300 N. The interval lasts 2.64 s.
Question 4
10 marksA proposed Earth-observation satellite is modelled in a circular orbit for repeated imaging of Tasmania. The survey satellite follows a circular orbit of radius 7.10 × 10^6 m around Earth, whose mass is 5.97 × 10^24 kg. Treat Earth as spherical.
Question 5
12 marksA winch pulls a turbine component at constant applied force up an inclined hydro-electric service rail. A 180 kg turbine component starts from rest and is pulled 25.0 m up a 12.0 degree incline by a constant 900 N force parallel to the slope. Friction is 120 N and the motion lasts 4.67 s.
Section B - Electromagnetism
5 compulsory questions, 45 marks. Suggested working time: 45 minutes. This section assesses Criterion 6.
Question 6
6 marksTwo positively charged fog droplets are held at fixed positions inside an atmospheric sampling chamber. Two fixed positive point charges, +2.00 nC and +8.00 nC, are 0.300 m apart.
Question 7
8 marksA proton beam enters horizontally midway between parallel deflection plates in a compact medical-beam demonstrator. A positive proton of charge 1.60 × 10^-19 C and mass 1.67 × 10^-27 kg enters horizontally at 3.00 × 10^5 m s^-1 between plates 0.0400 m apart and 0.0500 m long. The potential difference is 80.0 V; ignore gravity.
Question 8
9 marksA velocity-selected proton enters a uniform magnetic analyser field perpendicular to its motion. The proton, of mass 1.67 × 10^-27 kg and charge +1e, travels right at 1.80 × 10^6 m s^-1 into a uniform 0.220 T magnetic field directed into the page.
Question 9
10 marksTwo long parallel substation busbars carry steady currents in the same direction and are mechanically supported over a known length. Two long parallel conductors are 0.180 m apart and carry currents 35.0 A and 55.0 A in the same direction. Consider an active parallel length of 1.20 m.
Question 10
12 marksA conducting rail on a harbour wave monitor slides through a uniform field while completing a resistive circuit. A 0.450 m conducting rail moves right at 4.00 m s^-1 through a uniform 0.600 T field into the page. The rail-and-load circuit resistance is 1.80 ohm. The graph shows the ideal emf-speed relationship.
Section C - Wave Motion
5 compulsory questions, 45 marks. Suggested working time: 45 minutes. This section assesses Criterion 7.
Question 11
6 marksA sonar pulse crosses from one uniform seawater layer into a slower layer at an oblique boundary. A wave of frequency 500 Hz travels from upper seawater, where its speed is 1500 m s^-1, into lower seawater, where its speed is 1100 m s^-1. Its incident angle is 35.0 degrees to the normal.
Question 12
8 marksA taut monitoring cable fixed at both ends is driven in a stable standing-wave pattern. The monitoring cable has vibrating length 0.750 m, tension 120 N and linear density 4.00 × 10^-3 kg m^-1. It is fixed at both ends and vibrates in harmonic 3.
Question 13
9 marksTwo coherent loudspeakers form an interference pattern across a straight listening line in a test hall. Two in-phase coherent sources, identified as loudspeaker pair, have wavelength 0.340 m and separation 1.20 m. They lie at transverse coordinates +d/2 and -d/2. A parallel observation line is 8.00 m from the source line; y is measured from the central axis. Use the small-angle approximation only for the central band width.
Question 14
10 marksA laboratory adjustable oscillator is compared with a reference oscillator before a school acoustics investigation. The reference oscillator has a calibrated frequency of 440 Hz. When sounded with the adjustable oscillator, 12 beats are counted in 4.00 s. Increasing the adjustable oscillator frequency slightly makes the beats slower.
Question 15
12 marksStudents vary tension in the same uniform string and determine wave speed from repeated frequency and wavelength measurements. The data for this laboratory string are 20.0 N: 78.3 m s^-1, 40.0 N: 112 m s^-1, 60.0 N: 136 m s^-1, 80.0 N: 160 m s^-1. The accepted linear density is 3.20 × 10^-3 kg m^-1.
Section D - Light, Atomic and Nuclear Physics
5 compulsory questions, 45 marks. Suggested working time: 45 minutes. This section assesses Criterion 8.
Question 16
6 marksA solar-observation exercise links a thermal spectrum to electron emission from a clean photocathode. A thermal source has temperature 5800 K. Light of frequency 7.00 × 10^14 Hz from the system illuminates caesium photocathode, whose work function is 2.10 eV.
Question 17
8 marksIn a hospital X-ray tube, electrons are accelerated through 70.0 kV before striking a tungsten target.
Question 18
9 marksA simplified three-level model vapour atom with electron energies -10.0 eV, -5.00 eV and -2.00 eV produces discrete absorption and emission lines in a spectroscope.
Question 19
10 marksA short-lived medical tracer is monitored beside a detector with a stable background count rate. A detector records 250 counts s^-1 from diagnostic tracer plus a steady background of 10.0 counts s^-1. The half-life is 6.00 h and the sample is observed again after 15.0 h.
Question 20
12 marksA research team models the energy balance of light-nuclei fusion and compares it with continuous plant output. In this light-nuclei fusion process, light nuclei (mass number below 20) combine into a more tightly bound medium-light product. Total product mass is 0.215 u less than total reactant mass, and the system releases thermal power 120 MW.
Worked Solutions And Marking Guide
General marking principles
- Apply each question-specific criterion to the evidence shown and accept equivalent scientifically correct reasoning.
- Carry forward an earlier numerical error when later physics is internally consistent, unless a later criterion independently assesses the same step.
- Award numerical marks against the mapped external criterion and the evidence supplied for that question.
- Do not penalise a missing unit more than once in a linked calculation unless units are separately assessed.
Section A Question 1
(a) Vertical motion starts with zero vertical velocity, so 28.0 = 0.5(9.81)t^2 and t = 2.39 s.
(b) The range is vt = 33.4 m. The downward component is gt = 23.4 m s^-1.
(c) Combining perpendicular components gives 27.3 m s^-1 at 59.1 degrees below the horizontal.
Detailed marking criteria
Part a (1 mark)
Uses the vertical displacement relationship with zero initial vertical velocity.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part a (1 mark)
Calculates the time of flight as 2.39 s.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Calculates the horizontal range as 33.4 m.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Calculates the downward impact component as 23.4 m s^-1.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Combines the components to obtain an impact speed of 27.3 m s^-1.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Determines the direction as 59.1 degrees below horizontal.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Section A Question 2
(a) J_x = F_x delta t = -0.360 N s and J_y = F_y delta t = 0.960 N s.
(b) Initial momentum is 1.08 kg m s^-1 east. Final components are (0.720, 0.960) kg m s^-1, giving speed 6.67 m s^-1.
(c) The direction is tan^-1(p_y/p_x) = 53.1 degrees north of east. Impulse changes the momentum vector; scalar speed cannot retain the independent east and north changes.
Detailed marking criteria
Part a (1 mark)
Calculates the eastward impulse as -0.360 N s.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part a (1 mark)
Calculates the northward impulse as 0.960 N s.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Adds the impulse vector to the initial momentum vector.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (2 marks)
Obtains the final components and speed 6.67 m s^-1.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Calculates a direction of 53.1 degrees north of east.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (2 marks)
Explains that impulse and momentum are vectors whose perpendicular components must be combined.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Section A Question 3
(a) F_net = 900 N and W_net = F_net s = 1.62 × 10^4 J.
(b) K_i = 3360 J and K_f = K_i + W_net = 1.96 × 10^4 J, so v_f = 9.65 m s^-1. Independently, a = F_net/m = 2.14 m s^-2 and u + at gives the same final speed apart from rounding.
(c) Drive work is 2.16 × 10^4 J, so average drive power is 8191 W. The kinetic-energy fraction is W_net/W_drive = 75.0%.
Detailed marking criteria
Part a (1 mark)
Determines the net force as 900 N.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part a (2 marks)
Calculates net work of 1.62 × 10^4 J with its sign and unit.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Applies the work-energy theorem to the net work.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (2 marks)
Calculates the final speed as 9.65 m s^-1.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Calculates average drive power as 8191 W.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (2 marks)
Calculates and interprets the kinetic-energy fraction as 75.0%.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Section A Question 4
(a) g = GM/r^2 = 7.90 N kg^-1.
(b) Equating gravity and centripetal acceleration gives v = sqrt(GM/r) = 7489 m s^-1.
(c) T = 2pir/v = 5957 s. Kepler's third law gives T2/T1 = (1.30)^(3/2) = 1.48.
Detailed marking criteria
Part a (1 mark)
Uses the distance from the central body's centre in g = GM/r^2.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part a (2 marks)
Calculates the field strength as 7.90 N kg^-1.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Equates gravitational acceleration with centripetal acceleration.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (2 marks)
Calculates the orbital speed as 7489 m s^-1.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (2 marks)
Calculates the first orbital period as 5957 s.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (2 marks)
Applies Kepler's third law to obtain a period ratio of 1.48.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Section A Question 5
(a) The downhill weight component is mg sin(theta) = 367 N. Net uphill force is 413 N, giving a = 2.29 m s^-2.
(b) v = sqrt(2as) = 10.7 m s^-1. Input work 2.25 × 10^4 J equals kinetic gain 1.03 × 10^4 J plus potential gain 9178 J and friction loss 3000 J, apart from rounding.
(c) Average applied power is 4819 W and the kinetic-energy percentage is 45.9%. The remaining input raises gravitational potential energy and is dissipated thermally by friction.
Detailed marking criteria
Part a (1 mark)
Calculates the downhill weight component as 367 N.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part a (2 marks)
Combines all parallel forces and calculates acceleration 2.29 m s^-2.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (2 marks)
Calculates the final speed as 10.7 m s^-1.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (2 marks)
Accounts quantitatively for kinetic, gravitational-potential and frictional energy changes.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (2 marks)
Calculates average applied power as 4819 W.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Calculates the kinetic-energy percentage as 45.9%.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (2 marks)
Explains the gravitational-potential and frictional destinations of the remaining energy.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Section B Question 6
(a) Coulomb's law gives F = 1.60 × 10^-6 N. The like charges repel, so the forces are equal in magnitude and opposite along their joining line.
(b) The fields oppose at the midpoint. Their difference is 2400 N C^-1, directed from the larger charge towards the smaller charge.
(c) Equating kq1/x^2 and kq2/(d-x)^2 gives x = 0.100 m from the +2.00 nC charge.
Detailed marking criteria
Part a (1 mark)
Calculates the Coulomb-force magnitude as 1.60 × 10^-6 N.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part a (1 mark)
States that the equal and opposite forces are repulsive.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Calculates the midpoint field magnitude as 2400 N C^-1.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Gives the net direction from the larger positive charge towards the smaller one.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Equates the two inverse-square field magnitudes in the region between the charges.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Locates the point 0.100 m from the smaller charge.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Section B Question 7
(a) E = V/d = 2000 V m^-1 and F = qE = 3.20 × 10^-16 N towards the negative plate.
(b) a = F/m = 1.92 × 10^11 m s^-2 and t = L/v_x = 1.67 × 10^-7 s.
(c) y = 0.5at^2 = 2.66 × 10^-3 m. The vertical exit component is 3.19 × 10^4 m s^-1, so the direction is 6.08 degrees towards the negative plate from the horizontal.
Detailed marking criteria
Part a (1 mark)
Calculates the uniform field as 2000 V m^-1.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part a (1 mark)
Calculates force 3.20 × 10^-16 N towards the negative plate.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Calculates acceleration 1.92 × 10^11 m s^-2.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (2 marks)
Uses the unchanged horizontal speed to calculate transit time 1.67 × 10^-7 s.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Calculates deflection 2.66 × 10^-3 m.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (2 marks)
Calculates the exit component and direction 6.08 degrees from horizontal.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Section B Question 8
(a) F = qvB = 6.34 × 10^-14 N. For a positive charge moving right with B into the page, the force is upward.
(b) r = mv/(qB) = 0.0854 m. Magnetic force remains perpendicular to velocity, so it changes direction but does no work and does not change speed.
(c) T = 2pim/(qB) = 2.98 × 10^-7 s. Doubling speed doubles radius, while the non-relativistic period is unchanged.
Detailed marking criteria
Part a (1 mark)
Converts the stated charge multiple to coulombs.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part a (1 mark)
Calculates magnetic force 6.34 × 10^-14 N.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part a (1 mark)
Uses the positive-charge rule to give an upward force.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (2 marks)
Calculates the circular-path radius as 0.0854 m.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Explains that perpendicular magnetic force changes direction without doing work.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Calculates period 2.98 × 10^-7 s.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (2 marks)
Correctly predicts doubled radius and unchanged period when speed doubles.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Section B Question 9
(a) The fields are 7.78 × 10^-5 T and 1.22 × 10^-4 T. They oppose between same-direction currents, giving resultant magnitude 4.44 × 10^-5 T in the direction of the stronger field.
(b) F/l = k_BI1I2/r = 2.14 × 10^-3 N m^-1, so F = 2.57 × 10^-3 N. Same-direction currents attract.
(c) Each wire's field acts on the current in the other, producing equal-magnitude opposite forces on different conductors. Reversing one current makes the forces repulsive and makes the two midpoint field contributions reinforce rather than oppose.
Detailed marking criteria
Part a (1 mark)
Uses B = k_BI/r with the midpoint distance.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part a (2 marks)
Calculates both contributions and resultant 4.44 × 10^-5 T.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Applies the parallel-current force relationship.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (2 marks)
Calculates force 2.57 × 10^-3 N and identifies attraction.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (2 marks)
Identifies equal and opposite forces acting on different conductors.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (2 marks)
Predicts repulsion and reinforcement of the midpoint fields after one current reverses.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Section B Question 10
(a) emf = vlB = 1.08 V and I = emf/R = 0.600 A.
(b) The gradient is lB = 0.270 V s m^-1, so B = gradient/l = 0.600 T. At zero speed there is no conductor motion across field lines and hence zero motional emf.
(c) The induced current is upward in the moving conductor, making the magnetic force leftward and opposing the imposed motion. F = IlB = 0.162 N. Mechanical power Fv = 0.648 W equals I^2R = 0.648 W.
Detailed marking criteria
Part a (1 mark)
Applies the perpendicular motional-emf relationship.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part a (2 marks)
Calculates emf 1.08 V and current 0.600 A.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (2 marks)
Determines the graph gradient as 0.270 V s m^-1.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Uses the conductor length to recover B = 0.600 T.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Explains the physically required zero intercept.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Uses Lenz's law and the force rule to obtain a force opposite the motion.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (2 marks)
Calculates the opposing force as 0.162 N.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (2 marks)
Shows mechanical and resistive powers both equal 0.648 W.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Section C Question 11
(a) lambda_1 = v_1/f = 3.00 m and lambda_2 = v_2/f = 2.20 m.
(b) sin(theta_2) = (v_2/v_1)sin(theta_1), giving theta_2 = 24.9 degrees.
(c) The source fixes the frequency, so frequency and period remain unchanged. Wave speed and wavelength decrease in the slower second medium, and the ray bends towards the normal.
Detailed marking criteria
Part a (1 mark)
Calculates the first wavelength as 3.00 m.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part a (1 mark)
Calculates the second wavelength as 2.20 m.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Applies Snell's law using angles measured from the normal.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Calculates the refracted angle as 24.9 degrees.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
States that frequency and period remain unchanged across the boundary.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Connects reduced speed with reduced wavelength and bending towards the normal.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Section C Question 12
(a) v = sqrt(T/mu) = 173 m s^-1.
(b) f_1 = v/(2L) = 115 Hz and f_3 = 3f_1 = 346 Hz.
(c) lambda = 2L/3 = 0.500 m. There are 4 nodes including both ends and 3 antinodes, one in each loop.
Detailed marking criteria
Part a (1 mark)
Uses the tension and linear-density relationship.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part a (1 mark)
Calculates wave speed 173 m s^-1.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Uses the fixed-string fundamental relationship.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (2 marks)
Calculates f1 = 115 Hz and the harmonic frequency 346 Hz.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Calculates wavelength 0.500 m.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (2 marks)
Correctly identifies 4 nodes and 3 antinodes.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Section C Question 13
(a) W = lambda x/d = 2.27 m.
(b) The distances are sqrt(x^2 + (y-d/2)^2) = 8.91319 m and sqrt(x^2 + (y+d/2)^2) = 9.50352 m. Their difference is 0.5903 m = 1.736 wavelengths. This is not an integer number of wavelengths, so it is not a constructive maximum. The off-axis small-angle estimate is inadequate here.
(c) Both changes increase W because band width is proportional to wavelength and inversely proportional to source separation. The interference features therefore spread farther apart.
Detailed marking criteria
Part a (1 mark)
Uses the prescribed two-source band-width relationship.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part a (1 mark)
Uses consistent SI units.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part a (1 mark)
Calculates band width 2.27 m.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Calculates path difference 0.590 m.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (2 marks)
Compares the exact difference of 1.736 wavelengths with 2 and judges the maximum claim with appropriate precision.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Predicts wider spacing for increased wavelength.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Predicts wider spacing for reduced source separation.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Justifies both predictions using the proportional relationship.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Section C Question 14
(a) f_B = 12/4.00 = 3.00 Hz. Alternating constructive and destructive superposition makes the resultant amplitude, and hence loudness, rise and fall.
(b) Because |f_2 - f_1| = f_B, the possibilities are 437 Hz and 443 Hz.
(c) The adjustment raises the adjustable oscillator's frequency towards the reference frequency, so it was initially below the reference. The unknown frequency was 437 Hz. At exact tuning the beat frequency becomes zero and a steady resultant tone replaces the periodic loudness variation.
Detailed marking criteria
Part a (1 mark)
Calculates beat frequency 3.00 Hz.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part a (2 marks)
Explains loudness variation through alternating constructive and destructive superposition.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Uses the absolute frequency-difference relationship.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (2 marks)
Obtains both 437 Hz and 443 Hz.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (2 marks)
The adjustment raises the adjustable oscillator's frequency towards the reference frequency, so it was initially below the reference.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Identifies the unknown frequency as 437 Hz.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Explains that exact tuning removes beats and produces steady amplitude.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Section C Question 15
(a) Since v = sqrt(T/mu), v^2 = (1/mu)T. Squaring the printed speeds gives the plotted points. A least-squares fit constrained through the model origin gives 315 kg^-1 m; graphical estimates from 299 to 331 kg^-1 m are reasonable.
(b) mu = 1/gradient = 3.17 × 10^-3 kg m^-1. This is close to 3.20 × 10^-3 kg m^-1; the difference is consistent with the small measurement scatter.
(c) The model gives v proportional to sqrt(T), so doubling tension multiplies speed by sqrt(2), not 2. The measured speed ratios are close to sqrt(2); the scatter does not establish exact agreement. At 160 N the accepted-density model predicts v = sqrt(T/mu) = 224 m s^-1, assuming the string retains the same linear density.
Detailed marking criteria
Part a (2 marks)
Linearises v = sqrt(T/mu) to v squared = T/mu.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part a (2 marks)
Draws or uses a reasonable best-fit line and determines a gradient from 299 to 331 kg^-1 m.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (2 marks)
Recognises that linear density is the reciprocal of the gradient.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Calculates an inferred density near 3.17 × 10^-3 kg m^-1.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Makes a justified comparison with the accepted value.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Rejects the linear-speed claim using the square-root dependence on tension.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Uses a pair of data values or their squared speeds to support the model within scatter.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Predicts speed 224 m s^-1 at twice the largest tension.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
States that the prediction assumes unchanged linear density.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Section D Question 16
(a) Wien's law gives lambda_peak = 2.90 × 10^-3/T = 5.00 × 10^-7 m.
(b) Photon energy is 2.90 eV, so K_max = 0.801 eV = 1.28 × 10^-19 J and the stopping potential is 0.801 V.
(c) Greater intensity supplies more photons per second, increasing the emission rate and photocurrent. Photon energy hf is unchanged, so maximum kinetic energy and stopping potential do not change.
Detailed marking criteria
Part a (1 mark)
Applies Wien's displacement law with absolute temperature.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part a (1 mark)
Calculates peak wavelength 5.00 × 10^-7 m.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Subtracts the work function from the photon energy.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Obtains K_max 0.801 eV and stopping potential 0.801 V.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Predicts increased photocurrent from the increased photon arrival rate.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Explains why maximum kinetic energy and stopping potential remain unchanged.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Section D Question 17
(a) The maximum photon energy is eV = 1.12 × 10^-14 J. Therefore f_max = 1.69 × 10^19 Hz and lambda_min = 1.78 × 10^-11 m.
(b) p = h/lambda = E/c = 3.73 × 10^-23 kg m s^-1.
(c) The continuous spectrum comes from electron deceleration through a range of energy losses. Characteristic lines come from target-atom inner-shell vacancies and fixed energy-level differences. Increasing tube current raises photon rate and intensity but does not change the cutoff frequency set by accelerating voltage.
Detailed marking criteria
Part a (1 mark)
Equates maximum photon energy with the accelerated electron energy eV.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part a (2 marks)
Calculates f_max 1.69 × 10^19 Hz and lambda_min 1.78 × 10^-11 m.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Uses a valid photon-momentum relationship.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Calculates photon momentum 3.73 × 10^-23 kg m s^-1.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Explains the continuous spectrum through electron deceleration.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Explains characteristic lines through target-atom energy-level differences.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Predicts increased intensity but unchanged cutoff frequency when tube current increases.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Section D Question 18
(a) Delta E = 8.00 eV, so f = Delta E/h = 1.93 × 10^15 Hz.
(b) The smallest energy step is 3.00 eV from the upper to middle level. It gives f = 7.24 × 10^14 Hz and lambda = 4.14 × 10^-7 m.
(c) Only discrete electron energies are allowed, so photons can carry only differences between levels. Absorption raises an electron when photon energy matches an allowed difference; emission occurs when an electron falls and releases that difference as a photon.
Detailed marking criteria
Part a (1 mark)
Determines excitation energy 8.00 eV.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part a (2 marks)
Converts energy and calculates frequency 1.93 × 10^15 Hz.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Identifies the smallest allowed energy difference.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (2 marks)
Calculates the corresponding wavelength 4.14 × 10^-7 m.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Links discrete spectral lines with quantised electron energies.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Correctly describes resonant photon absorption.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Correctly describes photon emission during a downward transition.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Section D Question 19
(a) A0 = 240 counts s^-1. The corrected later rate is 42.4 counts s^-1, so the detector reads 52.4 counts s^-1 after background is restored.
(b) lambda = 0.693/T1/2 = 0.115 h^-1 and N/N0 = 0.177.
(c) Only the source contribution halves. The total approaches the non-zero background as the source contribution decays. Decay is random, so predicted rates are statistical expectations; individual finite counts fluctuate around them.
Detailed marking criteria
Part a (1 mark)
Subtracts background to obtain 240 counts s^-1.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part a (2 marks)
Calculates the later detector reading as 52.4 counts s^-1.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Calculates decay constant 0.115 h^-1.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (2 marks)
Calculates remaining fraction 0.177.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Distinguishes halving of the source rate from the source-plus-background total.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Predicts the background rate as the long-time limit.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Identifies spontaneous random decay of individual nuclei.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Distinguishes an expected count from an individual fluctuating observation.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Section D Question 20
(a) Delta E = 0.215 × 931 = 200 MeV = 3.20 × 10^-11 J.
(b) Rate = P/Delta E = 1.20 × 10^8/3.20 × 10^-11 = 3.75 × 10^18 reactions s^-1.
(c) The fusion products lie to the right of the light reactants and closer to the iron-region maximum, so their binding energy per nucleon is greater. Their lower total mass corresponds to increased binding energy released by the reaction. The short-range strong force binds neighbouring nucleons.
(d) A proton is uud and a neutron is udd. The strong nuclear force is significant only over nuclear distances, so it is negligible for well-separated nuclei.
Detailed marking criteria
Part a (1 mark)
Uses the mass-to-energy conversion 931 MeV per atomic mass unit.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part a (1 mark)
Calculates 200 MeV per reaction.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part a (2 marks)
Converts the result to 3.20 × 10^-11 J.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (1 mark)
Converts megawatts to watts and relates power to energy per reaction.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part b (2 marks)
Calculates reaction rate 3.75 × 10^18 s^-1.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Uses the supplied curve to place light reactants left of their more tightly bound fusion products.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
Relates greater binding to lower product mass and released energy.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part c (1 mark)
States the short-range nucleon-binding role of the strong nuclear force.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part d (1 mark)
Gives proton uud and neutron udd compositions.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Part d (1 mark)
Uses the short range of the strong nuclear force to explain negligible attraction at large separation.
Acceptable alternatives: Accept equivalent scientifically accurate reasoning or working applied to the supplied context.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Criterion 5 - Newtonian mechanics and gravitational fields | Q1, Q2, Q3, Q4, Q5 | 45 | ___ | Review vectors, motion, momentum, Newton's laws, energy, circular motion and gravitation. |
| Criterion 6 - Electricity and magnetism | Q6, Q7, Q8, Q9, Q10 | 45 | ___ | Review electric and magnetic fields, charged-particle motion, currents and motional induction. |
| Criterion 7 - General principles of wave motion | Q11, Q12, Q13, Q14, Q15 | 45 | ___ | Review wave quantities, refraction, superposition, resonance, standing waves, interference and diffraction. |
| Criterion 8 - Light, atomic and nuclear physics | Q16, Q17, Q18, Q19, Q20 | 45 | ___ | Review photons, spectra, wave-particle duality, radioactivity, mass defect and nuclear processes. |