2025 IB DP Physics 模擬試題連答案詳解

For a pipe closed at one end, the harmonic frequencies are given by:
\(f_n = \frac{n v}{4L}\) where \(n = 1, 3, 5, \dots\)
The frequency of the third harmonic (\(n=3\)) is:
\(f_{3,\text{closed}} = \frac{3v}{4L}\)

For a pipe open at both ends, the harmonic frequencies are given by:
\(f_m = \frac{m v}{2 L_{\text{open}}}\) where \(m = 1, 2, 3, \dots\)
The frequency of the second harmonic (\(m=2\)) is:
\(f_{2,\text{open}} = \frac{2v}{2 L_{\text{open}}} = \frac{v}{L_{\text{open}}}\)

Equating the two frequencies because they are stated to be equal:
\(\frac{3v}{4L} = \frac{v}{L_{\text{open}}}\)

Rearranging to find the ratio \(\frac{L_{\text{open}}}{L}\):
\(\frac{L_{\text{open}}}{L} = \frac{4}{3}\)

[1 mark] award for identifying correct expressions for both frequencies, equating them, and solving correctly to obtain option C.

The gravitational potential energy is given by:
\(E_p = -\frac{GMm}{r}\)

Initial potential energy: \(E_{p,i} = -\frac{GMm}{R}\)
Final potential energy: \(E_{p,f} = -\frac{GMm}{3R}\)
Change in potential energy:
\(\Delta E_p = E_{p,f} - E_{p,i} = -\frac{GMm}{3R} - \left(-\frac{GMm}{R}\right) = \frac{2GMm}{3R}\)

The kinetic energy in a stable circular orbit is given by:
\(E_k = \frac{GMm}{2r}\)

Initial kinetic energy: \(E_{k,i} = \frac{GMm}{2R}\)
Final kinetic energy: \(E_{k,f} = \frac{GMm}{6R}\)
Change in kinetic energy:
\(\Delta E_k = E_{k,f} - E_{k,i} = \frac{GMm}{6R} - \frac{GMm}{2R} = -\frac{GMm}{3R}\)

Therefore, the potential energy increases by \(\frac{2GMm}{3R}\) and the kinetic energy decreases by \(\frac{GMm}{3R}\).

[1 mark] award for calculating both change in potential energy and change in kinetic energy correctly to find option B.

For the first ball, taking the direction away from the wall as positive, the change in momentum is:
\(\Delta p = mv - (-mv) = 2mv\)
The average force \(F\) is:
\(F = \frac{\Delta p}{t} = \frac{2mv}{t}\)

For the second ball, the change in momentum is:
\(\Delta p' = (2m)(2v) - (-(2m)(2v)) = 8mv\)
The contact time is \(t' = \frac{t}{2}\).
The new average force \(F'\) is:
\(F' = \frac{\Delta p'}{t'} = \frac{8mv}{t/2} = \frac{16mv}{t}\)

Since \(F = \frac{2mv}{t}\), we can substitute this to get:
\(F' = 8 \left(\frac{2mv}{t}\right) = 8F\)

[1 mark] award for finding the ratio of momentum changes and using the halved contact time to determine that the force is 8 times the original force, corresponding to option C.

After three half-lives, the remaining fraction of undecayed nuclei is:
\(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\)

Therefore, the fraction of the initial nuclei that have decayed is:
\(1 - \frac{1}{8} = \frac{7}{8}\)

Since activity is directly proportional to the number of remaining undecayed nuclei, the activity of the sample is:
\(A = \frac{A_0}{8}\)

[1 mark] award for calculating the decayed fraction as 7/8 and the remaining activity as 1/8 of the original value, which yields option B.

According to Einstein's photoelectric equation, the maximum kinetic energy for the first frequency \(f\) is:
\(E_{\text{max}} = hf - \Phi \implies hf = E_{\text{max}} + \Phi\)

For the second frequency \(2f\), the maximum kinetic energy \(E'\) is:
\(E' = h(2f) - \Phi = 2(hf) - \Phi\)

Substitute the expression for \(hf\) into the second equation:
\(E' = 2(E_{\text{max}} + \Phi) - \Phi = 2E_{\text{max}} + 2\Phi - \Phi = 2E_{\text{max}} + \Phi\)

[1 mark] award for correctly substituting Einstein's photoelectric equation into the frequency-doubled equation and obtaining option C.

The internal energy \(U\) of an ideal gas depends only on its absolute temperature. Since the gas returns to its initial temperature at the end of the two-step process, the total change in internal energy is:
\(\Delta U_{\text{total}} = 0\)

By the first law of thermodynamics applied to the entire process:
\(\Delta U_{\text{total}} = Q_{\text{net}} - W_{\text{net, by}}\)

During the adiabatic compression (step 1), work \(W\) is done ON the gas, so \(W_{\text{step 1, by}} = -W\). During the constant volume cooling (step 2), no work is done because \(\Delta V = 0\), so \(W_{\text{step 2, by}} = 0\).
The net work done by the gas is:
\(W_{\text{net, by}} = -W + 0 = -W\)

Substituting this into the first law equation:
\(0 = Q_{\text{net}} - (-W) \implies Q_{\text{net}} = -W\)

This negative sign indicates that a net quantity of thermal energy \(W\) is transferred from the gas to its surroundings.

Hence, the total change in internal energy is \(0\) and the net thermal energy transferred to the surroundings is \(W\).

[1 mark] award for identifying that change in internal energy is zero because final and initial temperatures are equal, and that the net heat output must equal the work input, leading to option B.

Let \(T_f\) be the final equilibrium temperature. Since the container is insulated, the thermal energy lost by the metal block equals the thermal energy gained by the liquid:
\(Q_{\text{lost}} = Q_{\text{gained}}\)

\(m \cdot c \cdot (2T - T_f) = (2m) \cdot (2c) \cdot (T_f - T)\)

Dividing both sides of the equation by the common factor \(m \cdot c\):
\(2T - T_f = 4(T_f - T)\)

Expanding the right side:
\(2T - T_f = 4T_f - 4T\)

Grouping the temperature terms:
\(6T = 5T_f \implies T_f = 1.2 T\)

[1 mark] award for setting up the conservation of energy equation correctly and solving for the final temperature to get 1.2T, which is option A.

By conservation of mechanical energy, the initial gravitational potential energy at the top of the incline is converted into both translational and rotational kinetic energy at the bottom:
\(Mgh = E_{k,\text{trans}} + E_{k,\text{rot}}\)

\(Mgh = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2\)

Because the cylinder rolls without slipping, the angular speed \(\omega\) is related to the linear speed \(v\) by \(\omega = \frac{v}{R}\). Substituting this and \(I = \frac{1}{2} M R^2\) into the energy equation:
\(Mgh = \frac{1}{2} M v^2 + \frac{1}{2} \left(\frac{1}{2} M R^2\right) \left(\frac{v}{R}\right)^2\)

\(Mgh = \frac{1}{2} M v^2 + \frac{1}{4} M v^2 = \frac{3}{4} M v^2\)

Canceling \(M\) from both sides and solving for \(v\):
\(gh = \frac{3}{4} v^2 \implies v^2 = \frac{4}{3} gh \implies v = \sqrt{\frac{4}{3} gh}\)

[1 mark] award for setting up conservation of energy including both rotational and translational kinetic energy terms, using the rolling without slipping condition, and solving for v to obtain option B.

The total energy \(E\) of a satellite in a circular orbit of radius \(r\) is given by:

\(E = -\frac{GMm}{2r}\)

The initial total energy is:

\(E_i = -\frac{GMm}{2r}\)

The final total energy at the new radius \(2r\) is:

\(E_f = -\frac{GMm}{2(2r)} = -\frac{GMm}{4r}\)

The change in total energy \(\Delta E\) is:

\(\Delta E = E_f - E_i = -\frac{GMm}{4r} - \left(-\frac{GMm}{2r}\right) = +\frac{GMm}{4r}\)

Therefore, the correct option is B.

[1] Award mark for the correct answer B.

For Pipe A (open at both ends), the fundamental frequency (first harmonic) is:
\(f_{A,1} = \frac{v}{2L_A}\)

For Pipe B (closed at one end), the harmonic frequencies are given by \(f_{B,n} = \frac{nv}{4L_B}\) for odd integers \(n = 1, 3, 5, \dots\). The frequency of the third harmonic (\(n=3\)) is:
\(f_{B,3} = \frac{3v}{4L_B}\)

We are given that \(f_{A,1} = f_{B,3}\):
\(\frac{v}{2L_A} = \frac{3v}{4L_B}\)

Cancelling \(v\) and rearranging for the ratio \(\frac{L_A}{L_B}\):
\(\frac{1}{2L_A} = \frac{3}{4L_B} \implies \frac{L_A}{L_B} = \frac{4}{6} = \frac{2}{3}\)

[1] Award mark for the correct answer A.

Using conservation of linear momentum:
\(m v = (m + 2m) v_f \implies v_f = \frac{v}{3}\)

The initial kinetic energy is:
\(E_{k,i} = \frac{1}{2} m v^2\)

The final kinetic energy is:
\(E_{k,f} = \frac{1}{2} (3m) v_f^2 = \frac{3}{2} m \left(\frac{v}{3}\right)^2 = \frac{1}{6} m v^2 = \frac{1}{3} E_{k,i}\)

The fraction of kinetic energy lost is:
\(\frac{E_{k,i} - E_{k,f}}{E_{k,i}} = 1 - \frac{1}{3} = \frac{2}{3}\)

[1] Award mark for the correct answer C.

After each half-life, the number of remaining active nuclei is halved.
After 1 half-life: \(\frac{1}{2} N_0\) remains.
After 2 half-lives: \(\frac{1}{4} N_0\) remains.
After 3 half-lives: \(\frac{1}{8} N_0\) remains.

The fraction of the initial active nuclei remaining is \(\frac{1}{8}\).
Therefore, the fraction of nuclei that has decayed is:
\(1 - \frac{1}{8} = \frac{7}{8}\)

[1] Award mark for the correct answer C.

From Einstein's photoelectric equation:
\(h f = \Phi + E_k \implies h f - \Phi = E_k\)

For light of frequency \(2f\), the new maximum kinetic energy \(E_k'\) is:
\(E_k' = h (2f) - \Phi = 2(h f) - \Phi\)

Substitute \(hf = E_k + \Phi\) into the equation:
\(E_k' = 2(E_k + \Phi) - \Phi = 2E_k + \Phi\)

[1] Award mark for the correct answer B.

According to the first law of thermodynamics:
\(Q = \Delta U + W\)
where:
- \(Q\) is the thermal energy added to the system. Since energy is removed, \(Q = -150\text{ J}\).
- \(W\) is the work done by the system on its surroundings. Since the gas does work, \(W = +400\text{ J}\).

Rearranging for \(\Delta U\):
\(\Delta U = Q - W = -150\text{ J} - 400\text{ J} = -550\text{ J}\)

Thus, the internal energy of the gas decreases by \(550\text{ J}\).

[1] Award mark for the correct answer D.

According to the Stefan-Boltzmann law, the power radiated by a black body is:
\(P = \sigma A T^4\)

If the new absolute temperature is \(T' = 2T\) and the new surface area is \(A' = \frac{A}{2}\), the new power \(P'\) is:
\(P' = \sigma A' (T')^4 = \sigma \left(\frac{A}{2}\right) (2T)^4 = \sigma \left(\frac{A}{2}\right) (16 T^4) = 8 \sigma A T^4 = 8 P\)

[1] Award mark for the correct answer C.

The radius \(r\) of a charged particle in a magnetic field is:
\(r = \frac{mv}{qB} = \frac{p}{qB}\)

Since kinetic energy \(E_k = \frac{p^2}{2m}\), the momentum is \(p = \sqrt{2mE_k}\). Therefore, we can write:
\(r = \frac{\sqrt{2mE_k}}{qB}\)

For the proton:
\(R = \frac{\sqrt{2mE_k}}{qB}\)

For the alpha particle with the same kinetic energy \(E_k\), mass \(4m\), and charge \(2q\):
\(R_{\alpha} = \frac{\sqrt{2(4m)E_k}}{(2q)B} = \frac{2\sqrt{2mE_k}}{2qB} = \frac{\sqrt{2mE_k}}{qB} = R\)

[1] Award mark for the correct answer B.

For a pipe closed at one end, the fundamental wavelength is \(\lambda_1 = 4L\), so the fundamental frequency is \(f_1 = \frac{v}{\lambda_1} = \frac{v}{4L}\). For a pipe open at both ends, the fundamental wavelength is \(\lambda'_1 = 2L\), so the fundamental frequency is \(f'_1 = \frac{v}{\lambda'_1} = \frac{v}{2L} = 2 f_1\).

Award [1] for the correct answer D. Award [0] for incorrect options.

The total mechanical energy \(E\) of an orbiting satellite in a circular orbit of radius \(r\) is \(E = -\frac{GMm}{2r}\). For the new orbit of radius \(2r\), the energy is \(E' = -\frac{GMm}{4r}\). The change in mechanical energy is \(\Delta E = E' - E = -\frac{GMm}{4r} - \left(-\frac{GMm}{2r}\right) = +\frac{GMm}{4r}\).

Award [1] for the correct answer B. Award [0] for incorrect options.

The de Broglie wavelength is given by \(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE_k}}\). Since the kinetic energy \(E_k\) is the same for both particles, the wavelength is inversely proportional to the square root of the mass: \(\frac{\lambda_e}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha}}{m_e}} = \sqrt{7300} \approx 85\).

Award [1] for the correct answer C. Award [0] for incorrect options.

Taking the initial direction of the ball as positive: \(v_i = +15\text{ m s}^{-1}\) and \(v_f = -10\text{ m s}^{-1}\). The change in momentum is \(\Delta p = m(v_f - v_i) = 0.20 \times (-10 - 15) = -5.0\text{ N s}\). The magnitude of the average force is \(F_{\text{avg}} = \frac{|\Delta p|}{\Delta t} = \frac{5.0}{0.050} = 100\text{ N}\).

Award [1] for the correct answer C. Award [0] for incorrect options.

The number of remaining nuclei at time \(t\) is given by \(N(t) = N_0 e^{-\lambda t}\). Substituting \(t = \frac{2}{\lambda}\) yields \(N = N_0 e^{-2}\). The activity of the sample is \(A(t) = \lambda N(t) = \lambda N_0 e^{-2}\).

Award [1] for the correct answer A. Award [0] for incorrect options.

For an adiabatic process, there is no heat transfer, so \(Q = 0\). In an expansion, work is done BY the gas, hence \(W > 0\). According to the first law of thermodynamics, \(Q = \Delta U + W\), so \(0 = \Delta U + W\), which means \(\Delta U = -W\). Since \(W > 0\), it follows that \(\Delta U < 0\).

Award [1] for the correct answer A. Award [0] for incorrect options.

The initial magnetic flux linkage through the coil is \(\Phi_{\text{linkage}} = N B A\). When the magnetic field becomes zero, the final flux linkage is zero. Therefore, the change in flux linkage is \(\Delta \Phi_{\text{linkage}} = N B A\). By Faraday's law of induction, the average induced emf is \(\epsilon = \frac{\Delta \Phi_{\text{linkage}}}{\Delta t} = \frac{N B A}{\Delta t}\).

Award [1] for the correct answer B. Award [0] for incorrect options.

The initial vertical component of velocity is \(u_y = v \sin\theta\). At maximum height, the vertical velocity component becomes \(v_y = 0\). Using the kinematic equation \(v_y^2 = u_y^2 - 2gH\), we get \(0 = (v \sin\theta)^2 - 2gH\), which yields \(H = \frac{v^2 \sin^2\theta}{2g}\).

Award [1] for the correct answer B. Award [0] for incorrect options.

The fundamental frequency of a closed pipe of length \(L_c\) is given by:
\(f = \frac{v}{4L_c}\)

For an open pipe of length \(L_o\), the frequency of the \(n\)-th harmonic is given by:
\(f_n = n \frac{v}{2L_o}\)

We are given \(L_o = 2L_c\). Substituting this into the equation for the third harmonic (\(n = 3\)) of the open pipe:
\(f_3 = 3 \frac{v}{2(2L_c)} = 3 \frac{v}{4L_c} = 3f\)

Award [1] for the correct answer B.

The kinetic energy \(E_k\) of a satellite of mass \(m\) in a stable circular orbit of radius \(r\) around a planet of mass \(M\) is given by:
\(E_k = \frac{GMm}{2r}\)

The initial kinetic energy is \(E_{k,i} = \frac{GMm}{2r}\).

The final kinetic energy in the new orbit of radius \(2r\) is:
\(E_{k,f} = \frac{GMm}{2(2r)} = \frac{GMm}{4r}\)

The change in kinetic energy is:
\(\Delta E_k = E_{k,f} - E_{k,i} = \frac{GMm}{4r} - \frac{GMm}{2r} = -\frac{GMm}{4r}\)

Award [1] for the correct answer A.

When the block slides down the incline at constant velocity, the net force is zero. Thus, the component of gravity pulling it down the slope is balanced by friction:
\(mg \sin\theta - f = 0 \implies f = mg \sin\theta\)

When the block is projected up the incline, both the component of gravity down the slope and the frictional force (which now opposes the upward motion) act in the downward direction along the slope:
\(F_{\text{net}} = mg \sin\theta + f = mg \sin\theta + mg \sin\theta = 2mg \sin\theta\)

Using Newton's second law, the magnitude of deceleration \(a\) is:
\(a = \frac{F_{\text{net}}}{m} = 2g \sin\theta\)

Award [1] for the correct answer C.

Let \(t\) be the elapsed time.
The number of remaining nuclei of X is given by:
\(N_X(t) = 8N_0 \left(\frac{1}{2}\right)^{t/T}\)

The number of remaining nuclei of Y is given by:
\(N_Y(t) = N_0 \left(\frac{1}{2}\right)^{t/(2T)}\)

Setting \(N_X(t) = N_Y(t)\):
\(8N_0 \left(\frac{1}{2}\right)^{t/T} = N_0 \left(\frac{1}{2}\right)^{t/(2T)}\)

Dividing both sides by \(N_0\):
\(8 \left(\frac{1}{2}\right)^{t/T} = \left(\frac{1}{2}\right)^{t/(2T)}\)

Let \(x = \left(\frac{1}{2}\right)^{t/(2T)}\). Since \(\left(\frac{1}{2}\right)^{t/T} = x^2\), we can write:
\(8x^2 = x \implies 8x = 1 \implies x = \frac{1}{8}\)

Thus:
\(\left(\frac{1}{2}\right)^{t/(2T)} = \frac{1}{8} = \left(\frac{1}{2}\right)^3\)

Equating exponents:
\(\frac{t}{2T} = 3 \implies t = 6T\)

Award [1] for the correct answer C.

From Einstein's photoelectric equation:
\(E_{\text{max}} = hf - \Phi \implies hf = E_{\text{max}} + \Phi\)

When the incident frequency is doubled to \(2f\), the new maximum kinetic energy \(E'_{\text{max}}\) is:
\(E'_{\text{max}} = h(2f) - \Phi = 2(hf) - \Phi\)

Substitute the expression for \(hf\):
\(E'_{\text{max}} = 2(E_{\text{max}} + \Phi) - \Phi = 2E_{\text{max}} + \Phi\)

Award [1] for the correct answer B.

For an adiabatic process, the relation between temperature and volume is given by:
\(T V^{\gamma-1} = \text{constant}\)

Therefore:
\(T_0 V_0^{\gamma-1} = T (2V_0)^{\gamma-1}\)

Solving for the final temperature \(T\):
\(T = T_0 \left(\frac{V_0}{2V_0}\right)^{\gamma-1} = T_0 \left(\frac{1}{2}\right)^{\gamma-1} = T_0 2^{1-\gamma}\)

Award [1] for the correct answer B.

Let \(\theta = T_{\text{block}} - T\). The rate of heat loss from the block is given by:
\(P = -mc \frac{dT_{\text{block}}}{dt} = -mc \frac{d\theta}{dt}\)

We are given \(P = k\theta\), so:
\(-mc \frac{d\theta}{dt} = k\theta \implies \frac{d\theta}{\theta} = -\frac{k}{mc} dt\)

Integrating from the initial state (where \(T_{\text{block}} = 3T \implies \theta_0 = 2T\) at \(t=0\)) to the final state (where \(T_{\text{block}} = 2T \implies \theta_f = T\) at \(t\)):
\(\int_{2T}^{T} \frac{d\theta}{\theta} = -\frac{k}{mc} \int_{0}^{t} dt\)

\(\ln\left(\frac{T}{2T}\right) = -\frac{k}{mc} t \implies \ln\left(\frac{1}{2}\right) = -\frac{k}{mc} t\)

\(-\ln(2) = -\frac{k}{mc} t \implies t = \frac{mc \ln(2)}{k}\)

Award [1] for the correct answer B.

As the loop is pulled out of the magnetic field, the rate of change of magnetic flux produces an electromotive force (EMF) given by:
\(\varepsilon = B L v\)

The rate of electrical energy dissipation (power) in the loop of resistance \(R\) is:
\(P = \frac{\varepsilon^2}{R} = \frac{B^2 L^2 v^2}{R}\)

The time \(t\) taken for the square loop of side \(L\) to completely leave the field is:
\(t = \frac{L}{v}\)

Therefore, the total electrical energy dissipated is:
\(E = P \cdot t = \frac{B^2 L^2 v^2}{R} \cdot \frac{L}{v} = \frac{B^2 L^3 v}{R}\)

Award [1] for the correct answer B.

For a tube closed at one end and open at the other, only odd harmonics exist. The fundamental frequency is \(f_1 = \frac{v}{4L}\). The third harmonic has a frequency of \(f = 3 f_1 = \frac{3v}{4L}\). Rearranging this formula for length gives \(L = \frac{3v}{4f}\).

Award 1 mark for the correct answer B. Award 0 marks for incorrect options. Method involves identifying that the third harmonic in a closed-open tube corresponds to three quarter-wavelengths fitting in the tube, leading to \(L = \frac{3\lambda}{4} = \frac{3v}{4f}\).

The total mechanical energy of a satellite in a circular orbit of radius \(r\) is given by \(E = -\frac{GMm}{2r}\). The initial energy is \(E_1 = -\frac{GMm}{2R}\). The final energy is \(E_2 = -\frac{GMm}{6R}\). The change in mechanical energy is \(\Delta E = E_2 - E_1 = -\frac{GMm}{6R} - \left(-\frac{GMm}{2R}\right) = \frac{GMm}{2R}\left(1 - \frac{1}{3}\right) = \frac{GMm}{3R}\).

Award 1 mark for the correct option B. Method: Recall the formula for the mechanical energy of an orbiting body, calculate initial and final energy, and find the positive difference (since energy increases as orbit radius increases).

By conservation of momentum: \(m \cdot u + 0 = (m + 2m) \cdot v\), which gives the final velocity \(v = \frac{u}{3}\). The initial kinetic energy is \(E_{ki} = \frac{1}{2} m u^2\). The final kinetic energy is \(E_{kf} = \frac{1}{2} (3m) v^2 = \frac{3}{2} m \left(\frac{u}{3}\right)^2 = \frac{1}{6} m u^2\). The kinetic energy lost is \(\Delta E_k = E_{ki} - E_{kf} = \frac{1}{2} m u^2 - \frac{1}{6} m u^2 = \frac{1}{3} m u^2\). The fraction of initial kinetic energy lost is \(\frac{\Delta E_k}{E_{ki}} = \frac{\frac{1}{3} m u^2}{\frac{1}{2} m u^2} = \frac{2}{3}\).

Award 1 mark for the correct option C. Method: Apply conservation of momentum to find the final velocity, compute initial and final kinetic energies, and calculate the ratio of energy lost to initial energy.

Let \(N_0\) be the initial number of nuclei. After three half-lives, the number of remaining nuclei is \(N = N_0 \left(\frac{1}{2}\right)^3 = \frac{1}{8} N_0\). The number of decayed nuclei is \(N_{\text{decayed}} = N_0 - N = \frac{7}{8} N_0\). The ratio of decayed nuclei to remaining nuclei is \(\frac{N_{\text{decayed}}}{N} = \frac{\frac{7}{8} N_0}{\frac{1}{8} N_0} = 7\).

Award 1 mark for the correct option B. Method: Determine the fraction of remaining nuclei after three half-lives, deduce the fraction of decayed nuclei, and calculate their ratio.

From Einstein's photoelectric equation, we have: \(\frac{hc}{\lambda} = \Phi + E_{\max}\) and \(\frac{hc}{\lambda/2} = 2 \frac{hc}{\lambda} = \Phi + 3 E_{\max}\). Substituting the first equation into the second gives: \(2 (\Phi + E_{\max}) = \Phi + 3 E_{\max}\). Expanding and simplifying: \(2 \Phi + 2 E_{\max} = \Phi + 3 E_{\max} \implies \Phi = E_{\max}\). Therefore, the work function is equal to \(1.0 E_{\max}\).

Award 1 mark for the correct option B. Method: Write down the two photoelectric equations, eliminate the photon energy term, and solve for the work function in terms of maximum kinetic energy.

Since the gas returns to its initial state at the end of the cycle, internal energy, being a state function, has a total change of \(\Delta U = 0\). On a pressure-volume (\(P\)-\(V\)) diagram, the cycle proceeds clockwise: isobaric expansion at high pressure, isochoric cooling to low pressure, and isothermal compression at low temperature back to the initial state. The area enclosed by a clockwise cycle on a \(P\)-\(V\) diagram is positive, meaning the net work done by the gas is positive (\(W_{\text{net}} > 0\)).

Award 1 mark for the correct option A. Method: Identify that change in internal energy is zero for any complete thermodynamic cycle. Recognize that the clockwise cycle path results in positive net work.

According to Wien's displacement law, \(\lambda_{\max} \cdot T = \text{constant}\), so when temperature is doubled, the peak wavelength is halved (\(\frac{\lambda_{\max}}{2}\)). According to the Stefan-Boltzmann law, the total power radiated is proportional to \(T^4\) when the surface area is constant. Since the temperature is doubled, the power increases by a factor of \(2^4 = 16\), so the new power is \(16P\).

Award 1 mark for the correct option C. Method: Apply Wien's displacement law to find the new wavelength, and apply the Stefan-Boltzmann law to calculate the new power.

As the loop is pulled out of the magnetic field, the magnetic flux through it decreases. The magnitude of the induced electromotive force (emf) is given by Faraday's law: \(\varepsilon = \frac{\Delta \Phi}{\Delta t} = B \cdot s \cdot v\). The electrical power dissipated in the loop is given by \(P = \frac{\varepsilon^2}{R} = \frac{(Bsv)^2}{R} = \frac{B^2 s^2 v^2}{R}\).

Award 1 mark for the correct option B. Method: Use Faraday's law of induction to find the induced electromotive force, and then use the electrical power formula \(P = \frac{V^2}{R}\) to find the dissipated power.

\( \textbf{(a)} \) The fundamental frequency of a standing wave on a string fixed at both ends is given by:
\( f = \frac{v}{2L} \)
Since the wave speed is \( v = \sqrt{\frac{T}{\mu}} \), substituting this gives:
\( f = \frac{1}{2L}\sqrt{\frac{T}{\mu}} \implies f^2 = \left( \frac{1}{4L^2\mu} \right) T \)
Since the length \( L \) and mass per unit length \( \mu \) are constant, \( f^2 \) is directly proportional to \( T \). Thus, a graph of \( f^2 \) against \( T \) is a straight line through the origin with a gradient of \( \frac{1}{4L^2\mu} \).

\( \textbf{(b)} \) First, calculate \( f^2 \):
\( f^2 = (150)^2 = 22500 \text{ Hz}^2 \)
Next, calculate the percentage uncertainty in \( f \):
\( \frac{\Delta f}{f} = \frac{3}{150} = 0.02 = 2.0\% \)
The percentage uncertainty in \( f^2 \) is twice the percentage uncertainty in \( f \):
\( \%\Delta (f^2) = 2 \times 2.0\% = 4.0\% \)
Now, calculate the absolute uncertainty in \( f^2 \):
\( \Delta (f^2) = 22500 \times 0.040 = 900 \text{ Hz}^2 \)
Thus, \( f^2 = 22500 \pm 900 \text{ Hz}^2 \) (or \( (2.25 \pm 0.09) \times 10^4 \text{ Hz}^2 \)).

\( \textbf{(c)} \) The gradient \( S \) of the graph is given by \( S = \frac{1}{4L^2\mu} \).
Rearranging for \( \mu \):
\( \mu = \frac{1}{4L^2 S} \)
Substituting the values:
\( \mu = \frac{1}{4 \times (0.600)^2 \times 1125} = \frac{1}{4 \times 0.360 \times 1125} = \frac{1}{1620} \approx 6.17 \times 10^{-4} \text{ kg m}^{-1} \).

\( \textbf{(d)} \) The relation for \( \mu \) is \( \mu = \frac{1}{4L^2 S} \).
Therefore, the fractional uncertainty in \( \mu \) is:
\( \frac{\Delta \mu}{\mu} = 2\frac{\Delta L}{L} + \frac{\Delta S}{S} \)
We are given the percentage uncertainty in \( L \):
\( \frac{\Delta L}{L} = \frac{0.002}{0.600} \approx 0.00333 \) (or \( 0.33\% \))
Therefore:
\( \%\Delta \mu = 2 \times (0.333\%) + 4.0\% = 0.67\% + 4.0\% = 4.67\% \approx 4.7\% \) (or \( 5\% \) to 1 significant figure).

\( \textbf{(a)} \)
- Show derivation leading to \( f^2 = \frac{T}{4L^2\mu} \) [1]
- Explicitly state that \( f^2 \propto T \) with gradient \( \frac{1}{4L^2\mu} \) [1]

\( \textbf{(b)} \)
- Correct calculation of \( f^2 = 22500 \text{ Hz}^2 \) [1]
- Correct determination of percentage uncertainty in \( f^2 \) as \( 4\% \) [1]
- Correct absolute uncertainty of \( \pm 900 \text{ Hz}^2 \) with consistent units [1]

\( \textbf{(c)} \)
- Correct rearrangement of equation for \( \mu \) [1]
- Correct substitution of values [1]
- Final answer \( 6.17 \times 10^{-4} \text{ kg m}^{-1} \) (accept \( 6.2 \times 10^{-4} \text{ kg m}^{-1} \); must include correct unit) [1]

\( \textbf{(d)} \)
- Correct doubling of the percentage uncertainty in \( L \) to yield \( 0.67\% \) [1]
- Summing the uncertainties to find \( 4.7\% \) (accept \( 4.67\% \) or \( 5\% \)) [1]

\( \textbf{(a)} \) The dry air is trapped in a capillary tube of uniform cross-section by a concentrated sulfuric acid index. The capillary tube is placed vertically in a water bath alongside a thermometer and a ruler. The temperature \( \theta \) of the water is measured using the thermometer. The length \( L \) of the air column is measured using the ruler. Since the capillary tube has a uniform cross-sectional area \( A \), the volume \( V \) is directly proportional to the length of the column \( L \) (\( V = A L \)).

\( \textbf{(b)} \) According to Charles's Law, the volume of an ideal gas decreases linearly with temperature and extrapolates to zero at absolute zero. By plotting volume \( V \) against temperature \( \theta \), the straight line of best fit can be extrapolated backwards to the horizontal axis (where \( V = 0 \)). The intercept on this axis yields the value of absolute zero in \( ^\circ\text{C} \).

\( \textbf{(c) (i)} \) Set \( V = 0 \) in the equation of the line of best fit:
\( 0 = 0.150\theta + 41.2 \implies 0.150\theta = -41.2 \implies \theta = -\frac{41.2}{0.150} \approx -274.67^\circ\text{C} \)
To an appropriate number of significant figures, this is \( -275^\circ\text{C} \).

\( \textbf{(c) (ii)} \) Acceptable systematic errors include:
- Thermometer calibration error (constant temperature offset).
- The trapped air was not completely dry (water vapor present in the gas does not behave as an ideal gas and condenses at lower temperatures).
- The capillary tube does not have a perfectly uniform cross-sectional area, meaning the measured column length is not perfectly proportional to the volume.

\( \textbf{(d)} \) The part of the capillary tube outside the water bath is exposed to room temperature, which is lower than the water bath temperature. This portion of the trapped air is cooler than the measured bath temperature, meaning it expands less than expected. Consequently, the total length of the column (and hence the recorded volume) is smaller than it would be if the entire gas column were at the uniform temperature of the water bath. This effect becomes more significant at higher water bath temperatures as the temperature gradient between the bath and room air increases.

\( \textbf{(a)} \)
- Identifies the use of a uniform capillary tube with a liquid plug (e.g., sulfuric acid) to trap a fixed mass of dry air [1]
- Mentions immersing the tube in a water bath and measuring temperature with a thermometer [1]
- Mentions measuring the length of the trapped air column with a scale/ruler to represent volume [1]

\( \textbf{(b)} \)
- Recognizes that the volume of an ideal gas reduces to zero at absolute zero [1]
- States that absolute zero is found by extrapolating the line of best fit to find the horizontal intercept where \( V = 0 \) [1]

\( \textbf{(c) (i)} \)
- Sets \( V = 0 \) in the given equation [1]
- Obtains \( -275^\circ\text{C} \) (accept \( -274.7^\circ\text{C} \)) [1]

\( \textbf{(c) (ii)} \)
- Suggests a valid systematic error (e.g., wet/moist trapped air, thermometer calibration error, thermal expansion of the glass) [1]

\( \textbf{(d)} \)
- Identifies that the unsubmerged air is cooler than the water bath temperature [1]
- Explains that this causes the gas to expand less, leading to an underestimate of the true volume at higher temperatures [1]

(a) A wave from the sound source travels down the tube and undergoes reflection at the closed boundary. The incident and reflected waves, which have the same frequency and amplitude but travel in opposite directions, superpose (interfere). This interference creates stationary points of zero displacement (nodes) and maximum displacement (antinodes), resulting in a standing wave.

(b) For a tube closed at one end, the fundamental mode occurs when the length of the tube represents a quarter of a wavelength: \( L = \frac{\lambda}{4} \). Thus, the fundamental wavelength is \( \lambda = 4L = 4 \times 0.68\text{ m} = 2.72\text{ m} \). The fundamental frequency is then \( f_1 = \frac{v}{\lambda} = \frac{340\text{ m s}^{-1}}{2.72\text{ m}} = 125\text{ Hz} \).

(c) In a tube closed at one end, only odd harmonics are produced. The first overtone corresponds to the third harmonic (\( n = 3 \)). Using the formula for the third harmonic frequency: \( f_3 = 3 \times f_1' = \frac{3 v'}{4 L} \). Substituting the speed of sound in helium: \( f_3 = \frac{3 \times 970\text{ m s}^{-1}}{4 \times 0.68\text{ m}} = \frac{2910}{2.72} \approx 1069.85\text{ Hz} \approx 1070\text{ Hz} \) (or \( 1.1\text{ kHz} \) to two significant figures).

Part (a):
- Award [1] for noting reflection at the closed boundary / end of the tube.
- Award [1] for identifying that incident and reflected waves travel in opposite directions and superpose/interfere.
- Award [1] for describing the resulting pattern of nodes and antinodes.

Part (b):
- Award [1] for correctly stating or calculating \( \lambda = 4 L = 2.72\text{ m} \).
- Award [1] for showing \( f = \frac{340}{2.72} = 125\text{ Hz} \).

Part (c):
- Award [1] for identifying that the first overtone/third harmonic has frequency \( f_3 = 3 f_1 = \frac{3 v}{4 L} \).
- Award [1] for substitution of values: \( \frac{3 \times 970}{4 \times 0.68} \).
- Award [1] for correct final calculation (\( 1070\text{ Hz} \) or \( 1.1\text{ kHz} \)).
- Award [1] for correct unit (\( \text{Hz} \) or \( \text{kHz} \)).

(a) Gravitational field strength at a point is defined as the gravitational force per unit mass experienced by a small test mass placed at that point.

(b) For a circular orbit, the gravitational force acts as the centripetal force: \( \frac{G M m}{r^2} = \frac{m v^2}{r} \). Solving for \( v \) gives \( v = \sqrt{\frac{G M}{r}} \). Substituting the values: \( v = \sqrt{\frac{6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2} \times 3.8 \times 10^{24}\text{ kg}}{9.2 \times 10^6\text{ m}}} = \sqrt{\frac{2.5346 \times 10^{14}}{9.2 \times 10^6}} = \sqrt{2.755 \times 10^7} \approx 5249\text{ m s}^{-1} \approx 5.2 \times 10^3\text{ m s}^{-1} \) (or \( 5.25 \times 10^3\text{ m s}^{-1} \)).

(c) The total mechanical energy of a satellite in orbit of radius \( r \) is \( E = -\frac{G M m}{2r} \).
- Initial total energy: \( E_i = -\frac{6.67 \times 10^{-11} \times 3.8 \times 10^{24} \times 450}{2 \times 9.2 \times 10^6} \approx -6.20 \times 10^9\text{ J} \).
- Final total energy: \( E_f = -\frac{6.67 \times 10^{-11} \times 3.8 \times 10^{24} \times 450}{2 \times 1.5 \times 10^7} \approx -3.80 \times 10^9\text{ J} \).
- Work done: \( W = E_f - E_i = -3.80 \times 10^9\text{ J} - (-6.20 \times 10^9\text{ J}) = 2.40 \times 10^9\text{ J} \) (or \( 2.4 \times 10^9\text{ J} \)).

Part (a):
- Award [1] for stating 'force per unit mass'.
- Award [1] for specifying that this is experienced by a 'small' or 'test' mass placed at that point.

Part (b):
- Award [1] for equating gravitational force to centripetal force: \( \frac{G M m}{r^2} = \frac{m v^2}{r} \).
- Award [1] for rearranging to obtain \( v = \sqrt{\frac{G M}{r}} \).
- Award [1] for correct substitution of values: \( \sqrt{\frac{6.67 \times 10^{-11} \times 3.8 \times 10^{24}}{9.2 \times 10^6}} \).
- Award [1] for final speed value and unit: \( 5.2 \times 10^3\text{ m s}^{-1} \) (accept \( 5200 \text{ to } 5300\text{ m s}^{-1} \)).

Part (c):
- Award [1] for using the total energy formula \( E = -\frac{G M m}{2r} \) (or calculating kinetic and potential energy separately).
- Award [1] for setting up the difference \( \Delta E = E_f - E_i \) with values.
- Award [1] for correct final answer: \( 2.4 \times 10^9\text{ J} \) (accept range \( 2.39 \times 10^9\text{ J} \) to \( 2.41 \times 10^9\text{ J} \)).

(a) The total linear momentum of a closed (isolated) system remains constant, provided no external forces act on the system.

(b) By conservation of momentum:
\( p_{\text{initial}} = p_{\text{final}} \)
\( m_1 v_1 = (m_1 + m_2) v_f \)
\( 1.2 \times 4.5 = (1.2 + 2.8) v_f \)
\( 5.4 = 4.0 v_f \)
\( v_f = 1.35\text{ m s}^{-1} \) (or \( 1.4\text{ m s}^{-1} \) to 2 significant figures).

(c)
- Initial kinetic energy: \( E_{k, i} = \frac{1}{2} m_1 v_1^2 = 0.5 \times 1.2 \times 4.5^2 = 12.15\text{ J} \).
- Final kinetic energy: \( E_{k, f} = \frac{1}{2} (m_1 + m_2) v_f^2 = 0.5 \times 4.0 \times 1.35^2 = 3.645\text{ J} \).
- Ratio: \( \frac{E_{k, f}}{E_{k, i}} = \frac{3.645}{12.15} = 0.30 \) (exactly 30%).
- Discussion: The remaining 70% of the mechanical energy is lost as it is transformed into internal (thermal) energy of the blocks and acoustic (sound) energy during the impact.

Part (a):
- Award [1] for stating that the total momentum remains constant/conserved.
- Award [1] for specifying that this holds for a closed/isolated system or in the absence of external forces.

Part (b):
- Award [1] for writing a correct momentum conservation expression: \( m_1 v_1 = (m_1 + m_2) v_f \).
- Award [1] for substitution: \( 1.2 \times 4.5 = 4.0 v_f \).
- Award [1] for correct final speed: \( 1.35\text{ m s}^{-1} \) (accept \( 1.4\text{ m s}^{-1} \)).

Part (c):
- Award [1] for calculating initial kinetic energy: \( 12.15\text{ J} \) (or \( 12\text{ J} \)).
- Award [1] for calculating final kinetic energy: \( 3.65\text{ J} \) (or \( 3.6\text{ J} \)).
- Award [1] for calculating the ratio \( 0.30 \).
- Award [1] for explaining that the 'lost' mechanical energy is converted to thermal/internal energy and sound energy.

(a) Half-life is the time taken for the activity of a radioactive sample (or the number of unstable nuclei) to decrease to half of its original value.

(b) Using the decay law for activity:
\( A(t) = A_0 e^{-\lambda t} \)
\( 2.5 \times 10^5 = 8.0 \times 10^5 e^{-12 \lambda} \)
\( \frac{2.5}{8.0} = e^{-12 \lambda} \)
\( 0.3125 = e^{-12 \lambda} \)
Taking the natural logarithm of both sides:
\( \ln(0.3125) = -12 \lambda \)
\( -1.16315 = -12 \lambda \)
\( \lambda = \frac{1.16315}{12} \approx 0.0969\text{ hour}^{-1} \approx 0.097\text{ hour}^{-1} \) (or \( 9.7 \times 10^{-2}\text{ hr}^{-1} \)).

(c) First, convert the decay constant to SI units (\text{s}^{-1}):
\( \lambda = \frac{0.096929\text{ hour}^{-1}}{3600\text{ s/hour}} \approx 2.692 \times 10^{-5}\text{ s}^{-1} \).
Using the relationship between activity and number of nuclei:
\( A = \lambda N \)
At \( t = 12\text{ hours} \), the activity is \( A = 2.5 \times 10^5\text{ Bq} \).
\( N = \frac{A}{\lambda} = \frac{2.5 \times 10^5\text{ Bq}}{2.692 \times 10^{-5}\text{ s}^{-1}} \approx 9.287 \times 10^9 \approx 9.3 \times 10^9 \) nuclei.

Part (a):
- Award [2] for a complete definition mentioning both 'time taken' and 'activity/number of nuclei to halve' (or award [1] for a partial/less clear definition).

Part (b):
- Award [1] for recall of the decay law: \( A = A_0 e^{-\lambda t} \).
- Award [1] for correct substitution: \( 2.5 \times 10^5 = 8.0 \times 10^5 e^{-12 \lambda} \).
- Award [1] for taking natural logarithm correctly: \( -1.16 = -12 \lambda \).
- Award [1] for correct final value: \( 0.097\text{ hour}^{-1} \) (accept range \( 0.096 \text{ to } 0.098\text{ hour}^{-1} \)).

Part (c):
- Award [1] for converting \( \lambda \) to \( \text{s}^{-1} \): \( 2.7 \times 10^{-5}\text{ s}^{-1} \) (accept error carried forward from b).
- Award [1] for using \( A = \lambda N \).
- Award [1] for correct calculation of \( N = 9.3 \times 10^9 \) (accept range \( 9.2 \times 10^9 \text{ to } 9.4 \times 10^9 \)).

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

(b) According to Einstein's photoelectric equation:
\( E_{k,\text{max}} = h f - \Phi = \frac{h c}{\lambda} - \Phi \)
- Energy of the incident photons:
\( E_{\text{photon}} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{380 \times 10^{-9}\text{ m}} = 5.234 \times 10^{-19}\text{ J} \).
- Converting work function to Joules:
\( \Phi = 2.14\text{ eV} \times 1.60 \times 10^{-19}\text{ J/eV} = 3.424 \times 10^{-19}\text{ J} \).
- Calculating maximum kinetic energy:
\( E_{k,\text{max}} = 5.234 \times 10^{-19}\text{ J} - 3.424 \times 10^{-19}\text{ J} = 1.81 \times 10^{-19}\text{ J} \).

(c) The stopping voltage \( V_s \) is related to the maximum kinetic energy of the electrons by:
\( e V_s = E_{k,\text{max}} \)
In electron-volts, the maximum kinetic energy is:
\( E_{k,\text{max}} = \frac{1.81 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ J/eV}} = 1.131\text{ eV} \).
Therefore, the stopping voltage is \( V_s = 1.13\text{ V} \) (or \( 1.1\text{ V} \) to 2 significant figures).

Part (a):
- Award [1] for 'minimum energy required'.
- Award [1] for 'to liberate/remove/emit an electron from the surface of a metal'.

Part (b):
- Award [1] for using Einstein's photoelectric equation: \( E_{k,\text{max}} = \frac{h c}{\lambda} - \Phi \).
- Award [1] for calculating photon energy: \( 5.23 \times 10^{-19}\text{ J} \).
- Award [1] for converting work function to Joules: \( 3.42 \times 10^{-19}\text{ J} \).
- Award [1] for correct final subtraction and answer: \( 1.81 \times 10^{-19}\text{ J} \) (accept \( 1.8 \times 10^{-19}\text{ J} \)).

Part (c):
- Award [1] for recalling the relationship \( e V_s = E_{k,\text{max}} \).
- Award [1] for dividing \( E_{k,\text{max}} \) by the elementary charge \( e \).
- Award [1] for correct value and unit: \( 1.13\text{ V} \) (or \( 1.1\text{ V} \)).

(a) For an ideal gas, internal energy is proportional only to its absolute temperature (as there are no intermolecular potential energies). Because the expansion is isothermal, the temperature remains constant, and therefore the change in internal energy (\( \Delta U \)) is zero.

(b) By the first law of thermodynamics: \( Q = \Delta U + W \).
Since \( \Delta U = 0 \) during this isothermal process, we get \( W = Q \).
Given that \( 420\text{ J} \) of heat is added (\( Q = +420\text{ J} \)), the work done by the gas must also be \( 420\text{ J} \).

(c) The work done during an isobaric process is \( W = P \Delta V \).
Here, the volume decreases, so \( \Delta V = V_{\text{final}} - V_{\text{initial}} = 2.0 \times 10^{-3}\text{ m}^3 - 5.0 \times 10^{-3}\text{ m}^3 = -3.0 \times 10^{-3}\text{ m}^3 \).
Thus, the work done by the gas is:
\( W_{\text{by}} = 1.2 \times 10^5\text{ Pa} \times (-3.0 \times 10^{-3}\text{ m}^3) = -360\text{ J} \).
Since work done on the gas is defined as the negative of the work done by the gas:
\( W_{\text{on}} = -W_{\text{by}} = 360\text{ J} \).

Part (a):
- Award [1] for stating that for an ideal gas, internal energy depends only on temperature.
- Award [1] for concluding that constant temperature (isothermal) means change in internal energy is zero.

Part (b):
- Award [1] for stating the first law of thermodynamics: \( Q = \Delta U + W \).
- Award [1] for explaining that \( \Delta U = 0 \) implies \( W = Q \).
- Award [1] for stating the final work done is \( 420\text{ J} \).

Part (c):
- Award [1] for using the isobaric work formula: \( W = P \Delta V \).
- Award [1] for calculating change in volume correctly: \( -3.0 \times 10^{-3}\text{ m}^3 \) (or taking the magnitude as \( 3.0 \times 10^{-3}\text{ m}^3 \)).
- Award [1] for substitution of values: \( 1.2 \times 10^5 \times 3.0 \times 10^{-3} \).
- Award [1] for correct final answer: \( 360\text{ J} \) with the correct sign/direction (positive since work is done *on* the gas).

(a) Albedo is the ratio of reflected radiation power (or intensity) to the total incident radiation power (or intensity) on a surface.

(b) First, calculate the total incident power on the solar panel:
\( P_{\text{incident}} = I \times A = 850\text{ W m}^{-2} \times 2.4\text{ m}^2 = 2040\text{ W} \).
Using the albedo \( \alpha = 0.15 \), the fraction of energy absorbed is \( 1 - \alpha = 1 - 0.15 = 0.85 \).
Thus, the absorbed power is:
\( P_{\text{absorbed}} = 0.85 \times 2040\text{ W} = 1734\text{ W} \approx 1730\text{ W} \) (or \( 1.7\text{ kW} \)).

(c) Assuming all absorbed solar power goes into heating the water:
\( P_{\text{absorbed}} = \frac{\Delta Q}{\Delta t} = \frac{\Delta m}{\Delta t} \times c \times \Delta T \)
where \( \frac{\Delta m}{\Delta t} = 0.035\text{ kg s}^{-1} \) is the mass flow rate of water.
Substitute the known values:
\( 1734 = 0.035 \times 4180 \times \Delta T \)
\( 1734 = 146.3 \times \Delta T \)
\( \Delta T = \frac{1734}{146.3} \approx 11.85\text{ K} \approx 12\text{ K} \) (or \( 12\text{ }^\circ\text{C} \)).

Part (a):
- Award [1] for stating 'ratio of reflected power/intensity to incident power/intensity'.
- Award [1] for clarifying that it applies to radiation incident on a surface.

Part (b):
- Award [1] for calculating incident power: \( 2040\text{ W} \).
- Award [1] for identifying that absorbed fraction is \( 1 - \alpha \) (or 0.85).
- Award [1] for final absorbed power: \( 1734\text{ W} \) (accept \( 1730\text{ W} \) or \( 1.7\text{ kW} \)).

Part (c):
- Award [1] for recalling the heating formula in terms of flow rate: \( P = \frac{\Delta m}{\Delta t} c \Delta T \).
- Award [1] for correct substitution of values: \( 1734 = 0.035 \times 4180 \times \Delta T \) (allow error carried forward from b).
- Award [1] for calculating \( \Delta T \approx 11.9\text{ K} \) or \( 12\text{ K} \).
- Award [1] for correct unit (\( \text{K} \) or \( ^\circ\text{C} \)).

(a)
- The electric field exerts a force on the ions of magnitude \( F_E = qE \).
- The magnetic field exerts a magnetic force of magnitude \( F_B = qvB \) in the opposite direction (since fields are perpendicular).
- For ions to pass through undeflected, the net force must be zero, meaning these two forces must balance: \( qE = qvB \). This simplifies to \( v = \frac{E}{B} \). Only ions with this specific velocity will have equal opposing forces and thus remain undeflected.

(b) Using the balance condition:
\( v = \frac{E}{B} = \frac{1.2 \times 10^5\text{ V m}^{-1}}{0.45\text{ T}} \approx 2.67 \times 10^5\text{ m s}^{-1} \approx 2.7 \times 10^5\text{ m s}^{-1} \).

(c) In the magnetic field, the magnetic force provides the centripetal force:
\( q v B = \frac{m v^2}{r} \)
Solving for the radius \( r \):
\( r = \frac{m v}{q B} \)
Substituting the values:
\( r = \frac{3.3 \times 10^{-27}\text{ kg} \times 2.667 \times 10^5\text{ m s}^{-1}}{1.6 \times 10^{-19}\text{ C} \times 0.45\text{ T}} \)
\( r = \frac{8.8 \times 10^{-22}}{7.2 \times 10^{-20}} \approx 0.0122\text{ m} \approx 1.2\text{ cm} \) (or \( 0.012\text{ m} \)).

Part (a):
- Award [1] for identifying electric force \( F_E = qE \) and magnetic force \( F_B = qvB \).
- Award [1] for stating that the forces must act in opposite directions.
- Award [1] for stating that only when the forces balance (\( qE = qvB \)) does the net force equal zero, which determines a unique undeflected velocity \( v = \frac{E}{B} \).

Part (b):
- Award [1] for substitution: \( v = \frac{1.2 \times 10^5}{0.45} \).
- Award [1] for correct answer with unit: \( 2.7 \times 10^5\text{ m s}^{-1} \) (accept range \( 2.6 \times 10^5 \text{ to } 2.7 \times 10^5\text{ m s}^{-1} \)).

Part (c):
- Award [1] for equating magnetic force to centripetal force: \( q v B = \frac{m v^2}{r} \).
- Award [1] for rearranging to find \( r = \frac{m v}{q B} \).
- Award [1] for correct calculation of radius value: \( 0.012\text{ m} \) (or \( 1.2\text{ cm} \)) (allow error carried forward from b).
- Award [1] for correct units.

(a) The orbital radius of the satellite is \( r = R + h = 6.37 \times 10^6 \text{ m} + 1.20 \times 10^6 \text{ m} = 7.57 \times 10^6 \text{ m} \). Equating the gravitational force to the centripetal force gives: \( \frac{G M m}{r^2} = \frac{m v^2}{r} \). Solving for speed: \( v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{7.57 \times 10^6}} \approx 7253 \text{ m s}^{-1} \approx 7.3 \times 10^3 \text{ m s}^{-1} \).

(b) The total mechanical energy \( E \) is the sum of kinetic and gravitational potential energy: \( E = E_k + E_p = \frac{1}{2}mv^2 - \frac{GMm}{r} = -\frac{GMm}{2r} \). Substituting the values: \( E = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 4.2 \times 10^3}{2 \times 7.57 \times 10^6} = -1.105 \times 10^{11} \text{ J} \approx -1.1 \times 10^{11} \text{ J} \).

(c) From Kepler's Third Law, \( T^2 \propto r^3 \), which can be written as \( \left(\frac{T_2}{T_1}\right)^2 = \left(\frac{r_2}{r_1}\right)^3 \). Since the period is doubled, \( T_2 = 2T_1 \), so \( 2^2 = 4 = \left(\frac{r_2}{r_1}\right)^3 \). Thus, \( r_2 = 4^{1/3} r_1 \approx 1.587 \times 7.57 \times 10^6 \text{ m} \approx 1.20 \times 10^7 \text{ m} \).

(a) [3 marks]
• Calculates the orbital radius \( r = 7.57 \times 10^6 \text{ m} \) [1]
• Equates centripetal force to gravitational force, showing the algebraic step: \( v = \sqrt{\frac{GM}{r}} \) [1]
• Obtains a calculated value of \( 7.25 \times 10^3 \text{ m s}^{-1} \) or \( 7253 \text{ m s}^{-1} \) [1]

(b) [3 marks]
• States formula for total orbital energy: \( E = -\frac{GMm}{2r} \) or correctly calculates \( E_k \) and \( E_p \) separately [1]
• Correct substitution of values [1]
• Correct final value with negative sign and units: \( -1.1 \times 10^{11} \text{ J} \) (accept range \( -1.10 \times 10^{11} \text{ J} \) to \( -1.11 \times 10^{11} \text{ J} \)) [1]

(c) [3 marks]
• Recalls or derives Kepler's Third Law relation \( T^2 \propto r^3 \) [1]
• Sets up correct ratio \( r_2^3 = 4 r_1^3 \) or \( r_2 = 1.59 r_1 \) [1]
• Obtains final radius of \( 1.2 \times 10^7 \text{ m} \) (accept \( 1.20 \times 10^7 \text{ m} \)) [1]

(a) Energy of the incident photon: \( E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{380 \times 10^{-9}} = 5.234 \times 10^{-19} \text{ J} \). Work function of sodium in joules: \( \Phi = 2.28 \times 1.60 \times 10^{-19} \text{ J} = 3.648 \times 10^{-19} \text{ J} \). Using Einstein's photoelectric equation: \( E_{k,\text{max}} = E - \Phi = 5.234 \times 10^{-19} - 3.648 \times 10^{-19} = 1.586 \times 10^{-19} \text{ J} \approx 1.59 \times 10^{-19} \text{ J} \).

(b) (i) The maximum kinetic energy remains unchanged because the energy of individual photons \( E = hf \) depends only on the wavelength, which is kept constant. (ii) The rate of electron emission increases because higher intensity means more photons are incident per second, leading to a greater number of photoelectron emissions per second.

(c) First, calculate the photon energy for \( \lambda = 310 \text{ nm} \): \( E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{310 \times 10^{-9}} = 6.416 \times 10^{-19} \text{ J} = 4.01 \text{ eV} \). Max kinetic energy is: \( E_{k,\text{max}} = E - \Phi = 4.01 \text{ eV} - 2.28 \text{ eV} = 1.73 \text{ eV} \). Since \( e V_s = E_{k,\text{max}} \\, the stopping potential \) V_s = 1.73 \\text{ V} \) (or using joules: \( V_s = \frac{6.416 \times 10^{-19} - 3.648 \times 10^{-19}}{1.60 \times 10^{-19}} = 1.73 \text{ V} \)).

(a) [3 marks]
• Calculates incident photon energy: \( 5.23 \times 10^{-19} \text{ J} \) [1]
• Calculates work function in joules: \( 3.65 \times 10^{-19} \text{ J} \) [1]
• Subtracts values to find max kinetic energy: \( 1.59 \times 10^{-19} \text{ J} \) (accept \( 1.6 \times 10^{-19} \text{ J} \)) [1]

(b) [3 marks]
• (i) States that maximum kinetic energy does not change [1]
• (ii) States that the rate of emission increases [1]
• Explains that higher intensity means more photons per unit time, but each photon has the same energy [1]

(c) [3 marks]
• Calculates photon energy for new wavelength: \( 6.42 \times 10^{-19} \text{ J} \) or \( 4.01 \text{ eV} \) [1]
• Relates stopping potential to kinetic energy: \( e V_s = E_{k,\text{max}} \) [1]
• Calculates stopping potential: \( 1.73 \text{ V} \) (accept \( 1.7 \text{ V} \)) [1]