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As the block slides up the slope, both the component of gravity along the slope and the kinetic friction force act down the slope (opposing the motion). The component of gravity is \( mg \sin\theta \). The kinetic friction is \( f_k = \mu N = \mu mg \cos\theta \). The net force down the slope is \( F_{\text{net}} = mg \sin\theta + \mu mg \cos\theta \). By Newton's second law, the deceleration \( a \) is \( \frac{F_{\text{net}}}{m} = g(\sin\theta + \mu\cos\theta) \).

[1 mark] award for identifying the correct deceleration expression: \( a = g(\sin\theta + \mu\cos\theta) \). No partial credit.

The parallel combination of two resistors of resistance \( R \) has an equivalent resistance of \( R_p = R/2 \). The total external resistance of the circuit is \( R_{\text{ext}} = R + R_p = 1.5R = \frac{3R}{2} \). The total circuit resistance is \( R_{\text{total}} = R_{\text{ext}} + r = \frac{3R}{2} + \frac{R}{3} = \frac{11R}{6} \). The total current is \( I = \frac{\mathcal{E}}{R_{\text{total}}} = \frac{6\mathcal{E}}{11R} \). The terminal potential difference is the potential difference across the external circuit: \( V = I R_{\text{ext}} = \left(\frac{6\mathcal{E}}{11R}\right) \left(\frac{3R}{2}\right) = \frac{9}{11}\mathcal{E} \).

[1 mark] award for correct analytical derivation of terminal potential difference as \( \frac{9}{11}\mathcal{E} \).

According to the kinetic theory of gases, the average translational kinetic energy of gas molecules is directly proportional to the absolute temperature \( T \) (expressed as \( E_k = \frac{3}{2} k_B T \)). Therefore, if the absolute temperature is doubled, the average kinetic energy is also doubled. The average speed of the molecules increases by a factor of \( \sqrt{2} \), not 2. The collision rate with the walls also increases, and the average momentum change per collision increases by \( \sqrt{2} \).

[1 mark] award for identifying that average kinetic energy is directly proportional to absolute temperature.

The photons of specific energies (corresponding to energy differences between electron shells in the gas atoms) are absorbed by the electrons, promoting them to higher energy levels. When these electrons transition back to lower energy levels, they re-emit the photons of identical energy in all directions. As a result, very few of the re-emitted photons continue along the original line of sight, which creates the appearance of dark absorption lines in the transmitted spectrum.

[1 mark] award for identifying the process of absorption and isotropic re-emission of photons by electrons.

The solar radiation intercepted by the planet's cross-sectional area \( \pi R^2 \) is \( P_{\text{incident}} = S \pi R^2 \). A fraction \( \alpha \) is reflected, so the power absorbed is \( P_{\text{absorbed}} = S(1 - \alpha)\pi R^2 \). This absorbed power is distributed over the entire surface area of the spherical planet \( 4\pi R^2 \). Therefore, the average absorbed intensity is \( I_{\text{absorbed}} = \frac{S(1 - \alpha)\pi R^2}{4\pi R^2} = \frac{S(1 - \alpha)}{4} \).

[1 mark] award for showing correct division of intercepted power by the total spherical surface area to obtain \( \frac{S(1 - \alpha)}{4} \).

For SHM starting at the equilibrium position at \( t=0 \), the displacement is described by \( x(t) = A \sin(\omega t) \), where \( \omega = \frac{2\pi}{T} \). To find the time for the displacement to reach \( \frac{A}{2} \), we set: \( \frac{A}{2} = A \sin\left(\frac{2\pi t}{T}\right) \implies \sin\left(\frac{2\pi t}{T}\right) = \frac{1}{2} \). The smallest positive angle for this is \( \frac{2\pi t}{T} = \frac{\pi}{6} \), which yields \( t = \frac{T}{12} \).

[1 mark] award for calculating the correct time interval of \( T/12 \) using the sine equation for SHM.

Efficiency \( \eta \) is defined as \( \frac{P_{\text{out}}}{P_{\text{in}}} \). The power input is the supplied electrical power \( P \). The useful power output is the rate of doing work against gravity: \( P_{\text{out}} = F v = mgv \) (since the load is lifted at constant speed, the tension force equals the weight \( mg \)). Therefore, the efficiency is \( \eta = \frac{mgv}{P} \).

[1 mark] award for identifying power output as \( mgv \) and dividing by input power \( P \) to obtain \( \frac{mgv}{P} \).

After three half-lives, the number of remaining active nuclei is \( N = N_0 \left(\frac{1}{2}\right)^3 = \frac{1}{8} N_0 \). The number of nuclei that have decayed is the difference between the initial number and the remaining number: \( N_{\text{decayed}} = N_0 - N = N_0 - \frac{1}{8} N_0 = \frac{7}{8} N_0 \).

[1 mark] award for determining the fraction of decayed nuclei is \( 7/8 \).

The relationship between pressure and absolute temperature for a gas at constant volume is given by Gay-Lussac's Law: \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\). First, convert temperatures from Celsius to Kelvin: \(T_1 = 27 + 273 = 300\text{ K}\) and \(T_2 = 327 + 273 = 600\text{ K}\). Now, substitute the values into the equation: \(P_2 = P_1 \times \frac{T_2}{T_1} = P \times \frac{600}{300} = 2P\). Therefore, the correct option is A.

[1 mark] award for identifying that temperature must be in Kelvin and doubling the absolute temperature leads to doubling the pressure, hence option A is correct.

First, apply the law of conservation of momentum to find the common final velocity \(V_f\) of the two trolleys after the inelastic collision: \(m v + 2m(0) = (m + 2m) V_f \implies V_f = \frac{v}{3}\). Next, determine the initial total kinetic energy of the system: \(E_{ki} = \frac{1}{2} m v^2\). Determine the final total kinetic energy of the system: \(E_{kf} = \frac{1}{2} (3m) V_f^2 = \frac{3}{2} m \left(\frac{v}{3}\right)^2 = \frac{1}{6} m v^2\). Find the ratio of final kinetic energy to initial kinetic energy: \(\frac{E_{kf}}{E_{ki}} = \frac{\frac{1}{6} m v^2}{\frac{1}{2} m v^2} = \frac{1}{3}\). Therefore, the correct option is A.

[1 mark] award for using conservation of momentum to find the final velocity and correctly calculating the ratio of the final kinetic energy to the initial kinetic energy, resulting in option A.

The useful power output of the motor is the power required to lift the mass vertically at constant speed: \(P_{\text{out}} = F v = m g v = 50 \times 10 \times 1.2 = 600\text{ W}\). Using the definition of efficiency \(\eta\): \(\eta = \frac{P_{\text{out}}}{P_{\text{in}}}\). Rearranging for the power input: \(P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{600}{0.60} = 1000\text{ W}\). Therefore, the correct option is C.

[1 mark] award for calculating the useful output power as 600 W, and dividing by the efficiency of 0.60 to obtain an input power of 1000 W, which corresponds to option C.

For the two parallel resistors, the equivalent resistance \(R_p\) is: \(\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \implies R_p = \frac{R}{2}\). This combination is in series with the third resistor of resistance \(R\), so the total equivalent resistance of the circuit \(R_{\text{eq}}\) is: \(R_{\text{eq}} = R + \frac{R}{2} = \frac{3R}{2}\). The total power dissipated in the circuit is: \(P = \frac{V^2}{R_{\text{eq}}} = \frac{V^2}{\frac{3R}{2}} = \frac{2V^2}{3R}\). Therefore, the correct option is A.

[1 mark] award for finding the equivalent resistance of the circuit as \(\frac{3R}{2}\) and calculating the power dissipated as \(\frac{2V^2}{3R}\), which corresponds to option A.

The total mechanical energy in simple harmonic motion is: \(E_T = \frac{1}{2} k A^2\). The potential energy at displacement \(x\) is: \(E_p = \frac{1}{2} k x^2\). The kinetic energy is the difference between total energy and potential energy: \(E_k = E_T - E_p = \frac{1}{2} k (A^2 - x^2)\). We are looking for the displacement where \(E_k = E_p\): \(\frac{1}{2} k (A^2 - x^2) = \frac{1}{2} k x^2 \implies A^2 - x^2 = x^2 \implies A^2 = 2x^2 \implies x = \frac{A}{\sqrt{2}}\). Therefore, the correct option is B.

[1 mark] award for equating the kinetic energy and potential energy expressions, solving for the displacement in terms of the amplitude to find \(x = \frac{A}{\sqrt{2}}\), corresponding to option B.

The intensity \(I\) of spherical waves emitted from a point source decreases as the inverse square of the distance \(r\) from the source: \(I \propto \frac{1}{r^2}\). Let \(I_1 = I\) at distance \(r_1 = d\), and let \(I_2\) be the intensity at distance \(r_2 = 3d\). Therefore, \(\frac{I_2}{I_1} = \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{d}{3d}\right)^2 = \frac{1}{9}\), meaning \(I_2 = \frac{I}{9}\). Hence, the correct option is B.

[1 mark] award for identifying and applying the inverse square law for wave intensity to show that tripling the distance reduces the intensity by a factor of 9, corresponding to option B.

After three half-lives, the activity of the sample is reduced by a factor of \(2^3 = 8\). Therefore, the new activity is: \(A = \frac{A_0}{8}\). Similarly, the number of active nuclei remaining is \(\frac{1}{8}\) of the original number \(N_0\). The fraction of the original active nuclei that have decayed is: \(1 - \frac{1}{8} = \frac{7}{8}\). Thus, the activity is \(\frac{A_0}{8}\) and the fraction decayed is \(\frac{7}{8}\). Therefore, the correct option is A.

[1 mark] award for identifying that after three half-lives the activity is \(\frac{A_0}{8}\) and correctly calculating that \(\frac{7}{8}\) of the original nuclei have decayed, corresponding to option A.

The solar constant \(S\) represents the power per unit area of solar radiation incident perpendicular to a surface: \(S = 1400\text{ W m}^{-2}\). The total solar power intercepted by a planet of radius \(R\) is \(P_{\text{intercepted}} = S \times \pi R^2\). Since the average albedo of the planet is \(\alpha = 0.30\), the fraction of solar power reflected is \(0.30\), and the fraction absorbed is \(1 - \alpha = 0.70\). The total power absorbed is: \(P_{\text{absorbed}} = 0.70 \times S \times \pi R^2\). This total absorbed power is distributed over the entire spherical surface area of the planet, which is \(4 \pi R^2\). Therefore, the average power absorbed per unit area is: \(I_{\text{avg}} = \frac{0.70 \times S \times \pi R^2}{4 \pi R^2} = \frac{0.70 \times S}{4} = \frac{0.70 \times 1400}{4} = 245\text{ W m}^{-2}\). Therefore, the correct option is B.

[1 mark] award for calculating the average power per unit area using the formula \(\frac{(1-\alpha)S}{4}\) to obtain \(245\text{ W m}^{-2}\), which corresponds to option B.

The work done by the motor in time \(t\) is given by:
\(W = P \times t\)

According to the work-energy theorem, since resistive forces are negligible, this work is entirely converted into the kinetic energy of the vehicle:
\(W = \Delta E_k\)
\(P t = \frac{1}{2} m v^2 - 0\)

Solving for the final speed \(v\):
\(v^2 = \frac{2Pt}{m}\)
\(v = \sqrt{\frac{2Pt}{m}}\)

Therefore, the correct option is C.

Award [1] mark for the correct answer C.
- Award [1] for realizing that work done \(W = Pt\) equals kinetic energy \(\frac{1}{2}mv^2\), and correctly solving for \(v\).

The ideal gas equation can be written in terms of the number of molecules \(N\) as:
\(P V = N k_B T\)

Since the volume \(V\) is fixed and Boltzmann's constant \(k_B\) is constant, the pressure \(P\) is directly proportional to the product of the number of molecules and the absolute temperature:
\(P \propto N T\)

Let the initial state be described by:
\(P_1 = C N_1 T_1\) (where \(C\) is a constant)

In the final state:
- The number of molecules is \(N_2 = \frac{1}{2} N_1\)
- The absolute temperature is \(T_2 = 3 T_1\)

Substituting these values into the proportional relation:
\(P_2 = C N_2 T_2 = C \left(\frac{1}{2} N_1\right) (3 T_1) = \frac{3}{2} C N_1 T_1 = \frac{3}{2} P_1\)

Thus, the new pressure is \(\frac{3}{2}P\).

Award [1] mark for the correct answer B.
- Award [1] for identifying that pressure is proportional to the number of molecules and absolute temperature when volume is constant, and correctly finding the scale factor of \(\frac{3}{2}\).

The initial kinetic energy of the bullet is:
\(E_i = \frac{1}{2} m v^2\)

By conservation of linear momentum, the common final speed \(V\) of the block and the embedded bullet is:
\(m v = (M + m) V \implies V = \frac{m}{M + m} v\)

The final kinetic energy of the combined system is:
\(E_f = \frac{1}{2} (M + m) V^2 = \frac{1}{2} (M + m) \left(\frac{m}{M+m} v\right)^2 = \frac{1}{2} \frac{m^2}{M + m} v^2\)

The fraction of the initial kinetic energy remaining is:
\(\frac{E_f}{E_i} = \frac{\frac{1}{2} \frac{m^2}{M+m} v^2}{\frac{1}{2} m v^2} = \frac{m}{M + m}\)

The fraction of initial kinetic energy dissipated (lost) as thermal energy is:
\(1 - \frac{E_f}{E_i} = 1 - \frac{m}{M + m} = \frac{M}{M + m}\)

Award [1] mark for the correct answer B.
- Award [1] for showing conservation of momentum to find the final velocity, calculating the ratio of final kinetic energy to initial kinetic energy, and subtracting from 1 to find the fraction of energy dissipated.

First, calculate the equivalent resistance of the two parallel resistors:
\(R_p = \frac{R \times R}{R + R} = \frac{R}{2}\)

The total equivalent resistance of the circuit is the sum of the parallel combination and the third resistor in series:
\(R_{\text{total}} = R_p + R = \frac{R}{2} + R = \frac{3}{2}R\)

The total current \(I\) flowing from the cell is:
\(I = \frac{V}{R_{\text{total}}} = \frac{V}{\frac{3}{2}R} = \frac{2V}{3R}\)

Since the third resistor is connected in series with the rest of the circuit, the entire total current \(I\) flows through it. The power \(P\) dissipated in the third resistor is:
\(P = I^2 R = \left(\frac{2V}{3R}\right)^2 R = \frac{4V^2}{9R^2} R = \frac{4V^2}{9R}\)

Therefore, the correct option is C.

Award [1] mark for the correct answer C.
- Award [1] for calculating the correct circuit total resistance, the total current flowing through the series resistor, and using \(P = I^2 R\) to determine the power.

First, we calculate the number of half-lives that have elapsed for each nuclide after 8 hours:

For nuclide X:
\(n_X = \frac{8\text{ hours}}{2\text{ hours}} = 4\text{ half-lives}\)
The remaining number of X nuclei is:
\(N_X = N_0 \left(\frac{1}{2}\right)^4 = \frac{N_0}{16}\)

For nuclide Y:
\(n_Y = \frac{8\text{ hours}}{4\text{ hours}} = 2\text{ half-lives}\)
The remaining number of Y nuclei is:
\(N_Y = N_0 \left(\frac{1}{2}\right)^2 = \frac{N_0}{4}\)

Now, we find the ratio of the remaining nuclei of X to Y:
\(\frac{N_X}{N_Y} = \frac{\frac{N_0}{16}}{\frac{N_0}{4}} = \frac{4}{16} = \frac{1}{4}\)

Therefore, the correct option is D.

Award [1] mark for the correct answer D.
- Award [1] for correctly determining the remaining fractions of both nuclides and computing their ratio after 8 hours.

When unpolarized light of intensity \(I_0\) passes through the first polarizing filter, its intensity is reduced to half because only the component parallel to the transmission axis is transmitted:
\(I_1 = \frac{1}{2} I_0\)

The light emerging from the first filter is now vertically polarized.

Next, it passes through the second filter whose transmission axis makes an angle of \(\theta = 60^\circ\) with the vertical axis of polarization. Applying Malus's Law:
\(I_2 = I_1 \cos^2(\theta)\)
\(I_2 = \left(\frac{1}{2} I_0\right) \cos^2(60^\circ)\)

Since \(\cos(60^\circ) = \frac{1}{2}\), we have \(\cos^2(60^\circ) = \frac{1}{4}\).

Substituting this in:
\(I_2 = \frac{1}{2} I_0 \times \frac{1}{4} = \frac{1}{8} I_0\)

Therefore, the correct option is B.

Award [1] mark for the correct answer B.
- Award [1] for recalling that passing unpolarized light through a polarizer halves the intensity, and correctly applying Malus's Law with \(\cos(60^\circ) = 0.5\).

(a) From conservation of linear momentum: \(m_1 v_1 = m_1 v_{1f} + m_2 v_{2f}\). Substituting the given values: \(0.40 \times 3.0 = 0.40 \times (-0.60) + 0.60 \times v_{2f}\). This gives \(1.20 = -0.24 + 0.60 v_{2f}\), which simplifies to \(1.44 = 0.60 v_{2f}\), yielding \(v_{2f} = 2.4\text{ m s}^{-1}\). (b) Initial kinetic energy: \(E_{ki} = \frac{1}{2} m_1 v_1^2 = 0.5 \times 0.40 \times 3.0^2 = 1.8\text{ J}\). Final kinetic energy: \(E_{kf} = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 = 0.5 \times 0.40 \times (-0.60)^2 + 0.5 \times 0.60 \times 2.4^2 = 0.072 + 1.728 = 1.8\text{ J}\). Since \(E_{ki} = E_{kf}\), kinetic energy is conserved, indicating that the collision is elastic. (c) The average force on the second block is \(F = \frac{\Delta p}{\Delta t} = \frac{m_2 v_{2f} - 0}{\Delta t} = \frac{0.60 \times 2.4}{0.12} = 12\text{ N}\).

(a) 1 mark for applying conservation of momentum equation with correct sign for recoil velocity. 1 mark for showing correct algebraic steps leading to 2.4 m/s. (b) 1 mark for calculating initial kinetic energy (1.8 J). 1 mark for calculating final kinetic energy (1.8 J). 1 mark for stating that kinetic energy is conserved, hence the collision is elastic. (c) 1 mark for stating Newton's second law \(F = \frac{\Delta p}{\Delta t}\). 1 mark for calculating momentum change of 1.44 kg m/s. 1.33 marks for the final force of 12 N with units.

(a) Total solar power incident on the planet is \(P_{\text{incident}} = I \times \pi R^2\). The absorbed power is \(P_{\text{absorbed}} = I \times \pi R^2 \times (1 - \alpha) = 0.70 I \pi R^2\). This power is distributed over the entire surface area of the spherical planet \(4\pi R^2\). Thus, the average absorbed intensity is \(I_{\text{absorbed}} = \frac{P_{\text{absorbed}}}{4\pi R^2} = \frac{0.70 I}{4} = \frac{0.70 \times 1000}{4} = 175\text{ W m}^{-2}\). (b) For thermal equilibrium, the power radiated per unit area must equal the average power absorbed per unit area: \(\sigma T^4 = 175\text{ W m}^{-2}\). Substituting \(\sigma = 5.67 \times 10^{-8}\text{ W m}^{-2}\text{ K}^{-4}\): \(T = \left(\frac{175}{5.67 \times 10^{-8}}\right)^{0.25} = 235.9\text{ K} \approx 236\text{ K}\). (c) The atmosphere contains greenhouse gases which are transparent to incoming short-wavelength solar radiation but absorb outgoing long-wavelength infrared radiation from the planet's surface. These gases then re-emit the infrared radiation in all directions, including back towards the planet's surface (back-radiation), which further warms the surface.

(a) 1 mark for finding absorbed power fraction (0.70). 1 mark for writing the ratio of cross-sectional area to surface area (1/4). 1 mark for obtaining 175 W m^-2. (b) 1 mark for setting up the equilibrium equation \(\sigma T^4 = 175\). 1 mark for substituting \(\sigma = 5.67 \times 10^{-8}\). 1 mark for the final temperature of 236 K (accept 235.9 K). (c) 1 mark for mentioning that greenhouse gases absorb infrared/long-wavelength radiation emitted by the surface. 1.33 marks for explaining that these gases re-emit the radiation, with some returning to the surface (back-radiation) and causing further warming.

(a)(i) Total resistance of the circuit is \(R_{\text{total}} = R + r = 4.50 + 1.50 = 6.00\ \Omega\). The current is \(I = \frac{\mathcal{E}}{R_{\text{total}}} = \frac{12.0}{6.00} = 2.00\text{ A}\). (a)(ii) Terminal potential difference is \(V = \mathcal{E} - I r = 12.0 - 2.00 \times 1.50 = 9.00\text{ V}\). (b)(i) Maximum power transfer occurs when the external load resistance equals the internal resistance of the battery, so \(R = r = 1.50\ \Omega\). (b)(ii) The current at maximum power is \(I_{\text{max}} = \frac{\mathcal{E}}{R+r} = \frac{12.0}{1.50+1.50} = 4.00\text{ A}\). The maximum power dissipated is \(P_{\text{max}} = I_{\text{max}}^2 R = 4.00^2 \times 1.50 = 24.0\text{ W}\).

(a)(i) 1 mark for calculating total resistance (6.00 Ohm). 1 mark for final current of 2.00 A. (a)(ii) 1 mark for using \(V = \mathcal{E} - Ir\) or \(V = IR\). 1 mark for final terminal voltage of 9.00 V. (b)(i) 1 mark for stating \(R = 1.50\ \Omega\). (b)(ii) 1 mark for finding new current of 4.00 A. 1 mark for using \(P = I^2 R\). 1.33 marks for final power of 24.0 W.

(a) From the ideal gas equation: \(P = \frac{n R T}{V}\). Substituting the values: \(P = \frac{0.080 \times 8.31 \times 300}{2.0 \times 10^{-3}} = 9.972 \times 10^4\text{ Pa} \approx 1.0 \times 10^5\text{ Pa}\). (b)(i) Since volume and amount of gas are constant, we can use Gay-Lussac's Law: \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\). Thus, \(T_2 = T_1 \times \frac{P_2}{P_1} = 300 \times \frac{1.5 \times 10^5}{9.972 \times 10^4} \approx 451.3\text{ K} \approx 451\text{ K}\). (b)(ii) The average kinetic energy of a gas molecule is given by \(E_k = \frac{3}{2} k_B T\), where \(k_B = 1.38 \times 10^{-23}\text{ J K}^{-1}\). Substituting \(T_2 = 451.3\text{ K}\): \(E_k = 1.5 \times 1.38 \times 10^{-23} \times 451.3 = 9.34 \times 10^{-21}\text{ J}\).

(a) 1 mark for stating the ideal gas law formula rearranged for P. 1 mark for correct calculation of \(1.0 \times 10^5\text{ Pa}\) (accept \(9.97 \times 10^4\text{ Pa}\)). (b)(i) 1 mark for recognizing volume is constant and using \(P/T = \text{constant}\). 1 mark for substituting values correctly. 1 mark for correct final temperature of 451 K. (b)(ii) 1 mark for stating \(E_k = 1.5 k_B T\). 1 mark for substituting Boltzmann constant and temperature. 1.33 marks for final correct energy of \(9.34 \times 10^{-21}\text{ J}\).

(a)(i) The energy difference between the levels is \(\Delta E = E_3 - E_2 = -1.51\text{ eV} - (-3.40\text{ eV}) = 1.89\text{ eV}\). To convert to Joules: \(\Delta E = 1.89 \times 1.60 \times 10^{-19}\text{ J eV}^{-1} = 3.024 \times 10^{-19}\text{ J} \approx 3.02 \times 10^{-19}\text{ J}\). (a)(ii) The wavelength of the emitted photon is given by \(\lambda = \frac{h c}{E}\). Substituting values: \(\lambda = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{3.024 \times 10^{-19}} = 6.577 \times 10^{-7}\text{ m} \approx 6.58 \times 10^{-7}\text{ m}\) (or 658 nm). (b) The energy levels of the electrons in a hydrogen atom are discrete (quantized). Electrons can only exist in these specific, discrete states. When an electron falls from a higher energy level to a lower one, it emits a single photon whose energy is exactly equal to the difference between these two discrete levels. Since the energy differences are discrete, the wavelengths of the emitted photons can only have specific, discrete values, resulting in a line spectrum instead of a continuous spectrum.

(a)(i) 1 mark for calculating the energy change in eV (1.89 eV). 1 mark for using the correct conversion factor \(1.6 \times 10^{-19}\). 1 mark for final value of \(3.02 \times 10^{-19}\text{ J}\). (a)(ii) 1 mark for stating formula \(\lambda = \frac{hc}{E}\). 1.33 marks for correct wavelength \(6.58 \times 10^{-7}\text{ m}\). (b) 1 mark for stating that atomic energy levels are quantized/discrete. 1 mark for explaining that photon energy equals the difference between two discrete energy states. 1 mark for linking the discrete energy differences to specific wavelengths (via \(E = \frac{hc}{\lambda}\)).

(a) The period of a mass-spring system is given by \(T = 2\pi \sqrt{\frac{m}{k}}\). Substituting the given values: \(T = 2\pi \sqrt{\frac{0.250}{40.0}} = 2\pi \sqrt{0.00625} \approx 0.4967\text{ s}\), which rounds to approximately 0.50 s. (b) The maximum velocity is given by \(v_{\text{max}} = \omega A\), where \(\omega = \sqrt{\frac{k}{m}}\). \(\omega = \sqrt{\frac{40.0}{0.250}} = \sqrt{160} \approx 12.65\text{ rad s}^{-1}\). Thus, \(v_{\text{max}} = 12.65 \times 0.080 = 1.012\text{ m s}^{-1} \approx 1.01\text{ m s}^{-1}\). (c) The magnitude of the acceleration in SHM is given by \(a = \omega^2 x\). Since \(\omega^2 = \frac{k}{m} = 160\text{ s}^{-2}\) and the displacement \(x = 0.040\text{ m}\), we have: \(a = 160 \times 0.040 = 6.4\text{ m s}^{-2}\).

(a) 1 mark for stating the period formula \(T = 2\pi \sqrt{\frac{m}{k}}\). 1 mark for substituting values correctly and calculating 0.497 s or showing it rounds to 0.50 s. (b) 1 mark for calculating angular frequency \(\omega = 12.65\text{ rad s}^{-1}\). 1 mark for stating \(v_{\text{max}} = \omega A\). 1 mark for final value \(1.01\text{ m s}^{-1}\). (c) 1 mark for recognizing \(a = \omega^2 x\) or \(a = \frac{kx}{m}\). 1 mark for substituting correct values. 1.33 marks for final correct acceleration of \(6.4\text{ m s}^{-2}\).

(a)(i) The period is the total time divided by the number of oscillations: \(T = \frac{t}{20} = \frac{36.0}{20} = 1.80\text{ s}\). The absolute uncertainty in the period is \(\Delta T = \frac{\Delta t}{20} = \frac{0.2}{20} = 0.01\text{ s}\). Thus, \(T = 1.80 \pm 0.01\text{ s}\). (a)(ii) The percentage uncertainty in \(T\) is \(\frac{\Delta T}{T} \times 100\% = \frac{0.01}{1.80} \times 100\% \approx 0.556\%\). For \(T^2\), we double the percentage uncertainty: \(2 \times 0.556\% = 1.11\% \approx 1.1\%\). (b)(i) Squaring the pendulum period equation \(T = 2\pi\sqrt{\frac{L}{g}}\ yards \)T^2 = \frac{4\pi^2}{g} L\). This corresponds to the linear form \(y = mx\) where \(y = T^2\), \(x = L\), and the gradient is \(m = \frac{4\pi^2}{g}\). Since there is no y-intercept constant, the line passes through the origin. (b)(ii) Substituting the values: \(g = \frac{4\pi^2 L}{T^2} = \frac{4\pi^2 \times 0.800}{1.80^2} \approx 9.748\text{ m s}^{-2}\). The fractional uncertainty in \(g\) is \(\frac{\Delta g}{g} = \frac{\Delta L}{L} + \frac{\Delta(T^2)}{T^2} = \frac{0.005}{0.800} + 0.0111 = 0.00625 + 0.01111 = 0.01736\). The absolute uncertainty in \(g\) is \(\Delta g = 9.748 \times 0.01736 \approx 0.17\text{ m s}^{-2}\). Therefore, \(g = 9.7 \pm 0.2\text{ m s}^{-2}\) (or \(9.75 \pm 0.17\text{ m s}^{-2}\)).

(a)(i) Award [1] for calculating T = 1.80 s. Award [1] for absolute uncertainty ΔT = ±0.01 s. (a)(ii) Award [1] for doubling the percentage uncertainty of T. Award [1] for correct value of 1.1% (or 1.11%). (b)(i) Award [1] for deriving T² = (4π²/g)L. Award [1] for stating gradient = 4π²/g and explaining the zero intercept. (b)(ii) Award [1] for g = 9.75 m s⁻² (accept 9.7 m s⁻²). Award [1] for absolute uncertainty Δg = ±0.17 m s⁻² or ±0.2 m s⁻².

(a) Starting with \(P(V + V_{\text{dead}}) = C\), we divide both sides by \(P\) and \(C\) to get: \(\frac{V + V_{\text{dead}}}{C} = \frac{1}{P}\), which rearranges to \(\frac{1}{P} = \frac{1}{C}V + \frac{V_{\text{dead}}}{C}\). This is of the linear form \(y = mx + c\) where \(y = \frac{1}{P}\) and \(x = V\). The gradient \(m = \frac{1}{C}\) and the vertical intercept is \(\frac{V_{\text{dead}}}{C}\). (b)(i) From the best-fit equation, the gradient is \(0.0120\text{ atm}^{-1}\text{ cm}^{-3}\). Since \(m = \frac{1}{C}\), we find \(C = \frac{1}{0.0120} = 83.3\text{ atm cm}^3\). (b)(ii) The vertical intercept is \(0.0360\text{ atm}^{-1}\). Since the intercept is \(\frac{V_{\text{dead}}}{C}\), we find \(V_{\text{dead}} = 0.0360 \times C = 0.0360 \times 83.33 = 3.00\text{ cm}^3\). Alternatively, \(V_{\text{dead}} = \frac{\text{intercept}}{\text{gradient}} = \frac{0.0360}{0.0120} = 3.00\text{ cm}^3\). (c) A possible systematic error is that the temperature of the gas does not remain constant during compression or expansion (violating Boyle's Law), or there is a slow leak of gas from the syringe.

(a) Award [1] for correct algebraic steps to obtain \(\frac{1}{P} = \frac{1}{C}V + \frac{V_{\text{dead}}}{C}\). Award [1] for identifying gradient = \(1/C\) and intercept = \(V_{\text{dead}}/C\). (b)(i) Award [1] for calculating C = 83.3 (or 83). Award [1] for correct units: atm cm³ (or equivalent). (b)(ii) Award [1] for setting up the calculation using the intercept and C, or the ratio of intercept to gradient. Award [1] for V_dead = 3.00 cm³. (c) Award [1] for identifying a valid systematic error: e.g., temperature changes of the gas during compression/expansion, or leakage of air from the apparatus.

(a) Stellar parallax involves measuring the apparent angular shift in position of a nearby star against a background of very distant stars as the Earth orbits the Sun. Measurements are taken six months apart, and the parallax angle is half of the total angular shift.

(b) (i) Using the relationship \(d = \frac{1}{p}\), where \(p\) is in arcseconds and \(d\) is in parsecs: \(d = \frac{1}{0.125} = 8.0\text{ pc}\).
(ii) Since \(1\text{ pc} = 3.26\text{ ly}\): \(d = 8.0 \times 3.26 = 26.1\text{ ly}\).

(c) Convert distance to meters: \(d = 8.0\text{ pc} \times 3.09 \times 10^{16}\text{ m pc}^{-1} = 2.47 \times 10^{17}\text{ m}\).
From the apparent brightness formula \(b = \frac{L}{4\pi d^2}\), we find \(L = 4\pi d^2 b = 4\pi \times (2.47 \times 10^{17})^2 \times 1.3 \times 10^{-10} = 9.98 \times 10^{25}\text{ W} \approx 1.0 \times 10^{26}\text{ W}\).

(a)
- Measurement of apparent angular shift against background stars at six-month intervals [1]
- Parallax angle is defined as half of this total shift / distance is inversely proportional to this angle [1]

(b)
- (i) 8.0 [pc] [1]
- (ii) 26 [ly] or 26.1 [ly] (allow ECF from b(i)) [1]

(c)
- Correct conversion of pc to meters: \(2.47 \times 10^{17}\text{ m}\) [1]
- Correct rearrangement/use of \(L = 4\pi d^2 b\) [1]
- Correct final answer in range \(9.9 \times 10^{25}\text{ W}\) to \(1.0 \times 10^{26}\text{ W}\) [1]

(a) Wien's displacement law is \(\lambda_{\text{max}} T = 2.90 \times 10^{-3}\text{ m K}\).
\(T = \frac{2.90 \times 10^{-3}}{450 \times 10^{-9}\text{ m}} = 6444\text{ K}\), which is approximately \(6400\text{ K}\).

(b) The radius of the star is \(R = 1.2 \times 7.0 \times 10^8\text{ m} = 8.4 \times 10^8\text{ m}\).
Using Stefan-Boltzmann law: \(L = 4\pi R^2 \sigma T^4\).
\(L = 4\pi \times (8.4 \times 10^8)^2 \times (5.67 \times 10^{-8}) \times (6444)^4 = 8.7 \times 10^{26}\text{ W}\). (Using \(6400\text{ K}\), \(L = 8.4 \times 10^{26}\text{ W}\)).

(c) The Sun has a surface temperature of around \(5800\text{ K}\). Since the star has a higher surface temperature (\(6400\text{ K}\)) and higher luminosity than the Sun, it lies to the left (higher temperature) and above (higher luminosity) the Sun on the main sequence of the HR diagram.

(a)
- States or uses Wien's displacement law: \(\lambda_{\text{max}} T = 2.90 \times 10^{-3}\text{ m K}\) [1]
- Substitutes values to obtain \(6444\text{ K}\), showing calculation with at least 3 significant figures [1]

(b)
- Calculates radius of the star: \(R = 8.4 \times 10^8\text{ m}\) [1]
- Recalls/uses Stefan-Boltzmann law: \(L = 4\pi R^2 \sigma T^4\) [1]
- Substitutes and calculates \(L = 8.4 \times 10^{26}\text{ W}\) or \(8.7 \times 10^{26}\text{ W}\) [1]

(c)
- Mentions that the star is located to the left of the Sun (due to higher surface temperature) [1]
- Mentions that the star is located above the Sun (due to higher luminosity) [1]
*Award [1 max] if candidate states it is on the Main Sequence but does not specify both directions relative to the Sun.*

(a) The primary fusion process is the proton-proton (p-p) chain where hydrogen is converted into helium.

(b) Hydrostatic equilibrium is maintained because the inward gravitational force (which pulls stellar matter toward the center) is precisely balanced by the outward thermal and radiation pressure generated by core fusion reactions. This balance prevents further collapse or expansion.

(c) Electron degeneracy pressure, which arises from the Pauli exclusion principle, resists further gravitational collapse. The Chandrasekhar limit (approximately \(1.4 M_\odot\)) represents the maximum mass of a white dwarf. If a stellar remnant's core mass exceeds this limit, electron degeneracy pressure is overcome, and it must collapse further into a neutron star or black hole.

(a)
- Proton-proton (p-p) chain / hydrogen fusion (into helium) [1]

(b)
- Inward gravitational force is balanced by outward radiation pressure / thermal gas pressure [1]
- The net radial force is zero, resulting in a stable star radius [1]

(c)
- Electron degeneracy pressure prevents further collapse [1]
- Chandrasekhar limit is the maximum mass of a stable white dwarf / is approximately \(1.4 M_\odot\) [1]
- If mass exceeds this limit, the star collapses further (into a neutron star) / electron degeneracy pressure is insufficient to halt gravity [1]