2023 IB DP Physics 模擬試題連答案詳解

Since the block moves at a constant speed along a straight line, it is in dynamic equilibrium. This means the net force acting on the block is zero. The forces acting on the block are the gravitational force \(mg\) acting vertically downwards, and the contact force exerted by the slope (which is the vector sum of the normal reaction force and the frictional force). For the net force to be zero, the contact force must be equal in magnitude and opposite in direction to the gravitational force. Therefore, the magnitude of the contact force is \(mg\).

1 mark for identifying that the net force is zero and thus the contact force must balance the gravitational force \(mg\) exactly.

According to the Stefan-Boltzmann law, the total power \(P\) emitted by a black-body star of radius \(R\) and temperature \(T\) is given by \(P = \sigma A T^4 = \sigma (4\pi R^2) T^4\). Thus, the power is proportional to \(R^2 T^4\). For star X, we have \(P_X \propto R^2 T^4\). For star Y, we have \(P_Y \propto (2R)^2 (T/2)^4 = 4 R^2 (T^4 / 16) = 0.25 R^2 T^4\). Comparing the two values, the ratio \(P_X / P_Y\) is equal to \(1 / 0.25 = 4\).

1 mark for using the Stefan-Boltzmann law to find that the ratio is 4.

Since the two resistors are connected in parallel, they both experience the same potential difference \(V\). The power dissipated in the first resistor is \(P = V^2 / R\). The power dissipated in the second resistor is \(P_2 = V^2 / (2R) = 0.5 P\). Therefore, the total power dissipated in the circuit is the sum of the individual powers: \(P_{\text{total}} = P + P_2 = P + 0.5 P = 1.5 P\), which can be written as \(\frac{3}{2}P\).

1 mark for correctly determining that the voltage is the same and summing the individual powers to get \(\frac{3}{2}P\).

The energy of a photon emitted during a transition is given by \(E = hc / \lambda\). By conservation of energy, the energy difference for the transition from level 3 to level 1 is the sum of the energy differences from 3 to 2 and from 2 to 1. This gives: \(E_{3 \to 1} = E_{3 \to 2} + E_{2 \to 1}\). Substituting the expressions for photon energy: \(hc / \lambda_3 = hc / \lambda_1 + hc / \lambda_2\). Dividing both sides by \(hc\) gives: \(1 / \lambda_3 = 1 / \lambda_1 + 1 / \lambda_2 = (\lambda_1 + \lambda_2) / (\lambda_1 \lambda_2)\). Inverting this expression yields \(\lambda_3 = (\lambda_1 \lambda_2) / (\lambda_1 + \lambda_2)\).

1 mark for using conservation of energy and the photon energy equation to arrive at the correct fractional relation.

Let the length of the closed pipe be \(L\). For a pipe closed at one end, the fundamental wavelength is \(\lambda_{\text{closed}} = 4L\), and its fundamental frequency is \(f = v / (4L)\). The second pipe is open at both ends and has a length \(L' = 2L\). For a pipe open at both ends, the fundamental wavelength is \(\lambda_{\text{open}} = 2L' = 2(2L) = 4L\). Therefore, the fundamental frequency of the second pipe is \(f' = v / (4L) = f\).

1 mark for establishing the relationship between the fundamental frequencies and lengths to show that the new frequency is equal to the original frequency.

The average force is given by the rate of change of momentum: \(F = \Delta p / \Delta t\). Taking the initial direction of motion as positive, the initial momentum is \(p_i = mv\) and the final momentum is \(p_f = -m(v/2)\) since it rebounds in the opposite direction. The change in momentum is \(\Delta p = p_f - p_i = -0.5mv - mv = -1.5mv\). The magnitude of this change in momentum is \(1.5mv = \frac{3}{2}mv\). Thus, the magnitude of the average force is \(\frac{3mv}{2\Delta t}\).

1 mark for calculating the magnitude of the change in momentum as 1.5mv and dividing by the collision time.

In the core of a main sequence star, lighter hydrogen nuclei fuse to form heavier helium nuclei. The total mass of the resulting helium nucleus is slightly less than the sum of the masses of the initial hydrogen reactants. This difference is known as the mass defect, and it is converted directly into energy according to Einstein's mass-energy equivalence equation \(\Delta E = \Delta m c^2\).

1 mark for identifying that mass decreases because of the mass defect, which is converted to energy.

The electric current in a wire is related to the drift speed by \(I = n A q v\), where \(n\) is the number density of charge carriers, \(A\) is the cross-sectional area, \(q\) is the charge of an electron, and \(v\) is the drift speed. The cross-sectional area of a wire of diameter \(d\) is proportional to \(d^2\). Since the current, number density, and charge are constant, the drift speed is inversely proportional to the square of the diameter: \(v \propto 1/d^2\). When the diameter is doubled, the drift speed becomes \(1/2^2 = 1/4\) of its original value, which is \(\frac{v}{4\)}.

1 mark for using the drift speed formula to deduce that drift speed is inversely proportional to the square of the diameter, leading to a factor of 1/4.

The mass flow rate of the liquid striking the wall is given by \(\frac{\Delta m}{\Delta t} = \rho A v\). The initial velocity is \(u = v\) and the final velocity is \(v' = -0.2v\) (taking the initial direction as positive). The change in velocity is \(\Delta v = v' - u = -0.2v - v = -1.2v\). The average force on the liquid is \(F = \frac{\Delta m}{\Delta t} \Delta v = (\rho A v)(-1.2v) = -1.2 \rho A v^2\). By Newton's third law, the force exerted on the wall is equal in magnitude and opposite in direction, which is \(1.2 \rho A v^2\).

1 mark for the correct answer (C).

Let the acceleration of the system be \(a\) and the tension in the string be \(T\). For the block of mass \(3M\), the equation of motion is: \(3Mg - T = 3Ma\). For the block of mass \(M\), the equation of motion is: \(T - Mg = Ma\). Adding these two equations gives: \(2Mg = 4Ma\), which simplifies to \(a = 0.5g\). Substituting \(a\) back into the equation for the smaller block gives: \(T = M(g + a) = M(g + 0.5g) = 1.5Mg = \frac{3}{2}Mg\).

1 mark for the correct answer (D).

The rate at which energy is delivered to the liquid is \(P\), and the rate of energy loss is \(Q\). Therefore, the net rate of energy gain by the liquid is \(P - Q\). The total net thermal energy supplied to the liquid in time \(t\) is \(E = (P - Q)t\). Using the thermal energy equation \(E = mc\Delta T\), we have \((P - Q)t = mc\Delta T\), which rearranges to \(\Delta T = \frac{(P - Q)t}{mc\)}.

1 mark for the correct answer (A).

The relationship between current and drift velocity is \(I = nAvq\), where \(n\) is the number density of conduction electrons and \(q\) is the elementary charge. Since both wires are made of copper, \(n\) is the same. For the second wire, \(3I = n(2A)v_2q\). Dividing this equation by the first equation gives: \(\frac{3I}{I} = \frac{n(2A)v_2q}{nAvq}\), which simplifies to \(3 = 2\frac{v_2}{v}\). Solving for \(v_2\) gives \(v_2 = 1.5v\).

1 mark for the correct answer (B).

The energy levels of a hydrogen atom are given by \(E_n = -\frac{E_0}{n^2}\). The energy of the photon is equal to the transition energy: \(\Delta E = \frac{hc}{\lambda}\). For the \(3 \to 2\) transition: \(\Delta E_{3\to2} = E_3 - E_2 = -\frac{E_0}{9} - (-\frac{E_0}{4}) = \frac{5}{36}E_0\), so \(\lambda = \frac{36hc}{5E_0}\). For the \(4 \to 2\) transition: \(\Delta E_{4\to2} = E_4 - E_2 = -\frac{E_0}{16} - (-\frac{E_0}{4}) = \frac{3}{16}E_0\), so \(\lambda' = \frac{16hc}{3E_0}\). Taking the ratio gives: \(\frac{\lambda'}{\lambda} = \frac{16/3}{36/5} = \frac{16}{3} \times \frac{5}{36} = \frac{80}{108} = \frac{20}{27}\). Therefore, \(\lambda' = \frac{20}{27}\lambda\).

1 mark for the correct answer (B).

For a pipe of length \(L\) open at one end and closed at the other, the fundamental wavelength is \(\lambda_1 = 4L\), and its fundamental frequency is \(f = \frac{v}{\lambda_1} = \frac{v}{4L}\), where \(v\) is the speed of sound. For a pipe of length \(2L\) open at both ends, the fundamental wavelength is \(\lambda_2 = 2 \times (2L) = 4L\). Its fundamental frequency is therefore \(f_2 = \frac{v}{\lambda_2} = \frac{v}{4L} = f\).

1 mark for the correct answer (B).

According to Wien's displacement law, \(\lambda_{\text{max}} T = \text{constant}\), so the surface temperature of the stars is inversely proportional to their peak wavelength. Thus, \(\frac{T_X}{T_Y} = \frac{\lambda_Y}{\lambda_X} = \frac{800}{400} = 2\). According to the Stefan-Boltzmann law, the luminosity of a star is given by \(L = 4\pi R^2 \sigma T^4\). Comparing the luminosities of X and Y: \(\frac{L_X}{L_Y} = \left(\frac{R_X}{R_Y}\right)^2 \left(\frac{T_X}{T_Y}\right)^4\). Substituting the given values: \(16 = \left(\frac{R_X}{R_Y}\right)^2 (2)^4 = 16 \left(\frac{R_X}{R_Y}\right)^2\). This simplifies to \(\left(\frac{R_X}{R_Y}\right)^2 = 1\), which gives \(\frac{R_X}{R_Y} = 1\).

1 mark for the correct answer (C).

The distance between consecutive nodes in a standing wave is half a wavelength (\(\frac{\lambda}{2}\)). The distance from the first node to the fourth node corresponds to three of these intervals: \(3 \times \frac{\lambda}{2} = 1.5\lambda\). We are given that this distance is \(30\text{ cm} = 0.30\text{ m}\). Therefore, \(1.5\lambda = 0.30\text{ m}\), which gives \(\lambda = 0.20\text{ m}\). Using the wave equation \(v = f\lambda\), we find the frequency: \(f = \frac{v}{\lambda} = \frac{340}{0.20} = 1700\text{ Hz}\).

1 mark for the correct answer (A).

For an elastic collision, the change in momentum is \(\Delta p = mv - (-mv) = 2mv\). The impulse \(J\) delivered to the object is equal to the area under the force-time graph. For a triangular force profile with peak force \(F_0\) and duration \(T\), the impulse is \(J = \frac{1}{2} F_0 T\). Equating the impulse to the change in momentum gives \(\frac{1}{2} F_0 T = 2mv\), which simplifies to \(F_0 = \frac{4mv}{T}\).

Award 1 mark for the correct option C.

For a closed-open tube of length \(L\), the fundamental wavelength is \(\lambda_1 = 4L\), so its fundamental frequency is \(f_1 = \frac{v}{4L}\). For an open-open tube of length \(L' = 2L\), the fundamental wavelength is \(\lambda'_1 = 2L' = 4L\), meaning its fundamental frequency is \(f'_1 = \frac{v}{4L} = f_1\). The harmonics of an open-open tube are integer multiples of its fundamental frequency, so the frequency of the third harmonic is \(f'_3 = 3 f'_1 = 3 f_1\).

Award 1 mark for the correct option B.

The energy emitted during the transition is the difference between the initial and final energy states: \(\Delta E = E_{\text{initial}} - E_{\text{final}} = -E_0 - (-4E_0) = 3E_0\). Since the energy of a photon is given by \(\Delta E = h f\), where \(f\) is the frequency, we have \(3E_0 = h f\). Solving for \(f\) yields \(f = \frac{3E_0}{h}\).

Award 1 mark for the correct option B.

In the first configuration, the total resistance is \(R_1 = (R \parallel R) + R = \frac{R}{2} + R = \frac{3}{2} R\). The power dissipated is \(P_1 = \frac{V^2}{R_1} = \frac{2V^2}{3R}\). In the second configuration, the total resistance is \(R_2 = (R + R) \parallel R = 2R \parallel R = \frac{2R \cdot R}{2R + R} = \frac{2}{3} R\). The power dissipated is \(P_2 = \frac{V^2}{R_2} = \frac{3V^2}{2R}\). The ratio of the powers is \(\frac{P_2}{P_1} = \frac{3/2}{2/3} = \frac{9}{4}\).

Award 1 mark for the correct option D.

According to the Stefan-Boltzmann law, the power radiated by a blackbody is \(P = \sigma A T^4\). For the new state, the surface area becomes \(A' = 0.5A\) and the temperature becomes \(T' = 1.5T\). The new power radiated is \(P' = \sigma A' (T')^4 = \sigma (0.5A) (1.5T)^4 = 0.5 \times (1.5)^4 \times \sigma A T^4 = 0.5 \times \frac{81}{16} P = \frac{81}{32} P\).

Award 1 mark for the correct option A.

Apparent brightness is given by \(b = \frac{L}{4\pi d^2}\). For the first star, \(b_1 = \frac{L}{4\pi d^2}\). For the second star, \(b_2 = \frac{4L}{4\pi (3d)^2} = \frac{4}{9} \frac{L}{4\pi d^2} = \frac{4}{9} b_1\). The ratio of the apparent brightness of the first star to the second star is \(\frac{b_1}{b_2} = \frac{9}{4}\).

Award 1 mark for the correct option C.

In a vacuum, the only force acting is gravity, so the deceleration is \(g\) and the time to reach maximum height is \(t_{\text{vac}} = \frac{u}{g}\). In the medium, both gravity and air resistance act downwards during the upward motion. The total downward force is \(mg + \frac{1}{4}mg = \frac{5}{4}mg\), which results in a deceleration of \(\frac{5}{4}g\). The time to reach maximum height in the medium is \(t_{\text{medium}} = \frac{u}{\frac{5}{4}g} = \frac{4}{5} \frac{u}{g}\). The ratio of the times is \(\frac{t_{\text{medium}}}{t_{\text{vac}}} = \frac{4}{5}\).

Award 1 mark for the correct option A.

The terminal potential difference across the cell is \(V = \varepsilon \frac{R}{R+r}\). As \(R\) increases from a very small value (approaching 0) to a very large value, \(V\) increases monotonically from 0 towards \(\varepsilon\). The power dissipated in the external resistor is \(P = I^2 R = \frac{\varepsilon^2 R}{(R+r)^2}\). According to the maximum power theorem, \(P\) is maximized when the external resistance equals the internal resistance (\(R = r\)). Thus, as \(R\) increases, \(P\) first increases to a maximum and then decreases.

Award 1 mark for the correct option C.

Let \(E_i\) be the initial kinetic energy:

\(E_i = \frac{1}{2}mv^2\)

From the conservation of linear momentum, the final velocity \(v_f\) of the combined system of mass \(3m\) is given by:

\(m v = (m + 2m)v_f \implies v_f = \frac{v}{3}\)

The final kinetic energy \(E_f\) is:

\(E_f = \frac{1}{2}(3m)v_f^2 = \frac{1}{2}(3m)\left(\frac{v}{3}\right)^2 = \frac{1}{6}mv^2\)

The kinetic energy lost is:

\(\Delta E = E_i - E_f = \frac{1}{2}mv^2 - \frac{1}{6}mv^2 = \frac{1}{3}mv^2\)

The fraction of the initial kinetic energy lost is:

\Fraction = \frac{\Delta E}{E_i} = \frac{\frac{1}{3}mv^2}{\frac{1}{2}mv^2} = \frac{2}{3}\)

[1 mark] Award for correct option C.

For a pipe closed at one end and of length \(L\), the fundamental wavelength is \(\lambda = 4L\).

Its fundamental frequency \(f\) is:

\(f = \frac{v}{4L} \implies v = 4Lf\)

For a pipe open at both ends of length \(L' = \frac{L}{2}\), the frequencies of the harmonics are:

\(f_n = n \frac{v}{2L'} = n \frac{v}{2(L/2)} = n \frac{v}{L}\)

The first overtone of an open pipe is the second harmonic (\(n = 2\)).

Therefore, its frequency is:

\(f_2 = 2 \frac{v}{L} = 2 \frac{4Lf}{L} = 8f\)

[1 mark] Award for correct option C.

Let the original wire have length \(L\) and cross-sectional area \(A\). Its resistance is:

\(R = \rho \frac{L}{A}\)

When stretched to twice its original length, its new length is \(L' = 2L\). Since the volume \(V = A L\) is kept constant, its cross-sectional area becomes:

\(A' = \frac{A}{2}\)

The resistance of the entire stretched wire is:

\(R' = \rho \frac{L'}{A'} = \rho \frac{2L}{A/2} = 4\rho \frac{L}{A} = 4R\)

Cutting the wire into two equal halves gives two pieces, each with half the length of the stretched wire, and therefore each having a resistance of:

\(R_{\text{half}} = \frac{4R}{2} = 2R\)

Connecting these two halves in parallel yields an equivalent resistance of:

\(R_{\text{eq}} = \frac{R_{\text{half}}}{2} = \frac{2R}{2} = R\)

[1 mark] Award for correct option C.

The energy of level \(n\) in a hydrogen-like atom is given by \(E_n = -\frac{E_0}{n^2}\).

For the first transition from \(n=4\) to \(n=2\), the energy of the emitted photon is:

\(\Delta E_{4\rightarrow 2} = E_4 - E_2 = E_0\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3}{16}E_0\)

This corresponds to wavelength \(\lambda\):

\(\lambda = \frac{hc}{\Delta E_{4\rightarrow 2}} = \frac{16hc}{3E_0}\)

For the second transition from \(n=4\) to \(n=3\), the energy of the emitted photon is:

\(\Delta E_{4\rightarrow 3} = E_4 - E_3 = E_0\left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7}{144}E_0\)

This corresponds to wavelength \(\lambda'\):

\(\lambda' = \frac{hc}{\Delta E_{4\rightarrow 3}} = \frac{144hc}{7E_0}\)

Finding the ratio \(\frac{\lambda'}{\lambda}\):

\(\frac{\lambda'}{\lambda} = \frac{144/7}{16/3} = \frac{144}{7} \times \frac{3}{16} = \frac{9 \times 3}{7} = \frac{27}{7}\)

Therefore, \(\lambda' = \frac{27}{7}\lambda\).

[1 mark] Award for correct option C.

By Wien's displacement law, the temperature \(T\) of a star is inversely proportional to its peak wavelength \(\lambda\):

\(\frac{T_X}{T_Y} = \frac{\lambda_Y}{\lambda_X} = \frac{800}{400} = 2\)

By Stefan-Boltzmann law, the luminosity of a spherical star is given by:

\(L = 4\pi R^2 \sigma T^4 \implies L \propto R^2 T^4\)

Taking the ratio of luminosities for X and Y:

\(\frac{L_X}{L_Y} = \left(\frac{R_X}{R_Y}\right)^2 \left(\frac{T_X}{T_Y}\right)^4\)

Substitute the given values into the ratio:

\(64 = \left(\frac{R_X}{R_Y}\right)^2 (2)^4\)

\(64 = \left(\frac{R_X}{R_Y}\right)^2 \times 16\)

\(\left(\frac{R_X}{R_Y}\right)^2 = \frac{64}{16} = 4 \implies \frac{R_X}{R_Y} = 2\)

[1 mark] Award for correct option B.

The lifetime \(\tau\) on the main sequence is given by:

\(\tau \propto \frac{\text{Mass of Fuel}}{L}\)

We are given that the mass of fuel is proportional to the total mass of the star, \(M\):

\(\tau \propto \frac{M}{L}\)

Substituting the luminosity relationship \(L \propto M^3\) into the equation:

\(\tau \propto \frac{M}{M^3} = \frac{1}{M^2} = M^{-2}\)

[1 mark] Award for correct option C.

(a) Using conservation of linear momentum:
\(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\)
\((2.0 \times 6.0) + 0 = (2.0 \times 1.0) + (4.0 \times v_2)\)
\(12.0 = 2.0 + 4.0 v_2\)
\(4.0 v_2 = 10.0 \implies v_2 = 2.5\text{ m s}^{-1}\)

(b) Initial kinetic energy:
\(E_{k,i} = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} (2.0) (6.0)^2 = 36\text{ J}\)

Final kinetic energy:
\(E_{k,f} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} (2.0) (1.0)^2 + \frac{1}{2} (4.0) (2.5)^2\)
\(E_{k,f} = 1.0 + 2.0 (6.25) = 1.0 + 12.5 = 13.5\text{ J}\)

Loss in kinetic energy:
\(\Delta E_k = E_{k,i} - E_{k,f} = 36 - 13.5 = 22.5\text{ J}\)
Since \(\Delta E_k > 0\), kinetic energy is lost, proving the collision is inelastic.

(c) The average force is equal to the rate of change of momentum of the second block:
\(F_{\text{avg}} = \frac{\Delta p_2}{\Delta t} = \frac{m_2 v_2 - 0}{\Delta t}\)
\(F_{\text{avg}} = \frac{4.0 \times 2.5}{0.12} = \frac{10.0}{0.12} \approx 83.3\text{ N}\)

(d) Momentum is not conserved. For momentum to be conserved, the net external force on the system must be zero. Friction represents a non-zero net external force acting on the blocks, which decreases the total momentum of the system over time.

(a)
- [1 mark] for utilizing momentum conservation formula.
- [1 mark] for correct substitution of values.
- [1 mark] for final answer of \(2.5\text{ m s}^{-1}\).

(b)
- [1 mark] for calculating initial kinetic energy (\(36\text{ J}\)).
- [1 mark] for calculating final kinetic energy (\(13.5\text{ J}\)).
- [1 mark] for finding the difference (\(22.5\text{ J}\)) and explicitly stating that since KE is not conserved, the collision is inelastic.

(c)
- [1 mark] for setting up the impulse-momentum equation (\(F \Delta t = \Delta p\)).
- [1 mark] for calculating the force as \(83\text{ N}\) (accept \(83.3\text{ N}\)).

(d)
- [1 mark] for stating "No/Not conserved".
- [1 mark] for explaining that friction acts as an external force (or violating the closed/isolated system condition).

(a) In a \(1.5 M_{\odot}\) star, the primary fusion process is the proton-proton chain (or the CNO cycle, which dominates in stars slightly more massive than the Sun).
Gravity pulls the star's mass inward, creating immense pressure and temperature in the core. This thermal energy provides the protons with high enough kinetic energy to overcome their mutual electrostatic repulsion (Coulomb barrier) and undergo nuclear fusion.

(b) The Jeans criterion states that for an interstellar gas cloud to collapse under gravity, its gravitational potential energy must exceed the total random kinetic energy of its particles.
This occurs when the total mass \(M\) of the cloud exceeds a threshold called the Jeans mass (\(M > M_J\)), which typically happens in clouds that are very cold (low kinetic energy) and dense (stronger gravitational forces).

(c) When hydrogen in the core is exhausted, the core contracts and heats up while the outer layers expand and cool, turning the star into a Red Giant.
Helium fusion begins in the core once the temperature is high enough.
After helium is depleted, the core cannot reach the temperatures required to fuse carbon. The outer layers are ejected into space as a planetary nebula.
The remaining carbon-oxygen core contracts under gravity to form a White Dwarf, which is prevented from further collapse by electron degeneracy pressure.

(a)
- [1 mark] for identifying either "proton-proton (p-p) chain" or "CNO cycle".
- [1 mark] for linking gravity to high core density/temperature/pressure.
- [1 mark] for explaining that this extreme environment allows protons to overcome electrostatic repulsion.

(b)
- [1 mark] for stating that gravitational potential energy must be greater than thermal kinetic energy.
- [1 mark] for referencing the Jeans mass (\(M > M_J\)).
- [1 mark] for identifying that low temperature and/or high density promote collapse.

(c)
- [1 mark] for mentioning the expansion into a red giant after hydrogen exhaustion.
- [1 mark] for mentioning core helium fusion.
- [1 mark] for describing the ejection of outer layers as a planetary nebula.
- [1 mark] for identifying the final remnant as a white dwarf supported by electron degeneracy pressure.

(a) Incident power:
\(P_{\text{inc}} = I \times A = 800\text{ W m}^{-2} \times 2.5\text{ m}^2 = 2000\text{ W}\)

Useful absorbed power:
\(P_{\text{useful}} = \eta \times P_{\text{inc}} = 0.65 \times 2000 = 1300\text{ W}\)

(b) Using the thermal energy flow equation:
\(P_{\text{useful}} = \frac{Q}{\Delta t} = \left(\frac{\Delta m}{\Delta t}\right) c \Delta T\)

Where \(\frac{\Delta m}{\Delta t}\) is the mass flow rate:
\(\frac{\Delta m}{\Delta t} = \frac{P_{\text{useful}}}{c \Delta T} = \frac{1300}{4180 \times 15}\)
\(\frac{\Delta m}{\Delta t} = \frac{1300}{62700} \approx 0.0207\text{ kg s}^{-1}\) (or \(0.021\text{ kg s}^{-1}\))

(c) Convert temperatures to Kelvin:
\(T_{\text{panel}} = 12 + 273.15 = 285.15\text{ K}\)
\(T_{\text{surr}} = 2 + 273.15 = 275.15\text{ K}\)

Using the net Stefan-Boltzmann radiation equation:
\(P_{\text{net}} = e \sigma A (T_{\text{panel}}^4 - T_{\text{surr}}^4)\)
\(P_{\text{net}} = 0.90 \times (5.67 \times 10^{-8}\text{ W m}^{-2}\text{ K}^{-4}) \times 2.5 \times (285.15^4 - 275.15^4)\)
\(P_{\text{net}} = 1.27575 \times 10^{-7} \times (6.6115 \times 10^9 - 5.7319 \times 10^9)\)
\(P_{\text{net}} = 1.27575 \times 10^{-7} \times (8.796 \times 10^8) \approx 112.2\text{ W}\) (Accept \(110\text{ W}\) to \(112\text{ W}\))

(d) Two other factors:
1. Convection (wind currents carrying thermal energy away from the surface).
2. Evaporation (of condensation or rain water resting on the panel surface).

(a)
- [1 mark] for calculating incident power (\(2000\text{ W}\)).
- [1 mark] for applying efficiency to find useful power (\(1300\text{ W}\)).

(b)
- [1 mark] for referencing \(Q = mc\Delta T\) or \(P = \frac{\Delta m}{\Delta t} c \Delta T\).
- [1 mark] for correct substitution of values.
- [1 mark] for final answer of \(0.021\text{ kg s}^{-1}\) (or \(0.0207\text{ kg s}^{-1}\)).

(c)
- [1 mark] for converting temperatures to Kelvin (\(285\text{ K}\) and \(275\text{ K}\)).
- [1 mark] for correct formula substitution using Stefan-Boltzmann constant.
- [1 mark] for final answer of \(110\text{ W}\) to \(112\text{ W}\).

(d)
- [1 mark] for naming convection.
- [1 mark] for naming evaporation (or heat loss to mounting brackets/wind currents).

(a) The diagram must show:
- The cell and variable resistor \(R\) connected in a single loop (series path).
- An ammeter connected in series anywhere in that main loop.
- A voltmeter connected in parallel across either the poles of the cell or across the variable resistor \(R\).

(b) The terminal potential difference is given by \(V = \varepsilon - Ir\). As the current \(I\) increases, the potential difference across the internal resistance (the 'lost volts', \(Ir\)) increases. Since the emf \(\varepsilon\) remains constant, the remaining potential difference available to the external circuit (\(V\)) must decrease.

(c) From the linear equation \(V = -rI + \varepsilon\):
- The emf \(\varepsilon\) is equal to the y-intercept (the value of \(V\) when \(I = 0\)).
- The internal resistance \(r\) is equal to the negative gradient (or the magnitude of the slope) of the line.

(d)
(i) Using Ohm's law for the complete circuit:
\(\varepsilon = I(R + r)\)
\(6.0 = 0.60(8.0 + r)\)
\(10.0 = 8.0 + r \implies r = 2.0\text{ }\Omega\)

(ii) Power dissipated in the internal resistance:
\(P = I^2 r = (0.60)^2 \times 2.0 = 0.36 \times 2.0 = 0.72\text{ W}\)

(a)
- [1 mark] for ammeter placed correctly in series.
- [1 mark] for voltmeter placed correctly in parallel across the cell or variable resistor.

(b)
- [1 mark] for identifying the equation \(V = \varepsilon - Ir\) or referencing 'lost volts' (\(Ir\)).
- [1 mark] for explaining that as current increases, more potential is dropped inside the cell, reducing terminal PD.

(c)
- [1 mark] for identifying emf as the y-intercept.
- [1 mark] for identifying internal resistance as the magnitude of the gradient.

(d)(i)
- [1 mark] for setting up \(\varepsilon = I(R + r)\) or equivalent.
- [1 mark] for final answer of \(2.0\text{ }\Omega\).

(d)(ii)
- [1 mark] for using \(P = I^2 r\).
- [1 mark] for final answer of \(0.72\text{ W}\) (allow ECF from d(i)).

(a)
- Energy transfer: A travelling wave propagates energy through the medium, whereas a standing wave stores energy (there is no net transfer of energy along the wave).
- Phase: In a travelling wave, all neighboring particles have different phases (phase changes continuously with distance). In a standing wave, all particles between two adjacent nodes vibrate in phase with each other (and are \(180^\circ\) or \(\pi\text{ rad}\) out of phase with particles in adjacent loops).

(b) The sketch should show:
- A displacement node (zero amplitude) at the closed end (left).
- A displacement antinode (maximum amplitude) at the open end (right).
- A smooth curve starting at zero at the closed end and opening up to maximum width at the open end (a quarter of a sine wave pattern).
- Clear labels: "Node" or "N" at the closed end, "Antinode" or "A" at the open end.

(c) For a tube closed at one end, the fundamental wavelength \(\lambda_1\) is:
\(\lambda_1 = 4L = 4 \times 0.85\text{ m} = 3.4\text{ m}\)

Using the wave equation \(v = f \lambda\):
\(f_1 = \frac{v}{\lambda_1} = \frac{340}{3.4} = 100\text{ Hz}\)

(d) For a tuning fork of frequency \(f = 300\text{ Hz}\), the wavelength of the sound wave produced is:
\(\lambda = \frac{v}{f} = \frac{340}{300} \approx 1.133\text{ m}\)

Resonance in a closed-open tube occurs when the length of the tube corresponds to a boundary condition with a node at one end and an antinode at the other.
The shortest such length is the fundamental resonance:
\(L_{\text{min}} = \frac{\lambda}{4} = \frac{1.133}{4} \approx 0.283\text{ m}\) (or \(0.28\text{ m}\))

(a)
- [1 mark] for stating standing waves store energy/no net transfer while travelling waves transfer energy.
- [1 mark] for describing travelling wave phase (varies continuously with distance).
- [1 mark] for describing standing wave phase (all particles between adjacent nodes are in phase).

(b)
- [1 mark] for correct shape (quarter-wavelength profile, zero at closed end, max at open end).
- [1 mark] for correct labels of node (closed end) and antinode (open end).

(c)
- [1 mark] for identifying \(\lambda = 4L = 3.4\text{ m}\).
- [1 mark] for calculating \(f = 100\text{ Hz}\).

(d)
- [1 mark] for calculating wavelength \(\lambda = 1.13\text{ m}\).
- [1 mark] for identifying that the shortest resonant length is \(L = \frac{\lambda}{4}\).
- [1 mark] for finding the final length of \(0.28\text{ m}\) (accept \(0.283\text{ m}\)).

(a) Independent variable: frequency \(f\) (or \(1/f\)). Dependent variable: length of the air column \(L\). (b) Taking multiple measurements reduces the impact of random errors through graphing/averaging, and helps identify systematic errors (such as the end correction) from the intercept. (c)(i) Rearranging \(L + c = \frac{v}{4f}\) gives \(L = \frac{v}{4}\left(\frac{1}{f}\right) - c\). Comparing with \(y = mx + c\), where \(y = L\) and \(x = \frac{1}{f}\), the gradient \(m = \frac{v}{4}\). (c)(ii) Gradient is \(\frac{\Delta L}{\Delta (1/f)}\). The unit of \(L\) is meters (\(\text{m}\)) and the unit of \(1/f\) is seconds (\(\text{s}\)), so the unit of the gradient is \(\text{m s}^{-1}\). (d)(i) \(v = 4 \times \text{gradient} = 4 \times 82.5 = 330\text{ m s}^{-1}\). (d)(ii) Absolute uncertainty in \(v\) is \(\Delta v = 4 \times \Delta (\text{gradient}) = 4 \times 2.5 = 10\text{ m s}^{-1}\). (e) The y-intercept is \(-1.4 \pm 0.3\text{ cm}\). Since the value of 0 is outside the uncertainty range of the intercept (\([-1.7, -1.1]\text{ cm}\)), the end correction is statistically significant, confirming a non-zero systematic error.

(a) [1] Award [1] for correctly identifying both variables: independent = frequency \(f\) (or \(1/f\)) AND dependent = length \(L\). (b) [1] Award [1] for stating that a graph reduces random error or allows determination of systematic error/intercept. (c)(i) [1] Award [1] for rearranging the formula to \(L = \frac{v}{4}\left(\frac{1}{f}\right) - c\) and comparing to the equation of a straight line \(y = mx + c\). (c)(ii) [1] Award [1] for \(\text{m s}^{-1}\). (d)(i) [1] Award [1] for \(330\text{ m s}^{-1}\). (d)(ii) [2] Award [1] for realizing that absolute uncertainty is multiplied by 4: \(\Delta v = 4 \times \Delta m\). Award [1] for correct final value of \(10\text{ m s}^{-1}\). (e) [1] Award [1] for stating that it is significant because \(0\) is outside the uncertainty range (or equivalent reasoning).

(a) Stir the water continuously or immediately before taking each temperature reading to prevent thermal gradients. (b)(i) \(\theta = T - T_s = 50.0 - 20.0 = 30.0\ ^\circ\text{C}\) (or \(30.0\text{ K}\)). (b)(ii) For addition/subtraction, absolute uncertainties are added: \(\Delta \theta = \Delta T + \Delta T_s = 0.1 + 0.5 = 0.6\ ^\circ\text{C}\) (or \(0.6\text{ K}\)). (b)(iii) Percentage uncertainty = \(\frac{0.6}{30.0} \times 100\% = 2.0\%\). (c)(i) Temperature change \(\Delta T = T_1 - T_2 = 51.5 - 48.5 = 3.0\ ^\circ\text{C}\). Cooling rate \(R = \frac{\Delta T}{\Delta t} = \frac{3.0}{60.0} = 0.050\ ^\circ\text{C s}^{-1}\). (c)(ii) Absolute uncertainty in \(\Delta T\) is \(\Delta (\Delta T) = 0.1 + 0.1 = 0.2\ ^\circ\text{C}\). Fractional uncertainty in \(\Delta T\) is \(\frac{0.2}{3.0} \approx 0.0667\). Fractional uncertainty in \(\Delta t\) is \(\frac{0.5}{60.0} \approx 0.0083\). Fractional uncertainty in \(R\) is the sum of these fractional uncertainties: \(\frac{\Delta R}{R} = \frac{0.2}{3.0} + \frac{0.5}{60.0} \approx 0.075\) (or \(7.5\%\)).

(a) [1] Award [1] for mentioning stirring the water. (b)(i) [1] Award [1] for \(30.0\ ^\circ\text{C}\) (accept \(30\) or with unit \(\text{K}\)). (b)(ii) [1] Award [1] for adding absolute uncertainties to get \(0.6\ ^\circ\text{C}\) (or \(0.6\text{ K}\)). (b)(iii) [1] Award [1] for \(2.0\%\). (c)(i) [1] Award [1] for \(0.050\ ^\circ\text{C s}^{-1}\) (or \(\text{K s}^{-1}\)). (c)(ii) [2] Award [1] for finding absolute uncertainty of \(\Delta T\) as \(0.2\ ^\circ\text{C}\) and adding fractional/percentage uncertainties: \(\frac{0.2}{3.0} + \frac{0.5}{60.0}\). Award [1] for correct final fractional uncertainty of \(0.075\) (or \(7.5\%\)).

a) Distance in parsecs: \( d = \frac{1}{p} = \frac{1}{0.25} = 4.0 \text{ pc} \).
Distance in meters: \( d = 4.0 \times 3.09 \times 10^{16} \text{ m} = 1.24 \times 10^{17} \text{ m} \).

b) Luminosity: \( L = b \cdot 4\pi d^2 \).
\( L = (2.5 \times 10^{-11}) \times 4\pi \times (1.236 \times 10^{17})^2 = 4.8 \times 10^{24} \text{ W} \).

a)
* \( d = 4.0 \text{ pc} \) [1]
* \( d = 1.2 \times 10^{17} \text{ m} \) [1] (Accept \( 1.24 \times 10^{17} \text{ m} \))

b)
* Recall or use formula \( L = 4\pi d^2 b \) [1]
* Correct substitution of values [1]
* Correct final value: \( L = 4.8 \times 10^{24} \text{ W} \) (Accept range \( 4.7 \times 10^{24} \) to \( 4.8 \times 10^{24} \text{ W} \)) [1]

a) Cepheid variables have a known relationship between their period of pulsation and their average absolute luminosity. By measuring their period, their luminosity is known. Comparing this to their measured apparent brightness allows their distance to be calculated.

b) Using the inverse square law for brightness: \( b = \frac{L}{4\pi d^2} \Rightarrow d = \sqrt{\frac{L}{4\pi b}} \).
\( d = \sqrt{\frac{3.5 \times 10^{30}}{4\pi \times 1.2 \times 10^{-16}}} = 4.82 \times 10^{22} \text{ m} \).
Convert to parsecs: \( d = \frac{4.82 \times 10^{22}}{3.09 \times 10^{16}} = 1.56 \times 10^6 \text{ pc} = 1.56 \text{ Mpc} \).

a)
* There is a known period-luminosity relationship for Cepheids (allowing luminosity to be found from period) [1]
* Comparing known luminosity with measured apparent brightness yields distance (acting as a distance indicator / standard candle) [1]

b)
* Correct substitution into inverse square law to find \( d = 4.8 \times 10^{22} \text{ m} \) [1]
* Division by \( 3.09 \times 10^{16} \text{ m} \) to convert to parsecs (yields \( 1.56 \times 10^6 \text{ pc} \)) [1]
* Correct conversion to Mpc: \( 1.6 \text{ Mpc} \) (Accept \( 1.5 \text{ Mpc} \) to \( 1.6 \text{ Mpc} \)) [1]

a) From Wien's displacement law, \( \lambda_{\max} T = \text{constant} \), so \( \lambda_{\max} \propto \frac{1}{T} \).
\( \frac{\lambda_{\text{giant}}}{\lambda_{\text{MS}}} = \frac{T_{\text{MS}}}{T_{\text{giant}}} = \frac{7000}{3500} = 2.0 \).

b) From Stefan-Boltzmann law, \( L = 4\pi R^2 \sigma T^4 \), so \( R \propto \frac{\sqrt{L}}{T^2} \).
\( \frac{R_{\text{giant}}}{R_{\text{MS}}} = \sqrt{\frac{L_{\text{giant}}}{L_{\text{MS}}}} \times \left(\frac{T_{\text{MS}}}{T_{\text{giant}}}\right)^2 = \sqrt{800} \times 2^2 \approx 28.28 \times 4 = 113.1 \approx 110 \).

a)
* State Wien's displacement law / inverse relationship: \( \lambda_{\max} \propto \frac{1}{T} \) [1]
* Obtain ratio of \( 2.0 \) [1]

b)
* State relation \( L \propto R^2 T^4 \) [1]
* Express radius ratio in terms of luminosity and temperature: \( \frac{R_{\text{giant}}}{R_{\text{MS}}} = \sqrt{800} \times \left(\frac{7000}{3500}\right)^2 \) [1]
* Obtain correct final ratio: \( 110 \) (Accept \( 113 \)) [1]

a) Redshift: \( z = \frac{\lambda_{\text{observed}} - \lambda_{\text{rest}}}{\lambda_{\text{rest}}} \).
\( z = \frac{787.6 - 656.3}{656.3} = \frac{131.3}{656.3} = 0.200 \).

b) The relationship between scale factor and redshift is: \( \frac{R_0}{R} = 1 + z \).
Given \( R_0 = 1 \), \( R = \frac{1}{1 + z} = \frac{1}{1 + 0.20} = 0.833 \approx 0.83 \).

a)
* State redshift formula: \( z = \frac{\Delta\lambda}{\lambda_0} \) [1]
* Calculate redshift: \( z = 0.20 \) (Accept \( 0.200 \)) [1]

b)
* State scale factor relationship: \( \frac{R}{R_0} = \frac{1}{1+z} \) [1]
* Substitute values: \( R = \frac{1}{1.20} \) [1]
* Calculate scale factor: \( 0.83 \) (Accept \( 0.833 \)) [1]