2024 IB DP Physics 模擬試題連答案詳解

The current in the circuit is given by \(I = \frac{V_0}{R + r}\), which can be rearranged to give the external resistance: \(R = \frac{V_0}{I} - r\). The power dissipated in the external resistor is \(P = I^2 R = I^2 \left(\frac{V_0}{I} - r\right) = V_0 I - r I^2\). This equation represents a parabola that opens downwards. The maximum power occurs at the vertex where \(\frac{dP}{dI} = V_0 - 2rI = 0\), which corresponds to a current of \(I = \frac{V_0}{2r}\).

Award [1] for identifying the correct equation relating power and current, and identifying the correct parabolic shape with its maximum.

By the Rayleigh criterion, the minimum angular separation required for resolution is \(\theta = 1.22 \frac{\lambda}{D}\). Substituting the given values: \(\theta = 1.22 \times \frac{500 \times 10^{-9}\text{ m}}{2.44 \times 10^{-3}\text{ m}} = 0.5 \times 5.0 \times 10^{-4} = 2.5 \times 10^{-4}\text{ rad}\). For a headlight separation \(d = 1.5\text{ m}\), the maximum distance \(L\) is given by \(L = \frac{d}{\theta} = \frac{1.5}{2.5 \times 10^{-4}} = 6000\text{ m} = 6.0\text{ km}\).

Award [1] for calculating the angular separation and resolving it to find the maximum distance.

The photoelectric equation for the first case is \(E_k = \frac{hc}{\lambda} - \Phi\), which gives \(\frac{hc}{\lambda} = E_k + \Phi\), where \(\Phi\) is the work function. For the second case, \(E_k' = \frac{hc}{\lambda/2} - \Phi = 2\frac{hc}{\lambda} - \Phi\). Substituting \(\frac{hc}{\lambda}\) into this equation yields \(E_k' = 2(E_k + \Phi) - \Phi = 2E_k + \Phi\). Since the work function \(\Phi\) is a positive physical constant, it follows that \(E_k' > 2E_k\).

Award [1] for establishing the relationship between the two maximum kinetic energies and correctly concluding that the second is greater than twice the first.

The total energy of the system is \(E = \frac{1}{2} k A^2\). The potential energy is \(E_p = \frac{1}{2} k x^2\) and the kinetic energy is \(E_k = E - E_p = \frac{1}{2} k (A^2 - x^2)\). We are given that \(E_k = 3 E_p\), so \(\frac{1}{2} k (A^2 - x^2) = 3 \left(\frac{1}{2} k x^2\right)\). Simplifying this yields \(A^2 - x^2 = 3 x^2\), which gives \(4 x^2 = A^2\). Taking the square root gives \(x = \pm \frac{A}{2}\).

Award [1] for setting up the energy relation and correctly solving for the displacement.

By conservation of momentum, the initial momentum is zero, so the final momenta must be equal and opposite: \(p_1 = -p_2\), which means their magnitudes are equal: \(|p_1| = |p_2| = p\). The kinetic energy of an object can be written in terms of momentum as \(E = \frac{p^2}{2m}\). Therefore, the ratio of the kinetic energies is \(\frac{E_1}{E_2} = \frac{p^2 / (2m_1)}{p^2 / (2m_2)} = \frac{m_2}{m_1}\). Since \(m_1 = 2 m_2\), we find \(\frac{E_1}{E_2} = \frac{m_2}{2 m_2} = \frac{1}{2}\).

Award [1] for using momentum conservation and the relationship between kinetic energy and momentum to find the ratio.

The total energy \(E_T\) of a satellite in a circular orbit of radius \(r\) is \(E_T = -\frac{GMm}{2r}\). The initial total energy is \(E_{T,i} = -\frac{GMm}{2R}\), and the final total energy is \(E_{T,f} = -\frac{GMm}{2(2R)} = -\frac{GMm}{4R}\). The change in total energy is \(\Delta E_T = E_{T,f} - E_{T,i} = -\frac{GMm}{4R} - \left(-\frac{GMm}{2R}\right) = +\frac{GMm}{4R}\).

Award [1] for calculating the change in total orbital energy using the correct orbital energy formula.

From the ideal gas equation \(P V = n R T\), since the volume \(V\) and pressure \(P\) are held constant, the product \(n T\) must remain constant: \(n_i T_i = n_f T_f\). Given that the temperature is tripled (\(T_f = 3T_i\)), the final number of moles is \(n_f = \frac{1}{3} n_i\). Because the mass of the gas is directly proportional to the number of moles, the remaining mass is \(m_f = \frac{1}{3} m_i\). The fraction of the initial mass that has leaked out is \(1 - \frac{1}{3} = \frac{2}{3}\).

Award [1] for using the gas law to find the remaining fraction of gas and calculating the correct leaked fraction.

The displacement of the accelerating car is \(s_c = \frac{1}{2} a t^2\), and the displacement of the truck moving at constant speed is \(s_t = u t\). When the car overtakes the truck, their displacements are equal: \(\frac{1}{2} a t^2 = u t\). Solving for \(t\) (since \(t > 0\)) yields \(t = \frac{2u}{a}\). The speed of the car at this instant is \(v_c = a t = a \left(\frac{2u}{a}\right) = 2u\).

Award [1] for equating displacements, finding the time of overtake, and determining the final speed of the car.

The two identical external resistors are in parallel, so their equivalent resistance is \( R_{eq} = \frac{R}{2} \). The total resistance of the circuit (including the internal resistance of the battery) is \( R_{total} = R_{eq} + r = \frac{R}{2} + r = \frac{R+2r}{2} \). The current \( I \) flowing through the circuit is \( I = \frac{E}{R_{total}} = \frac{2E}{R+2r} \). The voltmeter measures the terminal potential difference \( V \) of the battery, which is equal to the voltage across the external load: \( V = I R_{eq} = \left( \frac{2E}{R+2r} \right) \left( \frac{R}{2} \right) = \frac{ER}{R+2r} \).

Award 1 mark for the correct option B. Only one option can be selected. No partial credit is given.

The first minimum of the single-slit diffraction pattern occurs at an angle \( \theta \) given by \( \sin\theta \approx \theta = \frac{\lambda}{b} \). Double-slit interference maxima occur at angles satisfying \( d \sin\theta = m\lambda \), or \( \theta \approx \frac{m\lambda}{d} \). The interference maximum that coincides with the first diffraction minimum occurs when \( \frac{m\lambda}{d} = \frac{\lambda}{b} \implies m = \frac{d}{b} = \frac{4.0b}{b} = 4 \). Because this interference maximum coincides with the diffraction minimum (where intensity is zero), the \( m = \pm 4 \) interference fringes are completely suppressed (missing). Therefore, the visible interference maxima within the central diffraction envelope correspond to \( m = 0, \pm 1, \pm 2, \pm 3 \). Counting these values, we have \( 1 + 2 \times 3 = 7 \) fringes.

Award 1 mark for the correct option B. Only one option can be selected. No partial credit is given.

The kinetic energy \( E_k \) gained by the electron accelerated through a potential difference \( V \) is \( E_k = eV \). Since \( E_k = \frac{p^2}{2m} \), the momentum \( p \) is \( p = \sqrt{2m eV} \). The de Broglie wavelength is \( \lambda = \frac{h}{p} = \frac{h}{\sqrt{2m eV}} \). Since \( \lambda \propto \frac{1}{\sqrt{V}} \), if the potential difference increases by a factor of 4, the wavelength decreases by a factor of \( \sqrt{4} = 2 \). Thus, the new wavelength is \( \frac{\lambda}{2} \).

Award 1 mark for the correct option C. Only one option can be selected. No partial credit is given.

For a circular orbit of radius \( r \), the gravitational force provides the centripetal acceleration: \( \frac{GMm}{r^2} = \frac{m v^2}{r} \). This simplifies to \( m v^2 = \frac{GMm}{r} \). The kinetic energy \( E_k \) is \( E_k = \frac{1}{2}mv^2 = \frac{GMm}{2r} \). Therefore, the kinetic energy is inversely proportional to the radius of the orbit: \( E_k \propto \frac{1}{r} \). When the radius increases from \( R \) to \( 3R \), the kinetic energy decreases by a factor of 3. Thus, the ratio \( \frac{E_{k,\text{new}}}{E_{k,\text{initial}}} = \frac{1}{3} \).

Award 1 mark for the correct option D. Only one option can be selected. No partial credit is given.

The change in momentum (impulse) is equal to the area under the force-time graph. The force-time graph forms a triangle with a base of \( 0.50 \text{ s} \) and a height of \( 12 \text{ N} \). The area of this triangle is: \( \text{Impulse} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 0.50 \text{ s} \times 12 \text{ N} = 3.0 \text{ N s} \). The impulse is also equal to \( m \Delta v \). Therefore: \( 3.0 \text{ N s} = 2.0 \text{ kg} \times \Delta v \implies \Delta v = 1.5 \text{ m s}^{-1} \). Since the force acts in the direction of motion, the final speed is \( v = v_{\text{initial}} + \Delta v = 4.0 \text{ m s}^{-1} + 1.5 \text{ m s}^{-1} = 5.5 \text{ m s}^{-1} \).

Award 1 mark for the correct option B. Only one option can be selected. No partial credit is given.

Using the equation of state for an ideal gas, \( PV = nRT \). Since the number of moles \( n \) is constant, we can use the relation: \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \). Substituting the given values: \( \frac{P_0 V_0}{T_0} = \frac{P_2 \left(\frac{V_0}{3}\right)}{2T_0} \). Solving for \( P_2 \): \( P_2 = P_0 \times \left(\frac{V_0}{\frac{V_0}{3}}\right) \times \left(\frac{2T_0}{T_0}\right) = P_0 \times 3 \times 2 = 6P_0 \).

Award 1 mark for the correct option D. Only one option can be selected. No partial credit is given.

Let \( N_0 \) be the initial number of nuclei for both isotopes. For isotope X, after \( t = 18 \text{ hours} \), the number of half-lives elapsed is \( n_X = \frac{18}{3.0} = 6 \). The remaining number of active nuclei is \( N_X = N_0 \left(\frac{1}{2}\right)^6 = \frac{N_0}{64} \). For isotope Y, after \( t = 18 \text{ hours} \), the number of half-lives elapsed is \( n_Y = \frac{18}{9.0} = 2 \). The remaining number of active nuclei is \( N_Y = N_0 \left(\frac{1}{2}\right)^2 = \frac{N_0}{4} \). The ratio of remaining nuclei is \( \frac{N_X}{N_Y} = \frac{N_0/64}{N_0/4} = \frac{4}{64} = \frac{1}{16} \).

Award 1 mark for the correct option C. Only one option can be selected. No partial credit is given.

The total power incident on the planet of radius \( R \) is determined by its projected cross-sectional area: \( P_{\text{incident}} = S \times \pi R^2 \). Since the planet's albedo is \( \alpha \), the fraction of radiation reflected is \( \alpha \), and the fraction absorbed is \( (1-\alpha) \). Thus, the total power absorbed by the planet is \( P_{\text{absorbed}} = S(1-\alpha) \times \pi R^2 \). This absorbed power is distributed and averaged over the entire spherical surface area of the planet, which is \( 4\pi R^2 \). The average intensity of absorbed radiation is \( I_{\text{average}} = \frac{P_{\text{absorbed}}}{4\pi R^2} = \frac{S(1-\alpha)\pi R^2}{4\pi R^2} = \frac{S(1-\alpha)}{4} \).

Award 1 mark for the correct option C. Only one option can be selected. No partial credit is given.

The drift speed \(v\) of free electrons in a conductor is related to the electric current \(I\) by:
\(I = n A e v\)
where \(n\) is the number density of free electrons, \(A\) is the cross-sectional area, and \(e\) is the elementary charge.

From Ohm's law, \(I = \frac{V}{R}\), where the resistance \(R\) is given by:
\(R = \rho \frac{L}{A}\)

Substituting this into the expression for current gives:
\(I = \frac{V A}{\rho L}\)

We can now substitute \(I\) back into the drift speed equation:
\(v = \frac{I}{nAe} = \frac{V A}{\rho L n A e} = \frac{V}{\rho n e L}\)

Since \(V\), \(\rho\), \(n\), and \(e\) remain constant, the drift speed is inversely proportional to the length of the conductor:
\(v \propto \frac{1}{L}\)

Therefore, when the length is doubled to \(2L\), the drift speed is halved to \(\frac{v}{2}\).

Award 1 mark for the correct answer of B.
- 1 mark for identifying that drift speed is inversely proportional to length, leading to half the initial value.

Let the amplitude of the electric field on the screen due to a single slit be \(E_0\).

When both slits are open, the two waves interfere constructively at the central maximum, resulting in a combined amplitude of:
\(E_{\text{total}} = E_0 + E_0 = 2E_0\)

Since intensity \(I\) is proportional to the square of the amplitude (\(I \propto E^2\)), the original intensity \(I_0\) is:
\(I_0 \propto (2E_0)^2 = 4E_0^2\)

When one slit is covered, only one wave reaches the center of the screen, so the new amplitude is \(E_{\text{new}} = E_0\). The new intensity \(I_{\text{new}}\) is:
\(I_{\text{new}} \propto E_0^2\)

Comparing the two expressions:
\(I_{\text{new}} = \frac{I_0}{4}\)

Award 1 mark for the correct answer of A.
- 1 mark for recognizing that closing one slit reduces the electric field amplitude to half, and since intensity is proportional to amplitude squared, intensity reduces to one-quarter.

For an electron confined in a one-dimensional infinite potential well of width \(L\), the wave function forms standing waves where the boundaries must be nodes.

The relationship between the length \(L\) and the de Broglie wavelength \(\lambda\) is given by:
\(L = n \frac{\lambda}{2}\)
where \(n\) is the quantum number representing the energy state.

For the ground state, \(n=1\). For the first excited state, \(n=2\).

Substituting \(n=2\) into the formula:
\(L = 2 \left(\frac{\lambda}{2}\right) = \lambda\)

Therefore, the de Broglie wavelength \(\lambda\) is equal to \(L\).

Award 1 mark for the correct answer of B.
- 1 mark for correctly identifying that the first excited state corresponds to \(n=2\) and solving for \(\lambda = L\).

The total energy of the particle in SHM is:
\(E_{\text{total}} = E_k + E_p = \frac{1}{2} k A^2\)

When kinetic energy equals potential energy, \(E_k = E_p\), which means:
\(E_p = \frac{1}{2} E_{\text{total}} \implies \frac{1}{2} k x^2 = \frac{1}{2} \left(\frac{1}{2} k A^2\right) \implies x = \frac{A}{\sqrt{2}}\)

Since the motion starts at maximum positive displacement, the displacement as a function of time is:
\(x(t) = A \cos(\omega t)\)

Setting \(x(t) = \frac{A}{\sqrt{2}}\):
\(A \cos(\omega t) = \frac{A}{\sqrt{2}} \implies \cos(\omega t) = \frac{1}{\sqrt{2}}\)

The minimum positive value of \(t\) satisfies:
\(\omega t = \frac{\pi}{4}\)

Substitute \(\omega = \frac{2\pi}{T}\):
\(\left(\frac{2\pi}{T}\right) t = \frac{\pi}{4} \implies t = \frac{T}{8}\)

Award 1 mark for the correct answer of A.
- 1 mark for determining that \(x = \frac{A}{\sqrt{2}}\) when energy is shared equally and finding the time \(t = \frac{T}{8}\).

The rate of generation of thermal energy is equal to the power dissipated by the force of friction.

Since the velocity is constant, the net force is zero. Therefore, the pulling force \(F\) is equal in magnitude to the frictional force \(F_f\):
\(F_f = F = \mu m g\)

The power dissipated by the frictional force is:
\(P = F_f v = F v\)

Thus, the correct option represents the power delivered by the pulling force (or equivalently, the thermal energy generation rate), which is \(F v\).

Award 1 mark for the correct answer of B.
- 1 mark for identifying that the rate of thermal energy generation is power, which equals the product of the frictional force (equal to \(F\)) and velocity \(v\).

For circular orbit, the centripetal force is provided by the gravitational attraction:
\(\frac{G M m}{r^2} = \frac{m v^2}{r} \implies m v^2 = \frac{G M m}{r}\)

The kinetic energy \(E_k\) of the satellite is:
\(E_k = \frac{1}{2} m v^2 = \frac{G M m}{2r}\)

The gravitational potential energy \(E_p\) is:
\(E_p = -\frac{G M m}{r}\)

The total mechanical energy \(E\) is the sum of kinetic and potential energies:
\(E = E_k + E_p = \frac{G M m}{2r} - \frac{G M m}{r} = -\frac{G M m}{2r}\)

Award 1 mark for the correct answer of B.
- 1 mark for correctly showing that total mechanical energy is negative and equal to half the potential energy.

The average kinetic energy of gas molecules in an ideal gas is directly proportional to the absolute temperature \(T\):
\(\frac{1}{2} m \langle v^2 \rangle = \frac{3}{2} k_B T\)

This implies that the molecular speed \(v\) is proportional to the square root of the absolute temperature:
\(v \propto \sqrt{T}\)

Therefore, the ratio of the new speed to the initial speed is:
\(\frac{v_{\text{new}}}{v_{\text{initial}}} = \sqrt{\frac{T_{\text{new}}}{T_{\text{initial}}}} = \sqrt{1.44} = 1.2\)

Award 1 mark for the correct answer of A.
- 1 mark for identifying the proportional relationship \(v \propto \sqrt{T}\) and correctly evaluating \(\sqrt{1.44} = 1.2\).

Let \(S\) be the incoming solar power per unit area absorbed by the planet. Without an atmosphere, radiative equilibrium requires:
\(\sigma T_0^4 = S\)

With the atmospheric greenhouse layer:
Let \(T_s\) be the new surface temperature and \(T_a\) be the atmospheric temperature.

For the atmosphere to be in equilibrium, the energy it absorbs from the surface must equal the total energy it emits (both upwards and downwards):
\(\sigma T_s^4 = 2\sigma T_a^4 \implies \sigma T_a^4 = \frac{1}{2}\sigma T_s^4\)

For the surface to be in equilibrium, the incoming energy (solar radiation \(S\) plus downward atmospheric emission \(\sigma T_a^4\)) must equal the radiation emitted by the surface:
\(S + \sigma T_a^4 = \sigma T_s^4\)

Substitute \(\sigma T_a^4 = \frac{1}{2}\sigma T_s^4\) into the surface equation:
\(S + \frac{1}{2}\sigma T_s^4 = \sigma T_s^4 \implies S = \frac{1}{2}\sigma T_s^4 \implies \sigma T_s^4 = 2S\)

Since \(\sigma T_0^4 = S\):
\(\sigma T_s^4 = 2\sigma T_0^4 \implies T_s = 2^{1/4} T_0\)

Award 1 mark for the correct answer of B.
- 1 mark for setting up the heat balance equations for both the surface and atmosphere layer, obtaining \(T_s^4 = 2 T_0^4\), and solving for \(T_s = 2^{1/4} T_0\).

Using the terminal potential difference equation for a cell: \(V = E - I r\). Since \(I = \frac{V}{R}\), we can write this as \(E = V + \frac{V}{R} r\). Setting this up for both states: \(E = V_1 + \frac{V_1}{R_1} r\) and \(E = V_2 + \frac{V_2}{R_2} r\). Equating the two expressions for \(E\): \(V_1 + \frac{V_1}{R_1} r = V_2 + \frac{V_2}{R_2} r\). Rearranging to solve for \(r\): \(V_1 - V_2 = r \left( \frac{V_2}{R_2} - \frac{V_1}{R_1} \right) = r \left( \frac{V_2 R_1 - V_1 R_2}{R_1 R_2} \right)\). Therefore, \(r = \frac{R_1 R_2 (V_1 - V_2)}{V_2 R_1 - V_1 R_2} = \frac{R_1 R_2 (V_2 - V_1)}{V_1 R_2 - V_2 R_1}\).

[1 mark] for the correct derivation equating the emf expressions or terminal voltages and arriving at option A. Reject other options due to algebraic sign errors or dimensional inconsistency.

In air, the fringe spacing is given by \(s_1 = \frac{\lambda D}{d}\). In the liquid of refractive index \(n\), the wavelength of light becomes \(\lambda' = \frac{\lambda}{n}\). The new fringe spacing is \(s_2 = \frac{\lambda' D_2}{d} = \frac{\lambda D_2}{n d}\). For the fringe spacing to remain unchanged, we require \(s_2 = s_1\), which means \(\frac{\lambda D_2}{n d} = \frac{\lambda D}{d}\). Solving for the ratio gives \(\frac{D_2}{D} = n\).

[1 mark] for relating the change in wavelength in a medium to refractive index and using the double slit formula to solve for the correct ratio of distances, matching option A.

The kinetic energy gained by a charged particle accelerated through a potential difference is \(E_k = qV\). The momentum is related to kinetic energy by \(p = \sqrt{2mE_k} = \sqrt{2mqV}\). The de Broglie wavelength is \(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}\). For particle \(Y\), \(\lambda_Y = \frac{h}{\sqrt{2(4m)(2q)(2V)}} = \frac{h}{\sqrt{16 \cdot 2mqV}} = \frac{1}{4} \frac{h}{\sqrt{2mqV}} = \frac{\lambda_X}{4}\).

[1 mark] for relating de Broglie wavelength to mass, charge, and voltage, and calculating the correct scaling factor of 1/4, matching option B.

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 \(E_K = E_T - E_P = \frac{1}{2} k (A^2 - x^2)\). We are given that \(E_K = 3 E_P\), so \(\frac{1}{2} k (A^2 - x^2) = 3 \left( \frac{1}{2} k x^2 \right)\). Simplifying this gives \(A^2 - x^2 = 3 x^2 \implies A^2 = 4 x^2 \implies x = \pm \frac{A}{2}\).

[1 mark] for setting up the energy conservation equation with SHM expressions and solving for the displacement in terms of amplitude, matching option B.

We use the conservation of momentum in two dimensions. Let the initial motion of the mass \(2m\) be along the positive x-axis, so \(\vec{p}_{initial} = 2mv \hat{i}\). After the collision, the mass \(2m\) moves perpendicular to this, say along the positive y-axis: \(\vec{p}_{2m, final} = 2m \left(\frac{v}{2}\right) \hat{j} = mv \hat{j}\). Let the final momentum of the mass \(m\) be \(\vec{p}_{m, final} = p_x \hat{i} + p_y \hat{j}\). By conservation of momentum, \(\vec{p}_{m, final} = \vec{p}_{initial} - \vec{p}_{2m, final} = 2mv \hat{i} - mv \hat{j}\). The magnitude of this momentum is \(p = \sqrt{(2mv)^2 + (-mv)^2} = \sqrt{5} mv\). Since momentum \(p = m v_{final}\), the speed of the mass \(m\) is \(v_{final} = \sqrt{5}v\).

[1 mark] for resolving momentum into perpendicular components, applying conservation of momentum, and finding the resultant speed using Pythagoras' theorem, matching option C.

For a circular orbit of radius \(r\), the gravitational force provides the centripetal acceleration: \(\frac{GMm}{r^2} = \frac{mv^2}{r} \implies \frac{GM}{r} = v^2\). The gravitational potential energy is given by \(E_p = -\frac{GMm}{r}\). For the initial orbit of radius \(R\), we have \(E_{p1} = -\frac{GMm}{R} = -m v^2\). For the final orbit of radius \(4R\), the potential energy is \(E_{p2} = -\frac{GMm}{4R} = -\frac{1}{4} m v^2\). The increase in potential energy is \(\Delta E_p = E_{p2} - E_{p1} = -\frac{1}{4} m v^2 - (-m v^2) = \frac{3}{4} m v^2\).

[1 mark] for relating orbital speed to gravitational variables, writing down correct initial and final potential energies in terms of \(mv^2\), and finding the difference, matching option B.

According to Gay-Lussac's Law for an ideal gas at constant volume, pressure is directly proportional to absolute temperature: \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\). The temperature must be in Kelvin. The initial temperature is \(T_1 = 27 + 273 = 300\text{ K}\). Since the pressure doubles to \(2P\), the absolute temperature must also double: \(T_2 = 2 T_1 = 600\text{ K}\). Converting this back to Celsius: \(T_2 = 600 - 273 = 327^\circ\text{C}\).

[1 mark] for converting Celsius to Kelvin, applying the ideal gas law relationship, and converting the final Kelvin temperature back to Celsius, matching option C.

Let the point of projection at the top of the cliff be the origin, with the upward direction chosen as positive. The displacement of the ball when it hits the ground is \(s = -h\). The initial velocity is \(v_0 = +u\), and the acceleration is \(a = -g\). Using the kinematic equation \(s = v_0 t + \frac{1}{2} a t^2\), we substitute these values: \(-h = u T - \frac{1}{2} g T^2\). Multiplying both sides by \(-1\) yields \(h = \frac{1}{2} g T^2 - u T\).

[1 mark] for applying kinematic equations with correct sign conventions for displacement, initial velocity, and acceleration, and rearranging to match option C.

Let the total current flowing from the source be I. The current passes entirely through the series resistor \(R_1\), so the power dissipated in it is \(P_1 = I^2 R\). The current splits equally between the two identical parallel resistors, so the current through each is \(I/2\). The power dissipated in one of these parallel resistors (say \(R_2\)) is \(P_2 = (I/2)^2 R = \frac{I^2 R}{4}\). Therefore, the ratio \(\frac{P_1}{P_2} = \frac{I^2 R}{I^2 R / 4} = 4\).

Award 1 mark for the correct calculation of currents and power ratios leading to option C.

The angular half-width of the central maximum in single-slit diffraction is given by \(\theta \approx \frac{\lambda}{b}\). The full angular width of the central maximum is \(2\theta = \frac{2\lambda}{b}\). If the wavelength becomes \(2\lambda\) and the slit width becomes \(b/2\), the new angular width is \(\frac{2(2\lambda)}{b/2} = 4 \times \frac{2\lambda}{b}\). This is 4 times the original angular width.

Award 1 mark for identifying the proportional relationship between angular width, wavelength, and slit width to obtain 4.

From Einstein's photoelectric equation, \(e V_s = h f - \Phi\), which gives \(h f = e V_s + \Phi\). When the frequency is doubled, the new stopping potential \(V'_s\) satisfies \(e V'_s = h(2f) - \Phi = 2(h f) - \Phi\). Substituting \(h f\) into this equation yields \(e V'_s = 2(e V_s + \Phi) - \Phi = 2e V_s + \Phi\). Dividing by \(e\) gives \(V'_s = 2 V_s + \frac{\Phi}{e}\).

Award 1 mark for setting up the photoelectric equation for both cases and correctly solving for the new stopping potential.

In simple harmonic motion, the maximum acceleration is given by \(a_{\text{max}} = \omega^2 A\), where \(\omega\) is the angular frequency. Since \(\omega = \frac{2\pi}{T}\), substituting this gives \(a_{\text{max}} = \left(\frac{2\pi}{T}\right)^2 A = \frac{4\pi^2 A}{T^2}\).

Award 1 mark for combining the angular frequency and maximum acceleration formulas correctly.

The impulse-momentum theorem states that Impulse = \(F \Delta t = \Delta p\), where \(\Delta p\) is the change in momentum. The magnitude of the change in momentum is \(|p_f - p_i| = |m(v/3) - mv| = \frac{2}{3} mv\). Therefore, \(F \Delta t = \frac{2}{3} mv\), which gives \(F = \frac{2mv}{3\Delta t}\).

Award 1 mark for applying the impulse-momentum theorem to determine the magnitude of the resisting force.

For a circular orbit, the centripetal force is provided by gravity: \(\frac{m v^2}{R} = \frac{G M m}{R^2}\), which gives the kinetic energy \(E_k = \frac{1}{2} m v^2 = \frac{G M m}{2 R}\). The magnitude of the gravitational potential energy is \(|E_p| = \frac{G M m}{R}\). Therefore, the ratio \(\frac{E_k}{|E_p|} = \frac{G M m / 2R}{G M m / R} = \frac{1}{2}\).

Award 1 mark for expressing kinetic and potential energy in terms of G, M, m, and R, and computing their ratio.

The root-mean-square speed of molecules in an ideal gas is given by \(v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}}\), which means \(v_{\text{rms}} \propto \sqrt{T}\). When the absolute temperature is scaled by a factor of 1.5, the rms speed is scaled by a factor of \(\sqrt{1.5}\). Thus, the new rms speed is \(\sqrt{1.5} v\).

Award 1 mark for recognizing that the rms speed is proportional to the square root of the absolute temperature.

The lifetime \(\tau\) of a star is proportional to the fuel available (mass \(M\)) divided by the rate of consumption (luminosity \(L\)), so \(\tau \propto \frac{M}{L}\). Given that \(L \propto M^4\), the lifetime scales as \(\tau \propto \frac{M}{M^4} = M^{-3}\). Therefore, the ratio of lifetimes is \(\frac{\tau_Y}{\tau_X} = \left(\frac{M_X}{M_Y}\right)^3 = 2^3 = 8\).

Award 1 mark for relating lifetime to mass and luminosity, establishing the inverse-cube relationship, and calculating the ratio of 8.

(a) The electromotive force (emf) is defined as the total work done per unit charge by the source in moving a charge around a complete circuit. (b)(i) Using the circuit equation: \(\varepsilon = I(R + r) + V_{\text{lamp}}\). Substituting the given values: \(6.0 = 0.40(R + 1.5) + 3.2\). This simplifies to \(2.8 = 0.40(R + 1.5)\), leading to \(R + 1.5 = 7.0\), which gives \(R = 5.5\ \Omega\). (b)(ii) The power dissipated in the internal resistance is \(P = I^2 r = (0.40)^2 \times 1.5 = 0.16 \times 1.5 = 0.24\text{ W}\). (c) No. If \(R\) is reduced to zero, the current increases. An increase in current causes the temperature of the filament lamp to rise. Since the resistance of a metal filament increases with temperature, the total resistance of the circuit will be greater than \(r + R_{\text{lamp, initial}}\). Therefore, the current will be less than the expected \(0.80\text{ A}\).

(a) 1 mark for stating work done per unit charge supplied by the source. (b)(i) 1 mark for recalling \(\varepsilon = I(R+r) + V_{\text{lamp}}\), 1 mark for substitution, 1 mark for correct final answer of \(5.5\ \Omega\). (b)(ii) 1 mark for \(0.24\text{ W}\). (c) 1 mark for stating that current will be less than \(0.80\text{ A}\), 1 mark for relating increased current to increased temperature, 1 mark for explaining that higher temperature increases filament resistance.

(a) The angle to the first minimum in a single-slit diffraction pattern is given by \(\theta = \frac{\lambda}{b}\). Substituting the values: \(\theta = \frac{6.33 \times 10^{-7}\text{ m}}{1.20 \times 10^{-4}\text{ m}} = 5.275 \times 10^{-3}\text{ rad}\). The angular width of the central maximum is the angle between the first minima on either side, which is \(2\theta = 2 \times 5.275 \times 10^{-3}\text{ rad} = 1.055 \times 10^{-2}\text{ rad} \approx 0.0106\text{ rad}\). (b) The distance on the screen between the two minima is \(Y = 2\theta \times D = 1.055 \times 10^{-2}\text{ rad} \times 2.00\text{ m} = 0.0211\text{ m} = 2.11\text{ cm}\). (c) According to the Rayleigh criterion for a circular aperture, the minimum angular separation is \(\theta_{\min} = 1.22 \frac{\lambda}{d}\). Calculating this gives \(\theta_{\min} = 1.22 \times \frac{6.33 \times 10^{-7}\text{ m}}{2.50 \times 10^{-3}\text{ m}} = 3.089 \times 10^{-4}\text{ rad}\). The minimum spatial separation is \(s = \theta_{\min} \times L = 3.089 \times 10^{-4}\text{ rad} \times 15.0\text{ m} = 4.63 \times 10^{-3}\text{ m} = 4.63\text{ mm}\).

(a) 1 mark for finding the angle to the first minimum \(\theta = 5.275 \times 10^{-3}\text{ rad}\). 1 mark for doubling to obtain \(0.0106\text{ rad}\). (b) 1 mark for using \(Y = 2\theta D\) or equivalent. 1 mark for the correct answer \(0.0211\text{ m}\) (or \(2.11\text{ cm}\)). (c) 1 mark for Rayleigh criterion formula, 1 mark for calculating \(\theta_{\min} = 3.09 \times 10^{-4}\text{ rad}\), 1 mark for using \(s = L \theta\), 1 mark for correct final answer \(4.63\text{ mm}\).

(a) The work function is the minimum energy required to eject an electron from the surface of a metal. (b) The energy of the incident photon is \(E = h f = 6.63 \times 10^{-34}\text{ J s} \times 7.20 \times 10^{14}\text{ Hz} = 4.774 \times 10^{-19}\text{ J}\item. The work function in joules is \)\Phi = 2.10\text{ eV} \times 1.60 \times 10^{-19}\text{ J eV}^{-1} = 3.360 \times 10^{-19}\text{ J}\). The photoelectric equation gives \(E_{k,\max} = h f - \Phi = 4.774 \times 10^{-19}\text{ J} - 3.360 \times 10^{-19}\text{ J} = 1.414 \times 10^{-19}\text{ J} \approx 1.41 \times 10^{-19}\text{ J}\). (c) The momentum \(p\) of the electron is related to its kinetic energy by \(p = \sqrt{2 m_e E_k}\). Substituting values: \(p = \sqrt{2 \times 9.11 \times 10^{-31}\text{ kg} \times 1.414 \times 10^{-19}\text{ J}} = \sqrt{2.577 \times 10^{-49}} = 5.076 \times 10^{-25}\text{ kg m s}^{-1}\). The de Broglie wavelength is \(\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}\text{ J s}}{5.076 \times 10^{-25}\text{ kg m s}^{-1}} = 1.306 \times 10^{-9}\text{ m} \approx 1.31\text{ nm}\).

(a) 1 mark for correct definition. (b) 1 mark for calculating photon energy in J. 1 mark for converting work function to J. 1 mark for correct subtraction and result \(1.41 \times 10^{-19}\text{ J}\). (c) 1 mark for expressing relation between momentum and kinetic energy. 1 mark for finding momentum \(5.08 \times 10^{-25}\text{ kg m s}^{-1}\). 1 mark for de Broglie formula. 1 mark for correct final wavelength \(1.31\text{ nm}\).

(a) The period of a mass-spring system is \(T = 2\pi \sqrt{\frac{m}{k}}\). Squaring and rearranging for \(k\) gives: \(k = \frac{4\pi^2 m}{T^2} = \frac{4\pi^2 \times 0.350}{1.20^2} = \frac{13.817}{1.44} = 9.595\text{ N m}^{-1} \approx 9.6\text{ N m}^{-1}\). (b) The angular frequency is \(\omega = \frac{2\pi}{T} = \frac{2\pi}{1.20} = 5.236\text{ rad s}^{-1}\). The maximum acceleration is \(a_{\max} = \omega^2 x_0 = (5.236)^2 \times 0.0850\text{ m} = 2.33\text{ m s}^{-2}\). (c) The kinetic energy at displacement \(x\) is given by \(E_k = \frac{1}{2} m \omega^2 (x_0^2 - x^2)\). Substituting the values: \(E_k = 0.5 \times 0.350 \times (5.236)^2 \times (0.0850^2 - 0.0400^2) = 0.5 \times 0.350 \times 27.416 \times (0.007225 - 0.001600) = 4.798 \times 0.005625 = 0.0270\text{ J}\).

(a) 1 mark for writing/rearranging \(T = 2\pi \sqrt{m/k}\). 1 mark for showing value of \(9.60\text{ N m}^{-1}\). (b) 1 mark for finding \(\omega = 5.24\text{ rad s}^{-1}\). 1 mark for correct maximum acceleration \(2.33\text{ m s}^{-2}\). (c) 1 mark for recalling \(E_k = \frac{1}{2} m \omega^2 (x_0^2 - x^2)\) (or conservation of energy equivalent). 1 mark for substituting values. 1 mark for evaluating terms. 1 mark for final correct value of \(0.0270\text{ J}\).

(a) Impulse is defined as the change in momentum (or the product of the average force and the time interval during which it acts). (b) The impulse is \(\Delta p = m v - m u = 0.045 \times 68 - 0 = 3.06\text{ N s}\). The average force is \(F_{\text{average}} = \frac{\Delta p}{\Delta t} = \frac{3.06}{0.82 \times 10^{-3}\text{ s}} = 3732\text{ N} \approx 3.73 \times 10^3\text{ N}\item. (c) The graph should show force on the y-axis and time on the x-axis, with a smooth bell-shaped or triangular curve starting at \)t = 0\), peaking at some value \(F_{\max}\), and returning to zero at \(t = 0.82\text{ ms}\). The area under the force-time curve is the total impulse. For a symmetric peak (like a triangle), the area is roughly \(\frac{1}{2} F_{\max} \Delta t\). Since the average force is defined by \(F_{\text{average}} \Delta t\), the peak force \(F_{\max}\) must be approximately double the average force to yield the same total impulse (area under the curve).

(a) 1 mark for defining impulse as change in momentum. (b) 1 mark for finding the change in momentum (3.06 N s). 1 mark for using \(F = \Delta p / \Delta t\). 1 mark for correct final answer of \(3.73 \times 10^3\text{ N}\). (c) 1 mark for a sketch showing force starting and ending at zero with a single peak. 1 mark for correct labels of \(F_{\max}\) and \(0.82\text{ ms}\). 1 mark for explaining that the area represents impulse. 1 mark for explaining why the peak force must be greater (typically double) than the average force.

(a) The radius of the orbit is \(r = R + h = 6.37 \times 10^6\text{ m} + 3.60 \times 10^5\text{ m} = 6.73 \times 10^6\text{ m}\). The centripetal force is provided by gravity: \(\frac{m v^2}{r} = \frac{G M m}{r^2}\), which gives \(v = \sqrt{\frac{G M}{r}}\). Substituting values: \(v = \sqrt{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.73 \times 10^6}} = \sqrt{5.917 \times 10^7} = 7692\text{ m s}^{-1} \approx 7.69 \times 10^3\text{ m s}^{-1}\). (b) Gravitational potential energy is \(E_p = -\frac{G M m}{r}\). Substituting values: \(E_p = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 1200}{6.73 \times 10^6} = -7.101 \times 10^{10}\text{ J} \approx -7.10 \times 10^{10}\text{ J}\). (c) The total energy of the satellite in circular orbit is \(E_T = E_k + E_p = \frac{1}{2} m v^2 - \frac{G M m}{r} = -\frac{G M m}{2r} = -3.55 \times 10^{10}\text{ J}\). To escape, the total energy must be at least zero. The additional kinetic energy required is therefore equal to the magnitude of the total energy: \(\Delta E_k = 0 - E_T = 3.55 \times 10^{10}\text{ J}\).

(a) 1 mark for finding \(r = 6.73 \times 10^6\text{ m}\). 1 mark for equating centripetal and gravitational forces. 1 mark for correct orbital speed \(7.69 \times 10^3\text{ m s}^{-1}\). (b) 1 mark for using potential energy formula. 1 mark for correct negative answer \(-7.10 \times 10^{10}\text{ J}\). (c) 1 mark for realizing total orbit energy is half the potential energy. 1 mark for stating escape total energy is zero. 1 mark for correct additional kinetic energy \(3.55 \times 10^{10}\text{ J}\).

(a) From the ideal gas equation: \(p V = n R T\). Temperature \(T = 22 + 273.15 = 295.15\text{ K}\). Pressure is \(p = \frac{n R T}{V} = \frac{2.4 \times 8.31 \times 295.15}{0.045} = 1.308 \times 10^5\text{ Pa} \approx 1.31 \times 10^5\text{ Pa}\). (b) The average translational kinetic energy of an ideal gas atom is \(\bar{E}_k = \frac{3}{2} k_B T\). Substituting values: \(\bar{E}_k = 1.5 \times 1.38 \times 10^{-23}\text{ J K}^{-1} \times 295.15\text{ K} = 6.11 \times 10^{-21}\text{ J}\). (c) At constant volume, the number density of the gas is constant. Pressure is determined by the rate of molecular collisions with the walls and the change in momentum per collision. If pressure doubles, the average force exerted during collisions must double, which implies the average kinetic energy of the molecules must double. Since temperature is directly proportional to average kinetic energy, absolute temperature must double. Since kinetic energy is proportional to \(v^2\), doubling the temperature increases the average speed (root-mean-square speed) by a factor of \(\sqrt{2} \approx 1.41\).

(a) 1 mark for converting temperature to Kelvin. 1 mark for correct pressure \(1.31 \times 10^5\text{ Pa}\). (b) 1 mark for using \(\bar{E}_k = \frac{3}{2} k_B T\). 1 mark for substitution of Boltzmann's constant and temperature. 1 mark for correct result \(6.11 \times 10^{-21}\text{ J}\). (c) 1 mark for explaining that pressure is proportional to average kinetic energy at constant volume. 1 mark for stating that temperature is proportional to average kinetic energy. 1 mark for stating that the average speed increases by a factor of \(\sqrt{2}\).

(a) The horizontal component is \(u_x = u \cos\theta = 24.0 \cos(35.0^\circ) = 19.66\text{ m s}^{-1} \approx 19.7\text{ m s}^{-1}\). The vertical component is \(u_y = u \sin\theta = 24.0 \sin(35.0^\circ) = 13.77\text{ m s}^{-1} \approx 13.8\text{ m s}^{-1}\). (b) At the peak of flight, the vertical velocity component is zero (\(v_y = 0\)). Using \(v_y^2 = u_y^2 - 2 g h\): \(0 = (13.77)^2 - 2 \times 9.81 \times h\), leading to \(189.6 = 19.62 h\), which gives \(h = 9.66\text{ m}\). (c) The total time of flight \(t\) can be found using the vertical equation \(s_y = u_y t - \frac{1}{2} g t^2 = 0\). Thus, \(t = \frac{2 u_y}{g} = \frac{2 \times 13.77}{9.81} = 2.807\text{ s}\). The horizontal range is \(R = u_x \times t = 19.66\text{ m s}^{-1} \times 2.807\text{ s} = 55.2\text{ m}\).

(a) 1 mark for horizontal component \(19.7\text{ m s}^{-1}\). 1 mark for vertical component \(13.8\text{ m s}^{-1}\). (b) 1 mark for recalling kinematic relation with \(v_y = 0\). 1 mark for substitution of \(g = 9.81\text{ m s}^{-2}\). 1 mark for correct maximum height \(9.66\text{ m}\). (c) 1 mark for calculating time of flight \(2.81\text{ s}\). 1 mark for range formula \(R = u_x t\). 1 mark for correct range \(55.2\text{ m}\).

(a) The Rayleigh criterion states that two point sources are just resolved when the central maximum of the diffraction pattern of one source falls directly on the first minimum of the diffraction pattern of the other source.

(b) For a circular aperture, the minimum angular separation is given by:
\(\theta_{\text{min}} = 1.22 \frac{\lambda}{D}\)
\(\theta_{\text{min}} = 1.22 \times \frac{550 \times 10^{-9}\text{ m}}{2.4\text{ m}}\)
\(\theta_{\text{min}} = 2.80 \times 10^{-7}\text{ rad}\) (or \(2.8 \times 10^{-7}\text{ rad}\))

(c) The minimum physical separation \(d\) is:
\(d = s \theta_{\text{min}} = 4.2 \times 10^{16}\text{ m} \times 2.80 \times 10^{-7}\text{ rad}\)
\(d = 1.18 \times 10^{10}\text{ m} \approx 1.2 \times 10^{10}\text{ m}\)

(d) Decreasing the wavelength from \(550\text{ nm}\) to \(450\text{ nm}\) decreases the minimum resolvable angular separation \(\theta_{\text{min}}\) since \(\theta_{\text{min}} \propto \lambda\). A smaller \(\theta_{\text{min}}\) means the telescope can resolve smaller details, thereby increasing its resolving power.

(a)
- Central maximum of one diffraction pattern coincides with the first minimum of the other; [1]
- Mentions diffraction/Airy disc patterns of two sources; [1]

(b)
- Recall of \(\theta = 1.22 \frac{\lambda}{D}\); [1]
- Correct calculation leading to \(2.8 \times 10^{-7}\text{ rad}\); [1]

(c)
- Use of \(d = s \theta\); [1]
- Correct calculation leading to \(1.2 \times 10^{10}\text{ m}\) (accept \(1.18 \times 10^{10}\text{ m}\)); [1]

(d)
- States that a shorter wavelength \(\lambda\) decreases \(\theta_{\text{min}}\); [1]
- Explains that a smaller minimum angular separation means higher resolving power / ability to see finer details; [1.18]

(a) The orbital radius is \(r = R + h = 5.2 \times 10^6\text{ m} + 1.8 \times 10^6\text{ m} = 7.0 \times 10^6\text{ m}\).
Equating the gravitational force to the centripetal force:
\(\frac{GMm}{r^2} = \frac{mv^2}{r} \implies v = \sqrt{\frac{GM}{r}}\)
\(v = \sqrt{\frac{6.67 \times 10^{-11} \times 4.8 \times 10^{24}}{7.0 \times 10^6}}\)
\(v = \sqrt{4.57 \times 10^7} \approx 6.76 \times 10^3\text{ m s}^{-1} \approx 6.8 \times 10^3\text{ m s}^{-1}\).

(b) The gravitational potential energy \(E_p\) is:
\(E_p = -\frac{GMm}{r}\)
\(E_p = -\frac{6.67 \times 10^{-11} \times 4.8 \times 10^{24} \times 650}{7.0 \times 10^6}\)
\(E_p = -2.97 \times 10^{10}\text{ J} \approx -3.0 \times 10^{10}\text{ J}\).

(c) The total mechanical energy is:
\(E_{\text{total}} = E_k + E_p = \frac{1}{2}mv^2 - \frac{GMm}{r} = -\frac{GMm}{2r} = \frac{E_p}{2}\)
\(E_{\text{total}} = \frac{-2.97 \times 10^{10}\text{ J}}{2} = -1.49 \times 10^{10}\text{ J} \approx -1.5 \times 10^{10}\text{ J}\). (Alternative: using the shown value of \(v = 6.76 \times 10^3\text{ m s}^{-1}\), \(E_k \approx 1.49 \times 10^{10}\text{ J}\), yielding \(E_{\text{total}} = -1.48 \times 10^{10}\text{ J}\)).

(d) The resistive drag force does work against the satellite's motion, causing its total mechanical energy to decrease (become more negative). Since \(E_{\text{total}} = -\frac{GMm}{2r}\), a decrease in total energy means \(r\) must decrease, so the orbit spirals inwards. As the orbital radius \(r\) decreases, the potential energy decreases twice as fast as the total energy, causing the kinetic energy to increase. Thus, the orbital speed \(v = \sqrt{\frac{GM}{r}}\) increases.

(a)
- Calculates orbital radius \(r = 7.0 \times 10^6\text{ m}\); [1]
- Equates centripetal force to gravitational force and shows \(v \approx 6.8 \times 10^3\text{ m s}^{-1}\) (must show substitution of values); [1]

(b)
- Uses \(E_p = -\frac{GMm}{r}\) with correct negative sign; [1]
- Obtains \(-3.0 \times 10^{10}\text{ J}\) (accept \(-2.97 \times 10^{10}\text{ J}\)); [1]

(c)
- Recognizes \(E_{\text{total}} = \frac{1}{2} E_p\) or calculates \(E_k + E_p\); [1]
- Obtains \(-1.5 \times 10^{10}\text{ J}\) (accept \(-1.49 \times 10^{10}\text{ J}\) or \(-1.48 \times 10^{10}\text{ J}\)); [1]

(d)
- Drag force does work, reducing the total mechanical energy of the satellite; [1]
- Lower total energy results in a smaller orbital radius \(r\) (spiraling inwards); [1]
- Because \(v = \sqrt{\frac{GM}{r}}\), a smaller \(r\) leads to a higher orbital speed \(v\); [0.18]

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

(b) The energy of an incident photon is:
\(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{320 \times 10^{-9}} = 6.216 \times 10^{-19}\text{ J}\)
Converting this energy to \(\text{eV}\):
\(E = \frac{6.216 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ J/eV}} = 3.88\text{ eV}\)
Using Einstein's photoelectric equation:
\(E_{k,\text{max}} = E - \Phi = 3.88\text{ eV} - 2.28\text{ eV} = 1.60\text{ eV}\).

(c) The maximum kinetic energy in Joules is:
\(E_{k,\text{max}} = 1.60 \times 1.60 \times 10^{-19}\text{ J} = 2.56 \times 10^{-19}\text{ J}\)
The momentum \(p\) of the electron is:
\(p = \sqrt{2 m_e E_k} = \sqrt{2 \times 9.11 \times 10^{-31}\text{ kg} \times 2.56 \times 10^{-19}\text{ J}} = 6.83 \times 10^{-25}\text{ kg m s}^{-1}\)
The de Broglie wavelength \(\lambda_{\text{dB}}\) is:
\(\lambda_{\text{dB}} = \frac{h}{p} = \frac{6.63 \times 10^{-34}\text{ J s}}{6.83 \times 10^{-25}\text{ kg m s}^{-1}} = 9.71 \times 10^{-10}\text{ m}\) (or \(0.971\text{ nm}\)).

(d) Classical wave theory states that the energy of a wave is determined by its intensity (amplitude squared), and is independent of its frequency. Therefore, classical theory predicts that:
- Any frequency of light should cause emission if the intensity is high enough.
- At low intensities, there should be a measurable time delay while the electron continuously absorbs energy until it has enough to escape.
Neither prediction is observed; electron emission is instantaneous and only occurs if the frequency is above a specific threshold value, which is explained only by the photon model of light where energy is delivered in discrete packets (\(E = hf\)).

(a)
- Minimum energy; [1]
- Required to remove/liberate an electron from the surface (of a metal); [1]

(b)
- Calculates photon energy as \(3.88\text{ eV}\) or \(6.2 \times 10^{-19}\text{ J}\); [1]
- Subtracts work function to get \(1.60\text{ eV}\); [1]

(c)
- Uses \(p = \sqrt{2mE_k}\) or calculates speed first (\(v \approx 7.5 \times 10^5\text{ m s}^{-1}\)); [1]
- Uses \(\lambda = \frac{h}{p}\) to obtain \(9.7 \times 10^{-10}\text{ m}\) (accept \(9.5 \times 10^{-10}\text{ m}\) to \(9.9 \times 10^{-10}\text{ m}\)); [1]

(d)
- Classical theory states wave energy depends on intensity / does not depend on frequency; [1]
- Classical theory predicts any frequency should cause emission if intensity is high or given enough time; [1]
- Experimental results show instantaneous emission only above a threshold frequency (which is independent of intensity); [0.18]

a) To collect the data, the student must measure the terminal potential difference across the cell (or across the external circuit) and the current flowing through the circuit. The voltmeter must be connected in parallel with either the cell or the variable resistor. The ammeter must be connected in series with the cell and the variable resistor.

b) The relationship between terminal potential difference \(V\), emf \(E\), current \(I\), and internal resistance \(r\) is given by:
\(V = E - Ir \implies V = -rI + E\)
Comparing this to the equation of a straight line, \(y = mx + c\):
- The vertical intercept (on the \(V\)-axis) represents the electromotive force, \(E\).
- The gradient of the graph represents the negative internal resistance, \(-r\).

c) The power delivered to the external resistor is given by:
\(P = VI = 1.26 \times 0.45 = 0.567 \text{ W}\)

To find the absolute uncertainty in \(P\), first sum the fractional uncertainties of \(V\) and \(I\):
\(\frac{\Delta P}{P} = \frac{\Delta V}{V} + \frac{\Delta I}{I} = \frac{0.04}{1.26} + \frac{0.02}{0.45} \approx 0.0317 + 0.0444 = 0.0761\)

Now, calculate the absolute uncertainty \(\Delta P\):
\(\Delta P = 0.0761 \times 0.567 \approx 0.043 \text{ W}\)

Applying proper significant figures (uncertainties are typically stated to 1 significant figure):
\(\Delta P = 0.04 \text{ W}\) (giving \(P = (0.57 \pm 0.04) \text{ W}\)).

d) A systematic error that shifts all voltmeter readings upwards by \(0.10 \text{ V}\) will shift the entire line of best fit upwards along the vertical axis by \(0.10 \text{ V}\). However, the slope (gradient) of the line remains unchanged because the difference between any two voltage measurements remains constant. Since the internal resistance \(r\) is derived solely from the gradient of the graph, this systematic error has no effect on the determined value of \(r\).

a) [2 marks]
- Correct circuit symbols for cell, variable resistor, ammeter, and voltmeter connected in a single loop [1]
- Voltmeter connected correctly in parallel across either the cell or the variable resistor [1]

b) [2 marks]
- Identifies the vertical intercept as the emf \(E\) [1]
- Identifies the gradient as the negative of the internal resistance \(-r\) (accept 'magnitude of the gradient is the internal resistance') [1]

c) [2 marks]
- Calculates the sum of fractional uncertainties: \(\frac{0.04}{1.26} + \frac{0.02}{0.45} \approx 0.076\) [1]
- Determines the absolute uncertainty to 1 or 2 sig figs: \(\Delta P \approx 0.04 \text{ W}\) (or \(0.043 \text{ W}\)) [1]

d) [2 marks]
- States that a constant systematic shift does not alter the gradient of the graph [1]
- Concludes that the experimental value for \(r\) remains unchanged [1]

a) The independent variable is the length of the pendulum, \(L\) (which is directly varied by the experimenter). The dependent variable is the period of oscillation, \(T\) (which is measured as a consequence of changing the length).

b) Squaring both sides of the equation yields:
\(T^2 = \frac{4\pi^2}{g} L\)
This is in the form of a linear equation \(y = mx\), where \(y = T^2\) and \(x = L\).
Therefore, the gradient \(m\) is given by:
\(m = \frac{4\pi^2}{g}\)
Rearranging this gives:
\(g = \frac{4\pi^2}{m}\)

c) First, calculate the value of \(g\):
\(g = \frac{4\pi^2}{4.02} \approx 9.820 \text{ m s}^{-2}\)

Since \(g = \frac{4\pi^2}{m}\), the fractional uncertainty in \(g\) is equal to the fractional uncertainty in the gradient \(m\):
\(\frac{\Delta g}{g} = \frac{\Delta m}{m} = \frac{0.08}{4.02} \approx 0.0199\)

Now, calculate the absolute uncertainty \(\Delta g\):
\(\Delta g = 0.0199 \times 9.820 \approx 0.20 \text{ m s}^{-2}\)

Reporting the final value with its uncertainty:
\(g = (9.8 \pm 0.2) \text{ m s}^{-2}\) (or \(g = (9.82 \pm 0.20) \text{ m s}^{-2}\))

d) The absolute uncertainty introduced by human reaction time when starting and stopping the stopwatch (typically around \(\pm 0.2 \text{ s}\)) remains roughly constant regardless of the duration measured. By timing 20 oscillations, this total absolute uncertainty is divided by 20 to find the period, thereby reducing the random uncertainty in the period of a single oscillation.

a) [1 mark]
- Correctly identifies both the independent variable (length \(L\)) and the dependent variable (period \(T\) or time \(t\)) [1]

b) [2 marks]
- Shows the linearised relationship \(T^2 = \left(\frac{4\pi^2}{g}\right)L\) [1]
- States that the gradient is \(m = \frac{4\pi^2}{g}\) and hence \(g = \frac{4\pi^2}{m}\) [1]

c) [3 marks]
- Correctly calculates \(g \approx 9.82 \text{ m s}^{-2}\) (accept \(9.8\)) [1]
- Identifies that \(\frac{\Delta g}{g} = \frac{\Delta m}{m}\) (or equivalent fractional uncertainty step) [1]
- Calculates the absolute uncertainty \(\Delta g = 0.2 \text{ m s}^{-2}\) (accept range \(0.19\) to \(0.20\)) [1]

d) [1 mark]
- Mentions that human reaction time uncertainty is constant, so dividing the total time by 20 divides the absolute uncertainty by 20 / reduces the percentage uncertainty in \(T\) [1]

(a) The parallax angle is defined as the angle subtended at a star by the semi-major axis of the Earth's orbit (1 AU).

(b) Using the parallax formula:
\(d = \frac{1}{p}\)
where \(p = 0.310''\).
\(d = \frac{1}{0.310} = 3.226 \text{ pc} \approx 3.2 \text{ pc}\).

(c) First, convert the distance \(d\) from parsecs to meters:
\(d = 3.226 \text{ pc} \times (3.08 \times 10^{16} \text{ m pc}^{-1}) = 9.94 \times 10^{16} \text{ m}\).
(If using \(d = 3.2 \text{ pc}\), then \(d = 3.2 \times 3.08 \times 10^{16} = 9.86 \times 10^{16} \text{ m}\)).

Using the apparent brightness formula:
\(b = \frac{L}{4\pi d^2} \implies L = 4\pi d^2 b\)
\(L = 4 \pi (9.94 \times 10^{16})^2 \times 1.1 \times 10^{-8} = 1.36 \times 10^{27} \text{ W}\).
(If using \(d = 3.2 \text{ pc}\), \(L = 4 \pi (9.86 \times 10^{16})^2 \times 1.1 \times 10^{-8} = 1.34 \times 10^{27} \text{ W}\)).

(d) The spectral class is determined by analyzing the specific absorption lines present in the star's spectrum. Since different elements are ionized or excited at different temperatures, the presence and strength of particular spectral lines indicate the surface temperature of the star, classifying it into O, B, A, F, G, K, or M.

(a) [1] Award [1] for a definition of parallax angle as the angle subtended at the star by the radius of Earth's orbit, or half the angular shift over 6 months.

(b) [2]
[1] for stating \(d = 1/p\) and substituting \(p = 0.310\).
[1] for showing calculation to get \(3.2\) or \(3.23 \text{ pc}\).

(c) [3]
[1] for converting \(3.23 \text{ pc}\) to meters (accept range \(9.8 \times 10^{16}\) to \(1.0 \times 10^{17} \text{ m}\)).
[1] for correctly rearranging \(L = 4\pi d^2 b\).
[1] for final answer of \(1.3 \times 10^{27} \text{ W}\) or \(1.4 \times 10^{27} \text{ W}\) (allow ECF from incorrect distance conversion).

(d) [2]
[1] for mentioning absorption lines in the stellar spectrum.
[1] for connecting these lines to temperature or specific elements/ions excited at that temperature.

(a) Cepheid variables undergo periodic changes in luminosity due to outer-layer pulsations. There is a well-established period-luminosity relation. By measuring the period, astronomers can determine its absolute luminosity, which makes it a 'standard candle' to calculate distance.

(b) Find the luminosity \(L\) in watts:
\(L = 1.2 \times 10^4 \times 3.8 \times 10^{26} \text{ W} = 4.56 \times 10^{30} \text{ W}\).

Use the apparent brightness formula to find the distance \(d\) in meters:
\(b = \frac{L}{4\pi d^2} \implies d = \sqrt{\frac{L}{4\pi b}}\)
\(d = \sqrt{\frac{4.56 \times 10^{30}}{4\pi \times 4.8 \times 10^{-16}}} = \sqrt{7.56 \times 10^{44}} = 2.75 \times 10^{22} \text{ m}\).

Convert distance to Mpc:
\(1 \text{ pc} = 3.08 \times 10^{16} \text{ m}\)
\(1 \text{ Mpc} = 3.08 \times 10^{22} \text{ m}\)
\(d = \frac{2.75 \times 10^{22}}{3.08 \times 10^{22}} = 0.89 \text{ Mpc}\) (or \(0.893 \text{ Mpc}\)).

(c) At very large distances (e.g., beyond 50-100 Mpc), individual stars cannot be resolved because they are too faint and blur into the general galaxy light.

(a) [2]
[1] for stating there is a known relation between period and luminosity.
[1] for explaining that measuring the period gives luminosity, and together with apparent brightness, distance can be found.

(b) [4]
[1] for calculating \(L = 4.56 \times 10^{30} \text{ W}\).
[1] for using the formula \(d = \sqrt{\frac{L}{4\pi b}}\).
[1] for calculating distance in meters as \(2.75 \times 10^{22} \text{ m}\) (allow range \(2.7 \times 10^{22} \text{ m}\) to \(2.8 \times 10^{22} \text{ m}\)).
[1] for final answer of \(0.89 \text{ Mpc}\) (allow range \(0.88\) to \(0.91 \text{ Mpc}\)).

(c) [1] Award [1] for noting that at extreme distances individual stars are too faint to be resolved / detected.

(a) Using the relation:
\(\frac{L}{L_{\odot}} = \left(\frac{M}{M_{\odot}}\right)^{3.5} = 12^{3.5}\)
\(12^{3.5} = 12^3 \times \sqrt{12} = 1728 \times 3.464 = 5986 \approx 6000\).

(b) Post-main sequence evolution:
1. The star leaves the main sequence and expands to become a red supergiant as helium fusion begins in a shell around the core.
2. Heavy element fusion occurs in concentric shells, culminating in an iron core.
3. The core undergoes rapid gravitational collapse when it exceeds the Chandrasekhar limit, leading to a supernova explosion.
4. The remaining core is compressed into a neutron star (since the initial mass is between \(8M_{\odot}\) and about \(20 M_{\odot}\)).

(c) A white dwarf is supported against gravity by electron degeneracy pressure, which arises from the Pauli exclusion principle preventing electrons from occupying the same quantum state. A neutron star is supported by neutron degeneracy pressure, which arises because neutrons are squeezed so close together that they provide a similar degeneracy pressure.

(a) [2]
[1] for identifying the ratio formula \(L/L_{\odot} = (12/1)^{3.5}\).
[1] for showing that this value is \(5986\) or \(6000\).

(b) [4]
[1] for transition to red supergiant phase.
[1] for noting shell fusion of heavier elements ending with an iron core.
[1] for describing gravitational collapse resulting in a supernova.
[1] for identifying the final remnant as a neutron star.

(c) [2]
[1] for stating that white dwarfs are supported by electron degeneracy pressure.
[1] for stating that neutron stars are supported by neutron degeneracy pressure (both relate to Pauli exclusion principle for different particles).

(a) Wien's displacement law relates peak wavelength and temperature:
\(\lambda_{\text{max}} T = 2.90 \times 10^{-3} \text{ m K}\)
\(T = \frac{2.90 \times 10^{-3}}{1.06 \times 10^{-3}} = 2.736 \text{ K} \approx 2.7 \text{ K}\).

(b) The temperature of the cosmic background radiation is inversely proportional to the scale factor \(R\) of the universe:
\(T \propto \frac{1}{R}\)
Therefore, the expansion factor is:
\(\frac{R_{\text{today}}}{R_{\text{recomb}}} = \frac{T_{\text{recomb}}}{T_{\text{today}}} = \frac{3000 \text{ K}}{2.736 \text{ K}} \approx 1100\).

(c) The redshift \(z\) of the galaxy is:
\(z = \frac{\Delta \lambda}{\lambda_0} = \frac{702.2 - 656.3}{656.3} = \frac{45.9}{656.3} = 0.06994\).

Since \(z \ll 1\), we can use the linear Doppler relation:
\(z = \frac{v}{c} \implies v = z c\)
\(v = 0.06994 \times 3.00 \times 10^8 \text{ m s}^{-1} = 2.10 \times 10^7 \text{ m s}^{-1}\) (or \(2.1 \times 10^4 \text{ km s}^{-1}\)).

(a) [2]
[1] for stating Wien's displacement law formula.
[1] for substituting and obtaining \(2.74 \text{ K}\) or \(2.7 \text{ K}\).

(b) [2]
[1] for stating that scale factor \(R\) is inversely proportional to temperature (or equivalent relation \(T \propto 1/R\)).
[1] for calculating the expansion factor as \(1100\) (allow range \(1090\) to \(1110\)).

(c) [3]
[1] for calculating \(z = 0.070\) (or \(0.0699\)).
[1] for stating/using \(v = z c\).
[1] for final answer of \(2.10 \times 10^7 \text{ m s}^{-1}\) (or \(2.1 \times 10^4 \text{ km s}^{-1}\)).