2023 IB DP Physics 模擬試題連答案詳解

The induced electromotive force (emf) is given by Faraday's Law: \(\varepsilon = BLv\).
The induced current in the loop is \(I = \frac{\varepsilon}{R} = \frac{BLv}{R}\).
The rate of thermal energy dissipation (power) is \(P = I^2 R = \frac{B^2 L^2 v^2}{R}\).
The time taken for the square loop of length \(L\) to completely exit the magnetic field at velocity \(v\) is \(\Delta t = \frac{L}{v}\).
Therefore, the total thermal energy dissipated is: \(E = P \Delta t = \frac{B^2 L^2 v^2}{R} \cdot \frac{L}{v} = \frac{B^2 L^3 v}{R}\).

Award 1 mark for the correct option B.
[1 mark] for correctly showing the energy calculation: \(E = P \Delta t\) with correct substitutions.

The de Broglie wavelength is \(\lambda = \frac{h}{p}\). Since both particles have the same wavelength, they must have the same momentum: \(p_e = p_p = p\).
The kinetic energy in terms of momentum is given by \(E = \frac{p^2}{2mass}\).
Thus, for the electron, \(E_e = \frac{p^2}{2m}\), and for the proton, \(E_p = \frac{p^2}{2M}\).
Taking the ratio: \(\frac{E_e}{E_p} = \frac{p^2 / (2m)}{p^2 / (2M)} = \frac{M}{m}\).

Award 1 mark for the correct option C.
[1 mark] for recognizing that equal wavelengths imply equal momentum and using \(E_k = \frac{p^2}{2m}\) to find the correct inverse ratio.

By conservation of energy, the gravitational potential energy at the top is converted into both translational and rotational kinetic energy at the bottom:
\(Mgh = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2\).
Since the sphere rolls without slipping, \(\omega = \frac{v}{R}\).
Substituting the moment of inertia and \(\omega\):
\(Mgh = \frac{1}{2} M v^2 + \frac{1}{2} \left(\frac{2}{5}MR^2\right)\left(\frac{v}{R}\right)^2\)
\(Mgh = \frac{1}{2} M v^2 + \frac{1}{5} M v^2 = \frac{7}{10} M v^2\).
Solving for \(v\):
\(v = \sqrt{\frac{10}{7}gh}\).

Award 1 mark for the correct option B.
[1 mark] for applying conservation of energy including both translational and rotational kinetic energy terms with correct substitution of \(\omega = v/R\).

The angular half-width \(\theta\) of the central maximum in single-slit diffraction is given by \(\theta \approx \frac{\lambda}{b}\).
The linear width \(W\) of the central maximum on a screen at distance \(D\) is proportional to \(\frac{\lambda}{b}\).
Initially, \(W_1 \propto \frac{\lambda}{b}\).
With the new parameters, \(\lambda' = 2\lambda\) and \(b' = \frac{b}{2}\):
\(W_2 \propto \frac{2\lambda}{b/2} = 4 \left(\frac{\lambda}{b}\right)\).
Therefore, the ratio \(\frac{W_2}{W_1} = 4\).

Award 1 mark for the correct option C.
[1 mark] for recognizing that the width of the central maximum is proportional to \(\frac{\lambda}{b}\) and calculating the factor of 4.

Using Ohm's law, the current in each case can be found from the terminal potential difference and external resistance:
Case 1: \(I_1 = \frac{V_1}{R_1} = \frac{6.0\text{ V}}{3.0\ \Omega} = 2.0\text{ A}\).
Case 2: \(I_2 = \frac{V_2}{R_2} = \frac{8.0\text{ V}}{8.0\ \Omega} = 1.0\text{ A}\).
Using the equation for terminal voltage \(E = V + Ir\):
\(E = 6.0 + 2.0r\)
\(E = 8.0 + 1.0r\)
Equating the two expressions:
\(6.0 + 2.0r = 8.0 + 1.0r \implies r = 2.0\ \Omega\).

Award 1 mark for the correct option B.
[1 mark] for determining the current in each state and solving the simultaneous equations to find \(r = 2.0\ \Omega\).

Let the charge \(+q\) be at the origin \(x = 0\) and \(-4q\) be at \(x = d\).
We look for a point between them at distance \(x\) (where \(0 < x < d\)) such that the potential \(V = 0\).
The potential at position \(x\) is:
\(V = \frac{kq}{x} + \frac{k(-4q)}{d-x} = 0\).
This simplifies to:
\(\frac{1}{x} = \frac{4}{d-x}\)
\(d-x = 4x \implies 5x = d \implies x = \frac{d}{5}\).
Thus, the potential is zero at a distance of \(\frac{d}{5}\) from the charge \(+q\).

Award 1 mark for the correct option A.
[1 mark] for setting up the electric potential equation \(V_1 + V_2 = 0\) and correctly solving for the distance from the positive charge.

The component of velocity parallel to the wall, \(v \sin\theta\), is unchanged during the elastic collision.
The component of velocity perpendicular to the wall changes from \(v \cos\theta\) (towards the wall) to \(-v \cos\theta\) (away from the wall).
The change in momentum of the block is:
\(\Delta p = p_f - p_i = -mv\cos\theta - mv\cos\theta = -2mv\cos\theta\).
The impulse delivered to the wall is equal in magnitude and opposite in direction to the change in momentum of the block, so:
\(J = 2mv\cos\theta\).

Award 1 mark for the correct option B.
[1 mark] for using vector components to calculate the change in perpendicular momentum and obtaining the correct expression.

In an adiabatic expansion, there is no thermal energy transfer: \(Q = 0\).
Since the gas expands, it does positive work on the surroundings: \(W > 0\).
According to the First Law of Thermodynamics, \(\Delta U = Q - W = -W < 0\).
Since internal energy decreases, the temperature of the ideal gas must decrease.
For a reversible adiabatic process, the change in entropy is defined by \(\Delta S = \int \frac{dQ_{rev}}{T} = 0\), so entropy remains constant.

Award 1 mark for the correct option A.
[1 mark] for relating adiabatic work to internal energy change to find the temperature decrease, and noting that entropy change is zero for a reversible adiabatic process.

As the loop moves out of the magnetic field at constant speed \(v\), the magnetic flux \(\Phi\) through the loop changes. In a time interval \(\Delta t\), the loop moves a distance \(\Delta x = v \Delta t\), reducing the area inside the field by \(\Delta A = L \Delta x = L v \Delta t\). According to Faraday's law of induction, the magnitude of the induced electromotive force (emf) \(\varepsilon\) is: \(\varepsilon = \frac{\Delta \Phi}{\Delta t} = B \frac{\Delta A}{\Delta t} = B L v\). The rate of electrical energy dissipation (power \(P\)) in the loop of resistance \(R\) is given by Joule's law: \(P = \frac{\varepsilon^2}{R} = \frac{(B L v)^2}{R} = \frac{B^2 L^2 v^2}{R}\).

Award [1] for the correct answer B. Award [0] for other options. Correct identification of induced emf \(\varepsilon = B L v\) and its relationship to power dissipation \(P = \frac{\varepsilon^2}{R}\) leads to Option B.

The kinetic energy \(E_k\) of a particle of charge \(q\) accelerated through a potential difference \(V\) is \(E_k = qV\). The momentum \(p\) of the particle is related to its kinetic energy by: \(p = \sqrt{2 m E_k} = \sqrt{2 m q V}\). The de Broglie wavelength is: \(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m q V}}\). For the electron: \(\lambda_{\text{electron}} = \frac{h}{\sqrt{2 m e V}}\). For the proton (mass \(M\), charge \(e\), potential difference \(4V\)): \(\lambda_{\text{proton}} = \frac{h}{\sqrt{2 M e (4V)}} = \frac{h}{2\sqrt{2 M e V}}\). Taking the ratio: \(\frac{\lambda_{\text{electron}}}{\lambda_{\text{proton}}} = 2 \sqrt{\frac{M}{m}}\).

Award [1] for the correct answer A. Award [0] for other options. Correctly using the de Broglie wavelength formula and finding the ratio of 2 times \(\sqrt{M/m}\) leads to Option A.

For a single slit diffraction, the angle \(\theta\) of the first minimum is given by the single-slit diffraction equation: \(\theta = \frac{\lambda}{b}\). When the wavelength is changed to \(\lambda' = 1.5\lambda\) and the slit width is changed to \(b' = 0.5b\), the new angular position \(\theta'\) is: \(\theta' = \frac{1.5\lambda}{0.5b} = 3.0 \left(\frac{\lambda}{b}\right) = 3.0\theta\).

Award [1] for the correct answer C. Award [0] for other options. Identification of \(\theta \propto \frac{\lambda}{b}\) and finding the ratio \(\frac{1.5}{0.5} = 3\) leads to Option C.

Let the charge \(Q\) be placed at a distance \(x\) from the charge \(+q\). Its distance from the charge \(+4q\) is \(d - x\). For the net electrostatic force on \(Q\) to be zero, the magnitude of the repulsive force from \(+q\) must equal the magnitude of the repulsive force from \(+4q\): \(k \frac{q Q}{x^2} = k \frac{4q Q}{(d - x)^2}\). Simplifying this expression by dividing both sides by \(k q Q\) yields: \(\frac{1}{x^2} = \frac{4}{(d - x)^2}\). Taking the positive square root of both sides (since \(x\) lies between \(0\) and \(d\)) gives: \(\frac{1}{x} = \frac{2}{d - x} \implies d - x = 2x \implies 3x = d \implies x = \frac{d}{3}\).

Award [1] for the correct answer C. Award [0] for other options. Setting up Coulomb's law force balance and solving the square root equation leads to Option C.

Let the direction of the initial velocity be positive. Initial momentum is \(p_i = m v\) and final momentum is \(p_f = -m \left(\frac{v}{2}\right)\) since it moves in the opposite direction. The change in momentum of the ball is: \(\Delta p = p_f - p_i = -\frac{1}{2} m v - m v = -\frac{3}{2} m v\). The magnitude of the change in momentum is \(|\Delta p| = \frac{3}{2} m v\). According to Newton's second law, the magnitude of the average force \(F\) is: \(F = \frac{|\Delta p|}{\Delta t} = \frac{3 m v}{2 \Delta t}\).

Award [1] for the correct answer C. Award [0] for other options. Correctly determining the magnitude of the momentum change as \(\frac{3}{2}mv\) and dividing by \(\Delta t\) leads to Option C.

Let the current passing through the series resistor be \(I\). Since the other two resistors are identical and in parallel, the current \(I\) splits equally between them. Therefore, the current through each of the parallel resistors is \(\frac{I}{2}\). The power dissipated in the series resistor is \(P_{\text{series}} = I^2 R\). The power dissipated in one of the parallel resistors is \(P_{\text{parallel}} = \left(\frac{I}{2}\right)^2 R = \frac{I^2 R}{4}\). The ratio is \(\frac{P_{\text{parallel}}}{P_{\text{series}}} = 0.25\).

Award [1] for the correct answer A. Award [0] for other options. Correct identification that current in parallel branches is half the main current and power is proportional to current squared leads to Option A.

The luminosity \(L\) of a star modeled as a black-body radiator is given by the Stefan-Boltzmann law: \(L = 4 \pi R^2 \sigma T^4\), where \(R\) is the radius and \(T\) is the absolute surface temperature. Let \(R_X\) and \(T_X\) be the radius and temperature of Star X. For Star Y, \(R_Y = 2 R_X\) and \(T_Y = 0.5 T_X\). The luminosity of Star Y is: \(L_Y = 4 \pi (2 R_X)^2 \sigma (0.5 T_X)^4 = 4 \pi (4 R_X^2) \sigma (0.0625 T_X^4) = 0.25 L_X\). Therefore, the ratio is: \(\frac{L_X}{L_Y} = \frac{1}{0.25} = 4\).

Award [1] for the correct answer D. Award [0] for other options. Correctly applying the proportional relationship and showing that \(L_Y = 0.25 L_X\) leads to Option D.

When unpolarized light of intensity \(I_0\) passes through the first polarizing filter, it becomes polarized and its intensity is reduced by half: \(I_1 = \frac{1}{2} I_0\). When this polarized light passes through the second filter, the intensity of the transmitted light is given by Malus's law: \(I_2 = I_1 \cos^2(\theta)\). Given \(\theta = 60^\circ\) and \(\cos(60^\circ) = 0.5\): \(I_2 = \left(\frac{1}{2} I_0\right) \cos^2(60^\circ) = \left(\frac{1}{2} I_0\right) (0.5)^2 = \frac{1}{8} I_0\).

Award [1] for the correct answer C. Award [0] for other options. Correct application of the unpolarized-to-polarized intensity reduction factor of half and Malus's law with \(\cos^2(60^\circ) = \frac{1}{4}\) leads to Option C.

The magnetic flux through the square loop is initially \(\Phi = B A = B L^2\), and finally zero. The rate of change of magnetic flux is \(\frac{\Delta \Phi}{\Delta t} = \frac{B L^2}{\Delta t}\). According to Faraday's law, the magnitude of the induced electromotive force (emf) is \(\varepsilon = \frac{\Delta \Phi}{\Delta t} = \frac{B L^2}{\Delta t}\). The power dissipated as thermal energy in the loop of resistance \(R\) is given by \(P = \frac{\varepsilon^2}{R} = \frac{B^2 L^4}{R (\Delta t)^2}\). The total thermal energy dissipated in the time interval \(\Delta t\) is \(E = P \Delta t = \frac{B^2 L^4}{R \Delta t}\).

Award [1] for the correct answer (B). Award [0] for other options.

From Einstein's photoelectric equation, the maximum kinetic energy is given by: \(E_{K, \text{max}} = \frac{hc}{\lambda} - \Phi\). For the first case: \(E_K = \frac{hc}{\lambda} - \Phi \implies \frac{hc}{\lambda} = E_K + \Phi\). For the second case: \(3E_K = \frac{hc}{\lambda/2} - \Phi = \frac{2hc}{\lambda} - \Phi\). Substitute the expression for \(\frac{hc}{\lambda}\) from the first equation into the second equation: \(3E_K = 2(E_K + \Phi) - \Phi \implies 3E_K = 2E_K + 2\Phi - \Phi \implies \Phi = E_K\).

Award [1] for the correct answer (B). Award [0] for other options.

The diffraction grating equation is \(d \sin\theta = n \lambda\), where \(d\) is the slit spacing in metres. Given \(n = 2\) and \(\theta = 30^\circ\), we have: \(d \sin(30^\circ) = 2 \lambda \implies d (0.5) = 2 \lambda \implies d = 4\lambda\text{ m}\). The number of lines per metre is \(N_{\text{metre}} = \frac{1}{d} = \frac{1}{4\lambda}\). Since there are \(10^{-3}\) metres in a millimetre, the number of lines per millimetre is: \(N_{\text{mm}} = N_{\text{metre}} \times 10^{-3} = \frac{10^{-3}}{4\lambda} = \frac{1}{4\lambda \times 10^3}\).

Award [1] for the correct answer (A). Award [0] for other options.

The equivalent resistance of the parallel combination of \(R_2\) and \(R_3\) is \(R_{23} = \frac{R \cdot R}{R + R} = \frac{R}{2}\). The total resistance of the circuit is \(R_{\text{total}} = R_1 + R_{23} = R + \frac{R}{2} = \frac{3R}{2}\). The total current drawn from the source is \(I = \frac{V}{R_{\text{total}}} = \frac{V}{3R/2} = \frac{2V}{3R}\). The potential difference across the parallel section (the voltmeter reading across \(R_2\)) is \(V_2 = I \cdot R_{23} = \left(\frac{2V}{3R}\right) \cdot \left(\frac{R}{2}\right) = \frac{V}{3}\).

Award [1] for the correct answer (B). Award [0] for other options.

Let \(v_1\) be the final velocity of the block of mass \(m\) and \(v_2\) be the final velocity of the block of mass \(3m\). From 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 separation equals the relative speed of approach: \(v_2 - v_1 = v\). Adding these two equations gives: \(4v_2 = 2v \implies v_2 = +\frac{v}{2}\). Substituting this into the separation equation gives: \(v_1 = v_2 - v = \frac{v}{2} - v = -\frac{v}{2}\).

Award [1] for the correct answer (A). Award [0] for other options.

By the Stefan-Boltzmann law, the total power radiated by a star of radius \(R\) and surface temperature \(T\) is \(P = \sigma A T^4 = \sigma (4\pi R^2) T^4 \propto R^2 T^4\). Therefore: \(\frac{P_Y}{P_X} = \frac{R_Y^2 T_Y^4}{R_X^2 T_X^4} = \frac{(4R)^2 (2T)^4}{R^2 T^4} = 16 \times 16 = 256\).

Award [1] for the correct answer (C). Award [0] for other options.

The kinetic energy gained by the electron is equal to the work done by the electric field: \(e V = \frac{1}{2} m_e v^2 \implies v = \sqrt{\frac{2eV}{m_e}}\). Inside the magnetic field, the magnetic force provides the centripetal force: \(e v B = \frac{m_e v^2}{r} \implies r = \frac{m_e v}{e B}\). Substituting \(v\) gives: \(r = \frac{m_e}{e B} \sqrt{\frac{2eV}{m_e}} = \sqrt{\frac{m_e^2}{e^2 B^2} \cdot \frac{2eV}{m_e}} = \sqrt{\frac{2 m_e V}{e B^2}}\).

Award [1] for the correct answer (A). Award [0] for other options.

The energy of an emitted photon is related to the wavelength by \(E = \frac{hc}{\lambda}\). For the transition from \(E_4\) to \(E_2\), we have \(E_4 - E_2 = \frac{hc}{\lambda} \implies hc = \lambda (E_4 - E_2)\). For the transition from \(E_3\) to \(E_1\), the wavelength \(\lambda'\) is given by: \(E_3 - E_1 = \frac{hc}{\lambda'}\). Substituting \(hc\) gives: \(E_3 - E_1 = \frac{\lambda (E_4 - E_2)}{\lambda'} \implies \lambda' = \lambda \frac{E_4 - E_2}{E_3 - E_1}\).

Award [1] for the correct answer (A). Award [0] for other options.

The magnetic flux through a single loop of the coil is given by \(\Phi = B A\). Since the coil is square of side length \(s\), the area is \(A = s^2\). The total magnetic flux linkage through the coil of \(N\) turns is \(N \Phi = N B s^2\). According to Faraday's law, the magnitude of the average induced emf is given by the rate of change of magnetic flux linkage: \(\mathcal{E} = \frac{\Delta (N \Phi)}{\Delta t} = \frac{N B s^2}{t}\).

[1 mark] awarded for the correct option A. [0 marks] for incorrect options.

The wavelength of a photon of energy \(E\) is given by \(\lambda = \frac{h c}{E}\), which means its momentum is \(p = \frac{h}{\lambda} = \frac{E}{c}\). Since the electron has the same de Broglie wavelength \(\lambda\), its momentum is also given by the de Broglie relation \(p = \frac{h}{\lambda}\). Therefore, the momentum of the electron is also equal to \(\frac{E}{c}\).

[1 mark] awarded for the correct option A. [0 marks] for incorrect options.

By conservation of energy, the gravitational potential energy lost equals the total kinetic energy gained (translational plus rotational): \(Mgh = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2\). Since it rolls without slipping, \(\omega = \frac{v}{R}\). Substituting this and \(I = \frac{1}{2} M R^2\) into the equation: \(Mgh = \frac{1}{2} M v^2 + \frac{1}{2} (\frac{1}{2} M R^2) (\frac{v}{R})^2 = \frac{1}{2} M v^2 + \frac{1}{4} M v^2 = \frac{3}{4} M v^2\). Solving for \(v\) gives: \(v = \sqrt{\frac{4}{3}gh}\).

[1 mark] awarded for the correct option B. [0 marks] for incorrect options.

By conservation of momentum, \(m u = (m + 3m) v\), where \(v\) is the common velocity after the collision. This gives \(v = \frac{u}{4}\). The initial kinetic energy is \(E_k = \frac{1}{2} m u^2\). The final kinetic energy is \(E_f = \frac{1}{2} (4m) v^2 = \frac{1}{2} (4m) (\frac{u}{4})^2 = \frac{1}{8} m u^2\). The energy lost is \(\Delta E_k = E_k - E_f = \frac{1}{2} m u^2 - \frac{1}{8} m u^2 = \frac{3}{8} m u^2\). The fraction of initial kinetic energy lost is \(\frac{\Delta E_k}{E_k} = \frac{3}{4}\).

[1 mark] awarded for the correct option C. [0 marks] for incorrect options.

The formula for double-slit fringe spacing is \(s = \frac{\lambda D}{d}\). If the new slit separation is \(d' = 2d\) and the new screen distance is \(D' = \frac{D}{2}\), the new fringe spacing is \(s' = \frac{\lambda (D/2)}{2d} = \frac{1}{4} \frac{\lambda D}{d} = \frac{s}{4}\).

[1 mark] for the correct option D. [0 marks] for incorrect options.

Let \(+q\) be at \(x = 0\) and \(-4q\) at \(x = L\). For the net field to be zero, the position must be outside the two charges and closer to the smaller charge in magnitude (to the left of \(+q\) at some distance \(d\)). Setting the field magnitudes equal: \(\frac{k q}{d^2} = \frac{k (4q)}{(L+d)^2}\) gives \(\frac{1}{d} = \frac{2}{L+d}\). Solving this yields \(d = L\). Therefore, the point is at a distance \(L\) to the left of \(+q\).

[1 mark] for the correct option B. [0 marks] for incorrect options.

The parallel combination has resistance \(R_p = \frac{R}{2}\). The total resistance of the circuit is \(R_{eq} = R + \frac{R}{2} = \frac{3}{2} R\). The total current is \(I = \frac{V}{R_{eq}} = \frac{2V}{3R}\). The current divides equally in the parallel branches, so the current through one parallel resistor is \(I_p = \frac{I}{2} = \frac{V}{3R}\). The power dissipated in it is \(P = I_p^2 R = (\frac{V}{3R})^2 R = \frac{V^2}{9R}\).

[1 mark] for the correct option A. [0 marks] for incorrect options.

The useful power output of the motor is the rate of work done in lifting the load: \(P_{out} = F v = m g v\). The efficiency is given by \(\eta = \frac{P_{out}}{P_{in}}\). Rearranging for the electrical power input gives \(P_{in} = \frac{P_{out}}{\eta} = \frac{m g v}{\eta}\).

[1 mark] for the correct option B. [0 marks] for incorrect options.

When the loop falls at terminal velocity, the magnetic force acting upwards on the loop balances the downward gravitational force:

\(F_M = mg\)

The induced electromotive force (emf) in the bottom segment of the loop moving at speed \(v\) is:

\(V = BLv\)

The induced current \(I\) flowing through the loop of resistance \(R\) is:

\(I = \frac{V}{R} = \frac{BLv}{R}\)

The upward magnetic force acting on this current-carrying wire of length \(L\) is:

\(F_M = I L B = \left(\frac{BLv}{R}\right) L B = \frac{B^2 L^2 v}{R}\)

Equating this to the gravitational force:

\(\frac{B^2 L^2 v}{R} = mg \implies v = \frac{mgR}{B^2 L^2}\)

Award [1] for the correct answer A.
- [1] mark for identifying that gravitational force balances magnetic force at terminal speed.
- [1] mark for correctly substituting the expression for induced current into the magnetic force formula to solve for velocity.

The kinetic energy of the accelerated electron is given by:

\(E_k = eV\)

Using the relationship between kinetic energy and momentum:

\(p = \sqrt{2m E_k} = \sqrt{2meV\)}

The de Broglie wavelength is:

\(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV\)}}

This shows that the wavelength is inversely proportional to the square root of the potential difference:

\(\lambda \propto \frac{1}{\sqrt{V}}\)

Therefore:

\(\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}}\)

Given that \(\lambda_2 = \frac{1}{3}\lambda_1\), we have:

\(\frac{1}{3} = \sqrt{\frac{V_1}{V_2}} \implies \frac{V_1}{V_2} = \frac{1}{9} \implies \frac{V_2}{V_1} = 9\)

Award [1] mark for the correct answer B.
- [1] mark for showing that the de Broglie wavelength is inversely proportional to the square root of potential difference.
- [1] mark for calculating the ratio of potential differences correctly.

According to the Rayleigh criterion for a circular aperture, the minimum angular separation \(\theta\) required to resolve two point sources is:

\(\theta = 1.22 \frac{\lambda}{D}\)

Let the new wavelength be \(\lambda' = 2\lambda\) and the new diameter be \(D' = \frac{D}{2}\).

The new minimum angular separation \(\theta'\) is:

\(\theta' = 1.22 \frac{\lambda'}{D'} = 1.22 \frac{2\lambda}{\frac{D}{2}} = 4 \left(1.22 \frac{\lambda}{D}\right) = 4\theta\)

Award [1] mark for the correct answer C.
- [1] mark for applying the Rayleigh criterion formula.
- [1] mark for correctly evaluating the effect of both changes to find the factor of 4.

The moment of inertia of the uniform disc is:

\(I_{\text{disc}} = \frac{1}{2} M R^2\)

Since the clay is dropped vertically onto the outer edge at distance \(R\), the moment of inertia of the clay is:

\(I_{\text{clay}} = m R^2 = \left(\frac{M}{2}\right) R^2 = \frac{1}{2} M R^2\)

The total final moment of inertia of the system is:

\(I_f = I_{\text{disc}} + I_{\text{clay}} = \frac{1}{2} M R^2 + \frac{1}{2} M R^2 = M R^2\)

By conservation of angular momentum (since no external torques act on the system):

\(I_i \omega = I_f \omega_f\)

\(\left(\frac{1}{2} M R^2\right) \omega = (M R^2) \omega_f \implies \omega_f = \frac{1}{2}\omega\)

Award [1] mark for the correct answer B.
- [1] mark for calculating the final moment of inertia as twice the initial moment of inertia.
- [1] mark for applying the conservation of angular momentum to find the new angular speed.

The net work done during a thermodynamic cycle is represented by the area enclosed by the path on a \(P\)-\(V\) diagram.

The path forms a right-angled triangle on the \(P\)-\(V\) plane with vertices at:
- \(A(V, P)\)
- \(B(2V, P)\)
- \(C(2V, \frac{1}{2}P)\)

The base of this triangle (along the constant pressure line) has length:

\(\Delta V = 2V - V = V\)

The height of the triangle (along the constant volume line) is:

\(\Delta P = P - \frac{1}{2}P = \frac{1}{2}P\)

The area of the triangle is:

\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times V \times \frac{1}{2}P = \frac{1}{4}PV\)

Since the cycle is clockwise (\(A \to B\) expansion, then compression), the net work done by the gas is positive: \(W_{\text{net}} = \frac{1}{4}PV\).

Award [1] mark for the correct answer A.
- [1] mark for recognizing that the net work done is the enclosed area of the triangle on the P-V diagram.
- [1] mark for evaluating the area formula correctly.

The total useful output power \(P_{\text{out}}\) of the pump must provide both the gravitational potential energy per second and the kinetic energy per second to the water.

1. Gravitational potential power:

\(P_p = \frac{\Delta m}{\Delta t} g h = 15 \times 10 \times 20 = 3000\text{ W}\)

2. Kinetic power:

\(P_k = \frac{1}{2} \frac{\Delta m}{\Delta t} v^2 = \frac{1}{2} \times 15 \times (10)^2 = 750\text{ W}\)

Total output power:

\(P_{\text{out}} = P_p + P_k = 3000 + 750 = 3750\text{ W}\)

Since the efficiency is \(\eta = 0.60\):

\(P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{3750}{0.60} = 6250\text{ W} = 6.25\text{ kW}\)

Award [1] mark for the correct answer C.
- [1] mark for calculating both gravitational and kinetic energy transfers per unit time.
- [1] mark for applying the efficiency formula to determine the correct input power.

The total external resistance \(R_{\text{ext}}\) of the combination is:
- The parallel combination of two resistors \(R\) has resistance:

\(R_p = \frac{R}{2} = 0.5R\)

- In series with the third resistor \(R\), the total resistance is:

\(R_{\text{ext}} = R_p + R = 0.5R + R = 1.5R\)

According to the Maximum Power Transfer Theorem, the power dissipated in the external circuit is maximized when the external resistance equals the internal resistance of the power source:

\(R_{\text{ext}} = r\)

\(1.5R = 3.0\ \Omega \implies R = 2.0\ \Omega\)

Award [1] mark for the correct answer B.
- [1] mark for finding the total equivalent resistance of the network in terms of R.
- [1] mark for applying the maximum power transfer condition \(R_{\text{ext}} = r\) to solve for R.

Let the charge \(+q\) be at \(x = 0\) and the charge \(-4q\) be at \(x = d\).

Let the point where potential is zero be at a distance \(x\) from \(+q\) (where \(0 < x < d\)).

The electric potential \(V\) at this point is the sum of the potentials due to each charge:

\(V = \frac{k q}{x} + \frac{k (-4q)}{d - x} = 0\)

This simplifies to:

\(\frac{k q}{x} = \frac{4 k q}{d - x}\)

\(\frac{1}{x} = \frac{4}{d - x}\)

\(d - x = 4x \implies 5x = d \implies x = \frac{d}{5}\)

Award [1] mark for the correct answer A.
- [1] mark for writing the total potential expression as the sum of scalar potentials.
- [1] mark for setting the potential equal to zero and solving for \(x\) algebraically.

a. Solar radiation (mostly short-wavelength visible light and ultraviolet) passes through the atmosphere and is absorbed by the Earth's surface. The surface re-emits this energy as long-wavelength infrared radiation. Greenhouse gas molecules in the atmosphere absorb specific wavelengths of this infrared radiation because the frequencies match the vibrational modes (resonance) of the gas molecules. The molecules then re-emit this radiation in all directions, including back towards the Earth's surface, which further warms the surface.

b. The solar constant (intensity \(I\)) is inversely proportional to the square of the distance \(r\) from the Sun:
\(I \propto \frac{1}{r^2}\)
\(I_{\text{Mars}} = I_{\text{Earth}} \times \left(\frac{1.00}{1.52}\right)^2\)
\(I_{\text{Mars}} = 1360 \times \frac{1}{2.3104} = 588.64\text{ W m}^{-2} \approx 589\text{ W m}^{-2}\)

c. In radiative equilibrium, the total power absorbed by Mars equals the total power radiated by Mars:
\(\text{Power absorbed} = (1 - \alpha) I \pi R^2\)
\(\text{Power radiated} = \sigma 4\pi R^2 T^4\)
Equating the two yields:
\((1 - \alpha) I = 4 \sigma T^4\)
\(T = \left[ \frac{(1 - \alpha) I}{4 \sigma} \right]^{1/4}\)
Substituting the values:
\(T = \left[ \frac{(1 - 0.250) \times 588.64}{4 \times 5.67 \times 10^{-8}} \right]^{1/4}\)
\(T = \left[ \frac{441.48}{2.268 \times 10^{-7}} \right]^{1/4} = (1.9466 \times 10^9)^{1/4} = 210.1\text{ K} \approx 210\text{ K}\)

a.
- Solar radiation (short wavelength) passes through atmosphere and warms Earth [1 mark]
- Earth re-emits longer wavelength infrared radiation [1 mark]
- Greenhouse gases absorb infrared radiation due to resonance / vibrational states matching IR frequency [1 mark]
- Gases re-emit IR in all directions, some back to Earth's surface, increasing temperature [1 mark]

b.
- Use of inverse-square law: \(I \propto \frac{1}{r^2}\) [1 mark]
- Substitution of values: \(1360 \times \left(\frac{1}{1.52}\right)^2\) [1 mark]
- Final answer: \(589\text{ W m}^{-2}\) (accept 590 or 588.6) [1 mark]

c.
- Equating absorbed power to emitted power: \((1 - \alpha) I = 4 \sigma T^4\) [1.25 marks]
- Correct substitution of values including \(\sigma = 5.67 \times 10^{-8}\) [1 mark]
- Rearranging for \(T\) correctly [1 mark]
- Final answer: \(210\text{ K}\) (accept \(210.1\text{ K}\); reject if in Celsius unless converted properly) [1 mark]

a. The total resistance of the circuit is \(R_{\text{total}} = R + r\). The current \(I\) is given by Ohm's law: \(I = \frac{\varepsilon}{R+r}\). The power \(P\) dissipated in the variable resistor \(R\) is \(P = I^2 R\). Substituting \(I\) gives: \(P = \left(\frac{\varepsilon}{R+r}\right)^2 R = \frac{\varepsilon^2 R}{(R+r)^2}\).
To find the condition for maximum power, differentiate \(P\) with respect to \(R\) and set equal to zero: \(\frac{dP}{dR} = \varepsilon^2 \frac{(R+r)^2 - 2R(R+r)}{(R+r)^4} = 0 \implies (R+r) - 2R = 0 \implies R = r\). Alternatively, qualitatively: if \(R\) is very small, \(I\) is high but \(R\) is low (so \(P\) is small); if \(R\) is very large, \(I\) is very small (so \(P\) is small). The peak occurs at the matching resistance, \(R = r\).

b. (i) Total resistance of the series circuit is:
\(R_{\text{total}} = r + R_1 + R_{\text{LDR}} = 1.50\ \Omega + 3.50\ \Omega + 5.00\ \Omega = 10.00\ \Omega\)
The current in the circuit is:
\(I = \frac{\varepsilon}{R_{\text{total}}} = \frac{6.00\text{ V}}{10.00\ \Omega} = 0.60\text{ A}\)

(ii) The terminal potential difference is:
\(V = \varepsilon - Ir = 6.00\text{ V} - (0.60\text{ A} \times 1.50\ \Omega) = 6.00\text{ V} - 0.90\text{ V} = 5.10\text{ V}\)

c. As light intensity decreases, the resistance of the LDR increases. This increases the total resistance of the circuit, which decreases the current \(I\). Since the potential difference across the fixed resistor is given by \(V_1 = I R_1\), a decrease in \(I\) leads to a decrease in the potential difference across \(R_1\).

a.
- Correct expression for current \(I = \frac{\varepsilon}{R+r}\) [1 mark]
- Derivation of power formula \(P = I^2 R = \frac{\varepsilon^2 R}{(R+r)^2}\) [1 mark]
- Outline of maximum power condition: showing \(\frac{dP}{dR} = 0\) leading to \(R = r\) OR explaining qualitative limits at \(R \to 0\) and \(R \to \infty\) with a peak at \(R=r\) [2 marks]

b.
(i)
- Summing resistances correctly: \(R_{\text{total}} = 10.0\ \Omega\) [1.25 marks]
- Correct current: \(0.60\text{ A}\) [1 mark]
(ii)
- Use of \(V = ε - Ir\) [1 mark]
- Correct calculation: \(5.10\text{ V}\) (accept 5.1 V) [1 mark]

c.
- States that LDR resistance increases [1 mark]
- Explains that total circuit resistance increases, reducing current [1 mark]
- Concludes that potential difference across fixed resistor decreases because \(V = I R_1\) [1 mark]

a. We use conservation of linear momentum in the \(x\) and \(y\) directions.

Initial momentum:
\(p_{ix} = M v_0 = 1200\text{ kg} \times 350\text{ m s}^{-1} = 420000\text{ kg m s}^{-1}\)
\(p_{iy} = 0\)

For piece A after the explosion:
\(p_{Ax} = m_A v_A \cos(30^\circ) = 400\text{ kg} \times 500\text{ m s}^{-1} \cos(30^\circ) = 173205\text{ kg m s}^{-1}\)
\(p_{Ay} = m_A v_A \sin(30^\circ) = 400\text{ kg} \times 500\text{ m s}^{-1} \sin(30^\circ) = 100000\text{ kg m s}^{-1}\)

By conservation of momentum for piece B:
\(p_{Bx} = p_{ix} - p_{Ax} = 420000 - 173205 = 246795\text{ kg m s}^{-1}\)
\(p_{By} = p_{iy} - p_{Ay} = 0 - 100000 = -100000\text{ kg m s}^{-1}\)

Velocity components of piece B:
\(v_{Bx} = \frac{p_{Bx}}{m_B} = \frac{246795}{800} = 308.49\text{ m s}^{-1}\)
\(v_{By} = \frac{p_{By}}{m_B} = \frac{-100000}{800} = -125.00\text{ m s}^{-1}\)

Magnitude of velocity \(v_B\):
\(v_B = \sqrt{v_{Bx}^2 + v_{By}^2} = \sqrt{308.49^2 + (-125.00)^2} = \sqrt{95166 + 15625} \approx 332.85\text{ m s}^{-1} \approx 333\text{ m s}^{-1}\)

Direction angle \(\theta\) below the positive \(x\)-axis:
\(\theta = \tan^{-1}\left(\frac{125.00}{308.49}\right) = 22.05^\circ \approx 22.1^\circ\)

b. Initial Kinetic Energy:
\(E_{ki} = \frac{1}{2} M v_0^2 = 0.5 \times 1200 \times 350^2 = 7.350 \times 10^7\text{ J}\)

Final Kinetic Energy:
\(E_{kf} = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 = 0.5 \times 400 \times 500^2 + 0.5 \times 800 \times 332.85^2\)
\(E_{kf} = 5.000 \times 10^7 + 4.431 \times 10^7 = 9.431 \times 10^7\text{ J}\)

Change in kinetic energy:
\(\Delta E_k = 9.431 \times 10^7 - 7.350 \times 10^7 = +2.081 \times 10^7\text{ J} \approx +2.08 \times 10^7\text{ J}\)

Since kinetic energy increases (is not conserved), the explosion is inelastic.

c. According to Newton's third law, the force piece A exerts on piece B (\(F_{AB}\)) is equal in magnitude and opposite in direction to the force piece B exerts on piece A (\(F_{BA}\)): \(F_{AB} = -F_{BA}\).
The time interval \(\Delta t\) over which these forces act is identical. Since impulse \(\Delta p = F \Delta t\), the change in momentum of A is equal and opposite to the change in momentum of B: \(\Delta p_A = -\Delta p_B\). Thus, the net change in momentum of the system is zero, meaning momentum is conserved.

a.
- Correct initial momentum components [1 mark]
- Correct momentum components for piece A [1 mark]
- Correct calculation of piece B's velocity components: \(v_{Bx} = 308.5\text{ m s}^{-1}\) and \(v_{By} = -125\text{ m s}^{-1}\) [1 mark]
- Magnitude of \(v_B = 333\text{ m s}^{-1}\) (accept range 332-333) [1 mark]
- Direction of \(v_B = 22.1^\circ\) below the positive x-axis (or \(-22.1^\circ\)) [1 mark]

b.
- Correct calculation of initial KE (\(7.35 \times 10^7\text{ J}\)) [1 mark]
- Correct calculation of final KE (\(9.43 \times 10^7\text{ J}\)) [1 mark]
- Calculates difference and identifies as "inelastic" (kinetic energy is not conserved) [1.25 marks]

c.
- States Newton's third law: Action and reaction forces are equal and opposite [1 mark]
- Connects forces to impulse: \(F \Delta t = \Delta p\) [1 mark]
- Concludes that equal and opposite impulses lead to zero net momentum change [1 mark]

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. (i) For a circular aperture, the minimum resolvable angular separation \(\theta_{\text{min}}\) is given by:
\(\theta_{\text{min}} = \frac{1.22 \lambda}{D}\)
\(\theta_{\text{min}} = \frac{1.22 \times 580 \times 10^{-9}\text{ m}}{1.50\text{ m}} = 4.717 \times 10^{-7}\text{ rad} \approx 4.72 \times 10^{-7}\text{ rad}\)

(ii) The physical separation \(s\) at distance \(d\) is:
\(s = d \times \theta_{\text{min}}\)
\(s = 8.50 \times 10^{16}\text{ m} \times 4.717 \times 10^{-7}\text{ rad} = 4.01 \times 10^{10}\text{ m}\)

c. Resolution:
Since \(\theta_{\text{min}} \propto \frac{1}{D}\), doubling the diameter of the lens halves the minimum angular separation (to \(2.36 \times 10^{-7}\text{ rad}\)). This improves the resolution (allowing smaller details to be distinguished).

Brightness:
The area of the aperture is proportional to \(D^2\). Doubling \(D\) increases the light-collecting area by a factor of \(2^2 = 4\). Therefore, the image becomes 4 times brighter.

a.
- Mentions diffraction patterns of the two sources [1 mark]
- Mentions "first minimum" of one pattern [1 mark]
- Aligns with "central maximum/principal maximum" of the other pattern [1 mark]

b.
(i)
- Use of \(\theta = \frac{1.22 \lambda}{D}\) [1 mark]
- Final answer: \(4.72 \times 10^{-7}\text{ rad}\) [1 mark]
(ii)
- Use of \(s = d \theta\) [1 mark]
- Substitution of correct values [1 mark]
- Final answer: \(4.01 \times 10^{10}\text{ m}\) (accept \(4.0 \times 10^{10}\text{ m}\)) [1 mark]

c.
- Resolution: Smaller angular limit \(\theta_{\text{min}}\) by a factor of 2, hence resolution improves / is doubled [2 marks]
- Brightness: Increases by a factor of 4, because area depends on \(D^2\) [1.25 marks]

a. Faraday's law of induction states that the magnitude of the induced electromotive force (emf) is directly proportional to the rate of change of magnetic flux linkage.
Lenz's law states that the direction of the induced current opposes the change in magnetic flux that created it. If the induced current assisted the change, the magnetic force would accelerate the magnet, increasing both its kinetic energy and generating electrical energy. This would create energy out of nothing, violating the law of conservation of energy. Thus, work must be done against the opposing force to convert mechanical energy into electrical energy.

b. First, find the area of the loop:
\(A = \pi r^2 = \pi \times (0.0400\text{ m})^2 = 5.027 \times 10^{-3}\text{ m}^2\)

The initial magnetic flux linkage \(\Phi_i\) is:
\(\Phi_i = N B A = 150 \times 0.250\text{ T} \times 5.027 \times 10^{-3}\text{ m}^2 = 0.1885\text{ Wb}\)

Since the field is reduced to zero, the final flux linkage is \(\Phi_f = 0\).
The change in flux linkage is \(\Delta \Phi = 0.1885\text{ Wb}\).

The average induced emf \(\varepsilon\) is:
\(\varepsilon = \frac{\Delta \Phi}{\Delta t} = \frac{0.1885\text{ Wb}}{0.120\text{ s}} = 1.571\text{ V} \approx 1.57\text{ V}\)

c. As the coil rotates, the angle \(\theta\) between the normal to the coil's surface and the magnetic field changes continuously as \(\theta = \omega t\). The flux linkage varies sinusoidally: \(\Phi = N B A \cos(\omega t)\). By Faraday's law, the rate of change of flux is \(\frac{d\Phi}{dt} = -N B A \omega \sin(\omega t)\), yielding an induced emf \(\varepsilon = N B A \omega \sin(\omega t)\). This output voltage changes direction (polarity) periodically, which produces an alternating current in the circuit.

a.
- Correct statement of Faraday's law [2 marks]
- Explanation of Lenz's law in terms of opposing work [1 mark]
- Linking opposing work to the preservation of conservation of energy [1 mark]

b.
- Correct area: \(5.03 \times 10^{-3}\text{ m}^2\) [1.25 marks]
- Correct calculation of initial flux linkage: \(0.189\text{ Wb}\) [1 mark]
- Use of Faraday's formula: \(\varepsilon = \frac{\Delta \Phi}{\Delta t}\) [1 mark]
- Final answer: \(1.57\text{ V}\) (accept range 1.56-1.58 V) [1 mark]

c.
- Recognizes that rotation changes the angle between the field and the normal to the coil [1 mark]
- States that the flux linkage varies sinusoidally (as a cosine/sine wave) [1 mark]
- Concludes that the induced emf is also sinusoidal and changes polarity, producing AC [1 mark]

a. (i) The energy \(E\) of the incoming photon is:
\(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{380 \times 10^{-9}\text{ m}} = 5.234 \times 10^{-19}\text{ J}\)

Converting this energy to electron-volts (eV):
\(E = \frac{5.234 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ J eV}^{-1}} = 3.271\text{ eV}\)

Using Einstein's photoelectric equation:
\(E_{k,\text{max}} = E - \Phi = 3.271\text{ eV} - 2.28\text{ eV} = 0.991\text{ eV} \approx 0.99\text{ eV}\)

(ii) The stopping voltage \(V_s\) is related to the maximum kinetic energy by \(E_{k,\text{max}} = e V_s\). Thus:
\(V_s = 0.99\text{ V}\)

b. Classical wave theory views light as a continuous wave where the energy carried is proportional to intensity (amplitude squared) and is independent of frequency. If classical theory were correct, light of any frequency should eventually deliver enough energy to electrons to liberate them if the light is left on long enough, or if the intensity is high enough. However, experiments show that if the frequency is below the threshold frequency, no electrons are emitted regardless of intensity. This can only be explained if light is quantized into packets (photons), where each electron absorbs only one photon at a time, requiring a minimum energy of \(hf \ge \Phi\).

c. The kinetic energy of the accelerated electron is:
\(E_k = q V = 1.60 \times 10^{-19}\text{ C} \times 120\text{ V} = 1.92 \times 10^{-17}\text{ J}\)

The momentum \(p\) of the electron is:
\(p = \sqrt{2 m_e E_k} = \sqrt{2 \times 9.11 \times 10^{-31}\text{ kg} \times 1.92 \times 10^{-17}\text{ J}} = \sqrt{3.498 \times 10^{-47}} = 5.914 \times 10^{-24}\text{ kg m s}^{-1}\)

The de Broglie wavelength \(\lambda\) is:
\(\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}\text{ J s}}{5.914 \times 10^{-24}\text{ kg m s}^{-1}} = 1.121 \times 10^{-10}\text{ m} \approx 0.112\text{ nm}\)

a.
(i)
- Correct calculation of photon energy in Joules or eV [1 mark]
- Conversion of Joules to eV (or vice versa) [0.25 marks]
- Correct subtraction of work function: \(0.99\text{ eV}\) [1 mark]
(ii)
- Identifies Stopping Voltage directly from KE: \(0.99\text{ V}\) [1 mark]

b.
- Classically, energy depends on intensity, not frequency [1 mark]
- Classically, low-frequency light should produce emission given enough time/intensity [1 mark]
- Particle theory states energy is delivered in discrete packets (photons) [1 mark]
- Single photon-electron interaction means if \(hf < \Phi\), no emission can occur [1 mark]

c.
- Calculates KE in Joules: \(1.92 \times 10^{-17}\text{ J}\) [1 mark]
- Relates KE to momentum: \(p = \sqrt{2mE_k}\) [1 mark]
- Computes correct momentum: \(5.91 \times 10^{-24}\text{ kg m s}^{-1}\) [1 mark]
- Computes wavelength: \(1.12 \times 10^{-10}\text{ m}\) (or \(1.1 \times 10^{-10}\text{ m}\)) [1 mark]

a. Atoms only emit light at discrete, specific wavelengths (line spectra) rather than a continuous range of colors. Since photon energy is linked to frequency/wavelength by \(E = hf\), this means emitted photons have specific energy values. Because these photons are produced when electrons fall from higher to lower energy states, the discrete energy of emitted photons implies that the energy differences between states are discrete. Therefore, the energy levels of electrons within an atom must be quantized.

b. (i) Calculate the energy levels:
\(E_4 = -\frac{13.6}{4^2} = -0.85\text{ eV}\)
\(E_2 = -\frac{13.6}{2^2} = -3.40\text{ eV}\)

The energy of the emitted photon \(\Delta E\) is:
\(\Delta E = E_4 - E_2 = -0.85 - (-3.40) = 2.55\text{ eV}\)

Converting to Joules:
\(\Delta E = 2.55 \times 1.60 \times 10^{-19}\text{ J} = 4.08 \times 10^{-19}\text{ J}\)

(ii) Frequency \(f\):
\(f = \frac{\Delta E}{h} = \frac{4.08 \times 10^{-19}\text{ J}}{6.63 \times 10^{-34}\text{ J s}} = 6.154 \times 10^{14}\text{ Hz} \approx 6.15 \times 10^{14}\text{ Hz}\)

Wavelength \(\lambda\):
\(\lambda = \frac{c}{f} = \frac{3.00 \times 10^8\text{ m s}^{-1}}{6.154 \times 10^{14}\text{ Hz}} = 4.875 \times 10^{-7}\text{ m} \approx 488\text{ nm}\)

c. Deflection of alpha particles through large angles (> 90 degrees) occurs because of the strong electrostatic repulsion between the positively charged alpha particle and the highly concentrated positive charge of the atom. This reveals that almost all of the mass and all of the positive charge of the atom are concentrated in an extremely small volume at the center (the nucleus), with the rest of the atom being mostly empty space.

a.
- Relates discrete emission lines to specific photon energies/frequencies (\(E=hf\)) [1.25 marks]
- Mentions that emissions correspond to electron transitions between states [1 mark]
- Concludes that discrete differences require discrete/quantized energy levels [1 mark]

b.
(i)
- Calculates the two energy levels correctly in eV (\(-0.85\text{ eV}\) and \(-3.40\text{ eV}\)) [1 mark]
- Finds difference and converts correctly to Joules (\(4.08 \times 10^{-19}\text{ J}\)) [1 mark]
(ii)
- Calculates frequency \(f = 6.15 \times 10^{14}\text{ Hz}\) [1.5 marks]
- Calculates wavelength \(\lambda = 4.88 \times 10^{-7}\text{ m}\) (accept 4.9 x 10^-7 m or 488 nm) [1.5 marks]

c.
- Identifies Coulomb/electrostatic repulsion as cause for deflection [1 mark]
- Concludes positive charge is concentrated in a tiny central region (nucleus) [1 mark]
- Concludes nearly all the mass is concentrated in this tiny nucleus [1 mark]

a. The electric force acting on the particle is given by \(F_E = qE\).
The magnetic force acting on the particle is given by \(F_B = qvB\) (since velocity is perpendicular to the magnetic field).
For the particle to travel in a straight line without deflecting, the electric and magnetic forces must be equal in magnitude and opposite in direction:
\(F_E = F_B \implies qE = qvB\)

Dividing both sides by the charge \(q\) gives:
\(E = vB \implies v = \frac{E}{B}\)

Since \(q\) cancels out, and mass \(m\) is not present in the equation, the selected velocity is independent of both the charge and mass of the particle.

b. (i) The speed of the selected ions is:
\(v = \frac{E}{B_1} = \frac{2.40 \times 10^4\text{ V m}^{-1}}{0.150\text{ T}} = 1.60 \times 10^5\text{ m s}^{-1}\)

(ii) In the second magnetic field \(B_2\), the centripetal force is provided by the magnetic force:
\(\frac{m v^2}{r} = q v B_2 \implies r = \frac{m v}{q B_2}\)

Calculate the radius of the circular path for both isotopes:
For Neon-20:
\(r_1 = \frac{3.32 \times 10^{-26}\text{ kg} \times 1.60 \times 10^5\text{ m s}^{-1}}{1.60 \times 10^{-19}\text{ C} \times 0.250\text{ T}} = 0.1328\text{ m}\)

For Neon-22:
\(r_2 = \frac{3.65 \times 10^{-26}\text{ kg} \times 1.60 \times 10^5\text{ m s}^{-1}}{1.60 \times 10^{-19}\text{ C} \times 0.250\text{ T}} = 0.1460\text{ m}\)

The separation of the two spots on the detector (which corresponds to the difference in the diameters of their semi-circular paths) is:
\(d = 2(r_2 - r_1) = 2(0.1460 - 0.1328) = 2(0.0132\text{ m}) = 0.0264\text{ m}\) (or \(2.64\text{ cm}\))

c. The magnetic force on a charged particle is given by \(\vec{F}_B = q (\vec{v} \times \vec{B})\). This means the force is always perpendicular to the velocity vector \(\vec{v}\) of the ion. The rate of doing work is \(P = \vec{F}_B \cdot \vec{v}\). Since the vectors are perpendicular, the dot product is zero, meaning the work done on the ion by the magnetic field is always zero. By the work-energy theorem, if no work is done on the particle, its kinetic energy (and thus its speed) remains constant, while only the direction of motion changes.

a.
- Correct expression for electric force \(F_E = qE\) [1 mark]
- Correct expression for magnetic force \(F_B = qvB\) [1 mark]
- Equates forces and solves for \(v\) [1 mark]
- Explains that the charge cancels and mass is absent, making it independent of both [1 mark]

b.
(i)
- Correct calculation of velocity: \(1.60 \times 10^5\text{ m s}^{-1}\) [1.25 marks]
(ii)
- Employs centripetal force formula: \(r = \frac{mv}{qB}\) [1 mark]
- Correct values for \(r_1 = 0.133\text{ m}\) and \(r_2 = 0.146\text{ m}\) [1 mark]
- Correct separation (difference of diameters): \(0.0264\text{ m}\) (or \(2.64\text{ cm}\)) [1 mark]

c.
- States that the magnetic force is perpendicular to the velocity [1.25 marks]
- Mentions that work done \(W = F d \cos(90^\circ) = 0\) [1 mark]
- Connects lack of work to conservation of kinetic energy via work-energy theorem [1 mark]

(a) The cross-sectional area is calculated as:
\(A = \frac{\pi d^2}{4} = \frac{\pi (0.38 \times 10^{-3}\text{ m})^2}{4} = 1.134 \times 10^{-7}\text{ m}^2 \approx 1.1 \times 10^{-7}\text{ m}^2\).

The percentage uncertainty in the diameter \(d\) is:
\(\frac{\Delta d}{d} \times 100\% = \frac{0.02}{0.38} \times 100\% = 5.26\%\).

Since \(A \propto d^2\), the percentage uncertainty in \(A\) is twice the percentage uncertainty in \(d\):
\(\frac{\Delta A}{A} = 2 \times 5.26\% = 10.5\% \approx 11\%\).

Thus, \(A = (1.1 \pm 0.1) \times 10^{-7}\text{ m}^2\).

(b) The relationship between resistance and length is:
\(R = \frac{\rho L}{A} \implies \frac{R}{L} = \frac{\rho}{A}\).

Since the gradient \(m = \frac{R}{L}\), we have:
\(\rho = m \times A = 12.5 \times 1.134 \times 10^{-7} = 1.4175 \times 10^{-6}\text{ }\Omega\text{ m} \approx 1.4 \times 10^{-6}\text{ }\Omega\text{ m}\).

The percentage uncertainty in \(\rho\) is the sum of the percentage uncertainties in \(m\) and \(A\):
\(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta A}{A} = \frac{0.5}{12.5} + 0.1053 = 0.04 + 0.1053 = 0.1453\) (or \(14.5\%\)).

The absolute uncertainty in \(\rho\) is:
\(\Delta \rho = 1.4175 \times 10^{-6} \times 0.1453 = 0.206 \times 10^{-6}\text{ }\Omega\text{ m} \approx 0.2 \times 10^{-6}\text{ }\Omega\text{ m}\).

Thus, \(\rho = (1.4 \pm 0.2) \times 10^{-6}\text{ }\Omega\text{ m}\).

(c) A positive y-intercept on the resistance axis indicates that when the length of the wire is extrapolated to zero, there is still a non-zero measured resistance. This systematic error is caused by the resistance of the connecting leads and/or the contact resistance between the measurement probes and the wire. This adds a constant offset to all measured resistance values.

Part (a) [3 marks]
- Award [1] for calculating correct area \(A \approx 1.1 \times 10^{-7}\text{ m}^2\).
- Award [1] for showing that the fractional/percentage uncertainty of area is twice that of the diameter: \(\frac{\Delta A}{A} = 2\frac{\Delta d}{d}\).
- Award [1] for calculating percentage uncertainty as \(11\%\) (accept \(10.5\%\) or \(10\%\)).

Part (b) [3 marks]
- Award [1] for identifying and calculating the value of resistivity \(\rho = m \times A \approx 1.4 \times 10^{-6}\text{ }\Omega\text{ m}\).
- Award [1] for adding the percentage uncertainties of \(m\) and \(A\): \(\%\Delta \rho = 4\% + 10.5\% = 14.5\%\) (accept \(14\%\) to \(15\%\)).
- Award [1] for stating the final answer with absolute uncertainty and unit: \((1.4 \pm 0.2) \times 10^{-6}\text{ }\Omega\text{ m}\).

Part (c) [2 marks]
- Award [1] for identifying the source of systematic error as lead resistance or contact resistance.
- Award [1] for explaining that this resistance is in series with the wire, thus shifting the entire line up by a constant amount.

(a) A graph of \(T^2\) against \(L\) is linear (yields a straight line through the origin), which makes it straightforward to determine the gradient and extract a value for \(g\). A graph of \(T\) against \(L\) would be a curve, which is more difficult to analyze accurately.

(b) Squaring the theoretical equation yields:
\(T^2 = \frac{4\pi^2}{g} L\).

Comparing this to the equation of a straight line, the gradient \(m\) is:
\(m = \frac{4\pi^2}{g}\).

Solving for \(g\):
\(g = \frac{4\pi^2}{m} = \frac{4\pi^2}{4.08} = 9.68\text{ m s}^{-2}\).

(c) Since \(g = \frac{4\pi^2}{m}\), the fractional uncertainty in \(g\) is equal to the fractional uncertainty in \(m\):
\(\frac{\Delta g}{g} = \frac{\Delta m}{m} = \frac{0.12}{4.08} \approx 0.0294\) (or \(2.94\%\)).

The absolute uncertainty in \(g\) is:
\(\Delta g = 9.68 \times 0.0294 = 0.284\text{ m s}^{-2} \approx 0.3\text{ m s}^{-2}\).

Thus, \(g = 9.7 \pm 0.3\text{ m s}^{-2}\) (or \(9.68 \pm 0.28\text{ m s}^{-2}\)).

(d) The range of the experimental value of \(g\) is:
\(9.7 - 0.3 = 9.4\text{ m s}^{-2}\) to \(9.7 + 0.3 = 10.0\text{ m s}^{-2}\) (or more precisely \(9.40\text{ m s}^{-2}\) to \(9.96\text{ m s}^{-2}\)).

Since the accepted value of \(9.81\text{ m s}^{-2}\) lies within this range of uncertainty, the experimental result is consistent with the accepted value.

Part (a) [1 mark]
- Award [1] for explaining that \(T^2\) against \(L\) is linear / yields a straight line, allowing for easy determination of the gradient / analysis of \(g\).

Part (b) [2 marks]
- Award [1] for correctly rearranging the theoretical equation to find \(m = \frac{4\pi^2}{g}\).
- Award [1] for calculating \(g = 9.68\text{ m s}^{-2}\) or \(9.7\text{ m s}^{-2}\).

Part (c) [2 marks]
- Award [1] for setting the fractional/percentage uncertainty of \(g\) equal to that of \(m\): \(\frac{\Delta g}{g} = \frac{\Delta m}{m}\).
- Award [1] for calculating \(\Delta g = 0.28\text{ m s}^{-2}\) or \(0.3\text{ m s}^{-2}\).

Part (d) [2 marks]
- Award [1] for identifying the experimental range of uncertainty: \([9.4, 10.0]\text{ m s}^{-2}\) (or \([9.40, 9.96]\text{ m s}^{-2}\)).
- Award [1] for stating that the accepted value of \(9.81\text{ m s}^{-2}\) lies within this range, indicating consistency.

(a) The total energy is given by \(E = \gamma m_0 c^2\). Since \(E = 3 m_0 c^2\), we have \(\gamma = 3\). Now, use the definition of the Lorentz factor: \(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = 3\). Squaring both sides: \(1 - \frac{v^2}{c^2} = \frac{1}{9}\), which leads to \(\frac{v^2}{c^2} = \frac{8}{9}\) and thus \(v = \sqrt{\frac{8}{9}} c \approx 0.943 c\).

(b) The relativistic momentum is given by \(p = \gamma m_0 v\). Substituting the known values: \(p = 3 \times (938\text{ MeV } c^{-2}) \times \sqrt{\frac{8}{9}} c = 938 \sqrt{8}\text{ MeV } c^{-1} \approx 2653\text{ MeV } c^{-1}\). Alternatively, using \(E^2 = p^2 c^2 + (m_0 c^2)^2\), we find \((3 m_0 c^2)^2 = p^2 c^2 + (m_0 c^2)^2\), which simplifies to \(8 (m_0 c^2)^2 = p^2 c^2\), giving \(p = \sqrt{8} m_0 c = \sqrt{8} \times 938 \approx 2653\text{ MeV } c^{-1}\).

(a)
- Identifies that \(\gamma = 3\). [1 mark]
- Substitutes into the Lorentz factor formula. [1 mark]
- Calculates the final speed \(v = 0.943c\) (or \(0.94c\)). [1 mark]

(b)
- Recalls a valid relativistic momentum equation, e.g., \(p = \gamma m_0 v\) or \(E^2 = p^2 c^2 + (m_0 c^2)^2\). [1 mark]
- Correct substitution of values. [1 mark]
- Correct final answer in the range \(2650\text{ MeV } c^{-1}\) to \(2653\text{ MeV } c^{-1}\). [1 mark]

(a) Using the moment of inertia formula: \(I = \frac{1}{2} M R^2 = 0.5 \times 2.5\text{ kg} \times (0.12\text{ m})^2 = 0.018\text{ kg m}^2\).

(b) First, calculate the torque: \(\tau = F R = 15\text{ N} \times 0.12\text{ m} = 1.8\text{ N m}\). Next, apply Newton's second law for rotation: \(\alpha = \frac{\tau}{I} = \frac{1.8\text{ N m}}{0.018\text{ kg m}^2} = 100\text{ rad s}^{-2}\).

(c) Using the rotational kinematic equation \(\omega^2 = \omega_0^2 + 2\alpha\theta\) with \(\omega_0 = 0\): \(\omega^2 = 2 \times 100\text{ rad s}^{-2} \times 5.0\text{ rad} = 1000\text{ rad}^2\text{ s}^{-2}\). Taking the square root gives \(\omega = \sqrt{1000} \approx 31.6\text{ rad s}^{-1}\).

(a)
- Correct substitution into the formula. [1 mark]
- Correct answer with appropriate units: \(0.018\text{ kg m}^2\). [1 mark]

(b)
- Correct calculation of torque as \(1.8\text{ N m}\). [1 mark]
- Divides torque by moment of inertia to get \(100\text{ rad s}^{-2}\). [1 mark]

(c)
- Selects a correct kinematic equation and substitutes values. [1 mark]
- Correctly calculates \(31.6\text{ rad s}^{-1}\) (or \(32\text{ rad s}^{-1}\)). [1 mark]

(a) Work done during an isobaric process is given by \(W = P \Delta V = P (V_2 - V_1) = (2.0 \times 10^5\text{ Pa}) \times (3.0 \times 10^{-3}\text{ m}^3 - 1.0 \times 10^{-3}\text{ m}^3) = 400\text{ J}\).

(b) Applying the First Law of Thermodynamics: \(Q = \Delta U + W\). Since \(\Delta U = 600\text{ J}\) and \(W = 400\text{ J}\), the heat added is \(Q = 600\text{ J} + 400\text{ J} = 1000\text{ J}\).

(c) During process C \(\to\) A, the gas is compressed, meaning its volume decreases (\(V_f < V_i\)). Since the displacement of the boundary is opposite to the pressure force exerted by the gas, the work done *by* the gas is negative (work is done *on* the gas).

(a)
- Uses \(W = P \Delta V\) with correct values. [1 mark]
- Obtains \(400\text{ J}\). [1 mark]

(b)
- Applies the first law \(Q = \Delta U + W\). [1 mark]
- Obtains \(1000\text{ J}\). [1 mark]

(c)
- Identifies the work as negative. [1 mark]
- Explains that the volume is decreasing (compression occurs). [1 mark]

(a) The Lorentz factor is: \(\gamma = \frac{1}{\sqrt{1 - v^2/c^2}} = \frac{1}{\sqrt{1 - 0.99^2}} = \frac{1}{\sqrt{1 - 0.9801}} \approx 7.09\).

(b) Length contraction means the distance in the muon's frame is contracted: \(L = \frac{L_0}{\gamma} = \frac{8.0\text{ km}}{7.09} \approx 1.13\text{ km} \approx 1.1\text{ km}\).

(c) Classically, \(d = v t_{1/2} = 0.99 \times (3.0 \times 10^8\text{ m s}^{-1}) \times (1.5 \times 10^{-6}\text{ s}) \approx 446\text{ m}\). Because of length contraction, they only need to travel \(1.1\text{ km}\) in their own frame to reach the ground, or equivalently because of time dilation, their half-life in the Earth's frame is extended to \(\gamma \times 1.5\text{ \mu s} \approx 10.6\text{ \mu s}\), allowing them to travel much further.

(a)
- Correct formula written down. [1 mark]
- Correct calculation yielding \(7.1\) or \(7.09\). [1 mark]

(b)
- Applies length contraction formula. [1 mark]
- Correct calculation yielding \(1.1\text{ km}\) (or \(1.13\text{ km}\)). [1 mark]

(c)
- Calculates classical distance \(\approx 450\text{ m}\). [1 mark]
- Explains how either length contraction or time dilation allows the muons to reach the ground. [1 mark]

(a) By conservation of angular momentum: \(I_i \omega_i = I_f \omega_f\). Thus: \(3.6\text{ kg m}^2 \times 4.0\text{ rad s}^{-1} = 1.2\text{ kg m}^2 \times \omega_f \implies \omega_f = 12\text{ rad s}^{-1}\).

(b) The kinetic energy ratio is \(\frac{E_{k,f}}{E_{k,i}} = \frac{0.5 I_f \omega_f^2}{0.5 I_i \omega_i^2} = \frac{1.2 \times 12^2}{3.6 \times 4^2} = \frac{172.8}{57.6} = 3.0\). Alternatively, since \(E_k = \frac{L^2}{2I}\) and \(L\) is constant, the ratio is \(\frac{I_i}{I_f} = \frac{3.6}{1.2} = 3.0\).

(c) The increase in kinetic energy is due to the work done by the skater's internal muscles as she pulls her arms inward against the centripetal/inertial resistance. This converts her internal chemical energy into mechanical rotational kinetic energy.

(a)
- Uses conservation of angular momentum \(I_i \omega_i = I_f \omega_f\). [1 mark]
- Obtains \(12\text{ rad s}^{-1}\). [1 mark]

(b)
- Substitutes into the rotational kinetic energy formula. [1 mark]
- Calculates the ratio as \(3.0\) (or \(3\)). [1 mark]

(c)
- States that work is done by the skater's muscles. [1 mark]
- Explains that chemical energy is converted to mechanical energy. [1 mark]