2023 IB DP Physics 模擬試題連答案詳解

According to the Stefan-Boltzmann law, luminosity \(L\) is proportional to \(R^2 T^4\). For star A, \(L_A \propto R^2 T^4\). For star B, \(L_B \propto (0.5R)^2 (2T)^4 = 0.25 R^2 \times 16 T^4 = 4 R^2 T^4\). Therefore, the ratio of the luminosity of star B to star A is \(L_B / L_A = 4\).

1 mark for correctly applying the Stefan-Boltzmann law to both stars and calculating the ratio of 4.

The total initial mass of four protons is \(4 \times 1.673 \times 10^{-27} \text{ kg} = 6.692 \times 10^{-27} \text{ kg}\). The final mass of the helium-4 nucleus is \(6.646 \times 10^{-27} \text{ kg}\). The mass defect is the difference: \(\Delta m = 6.692 \times 10^{-27} \text{ kg} - 6.646 \times 10^{-27} \text{ kg} = 0.046 \times 10^{-27} \text{ kg} = 4.6 \times 10^{-29} \text{ kg}\).

1 mark for calculating the mass defect as the difference between four proton masses and one helium-4 mass.

The rate of thermal conduction is given by \(Q = k A \frac{\Delta T}{L}\). For the second rod, the rate of conduction is \(Q_2 = k (3A) \frac{2\Delta T}{2L} = 3 \left( k A \frac{\Delta T}{L} \right) = 3Q\).

1 mark for substituting the new values into the conduction equation and identifying that the rate is tripled.

The fringe separation is given by \(s = \frac{\lambda D}{d}\). If the screen distance becomes \(2D\) and the slit separation becomes \(d/2\), the new fringe separation is \(s_{new} = \frac{\lambda (2D)}{d/2} = 4 \left( \frac{\lambda D}{d} \right) = 4s\).

1 mark for using the double-slit interference equation to deduce that the new fringe separation is four times the original.

The impulse is equal to the area under the force-time graph. The graph is a triangle with base \(6.0 \text{ s}\) and height \(10 \text{ N}\). Area = \(0.5 \times 6.0 \times 10 = 30 \text{ N s}\). Since impulse equals change in momentum, \(\Delta p = 30 \text{ kg m s}^{-1}\). Given the mass is \(2.0 \text{ kg}\) and it starts from rest, the final speed is \(30 / 2.0 = 15 \text{ m s}^{-1}\).

1 mark for calculating the total impulse as the area under the triangle and dividing by mass to find the final speed.

In series, the equivalent resistance is \(3R\), so the power is \(P = V^2 / (3R)\). In parallel, the equivalent resistance is \(R/3\), so the power is \(P_{parallel} = V^2 / (R/3) = 3 V^2 / R = 9 P\).

1 mark for expressing power in terms of resistance for both series and parallel configurations and finding the correct ratio of 9.

The average kinetic energy of the molecules in an ideal gas is directly proportional to its absolute temperature \(T\). Since kinetic energy is proportional to the mean square speed \(\overline{v^2}\), the mean square speed is directly proportional to \(T\). Therefore, doubling the absolute temperature doubles the mean square speed. The ratio is 2.0.

1 mark for identifying that the mean square speed is directly proportional to temperature and therefore doubles when temperature is doubled.

The energy difference for the transition from 3 to 1 is the sum of the energy differences from 3 to 2 and from 2 to 1: \(\Delta E_{31} = \Delta E_{32} + \Delta E_{21}\). Using \(E = hc/\lambda\), we get \(\frac{hc}{\lambda_{31}} = \frac{hc}{\lambda_{32}} + \frac{hc}{\lambda_{21}}\). Simplifying gives \(\frac{1}{\lambda_{31}} = \frac{1}{\lambda_{32}} + \frac{1}{\lambda_{21}} = \frac{\lambda_{21} + \lambda_{32}}{\lambda_{32} \lambda_{21}}\), which yields \(\lambda_{31} = \frac{\lambda_{32} \lambda_{21}}{\lambda_{32} + \lambda_{21}}\).

1 mark for relating the energy levels to wavelengths and solving for the combined wavelength correctly.

Luminosity is given by the Stefan-Boltzmann law as \(L = 4\pi R^2 \sigma T^4\), which means \(L \propto R^2 T^4\). Therefore, the ratio of the luminosities is \(\frac{L_X}{L_Y} = \left(\frac{R_X}{R_Y}\right)^2 \left(\frac{T_X}{T_Y}\right)^4 = \left(\frac{1}{2}\right)^2 (2)^4 = \frac{1}{4} \times 16 = 4\).

Award 1 mark for the correct application of the luminosity formula and finding the correct numerical ratio.

The total energy supplied during the constant temperature phase change is \(Q = P \Delta t = P(t_2 - t_1)\). Since the energy required for fusion is given by \(Q = m L_f\), we can rearrange this to express the specific latent heat of fusion as \(L_f = \frac{P(t_2 - t_1)}{m}\).

Award 1 mark for calculating the energy transferred during the phase change and relating it to specific latent heat of fusion.

Fringe spacing is given by \(s = \frac{\lambda D}{d}\). Under the new conditions, the new slit separation is \(d' = 2d\) and the new screen distance is \(D' = D/2\). Substituting these values into the formula gives \(s' = \frac{\lambda (D/2)}{2d} = \frac{\lambda D}{4d} = \frac{s}{4}\).

Award 1 mark for setting up the fringe spacing proportionalities and obtaining \(s/4\).

According to the conservation of momentum, \(mv = (m + 3m)v' = 4mv'\), which gives the final common speed as \(v' = v/4\). The initial kinetic energy of the system is \(E_k = \frac{1}{2}mv^2\). The final kinetic energy is \(E_k' = \frac{1}{2}(4m)(v/4)^2 = \frac{1}{2}(4m)\frac{v^2}{16} = \frac{1}{8}mv^2 = \frac{1}{4}E_k\). The fraction of kinetic energy lost is \(1 - \frac{E_k'}{E_k} = 1 - \frac{1}{4} = \frac{3}{4}\).

Award 1 mark for identifying the correct final velocity via momentum conservation and comparing the final and initial kinetic energies to find the loss ratio.

The total resistance of the circuit is \(R_{eq} = R + \frac{R}{2} = \frac{3}{2}R\). The total current flowing from the battery is \(I = \frac{V}{R_{eq}} = \frac{2V}{3R}\). Since the two resistors are connected in parallel, the current divides equally between them. The current through one parallel resistor is therefore \(I_p = \frac{I}{2} = \frac{V}{3R}\). The power dissipated in this resistor is \(P = I_p^2 R = \left(\frac{V}{3R}\right)^2 R = \frac{V^2}{9R}\).

Award 1 mark for the correct combination of total resistance, current, and calculation of individual power dissipation in the parallel branch.

Starting from \((p, V, T)\), an isobaric expansion to double the volume means the new volume is \(2V\) and the pressure remains \(p\). By the ideal gas law \(pV = nRT\), the temperature must also double to \(2T\). In the second step, the gas is cooled at a constant volume (isochorically), so its volume remains \(2V\) and its temperature returns to \(T\). Applying the gas law again for the final state: \(p_{final} (2V) = nRT = pV\), which yields \(p_{final} = \frac{p}{2}\).

Award 1 mark for analyzing the sequential state changes of the gas to determine the final pressure correctly.

The energy of an level in a hydrogen atom is proportional to \(-1/n^2\). Thus, the energy of a transition from \(n_i\) to \(n_f\) is proportional to \(\frac{1}{n_f^2} - \frac{1}{n_i^2}\). For \(3 \to 2\), the energy change is proportional to \(\frac{1}{4} - \frac{1}{9} = \frac{5}{36}\), so the wavelength is \(\lambda \propto \frac{36}{5}\). For \(4 \to 2\), the energy change is proportional to \(\frac{1}{4} - \frac{1}{16} = \frac{3}{16}\), so the wavelength is \(\lambda' \propto \frac{16}{3}\). Comparing the two gives \(\frac{\lambda'}{\lambda} = \frac{16/3}{36/5} = \frac{16}{3} \times \frac{5}{36} = \frac{20}{27}\), hence \(\lambda' = \frac{20}{27}\lambda\).

Award 1 mark for expressing the wavelength ratio in terms of the Rydberg-type energy differences and calculating the exact numerical factor.

The total power incident on the planet's circular cross-sectional area is \(P_{inc} = I \pi R^2\). With an albedo of \(\alpha\), a fraction \(\alpha\) of the incident radiation is reflected, meaning \(1-\alpha\) is absorbed. The total power absorbed is therefore \(P_{abs} = I (1-\alpha) \pi R^2\). To find the average power absorbed per unit area, we divide this power by the total surface area of the spherical planet, \(4\pi R^2\): \(I_{avg} = \frac{I(1-\alpha)\pi R^2}{4\pi R^2} = \frac{I(1-\alpha)}{4}\).

Award 1 mark for utilizing both the albedo factor and the geometric ratio of circular cross-section area to spherical surface area (1/4).

According to Wien's displacement law, the peak emission wavelength \(\lambda_{\text{max}}\) is inversely proportional to the absolute temperature \(T\). Thus, \(T_Y / T_X = \lambda_X / \lambda_Y = 1/2\). According to the Stefan-Boltzmann law, the power emitted per unit area is proportional to \(T^4\). Therefore, the ratio of the power emitted per unit area of star Y to that of star X is \((T_Y / T_X)^4 = (1/2)^4 = 1/16\).

[1 mark] award for identifying the correct ratio of temperatures as 1/2. [1 mark] award for applying the Stefan-Boltzmann law to find the power ratio of 1/16.

According to the Jeans criterion, a gas cloud collapses gravitationally when its gravitational potential energy exceeds the random kinetic energy of its particles. Low temperature decreases the average random kinetic energy of the gas particles, while high density increases the strength of the gravitational attraction between them, making collapse more likely.

[1 mark] award for identifying that low temperature decreases kinetic energy and high density increases gravitational potential energy, both favoring collapse.

The rate of conduction through a material is given by \(P = \frac{k A \Delta T}{d}\). For the second insulator, the rate is \(P' = \frac{(3k) A \Delta T}{2d} = \frac{3}{2} \left(\frac{k A \Delta T}{d}\right) = \frac{3}{2}P\).

[1 mark] award for applying the thermal conduction formula to find that the new rate is 1.5 times the original rate.

The condition for the first minimum in single-slit diffraction is \(b \sin\theta = \lambda\), which gives \(\sin\theta = \frac{\lambda}{b} = x\). For the new configuration, \(\sin\theta' = \frac{2\lambda}{b/2} = 4\frac{\lambda}{b} = 4x\).

[1 mark] award for substituting the new wavelength and slit width into the single-slit diffraction formula to obtain 4x.

By conservation of linear momentum, the total momentum before and after the decay must be equal. Since the nucleus is initially stationary, the initial momentum is zero. Therefore: \(0 = m v + (M - m) v_{\text{recoil}}\). Solving for the recoil velocity of the remaining nucleus (which has mass \(M - m\)) gives \(v_{\text{recoil}} = -\frac{m v}{M - m}\).

[1 mark] award for using conservation of momentum and correctly identifying the mass of the daughter nucleus as (M - m) to get the correct recoil velocity.

Because the wires are connected in series, the electrical current \(I\) is the same in both wires. The drift speed is given by \(v = \frac{I}{n A q}\). Since both wires are made of copper, the charge carrier density \(n\) is identical. Thus, the drift speed is inversely proportional to the cross-sectional area: \(v \propto \frac{1}{A} \propto \frac{1}{r^2}\). Given \(r_X = 2r_Y\), we find \(\frac{v_X}{v_Y} = \left(\frac{r_Y}{r_X}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\).

[1 mark] award for recognizing that the current is identical in series and that cross-sectional area depends on the square of the radius, leading to a ratio of 1/4.

The average kinetic energy of the molecules in an ideal gas is directly proportional to its absolute temperature: \(\frac{1}{2} m v_{\text{rms}}^2 = \frac{3}{2} k_B T\). This implies that the rms speed of the molecules is proportional to the square root of the absolute temperature: \(v_{\text{rms}} \propto \sqrt{T}\). Therefore, tripling the temperature results in an rms speed increase by a factor of \(\sqrt{3}\).

[1 mark] award for identifying that the rms speed is proportional to the square root of the absolute temperature, hence the ratio is root 3.

The energy of the emitted photon is equal to the difference in energy between the initial and final states: \(\Delta E = E_{\text{initial}} - E_{\text{final}} = -E_1 - (-E_2) = E_2 - E_1\). Since \(\Delta E = \frac{h c}{\lambda}\), solving for \(\lambda\) gives \(\lambda = \frac{h c}{E_2 - E_1}\).

[1 mark] award for finding the energy difference as E2 - E1 and setting it equal to hc/wavelength to solve for wavelength.

The ideal gas equation can be written as \( P V = N k_B T \). Since the volume \( V \) of the rigid cylinder is constant, we can express the pressure as \( P \propto N T \). Let the initial pressure, number of molecules, and temperature be \( P_1 \), \( N_1 \), and \( T_1 \) respectively. The final values are \( N_2 = 0.5 N_1 \) and \( T_2 = 2 T_1 \). The final pressure is \( P_2 \propto N_2 T_2 = (0.5 N_1) (2 T_1) = N_1 T_1 \). Therefore, the ratio of the final pressure to the initial pressure is \( \frac{P_2}{P_1} = 1 \).

[1 mark] C is correct. Award 1 mark for the correct application of the ideal gas law under changing conditions of both temperature and quantity of gas.

According to the Stefan-Boltzmann law, the total power (luminosity) radiated by a star is given by \( L = \sigma A T^4 = 4\pi R^2 \sigma T^4 \). Thus, luminosity is proportional to \( R^2 T^4 \). For Star X: \( L_X \propto R^2 T^4 \). For Star Y: \( L_Y \propto \left( \frac{R}{2} \right)^2 (2T)^4 = \frac{R^2}{4} \times 16 T^4 = 4 R^2 T^4 \). The ratio \( \frac{L_Y}{L_X} \) is therefore \( \frac{4 R^2 T^4}{R^2 T^4} = 4 \).

[1 mark] B is correct. Award 1 mark for applying the Stefan-Boltzmann law including both the radius and the temperature dependency.

For a single slit, the angle to the first minimum is given by \( \theta_1 \approx \frac{\lambda}{b} \). The angular width of the central maximum is \( 2\theta_1 \propto \frac{\lambda}{b} \). If the original angular width is \( \theta \propto \frac{\lambda}{b} \), then the new angular width \( \theta' \) with a slit width of \( 2b \) and wavelength of \( \frac{\lambda}{2} \) is: \( \theta' \propto \frac{\lambda / 2}{2b} = \frac{1}{4} \frac{\lambda}{b} = \frac{\theta}{4} \).

[1 mark] A is correct. Award 1 mark for relating the angular width of the central maximum to wavelength and slit width.

Let \( v_1 \) be the final velocity of mass \( m \) and \( v_2 \) be the final velocity of mass \( 3m \). By conservation of linear momentum: \( m v = m v_1 + 3m v_2 \implies v = v_1 + 3v_2 \). Since the collision is elastic, the relative speed of approach equals the relative speed of separation: \( v - 0 = v_2 - v_1 \implies v_1 = v_2 - v \). Substituting \( v_1 \) into the momentum equation: \( v = (v_2 - v) + 3v_2 \implies 2v = 4v_2 \implies v_2 = \frac{v}{2} \).

[1 mark] C is correct. Award 1 mark for simultaneous application of conservation of momentum and conservation of kinetic energy (or relative velocity of approach/separation) to solve for the final speed.

The equivalent resistance of the two parallel resistors is \( R_p = \frac{R}{2} \). The total resistance of the entire circuit is \( R_{\text{total}} = R_p + R = \frac{R}{2} + R = \frac{3}{2}R \). The total current flowing from the cell is \( I = \frac{E}{R_{\text{total}}} = \frac{2E}{3R} \). The current splits equally between the two parallel branches since the branches are identical. Thus, the current through each of the parallel resistors is \( I_p = \frac{I}{2} = \frac{E}{3R} \). The power dissipated in one of these parallel resistors is \( P = I_p^2 R = \left( \frac{E}{3R} \right)^2 R = \frac{E^2}{9R} \).

[1 mark] A is correct. Award 1 mark for correctly finding the total circuit resistance, total current, the current division in parallel branches, and the power in a single resistor.

The luminosity of a star is given by \( L = 4\pi R^2 \sigma T^4 \), where \( R \) is the radius and \( T \) is the surface temperature. Since both stars have the same surface temperature, their luminosity is proportional to the square of their radius: \( L \propto R^2 \). This means that \( R \propto \sqrt{L} \). Therefore, the ratio of the radii is \( \frac{R_P}{R_Q} = \sqrt{\frac{L_P}{L_Q}} = \sqrt{81} = 9 \).

[1 mark] B is correct. Award 1 mark for using the relationship between luminosity and radius at constant temperature to find the ratio.

(a) Hydrostatic equilibrium is the state where the inward gravitational force acting on the star's matter is exactly balanced by the outward radiation pressure and gas pressure arising from fusion reactions in the core.

(b)(i) Using Stefan-Boltzmann's Law:
\(L = \sigma A T^4 = 5.67 \times 10^{-8} \times 4 \pi R^2 \times T^4\)
\(L = 5.67 \times 10^{-8} \times 4 \pi \times (7.0 \times 10^8)^2 \times (5800)^4\)
\(L = 5.67 \times 10^{-8} \times 6.1575 \times 10^{18} \times 1.1316 \times 10^{15} \approx 3.95 \times 10^{26}\text{ W}\) (or \(4.0 \times 10^{26}\text{ W}\))

(b)(ii) The star lies on the Main Sequence region (similar to the Sun).

(c)(i) The fundamental force is the weak nuclear interaction (weak force). This is because a proton decays into a neutron, producing a positron and an electron neutrino, which is a process involving flavour change and leptons.

(c)(ii) The mass defect is \(0.7\%\) of \(1.0\text{ kg}\):
\(\Delta m = 0.007 \times 1.0\text{ kg} = 0.007\text{ kg}\)
Using Einstein's mass-energy equivalence:
\(E = \Delta m c^2 = 0.007 \times (3.0 \times 10^8)^2 = 0.007 \times 9.0 \times 10^{16} = 6.3 \times 10^{14}\text{ J}\)

(a)
- Inward force due to gravity is balanced by outward force due to radiation pressure [1]
- Mentions that forces must act on any layer / star is in a stable size state [1]

(b)(i)
- Correct formula usage: \(L = \sigma 4\pi R^2 T^4\) [1]
- Correct final answer: \(3.95 \times 10^{26}\text{ W}\) or \(4.0 \times 10^{26}\text{ W}\) [1]

(b)(ii)
- Main sequence [1]

(c)(i)
- Weak force/interaction [1]
- Justification: change of quark flavour (proton to neutron) OR presence of leptons / neutrino / positron emission [1]

(c)(ii)
- Correct calculation of mass defect: \(\Delta m = 0.007\text{ kg}\) [1]
- Substitution into \(E = \Delta m c^2\) [1]
- Correct final answer: \(6.3 \times 10^{14}\text{ J}\) [1]

(a)
- In conduction, thermal energy is transferred through direct atomic/molecular vibrations and collisions. Faster-moving particles transfer kinetic energy to adjacent, slower-moving particles. In metals, free electrons also assist in carrying energy [1].
- In convection, thermal energy is transferred through the physical bulk movement of fluid. Warmer fluid expands, becomes less dense, rises, and is replaced by cooler, denser fluid [2].

(b)(i)
- Total power incident: \(P_{\text{incident}} = I \times A = 850 \times 2.5 = 2125\text{ W}\) [1]
- Power absorbed: \(P_{\text{absorbed}} = 2125 \times 0.92 = 1955\text{ W}\) [1]
- This value is approximately \(2000\text{ W}\).

(b)(ii)
- Using the heating rate equation: \(\frac{Q}{\Delta t} = \frac{m}{\Delta t} c \Delta T\) [1]
- Here, \(P = \left(\frac{m}{\Delta t}\right) c \Delta T\), so:
\(1955 = 0.045 \times 4200 \times \Delta T\) [1]
\(1955 = 189 \times \Delta T\)
\(\Delta T = \frac{1955}{189} \approx 10.3\text{ K}\) (or \(10.3\text{ }^\circ\text{C}\)) [1]
*(If using the show-that value of \(2000\text{ W}\): \(\Delta T = \frac{2000}{189} \approx 10.6\text{ K}\)*)

(c)
- Thermal energy is lost from the hot absorber plate and tubes to the cooler surroundings via conduction/convection/radiation [1].
- Evaporation of water or reflection from the glass cover limits the efficiency [1].

(a)
- Conduction involves energy transfer via atomic/molecular vibrations/collisions (without bulk motion) [1]
- Mentions role of free/delocalized electrons in conduction [1]
- Convection involves bulk macroscopic movement of fluid driven by density differences [1]

(b)(i)
- Correct calculation of incident power: \(850 \times 2.5 = 2125\text{ W}\) [1]
- Correct calculation of absorbed power: \(2125 \times 0.92 = 1955\text{ W}\) (must show explicit calculation to 3 or 4 sig figs before stating \(\approx 2000\text{ W}\)) [1]

(b)(ii)
- Uses \(P = \frac{\Delta m}{\Delta t} c \Delta T\) [1]
- Correct substitution of values (either 1955 or 2000) [1]
- Correct final answer: \(10.3\text{ K}\) or \(10.6\text{ K}\) (accept \(^{\circ}\text{C}\)) [1]

(c)
- Award [1] for each sensible reason, up to [2], e.g.:
- Heat loss by conduction/convection/radiation to the surrounding air [1]
- Reflections of sunlight from outer glass cover/surfaces [1]
- Insufficient contact/thermal resistance between the plate and tubes [1]

(a)(i) For unpolarized light passing through a linear polarizer, the intensity is halved. Thus, \(I_1 = 0.5 I_0\).

(a)(ii) Malus's law states \(I_2 = I_1 \cos^2(\theta)\).
We require \(I_2 = 0.125 I_0\). Substituting \(I_1 = 0.5 I_0\):
\(0.125 I_0 = 0.5 I_0 \cos^2(\theta)\)
\(\cos^2(\theta) = \frac{0.125}{0.5} = 0.25\)
\(\cos(\theta) = \sqrt{0.25} = 0.5\)
\(\theta = \arccos(0.5) = 60^\circ\) (or \(\frac{\pi}{3}\text{ rad}\)).

(b)(i) For single slit diffraction, the angle of the first minimum \(\theta\) is given by:
\(\theta \approx \frac{\lambda}{b}\)
The angular width of the central maximum is \(2\theta = \frac{2\lambda}{b}\).
The linear width on the screen is \(W = 2\theta D = \frac{2\lambda D}{b}\).
Substituting the values:
\(W = \frac{2 \times 633 \times 10^{-9} \times 2.0}{0.15 \times 10^{-3}}\)
\(W = \frac{2.532 \times 10^{-6}}{0.15 \times 10^{-3}} = 1.688 \times 10^{-2}\text{ m} \approx 1.7 \times 10^{-2}\text{ m}\) (or \(1.7\text{ cm}\)).

(b)(ii) If the slit width \(b\) is decreased:
- The angular width \(\theta\) of the central maximum increases, so the pattern spreads out wider [1].
- The distance between consecutive minima/maxima increases [1].
- The peak intensity of the central and secondary maxima decreases because less light energy passes through the narrower slit [1].

(a)(i)
- \(0.5 I_0\) (or \(\frac{1}{2} I_0\)) [1]

(a)(ii)
- Uses Malus's law: \(I_2 = I_1 \cos^2(\theta)\) [1]
- Solves for \(\cos^2(\theta) = 0.25\) [1]
- Obtains \(\theta = 60^\circ\) or \(1.05\text{ rad}\) (or \(\frac{\pi}{3}\text{ rad}\)) [1]

(b)(i)
- Recalls or uses \(\theta = \frac{\lambda}{b}\) for first minimum [1]
- Realizes that central width is \(2 \theta D\) or \(\frac{2\lambda D}{b}\) [1]
- Correct calculation to yield \(1.7 \times 10^{-2}\text{ m}\) (accept \(1.69 \times 10^{-2}\text{ m}\)) [1]

(b)(ii)
- Central maximum becomes wider/broader [1]
- Secondary maxima move further apart/diffraction spacing increases [1]
- Overall intensity of the pattern decreases [1]

(a) Impulse is defined as the product of the average net force acting on an object and the time duration of this force, which is equivalent to the change in momentum of the object (\(J = F \Delta t = \Delta p\)).

(b)(i) Define the direction of \(m_1\)'s initial velocity as positive.
Using conservation of linear momentum:
\(p_{\text{initial}} = m_1 u_1 + m_2 u_2 = (2.5 \times 4.0) + (1.5 \times (-2.0))\)
\(p_{\text{initial}} = 10.0 - 3.0 = 7.0\text{ kg m s}^{-1}\)
Since they stick together, the final momentum is:
\(p_{\text{final}} = (m_1 + m_2) v = (2.5 + 1.5) v = 4.0 v\)
Setting \(p_{\text{initial}} = p_{\text{final}}\):
\(4.0 v = 7.0 \implies v = 1.75\text{ m s}^{-1}\)
Since \(v > 0\), the direction is in the direction of \(m_1\)'s initial motion.

(b)(ii) Calculate initial kinetic energy:
\(E_{\text{k,i}} = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2}(2.5)(4.0)^2 + \frac{1}{2}(1.5)(-2.0)^2 = 20.0 + 3.0 = 23.0\text{ J}\)
Calculate final kinetic energy:
\(E_{\text{k,f}} = \frac{1}{2}(m_1 + m_2) v^2 = \frac{1}{2}(4.0)(1.75)^2 = 2.0 \times 3.0625 = 6.125\text{ J}\)
Loss in kinetic energy:
\(\Delta E_{\text{k}} = 23.0 - 6.125 = 16.875\text{ J} \approx 17\text{ J}\).
Since kinetic energy is not conserved (energy is lost), the collision is inelastic.

(c) Impulse on \(m_2\):
\(\Delta p_2 = m_2 v - m_2 u_2 = 1.5 \times 1.75 - 1.5 \times (-2.0) = 2.625 + 3.0 = 5.625\text{ kg m s}^{-1}\)
Using the impulse-force relationship:
\(F_{\text{avg}} \Delta t = \Delta p\)
\(150 \times \Delta t = 5.625\)
\[\Delta t = \frac{5.625}{150} = 0.0375\text{ s} \approx 0.038\text{ s}\]

(a)
- Force multiplied by time / change in momentum [1]

(b)(i)
- Uses conservation of momentum with signs: \(m_1 u_1 + m_2 u_2 = (m_1 + m_2)v\) [1]
- Correct calculation: \(7.0 = 4.0 v \implies v = 1.75\text{ m s}^{-1}\) [1]
- Direction: same direction as 2.5 kg block's initial motion [1]

(b)(ii)
- Correct initial \(E_{\text{k}} = 23.0\text{ J}\) [1]
- Correct final \(E_{\text{k}} = 6.13\text{ J}\) [1]
- Compares energies to show they are different and concludes inelastic [1]

(c)
- Correct change in momentum for either block (magnitude \(5.625\text{ N s}\)) [1]
- Uses \(F \Delta t = \Delta p\) [1]
- Correct final time: \(0.038\text{ s}\) (or \(0.0375\text{ s}\)) [1]

(a) Electromotive force (emf) is the work done per unit charge (energy converted from chemical to electrical form) in moving charge completely around a closed circuit.

(b)(i) The relationship is \(V = \varepsilon - Ir\).

(b)(ii) Since \(V = -rI + \varepsilon\):
- The y-intercept of the graph representing \(V\) versus \(I\) is equal to the electromotive force \(\varepsilon\) [1].
- The absolute value of the gradient (slope) of the line is equal to the internal resistance \(r\) [1].

(c)(i) The total resistance in the circuit is:
\(R_{\text{total}} = R + r = 12.0 + 1.5 = 13.5\ \Omega\) [1]
The current \(I\) is:
\(I = \frac{\varepsilon}{R_{\text{total}}} = \frac{6.0}{13.5} = 0.444\text{ A}\)
The total power dissipated in the circuit is:
\(P_{\text{total}} = I \varepsilon = 0.444 \times 6.0 = 2.67\text{ W}\) (or \(2.7\text{ W}\)) [1]

(c)(ii) The useful power delivered to the external resistor is:
\(P_{\text{useful}} = I^2 R = (0.444)^2 \times 12.0 = 2.37\text{ W}\)
Efficiency \(\eta\):
\(\eta = \frac{P_{\text{useful}}}{P_{\text{total}}} = \frac{I^2 R}{I^2(R+r)} = \frac{R}{R+r}\) [1]
\(\eta = \frac{12.0}{13.5} \approx 0.889\) [1]
In percentage form, the efficiency is \(88.9\%\) (or \(89\%\)) [1].

(a)
- Work done per unit charge [1]
- in moving charge completely around a circuit [1]

(b)(i)
- \(V = \varepsilon - Ir\) [1]

(b)(ii)
- \(\varepsilon\) is the y-intercept (when \(I=0\)) [1]
- \(r\) is the negative of the gradient / magnitude of the slope [1]

(c)(i)
- Correct calculation of total resistance \(13.5\ \Omega\) and current \(0.44\text{ A}\) [1]
- Correct calculation of total power: \(2.67\text{ W}\) (or \(2.7\text{ W}\)) [1]

(c)(ii)
- Understands efficiency expression, e.g., \(\frac{R}{R+r}\) or \(\frac{P_{\text{useful}}}{P_{\text{total}}}\) [1]
- Correct substitution: \(\frac{12.0}{13.5}\) or equivalent power values [1]
- Correct final percentage: \(88.9\%\) or \(89\%\) (or \(0.89\)) [1]

(a) The thermal energy supplied per unit time is equal to the electrical power: \(P = m c \frac{\Delta T}{\Delta t}\).

Rearranging for \(c\):
\(c = \frac{P}{m \times \text{gradient}} = \frac{35.0}{0.850 \times 0.046} \approx 895.14\text{ J kg}^{-1}\text{ K}^{-1}\).

Rounding to an appropriate number of significant figures, \(c = 900\text{ J kg}^{-1}\text{ K}^{-1}\) or \(895\text{ J kg}^{-1}\text{ K}^{-1}\).

(b) The fractional uncertainty in \(c\) is the sum of the fractional uncertainties of the constituent quantities:
\(\frac{\Delta c}{c} = \frac{\Delta P}{P} + \frac{\Delta m}{m} + \frac{\Delta (\text{gradient})}{\text{gradient}}\)

\(\frac{\Delta c}{c} = \frac{1.0}{35.0} + \frac{0.005}{0.850} + \frac{0.002}{0.046}\)

\(\frac{\Delta c}{c} \approx 0.02857 + 0.00588 + 0.04348 = 0.07793\)

Percentage uncertainty = \(0.07793 \times 100\\% \approx 7.8\\%\) (or \(8\\%\)).

(c) The absolute uncertainty is:
\(\Delta c = 0.07793 \times 895.14 \approx 69.8\text{ J kg}^{-1}\text{ K}^{-1}\).

Rounding to 1 significant figure gives \(\Delta c = 70\text{ J kg}^{-1}\text{ K}^{-1}\).

Since the uncertainty is in the tens place, the value of \(c\) is rounded to the tens place:
\(c = 900 \pm 70\text{ J kg}^{-1}\text{ K}^{-1}\) (or \(890 \pm 70\text{ J kg}^{-1}\text{ K}^{-1}\)).

(d) One major systematic error is thermal energy loss to the surroundings (due to incomplete insulation).

Because some energy is lost, the rate of temperature rise (gradient) is smaller than it would be under ideal conditions.

Since \(c \propto \frac{1}{\text{gradient}}\), a smaller gradient results in an overestimated (higher) calculated value of \(c\).

(a) [2 marks]
- Correct calculation of \(c\): \(895\) or \(900\text{ J kg}^{-1}\text{ K}^{-1}\) [1]
- Correct unit: \(\text{J kg}^{-1}\text{ K}^{-1}\) or \(\text{J kg}^{-1}\text{ }^\circ\text{C}^{-1}\) [1]

(b) [2 marks]
- Sums fractional uncertainties: \(\frac{1.0}{35.0} + \frac{0.005}{0.850} + \frac{0.002}{0.046}\) [1]
- Obtains percentage uncertainty of \(7.8\%\) or \(8\%\) [1]

(c) [1.5 marks]
- Correctly calculates absolute uncertainty to be \(70\) (or \(69.8\)) [0.5]
- States the final value with consistent significant figures: \(900 \pm 70\) or \(890 \pm 70\) [1] (Accept \(895 \pm 70\))

(d) [2 marks]
- Identifies heat loss to the surroundings [1]
- Explains that this leads to a smaller gradient, which results in an overestimation of the specific heat capacity \(c\) [1]

(a) Measuring a single fringe spacing directly is very small and would lead to a very large percentage uncertainty. By measuring the cumulative width of 7 spacings, the absolute uncertainty in the ruler measurement remains constant (\(\pm 0.1\text{ cm}\)), but the total distance is much larger. This significantly reduces the percentage uncertainty in the determined fringe spacing \(s\).

(b) The fringe spacing is:
\(s = \frac{w}{7} = \frac{4.3\text{ cm}}{7} = 0.6143\text{ cm} \approx 0.61\text{ cm}\) (or \(6.1\text{ mm}\)).

The absolute uncertainty is:
\(\Delta s = \frac{\Delta w}{7} = \frac{0.1\text{ cm}}{7} \approx 0.014\text{ cm}\) (or \(0.14\text{ mm}\)).

Thus, \(s = 6.1 \pm 0.1\text{ mm}\) or \(0.61 \pm 0.01\text{ cm}\).

(c) Using the double-slit formula \(s = \frac{\lambda D}{d}\), we rearrange for \(\lambda\):
\(\lambda = \frac{s d}{D} = \frac{6.143 \times 10^{-3}\text{ m} \times 0.150 \times 10^{-3}\text{ m}}{1.45\text{ m}} \approx 6.355 \times 10^{-7}\text{ m} = 635.5\text{ nm}\).

Now, calculate the fractional uncertainty in \(\lambda\):
\(\frac{\Delta \lambda}{\lambda} = \frac{\Delta s}{s} + \frac{\Delta d}{d} + \frac{\Delta D}{D}\)

\(\frac{\Delta \lambda}{\lambda} = \frac{0.1}{4.3} + \frac{0.002}{0.150} + \frac{0.02}{1.45} \approx 0.02326 + 0.01333 + 0.01379 = 0.05038\) (or \(5.04\\%\)).

The absolute uncertainty in \(\lambda\) is:
\(\Delta \lambda = 0.05038 \times 635.5\text{ nm} \approx 32\text{ nm}\).

Rounding \(\lambda\) and its uncertainty consistently:
\(\lambda = 640 \pm 30\text{ nm}\) or \(636 \pm 32\text{ nm}\).

(d) To reduce the percentage uncertainty in \(s\), the student can:
- Increase the distance \(D\) from the slits to the screen, which increases the fringe spacing \(s\) and thus the total measured width \(w\).
- Use a double-slit with a smaller separation \(d\), which also increases the fringe spacing \(s\).
- Measure a larger number of fringes (e.g., across 15 fringes), which increases the total measured distance \(w\).

(a) [1.5 marks]
- State that measuring a small single fringe spacing results in a high percentage uncertainty [1]
- State that measuring multiple fringes increases the total measured length, thereby reducing the percentage uncertainty [0.5]

(b) [2 marks]
- Calculates fringe spacing: \(s = 6.1\text{ mm}\) (or \(0.61\text{ cm}\)) [1]
- Calculates absolute uncertainty: \(\pm 0.1\text{ mm}\) (or \(\pm 0.01\text{ cm}\)) [1]

(c) [3 marks]
- Correct calculation of wavelength: \(636\text{ nm}\) or \(640\text{ nm}\) [1]
- Correct calculation of fractional or percentage uncertainty: \(5.0\%\) [1]
- Correct final value and absolute uncertainty: \(640 \pm 30\text{ nm}\) or \(636 \pm 32\text{ nm}\) [1]

(d) [1 mark]
- Suggests increasing \(D\) / using smaller slit separation \(d\) / measuring across more fringes [1]

(a) A parsec is defined as the distance at which an object has a parallax angle of one arcsecond (or the distance at which a baseline of \(1\text{ AU}\) subtends an angle of \(1\text{ arcsecond}\)).

(b) First, calculate the distance in parsecs:
\(d = \frac{1}{p} = \frac{1}{0.125} = 8.0\text{ pc}\)
Convert to light-years:
\(d = 8.0 \times 3.26 = 26.1\text{ ly}\) (or \(26\text{ ly}\) to two significant figures).

(c) First, convert the distance to meters:
\(1\text{ pc} = 3.09 \times 10^{16}\text{ m}\)
\(d = 8.0 \times 3.09 \times 10^{16} = 2.472 \times 10^{17}\text{ m}\)
Using the apparent brightness formula:
\(b = \frac{L}{4\pi d^2} \implies L = 4\pi d^2 b\)
\(L = 4\pi (2.472 \times 10^{17})^2 \times 4.2 \times 10^{-11} = 3.225 \times 10^{25}\text{ W} \approx 3.2 \times 10^{25}\text{ W}\)

(d) (i) Using Wien's displacement law:
\(\lambda_{\text{max}} T = 2.90 \times 10^{-3}\text{ m K}\)
\[\lambda_{\text{max}} = \frac{2.90 \times 10^{-3}}{4500} = 6.44 \times 10^{-7}\text{ m} \approx 6.4 \times 10^{-7}\text{ m}\quad (\text{or } 640\text{ nm})\\]
(ii) Astra-9 has a lower surface temperature (4500 K) than the Sun (5800 K) and is less luminous, which indicates that it is a lower-mass main-sequence star. Because lower-mass stars have much lower core pressures and temperatures, they fuse hydrogen at a much slower rate. Consequently, its lifetime on the main sequence is significantly longer than that of the Sun.

(a) Award [1] for a clear definition stating the distance at which 1 AU subtends 1 arcsecond, OR the distance corresponding to a parallax of 1 arcsecond.

(b) Award [1] for calculating distance in pc: \(d = 8.0\text{ pc}\).
Award [1] for multiplying by 3.26 to obtain \(26\text{ ly}\) (accept \(26.1\text{ ly}\)).

(c) Award [1] for converting parsecs to meters: \(2.47 \times 10^{17}\text{ m}\).
Award [1] for quoting or using the formula \(L = 4\pi d^2 b\).
Award [1] for the final correct answer of \(3.2 \times 10^{25}\text{ W}\) (accept range \(3.2 \times 10^{25}\text{ W}\) to \(3.3 \times 10^{25}\text{ W}\)).

(d) (i) Award [1] for quoting Wien's displacement law \(\lambda_{\text{max}} T = 2.90 \times 10^{-3}\text{ m K}\).
Award [1] for final correct peak wavelength \(6.4 \times 10^{-7}\text{ m}\) (accept \(6.44 \times 10^{-7}\text{ m}\) or \(640\text{ nm}\)).
(ii) Award [1] for recognizing that Astra-9 has a lower mass than the Sun.
Award [1] for stating that the lower mass leads to a much slower rate of fusion, resulting in a longer main-sequence lifetime.

(a) Cepheid variables are post-main-sequence stars that cross a region on the HR diagram known as the instability strip. In this phase, the outer layers of the star become unstable and undergo periodic expansion and contraction (pulsations). This cycle of changing size, temperature, and opacity causes the star's overall luminosity to vary periodically.

(b) Given \(P = 10\text{ days}\):
\(\log_{10} \left( \frac{L}{L_{\odot}} \right) = 1.20 \log_{10}(10) + 1.10\)
Since \[\log_{10}(10) = 1\\]:
\(\log_{10} \left( \frac{L}{L_{\odot}} \right) = 1.20(1) + 1.10 = 2.30\)
To find the ratio:
\(\frac{L}{L_{\odot}} = 10^{2.30} = 199.5 \approx 200\text{ } (\text{or } 2.0 \times 10^2)\).

(c) First, calculate the luminosity \(L\) in watts:
\(L = 199.5 \times 3.8 \times 10^{26}\text{ W} = 7.58 \times 10^{28}\text{ W}\)
(If using \(200\), \(L = 7.60 \times 10^{28}\text{ W}\)).
Use the apparent brightness formula to solve for distance \(d\):
\(b = \frac{L}{4\pi d^2} \implies d = \sqrt{\frac{L}{4\pi b}}\)
\(d = \sqrt{\frac{7.58 \times 10^{28}}{4\pi \times 1.5 \times 10^{-12}}} = \sqrt{4.02 \times 10^{39}} = 2.01 \times 10^{20}\text{ m}\)
Convert to parsecs:
\(d = \frac{2.01 \times 10^{20}}{3.09 \times 10^{16}} = 6500\text{ pc}\) (or \(6.5 \times 10^3\text{ pc}\)).

(d) Because the star's initial mass (\(12 M_{\odot}\)) is greater than the Chandrasekhar limit/greater than 8 solar masses, it will evolve into a red supergiant where it fuses heavier elements up to iron. It will eventually experience a core-collapse (Type II) supernova, leaving behind a neutron star as the stellar remnant.

(a) Award [1] for noting that Cepheids are unstable/pulsating stars located in the instability strip of the HR diagram.
Award [1] for describing the mechanism of periodic contraction and expansion (pulsation) of outer layers.
Award [1] for linking this physical pulsation directly to the periodic variation of surface area/temperature/opacity, and thus luminosity.

(b) Award [1] for correctly substituting \(P = 10\) and finding the log value as \(2.30\).
Award [1] for calculating \(\frac{L}{L_{\odot}} = 200\) (accept range \(199\) to \(200\)).

(c) Award [1] for calculating the luminosity in watts: \(7.6 \times 10^{28}\text{ W}\) (or \(7.58 \times 10^{28}\text{ W}\)).
Award [1] for calculating distance in meters: \(2.0 \times 10^{20}\text{ m}\).
Award [1] for converting to parsecs and stating \(6500\text{ pc}\) (accept range \(6400\text{ pc}\) to \(6500\text{ pc}\)).

(d) Award [1] for identifying that it will become a red supergiant and end in a supernova (core collapse/Type II).
Award [1] for identifying that the remnant remaining will be a neutron star.