2023 IB DP Physics Practice Paper with Answers

By conservation of momentum, \(mv = (m + 3m)v'\), which gives the common final velocity \(v' = \frac{v}{4}\). The initial kinetic energy is \(E_k = \frac{1}{2}mv^2\). The final kinetic energy is \(E_k' = \frac{1}{2}(4m)(v')^2 = \frac{1}{2}(4m)\left(\frac{v}{4}\right)^2 = \frac{1}{8}mv^2 = \frac{1}{4}E_k\). The kinetic energy dissipated is \(E_k - E_k' = \frac{3}{4}E_k\). Therefore, the fraction of initial kinetic energy dissipated is \\frac{3}{4}.

Award [1] for the correct answer D.

Luminosity is given by Stefan-Boltzmann law: \(L = \sigma A T^4 = \sigma (4\pi R^2) T^4 \propto R^2 T^4\). For Star X, the luminosity is proportional to \(R^2 T^4\). For Star Y, the luminosity is proportional to \((0.5R)^2 (2T)^4 = 0.25 R^2 \times 16 T^4 = 4 R^2 T^4\). Therefore, the ratio is \\frac{L_X}{L_Y} = \\frac{R^2 T^4}{4 R^2 T^4} = \\frac{1}{4}.

Award [1] for the correct answer D.

Using the ideal gas equation, we have \\frac{P_1 V_1}{T_1} = \\frac{P_2 V_2}{T_2}\). Substituting \(P_1 = P\), \(V_1 = V\), \(T_1 = T\), \(P_2 = 2P\), and \(T_2 = \frac{T}{2}\), we get \\frac{PV}{T} = \\frac{2P \\times V_2}{T/2} = \\frac{4 P V_2}{T}\). Solving for \(V_2\) gives \(V_2 = \frac{V}{4}\).

Award [1] for the correct answer D.

When unpolarized light passes through the first polarizer, its intensity is halved: \(I_1 = \frac{1}{2}I_0\). By Malus's Law, the intensity after passing through the analyzer is \(I_2 = I_1 \cos^2(60^\circ) = \frac{1}{2}I_0 \times \left(\frac{1}{2}\right)^2 = \frac{1}{8}I_0\).

Award [1] for the correct answer C.

For a source moving towards a stationary observer, the Doppler shifted frequency is given by \(f' = f \left(\frac{v}{v - v_s}\right)\). Substituting the speed of the source \(v_s = 0.10 v\) gives \(f' = f \left(\frac{v}{v - 0.10 v}\right) = f \left(\frac{v}{0.90 v}\right) = \frac{10}{9} f\).

Award [1] for the correct answer D.

Let the point of zero potential be at a distance \(x\) from the charge \( +q \). Since this point is between the two charges, its distance from the charge \( -4q \) is \(d - x\). The total electric potential is the sum of the potentials due to each charge: \(V = \frac{kq}{x} + \frac{k(-4q)}{d - x} = 0\). Rearranging this gives \\frac{1}{x} = \\frac{4}{d - x}\), which yields \(d - x = 4x \implies 5x = d \implies x = \frac{d}{5}\).

Award [1] for the correct answer A.

According to Kepler's Third Law for circular orbits, \(T^2 \propto R^3\). Therefore, the relationship between the orbital periods and radii of the two satellites is given by \\left(\\frac{T_Y}{T_X}\\right)^2 = \\left(\\frac{R_Y}{R_X}\\right)^3 = \\left(\\frac{4R}{R}\\right)^3 = 64\). Taking the square root of both sides gives \\frac{T_Y}{T_X} = 8\), which means \(T_Y = 8T\).

Award [1] for the correct answer C.

The energy of an emitted photon is related to its wavelength by \(E = \frac{hc}{\lambda}\), meaning \\lambda \\propto \\frac{1}{\\Delta E}\). The energy differences for the transitions are: \\Delta E_{3 \\to 1} = E_0 \\left(1 - \\frac{1}{9}\\right) = \\frac{8}{9}E_0\) and \\Delta E_{2 \\to 1} = E_0 \\left(1 - \\frac{1}{4}\\right) = \\frac{3}{4}E_0\). The ratio of the wavelengths is \\frac{\\lambda_{2 \\to 1}}{\\lambda_{3 \\to 1}} = \\frac{\\Delta E_{3 \\to 1}}{\\Delta E_{2 \\to 1}} = \\frac{8/9}{3/4} = \\frac{8}{9} \\times \\frac{4}{3} = \\frac{32}{27}\).

Award [1] for the correct answer B.

The time of flight is given by \( t = \sqrt{\frac{2h}{g}} \). The horizontal distance is \( d = v t = v \sqrt{\frac{2h}{g}} \). For the new scenario, the new time of flight is \( t' = \sqrt{\frac{2(\frac{h}{2})}{g}} = \frac{t}{\sqrt{2}} \). The new horizontal distance is \( d' = (2v) t' = 2v \frac{t}{\sqrt{2}} = \sqrt{2} vt = \sqrt{2}d \).

Award 1 mark for identifying that the time of flight is proportional to the square root of height and the horizontal distance is the product of horizontal speed and time of flight, leading to \( \sqrt{2} d \).

The total mass of the system is \( M = m + 3m = 4m \). The acceleration of both blocks is \( a = \frac{F}{4m} \). The force responsible for accelerating the second block of mass \( 3m \) is the normal contact force \( N \) from the first block. Therefore, \( N = 3m \times a = 3m \times \frac{F}{4m} = \frac{3}{4}F \).

Award 1 mark for calculating the correct acceleration of the combined system and multiplying it by the mass of the second block to find the contact force of \( \frac{3}{4}F \).

According to the Stefan-Boltzmann law, the total power radiated is \( P = \sigma A T^4 = 4\pi R^2 \sigma T^4 \). Thus, \( P \propto R^2 T^4 \). For Star Y, the power is \( P_Y \propto \left(\frac{R}{2}\right)^2 (2T)^4 = \frac{R^2}{4} \times 16 T^4 = 4 R^2 T^4 \). The ratio is \( \frac{P_Y}{P_X} = 4 \).

Award 1 mark for applying the Stefan-Boltzmann law and correctly identifying that doubling the temperature increases power by a factor of 16, while halving the radius decreases power by a factor of 4, yielding a net factor of 4.

During the isobaric expansion, pressure remains \( P \). Since volume doubles to \( 2V \), the absolute temperature must also double to \( 2T \) to satisfy the ideal gas law. During the isochoric cooling, volume remains constant at \( 2V \) while the temperature is halved back to \( T \). Since pressure is directly proportional to temperature at constant volume, the pressure must be halved to \( \frac{P}{2} \).

Award 1 mark for applying the ideal gas equation through both processes to determine that the final pressure is \( \frac{P}{2} \).

The fringe spacing is given by \( s = \frac{\lambda D}{d} \). Under the new conditions, the new fringe spacing \( s' \) is \( s' = \frac{\lambda (2D)}{\frac{d}{2}} = 4 \frac{\lambda D}{d} = 4s \).

Award 1 mark for substituting the new parameters into the double-slit formula to obtain a fringe spacing that is 4 times the original value.

Let the position where the potential is zero be at a distance \( x \) from the charge \( +q \). Since the point lies between the charges, its distance from the charge \( -2q \) is \( d - x \). The total electric potential at this point is \( V = \frac{kq}{x} - \frac{k(2q)}{d-x} = 0 \). Solving for \( x \) gives \( \frac{1}{x} = \frac{2}{d-x} \implies d-x = 2x \implies 3x = d \implies x = \frac{d}{3} \).

Award 1 mark for expressing the total potential as the sum of the potentials due to individual charges and solving for the location between the charges where it equals zero.

The energy of the emitted photon is equal to the difference in energy levels. For the first transition, \( E_{\text{photon1}} = hf = -E - (-4E) = 3E \). For the second transition, \( E_{\text{photon2}} = hf' = -E - (-9E) = 8E \). Therefore, the new frequency \( f' \) is given by \( f' = \frac{8}{3} f \).

Award 1 mark for calculating the energy differences of both transitions and establishing the ratio of the emitted frequencies.

The escape speed is given by the formula \( v = \sqrt{\frac{2GM}{R}} \). For the second planet, the escape speed \( v' \) is \( v' = \sqrt{\frac{2G(2M)}{8R}} = \sqrt{\frac{4GM}{8R}} = \sqrt{\frac{GM}{2R}} = \frac{1}{2} \sqrt{\frac{2GM}{R}} = \frac{v}{2} \).

Award 1 mark for correctly substituting the new mass and radius into the escape speed formula and showing that the new speed is half the original escape speed.

Let upward be the positive direction. The maximum height reached by the ball above the cliff is \(H = \frac{v^2}{2g}\). The ball then falls from this maximum height to the ground, a total displacement of \(-(H + h)\). Using the kinematic formula \(v_f^2 = v_i^2 + 2as\) for the entire motion from the launch to the ground: \((-3v)^2 = v^2 + 2(-g)(-h)\), which simplifies to \(9v^2 = v^2 + 2gh\). This gives \(8v^2 = 2gh\), so \(v^2 = \frac{gh}{4}\). Substituting this back into the expression for \(H\), we get \(H = \frac{gh/4}{2g} = \frac{h}{8}\).

Correct answer is A. Award 1 mark for the correct derivation of the maximum height in terms of the cliff height.

The initial kinetic energy is \(E_k = \frac{1}{2}mv^2\). By conservation of linear momentum: \(m v = (m + 2m)v_f\), where \(v_f\) is the common final speed. This gives \(v_f = \frac{v}{3}\). The final kinetic energy is \(E_{kf} = \frac{1}{2}(3m)v_f^2 = \frac{1}{2}(3m)\left(\frac{v}{3}\right)^2 = \frac{1}{6}mv^2\). The energy dissipated is \(\Delta E = E_k - E_{kf} = \frac{1}{2}mv^2 - \frac{1}{6}mv^2 = \frac{1}{3}mv^2\). The fraction of initial kinetic energy dissipated is \(\frac{\Delta E}{E_k} = \frac{\frac{1}{3}mv^2}{\frac{1}{2}mv^2} = \frac{2}{3}\).

Correct answer is C. 1 mark for calculating the correct fraction of energy dissipated.

According to Wien's displacement law, the peak wavelength is inversely proportional to the absolute temperature: \(\lambda \propto \frac{1}{T}\). Thus, doubling the temperature halves the peak wavelength to \(\frac{\lambda}{2}\). According to the Stefan-Boltzmann law, the total radiated power is directly proportional to the fourth power of the absolute temperature: \(P \propto T^4\). Thus, doubling the temperature increases the power by a factor of \(2^4 = 16\), giving \(16P\).

Correct answer is A. 1 mark for applying both Wien's law and the Stefan-Boltzmann law correctly.

Using the ideal gas equation of state, \(\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}\). Substituting the given values: \(\frac{p V}{T} = \frac{p_2 (V/3)}{2T}\). Rearranging to solve for the final pressure \(p_2\) gives \(p_2 = 6p\).

Correct answer is D. 1 mark for the correct application of the ideal gas law.

When unpolarized light of intensity \(I_0\) passes through the first polarizer, its intensity is halved: \(I_1 = \frac{I_0}{2}\). When this vertically polarized light passes through the second polarizer, Malus's law applies: \(I_2 = I_1 \cos^2(\theta)\). Here, \(\theta = 30^\circ\), so \(I_2 = \frac{I_0}{2} \cos^2(30^\circ) = \frac{I_0}{2} \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{I_0}{2} \times \frac{3}{4} = \frac{3}{8}I_0\).

Correct answer is B. 1 mark for correctly applying the halving of intensity for unpolarized light followed by Malus's law.

Let the potential be zero at a distance \(x\) from the charge \(+q\) on the line connecting the two charges. The distance from the charge \(-2q\) is \(d - x\). The total electric potential is the sum of the potentials from both charges: \(V = \frac{kq}{x} + \frac{k(-2q)}{d-x} = 0\). Simplifying this equation: \(\frac{1}{x} = \frac{2}{d-x}\), which gives \(d - x = 2x\), and thus \(3x = d \implies x = \frac{d}{3}\).

Correct answer is A. 1 mark for the correct algebraic solution for the point of zero potential.

The terminal potential difference is given by \(V = E \frac{R}{R+r}\). As \(R\) increases from \(0\) to \(\infty\), the fraction \(\frac{R}{R+r}\) increases from \(0\) to \(1\), so \(V\) increases continuously. The power dissipated in the external resistor is \(P = I^2 R = \frac{E^2 R}{(R+r)^2}\). At \(R = 0\), \(P = 0\). As \(R \to \infty\), \(P \to 0\). According to the maximum power transfer theorem, \(P\) reaches a maximum when \(R = r\). Therefore, \(P\) first increases to a maximum and then decreases.

Correct answer is B. 1 mark for identifying the correct behavior of both terminal voltage and power output.

The energy of the emitted photon is equal to the difference in energy levels: \(\Delta E = E_i - E_f = -1.51\text{ eV} - (-3.40\text{ eV}) = 1.89\text{ eV}\). Convert this energy to joules: \(\Delta E = 1.89 \times 1.60 \times 10^{-19}\text{ J} = 3.024 \times 10^{-19}\text{ J}\). Using the relation \(E = hf\), the frequency of the photon is: \(f = \frac{\Delta E}{h} = \frac{3.024 \times 10^{-19}\text{ J}}{6.63 \times 10^{-34}\text{ J s}} \approx 4.56 \times 10^{14}\text{ Hz}\).

Correct answer is B. 1 mark for calculating the correct energy difference and converting it to find the frequency.

The impulse is equal to the area under the force-time graph. The graph is a triangle with base \( b = 5.0 \text{ s} \) and height \( h = 10 \text{ N} \).

\( \text{Impulse} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5.0 \times 10 = 25 \text{ N s} \)

Since impulse is equal to the change in momentum \( \Delta p = m \Delta v \):

\( 25 = 0.50 \times v \implies v = 50 \text{ m s}^{-1} \)

[1 mark] Award for identifying the correct speed by calculating the triangular area under the F-t curve and dividing by the mass.

According to the Stefan-Boltzmann law, the total power radiated by a black body star of radius \( R \) and temperature \( T \) is given by:

\( P = \sigma A T^4 = \sigma (4\pi R^2) T^4 \propto R^2 T^4 \)

Therefore, the ratio of the powers is:

\( \frac{P_Y}{P_X} = \left(\frac{R_Y}{R_X}\right)^2 \left(\frac{T_Y}{T_X}\right)^4 = (3)^2 \times (2)^4 = 9 \times 16 = 144 \)

[1 mark] Award for applying the Stefan-Boltzmann law and correctly raising the radius ratio to power 2 and the temperature ratio to power 4.

The gas laws require absolute temperature in Kelvin:

\( T_1 = 27 + 273 = 300 \text{ K} \)

\( T_2 = 327 + 273 = 600 \text{ K} \)

Since the volume is constant, the pressure is directly proportional to absolute temperature (Gay-Lussac's Law):

\( \frac{p_1}{T_1} = \frac{p_2}{T_2} \implies p_2 = p \times \frac{600}{300} = 2p \)

[1 mark] Award for converting temperatures to Kelvin and finding the correct pressure ratio.

When unpolarized light of intensity \( I_0 \) passes through the first polarizer, its intensity becomes halved:

\( I_1 = \frac{1}{2} I_0 \)

According to Malus's Law, when this polarized light passes through the analyzer:

\( I_2 = I_1 \cos^2(30^\circ) = \frac{1}{2} I_0 \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{2} I_0 \times \frac{3}{4} = \frac{3}{8} I_0 \)

[1 mark] Award for applying the unpolarized light reduction factor of 0.5 followed by Malus's Law.

For the droplet to remain stationary, the upward electric force must equal the downward gravitational force:

\( F_e = F_g \implies qE = mg \)

For parallel plates, the electric field strength is \( E = \frac{V}{d} \). Substituting this in:

\( q\left(\frac{V}{d}\right) = mg \implies V = \frac{mgd}{q} \)

[1 mark] Award for equating electric force to gravitational force and rearranging for V.

The total energy \( E \) is the sum of kinetic energy \( E_k \) and potential energy \( E_p \).

Potential energy is given by: \( E_p = -\frac{GMm}{r} \)

For a circular orbit, the centripetal force is provided by gravity:

\( \frac{GMm}{r^2} = \frac{mv^2}{r} \implies m v^2 = \frac{GMm}{r} \)

Thus, kinetic energy is: \( E_k = \frac{1}{2}mv^2 = \frac{GMm}{2r} \)

Total mechanical energy is:

\( E = E_k + E_p = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r} \)

[1 mark] Award for deriving kinetic energy from orbital parameters and summing with the potential energy.

The activity decreased from \( 800 \text{ Bq} \) to \( 100 \text{ Bq} \), which is a factor of:

\( \frac{100}{800} = \frac{1}{8} = \left(\frac{1}{2}\right)^3 \)

This reduction represents exactly \( 3 \) half-lives. Therefore:

\( 3 \times T_{1/2} = 24 \text{ hours} \implies T_{1/2} = 8 \text{ hours} \)

[1 mark] Award for finding the number of half-lives that occurred and dividing the total duration by that number.

The energy of the emitted photon is equal to the difference in energy levels:

\( \Delta E = -1.5 \text{ eV} - (-3.4 \text{ eV}) = 1.9 \text{ eV} \)

Convert this energy into Joules:

\( \Delta E = 1.9 \times 1.60 \times 10^{-19} \text{ J} = 3.04 \times 10^{-19} \text{ J} \)

Now, relate energy to wavelength \( \lambda \):

\( E = \frac{hc}{\lambda} \implies \lambda = \frac{hc}{E} \)

\( \lambda = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{3.04 \times 10^{-19}} \approx 6.54 \times 10^{-7} \text{ m} = 654 \text{ nm} \)

This is closest to \( 650 \text{ nm} \).

[1 mark] Award for calculating energy transition in Joules and applying de Broglie / photon energy wavelength relationship.

Initial momentum is \(p_i = mv\). By conservation of linear momentum, the final momentum is \(p_f = (m + 3m)v_f = 4mv_f\). Therefore, the final velocity is \(v_f = \frac{v}{4}\). The initial kinetic energy is \(E_{ki} = \frac{1}{2}mv^2\). The final kinetic energy is \(E_{kf} = \frac{1}{2}(4m)v_f^2 = \frac{1}{2}(4m)\left(\frac{v}{4}\right)^2 = \frac{1}{8}mv^2\). The kinetic energy dissipated is \(\Delta E_k = E_{ki} - E_{kf} = \frac{1}{2}mv^2 - \frac{1}{8}mv^2 = \frac{3}{8}mv^2\). The fraction dissipated is \(\frac{\Delta E_k}{E_{ki}} = \frac{\frac{3}{8}mv^2}{\frac{1}{2}mv^2} = \frac{3}{4} = 75\%\).

Award [1] for the correct option C. Reject other options because they do not correctly represent the ratio of energy dissipated to initial energy.

From the ideal gas law, \(PV = nRT\), which implies \(P = \frac{nRT}{V}\). If the initial pressure is \(P_1 = \frac{nRT_1}{V_1}\), the final pressure is \(P_2 = \frac{nR(2T_1)}{\frac{V_1}{2}} = 4 \left(\frac{nRT_1}{V_1}\right) = 4P_1\). Thus, the ratio of final to initial pressure is 4.

Award [1] for the correct option C. Incorrect options result from neglecting the volume change (giving 2) or inversing the relation.

The fringe spacing formula is given by \(s = \frac{\lambda D}{d}\). For the new conditions, the new fringe spacing is \(s' = \frac{\lambda' D'}{d'}\). Substituting the new values: \(s' = \frac{\left(\frac{\lambda}{2}\right) (2D)}{\frac{d}{2}} = 2 \left(\frac{\lambda D}{d}\right) = 2s\).

Award [1] for the correct option C.

The total mechanical energy of a satellite in a circular orbit of radius \(r\) is \(E_i = -\frac{GMm}{2r}\). Here, \(r = 3R\), so \(E_i = -\frac{GMm}{6R}\). To move the satellite to an infinite distance where its total mechanical energy is at least \(E_f = 0\), the energy required is \(\Delta E = E_f - E_i = \frac{GMm}{6R}\). Since the surface gravitational acceleration is \(g = \frac{GM}{R^2}\), we have \(GM = gR^2\). Substituting this gives \(\Delta E = \frac{mgR^2}{6R} = \frac{mgR}{6}\).

Award [1] for the correct option A.

After three half-lives, the fraction of active nuclei remaining is \(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\). Since activity \(A\) is directly proportional to the number of active nuclei \(N\), the remaining fraction of activity is also \(\frac{1}{8}\). The fraction of original nuclei that have decayed is given by \(1 - \text{fraction remaining} = 1 - \frac{1}{8} = \frac{7}{8}\).

Award [1] for the correct option A.

The terminal potential difference is given by \(V = E - Ir = E\left(\frac{R}{R+r}\right)\). As \(R\) increases, \(V\) increases steadily toward \(E\). The power dissipated in the external resistor is given by \(P = I^2 R = \frac{E^2 R}{(R+r)^2}\). According to the maximum power transfer theorem, this power reaches a maximum value when \(R = r\). Thus, as \(R\) increases from \(R \ll r\) to \(R \gg r\), \(P\) first increases to a maximum and then decreases.

Award [1] for the correct option B.

The photon energy is equal to the energy difference between levels: \(\Delta E = E_i - E_f = -1.51\text{ eV} - (-3.40\text{ eV}) = 1.89\text{ eV}\). In Joules, this is \(\Delta E = 1.89 \times 1.60 \times 10^{-19}\text{ J} = 3.024 \times 10^{-19}\text{ J}\). The frequency \(f\) of the photon is given by \(f = \frac{\Delta E}{h} = \frac{3.024 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 4.56 \times 10^{14}\text{ Hz}\), which is \(4.6 \times 10^{14}\text{ Hz}\) to two significant figures.

Award [1] for the correct option A.

Using Einstein's photoelectric equation: \(e V_s = hf - \Phi \implies V_s = \frac{hf}{e} - \frac{\Phi}{e}\). When the frequency is doubled, the new stopping potential is \(V_s' = \frac{h(2f)}{e} - \frac{\Phi}{e} = 2\left(\frac{hf}{e}\right) - \frac{\Phi}{e}\). We can rewrite this as \(V_s' = 2\left(V_s + \frac{\Phi}{e}\right) - \frac{\Phi}{e} = 2V_s + \frac{\Phi}{e}\). Since work function \(\Phi\) and elementary charge \(e\) are positive constants, \(\frac{\Phi}{e} > 0\), and therefore \(V_s' > 2V_s\).

Award [1] for the correct option C.

(a) Conduction (through the low-conductivity gas gap between plates) and convection (by restricting circulation in the narrow gap).

(b) Total solar power incident:
\( P_{\text{incident}} = I \times A = 850\text{ W m}^{-2} \times 4.2\text{ m}^2 = 3570\text{ W} \)
Useful thermal energy rate:
\( P_{\text{useful}} = P_{\text{incident}} \times \text{efficiency} = 3570 \times 0.65 = 2320.5\text{ W} \approx 2300\text{ W} \) (or \( 2.3\text{ kW} \)).

(c) Using \( Q = m c \Delta T \), the power is \( P = \frac{\Delta m}{\Delta t} c \Delta T \).
\( 2320.5 = \frac{\Delta m}{\Delta t} \times 4180 \times (42 - 18) \)
\( 2320.5 = \frac{\Delta m}{\Delta t} \times 4180 \times 24 = \frac{\Delta m}{\Delta t} \times 100320 \)
\( \frac{\Delta m}{\Delta t} = \frac{2320.5}{100320} \approx 0.0231\text{ kg s}^{-1} \approx 0.023\text{ kg s}^{-1} \).

(d) There are thermal energy losses to the surrounding atmosphere from the outer surfaces of the panel (due to conduction, convection, and thermal radiation). Therefore, not all useful power goes into heating the water, resulting in a lower exit temperature.

(a) [2 marks max]
- Conduction [1 mark]
- Convection [1 mark]

(b) [3 marks max]
- Calculation of total incident power (3570 W) [1 mark]
- Correct multiplication by efficiency (0.65) [1 mark]
- Final answer of 2300 W or 2.3 kW with correct unit [1 mark]

(c) [3 marks max]
- Recall of \( P = (\Delta m / \Delta t) c \Delta T \) [1 mark]
- Correct substitution of values including \( \Delta T = 24\text{ K} \) [1 mark]
- Correct calculation of flow rate (0.023) and unit (\( \text{kg s}^{-1} \)) [1 mark]

(d) [3 marks max]
- Identifies energy loss mechanism (e.g., radiation, wind convection) [1 mark]
- Explains how this loss reduces net energy transferred to water [1 mark]
- Relates this clearly to a lower-than-calculated temperature change [1 mark]

(a) Resolving forces parallel to the incline:
\( F_{\text{net}} = m_1 g \sin\theta - f \)
Where \( f = \mu_k R = \mu_k m_1 g \cos\theta \).
So, \( a = g \sin\theta - \mu_k g \cos\theta \)
\( a = 9.81 \sin(30^\circ) - 0.15 \times 9.81 \cos(30^\circ) \)
\( a = 4.905 - 1.274 = 3.631\text{ m s}^{-2} \approx 3.6\text{ m s}^{-2} \).

(b) Using the equations of motion:
\( v^2 = u^2 + 2as \)
\( v^2 = 0 + 2 \times 3.631 \times 3.0 = 21.786 \)
\( v = \sqrt{21.786} \approx 4.67\text{ m s}^{-1} \approx 4.7\text{ m s}^{-1} \).

(c) Applying conservation of momentum during the collision:
\( m_1 v = (m_1 + m_2) V \)
\( 2.5 \times 4.67 = (2.5 + 1.5) V \)
\( 11.675 = 4.0 V \)
\( V = 2.92\text{ m s}^{-1} \approx 2.9\text{ m s}^{-1} \).

(d) Momentum is conserved because there are no net external horizontal forces acting on the two-block system during the collision. Kinetic energy is not conserved because the collision is completely inelastic; mechanical energy is transformed into thermal energy and sound.

(a) [3 marks max]
- Correct force equation parallel to incline [1 mark]
- Correct substitution of normal reaction force / friction term [1 mark]
- Correct calculations showing the final answer is 3.6 m/s^2 [1 mark]

(b) [2 marks max]
- Correct use of SUVAT equation [1 mark]
- Correct final velocity (4.7 m/s) with unit [1 mark]

(c) [3 marks max]
- Correct statement of momentum conservation equation [1 mark]
- Correct substitution of masses and initial velocity [1 mark]
- Correct final velocity (2.9 m/s) [1 mark]

(d) [3 marks max]
- Correctly states momentum is conserved and kinetic energy is not [1 mark]
- Justifies momentum conservation (no external force) [1 mark]
- Justifies non-conservation of kinetic energy (inelastic collision/loss to heat/sound) [1 mark]

(a) Coherent sources have a constant phase difference and the same frequency.

(b) The distance to the fifth bright fringe is \( 5s = 4.75\text{ cm} \).
Therefore, the fringe separation \( s = \frac{4.75 \times 10^{-2}\text{ m}}{5} = 9.50 \times 10^{-3}\text{ m} \).
Using the double-slit formula: \( s = \frac{\lambda D}{d} \)
\( \lambda = \frac{s d}{D} = \frac{9.50 \times 10^{-3}\text{ m} \times 0.12 \times 10^{-3}\text{ m}}{1.8\text{ m}} \)
\( \lambda = \frac{1.14 \times 10^{-6}}{1.8} = 6.33 \times 10^{-7}\text{ m} \approx 6.3 \times 10^{-7}\text{ m} \) (or \( 633\text{ nm} \)).

(c) The wavelength in the liquid becomes \( \lambda' = \frac{\lambda}{n} = \frac{\lambda}{1.33} \).
Since fringe spacing is proportional to the wavelength (\( s \propto \lambda \)), the fringe spacing decreases to \( s' = \frac{s}{1.33} \approx 7.1\text{ mm} \).

(d) For the single slit, instead of multiple narrow fringes of equal intensity, there will be a single broad central maximum that is twice the width of the subsidiary maxima. The intensity of successive maxima decreases rapidly on either side of the central maximum.

(a) [2 marks max]
- Constant phase difference [1 mark]
- Same frequency / wavelength [1 mark]

(b) [3 marks max]
- Correctly identifies fringe spacing \( s = 9.5\text{ mm} \) [1 mark]
- Correct rearrangement of double slit formula [1 mark]
- Correct final wavelength calculation with unit [1 mark]

(c) [3 marks max]
- States that wavelength decreases inside liquid [1 mark]
- Relates decrease in wavelength to decrease in fringe spacing [1 mark]
- Quantifies change correctly (fringe spacing decreases to ~7.1 mm / by factor of 1.33) [1 mark]

(d) [3 marks max]
- Central maximum becomes broader than other maxima [1 mark]
- Intensity of outer maxima decreases progressively [1 mark]
- No closely spaced double-slit interference fringes are seen [1 mark]

(a) The work function is the minimum energy required to liberate an electron from the surface of a metal.

(b) Energy of the incident photon:
\( E = hf = 6.63 \times 10^{-34}\text{ J s} \times 7.50 \times 10^{14}\text{ Hz} = 4.9725 \times 10^{-19}\text{ J} \)
Converting to eV:
\( E = \frac{4.9725 \times 10^{-19}}{1.60 \times 10^{-19}} \approx 3.11\text{ eV} \)
Using Einstein's photoelectric equation:
\( E_k = hf - \Phi \implies 1.20\text{ eV} = 3.11\text{ eV} - \Phi \implies \Phi = 3.11 - 1.20 = 1.91\text{ eV} \).

(c)
(i) Since \( f \) is doubled, photon energy becomes \( 2 \times 3.11\text{ eV} = 6.22\text{ eV} \). The new maximum kinetic energy is \( 6.22 - 1.91 = 4.31\text{ eV} \). Thus, maximum kinetic energy increases by more than a factor of two.
(ii) Since the intensity is constant but each photon carries twice the energy, the number of incident photons per second is halved. Thus, the rate of emission of photoelectrons decreases.

(d) Electrons transition between specific discrete orbits. During a transition to a lower state, a photon of energy \( E = hf = E_{\text{initial}} - E_{\text{final}} \) is emitted. Since only certain discrete wavelengths are observed in the spectrum, the energy differences must be discrete, proving that the energy levels themselves are quantized.

(a) [2 marks max]
- Minimum energy [1 mark]
- To remove an electron from the metal surface [1 mark]

(b) [3 marks max]
- Correct photon energy in Joules or eV [1 mark]
- Correct application of photoelectric equation [1 mark]
- Correct final value of work function (1.91 eV) [1 mark]

(c) [3 marks max]
- Max kinetic energy increases (to 4.31 eV / by more than double) with correct reasoning [1 mark]
- Number of photons per second decreases (because intensity is constant but energy per photon increases) [1 mark]
- Hence, number of emitted photoelectrons per second decreases [1 mark]

(d) [3 marks max]
- Photons are emitted when electrons drop to lower energy levels [1 mark]
- Observed spectral lines correspond to specific, distinct photon energies [1 mark]
- This implies only specific transitions/energy levels are permitted [1 mark]

(a) Lenz's law states that the direction of an induced current is such that it opposes the change in magnetic flux that produced it. If the induced current did not oppose the change, the magnetic force would accelerate the magnet/coil, creating kinetic energy and electrical energy from nothing, which violates the law of conservation of energy.

(b) Area of the square coil: \( A = 0.050\text{ m} \times 0.050\text{ m} = 2.5 \times 10^{-3}\text{ m}^2 \).
Initial magnetic flux: \( \Phi = B A = 0.45\text{ T} \times 2.5 \times 10^{-3}\text{ m}^2 = 1.125 \times 10^{-3}\text{ Wb} \).
Initial flux linkage: \( N\Phi = 120 \times 1.125 \times 10^{-3} = 0.135\text{ Wb-turns} \).
Using Faraday's Law:
\( \mathcal{E} = \frac{\Delta (N\Phi)}{\Delta t} = \frac{0.135 - 0}{0.15\text{ s}} = 0.90\text{ V} \).

(c) Average power dissipated:
\( P = \frac{\mathcal{E}^2}{R} = \frac{(0.90)^2}{3.5} = \frac{0.81}{3.5} \approx 0.231\text{ W} \approx 0.23\text{ W} \).

(d) As the coil is pulled out, a current is induced in it. This induced current experiences a magnetic force from the magnetic field that opposes the motion (by Lenz's Law). To overcome this magnetic force and maintain constant velocity, an equal and opposite external force must be applied.

(a) [3 marks max]
- Defines Lenz's law (induced emf opposes change in flux) [1 mark]
- States that if it assisted, energy would be created from nothing [1 mark]
- Concludes opposition represents work done by external agent converted to electrical energy [1 mark]

(b) [3 marks max]
- Correct calculation of area (\( 2.5 \times 10^{-3}\text{ m}^2 \)) [1 mark]
- Correct calculation of flux linkage (0.135 Wb-turns) [1 mark]
- Correct application of Faraday's Law to obtain 0.90 V [1 mark]

(c) [3 marks max]
- Recalls \( P = \mathcal{E}^2/R \) (or calculates \( I = 0.257\text{ A} \) then uses \( I^2 R \)) [1 mark]
- Substitute values correctly [1 mark]
- Correct final power with unit (0.23 W) [1 mark]

(d) [2 marks max]
- Moving coil has induced current which experiences a magnetic force opposing motion [1 mark]
- External force is required to balance this magnetic drag force to maintain constant speed [1 mark]

(a) The centripetal force is provided by the gravitational force:
\( F_c = F_g \implies \frac{m v^2}{r} = \frac{G M m}{r^2} \)
Solving for \( v \):
\( v^2 = \frac{G M}{r} \implies v = \sqrt{\frac{G M}{r}} \).

(b) Radius of orbit: \( r = R_E + h = 6.37 \times 10^6\text{ m} + 3.50 \times 10^5\text{ m} = 6.72 \times 10^6\text{ m} \).
Orbital speed:
\( v = \sqrt{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.72 \times 10^6}} = \sqrt{5.9256 \times 10^7} \approx 7698\text{ m s}^{-1} \).
Orbital period:
\( T = \frac{2 \pi r}{v} = \frac{2 \pi \times 6.72 \times 10^6}{7698} \approx 5485\text{ s} \approx 5.49 \times 10^3\text{ s} \) (approx. 91.4 minutes).

(c) \( \Delta E_p = -\frac{G M m}{r_{\text{final}}} - \left( -\frac{G M m}{r_{\text{initial}}} \right) = G M m \left( \frac{1}{r_i} - \frac{1}{r_f} \right) \)
\( G M m = 6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 1200 = 4.778 \times 10^{17}\text{ J m} \).
\( \Delta E_p = 4.778 \times 10^{17} \left( \frac{1}{6.72 \times 10^6} - \frac{1}{4.22 \times 10^7} \right) \)
\( \Delta E_p = 4.778 \times 10^{17} \left( 1.488 \times 10^{-7} - 2.370 \times 10^{-8} \right) = 4.778 \times 10^{17} \times 1.251 \times 10^{-7} \approx 5.98 \times 10^{10}\text{ J} \approx 6.0 \times 10^{10}\text{ J} \).

(d) Kinetic energy is given by \( E_k = \frac{1}{2} m v^2 = \frac{G M m}{2 r} \). Since \( r \) increases, the kinetic energy must decrease.

(a) [2 marks max]
- Equates centripetal force to gravitational force [1 mark]
- Correct algebraic simplification to obtain final expression [1 mark]

(b) [4 marks max]
- Calculates correct orbital radius (\( 6.72 \times 10^6\text{ m} \)) [1 mark]
- Calculates orbital velocity (\( 7.7 \times 10^3\text{ m/s} \)) [1 mark]
- Expresses relation between T, r, and v correctly [1 mark]
- Obtains final correct period (\( 5.49 \times 10^3\text{ s} \)) [1 mark]

(c) [3 marks max]
- Recalls \( E_p = -\frac{GMm}{r} \) [1 mark]
- Substitutes initial and final values correctly [1 mark]
- Calculates change as positive value \( 6.0 \times 10^{10}\text{ J} \) (accept \( 5.98 \times 10^{10}\text{ J} \)) [1 mark]

(d) [2 marks max]
- Recalls \( v \propto \frac{1}{\sqrt{r}} \) or \( E_k \propto \frac{1}{r} \) [1 mark]
- Explains that as orbit radius increases, kinetic energy decreases [1 mark]

(a) As the ambulance moves, it travels in the direction of the emitted sound waves. This causes the wavefronts ahead of it to be compressed (crowded closer together), reducing the observed wavelength. Since the speed of sound in the medium is constant, this reduced wavelength results in a higher observed frequency.

(b) Using the Doppler effect formula for an approaching source:
\( f' = f \left( \frac{v}{v - v_s} \right) \)
\( f' = 440 \left( \frac{343}{343 - 28} \right) = 440 \left( \frac{343}{315} \right) = 440 \times 1.0889 \approx 479.1\text{ Hz} \approx 479\text{ Hz} \).

(c) For the departing source:
\( f'' = f \left( \frac{v}{v + v_s} \right) \)
\( f'' = 440 \left( \frac{343}{343 + 28} \right) = 440 \left( \frac{343}{371} \right) = 440 \times 0.9245 \approx 406.8\text{ Hz} \approx 407\text{ Hz} \).
Change in frequency:
\( \Delta f = f' - f'' = 479.1 - 406.8 = 72.3\text{ Hz} \approx 72\text{ Hz} \).

(d) Sound requires a material medium (air), and the effect depends on whether the source or observer is moving relative to this medium. Light is an electromagnetic wave that requires no medium, so its Doppler effect depends only on the relative speed between the source and observer (special relativity description).

(a) [3 marks max]
- Wavefronts are compressed / crowded ahead of the source [1 mark]
- Wavelength decreases [1 mark]
- Relates constant wave speed to increased observed frequency [1 mark]

(b) [3 marks max]
- Recalls Doppler formula for approaching source [1 mark]
- Substitutes values correctly (343 and 28) [1 mark]
- Correct final answer (479 Hz) [1 mark]

(c) [3 marks max]
- Recalls Doppler formula for receding source [1 mark]
- Correctly calculates receding frequency (407 Hz) [1 mark]
- Calculates difference correctly (72 Hz) [1 mark]

(d) [2 marks max]
- Relates sound Doppler to motion relative to medium (air) [1 mark]
- Relates light Doppler to relative motion of source and observer directly (independent of a medium) [1 mark]

(a) Assumptions include:
1. Molecules are point masses with negligible volume compared to the container volume.
2. No intermolecular forces except during collisions.
3. Collisions are perfectly elastic.
4. Molecules are in continuous random motion.

(b) Using the ideal gas equation:
\( P V = n R T \)
\( 1.20 \times 10^5 \times 5.00 \times 10^{-3} = 0.25 \times 8.31 \times T \)
\( 600 = 2.0775 \times T \)
\( T = \frac{600}{2.0775} \approx 288.8\text{ K} \approx 289\text{ K} \).

(c) For an isothermal process:
\( P_1 V_1 = P_2 V_2 \)
\( 1.20 \times 10^5 \times 5.00 \times 10^{-3} = P_2 \times 2.00 \times 10^{-3} \)
\( P_2 = 1.20 \times 10^5 \times \frac{5}{2} = 3.00 \times 10^5\text{ Pa} \).

(d) Since the temperature is constant, the average speed (or average kinetic energy) of the gas molecules remains the same. When the volume is reduced, the molecules are packed into a smaller space, meaning they travel shorter distances between collisions. This leads to a higher frequency of collisions with the walls of the cylinder, increasing the total rate of momentum change per unit area (pressure).

(a) [3 marks max]
- Award [1 mark] for each valid assumption (up to 3 marks) [3 marks]

(b) [3 marks max]
- Recalls \( PV = nRT \) [1 mark]
- Correct substitution of values [1 mark]
- Correct temperature calculation (289 K) [1 mark]

(c) [2 marks max]
- Recalls Boyle's law relation \( P_1V_1 = P_2V_2 \) [1 mark]
- Correct calculation to obtain \( 3.00 \times 10^5\text{ Pa} \) [1 mark]

(d) [3 marks max]
- Constant temperature means constant molecular speed / KE [1 mark]
- Smaller volume means higher frequency of wall collisions [1 mark]
- Connects increased rate of collisions to greater force/momentum change per unit area [1 mark]

\( \textbf{(a)(i)} \)
\( V = 10.0\text{ cm}^3 = 10.0 \times 10^{-6}\text{ m}^3 \).
Therefore, \( 1/V = \frac{1}{10.0 \times 10^{-6}} = 1.00 \times 10^5\text{ m}^{-3} \).

\( \textbf{(a)(ii)} \)
The fractional uncertainty in \( V \) is:
\( \frac{\Delta V}{V} = \frac{0.5}{10.0} = 0.05 \) (or 5%).
Since \( y = 1/V \), the fractional uncertainty in \( 1/V \) is also 5%.
Absolute uncertainty \( \Delta(1/V) = 0.05 \times 1.00 \times 10^5\text{ m}^{-3} = 5 \times 10^3\text{ m}^{-3} \).

\( \textbf{(b)} \)
The ideal gas equation is \( PV = nRT \), which can be written as \( P = (nRT) \frac{1}{V} \).
Since \( n \), \( R \), and \( T \) are constant, \( P \) is directly proportional to \( 1/V \). Thus, a graph of \( P \) against \( 1/V \) yields a straight line through the origin with gradient \( nRT \).

\( \textbf{(c)} \)
From \( P = (nRT) \frac{1}{V} \), the gradient is \( m = nRT \).
Therefore, \( n = \frac{m}{RT} = \frac{2.40}{8.31 \times 295} = 9.79 \times 10^{-4}\text{ mol} \).

To find the uncertainty in \( n \):
Fractional uncertainty in \( m \): \( \frac{\Delta m}{m} = \frac{0.08}{2.40} \approx 0.0333 \) (3.33%).
Fractional uncertainty in \( T \): \( \frac{\Delta T}{T} = \frac{1}{295} \approx 0.0034 \) (0.34%).

Using the sum of fractional uncertainties:
\( \frac{\Delta n}{n} = 0.0333 + 0.0034 = 0.0367 \) (3.67%).
\( \Delta n = 0.0367 \times 9.79 \times 10^{-4}\text{ mol} \approx 0.36 \times 10^{-4}\text{ mol} \approx 0.4 \times 10^{-4}\text{ mol} \).

Using the quadratic sum (quadrature):
\( \frac{\Delta n}{n} = \sqrt{0.0333^2 + 0.0034^2} \approx 0.0335 \) (3.35%).
\( \Delta n = 0.0335 \times 9.79 \times 10^{-4}\text{ mol} \approx 0.33 \times 10^{-4}\text{ mol} \approx 0.3 \times 10^{-4}\text{ mol} \).

Thus, \( n = (9.8 \pm 0.4) \times 10^{-4}\text{ mol} \) or \( (9.8 \pm 0.3) \times 10^{-4}\text{ mol} \).

\( \textbf{(d)} \)
Rapid compression is adiabatic, meaning work done on the gas increases its internal energy and temperature. As volume decreases (larger \( 1/V \)), the temperature rises above room temperature. This causes the pressure to be higher than expected at high values of \( 1/V \), causing the graph to curve upwards and resulting in an overestimated average gradient.

\( \textbf{(a)(i)} \)
[1 mark] \( 1.00 \times 10^5\text{ m}^{-3} \) (accept \( 1 \times 10^5 \) but award 0 if power of ten is incorrect).

\( \textbf{(a)(ii)} \)
[1 mark] \( 5 \times 10^3\text{ m}^{-3} \) (or \( 0.05 \times 10^5\text{ m}^{-3} \)). Allow ECF from (a)(i).

\( \textbf{(b)} \)
[1 mark] States \( PV = nRT \) and shows that \( P \propto 1/V \) because \( n, R, T \) are constant.

\( \textbf{(c)} \)
[1 mark] Correct calculation of \( n = 9.8 \times 10^{-4}\text{ mol} \) (accept \( 9.79 \times 10^{-4} \)).
[1 mark] Correct calculation of fractional uncertainty of \( n \) (either 3.4% to 3.7%).
[1 mark] Expresses final answer with absolute uncertainty to 1 or 2 sig figs: \( (9.8 \pm 0.4) \times 10^{-4}\text{ mol} \) or \( (9.8 \pm 0.3) \times 10^{-4}\text{ mol} \).

\( \textbf{(d)} \)
[1 mark] Identifies that temperature of the gas will rise during rapid compression.
[1 mark] Explains that this increases pressure at low volumes (high \( 1/V \)), leading to a larger average gradient.

\( \textbf{(a)} \)
Independent variable: Hanging mass \( m \).
Dependent variable: Acceleration \( a \).

\( \textbf{(b)} \)
To keep the total mass constant, masses are transferred between the trolley and the hanging mass hanger (rather than adding new mass from outside the system).

\( \textbf{(c)} \)
According to Newton's second law, the net force on the system is the weight of the hanging mass: \( F_{\text{net}} = mg \).
This force accelerates the entire system of mass \( M_T \):
\( mg = M_T a \implies a = \left(\frac{g}{M_T}\right) m \).
Since \( g \) and \( M_T \) are constant, \( a \) is directly proportional to \( m \), yielding a straight line with a gradient of \( \frac{g}{M_T} \).

\( \textbf{(d)} \)
Ratio \( R = \frac{a}{m} = \frac{2.04}{0.250} = 8.16\text{ m s}^{-2}\text{ kg}^{-1} \).

Percentage uncertainty in \( m \): \( \frac{0.001}{0.250} \times 100\% = 0.4\% \).
Percentage uncertainty in \( a \): \( \frac{0.05}{2.04} \times 100\% \approx 2.45\% \).

Total percentage uncertainty (by summing fractional uncertainties):
\( 0.4\% + 2.45\% = 2.85\% \approx 2.9\% \).
(By quadrature: \( \sqrt{0.4^2 + 2.45^2}\% \approx 2.48\% \approx 2.5\% \)).

Absolute uncertainty in \( R \):
- Using sum: \( 8.16 \times 0.0285 = 0.23\text{ m s}^{-2}\text{ kg}^{-1} \).
- Using quadrature: \( 8.16 \times 0.0248 = 0.20\text{ m s}^{-2}\text{ kg}^{-1} \).

\( \textbf{(e)} \)
Gradient \( = \frac{g}{M_T} \).
\( 8.16 = \frac{g}{1.20} \implies g = 8.16 \times 1.20 = 9.79\text{ m s}^{-2} \).

\( \textbf{(a)} \)
[1 mark] Correctly identifies both variables (Independent: hanging mass \( m \), Dependent: acceleration \( a \)).

\( \textbf{(b)} \)
[1 mark] States that masses are transferred between the trolley and the hanger.

\( \textbf{(c)} \)
[1 mark] Derives \( a = \left(\frac{g}{M_T}\right) m \) using \( F = ma \).
[1 mark] Identifies that the gradient is \( \frac{g}{M_T} \).

\( \textbf{(d)} \)
[1 mark] Correctly calculates percentage uncertainty of \( 2.9\% \) (or \( 2.5\% \) using quadrature).
[1 mark] Correctly calculates absolute uncertainty of \( 0.23\text{ m s}^{-2}\text{ kg}^{-1} \) (or \( 0.20\text{ m s}^{-2}\text{ kg}^{-1} \) using quadrature).

\( \textbf{(e)} \)
[1 mark] Correctly calculates \( g = 9.79\text{ m s}^{-2} \) (accept \( 9.8\text{ m s}^{-2} \)).

(a) Using the stellar parallax formula:
\( d = \frac{1}{p} = \frac{1}{0.040} = 25 \text{ pc} \)

Converting to meters using \( 1 \text{ pc} = 3.086 \times 10^{16} \text{ m} \):
\( d = 25 \times 3.086 \times 10^{16} \text{ m} = 7.715 \times 10^{17} \text{ m} \approx 7.7 \times 10^{17} \text{ m} \)

(b) Using the relationship between apparent brightness \( b \), luminosity \( L \), and distance \( d \):
\( b = \frac{L}{4\pi d^2} \Rightarrow L = 4\pi d^2 b \)
\( L = 4\pi \times (7.715 \times 10^{17} \text{ m})^2 \times (3.2 \times 10^{-11} \text{ W m}^{-2}) \)
\( L = 2.40 \times 10^{26} \text{ W} \approx 2.4 \times 10^{26} \text{ W} \)

(c) (i) According to Wien's displacement law, peak wavelength depends only on temperature: \( \lambda_{\text{max}} \propto \frac{1}{T} \). Since both stars have the same surface temperature, the peak emission wavelength is the same, so \( \frac{\lambda_{\text{B}}}{\lambda_{\text{A}}} = 1 \).
(ii) By Stefan-Boltzmann law, \( L = 4\pi R^2 \sigma T^4 \). Since temperature is the same, \( L \propto R^2 \):
\( \frac{L_{\text{B}}}{L_{\text{A}}} = \left(\frac{R_{\text{B}}}{R_{\text{A}}}\right)^2 = 3^2 = 9 \)

(d) Ground-based observations are limited by the Earth's atmosphere which causes light to refract randomly (atmospheric turbulence or "seeing"). This blurs the stellar image, making it impossible to measure extremely small parallax angles (for stars beyond approximately 100 pc) with high accuracy.

(a) [2 marks]
- Correct calculation of distance in parsecs: \( d = 25 \text{ pc} \) [1 mark]
- Correct conversion to meters: \( 7.7 \times 10^{17} \text{ m} \) (accept \( 7.72 \times 10^{17} \text{ m} \) or \( 7.73 \times 10^{17} \text{ m} \)) [1 mark]

(b) [2 marks]
- Correct substitution into luminosity formula: \( L = 4\pi \times (7.7 \times 10^{17})^2 \times 3.2 \times 10^{-11} \) [1 mark]
- Correct final value: \( 2.4 \times 10^{26} \text{ W} \) (accept range \( 2.38 \times 10^{26} \text{ W} \) to \( 2.42 \times 10^{26} \text{ W} \)) [1 mark]

(c) [2 marks]
- (i) State ratio of 1 [1 mark]
- (ii) Correct calculation of luminosity ratio as 9 [1 mark]

(d) [2 marks]
- Identifies atmospheric turbulence/blurring/"seeing" as the source of error [1 mark]
- Explains that this makes very small angles (or large distances) unresolvable from the ground [1 mark]

(a) The standard axes of an H-R diagram are:
- Vertical axis: Luminosity (or Absolute Magnitude), which increases upwards. [1 mark]
- Horizontal axis: Surface Temperature (or Spectral Class), which increases to the left (i.e., decreases to the right). [1 mark]

(b) (i) Using the mass-luminosity relation \( L \propto M^{3.5} \):
\( \frac{L}{L_{\odot}} = \left( \frac{M}{M_{\odot}} \right)^{3.5} = (4.0)^{3.5} = 128 \)
So, the luminosity is \( 128 \, L_{\odot} \). [1 mark]

(ii) Massive stars have a much greater mass, which creates much stronger gravitational forces in their cores. To remain in hydrostatic equilibrium, the core temperature and pressure must be much higher, leading to a vastly higher rate of fusion (as shown by \( L \propto M^{3.5} \)). Because they burn through their fuel supply at a rate that is disproportionately high compared to their mass, their lifetime on the main sequence is much shorter. [2 marks]

(c) Because the mass of the star is less than the Chandrasekhar limit (its initial mass \( 4.0 \, M_{\odot} < 8 \, M_{\odot} \)), it will not undergo a supernova. Instead, it will shed its outer layers as a planetary nebula, leaving behind a dense core. The final evolutionary state is a white dwarf, which is prevented from collapsing further by electron degeneracy pressure. [2 marks]

(a) [2 marks]
- Vertical axis identified as Luminosity / Absolute Magnitude AND specified as increasing upwards [1 mark]
- Horizontal axis identified as Temperature / Spectral Class AND specified as decreasing to the right (or increasing to the left) [1 mark]

(b) [3 marks]
- (i) \( 128 \, L_{\odot} \) (allow \( 130 \, L_{\odot} \) or using \( L \propto M^3 \) to give \( 64 \, L_{\odot} \) or \( L \propto M^4 \) to give \( 256 \, L_{\odot} \) if formula stated) [1 mark]
- (ii) Reference to much higher core temperatures/pressures needed for equilibrium [1 mark]
- (ii) Mentions that fuel consumption rate (fusion rate / luminosity) is vastly higher, outweighing the extra fuel mass [1 mark]

(c) [2 marks]
- Identifies the final state as a white dwarf [1 mark]
- Mentions that the outer layers are ejected (planetary nebula) OR states that it is supported by electron degeneracy pressure [1 mark]

(a) Redshift \( z \) is given by:
\( z = \frac{\Delta\lambda}{\lambda_0} = \frac{\lambda - \lambda_0}{\lambda_0} \)
\( z = \frac{682.6 \text{ nm} - 656.3 \text{ nm}}{656.3 \text{ nm}} = \frac{26.3}{656.3} \approx 0.04007 \approx 0.040 \)

(b) Using the relationship for relatively low speed recession (\( z < 0.1 \)):
\( z \approx \frac{v}{c} \)
\( v = z c = 0.04007 \times 3.00 \times 10^8 \text{ m s}^{-1} = 1.202 \times 10^7 \text{ m s}^{-1} \approx 1.2 \times 10^7 \text{ m s}^{-1} \)

(c) Convert the speed into \( \text{km s}^{-1} \):
\( v = 1.202 \times 10^4 \text{ km s}^{-1} \)
Using Hubble's Law:
\( v = H_0 d \Rightarrow d = \frac{v}{H_0} \)
\( d = \frac{1.202 \times 10^4 \text{ km s}^{-1}}{70 \text{ km s}^{-1} \text{ Mpc}^{-1}} \approx 171.7 \text{ Mpc} \approx 170 \text{ Mpc} \)

(d) Cosmological redshift occurs because space itself is expanding as light travels through it, stretching the wavelength of the light waves. Doppler redshift, on the other hand, is due to the physical movement of a source relative to the observer through a static space.

(a) [2 marks]
- Correct calculation of difference in wavelength: \( 26.3 \text{ nm} \) [1 mark]
- Correct calculation of redshift: \( 0.040 \) [1 mark]

(b) [2 marks]
- Correct substitution into \( v = zc \) [1 mark]
- Correct calculation of speed: \( 1.2 \times 10^7 \text{ m s}^{-1} \) (accept \( 1.20 \times 10^7 \text{ m s}^{-1} \) to \( 1.21 \times 10^7 \text{ m s}^{-1} \)) [1 mark]

(c) [2 marks]
- Correct conversion of speed to km/s (\( 1.2 \times 10^4 \text{ km s}^{-1} \)) and substitution into Hubble's law [1 mark]
- Correct distance: \( 170 \text{ Mpc} \) or \( 171 \text{ Mpc} \) (accept range \( 170 \) to \( 173 \)) [1 mark]

(d) [2 marks]
- Identifies cosmological redshift as being caused by the expansion of space itself stretching the light [1 mark]
- Identifies Doppler redshift as being caused by local motion of the emitter relative to the observer through space [1 mark]

(a) Two characteristics of the Cosmic Microwave Background (CMB) radiation are:
1. It is highly isotropic (extremely uniform in intensity and temperature across all directions in the sky).
2. Its spectrum is an almost perfect blackbody spectrum.

(b) (i) Wien's displacement law is:
\( \lambda_{\text{max}} T = 2.90 \times 10^{-3} \text{ m K} \)
Rearranging to solve for temperature \( T \):
\( T = \frac{2.90 \times 10^{-3} \text{ m K}}{\lambda_{\text{max}}} \)
\( T = \frac{2.90 \times 10^{-3} \text{ m K}}{1.06 \times 10^{-3} \text{ m}} \approx 2.74 \text{ K} \approx 2.7 \text{ K} \)
This shows that the current temperature is approximately \( 2.7 \text{ K} \).

(ii) The temperature \( T \) of cosmic radiation is inversely proportional to the scale factor \( R \) of the universe:
\( T \propto \frac{1}{R} \)
Therefore:
\( \frac{R_{\text{now}}}{R_{\text{rec}}} = \frac{T_{\text{rec}}}{T_{\text{now}}} \)
\( \frac{R_{\text{now}}}{R_{\text{rec}}} = \frac{3000 \text{ K}}{2.73 \text{ K}} \approx 1100 \) (or \( 1111 \) using \( 2.7 \text{ K} \))

(c) For a cloud of gas of a given mass \( M \) to begin gravitational collapse, the actual mass \( M \) of the gas cloud must be greater than the Jeans mass \( M_{\text{J}} \) (i.e., \( M > M_{\text{J}} \)).

(a) [2 marks]
- Identifies that CMB is isotropic / homogeneous / same in all directions (to high precision) [1 mark]
- Identifies that CMB has a blackbody / Planck spectrum (with peak in microwave region) [1 mark]

(b) [4 marks]
- (i) Clear rearrangement of Wien's law: \( T = \frac{2.90 \times 10^{-3}}{\lambda} \) [1 mark]
- (i) Correct calculation: \( 2.74 \text{ K} \) leading to shown value of \( 2.7 \text{ K} \) [1 mark]
- (ii) Identifies relationship \( T \propto \frac{1}{R} \) or \( R \cdot T = \text{constant} \) [1 mark]
- (ii) Correct calculation of ratio as approximately \( 1100 \) (accept range \( 1090 \) to \( 1111 \)) [1 mark]

(c) [1 mark]
- State that mass of gas cloud must be greater than the Jeans mass (or gravitational potential energy must exceed kinetic energy of particles) [1 mark]