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The radius of a charged particle moving in a perpendicular magnetic field is given by \(R = \frac{p}{qB}\), where \(p\) is the momentum. Since kinetic energy is \(E_k = \frac{p^2}{2m}\), we can write the momentum as \(p = \sqrt{2mE_k}\). Substituting this into the radius formula gives: \(R = \frac{\sqrt{2mE_k}}{qB}\). For the alpha particle, the mass is \(4m\) and the charge is \(2q\). Its radius \(R_\alpha\) is: \(R_\alpha = \frac{\sqrt{2(4m)E_k}}{(2q)B} = \frac{2\sqrt{2mE_k}}{2qB} = \frac{\sqrt{2mE_k}}{qB} = R\). Thus, the radius is the same as that of the proton.

[1 mark] for identifying the correct relationship between radius, kinetic energy, mass, and charge. [1 mark] for substituting the values for the alpha particle to show that the radius remains \(R\). Award 1 mark total for the correct option B.

The magnetic flux \(\Phi\) through the coil is given by \(\Phi = B A \cos(\theta)\). Since the coil rotates at a constant angular frequency \(\omega\), and is perpendicular to the field at \(t=0\), we have \(\theta = \omega t\), so \(\Phi = B A \cos(\omega t)\). The induced emf \(\varepsilon\) in a coil of \(N\) turns is given by Faraday's law: \(\varepsilon = -N \frac{d\Phi}{dt} = -N \frac{d}{dt}(B A \cos(\omega t)) = N B A \omega \sin(\omega t)\).

[1 mark] for correctly applying Faraday's law of induction and differentiating the flux expression with respect to time, including the negative sign from Lenz's law. Award 1 mark total for the correct option B.

According to Einstein's photoelectric equation, the maximum kinetic energy is given by \(E_{\text{max}} = hf - \Phi\), where \(\Phi\) is the work function of the metal. For an incident frequency of \(2f\), the new maximum kinetic energy \(E'_{\text{max}}\) is: \(E'_{\text{max}} = h(2f) - \Phi = 2hf - \Phi\). Since \(hf = E_{\text{max}} + \Phi\), we can substitute this back: \(E'_{\text{max}} = 2(E_{\text{max}} + \Phi) - \Phi = 2E_{\text{max}} + \Phi\).

[1 mark] for applying the photoelectric equation to both cases and solving for the new maximum kinetic energy in terms of the initial maximum kinetic energy and the work function. Award 1 mark total for the correct option B.

The total mechanical energy \(E\) of a satellite in a circular orbit of radius \(r\) is given by \(E = -\frac{GMm}{2r}\). The initial energy is \(E_i = -\frac{GMm}{2r}\). The final energy in the orbit of radius \(2r\) is \(E_f = -\frac{GMm}{2(2r)} = -\frac{GMm}{4r}\). The change in total mechanical energy is: \(\Delta E = E_f - E_i = -\frac{GMm}{4r} - \left(-\frac{GMm}{2r}\right) = +\frac{GMm}{4r}\).

[1 mark] for identifying the correct formula for the total energy of an orbiting satellite and calculating the difference between the final and initial states. Award 1 mark total for the correct option B.

Let the initial direction of motion be positive. The initial momentum is \(p_i = Mv\). The final momentum is \(p_f = -Mv\). The change in momentum is \(\Delta p = p_f - p_i = -Mv - Mv = -2Mv\). According to the impulse-momentum theorem, the impulse of the force is equal to the change in momentum: \(F \Delta t = \Delta p\). Taking the magnitude of both sides, we get: \(F \Delta t = 2Mv \implies F = \frac{2Mv}{\Delta t}\).

[1 mark] for determining the change in momentum taking direction into account, and using Newton's second law / impulse formula to solve for force. Award 1 mark total for the correct option B.

The power dissipated in the external resistor is \(P = I^2 R\), where \(I = \frac{\varepsilon}{R + r}\). Substituting \(I\) gives \(P = \frac{\varepsilon^2 R}{(R + r)^2}\). To find the maximum power transfer, we differentiate \(P\) with respect to \(R\) and set it to zero: \(\frac{dP}{dR} = \varepsilon^2 \frac{(R+r)^2 - 2R(R+r)}{(R+r)^4} = 0\). This simplifies to \((R+r) - 2R = 0\), which yields \(R = r\). This is the maximum power transfer theorem.

[1 mark] for utilizing the power equation for a circuit with internal resistance and applying the maximum power transfer theorem condition. Award 1 mark total for the correct option B.

According to the Stefan-Boltzmann law, the luminosity of a star is given by \(L = \sigma A T^4 = 4\pi R^2 \sigma T^4\), meaning \(L \propto R^2 T^4\). For Star \(X\): \(L_X \propto R^2 T^4\). For Star \(Y\): \(L_Y \propto (2R)^2 \left(\frac{T}{2}\right)^4 = 4R^2 \frac{T^4}{16} = \frac{1}{4} R^2 T^4\). Taking the ratio: \(\frac{L_X}{L_Y} = \frac{R^2 T^4}{\frac{1}{4} R^2 T^4} = 4\).

[1 mark] for applying the Stefan-Boltzmann law to both stars, factoring in both the radius and temperature changes, and correctly determining the ratio. Award 1 mark total for the correct option D.

The total energy in simple harmonic motion is the sum of kinetic and potential energy: \(E = E_k + E_p\). We are given that \(E_k = 3E_p\), so \(E = 3E_p + E_p = 4E_p\). The total energy is \(E = \frac{1}{2} k A^2\) and the potential energy at displacement \(x\) is \(E_p = \frac{1}{2} k x^2\). Substituting these into the energy equation gives: \(\frac{1}{2} k A^2 = 4 \left(\frac{1}{2} k x^2\right) \implies A^2 = 4x^2 \implies x = \pm \frac{A}{2}\).

[1 mark] for establishing the relationship between total energy and potential energy under the given condition, and solving for the displacement in terms of the amplitude. Award 1 mark total for the correct option B.

The impulse \(J\) delivered to the block is equal to the area under the force-time graph.

The graph is a triangle with base \(b = 6.0\text{ s}\) and peak height \(h = 12\text{ N}\).

\(J = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6.0\text{ s} \times 12\text{ N} = 36\text{ N s}\).

By the impulse-momentum theorem, the change in momentum is equal to the impulse:
\(\Delta p = m \Delta v = 36\text{ kg m s}^{-1}\).

Since the block starts from rest:
\(v_f = \frac{\Delta p}{m} = \frac{36\text{ kg m s}^{-1}}{2.0\text{ kg}} = 18\text{ m s}^{-1}\).

Award 1 mark for calculating the impulse as the area under the triangle: \(36\text{ N s}\). Award 1 mark for dividing by mass to get the correct speed of \(18\text{ m s}^{-1}\).

The escape speed from the surface of the planet is given by:
\(v_{\text{esc}} = \sqrt{\frac{2GM}{R}}\)

The orbital speed of a satellite in a circular orbit of radius \(r = 3R\) is given by:
\(v_{\text{orb}} = \sqrt{\frac{GM}{3R}}\)

Taking the ratio of these two speeds:
\(\frac{v_{\text{esc}}}{v_{\text{orb}}} = \frac{\sqrt{\frac{2GM}{R}}}{\sqrt{\frac{GM}{3R}}} = \sqrt{\frac{2GM}{R} \times \frac{3R}{GM}} = \sqrt{6}\).

Award 1 mark for the correct application of gravitational formulas and algebraic simplification leading to \(\sqrt{6}\).

The potential difference across \(R\) is \(V = I R = \frac{E}{R+r} R = \frac{E}{1 + \frac{r}{R}}\). As \(R\) increases from \(0\) to a very large value (approaching infinity), the term \(\frac{r}{R}\) approaches \(0\), meaning \(V\) increases from \(0\) and approaches \(E\).

The power dissipated in the external resistor is given by \(P = I^2 R = \frac{E^2 R}{(R+r)^2}\). At \(R = 0\), \(P = 0\). As \(R \to \infty\), \(P \to 0\). The power reaches a maximum when \(R = r\) (the maximum power transfer theorem). Therefore, \(P\) increases to a maximum and then decreases.

Award 1 mark for identifying that potential difference increases up to the emf, and that power peaks when external resistance matches internal resistance.

As the loop is pulled out of the magnetic field, the magnetic flux through the loop decreases. The rate of change of magnetic flux is:
\(\frac{\Delta \Phi}{\Delta t} = B L v\)

According to Faraday's law, this induces an emf in the loop:
\(\varepsilon = B L v\)

The induced current is:
\(I = \frac{\varepsilon}{R} = \frac{B L v}{R}\)

As the loop leaves the field, one of its sides of length \(L\) carries this current \(I\) and is perpendicular to the magnetic field. This experiences a magnetic force resisting the motion:
\(F_M = B I L = B \left(\frac{B L v}{R}\right) L = \frac{B^2 L^2 v}{R}\)

To keep the loop moving at a constant speed, the external force must balance this magnetic force, so:
\(F_{\text{ext}} = \frac{B^2 L^2 v}{R}\).

Award 1 mark for expressing the induced emf, current, and resolving the magnetic force equation to show the \(B^2 L^2 v / R\) dependence.

Einstein's photoelectric equation states:
\(E_{\text{max}} = hf - \Phi\) where \(\Phi\) is the work function of the metal.

For the first frequency:
\(E_1 = hf_1 - \Phi\)

For the doubled frequency:
\(E_2 = h(2f_1) - \Phi = 2hf_1 - \Phi\)

We can express \(2hf_1\) from the first equation as \(2(E_1 + \Phi)\):
\(E_2 = 2(E_1 + \Phi) - \Phi = 2E_1 + \Phi\)

Since the work function \(\Phi\) must be positive for photoemission to occur, it follows that:
\(E_2 > 2E_1\).

Award 1 mark for showing that the work function term does not double, resulting in \(E_2\) being strictly greater than \(2E_1\).

According to Wien's displacement law, the absolute temperature \(T\) of a black body is inversely proportional to its peak wavelength \(\lambda_{\text{max}}\):
\(T \propto \frac{1}{\lambda_{\text{max}}}\)

Since \(\lambda_X = 2\lambda_Y\), the absolute temperatures satisfy:
\(T_X = \frac{1}{2} T_Y\)

According to the Stefan-Boltzmann law, the total power radiated per unit area \(I\) is proportional to the fourth power of the absolute temperature:
\(I \propto T^4\)

Therefore, the ratio of the intensities is:
\(\frac{I_X}{I_Y} = \left(\frac{T_X}{T_Y}\right)^4 = \left(\frac{1}{2}\right)^4 = \frac{1}{16}\).

Award 1 mark for correctly relating temperature inversely to peak wavelength and utilizing the fourth-power dependence to find the ratio.

For a spherical wave emitted from a point source, the intensity \(I\) decreases as an inverse-square law with distance \(r\):
\(I \propto \frac{1}{r^2}\)

At a distance of \(3d\), the intensity is:
\(I = \frac{I_0}{3^2} = \frac{I_0}{9}\)

Since the intensity is proportional to the square of the amplitude (\(I \propto A^2\)), the amplitude \(A\) is inversely proportional to the distance \(r\):
\(A \propto \frac{1}{r}\)

At a distance of \(3d\), the amplitude is:
\(A = \frac{A_0}{3}\).

Award 1 mark for applying \(I \propto r^{-2}\) to find intensity and \(A \propto r^{-1}\) to find amplitude.

By the conservation of angular momentum, since there are no external torques:
\(L_i = L_f\)
\(I \omega_0 = (I + I) \omega = 2I \omega\)

This gives the final common angular speed as:
\(\omega = \frac{\omega_0}{2}\)

The initial rotational kinetic energy is:
\(E_i = \frac{1}{2} I \omega_0^2\)

The final rotational kinetic energy is:
\(E_f = \frac{1}{2} (2I) \omega^2 = I \left(\frac{\omega_0}{2}\right)^2 = \frac{1}{4} I \omega_0^2\)

The ratio of the final to initial rotational kinetic energy is:
\(\frac{E_f}{E_i} = \frac{\frac{1}{4} I \omega_0^2}{\frac{1}{2} I \omega_0^2} = \frac{1}{2}\).

Award 1 mark for applying conservation of angular momentum to find final angular velocity, and subsequently computing the kinetic energy ratio.

As the loop enters the magnetic field, the magnetic flux through the loop changes.

The rate of change of magnetic flux is given by:
\(\Phi = B \cdot A = B \cdot (L \cdot x)\) where \(x\) is the distance the loop has penetrated into the field.

Taking the time derivative to find the induced electromotive force (emf) \(\varepsilon\):
\(\varepsilon = \frac{d\Phi}{dt} = B L \frac{dx}{dt} = B L v\)

The power dissipated as thermal energy in the loop is given by:
\(P = \frac{\varepsilon^2}{R} = \frac{(B L v)^2}{R} = \frac{B^2 v^2 L^2}{R}\)

Award [1] for the correct answer B.
- Award [0] for other options.
- Method: Identifies emf \(\varepsilon = B L v\) and uses \(P = \frac{\varepsilon^2}{R}\) to arrive at option B.

The kinetic energy \(E_k\) acquired by the electron is given by:
\(E_k = e V\)

Since \(E_k = \frac{p^2}{2m_e}\), the momentum of the electron is:
\(p = \sqrt{2 m_e E_k} = \sqrt{2 m_e e V}\)

The de Broglie wavelength \(\lambda\) is:
\(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m_e e V}}\)

Therefore, the wavelength is inversely proportional to the square root of the potential difference:
\(\lambda \propto \frac{1}{\sqrt{V}}\)

When the potential difference is increased by a factor of 4, the new wavelength becomes:
\(\lambda' = \frac{\lambda}{\sqrt{4}} = \frac{\lambda}{2}\)

Award [1] for the correct answer B.
- Award [0] for other options.
- Method: Identifies that \(\lambda \propto V^{-1/2}\) and determines that scaling \(V\) by 4 scales \(\lambda\) by \(0.5\).

According to the Stefan-Boltzmann law, the total power \(P\) radiated by a black-body sphere of radius \(r\) and absolute temperature \(T\) is:
\(P = \sigma A T^4 = \sigma (4 \pi r^2) T^4\)

Thus, \(P \propto r^2 T^4\).

For star X:
\(P_X \propto R^2 T^4\)

For star Y:
\(P_Y \propto (2R)^2 (2T)^4 = 4 R^2 \cdot 16 T^4 = 64 R^2 T^4\)

The ratio is:
\(\frac{P_Y}{P_X} = 64\)

Award [1] for the correct answer D.
- Award [0] for other options.
- Method: Recalls \(P \propto A T^4 \propto r^2 T^4\) and applies the factors of 2 for radius and 2 for temperature.

The impulse \(J\) delivered to the object is equal to the area under the force-time graph.

The shape of the graph is a triangle with base \(b = 4.0\text{ s}\) and height \(h = 10\text{ N}\).

\(J = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0\text{ s} \times 10\text{ N} = 20\text{ N s}\)

By the impulse-momentum theorem:
\(J = \Delta p = m \Delta v\)

Since the object starts from rest (\(u = 0\)):
\(20\text{ N s} = 2.0\text{ kg} \times v\)
\(v = 10\text{ m s}^{-1}\)

Award [1] for the correct answer B.
- Award [0] for other options.
- Method: Calculates impulse as the area under the triangle, then divides by the mass to find the final speed.

For a satellite in a circular orbit of radius \(r\) around a body of mass \(M\), the gravitational force provides the centripetal force:
\(\frac{G M m}{r^2} = \frac{m v^2}{r}\)

Solving for the orbital speed \(v\):
\(v = \sqrt{\frac{G M}{r}}\)

This shows that the orbital speed is inversely proportional to the square root of the orbit radius:
\(v \propto \frac{1}{\sqrt{r}}\)

For satellite B, the orbit radius is \(4R\):
\(v_B = \sqrt{\frac{G M}{4R}} = \frac{1}{2} \sqrt{\frac{G M}{R}} = \frac{v}{2}\)

Award [1] for the correct answer B.
- Award [0] for other options.
- Method: Uses the relationship \(v = \sqrt{\frac{GM}{r}}\) and applies the radius ratio to find the speed ratio.

The current \(I\) in the circuit is given by Ohm's law:
\(I = \frac{E}{R + r}\)

The power dissipated in the external resistor \(R\) is:
\(P = I^2 R = \left( \frac{E}{R + r} \right)^2 R\)

When \(R = r\):
\(P = \left( \frac{E}{r + r} \right)^2 r = \left( \frac{E}{2r} \right)^2 r = \frac{E^2}{4r}\)

When \(R = 3r\):
\(P' = \left( \frac{E}{3r + r} \right)^2 (3r) = \left( \frac{E}{4r} \right)^2 (3r) = \frac{3 E^2}{16 r}\)

Expressing \(P'\) in terms of \(P\):
\(P' = \frac{3}{4} \left( \frac{E^2}{4r} \right) = \frac{3}{4} P\)

Award [1] for the correct answer C.
- Award [0] for other options.
- Method: Uses \(P = I^2 R\) combined with the total resistance to form expressions for both cases and divide them to find the ratio.

The total energy of the simple harmonic oscillator is:
\(E_T = \frac{1}{2} m \omega^2 x_0^2\)

The potential energy \(E_p\) at displacement \(x\) is:
\(E_p = \frac{1}{2} m \omega^2 x^2\)

The kinetic energy \(E_k\) at displacement \(x\) is:
\(E_k = E_T - E_p = \frac{1}{2} m \omega^2 (x_0^2 - x^2)\)

We are given that the kinetic energy is three times the potential energy:
\(E_k = 3 E_p\)

Substitute the energy expressions into this equation:
\(\frac{1}{2} m \omega^2 (x_0^2 - x^2) = 3 \left( \frac{1}{2} m \omega^2 x^2 \right)\)

Simplify by dividing both sides by \(\frac{1}{2} m \omega^2\):
\(x_0^2 - x^2 = 3 x^2\)
\(x_0^2 = 4 x^2\)
\(x^2 = \frac{x_0^2}{4}\)
\(x = \frac{x_0}{2}\)

Award [1] for the correct answer B.
- Award [0] for other options.
- Method: Sets up the equation \(E_k = 3 E_p\) using the standard formulas for energy in SHM and solves for \(x\).

Using the ideal gas equation:
\(P = \frac{N k_B T}{V}\) where \(N\) is the number of molecules.

Let the initial state be defined by \(N_1 = N\), \(T_1 = T\), and \(P_1 = P\).

In the final state, half of the molecules are removed:
\(N_2 = \frac{N}{2}\)

The absolute temperature is doubled:
\(T_2 = 2T\)

The volume remains constant: \(V_2 = V\).

Substitute these into the ideal gas equation for the final pressure \(P_2\):
\(P_2 = \frac{N_2 k_B T_2}{V} = \frac{\left(\frac{N}{2}\right) k_B (2T)}{V} = \frac{N k_B T}{V} = P\)

Thus, the pressure remains unchanged.

Award [1] for the correct answer C.
- Award [0] for other options.
- Method: Applies the ideal gas law \(P \propto N T\) under constant volume, noting that halving \(N\) and doubling \(T\) cancels out the effect on \(P\).

The total energy \(E\) of a satellite in a circular orbit of radius \(r\) is given by \(E = -\frac{GMm}{2r}\). The initial energy is \(E_i = -\frac{GMm}{2r}\). The final energy in the new orbit of radius \(2r\) is \(E_f = -\frac{GMm}{2(2r)} = -\frac{GMm}{4r}\). The change in the total energy is \(\Delta E = E_f - E_i = -\frac{GMm}{4r} - \left(-\frac{GMm}{2r}\right) = +\frac{GMm}{4r}\).

Award [1] for the correct answer B. Award [0] for any other response.

As the loop is pulled out, the magnetic flux through it decreases. The time taken to fully exit the field is \(\Delta t = \frac{L}{v}\). The change in magnetic flux is \(\Delta \Phi = B L^2\). According to Faraday's law, the magnitude of the induced electromotive force (EMF) is \(\varepsilon = \frac{\Delta \Phi}{\Delta t} = \frac{B L^2}{L/v} = B L v\). The thermal power dissipated in the loop is \(P = \frac{\varepsilon^2}{R} = \frac{B^2 L^2 v^2}{R}\).

Award [1] for the correct answer C. Award [0] for any other response.

The impulse \(J\) delivered to the block is equal to the area under the force-time graph. The graph is a triangle with base \(b = 4.0 \text{ s}\) and height \(h = 10 \text{ N}\). Thus, the impulse is \(J = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0 \times 10 = 20 \text{ N s}\). According to the impulse-momentum theorem, \(J = \Delta p = m \Delta v\). Since the block starts from rest, \(20 = 2.0 \times v\), which gives a final speed of \(v = 10 \text{ m s}^{-1}\).

Award [1] for the correct answer B. Award [0] for any other response.

The current in the circuit is given by \(I = \frac{E}{R+r}\). The power dissipated in the internal resistance of the cell is \(P_r = I^2 r = \left(\frac{E}{R+r}\right)^2 r\). Since \(E\) and \(r\) are constant, as the variable resistance \(R\) increases, the denominator \(R+r\) increases, which means the current \(I\) decreases. Therefore, the power dissipated in the internal resistance \(P_r\) must decrease continuously.

Award [1] for the correct answer A. Award [0] for any other response.

According to Einstein's photoelectric equation, \(E_{\text{max}} = hf - \Phi\), where \(\Phi\) is the work function of the metal. If the frequency is doubled, the new maximum kinetic energy is \(E_{\text{max}}' = h(2f) - \Phi = 2hf - \Phi\). We can rewrite this as \(E_{\text{max}}' = 2(E_{\text{max}} + \Phi) - \Phi = 2E_{\text{max}} + \Phi\). Since the work function \(\Phi\) is positive, \(2E_{\text{max}} + \Phi > 2E_{\text{max}}\), so the new maximum kinetic energy is more than \(2E_{\text{max}}\).

Award [1] for the correct answer C. Award [0] for any other response.

The Stefan-Boltzmann law states that luminosity \(L = \sigma A T^4\). Since the surface area of a spherical star is \(A = 4\pi R^2\), we have \(L \propto R^2 T^4\). Therefore, the ratio of the luminosities is \(\frac{L_X}{L_Y} = \frac{R_X^2 T_X^4}{R_Y^2 T_Y^4} = \frac{R^2 T^4}{(2R)^2 (2T)^4} = \frac{R^2 T^4}{4 R^2 \times 16 T^4} = \frac{1}{64}\).

Award [1] for the correct answer D. Award [0] for any other response.

For an ideal gas, the average kinetic energy of the molecules is directly proportional to the absolute temperature, \(\bar{E}_k = \frac{1}{2}m\bar{v^2} = \frac{3}{2}k_B T\). This implies that the mean square speed \(\bar{v^2}\) is directly proportional to \(T\). Thus, doubling \(T\) doubles the mean square speed. According to Gay-Lussac's law for a constant volume, pressure \(P\) is also directly proportional to the absolute temperature \(T\). Thus, doubling \(T\) doubles the pressure as well.

Award [1] for the correct answer A. Award [0] for any other response.

The intensity of a wave from a point source at distance \(r\) is proportional to the square of the amplitude \(A\) and inversely proportional to the square of the distance, so \(I \propto \frac{A^2}{r^2}\). Originally, \(I_1 = k\frac{A^2}{d^2} = I\). When amplitude is doubled to \(2A\) and distance is doubled to \(2d\), the new intensity is \(I_2 = k\frac{(2A)^2}{(2d)^2} = k\frac{4A^2}{4d^2} = k\frac{A^2}{d^2} = I\). Thus, the intensity remains unchanged.

Award [1] for the correct answer C. Award [0] for any other response.

By conservation of linear momentum, \(m v = (m + 3m) v_f\), which gives the final velocity \(v_f = \frac{v}{4}\). The initial kinetic energy is \(E_i = \frac{1}{2} m v^2\). The final kinetic energy is \(E_f = \frac{1}{2} (4m) v_f^2 = \frac{1}{2} (4m) \left(\frac{v}{4}\right)^2 = \frac{1}{8} m v^2\). The ratio of the final kinetic energy to the initial kinetic energy is \(\frac{E_f}{E_i} = \frac{\frac{1}{8} m v^2}{\frac{1}{2} m v^2} = \frac{1}{4}\).

[1 mark] Award for conservation of momentum to find final speed of \(v/4\). [1 mark] Award for calculating the ratio of kinetic energies to get \(1/4\).

The escape speed from a planet's surface is given by \(v_e = \sqrt{\frac{2GM}{R}}\). Since density \(\rho\) is uniform and the same for both planets, the mass of a planet is proportional to its volume, so \(M = \rho \frac{4}{3} \pi R^3 \propto R^3\). Substituting this into the escape speed formula gives \(v_e \propto \sqrt{\frac{R^3}{R}} = \sqrt{R^2} = R\). Since the escape speed is directly proportional to the radius, the ratio of the escape speeds is \(\frac{v_{e, Y}}{v_{e, X}} = \frac{3R}{R} = 3\).

[1 mark] Correctly relates mass to radius cubed using density. [1 mark] Correctly simplifies the escape speed relation to find that escape speed is proportional to radius, leading to the correct ratio of 3.

When the power dissipated in the external resistor is maximum, the load resistance \(R\) equals the internal resistance \(r\). The total resistance in the circuit is \(R_{total} = R + r = 2r\). The current in the circuit is \(I = \frac{E}{2r}\). The terminal potential difference \(V\) is the potential difference across the external resistor, which is \(V = I R = \left(\frac{E}{2r}\right) r = \frac{E}{2}\).

[1 mark] Recognizes that the total resistance is \(2r\) and calculates the current as \(I = E/(2r)\). [1 mark] Applies Ohm's law to the external resistor to find the potential difference as \(E/2\).

The initial magnetic flux linkage through the coil is \(\Phi_i = N B A \cos(0^\circ) = N B A\). After a rotation of \(90^\circ\), the plane of the coil is parallel to the magnetic field lines, so the final magnetic flux linkage is \(\Phi_f = N B A \cos(90^\circ) = 0\). According to Faraday's law of induction, the magnitude of the average induced emf is \(|\mathcal{E}| = \frac{|\Delta \Phi|}{\Delta t} = \frac{|0 - N B A|}{\Delta t} = \frac{N B A}{\Delta t}\).

[1 mark] Identifies initial and final flux linkage. [1 mark] Applies Faraday's law to find the magnitude of the average induced emf.

According to Einstein's photoelectric equation, the maximum kinetic energy is \(E_{max} = hf - \Phi\), where \(\Phi > 0\) is the work function of the metal. If the frequency is doubled to \(2f\), the new maximum kinetic energy is \(E'_{max} = h(2f) - \Phi = 2hf - \Phi\). Since \(hf = E_{max} + \Phi\), we can substitute this to get \(E'_{max} = 2(E_{max} + \Phi) - \Phi = 2E_{max} + \Phi\). Because the work function \(\Phi\) is always positive, the new maximum kinetic energy \(E'_{max}\) must be strictly greater than \(2E_{max}\).

[1 mark] Formulates the photoelectric equations for both cases. [1 mark] Rearranges and demonstrates that the new kinetic energy is \(2E_{max} + \Phi\), which is greater than \(2E_{max}\) since the work function is positive.

By the Stefan-Boltzmann law, the power radiated by a blackbody of surface area \(A\) at absolute temperature \(T\) is \(P = \sigma A T^4\). The ratio of the power radiated by A to that of B is \(\frac{P_A}{P_B} = \frac{\sigma A_A T_A^4}{\sigma A_B T_B^4} = \left(\frac{A_A}{A_B}\right) \left(\frac{T_A}{T_B}\right)^4\). Substituting the given values \(\frac{A_A}{A_B} = \frac{1}{2}\) and \(\frac{T_A}{T_B} = 2\), we find \(\frac{P_A}{P_B} = \frac{1}{2} \cdot 2^4 = \frac{1}{2} \cdot 16 = 8\).

[1 mark] Recalls and applies Stefan-Boltzmann law. [1 mark] Substitutes the temperature ratio and the area ratio to calculate the final power ratio of 8.

The total energy \(E_T\) of the simple harmonic oscillator is given by \(E_T = \frac{1}{2} m \omega^2 A^2\). The potential energy \(E_p\) at displacement \(x = \frac{A}{2}\) is \(E_p = \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} m \omega^2 \left(\frac{A}{2}\right)^2 = \frac{1}{8} m \omega^2 A^2\). The kinetic energy \(E_k\) is the difference between the total energy and potential energy: \(E_k = E_T - E_p = \frac{1}{2} m \omega^2 A^2 - \frac{1}{8} m \omega^2 A^2 = \frac{3}{8} m \omega^2 A^2\).

[1 mark] Identifies the total energy equation or the kinetic energy formula in terms of displacement. [1 mark] Substitutes \(x = A/2\) to arrive at the correct kinetic energy of \(\frac{3}{8} m \omega^2 A^2\).

When an object rolls down an incline without slipping, gravitational potential energy is converted into both translational kinetic energy and rotational kinetic energy. The solid cylinder has a moment of inertia of \(I_{solid} = \frac{1}{2} M R^2\) while the hollow cylinder has a moment of inertia of \(I_{hollow} = M R^2\). Since the solid cylinder has a smaller moment of inertia, it requires less torque and a smaller fraction of its energy to rotate at a given angular speed. Thus, more energy goes into its translational motion, resulting in a higher linear acceleration and speed, so it reaches the bottom first.

[1 mark] Compares the moments of inertia of the solid and hollow cylinders. [1 mark] Relates the lower moment of inertia to a greater share of energy going into translation and thus greater linear acceleration.

(a) Stellar parallax is the apparent shift in position of a nearby star against a background of distant stars when viewed from two opposite points on the Earth's orbit around the Sun. The distance in parsecs is calculated as \(d = \frac{1}{p}\), where \(p\) is the parallax angle in arcseconds.

(b) Distance in pc: \(d = \frac{1}{0.125} = 8.0\text{ pc}\). Distance in meters: \(d = 8.0 \times 3.09 \times 10^{16}\text{ m} = 2.47 \times 10^{17}\text{ m}\).

(c) Using the equation for apparent brightness: \(b = \frac{L}{4\pi d^2}\). Rearranging for luminosity: \(L = b \times 4\pi d^2 = 2.50 \times 10^{-10} \times 4 \pi \times (2.472 \times 10^{17})^2 = 1.92 \times 10^{26}\text{ W}\). Expressing this in solar units: \(\frac{L}{L_{\odot}} = \frac{1.92 \times 10^{26}}{3.83 \times 10^{26}} = 0.50 L_{\odot}\).

(a)
[1 mark] for definition of parallax as an apparent shift against distant background stars.
[1 mark] for linking observations to opposite positions in Earth's orbit (6 months apart).
[1 mark] for stating/showing \(d = 1/p\).

(b)
[1 mark] for correct parsecs calculation: \(8.0\text{ pc}\).
[1 mark] for using the conversion factor \(1\text{ pc} = 3.09 \times 10^{16}\text{ m}\).
[1 mark] for final answer of \(2.47 \times 10^{17}\text{ m}\) (accept \(2.5 \times 10^{17}\text{ m}\)).

(c)
[1 mark] for recalling/using \(L = 4\pi d^2 b\).
[1 mark] for substituting values correctly to find \(L = 1.92 \times 10^{26}\text{ W}\).
[1 mark] for finding the ratio to obtain \(0.50 L_{\odot}\).

(a) \(\frac{\tau}{\tau_{\odot}} = \left(\frac{M}{M_{\odot}}\right)^{-2.5} = (3.5)^{-2.5} = 0.0437\). Converting to a percentage: \(0.0437 \times 100\% \approx 4.4\%\).

(b) A star of mass \(3.5 M_{\odot}\) evolves into a red giant, sheds its outer layers in a planetary nebula, and leaves behind a white dwarf. A star of mass \(25 M_{\odot}\) evolves into a red supergiant, undergoes a core-collapse (Type II) supernova explosion, and leaves behind a neutron star or a black hole (since its remnant core mass exceeds the Tolman-Oppenheimer-Volkoff limit).

(c) A white dwarf is prevented from collapsing under gravity by electron degeneracy pressure. The Chandrasekhar limit (\(1.4 M_{\odot}\)) is the maximum mass for which electron degeneracy pressure can balance the gravitational forces. Above this mass, gravitational forces overcome electron degeneracy pressure, causing the star to collapse further.

(a)
[1 mark] for writing the ratio formula \(\frac{\tau}{\tau_{\odot}} = (3.5)^{-2.5}\).
[1 mark] for completing calculation to show \(4.37\%\) or \(4.4\%\).

(b)
[1 mark] for identifying the \(3.5 M_{\odot}\) star becomes a red giant and ends as a white dwarf.
[1 mark] for noting the loss of layers as planetary nebula for the lower-mass star.
[1 mark] for identifying the \(25 M_{\odot}\) star becomes a red supergiant and undergoes a supernova.
[1 mark] for stating it ends as a neutron star or black hole.

(c)
[1 mark] for identifying electron degeneracy pressure as the force supporting white dwarfs against gravity.
[1 mark] for stating that the Chandrasekhar limit is the maximum mass supported by this pressure.
[1 mark] for explaining that beyond \(1.4 M_{\odot}\), gravity overcomes this pressure, leading to further collapse.

(a) High temperature (approx. \(10^7\text{ K}\)) is needed to give nuclei high kinetic energy to overcome the electrostatic repulsion (Coulomb barrier) between positively charged nuclei. High density is required to ensure a high collision rate of nuclei.

(b) The mass deficit is \(\Delta m = [m({}^{1}_{1}\text{H}) + m({}^{2}_{1}\text{H})] - m({}^{3}_{2}\text{He})\) \(= (1.007825 + 2.014102) - 3.016029 = 3.021927 - 3.016029 = 0.005898\text{ u}\). The energy released is \(E = 0.005898 \times 931.5\text{ MeV} = 5.493987\text{ MeV} \approx 5.49\text{ MeV}\).

(c) From Einstein's mass-energy equivalence: \(E = \Delta m \cdot c^2\). Power (Luminosity \(L\)) is energy per unit time, so \(\frac{\Delta m}{\Delta t} = \frac{L}{c^2} = \frac{3.83 \times 10^{26}\text{ J s}^{-1}}{(3.00 \times 10^8\text{ m s}^{-1})^2} = 4.26 \times 10^9\text{ kg s}^{-1}\).

(a)
[1 mark] for specifying high temperature AND high density.
[1 mark] for explaining high temperature provides the necessary kinetic energy to overcome electrostatic repulsion.
[1 mark] for explaining high density ensures sufficient collision frequency.

(b)
[1 mark] for calculating mass difference \(\Delta m = 0.005898\text{ u}\).
[1 mark] for multiplying by \(931.5\text{ MeV u}^{-1}\).
[1 mark] for final answer of \(5.49\text{ MeV}\) (accept \(5.50\text{ MeV}\) depending on rounding precision).

(c)
[1 mark] for quoting \(E = \Delta m \cdot c^2\) or \(\Delta m/\Delta t = L/c^2\).
[1 mark] for substituting \(L = 3.83 \times 10^{26}\) and \(c = 3.00 \times 10^8\).
[1 mark] for final answer \(4.26 \times 10^9\text{ kg s}^{-1}\) (accept \(4.3 \times 10^9\text{ kg s}^{-1}\)).

(a) Let the initial direction of block \(A\) be positive.
Initial momentum: \(p_i = m_A u_A + m_B u_B = 2.0 \times 6.0 + 4.0 \times 0 = 12.0\text{ kg m s}^{-1}\).
Final momentum: \(p_f = m_A v_A + m_B v_B = 2.0 \times (-1.5) + 4.0 \times v_B = -3.0 + 4.0 v_B\).
By conservation of linear momentum: \(12.0 = -3.0 + 4.0 v_B \Rightarrow v_B = 3.75\text{ m s}^{-1}\) in the positive direction.

(b) Initial kinetic energy: \(E_{k,i} = \frac{1}{2} m_A u_A^2 = 0.5 \times 2.0 \times (6.0)^2 = 36.0\text{ J}\).
Final kinetic energy: \(E_{k,f} = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 = 0.5 \times 2.0 \times (-1.5)^2 + 0.5 \times 4.0 \times (3.75)^2 = 2.25 + 28.125 = 30.375\text{ J} \approx 30.4\text{ J}\).
Since \(E_{k,f} < E_{k,i}\), kinetic energy is not conserved, proving the collision is inelastic.

(c) Impulse on \(A\): \(\Delta p_A = m_A v_A - m_A u_A = 2.0 \times (-1.5) - 2.0 \times 6.0 = -15.0\text{ N s}\).
Average force on \(A\): \(F = \frac{\Delta p_A}{\Delta t} = \frac{-15.0}{0.080} = -187.5\text{ N} \approx -1.9 \times 10^2\text{ N}\).

(a)
[1 mark] for recognizing momentum is conserved and setting up the equation.
[1 mark] for applying correct signs: \(2.0 \times 6.0 = 2.0 \times (-1.5) + 4.0 v_B\).
[1 mark] for calculating \(v_B = 3.75\text{ m s}^{-1}\) (or \(3.8\text{ m s}^{-1}\)).

(b)
[1 mark] for calculating initial KE as \(36.0\text{ J}\).
[1 mark] for calculating final KE as \(30.4\text{ J}\) (or \(30.375\text{ J}\)).
[1 mark] for stating that since \(E_{k,f} < E_{k,i}\), the collision is inelastic.

(c)
[1 mark] for calculating momentum change of \(A\) as \(-15.0\text{ N s}\) (or magnitude \(15.0\text{ N s}\)).
[1 mark] for dividing momentum change by \(0.080\text{ s}\).
[1 mark] for final answer \(-190\text{ N}\) or \(190\text{ N}\) (accept \(-187.5\text{ N}\) or \(187.5\text{ N}\)).

(a) Period: \(T = \frac{t}{20} = \frac{35.8}{20} = 1.79\text{ s}\).
Absolute uncertainty in \(T\): \(\Delta T = \frac{\Delta t}{20} = \frac{0.4}{20} = 0.02\text{ s}\).
So, \(T = 1.79 \pm 0.02\text{ s}\).

(b) Rearranging \(T = 2\pi \sqrt{\frac{L}{g}}\): \(g = \frac{4\pi^2 L}{T^2} = \frac{4 \pi^2 \times 0.800}{1.79^2} = \frac{31.5827}{3.2041} = 9.857\text{ m s}^{-2} \approx 9.86\text{ m s}^{-2}\).

(c) The fractional uncertainty formula: \(\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}\).
\(\frac{\Delta L}{L} = \frac{0.005}{0.800} = 0.00625\).
\(\frac{\Delta T}{T} = \frac{0.02}{1.79} = 0.01117\).
\(\frac{\Delta g}{g} = 0.00625 + 2(0.01117) = 0.00625 + 0.02234 = 0.0286\) (or \(2.9\%\)).
Absolute uncertainty: \(\Delta g = 9.857 \times 0.0286 = 0.282\text{ m s}^{-2}\).
Rounding uncertainty to 1 significant figure gives \(\Delta g = 0.3\text{ m s}^{-2}\).
Expressing \(g\) to match the uncertainty decimal places: \(g = 9.9 \pm 0.3\text{ m s}^{-2}\).

(a)
[1 mark] for calculating \(T = 1.79\text{ s}\).
[1 mark] for stating/calculating \(\Delta T = \Delta t / 20\).
[1 mark] for final answer \(0.02\text{ s}\).

(b)
[1 mark] for correctly rearranging to make \(g\) the subject.
[1 mark] for calculating \(g = 9.86\text{ m s}^{-2}\) (accept \(9.9\text{ m s}^{-2}\)).

(c)
[1 mark] for using the correct fractional uncertainty relation (adding fractional uncertainties, doubling the power term).
[1 mark] for calculating fractional uncertainty in \(g\) as \(0.029\) or \(2.9\%\).
[1 mark] for calculating \(\Delta g = 0.28\text{ m s}^{-2}\).
[1 mark] for stating the final answer with appropriate significant figures: \(9.9 \pm 0.3\text{ m s}^{-2}\).

(a) Equating centripetal force and gravitational force: \(\frac{G M m}{r^2} = \frac{m v^2}{r} \Rightarrow v = \sqrt{\frac{G M}{r}}\).
Using \(G = 6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}\):
\(v = \sqrt{\frac{6.67 \times 10^{-11} \times 6.00 \times 10^{24}}{1.20 \times 10^7}} = \sqrt{3.335 \times 10^7} = 5.775 \times 10^3\text{ m s}^{-1} \approx 5.78 \times 10^3\text{ m s}^{-1}\).

(b) Total mechanical energy: \(E_T = E_k + E_p = \frac{1}{2}mv^2 - \frac{G M m}{r} = -\frac{G M m}{2r}\).
\(E_T = -\frac{6.67 \times 10^{-11} \times 6.00 \times 10^{24} \times 2.50 \times 10^3}{2 \times 1.20 \times 10^7} = -4.16875 \times 10^{10}\text{ J} \approx -4.17 \times 10^{10}\text{ J}\).

(c) Escape speed is the minimum speed required for an object at a given distance from a gravitational source to escape its gravitational pull and reach infinity with zero kinetic energy (without further propulsion).
By conservation of energy: \(E_{escape} = 0 \Rightarrow \frac{1}{2}m v_{escape}^2 - \frac{G M m}{r} = 0 \Rightarrow v_{escape} = \sqrt{\frac{2 G M}{r}}\).
\(v_{escape} = \sqrt{2} \times v_{orbital} = 1.414 \times 5.775 \times 10^3\text{ m s}^{-1} = 8.17 \times 10^3\text{ m s}^{-1}\).

(a)
[1 mark] for equating centripetal force to gravitational force.
[1 mark] for showing formula \(v = \sqrt{G M / r}\).
[1 mark] for calculating \(5.78 \times 10^3\text{ m s}^{-1}\) (accept \(5.8 \times 10^3\text{ m s}^{-1}\)).

(b)
[1 mark] for recalling/deriving total energy equation \(E_T = -GMm/(2r)\) or \(E_k + E_p\).
[1 mark] for correct substitution of values.
[1 mark] for final answer \(-4.17 \times 10^{10}\text{ J}\) (negative sign required; accept \(-4.2 \times 10^{10}\text{ J}\)).

(c)
[1 mark] for a correct definition of escape speed (must mention escaping to infinity / zero total energy).
[1 mark] for showing relationship \(v_{escape} = \sqrt{2} v_{orbital}\) or writing the energy equation \(\frac{1}{2} m v^2 = GMm/r\).
[1 mark] for calculating \(8.17 \times 10^3\text{ m s}^{-1}\) (accept \(8.2 \times 10^3\text{ m s}^{-1}\)).

(a) The diagram should depict a battery represented by an emf source in series with an internal resistor \(r\) inside a dashed box. A voltmeter must be connected in parallel across the terminals of this combination (or in parallel across the external load resistor).

(b) Using the potential divider equation \(V = \varepsilon \frac{R}{R+r}\) or \(\varepsilon = V + I r\):
Case 1: \(I_1 = \frac{10.0}{5.0} = 2.0\text{ A}\) \(\Rightarrow \varepsilon = 10.0 + 2.0 r\) --- (1)
Case 2: \(I_2 = \frac{11.0}{11.0} = 1.0\text{ A}\) \(\Rightarrow \varepsilon = 11.0 + 1.0 r\) --- (2)

(i) Equating (1) and (2):
\(10.0 + 2.0 r = 11.0 + 1.0 r \Rightarrow r = 1.0\ \Omega\).

(ii) Substituting \(r = 1.0\ \Omega\) into (2):
\(\varepsilon = 11.0 + 1.0(1.0) = 12.0\text{ V}\).

(c) Power dissipated: \(P = I^2 R = (1.0)^2 \times 11.0 = 11.0\text{ W}\) (or \(P = \frac{V^2}{R} = \frac{11.0^2}{11.0} = 11.0\text{ W}\)).

(a)
[1 mark] for drawing emf source and internal resistance in series (enclosed by dashed box).
[1 mark] for correctly showing voltmeter in parallel across the battery terminals or external resistor.

(b)(i)
[1 mark] for setting up two correct equations (either using \(V = I R\) and \(\varepsilon = I(R+r)\) or using potential divider relation).
[1 mark] for equating them to eliminate \(\varepsilon\).
[1 mark] for correct calculation of \(r = 1.0\ \Omega\).

(b)(ii)
[1 mark] for substituting \(r\) back into one of the equations.
[1 mark] for correct calculation of \(\varepsilon = 12.0\text{ V}\).

(c)
[1 mark] for choosing a valid power formula: \(P = I^2 R\) or \(P = V^2/R\).
[1 mark] for correct final value of \(11.0\text{ W}\).

(a) Magnetic flux linkage is the product of the magnetic flux passing through a single loop of a coil and the number of turns in the coil: \(\Phi = N B A \cos\theta\).

(b) Lenz's law states that the direction of the induced current is such that its magnetic field opposes the change in magnetic flux that created it. This represents the conservation of energy: if the induced field reinforced the flux change, the current would increase indefinitely, creating electrical energy without any mechanical work being done.

(c) The area of the coil is \(A = w \times l = 0.080 \times 0.120 = 0.0096\text{ m}^2\).
The angular velocity is \(\omega = 2\pi f = 2\pi \times 50.0 = 100\pi \approx 314.16\text{ rad s}^{-1}\).
The induced emf is given by \(\varepsilon = N B A \omega \sin(\omega t)\).
The maximum induced emf occurs when \(\sin(\omega t) = 1\):
\(\varepsilon_{max} = N B A \omega = 150 \times 0.45 \times 0.0096 \times 314.16 = 203.57\text{ V} \approx 204\text{ V}\).

(a)
[1 mark] for defining magnetic flux through a single loop (product of magnetic field strength and area perpendicular to field).
[1 mark] for multiplying by the number of turns \(N\).

(b)
[1 mark] for stating that the induced emf/current direction opposes the change in flux.
[1 mark] for explaining that mechanical work must be done against opposing magnetic forces to rotate the coil.
[1 mark] for linking this to energy conservation (otherwise electrical energy would be generated for free, violating conservation of energy).

(c)
[1 mark] for calculating area \(A = 0.0096\text{ m}^2\).
[1 mark] for calculating angular velocity \(\omega = 314\text{ rad s}^{-1}\).
[1 mark] for stating/using the equation for maximum induced emf \(\varepsilon_{max} = N B A \omega\).
[1 mark] for final answer of \(204\text{ V}\) (accept \(200\text{ V}\) if using \(3.14\) for \(\pi\)).

(a) As the loop is pulled out of the magnetic field, the area \(A\) of the loop remaining inside the field decreases. Since magnetic flux linkage is given by \(\Phi = B A \cos\theta\), this reduction in area causes a decrease in the magnetic flux linkage through the loop over time. According to Faraday's law of induction, the induced electromotive force (emf) is proportional to the rate of change of magnetic flux linkage, hence an emf is established in the loop.

(b) The rate of change of area is given by \(\frac{\Delta A}{\Delta t} = w v\), where \(w = 0.080\text{ m}\) is the width of the loop cutting the field boundary and \(v = 2.0\text{ m s}^{-1}\) is the speed.
Using Faraday's law, the magnitude of the induced emf is:
\(\varepsilon = B w v = 0.60 \times 0.080 \times 2.0 = 0.096\text{ V}\).
By Ohm's law, the induced current is:
\(I = \frac{\varepsilon}{R} = \frac{0.096}{0.40} = 0.24\text{ A}\).

(c) The magnetic force acting on the wire segment of width \(w\) carrying the induced current \(I\) is given by:
\(F_M = B I w = 0.60 \times 0.24 \times 0.080 = 0.01152\text{ N}\).
To maintain a constant speed, an equal and opposite external mechanical force \(F_{\text{ext}}\) must be applied.
The rate of doing mechanical work (mechanical power input) is:
\(P_{\text{mech}} = F_{\text{ext}} v = 0.01152 \times 2.0 = 0.02304\text{ W}\).
The rate of thermal energy dissipation (electrical power) in the resistor is:
\(P_{\text{elec}} = I^2 R = (0.24)^2 \times 0.40 = 0.02304\text{ W}\).
Since \(P_{\text{mech}} = P_{\text{elec}}\), the mechanical work done per second equals the rate of thermal energy dissipation.

(a)
- Recognizes that the area of the loop within the magnetic field decreases [1 mark]
- Relates this to a change in magnetic flux linkage over time [1 mark]
- Mentions Faraday's law (induced emf is proportional to the rate of change of magnetic flux linkage) [1 mark]

(b)
- Calculates the induced emf \(\varepsilon = Bwv = 0.096\text{ V}\) [1 mark]
- Calculates the current \(I = \frac{\varepsilon}{R} = 0.24\text{ A}\) [1 mark]
- Correct final units and significant figures [1 mark]

(c)
- Calculates the magnetic/external force \(F = B I w = 0.01152\text{ N}\) [1 mark]
- Calculates the mechanical power input \(P_{\text{mech}} = F v = 0.023\text{ W}\) [1 mark]
- Calculates the electrical power dissipation \(P_{\text{elec}} = I^2 R = 0.023\text{ W}\) and concludes they are equal [1 mark]
- Accept alternative proof using symbolic algebra: \(P_{\text{mech}} = Fv = (BIw)v = B\left(\frac{Bwv}{R}\right)wv = \frac{B^2w^2v^2}{R}\) and \(P_{\text{elec}} = I^2R = \left(\frac{Bwv}{R}\right)^2 R = \frac{B^2w^2v^2}{R}\).

(a) For a satellite in circular orbit, the gravitational force provides the necessary centripetal force:
\(\frac{G M m}{r^2} = \frac{m v^2}{r}\)
Simplifying this gives:
\(\frac{G M}{r} = v^2\)
The orbital speed \(v\) is related to the orbital period \(T\) by:
\(v = \frac{2\pi r}{T}\)
Substituting this expression for speed into the simplified force equation yields:
\(\frac{G M}{r} = \left(\frac{2\pi r}{T}\right)^2 = \frac{4\pi^2 r^2}{T^2}\)
Rearranging for \(T^2\) gives the desired Kepler's Third Law relation:
\(T^2 = \frac{4\pi^2 r^3}{G M}\)

(b) First, convert the period from hours to seconds:
\(T = 2.4 \times 3600\text{ s} = 8640\text{ s}\)
Next, rearrange the orbital period equation to solve for the mass of the planet \(M\):
\(M = \frac{4\pi^2 r^3}{G T^2}\)
Substitute the given values into the equation:
\(M = \frac{4\pi^2 \times (8.5 \times 10^6)^3}{(6.67 \times 10^{-11}) \times (8640)^2}\)
\(M = \frac{39.478 \times 6.141 \times 10^{20}}{(6.67 \times 10^{-11}) \times (7.465 \times 10^7)} = \frac{2.424 \times 10^{22}}{4.979 \times 10^{-3}} \approx 4.87 \times 10^{24}\text{ kg}\)
Rounding to 2 significant figures gives \(4.9 \times 10^{24}\text{ kg}\).

(c) The formula for the gravitational potential energy \(E_p\) of the spacecraft in orbit is:
\(E_p = -\frac{G M m}{r}\)
The change in gravitational potential energy \(\Delta E_p\) as it moves from \(r_1\) to \(r_2\) is:
\(\Delta E_p = E_{p,2} - E_{p,1} = -\frac{G M m}{r_2} - \left(-\frac{G M m}{r_1}\right) = G M m \left(\frac{1}{r_1} - \frac{1}{r_2}\right)\)
Substitute the values:
\(\Delta E_p = (6.67 \times 10^{-11}) \times (4.87 \times 10^{24}) \times 1200 \times \left(\frac{1}{8.5 \times 10^6} - \frac{1}{1.2 \times 10^7}\right)\)
\(\Delta E_p = 3.898 \times 10^{17} \times (1.176 \times 10^{-7} - 8.333 \times 10^{-8})\)
\(\Delta E_p = 3.898 \times 10^{17} \times 3.427 \times 10^{-8} \approx 1.34 \times 10^{10}\text{ J}\)
Rounding to 2 significant figures gives \(1.3 \times 10^{10}\text{ J}\).

(a)
- Equates gravitational force to centripetal force [1 mark]
- Expresses speed in terms of period, \(v = \frac{2\pi r}{T}\) [1 mark]
- Correctly substitutes and rearranges to get the final relation [1 mark]

(b)
- Converts orbital period to seconds (\(8640\text{ s}\)) [1 mark]
- Correctly rearranges the formula and substitutes values [1 mark]
- Obtains a final answer between \(4.8 \times 10^{24}\text{ kg}\) and \(4.9 \times 10^{24}\text{ kg}\) with correct unit [1 mark]

(c)
- Recognizes potential energy formula \(E_p = -\frac{GMm}{r}\) [1 mark]
- Sets up the correct subtraction structure for \(\Delta E_p\) [1 mark]
- Obtains a final value of \(1.3 \times 10^{10}\text{ J}\) (accept range \(1.3 \times 10^{10}\text{ J}\) to \(1.4 \times 10^{10}\text{ J}\) depending on rounding of mass \(M\)) [1 mark]

(a) The period \(T\) is the total time divided by the number of oscillations: \(T = \frac{35.8}{20} = 1.79\text{ s}\). The absolute uncertainty in \(T\) is \(\Delta T = \frac{\Delta t}{20} = \frac{0.2}{20} = 0.01\text{ s}\). Thus, \(T = (1.79 \pm 0.01)\text{ s}\).

(b) Using the formula for a simple pendulum \(T = 2\pi\sqrt{\frac{L}{g}}\), we rearrange for \(g\): \(g = \frac{4\pi^2 L}{T^2}\). Substituting the measured values: \(g = \frac{4\pi^2 \times 0.800}{1.79^2} \approx 9.856\text{ m s}^{-2} \approx 9.9\text{ m s}^{-2}\).
The fractional uncertainty in \(g\) is: \(\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T} = \frac{0.005}{0.800} + 2 \times \frac{0.01}{1.79} \approx 0.00625 + 0.01117 = 0.01742\).
The absolute uncertainty in \(g\) is \(\Delta g = 9.856 \times 0.01742 \approx 0.17\text{ m s}^{-2}\). Therefore, to one significant figure, \(\Delta g = 0.2\text{ m s}^{-2}\), giving \(g = (9.9 \pm 0.2)\text{ m s}^{-2}\).

(c) If there is a systematic error that overestimates the length \(L\) (for example, not measuring to the center of mass of the bob), the value of \(L\) used in the calculation will be too large. Since \(g \propto L\), this will result in a systematically overestimated (too large) calculated value of \(g\).

[1 mark] for calculating \(T = 1.79\text{ s}\).
[1 mark] for calculating \(\Delta T = 0.01\text{ s}\).
[1 mark] for correctly substituting values into \(g = \frac{4\pi^2 L}{T^2}\) to get \(9.9\text{ m s}^{-2}\).
[1 mark] for calculating the fractional uncertainty of \(g\) (approx 1.7% or 0.017).
[1.43 marks] for calculating the absolute uncertainty \(\Delta g = 0.2\text{ m s}^{-2}\) (accept \(0.17\text{ m s}^{-2}\)).
[1 mark] for explaining that an overestimated \(L\) leads to a larger calculated \(g\) (or vice-versa).

(a) For a circular orbit, the gravitational force provides the centripetal force: \(\frac{GMm}{R^2} = \frac{mv^2}{R}\), which simplifies to \(v^2 = \frac{GM}{R}\). Since orbital speed is \(v = \frac{2\pi R}{T}\), we substitute to get \(\left(\frac{2\pi R}{T}\right)^2 = \frac{GM}{R} \implies \frac{4\pi^2 R^2}{T^2} = \frac{GM}{R}\). Rearranging yields \(T^2 = \frac{4\pi^2}{GM}R^3\).

(b) Convert period \(T\) to seconds: \(T = 1.77\text{ days} = 1.77 \times 24 \times 3600\text{ s} = 1.529 \times 10^5\text{ s}\).
Rearrange Kepler's Third Law for \(M\): \(M = \frac{4\pi^2 R^3}{G T^2} = \frac{4\pi^2 \times (4.22 \times 10^8)^3}{6.67 \times 10^{-11} \times (1.529 \times 10^5)^2} \approx 1.90 \times 10^{27}\text{ kg}\).

(c) The escape speed is given by \(v_{\text{esc}} = \sqrt{\frac{2GM}{R_{\text{planet}}}}\). Substituting \(M = 1.90 \times 10^{27}\text{ kg}\) and \(R_{\text{planet}} = 6.99 \times 10^7\text{ m}\):
\(v_{\text{esc}} = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 1.90 \times 10^{27}}{6.99 \times 10^7}} \approx 6.02 \times 10^4\text{ m s}^{-1}\).

[2 marks] for deriving Kepler's third law (equating gravitational and centripetal forces [1] and substituting \(v = 2\pi R / T\) [1]).
[1 mark] for converting \(1.77\text{ days}\) to \(1.53 \times 10^5\text{ s}\).
[1.43 marks] for calculating mass \(M = 1.90 \times 10^{27}\text{ kg}\) (accept range \(1.88 \times 10^{27}\) to \(1.92 \times 10^{27}\)).
[1 mark] for using escape velocity formula \(v_{\text{esc}} = \sqrt{\frac{2GM}{R}}\).
[1 mark] for calculating \(v_{\text{esc}} \approx 6.02 \times 10^4\text{ m s}^{-1}\) (accept \(6.0 \times 10^4\text{ m s}^{-1}\) to \(6.1 \times 10^4\text{ m s}^{-1}\)).

(a) Using conservation of momentum: \(m_A u_A + m_B u_B = (m_A + m_B) v\). Substituting the given values: \(2.0 \times 4.0 + 3.0 \times 0 = (2.0 + 3.0) v \implies 8.0 = 5.0 v \implies v = 1.6\text{ m s}^{-1}\).

(b) Initial kinetic energy \(E_{k,i} = \frac{1}{2} m_A u_A^2 = \frac{1}{2} \times 2.0 \times 4.0^2 = 16\text{ J}\).
Final kinetic energy \(E_{k,f} = \frac{1}{2} (m_A + m_B) v^2 = \frac{1}{2} \times 5.0 \times 1.6^2 = 6.4\text{ J}\).
Loss in kinetic energy = \(16 - 6.4 = 9.6\text{ J}\). Since kinetic energy is lost, the collision is completely inelastic.

(c) If friction acts on the blocks, there is a net external force acting on the two-block system. Because \(\Delta p = F_{\text{net}} \Delta t\), a non-zero external force means that the total momentum of the system is not conserved during the collision.

[2 marks] for momentum conservation equation [1] and calculating \(v = 1.6\text{ m s}^{-1}\) [1].
[2.43 marks] for calculating initial \(E_k = 16\text{ J}\) and final \(E_k = 6.4\text{ J}\) [1], identifying that \(E_k\) is lost [1], and concluding it is inelastic [0.43].
[2 marks] for explaining that total momentum is only conserved in an isolated system/no external forces [1], and the presence of friction (external force) means total momentum decreases/is not conserved [1].

(a) The Stefan-Boltzmann law states that the total power (luminosity) radiated per unit surface area of a blackbody is directly proportional to the fourth power of its absolute temperature: \(P = \sigma A T^4\).

(b) Luminosity is \(L = 4\pi R^2 \sigma T^4\). Therefore: \(\frac{L_X}{L_\odot} = \left(\frac{R_X}{R_\odot}\right)^2 \left(\frac{T_X}{T_\odot}\right)^4 \implies 80 = \left(\frac{R_X}{R_\odot}\right)^2 \left(\frac{1.2 \times 10^4}{5.8 \times 10^3}\right)^4\).
\(\left(\frac{12000}{5800}\right)^4 \approx (2.069)^4 \approx 18.33\).
Thus, \(\left(\frac{R_X}{R_\odot}\right)^2 = \frac{80}{18.33} \approx 4.36 \implies \frac{R_X}{R_\odot} = \sqrt{4.36} \approx 2.1\).

(c) Star X has a high temperature (\(1.2 \times 10^4\text{ K}\)) and luminosity (\(80 L_\odot\)), placing it on the upper part of the Main Sequence. This indicates it is a massive star with a very hot core. In stars more massive than the Sun with high core temperatures, the CNO (Carbon-Nitrogen-Oxygen) cycle is the dominant hydrogen fusion pathway.

[1 mark] for stating the law/formula clearly with definitions of terms.
[3.43 marks] for setting up the ratio equation [1], calculating the temperature ratio to the fourth power (\(\approx 18.3\)) [1.43], and getting \(\frac{R_X}{R_\odot} \approx 2.1\) (accept \(2.0\) to \(2.2\)) [1].
[2 marks] for identifying its position in the upper-left of the Main Sequence [1] and stating that the CNO cycle is dominant because of the high core temperature associated with massive stars [1].

(a) Redshift \(z = 0.15\) means that the wavelength of the light observed from the galaxy has increased by \(15\%\) relative to its emitted wavelength: \(z = \frac{\Delta \lambda}{\lambda_0} = 0.15\).

(b) The relationship between redshift and cosmic scale factor \(R\) is given by: \(1 + z = \frac{R_{\text{now}}}{R_{\text{then}}}\). Therefore, \(\frac{R_{\text{then}}}{R_{\text{now}}} = \frac{1}{1 + z} = \frac{1}{1.15} \approx 0.87\). This means the universe was \(87\%\) of its current size when the light was emitted.

(c) From Hubble's Law, the recessional speed is \(v = z c\) (for \(z \ll 1\)).
\(v = 0.15 \times 3.0 \times 10^5\text{ km s}^{-1} = 4.5 \times 10^4\text{ km s}^{-1}\).
Hubble's law states \(v = H_0 d\). Thus, \(d = \frac{v}{H_0} = \frac{4.5 \times 10^4}{70} \approx 643\text{ Mpc} \approx 640\text{ Mpc}\).

[1 mark] for explaining redshift as the fractional increase in wavelength (\(\frac{\Delta\lambda}{\lambda_0}\)) equal to \(0.15\).
[2 marks] for using \(1 + z = \frac{R_0}{R}\) [1] and getting the ratio \(\approx 0.87\) (or \(\frac{1}{1.15}\)) [1].
[3.43 marks] for calculating recessional speed \(v = 4.5 \times 10^4\text{ km s}^{-1}\) [1], using \(d = v / H_0\) [1], and calculating \(d \approx 640\text{ Mpc}\) (accept range \(640 - 643\text{ Mpc}\)) [1.43].

(a) The energy of a photon of this light is: \(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.0 \times 10^8}{410 \times 10^{-9}} = 4.85 \times 10^{-19}\text{ J}\).
Converting to electron-volts: \(E = \frac{4.85 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.03\text{ eV}\).
Since the photon energy (\(3.03\text{ eV}\)) is greater than the work function (\(2.2\text{ eV}\)) of the potassium surface, photoelectric emission will occur.

(b) According to Einstein's photoelectric equation: \(E_{k,\text{max}} = hf - \Phi = 3.03\text{ eV} - 2.2\text{ eV} = 0.83\text{ eV}\).

(c) Doubling the intensity of the light means doubling the number of photons incident on the surface per second. Since the photon energy remains unchanged (wavelength is constant), the maximum kinetic energy of the individual photoelectrons remains unchanged at \(0.83\text{ eV}\). However, because there are twice as many photons, the number of photoelectrons emitted per second (rate of emission) will double.

[2 marks] for calculating photon energy in J or eV [1] and comparing it to the work function to show \(E > \Phi\) [1].
[1.43 marks] for calculating \(E_{k,\text{max}} = 0.83\text{ eV}\) (accept \(0.8\text{ eV}\) to \(0.83\text{ eV}\)).
[3 marks] for stating maximum kinetic energy is unchanged [1] and explaining that it depends only on frequency/wavelength [1]; stating rate of emission doubles [1] because number of incident photons per second doubles [1] (max 3 marks total).

(a) Magnetic flux linkage is the product of the magnetic flux passing through a single loop of a coil and the total number of turns of the coil: \(\Phi = N B A \cos\theta\).

(b) The area of the square coil is \(A = (0.050\text{ m})^2 = 2.5 \times 10^{-3}\text{ m}^2\).
The initial magnetic flux linkage is \(\Phi_i = N B A = 200 \times 0.40 \times 2.5 \times 10^{-3} = 0.20\text{ Wb-turns}\).
The final magnetic flux linkage is \(\Phi_f = 0\).
By Faraday's law, the magnitude of the induced emf is: \(\mathcal{E} = \frac{\Delta \Phi}{\Delta t} = \frac{0.20}{0.25} = 0.80\text{ V}\).

(c) The average induced current is \(I = \frac{\mathcal{E}}{R} = \frac{0.80}{8.0} = 0.10\text{ A}\).
The rate of thermal energy dissipation is \(P = I^2 R = (0.10)^2 \times 8.0 = 0.080\text{ W}\).
The total thermal energy dissipated is \(E = P \Delta t = 0.080 \times 0.25 = 0.020\text{ J}\) (or \(20\text{ mJ}\)).

[1 mark] for defining magnetic flux linkage as \(N B A \cos\theta\) or product of flux and number of turns.
[2 marks] for calculating the area \(A = 2.5 \times 10^{-3}\text{ m}^2\) [1] and using Faraday's law to find \(\mathcal{E} = 0.80\text{ V}\) [1].
[3.43 marks] for calculating current \(I = 0.10\text{ A}\) [1], using a correct power/energy formula (\(E = I^2 R t\) or \(E = \frac{V^2}{R} t\)) [1], and calculating energy dissipated \(E = 0.020\text{ J}\) [1.43].