OCR A-Level · PastPaper.sampleTitle

MetadataPastPaper.sampleTitle

Thinka Jun 2023 Cambridge OCR A Level-Style Mock — Physics B (Advancing Physics) - H557

100 PastPaper.marks135 PastPaper.minutes2023
PastPaper.analysisPastPaper.samplePaper
An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 Cambridge OCR A Level Physics B (Advancing Physics) - H557 paper. Not affiliated with or reproduced from Cambridge.

Section A

Answer all questions in this section. Questions focus on core concepts and basic applications.
3 PastPaper.question · 29.009999999999998 PastPaper.marks
PastPaper.question 1 · Short Structured and Calculation
9.67 PastPaper.marks
A semiconductor has a band gap of energy \(E_g = 1.1\text{ eV}\).

(a) Show that the Boltzmann factor \(e^{-E_g/kT}\) for this transition at a temperature \(T = 293\text{ K}\) is approximately \(1.2 \times 10^{-19}\).

(b) A temperature-sensitive circuit requires the Boltzmann factor to increase by exactly a factor of 10.0 to trigger a threshold alarm. Calculate the temperature, in kelvin, at which the Boltzmann factor is exactly 10.0 times its value at \(293\text{ K}\).

(Boltzmann constant \(k = 1.38 \times 10^{-23}\text{ J K}^{-1}\), elementary charge \(e = 1.60 \times 10^{-19}\text{ C}\))
PastPaper.showAnswers

PastPaper.workedSolution

(a) First, convert the band gap energy from eV to joules:
\(E_g = 1.1 \times 1.60 \times 10^{-19}\text{ J} = 1.76 \times 10^{-19}\text{ J}\).

At \(T = 293\text{ K}\), calculate the thermal energy term \(kT\):
\(kT = 1.38 \times 10^{-23}\text{ J K}^{-1} \times 293\text{ K} = 4.043 \times 10^{-21}\text{ J}\).

Calculate the exponent:
\(\frac{E_g}{kT} = \frac{1.76 \times 10^{-19}}{4.043 \times 10^{-21}} \approx 43.53\).

Calculate the Boltzmann factor:
\(e^{-43.53} \approx 1.25 \times 10^{-19}\), which is approximately \(1.2 \times 10^{-19}\).

(b) Let \(\text{BF}_1 = 1.248 \times 10^{-19}\) and the target factor \(\text{BF}_2 = 1.248 \times 10^{-18}\).

Set up the equation for the new temperature \(T_2\):
\(e^{-E_g / k T_2} = 1.248 \times 10^{-18}\)

Take the natural logarithm of both sides:
\(-\frac{E_g}{k T_2} = \ln(1.248 \times 10^{-18}) \approx -41.225\)

Solve for \(T_2\):
\(T_2 = \frac{E_g}{41.225 \times k} = \frac{1.76 \times 10^{-19}}{41.225 \times 1.38 \times 10^{-23}} \approx 309.37\text{ K}\).

Rounded to 3 significant figures, \(T_2 = 309\text{ K}\).

PastPaper.markingScheme

(a) [3 marks total]:
- Convert eV to J: \(1.76 \times 10^{-19}\text{ J}\) [1 mark].
- Calculate \(kT\) at \(293\text{ K}\): \(4.04 \times 10^{-21}\text{ J}\) [1 mark].
- Compute exponential factor to show value close to \(1.2 \times 10^{-19}\) [1 mark].

(b) [5 marks total]:
- Identify the new Boltzmann factor value \(= 1.25 \times 10^{-18}\) [1 mark].
- Set up logarithmic relationship: \(\ln(\text{BF}_2) = -E_g / k T_2\) [1 mark].
- Rearrange equation to make \(T_2\) the subject [1 mark].
- Substitute correct values including \(k\) and \(E_g\) [1 mark].
- Obtain final temperature in range \(309\text{ K}\) to \(310\text{ K}\) [1 mark].

[1.67 marks reserved for consistent unit usage and correct significant figures across both parts].
PastPaper.question 2 · Short Structured and Calculation
9.67 PastPaper.marks
Two coherent beams of light from a single laser source of wavelength \(\lambda = 600\text{ nm}\) travel along different paths and meet at a point on a screen.

Path 1 has a length of \(1.2003 \times 10^{-4}\text{ m}\) and Path 2 has a length of \(1.2018 \times 10^{-4}\text{ m}\).

(a) Calculate the path difference between the two waves.

(b) Determine the phase difference between the two phasor arrows representing these waves at the screen, expressing your answer in radians.

(c) If each wave individually has an amplitude of \(4.0\text{ V m}^{-1}\) at the screen, calculate the resultant amplitude of the combined waves using a phasor model.
PastPaper.showAnswers

PastPaper.workedSolution

(a) The path difference \(\Delta s\) is the absolute difference between the two path lengths:
\(\Delta s = 1.2018 \times 10^{-4}\text{ m} - 1.2003 \times 10^{-4}\text{ m} = 0.00015 \times 10^{-4}\text{ m} = 1.50 \times 10^{-7}\text{ m}\) (or \(150\text{ nm}\)).

(b) Express the path difference as a fraction of the wavelength:
\(\frac{\Delta s}{\lambda} = \frac{150\text{ nm}}{600\text{ nm}} = 0.25\).

The phase difference in radians is:
\(\Delta \phi = 2\pi \times 0.25 = \frac{\pi}{2}\text{ rad} \approx 1.57\text{ rad}\).

(c) Since the phase difference is \(\pi/2\text{ rad}\) (or \(90^\circ\)), the two phasor arrows of length \(4.0\text{ V m}^{-1}\) are perpendicular to each other.
Using Pythagoras' theorem, the magnitude of the resultant phasor \(A_R\) is:
\(A_R = \sqrt{(4.0)^2 + (4.0)^2} = \sqrt{16 + 16} = \sqrt{32} \approx 5.66\text{ V m}^{-1}\).

Rounding to 2 significant figures gives \(5.7\text{ V m}^{-1}\).

PastPaper.markingScheme

(a) [2 marks]:
- Calculate subtraction correctly: \(1.5 \times 10^{-7}\text{ m}\) [1 mark].
- Correct unit included [1 mark].

(b) [3 marks]:
- Relate path difference to wavelength: \(0.25\lambda\) [1 mark].
- Convert fraction of cycle to phase angle: \(\Delta \phi = 2\pi \times 0.25\) [1 mark].
- Final value \(1.57\text{ rad}\) or \(\pi/2\text{ rad}\) [1 mark].

(c) [3 marks]:
- Recognise phasors are perpendicular [1 mark].
- Apply Pythagoras' theorem correctly [1 mark].
- State final amplitude of \(5.7\text{ V m}^{-1}\) (allow \(5.66\text{ V m}^{-1}\)) [1 mark].

[0.67 marks reserved for consistent scientific notation and sig figs].
PastPaper.question 3 · Short Structured and Calculation
9.67 PastPaper.marks
A patient is treated with a radioactive dose containing iodine-131 (\(^{131}\text{I}\)), which has a half-life of 8.0 days.

(a) Calculate the decay constant \(\lambda\) of \(^{131}\text{I}\) in \(\text{s}^{-1}\).

(b) The activity of the dose at the moment of injection is \(3.5 \times 10^8\text{ Bq}\). Calculate the mass of the \(^{131}\text{I}\) atoms in this dose.

(c) Calculate the activity of this dose remaining in the patient's body after 20 days, assuming none has been biologically excreted.

(Avogadro constant \(N_A = 6.02 \times 10^{23}\text{ mol}^{-1}\), molar mass of \(^{131}\text{I} = 131\text{ g mol}^{-1}\))
PastPaper.showAnswers

PastPaper.workedSolution

(a) Convert the half-life from days to seconds:
\(T_{1/2} = 8.0\text{ days} \times 24\text{ h/day} \times 3600\text{ s/h} = 691,200\text{ s}\).

Calculate the decay constant \(\lambda\):
\(\lambda = \frac{\ln(2)}{T_{1/2}} = \frac{0.6931}{691,200\text{ s}} \approx 1.003 \times 10^{-6}\text{ s}^{-1}\).

(b) Relate activity \(A\) to the number of nuclei \(N\):
\(A = \lambda N \implies N = \frac{A}{\lambda} = \frac{3.5 \times 10^8\text{ Bq}}{1.003 \times 10^{-6}\text{ s}^{-1}} \approx 3.49 \times 10^{14}\text{ nuclei}\).

Convert the number of nuclei to moles \(n\):
\(n = \frac{N}{N_A} = \frac{3.49 \times 10^{14}}{6.02 \times 10^{23}\text{ mol}^{-1}} \approx 5.797 \times 10^{-10}\text{ mol}\).

Calculate the mass \(m\):
\(m = n \times 131\text{ g mol}^{-1} = 5.797 \times 10^{-10} \times 131 \approx 7.59 \times 10^{-8}\text{ g} = 7.6 \times 10^{-11}\text{ kg}\).

(c) Calculate the activity remaining after \(t = 20\text{ days}\):
\(A = A_0 e^{-\lambda t}\) or \(A = A_0 (0.5)^{t / T_{1/2}}\).

Using the half-life fraction approach:
\(n_{\text{half-lives}} = \frac{20}{8.0} = 2.5\).

\(A = 3.5 \times 10^8 \times (0.5)^{2.5} = 3.5 \times 10^8 \times 0.1768 \approx 6.19 \times 10^7\text{ Bq}\).

Rounded to 2 significant figures, \(A = 6.2 \times 10^7\text{ Bq}\).

PastPaper.markingScheme

(a) [2 marks]:
- Convert 8.0 days correctly to seconds (\(691,200\text{ s}\)) [1 mark].
- Correct decay constant value \(1.0 \times 10^{-6}\text{ s}^{-1}\) [1 mark].

(b) [4 marks]:
- State and use \(A = \lambda N\) to find \(N = 3.5 \times 10^{14}\) nuclei [1 mark].
- Use Avogadro's constant to find number of moles (\(5.8 \times 10^{-10}\text{ mol}\)) [1 mark].
- Convert moles to mass in grams (\(7.6 \times 10^{-8}\text{ g}\)) [1 mark].
- Correct conversion to kg: \(7.6 \times 10^{-11}\text{ kg}\) [1 mark].

(c) [3 marks]:
- Correct decay formula selected: \(A = A_0 e^{-\lambda t}\) or equivalent [1 mark].
- Number of half-lives identified as 2.5 [1 mark].
- Final activity of \(6.2 \times 10^7\text{ Bq}\) (accept \(6.19 \times 10^7\text{ Bq}\)) [1 mark].

[0.67 marks reserved for consistent units and rounding across the question].

Section B

Answer all questions in this section. Questions include a mix of mathematical synthesis, diagrams, and a structured Level of Response question.
3 PastPaper.question · 44.01 PastPaper.marks
PastPaper.question 1 · Extended Structured and LOR
14.67 PastPaper.marks

This question is about the thermal behavior of gases and the Boltzmann factor.

Part (a): A rigid container of volume \(0.025\text{ m}^3\) contains helium-4 gas (molar mass \(4.0\text{ g mol}^{-1}\)) at a pressure of \(1.2 \times 10^5\text{ Pa}\) and a temperature of \(290\text{ K}\).
(i) Calculate the number of helium atoms in the container. [2 marks]
(ii) Calculate the mean square speed \(\langle v^2 \rangle\) of the helium atoms. [3 marks]

Part (b): The activation energy for a certain chemical reaction is \(0.22\text{ eV}\).
(i) Calculate this activation energy in joules. [1 mark]
(ii) Calculate the ratio of the probability of a molecular collision having enough energy to react at \(350\text{ K}\) compared to \(290\text{ K}\). [3 marks]

Part (c)*: Describe how the kinetic theory of gases explains the pressure exerted by a gas on the walls of its container. Explain how this pressure changes when the temperature increases. Discuss how the Boltzmann factor is used to model processes such as chemical reactions or evaporation, and why these processes are highly sensitive to temperature changes. [6 marks]

PastPaper.showAnswers

PastPaper.workedSolution

Part (a):
(i) Using the ideal gas law: \(pV = N k_B T\)
\(N = \frac{pV}{k_B T} = \frac{1.2 \times 10^5 \times 0.025}{1.38 \times 10^{-23} \times 290} = 7.50 \times 10^{23}\) atoms.
(ii) Mean kinetic energy of a molecule is given by: \(\frac{1}{2} m \langle v^2 \rangle = \frac{3}{2} k_B T \implies \langle v^2 \rangle = \frac{3 k_B T}{m}\)
The mass of one helium-4 atom is \(m = \frac{4.0 \times 10^{-3}\text{ kg mol}^{-1}}{6.02 \times 10^{23}\text{ mol}^{-1}} = 6.64 \times 10^{-27}\text{ kg}\).
\(\langle v^2 \rangle = \frac{3 \times 1.38 \times 10^{-23} \times 290}{6.64 \times 10^{-27}} = 1.81 \times 10^6\text{ m}^2\text{ s}^{-2}\).

Part (b):
(i) \(E_a = 0.22\text{ eV} = 0.22 \times 1.60 \times 10^{-19}\text{ J} = 3.52 \times 10^{-20}\text{ J}\).
(ii) The Boltzmann factor is \(f = e^{-\frac{E_a}{k_B T}}\).
At \(290\text{ K}\): \(k_B T_1 = 1.38 \times 10^{-23} \times 290 = 4.002 \times 10^{-21}\text{ J}\)
\(f_{290} = e^{-\frac{3.52 \times 10^{-20}}{4.002 \times 10^{-21}}} = e^{-8.796} = 1.51 \times 10^{-4}\)
At \(350\text{ K}\): \(k_B T_2 = 1.38 \times 10^{-23} \times 350 = 4.83 \times 10^{-21}\text{ J}\)
\(f_{350} = e^{-\frac{3.52 \times 10^{-20}}{4.83 \times 10^{-21}}} = e^{-7.288} = 6.84 \times 10^{-4}\)
Ratio \(\frac{f_{350}}{f_{290}} = \frac{6.84 \times 10^{-4}}{1.51 \times 10^{-4}} = 4.53\) (accept 4.5).

Part (c)*:
Gas pressure is caused by elastic collisions of gas molecules with the container walls, producing a force determined by the rate of change of momentum (Newton's Second Law). Since \(P = F/A\), this force per unit area is the pressure.
When temperature increases, molecules have higher average kinetic energy and move faster. This causes: (1) a larger momentum change per collision, and (2) more frequent collisions per second. Both increase the average force, increasing pressure.
The Boltzmann factor represents the ratio of particles in an excited/activated state compared to the ground state. Because the absolute temperature appears in the denominator of a negative exponent, a small fractional increase in temperature leads to a massive exponential increase in the number of particles with energy exceeding the activation energy, making these rate-based processes highly temperature sensitive.

PastPaper.markingScheme

Part (a)(i):
• [1 mark] Correct rearrangement of ideal gas equation to find \(N\).
• [1 mark] Correct final answer: \(7.5 \times 10^{23}\).

Part (a)(ii):
• [1 mark] Correct calculation of the mass of a single helium atom (\(6.64 \times 10^{-27}\text{ kg}\)).
• [1 mark] Correct formulation for mean square speed \(\langle v^2 \rangle = \frac{3 k_B T}{m}\).
• [1 mark] Correct final answer: \(1.8 \times 10^6\text{ m}^2\text{ s}^{-2}\) (allow \(1.81 \times 10^6\)).

Part (b)(i):
• [1 mark] Correct conversion of eV to J (\(3.52 \times 10^{-20}\text{ J}\)).

Part (b)(ii):
• [1 mark] Correct calculation of at least one Boltzmann factor exponent.
• [1 mark] Correct values for both Boltzmann factors (\(1.5 \times 10^{-4}\) and \(6.8 \times 10^{-4}\)).
• [1 mark] Correct ratio (\(4.5\) or \(4.53\)).

Part (c)* Level of Response:
• Level 3 (5-6 marks): Comprehensive, logically structured explanation covering molecular origin of pressure, why pressure scales with temperature (addressing both collision frequency and force), and a detailed explanation of the exponential nature of the Boltzmann factor explaining temperature sensitivity.
• Level 2 (3-4 marks): Structured explanation covering most aspects, but may lack detail on either the dual reason for temperature-induced pressure change or the mathematical explanation of the Boltzmann factor.
• Level 1 (1-2 marks): Basic description of pressure as collision-based, or simple statement of temperature effect without deep physical justification.

PastPaper.question 2 · Extended Structured and LOR
14.67 PastPaper.marks

This question is about gravitational fields, orbits, and astronomical measurements.

Part (a): A star near the center of a distant galaxy is observed orbiting a central object, assumed to be a supermassive black hole. The star is in a circular orbit of radius \(1.2 \times 10^{14}\text{ m}\) and has an orbital period of \(15\text{ years}\).
(i) Calculate the orbital speed of the star. [2 marks]
(ii) Calculate the mass of the supermassive black hole. [3 marks]

Part (b): For this star-black hole system:
(i) Calculate the gravitational potential at the position of the star's orbit due to the black hole. [2 marks]
(ii) Calculate the escape velocity for any object located at this orbital distance from the black hole. [2 marks]

Part (c)*: Explain how astronomers can measure the orbital speeds and distances of stars near the galactic center. Discuss the evidence this provides for the existence of supermassive black holes and the cosmological expansion of space. [6 marks]

PastPaper.showAnswers

PastPaper.workedSolution

Part (a):
(i) Period \(T = 15 \times 365.25 \times 24 \times 3600 = 4.734 \times 10^8\text{ s}\).
Orbital speed \(v = \frac{2 \pi r}{T} = \frac{2 \pi \times 1.2 \times 10^{14}}{4.734 \times 10^8} = 1.59 \times 10^6\text{ m s}^{-1}\).
(ii) Equating gravitational and centripetal force: \(\frac{G M m}{r^2} = \frac{m v^2}{r} \implies M = \frac{v^2 r}{G}\).
\(M = \frac{(1.59 \times 10^6)^2 \times 1.2 \times 10^{14}}{6.67 \times 10^{-11}} = 4.56 \times 10^{36}\text{ kg}\).

Part (b):
(i) Gravitational potential \(V_g = -\frac{G M}{r}\).
\(V_g = -\frac{6.67 \times 10^{-11} \times 4.56 \times 10^{36}}{1.2 \times 10^{14}} = -2.53 \times 10^{12}\text{ J kg}^{-1}\).
(ii) Escape velocity \(v_{esc} = \sqrt{\frac{2 G M}{r}} = \sqrt{2 |V_g|}\).
\(v_{esc} = \sqrt{2 \times 2.53 \times 10^{12}} = 2.25 \times 10^6\text{ m s}^{-1}\).

Part (c)*:
Astronomers measure orbital speed using the Doppler effect: shifts in spectral lines of known elements (like Hydrogen) allow speed to be calculated from \(\frac{\Delta \lambda}{\lambda} \approx \frac{v}{c}\). Distances can be found using parallax (for nearby structures) or standard candles (like Cepheid variables or Type Ia supernovae) where luminosity \(L\) and flux \(F\) are linked by \(F = \frac{L}{4 \pi d^2}\).
Observing stars with massive velocities moving in very small orbits allows the mass of the central object to be calculated. The incredibly high mass density required is evidence for a supermassive black hole.
Cosmological expansion is proven by observing that light from distant galaxies is redshifted, with recession speed proportional to distance (Hubble's Law, \(v = H_0 d\)), showing the fabric of space itself is expanding.

PastPaper.markingScheme

Part (a)(i):
• [1 mark] Conversion of 15 years into seconds (\(4.73 \times 10^8\text{ s}\)).
• [1 mark] Correct speed calculated (\(1.59 \times 10^6\text{ m s}^{-1}\)).

Part (a)(ii):
• [1 mark] Derivation of \(M = \frac{v^2 r}{G}\).
• [1 mark] Correct substitution of values.
• [1 mark] Correct mass calculated (\(4.56 \times 10^{36}\text{ kg}\) or ecf from speed).

Part (b)(i):
• [1 mark] Correct formula for gravitational potential (negative sign required).
• [1 mark] Correct value (\(-2.53 \times 10^{12}\text{ J kg}^{-1}\)).

Part (b)(ii):
• [1 mark] Correct escape velocity expression \(v_{esc} = \sqrt{\frac{2 G M}{r}}\).
• [1 mark] Correct speed (\(2.25 \times 10^6\text{ m s}^{-1}\)).

Part (c)* Level of Response:
• Level 3 (5-6 marks): Full explanation of both measurement techniques (Doppler shift and standard candles) and deep physical connections to both black hole evidence (mass density) and cosmic expansion (Hubble's Law).
• Level 2 (3-4 marks): Explains how measurements are made, but lacks depth in discussing either black hole characteristics or cosmological expansion.
• Level 1 (1-2 marks): Mentions Doppler shift or parallax briefly, with little to no link to the physical/cosmological significance.

PastPaper.question 3 · Extended Structured and LOR
14.67 PastPaper.marks

This question is about wave-particle duality and the quantum nature of light and matter.

Part (a): Light of wavelength \(380\text{ nm}\) is shone onto a clean sodium surface in a vacuum. The work function of sodium is \(2.30\text{ eV}\).
(i) Calculate the energy of a single photon in the incident light, in eV. [2 marks]
(ii) Calculate the maximum kinetic energy of the emitted photoelectrons, and hence determine their maximum velocity. [3 marks]

Part (b): Calculate the de Broglie wavelength of the photoelectrons that are emitted with this maximum velocity. [3 marks]

Part (c)*: Discuss how the photoelectric effect and electron diffraction experiments provide contrasting evidence for the wave-like and particle-like nature of electromagnetic radiation and matter. Explain how classical wave theory failed to explain the photoelectric effect, and how Einstein’s photon model resolved these issues. [6 marks]

PastPaper.showAnswers

PastPaper.workedSolution

Part (a):
(i) \(E = \frac{h c}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{380 \times 10^{-9}} = 5.23 \times 10^{-19}\text{ J}\).
In eV: \(E = \frac{5.23 \times 10^{-19}}{1.60 \times 10^{-19}} = 3.27\text{ eV}\).
(ii) Maximum kinetic energy \(E_{k,\text{max}} = hf - \phi = 3.27\text{ eV} - 2.30\text{ eV} = 0.97\text{ eV}\).
In Joules: \(E_{k,\text{max}} = 0.97 \times 1.60 \times 10^{-19} = 1.55 \times 10^{-19}\text{ J}\).
Since \(E_k = \frac{1}{2} m v^2\):
\(v_{\text{max}} = \sqrt{\frac{2 \times 1.55 \times 10^{-19}}{9.11 \times 10^{-31}}} = 5.83 \times 10^5\text{ m s}^{-1}\).

Part (b):
The de Broglie wavelength is \(\lambda = \frac{h}{p} = \frac{h}{m v}\).
\(\lambda = \frac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \times 5.83 \times 10^5} = \frac{6.63 \times 10^{-34}}{5.31 \times 10^{-25}} = 1.25 \times 10^{-9}\text{ m} = 1.25\text{ nm}\).

Part (c)*:
The photoelectric effect provides evidence for the particle-like nature of light, whereas electron diffraction provides evidence for the wave-like nature of matter.
In the photoelectric effect, classical wave theory predicted that electron emission could happen at any frequency if intensity was high enough, and that there would be a time delay for energy to accumulate. It failed because experiments showed a threshold frequency and instantaneous emission. Einstein resolved this by proposing that light consists of discrete quanta (photons) of energy \(E = hf\). A single photon interacts with a single electron; if \(hf < \phi\), no emission occurs, regardless of intensity.
In electron diffraction, a beam of accelerated electrons passing through graphite produces a circular interference pattern. Interference is exclusively a wave phenomenon, which shows that matter (conventionally particles) propagates as waves with a wavelength given by \(\lambda = h/p\).

PastPaper.markingScheme

Part (a)(i):
• [1 mark] Correct use of \(E = \frac{hc}{\lambda}\).
• [1 mark] Correct conversion to eV to yield \(3.27\text{ eV}\) (accept \(3.3\text{ eV}\)).

Part (a)(ii):
• [1 mark] Correct calculation of \(E_{k,\text{max}} = 0.97\text{ eV}\) or \(1.55 \times 10^{-19}\text{ J}\).
• [1 mark] Correct rearrangement of kinetic energy equation for \(v\).
• [1 mark] Correct velocity calculated (\(5.83 \times 10^5\text{ m s}^{-1}\) or ecf from energy).

Part (b):
• [1 mark] Use of \(p = mv\) to calculate momentum (\(5.31 \times 10^{-25}\text{ kg m s}^{-1}\)).
• [1 mark] Correct substitution into de Broglie equation \(\lambda = \frac{h}{p}\).
• [1 mark] Correct wavelength (\(1.25 \times 10^{-9}\text{ m}\) or \(1.25\text{ nm}\)).

Part (c)* Level of Response:
• Level 3 (5-6 marks): Detailed explanation of both experiments. Fully explains the failures of classical wave theory (threshold frequency, intensity independence, instantaneous emission) and how the photon model solves them. Explains how electron diffraction exhibits wave phenomena for matter.
• Level 2 (3-4 marks): Explains both experiments, but lacks complete detail on the specific failures of classical theory or the mechanics of the photon model.
• Level 1 (1-2 marks): Mentions photoelectric effect and electron diffraction without clearly distinguishing why they represent wave or particle nature, or why classical theory failed.

Section C

Answer all questions based on the Advance Notice Article. Deep focus on mathematical modelling and experimental analysis.
3 PastPaper.question · 27 PastPaper.marks
PastPaper.question 1 · structured
9 PastPaper.marks
Based on the Advance Notice Article regarding the thermodynamic exploration of the atmosphere:

(a) At launch altitude (sea level, where \(h = 0\)), the temperature is \(290\text{ K}\) and the atmospheric pressure is \(101\text{ kPa}\). A weather balloon is inflated with helium to a volume of \(1.5\text{ m}^3\). As the balloon ascends to an altitude of \(15\text{ km}\), the pressure drops to \(12\text{ kPa}\) and the temperature falls to \(217\text{ K}\). Calculate the volume of the balloon at this altitude, assuming helium behaves as an ideal gas.

(b) The maximum volume the balloon can reach before bursting is \(10\text{ m}^3\). Explain, using mathematical models of gas behaviour and the atmosphere, how modelling the atmosphere as isothermal (constant temperature) compared to modelling it with the actual lapse rate (temperature decreasing with altitude) affects the prediction of the balloon's burst altitude.
PastPaper.showAnswers

PastPaper.workedSolution

(a) Using the ideal gas equation:
\(P_0 V_0 / T_0 = P_1 V_1 / T_1\)
Rearranging for \(V_1\):
\(V_1 = \frac{P_0 V_0 T_1}{P_1 T_0}\)
\(V_1 = \frac{101\text{ kPa} \times 1.5\text{ m}^3 \times 217\text{ K}}{12\text{ kPa} \times 290\text{ K}}\)
\(V_1 = \frac{32875.5}{3480} \approx 9.45\text{ m}^3\) (or \(9.4\text{ m}^3\) to 2 significant figures).

(b) Level of Response analysis:
- **Isothermal Model**: Since temperature \(T\) is constant, the volume of the balloon is given by \(V = \frac{nRT_0}{P}\). Therefore, \(V \propto \frac{1}{P}\). As the pressure drops exponentially with altitude \(P = P_0 e^{-h/H}\), the balloon's volume will expand exponentially.
- **Lapse Rate Model**: In reality, temperature decreases with altitude in the troposphere. The volume is given by \(V = \frac{nRT(h)}{P}\). Since both \(T\) and \(P\) decrease, the reduction in temperature acts to decrease the volume, partially counteracting the expansion due to the drop in external pressure.
- **Comparison**: At any given altitude \(h > 0\), the temperature in the lapse rate model is lower than \(T_0\). Thus, the actual volume of the balloon will be smaller than the volume predicted by the isothermal model at that same pressure. Consequently, the balloon will expand more slowly and will reach its critical burst volume of \(10\text{ m}^3\) at a higher altitude (lower pressure) than predicted by the isothermal model.

PastPaper.markingScheme

**Part (a) [3 Marks]:**
- **1 Mark**: Correct rearrangement of the ideal gas formula to make \(V_1\) the subject: \(V_1 = \frac{P_0 V_0 T_1}{P_1 T_0}\).
- **1 Mark**: Correct substitution of values with consistent units (either keeping kPa or converting both to Pa): \(V_1 = \frac{101 \times 10^3 \times 1.5 \times 217}{12 \times 10^3 \times 290}\).
- **1 Mark**: Correct final calculated volume of \(9.4\text{ m}^3\) or \(9.45\text{ m}^3\) (2 or 3 sig figs).

**Part (b) [6 Marks] - Level of Response:**
- **Detailed marking criteria**:
- **Level 3 (5-6 Marks)**: Detailed explanation of both models using equations (e.g., \(V = nRT/P\)). Correctly identifies that a lower temperature at altitude reduces gas expansion, leading to the conclusion that the isothermal model predicts a *lower* burst altitude (or that the real-world lapse rate model predicts a *higher* burst altitude). Explanations are clear, coherent, and highly logical.
- **Level 2 (3-4 Marks)**: Compares both models qualitatively. Explains that lower temperature leads to smaller volume. States which model predicts a higher/lower burst altitude, but the mathematical reasoning is incomplete or lacks formal link to \(PV=nRT\).
- **Level 1 (1-2 Marks)**: Basic statement that volume depends on pressure and temperature, or identifies that temperature drops with altitude. No clear comparison of burst altitudes.
- **0 Marks**: No creditworthy response.
PastPaper.question 2 · structured
9 PastPaper.marks
Based on the Advance Notice Article regarding electromagnetic braking systems in transit networks:

(a) Show that the induced electromotive force (emf) \(\varepsilon\) in a radial segment of a conducting disc of radius \(R\) rotating with angular velocity \(\omega\) in a uniform magnetic field \(B\) perpendicular to the disc is given by:
\(\varepsilon = \frac{1}{2} B \omega R^2\)

(b) An experimenter measures the braking torque \(\tau\) as a function of angular velocity \(\omega\) for a rotating copper disc. A simplified model predicts that the torque should be directly proportional to the angular velocity (\(\tau = k\omega\)). The experimenter collects the following data:
- At \(\omega = 50\text{ rad s}^{-1}\), \(\tau = 0.45\text{ N m}\)
- At \(\omega = 150\text{ rad s}^{-1}\), \(\tau = 1.15\text{ N m}\)

Explain how this data demonstrates a deviation from the simplified model, and suggest physical reasons for this deviation based on the electrical and magnetic properties of materials.
PastPaper.showAnswers

PastPaper.workedSolution

(a) Proof:
- A radial line of length \(R\) sweeps out a sector of area \(A = \frac{1}{2} R^2 \theta\) in time \(t\).
- The rate at which flux is cut (which equals the induced emf \(\varepsilon\) by Faraday's Law) is given by:
\(\varepsilon = \frac{d\Phi}{dt} = B \frac{dA}{dt}\)
- Since \(\frac{dA}{dt} = \frac{1}{2} R^2 \frac{d\theta}{dt}\) and \(\frac{d\theta}{dt} = \omega\):
\(\varepsilon = \frac{1}{2} B \omega R^2\) (as required).

(b) Level of Response analysis:
- **Data Analysis**: Under the simple model \(\tau = k\omega\), the ratio \(k = \tau/\omega\) should remain constant.
- For \(\omega = 50\text{ rad s}^{-1}\): \(k_1 = 0.45 / 50 = 0.0090\text{ N m s}\).
- For \(\omega = 150\text{ rad s}^{-1}\): \(k_2 = 1.15 / 150 \approx 0.0077\text{ N m s}\).
- Since \(k_2 < k_1\), the torque increases less than linearly with angular velocity at high speeds.
- **Physical Reasons**:
1. **Thermal Effects**: The large induced eddy currents cause significant Joule heating (\(P = I^2 R_{disc}\)) in the copper disc. As the temperature of the copper increases, its resistivity increases, which reduces the electrical conductivity. Consequently, the induced current \(I = \varepsilon / R_{disc}\) is lower than expected at high speeds, reducing the magnetic force and resulting braking torque.
2. **Inductive / Demagnetisation Effects**: At higher angular speeds, the magnetic field produced by the rapidly changing eddy currents themselves becomes significant. This induced field opposes the applied external magnetic field \(B\) (Lenz's Law), reducing the net magnetic flux through the disc and thus limiting the braking torque.

PastPaper.markingScheme

**Part (a) [3 Marks]:**
- **1 Mark**: Recognises that induced emf is rate of change of magnetic flux linkage, \(\varepsilon = \frac{d\Phi}{dt}\) or \(\varepsilon = B \frac{dA}{dt}\).
- **1 Mark**: Correctly relates the area of the sector to the angle: \(A = \frac{1}{2} R^2 \theta\).
- **1 Mark**: Substitutes \(\frac{d\theta}{dt} = \omega\) to obtain \(\varepsilon = \frac{1}{2} B \omega R^2\).

**Part (b) [6 Marks] - Level of Response:**
- **Detailed marking criteria**:
- **Level 3 (5-6 Marks)**: Calculates the ratio \(\tau/\omega\) for both data points to demonstrate a decreasing trend (proving non-linearity). Provides two distinct, physically accurate explanations for the deviation (temperature rise increasing resistance, and/or self-induction/Lenz's law limiting net flux at high frequencies).
- **Level 2 (3-4 Marks)**: Shows that the ratio \(\tau/\omega\) is not constant but lacks accuracy in calculations or provides only one physical explanation (such as temperature-induced resistance increase) with limited detail.
- **Level 1 (1-2 Marks)**: Identifies that the data does not show proportionality but fails to calculate ratios correctly. Offers generic physical reasons without linking them clearly to materials physics.
- **0 Marks**: No creditworthy response.
PastPaper.question 3 · structured
9 PastPaper.marks
Based on the Advance Notice Article regarding the dynamics of low Earth orbit (LEO) satellites:

(a) Show that the total mechanical energy \(E\) of a satellite of mass \(m\) in a stable circular orbit of radius \(r\) around a planet of mass \(M\) is given by:
\(E = -\frac{GMm}{2r}\)

(b) During orbital decay, atmospheric drag causes the satellite to lose energy continuously. Explain the apparent paradox that as the satellite loses mechanical energy, its orbital speed actually increases. Discuss how this effect, combined with atmospheric density variation, influences the mathematical modelling of the satellite's remaining lifetime.
PastPaper.showAnswers

PastPaper.workedSolution

(a) Proof:
- The total energy is the sum of kinetic energy (\(E_k\)) and gravitational potential energy (\(E_p\)):
\(E = E_k + E_p = \frac{1}{2}mv^2 - \frac{GMm}{r}\)
- For a stable circular orbit, the centripetal force is provided by gravity:
\(\frac{mv^2}{r} = \frac{GMm}{r^2} \implies mv^2 = \frac{GMm}{r}\)
- Substituting this into the kinetic energy term:
\(E_k = \frac{GMm}{2r}\)
- Adding the potential energy:
\(E = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}\) (as required).

(b) Level of Response analysis:
- **The Paradox explained**: As drag does negative work, total energy \(E\) decreases (becomes more negative). Since \(E = -\frac{GMm}{2r}\), a more negative \(E\) corresponds to a smaller orbital radius \(r\).
- Since the orbital velocity is given by \(v = \sqrt{\frac{GM}{r}}\), a smaller \(r\) results in a larger orbital speed \(v\). Physically, as the satellite descends, it loses gravitational potential energy; half of this lost GPE is converted into heat (work done against drag) and the other half is converted into kinetic energy, increasing the speed.
- **Modelling orbital lifetime**: A simple constant-drag model is highly inaccurate. As the radius \(r\) decreases, the satellite enters lower, denser layers of the atmosphere. The atmospheric density \(\rho\) increases exponentially with decreasing altitude (\(\rho \propto e^{-h/H}\)). Since drag force \(F_d \propto \rho v^2\), the rate of energy loss (\(P_{drag} = F_d v\)) accelerates dramatically. This results in a positive feedback loop (runaway decay), making the prediction of orbital lifetime highly sensitive to accurate density profiles and requiring numerical integration methods rather than static linear models.

PastPaper.markingScheme

**Part (a) [3 Marks]:**
- **1 Mark**: States correct expressions for kinetic energy (\(\frac{1}{2}mv^2\)) and gravitational potential energy (\(-\frac{GMm}{r}\)).
- **1 Mark**: Uses centripetal force equivalence to express \(mv^2\) or \(\frac{1}{2}mv^2\) in terms of \(G\), \(M\), \(m\), and \(r\).
- **1 Mark**: Adds the terms correctly to obtain \(E = -\frac{GMm}{2r}\).

**Part (b) [6 Marks] - Level of Response:**
- **Detailed marking criteria**:
- **Level 3 (5-6 Marks)**: Explains the energetic trade-off thoroughly (loss of GPE is twice the gain in KE, meaning speed increases as radius decreases). Connects this clearly to orbital decay modelling, noting that atmospheric density increases exponentially with decreasing altitude, which leads to a runaway decay rate. Identifies the need for non-linear mathematical models.
- **Level 2 (3-4 Marks)**: Explains the speed-energy paradox correctly (lower orbit requires higher speed). Mentions that atmospheric density increases at lower altitudes but fails to detail how this affects the mathematical model / decay rate mathematically.
- **Level 1 (1-2 Marks)**: Identifies that speed increases as radius decreases, but does not explain the energetic mechanism (GPE vs KE) or the exponential density challenge.
- **0 Marks**: No creditworthy response.

PastPaper.sampleCTATitle

PastPaper.sampleCTADescription

PastPaper.ctaPrimaryPastPaper.ctaSecondary

PastPaper.sampleStickyMessage

PastPaper.stickyCtaText