2025 IB DP Physics 模擬試題連答案詳解

By conservation of linear momentum, \(mv = (m + 3m)v_f\), which gives the final velocity of the combined mass as \(v_f = \frac{v}{4}\). The initial kinetic energy of the system is \(E_i = \frac{1}{2}mv^2\). The final kinetic energy is \(E_f = \frac{1}{2}(4m)(v_f)^2 = \frac{1}{2}(4m)\left(\frac{v}{4}\right)^2 = \frac{1}{8}mv^2\). Therefore, the ratio of the final kinetic energy to the initial kinetic energy is \(\frac{E_f}{E_i} = \frac{\frac{1}{8}mv^2}{\frac{1}{2}mv^2} = 0.25\).

Award 1 mark for correct answer A.

Using the equation of state for an ideal gas, we have \(\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}\). For the initial state, the values are \(p\), \(V\), and \(T\). For the final state, the volume is \(2V\) and the temperature is \(T\). Substituting these into the formula: \(\frac{pV}{T} = \frac{p_f(2V)}{T}\). Solving for the final pressure gives \(p_f = 0.50p\).

Award 1 mark for correct answer B.

The total mechanical energy of a satellite in a circular orbit of radius \(r\) is given by \(E = -\frac{GMm}{2r}\). The initial total energy is \(E_1 = -\frac{GMm}{2r}\). The final total energy at orbital radius \(2r\) is \(E_2 = -\frac{GMm}{2(2r)} = -\frac{GMm}{4r}\). The change in the total mechanical energy is \(\Delta E = E_2 - E_1 = -\frac{GMm}{4r} - \left(-\frac{GMm}{2r}\right) = +\frac{GMm}{4r}\).

Award 1 mark for correct answer B.

Power \(P\) is the rate of work done, so the total work done on the car in time \(t\) is \(W = P \cdot t\). Assuming no energy loss, this work is converted entirely into the kinetic energy of the car. Since the car starts from rest, \(W = \Delta E_k = \frac{1}{2}mv^2\). Thus, \(Pt = \frac{1}{2}mv^2\). Rearranging for speed \(v\) yields \(v = \sqrt{\frac{2Pt}{m}}\).

Award 1 mark for correct answer C.

For a single slit, the condition for the first diffraction minimum is given by \(\sin\theta = \frac{\lambda}{b}\). Under the new conditions, the wavelength becomes \(\lambda' = 2\lambda\) and the slit width becomes \(b' = \frac{b}{2}\). Therefore, the sine of the new angle is \(\sin\theta' = \frac{\lambda'}{b'} = \frac{2\lambda}{b/2} = 4\frac{\lambda}{b} = 4\sin\theta\).

Award 1 mark for correct answer D.

Let the charge \(+2q\) be at the origin \(x = 0\) and the charge \(-q\) be at \(x = d\). For a point between the charges at a distance \(x\) from the \(+2q\) charge (where \(0 < x < d\)), the net potential \(V\) is the sum of potentials due to both charges: \(V = \frac{k(2q)}{x} + \frac{k(-q)}{d - x}\). Setting \(V = 0\) gives \(\frac{2}{x} = \frac{1}{d - x}\). Cross-multiplying yields \(2d - 2x = x\), which simplifies to \(3x = 2d\) or \(x = \frac{2}{3}d\).

Award 1 mark for correct answer B.

Einstein's photoelectric equation is given by \(E_{\text{max}} = hf - \Phi\), which means \(hf = E_{\text{max}} + \Phi\). When the frequency is doubled, the new maximum kinetic energy \(E_{\text{max}}'\) is: \(E_{\text{max}}' = h(2f) - \Phi = 2(hf) - \Phi\). Substituting \(hf\) into this equation yields \(E_{\text{max}}' = 2(E_{\text{max}} + \Phi) - \Phi = 2E_{\text{max}} + \Phi\).

Award 1 mark for correct answer C.

The Stefan-Boltzmann law states that stellar luminosity is given by \(L = 4\pi R^2 \sigma T^4\). Since both stars have the same surface temperature \(T\), their luminosity ratio depends only on their radii: \(\frac{L_X}{L_Y} = \left(\frac{R_X}{R_Y}\right)^2\). Given \(\frac{L_X}{L_Y} = 16\), we have \(\left(\frac{R_X}{R_Y}\right)^2 = 16\), which simplifies to \(\frac{R_X}{R_Y} = 4\).

Award 1 mark for correct answer B.

The ideal gas law is given by \(PV = nRT\). Initially, the pressure is \(P = \frac{nRT}{V}\). In the final state, the volume \(V\) remains constant because the container is rigid. Since half of the gas molecules escape, the number of moles becomes \(n_2 = 0.5n\). The temperature becomes \(T_2 = 1.5T\). Substituting these values into the ideal gas equation gives the new pressure: \(P_2 = \frac{n_2 R T_2}{V} = \frac{(0.5n) R (1.5T)}{V} = 0.75 \frac{nRT}{V} = 0.75P\).

Award [1] for the correct answer of B.

The change in momentum of the block is equal to the impulse of the force, which is represented by the area under the force-time graph. The area under the graph from \(t = 0\) to \(t = 5.0\text{ s}\) can be split into a triangle and a rectangle: \(\text{Area} = \text{Area}_{\text{triangle}} + \text{Area}_{\text{rectangle}} = \frac{1}{2} \times 3.0\text{ s} \times 12\text{ N} + (5.0\text{ s} - 3.0\text{ s}) \times 12\text{ N} = 18\text{ N s} + 24\text{ N s} = 42\text{ N s}\). Since the block is initially at rest, the final momentum is \(p = m v = 42\text{ N s}\). Solving for speed: \(v = \frac{42\text{ N s}}{2.0\text{ kg}} = 21\text{ m s}^{-1}\).

Award [1] for the correct answer of C.

The rate of thermal energy transfer is the power \(P = \frac{Q}{\Delta t}\). In the solid phase, \(Q = m c_s \Delta \theta\), which gives \(P = m c_s \frac{\Delta \theta}{\Delta t} = m c_s S_1\), so \(c_s = \frac{P}{m S_1}\). Similarly, in the liquid phase, \(P = m c_l S_2\), which gives \(c_l = \frac{P}{m S_2}\). Taking the ratio: \(\frac{c_s}{c_l} = \frac{P / (m S_1)}{P / (m S_2)} = \frac{S_2}{S_1}\).

Award [1] for the correct answer of A.

The total mechanical energy of a satellite of mass \(m\) in a circular orbit of radius \(r\) is \(E = -\frac{GMm}{2r}\). The initial total energy is \(E_1 = -\frac{GMm}{2r}\). The final total energy is \(E_2 = -\frac{GMm}{2(2r)} = -\frac{GMm}{4r}\). The change in the total energy is \(\Delta E = E_2 - E_1 = -\frac{GMm}{4r} - \left(-\frac{GMm}{2r}\right) = +\frac{GMm}{4r}\).

Award [1] for the correct answer of C.

According to Einstein's photoelectric equation: \(hf = E_k + \Phi\) where \(hf\) is the energy of the incident photon. When the frequency is doubled, the equation becomes: \(h(2f) = 3E_k + \Phi\). Multiplying the first equation by 2 gives: \(2hf = 2E_k + 2\Phi\). Equating the expression for \(2hf\) from both equations: \(2E_k + 2\Phi = 3E_k + \Phi\). Solving for \(\Phi\) gives: \(\Phi = E_k\).

Award [1] for the correct answer of B.

Initial moment of inertia is \(I_i = \frac{1}{2}MR^2\) and initial angular momentum is \(L_i = I_i \omega = \frac{1}{2}MR^2 \omega\). Initial kinetic energy is \(E_i = \frac{1}{2} I_i \omega^2 = \frac{1}{4}MR^2 \omega^2\). When the clay sticks at the edge, the final moment of inertia is \(I_f = I_i + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2\). Since no external torque acts on the system, angular momentum is conserved: \(I_i \omega = I_f \omega_f \implies \frac{1}{2}MR^2 \omega = \frac{3}{2}MR^2 \omega_f \implies \omega_f = \frac{1}{3}\omega\). The final kinetic energy is \(E_f = \frac{1}{2} I_f \omega_f^2 = \frac{1}{2} \left(\frac{3}{2}MR^2\right) \left(\frac{1}{3}\omega\right)^2 = \frac{1}{12}MR^2 \omega^2\). The ratio of final to initial kinetic energy is \(\frac{E_f}{E_i} = \frac{\frac{1}{12}MR^2 \omega^2}{\frac{1}{4}MR^2 \omega^2} = \frac{1}{3}\).

Award [1] for the correct answer of B.

Let the current through the series resistor be \(I\). The power dissipated in this series resistor is \(P = I^2 R\). Because the other two resistors are identical and connected in parallel, the total current \(I\) splits equally between them, so the current through each parallel resistor is \(\frac{I}{2}\). The power dissipated in each of the parallel resistors is \(P_{\text{parallel}} = \left(\frac{I}{2}\right)^2 R = \frac{I^2 R}{4} = \frac{P}{4}\). The total power dissipated in the circuit is the sum of the power in all three resistors: \(P_{\text{total}} = P + \frac{P}{4} + \frac{P}{4} = 1.5 P\). Alternatively, the equivalent resistance of the circuit is \(R_{\text{eq}} = R + \frac{R}{2} = 1.5 R\), so the total power is \(I^2 R_{\text{eq}} = 1.5 I^2 R = 1.5 P\).

Award [1] for the correct answer of A.

First, find the mass of water lifted per second: \(\frac{\Delta m}{\Delta t} = \text{density} \times \frac{\text{volume}}{\text{time}} = 1000\text{ kg m}^{-3} \times \frac{0.20\text{ m}^3}{60\text{ s}} = \frac{10}{3}\text{ kg s}^{-1} \approx 3.33\text{ kg s}^{-1}\). The useful power output needed to lift this water is: \(P_{\text{out}} = \frac{\Delta m}{\Delta t} g h = \frac{10}{3}\text{ kg s}^{-1} \times 9.8\text{ m s}^{-2} \times 12\text{ m} = 392\text{ W}\). Since the efficiency of the pump is \(\eta = 0.80\), the input electrical power is: \(P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{392\text{ W}}{0.80} = 490\text{ W}\).

Award [1] for the correct answer of C.

The ideal gas equation is \(PV = N k_B T\), which can be written as \(P = \frac{N k_B T}{V}\). Initially, the pressure is \(P_1 = \frac{N_1 k_B T_1}{V_1} = P\). When half of the molecules escape, the new number of molecules is \(N_2 = 0.5 N_1\). The new temperature is \(T_2 = 1.5 T_1\), while the volume remains constant (\(V_2 = V_1\)). Substituting these into the formula for the new pressure: \(P_2 = \frac{(0.5 N_1) k_B (1.5 T_1)}{V_1} = 0.75 \frac{N_1 k_B T_1}{V_1} = 0.75 P\).

[1 mark] For identifying that pressure is proportional to both temperature and number of molecules. [1 mark] For calculating the new pressure as 0.75 times the initial pressure.

Conservation of momentum gives \(p_{\text{initial}} = m v + (3m)(-\frac{v}{3}) = 0\). Since momentum is conserved, the final velocity must be zero, and the final kinetic energy is zero. The initial kinetic energy is \(E_{\text{initial}} = \frac{1}{2} m v^2 + \frac{1}{2} (3m) (-\frac{v}{3})^2 = \frac{1}{2} m v^2 + \frac{1}{6} m v^2 = \frac{2}{3} m v^2\). Therefore, the entire initial kinetic energy is lost, which is \(\frac{2}{3} m v^2\).

Award 1 mark for calculating the correct initial kinetic energy and identifying that the final velocity is zero. Award 1 mark for selecting option B.

The root-mean-square speed of the gas molecules is proportional to the square root of the absolute temperature, \(v_{\text{rms}} \propto \sqrt{T}\). Thus, doubling \(T\) increases \(v_{\text{rms}}\) by a factor of \(\sqrt{2}\). According to the ideal gas law \(P V = n R T\), at constant volume, pressure is directly proportional to temperature, \(P \propto T\). Thus, doubling \(T\) doubles the pressure \(P\).

Award 1 mark for identifying the correct relationships for both rms speed and pressure. Award 1 mark for selecting option B.

The mass flow rate is \(\frac{\Delta m}{\Delta t} = \rho A v\). The total energy gained by the water per second consists of potential energy and kinetic energy: \(P = \frac{\Delta E_{\text{p}}}{\Delta t} + \frac{\Delta E_{\text{k}}}{\Delta t} = \left(\frac{\Delta m}{\Delta t}\right) g h + \frac{1}{2} \left(\frac{\Delta m}{\Delta t}\right) v^2 = \rho A v g h + \frac{1}{2} \rho A v^3 = \frac{1}{2} \rho A v (2gh + v^2)\).

Award 1 mark for establishing the mass flow rate and combining potential and kinetic energies. Award 1 mark for selecting option B.

For the sphere to be in equilibrium, the upward electric force must balance the downward gravitational force: \(F_{\text{electric}} = F_{\text{gravity}} \implies q E = m g\). Since the electric field between parallel plates is \(E = \frac{V}{d}\), we have \(q \frac{V}{d} = m g\). Solving for \(V\) yields \(V = \frac{m g d}{q}\).

Award 1 mark for equating the gravitational force and electric force. Award 1 mark for selecting option C.

The kinetic energy of the electron is \(E_{\text{k}} = e V\). Since \(E_{\text{k}} = \frac{p^2}{2m}\), the momentum is \(p = \sqrt{2 m e V}\). The de Broglie wavelength is \(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m e V}}\), meaning \(\lambda \propto \frac{1}{\sqrt{V}}\). When the potential is multiplied by 4, the wavelength decreases by a factor of \(\sqrt{4} = 2\), resulting in \(\frac{\lambda}{2}\).

Award 1 mark for showing that the de Broglie wavelength is inversely proportional to the square root of potential difference. Award 1 mark for selecting option B.

For a circular orbit, gravity provides the centripetal force: \(\frac{G M m}{R^2} = \frac{m v^2}{R} \implies v = \sqrt{\frac{G M}{R}}\). The escape speed from a distance \(R\) is defined as \(v_{\text{esc}} = \sqrt{\frac{2 G M}{R}}\). Comparing these expressions gives \(v_{\text{esc}} = \sqrt{2} v\).

Award 1 mark for setting up the equations for orbital speed and escape speed. Award 1 mark for selecting option B.

Thermal efficiency is defined as \(\eta = \frac{W}{Q_{\text{in}}} = \frac{40}{120} = \frac{1}{3} \approx 33\%\). By conservation of energy, the heat rejected to the cold reservoir is \(Q_{\text{out}} = Q_{\text{in}} - W = 120 - 40 = 80\text{ J}\).

Award 1 mark for calculating the correct efficiency and heat output. Award 1 mark for selecting option A.

The general equation for simple harmonic motion is \(a = -\omega^2 x\). Comparing this with the given equation, we find \(\omega^2 = 16\pi^2\), which gives \(\omega = 4\pi\text{ rad s}^{-1}\). Since angular frequency \(\omega = 2\pi f\), we have \(2\pi f = 4\pi\), which yields \(f = 2\text{ Hz}\).

Award 1 mark for determining the angular frequency \(\omega\). Award 1 mark for selecting option A.

**(a)(i)**
The independent variable is the temperature of the water bath \( \theta \) (or temperature of the gas). [1]

**(a)(ii)**
The dependent variable is the pressure of the gas \( P \). [1]

**(b)(i)**
At absolute zero of temperature, the pressure of an ideal gas is theoretically zero (\( P = 0 \)).
\( 0 = a\theta_0 + b \implies \theta_0 = -\frac{b}{a} \)
\( \theta_0 = -\frac{98.4}{0.362} = -271.82^\circ\text{C} \)
Rounding to 3 significant figures: \( \theta_0 \approx -272^\circ\text{C} \). [2]

**(b)(ii)**
Let \( Z = \frac{b}{a} \). The fractional uncertainty of \( Z \) is:
\( \frac{\Delta Z}{Z} = \frac{\Delta b}{b} + \frac{\Delta a}{a} \)
\( \frac{\Delta Z}{Z} = \frac{1.5}{98.4} + \frac{0.012}{0.362} \approx 0.01524 + 0.03315 = 0.04839 \)
The absolute uncertainty is:
\( \Delta Z = Z \times 0.04839 = 271.82 \times 0.04839 \approx 13.15^\circ\text{C} \)
Rounding to 2 significant figures gives \( \Delta \theta_0 = 13^\circ\text{C} \).
Thus, the experimental absolute zero is \( -272 \pm 13^\circ\text{C} \). [3]

**(c)**
*Option 1: Dead space error.*
- A portion of the gas (in the capillary tube connecting the flask to the sensor) is outside the water bath and remains near room temperature.
- At high temperatures, some of the heated gas expands into this cooler "dead space", meaning the pressure in the flask does not rise as much as it would if all the gas were heated.
- This decreases the slope \( a \) of the \( P\text{--}\theta \) graph, which leads to a more negative (colder) value of absolute zero (\( \theta_0 = -b/a \) becomes larger in magnitude). [3]

*Option 2: Flask thermal expansion.*
- As the temperature increases, the glass flask expands slightly, increasing the volume of the gas.
- An increased volume at higher temperatures leads to lower pressures than expected for constant volume.
- This decreases the slope \( a \), leading to a more negative absolute zero value. [3]

**(a)**
(i) Temperature (\( \theta \)) [1] (Accept "temperature of the water bath")
(ii) Pressure (\( P \)) [1] (Accept "gas pressure")

**(b)(i)**
Sets \( P = 0 \) to get \( \theta_0 = -b/a \) [1]
\( \theta_0 = -272^\circ\text{C} \) (accept negative sign implicit or explicit) [1]

**(b)(ii)**
Calculation of fractional uncertainties: \( \frac{1.5}{98.4} \approx 0.015 \) and \( \frac{0.012}{0.362} \approx 0.033 \) [1]
Sum of fractional uncertainties = \( 0.048 \) [1]
Absolute uncertainty \( \approx 271.82 \times 0.0484 = 13^\circ\text{C} \) (accept \( 13^\circ\text{C} \) or \( 13.1^\circ\text{C} \) or \( 10^\circ\text{C} \)) [1]
*(Award full marks for a correct propagation using min-max method)*

**(c)**
Identifies a valid systematic error (such as "dead space" in the connecting tube or "thermal expansion of the glass flask") [1]
Explains how this error affects the pressure measurements at different temperatures [1]
Deduces the effect on the gradient \( a \) (decreases it) and consequently makes the calculated absolute zero value more negative [1]

**(a)**
A resonance is heard because a standing wave is formed in the air column, which has a displacement node at the water surface (closed end) and an antinode near the open top. This reinforces the sound wave amplitude at its fundamental frequency. [1]
The relationship is:
\( L + e = \frac{\lambda}{4} \) [1]

**(b)(i)**
Comparing the given equation \( L = \left(\frac{v}{4}\right)\frac{1}{f} - e \) to the equation of a straight line \( y = mx + c \):
- \( y = L \)
- \( x = \frac{1}{f} \)
- \( m = \frac{v}{4} \) (gradient)
- \( c = -e \) (vertical intercept)
Therefore, the gradient of the graph of \( L \) against \( \frac{1}{f} \) is equal to \( \frac{v}{4} \). [2]

**(b)(ii)**
Since \( \text{gradient} = \frac{v}{4} \):
\( v = 4 \times \text{gradient} = 4 \times 81.5 = 326\text{ m s}^{-1} \) [1]
The absolute uncertainty in \( v \) is:
\( \Delta v = 4 \times \Delta(\text{gradient}) = 4 \times 2.5 = 10\text{ m s}^{-1} \) [1]
Thus, \( v = 326 \pm 10\text{ m s}^{-1} \). [1]

**(b)(iii)**
From the comparison with \( y = mx + c \), the vertical intercept is \( c = -e \).
Given \( c = -0.015 \pm 0.005\text{ m} \), we have:
\( -e = -0.015 \pm 0.005\text{ m} \implies e = 0.015 \pm 0.005\text{ m} \) (or \( 1.5 \pm 0.5\text{ cm} \)). [2]

**(c)**
Any one of:
- Strike the tuning fork against a soft rubber pad (not a hard surface) to avoid high-frequency overtones.
- Position the tuning fork as close to the open end of the tube as possible without touching it.
- Raise and lower the tube slowly to find the exact point of maximum loudness.
- Use a vertical ruler parallel to the tube and align the line of sight with the water meniscus to avoid parallax error. [1]

**(a)**
Standing wave is formed (with node at water surface and antinode at open end) causing maximum loudness [1]
\( L + e = \frac{\lambda}{4} \) [1]

**(b)(i)**
Identifies \( L \) as the vertical coordinate (y-axis) and \( \frac{1}{f} \) as the horizontal coordinate (x-axis) [1]
Shows that the coefficient of \( \frac{1}{f} \) is \( \frac{v}{4} \), which corresponds to the slope \( m \) in \( y = mx + c \) [1]

**(b)(ii)**
Calculates \( v = 326\text{ m s}^{-1} \) [1]
Calculates absolute uncertainty \( \Delta v = 4 \times 2.5 = 10\text{ m s}^{-1} \) [1]
Correct final presentation of speed with uncertainty: \( 326 \pm 10\text{ m s}^{-1} \) (or \( 330 \pm 10\text{ m s}^{-1} \)) [1]

**(b)(iii)**
Identifies \( e = -c \) [1]
States \( e = 0.015 \pm 0.005\text{ m} \) (or \( 1.5 \pm 0.5\text{ cm} \)) [1]

**(c)**
Any valid experimental precaution, e.g.:
- Bring the fork close to the tube but do not let it touch.
- Move the tube slowly up and down to isolate the peak loudness.
- Avoid parallax error when measuring the column length. [1]

(a) Using the conservation of linear momentum with the initial direction of block A as positive:
\(m_A v_0 = m_A v_A' + m_B v_B'\)
\((2.0 \times 6.0) = (2.0 \times -1.0) + (4.0 \times v_B')\)
\(12.0 = -2.0 + 4.0 v_B'\)
\(14.0 = 4.0 v_B'\)
\(v_B' = 3.5\text{ m s}^{-1}\)

(b) The average force is given by the rate of change of momentum of block B:
\(F_{\text{avg}} = \frac{\Delta p}{\Delta t} = \frac{m_B v_B' - 0}{\Delta t} = \frac{4.0 \times 3.5}{0.15} = \frac{14.0}{0.15} \approx 93.3\text{ N}\)

(c) The frictional force acting on B is \(F_f = \mu m_B g\).
This causes a deceleration of \(a = \mu g = 0.25 \times 9.81\text{ m s}^{-2} = 2.45\text{ m s}^{-2}\).
Using kinematics to find the stopping distance \(s\):
\(v^2 = u^2 + 2as \implies 0 = (3.5)^2 - 2(2.45)s\)
\(12.25 = 4.9 s \implies s = 2.5\text{ m}\)

(d) During the sliding phase after the collision, an external friction force from the table acts on both blocks. This force transfers momentum from the blocks to the table and the Earth. Thus, the momentum of the blocks-table-Earth system is conserved, but the momentum of the blocks alone is not conserved because they are acted on by an external net force.

(a)
- [1 mark] for statement of conservation of momentum
- [1 mark] for correct substitution, noting the negative sign for A's velocity
- [1 mark] for final answer of \(3.5\text{ m s}^{-1}\)

(b)
- [1 mark] for formula for impulse/force: \(F = \Delta p / \Delta t\)
- [1 mark] for calculating change in momentum \(\Delta p = 14\text{ N s}\)
- [1 mark] for final answer of \(93.3\text{ N}\)

(c)
- [1 mark] for calculating deceleration \(a = 2.45\text{ m s}^{-2}\) (accept \(2.5\text{ m s}^{-2}\) if using \(g = 10\))
- [1 mark] for using kinematics formula \(v^2 = u^2 + 2as\)
- [1.25 marks] for final answer of \(2.5\text{ m}\) (accept \(2.45\text{ m}\) for \(g = 10\))

(d)
- [1 mark] for identifying friction as an external force acting on the blocks
- [1 mark] for concluding that momentum is transferred to the table/Earth, conserving the overall system's momentum but not the blocks' alone

(a) For an isobaric process, \(\frac{V_A}{T_A} = \frac{V_B}{T_B}\).
Therefore, \(T_B = T_A \frac{V_B}{V_A} = 300 \times \frac{4.0 \times 10^{-3}}{2.0 \times 10^{-3}} = 600\text{ K}\).

(b) Work done during isobaric expansion:
\(W_{AB} = P \Delta V = (5.0 \times 10^5)\text{ Pa} \times (4.0 \times 10^{-3} - 2.0 \times 10^{-3})\text{ m}^3\)
\(W_{AB} = 5.0 \times 10^5 \times 2.0 \times 10^{-3} = 1000\text{ J}\) (or \(1.0\text{ kJ}\)).

(c) For the isothermal process \(C \to A\), the temperature is held constant at \(T_A = 300\text{ K}\).
The volume changes from \(V_C = V_B = 4.0 \times 10^{-3}\text{ m}^3\) to \(V_A = 2.0 \times 10^{-3}\text{ m}^3\).
\(W_{CA} = n R T \ln\left(\frac{V_A}{V_C}\right) = 0.40 \times 8.31 \times 300 \times \ln\left(\frac{2.0 \times 10^{-3}}{4.0 \times 10^{-3}}\right)\)
\(W_{CA} = 997.2 \times \ln(0.5) \approx -691.2\text{ J}\).
Since \(W\) is negative, the work done ON the gas is \(691.2\text{ J} \approx 6.9 \times 10^2\text{ J}\).

(d) For a complete thermodynamic cycle, the net change in internal energy is \(\Delta U_{\text{net}} = 0\).
The net work done by the gas is \(W_{\text{net}} = W_{AB} + W_{BC} + W_{CA}\).
Since \(B \to C\) is isochoric, \(W_{BC} = 0\).
\(W_{\text{net}} = 1000 - 691.2 = 308.8\text{ J}\).
Using the First Law of Thermodynamics, \(Q = \Delta U + W_{\text{net}}\):
\(Q = 0 + 308.8 = 308.8\text{ J} \approx 3.1 \times 10^2\text{ J}\).

(a)
- [1 mark] for recognizing \(V \propto T\) for constant pressure
- [1 mark] for correct substitution and final evaluation showing \(600\text{ K}\)

(b)
- [1 mark] for using \(W = P \Delta V\)
- [1 mark] for correct values substitution
- [1 mark] for \(1000\text{ J}\) (or \(1.0\text{ kJ}\))

(c)
- [1 mark] for identifying \(T_C = T_A = 300\text{ K}\) and \(V_C = 4.0 \times 10^{-3}\text{ m}^3\)
- [1 mark] for correct substitution into the given formula
- [1.25 marks] for final answer of \(6.9 \times 10^2\text{ J}\) or \(691\text{ J}\)

(d)
- [1 mark] for stating \(\Delta U_{\text{net}} = 0\)
- [1 mark] for summing the work terms correctly (recognizing \(W_{BC} = 0\))
- [1 mark] for final answer of \(3.1 \times 10^2\text{ J}\) (or \(309\text{ J}\))

(a) Using the proportional relation:
\(t_{\text{star}} = t_{\odot} \times \left(\frac{M_{\text{star}}}{M_{\odot}}\right)^{-2.5} = 1.0 \times 10^{10} \times (3.0)^{-2.5}\)
\(t_{\text{star}} = 1.0 \times 10^{10} \times 0.06415 \approx 6.4 \times 10^8\text{ years}\).

(b) First, compute the mass of the reactants:
\(m_{\text{initial}} = 4 \times m_p = 4 \times (1.67262 \times 10^{-27}\text{ kg}) = 6.69048 \times 10^{-27}\text{ kg}\).

Next, compute the mass of the products:
\(m_{\text{final}} = m_{\alpha} + 2 m_e = 6.64466 \times 10^{-27}\text{ kg} + 2 \times (9.10938 \times 10^{-31}\text{ kg})\)
\(m_{\text{final}} = 6.64466 \times 10^{-27}\text{ kg} + 0.00182188 \times 10^{-27}\text{ kg} = 6.64648 \times 10^{-27}\text{ kg}\).

The mass defect is:
\(\Delta m = m_{\text{initial}} - m_{\text{final}} = 6.69048 \times 10^{-27} - 6.64648 \times 10^{-27} = 4.40 \times 10^{-29}\text{ kg}\).

The energy released is:
\(E = \Delta m c^2 = 4.40 \times 10^{-29} \times (3.00 \times 10^8)^2 = 3.96 \times 10^{-12}\text{ J}\).

Convert this energy to eV and then MeV:
\(E = \frac{3.96 \times 10^{-12}\text{ J}}{1.60 \times 10^{-19}\text{ J eV}^{-1}} = 2.475 \times 10^7\text{ eV} \approx 24.8\text{ MeV}\).

(c) A main-sequence star is in a state of hydrostatic equilibrium. Gravity pulls the matter of the star inward, creating an inward pressure that tends to collapse the star. On the other hand, the energy released from nuclear fusion reactions in the core produces an outward radiation pressure (and thermal gas pressure). When these two pressures are balanced, the star remains stable and maintains a constant size.

(a)
- [1 mark] for recognizing the proportionality equation
- [1.25 marks] for the correct calculation and unit of \(6.4 \times 10^8\text{ years}\)

(b)
- [1 mark] for calculating the total initial mass
- [1 mark] for calculating the total final mass (including both positrons)
- [1 mark] for calculating the mass defect \(\Delta m = 4.40 \times 10^{-29}\text{ kg}\)
- [1 mark] for calculating energy in Joules using \(E = \Delta m c^2\)
- [1 mark] for correctly converting to MeV to obtain \(24.8\text{ MeV}\)

(c)
- [1 mark] for mentioning gravitational pressure acts inwards
- [1 mark] for mentioning radiation pressure acts outwards
- [1 mark] for stating that the radiation pressure is caused by nuclear fusion in the core
- [1 mark] for explaining that hydrostatic equilibrium is reached when these forces/pressures balance

(a) The energy of a photon is given by:
\(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{240 \times 10^{-9}\text{ m}} = 8.2875 \times 10^{-19}\text{ J}\).

Convert Joules to electronvolts:
\(E = \frac{8.2875 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ J eV}^{-1}} \approx 5.18\text{ eV}\).

(b) Using Einstein's photoelectric equation:
\(E_{k,\text{max}} = hf - \Phi = 5.18\text{ eV} - 2.28\text{ eV} = 2.90\text{ eV}\).

(c) The stopping potential \(V_s\) is related to the maximum kinetic energy by:
\(E_{k,\text{max}} = e V_s \implies 2.90\text{ eV} = e V_s \implies V_s = 2.90\text{ V}\).

(d) The maximum kinetic energy in Joules is:
\(E_k = 2.90 \times 1.60 \times 10^{-19}\text{ J} = 4.64 \times 10^{-19}\text{ J}\).

The momentum \(p\) of the photoelectrons is:
\(p = \sqrt{2 m_e E_k} = \sqrt{2 \times (9.11 \times 10^{-31}\text{ kg}) \times (4.64 \times 10^{-19}\text{ J})} = \sqrt{8.454 \times 10^{-49}} \approx 9.20 \times 10^{-25}\text{ kg m s}^{-1}\).

The de Broglie wavelength is:
\(\lambda_{\text{dB}} = \frac{h}{p} = \frac{6.63 \times 10^{-34}\text{ J s}}{9.20 \times 10^{-25}\text{ kg m s}^{-1}} \approx 7.21 \times 10^{-10}\text{ m}\) (or \(0.721\text{ nm}\)).

(a)
- [1 mark] for the photon energy formula \(E = hc/\lambda\)
- [1 mark] for correct substitution in SI units
- [1 mark] for converting correctly to \(5.18\text{ eV}\)

(b)
- [1 mark] for using \(E_{k,\text{max}} = hf - \Phi\)
- [1 mark] for correct substitution to yield \(2.90\text{ eV}\)

(c)
- [1 mark] for relating \(E_k\) to \(e V_s\)
- [1 mark] for answer \(2.90\text{ V}\) (must have unit V)

(d)
- [1 mark] for converting \(E_k\) back to Joules
- [1 mark] for using \(p = \sqrt{2mE}\) or equivalent method
- [1 mark] for using \(\lambda = h/p\)
- [1.25 marks] for final correct calculation giving \(7.21 \times 10^{-10}\text{ m}\)

(a) For a stable circular orbit, the centripetal force is provided by the gravitational attraction:
\(\frac{G M m}{R_0^2} = \frac{m v^2}{R_0} \implies v = \sqrt{\frac{G M}{R_0}}\)

Substitute values:
\(v = \sqrt{\frac{6.67 \times 10^{-11} \times 4.80 \times 10^{24}}{8.00 \times 10^6}} = \sqrt{\frac{3.2016 \times 10^{14}}{8.00 \times 10^6}} = \sqrt{4.002 \times 10^7} \approx 6.326 \times 10^3\text{ m s}^{-1} \approx 6.3\text{ km s}^{-1}\).

(b) The total energy of a spacecraft in a circular orbit is given by:
\(E = -\frac{G M m}{2 R_0}\)
\(E = -\frac{6.67 \times 10^{-11} \times 4.80 \times 10^{24} \times 1.50 \times 10^3}{2 \times 8.00 \times 10^6} = -\frac{4.8024 \times 10^{17}}{1.60 \times 10^7} = -3.0015 \times 10^{10}\text{ J} \approx -3.00 \times 10^{10}\text{ J}\).

(c) To escape the planet's gravitational pull, the total energy of the spacecraft must be at least zero (\(E_{\text{escape}} = 0\)).
The minimum additional energy \(\Delta E\) required is:
\(\Delta E = E_{\text{escape}} - E = 0 - (-3.00 \times 10^{10}\text{ J}) = 3.00 \times 10^{10}\text{ J}\).

(d) From Kepler's third law, the square of the orbital period is directly proportional to the cube of the orbital radius (\(T^2 \propto r^3\)). As the radius \(r\) increases, the orbital period \(T\) must also increase.

(a)
- [1 mark] for equating centripetal force to gravitational force
- [1 mark] for rearranging to find \(v = \sqrt{\frac{G M}{R_0}}\)
- [1 mark] for showing intermediate calculation and final result of \(6.3\text{ km s}^{-1}\)

(b)
- [1 mark] for the total energy formula \(E = -\frac{G M m}{2 R_0}\) (or by summing \(E_k + E_p\))
- [1 mark] for correct substitution
- [1.25 marks] for the final correct value \(-3.00 \times 10^{10}\text{ J}\) (the negative sign must be present)

(c)
- [1 mark] for stating that the total energy at infinity (escape) is \(0\)
- [1 mark] for stating \(\Delta E = E_{\text{final}} - E_{\text{initial}}\)
- [1 mark] for final calculation of \(3.00 \times 10^{10}\text{ J}\)

(d)
- [1 mark] for referencing Kepler's Third Law (or deriving \(T^2 \propto r^3\))
- [1 mark] for concluding that the period increases

(a) The slit spacing \(d\) is:
\(d = \frac{1}{N} = \frac{1}{5.00 \times 10^5\text{ lines m}^{-1}} = 2.00 \times 10^{-6}\text{ m}\).

Using the grating equation \(d \sin\theta = n \lambda\) for \(n = 2\):
\(2.00 \times 10^{-6} \sin\theta = 2 \times 633 \times 10^{-9}\)
\[\sin\theta = \frac{1.266 \times 10^{-6}}{2.00 \times 10^{-6}} = 0.633\\
\theta = \arcsin(0.633) \approx 39.3^{\circ}\]

(b) For the maximum observable order, the angle \(\theta\) cannot exceed \(90^{\circ}\), so \(\sin\theta \le 1\):
\(n \le \frac{d}{\lambda} = \frac{2.00 \times 10^{-6}}{633 \times 10^{-9}} \approx 3.16\).

Since \(n\) must be an integer, the highest observable order is \(n = 3\).

(c) For a double slit, the fringe spacing \(s\) is given by:
\(s = \frac{\lambda D}{d}\)
\(s = \frac{633 \times 10^{-9} \times 2.00}{0.250 \times 10^{-3}} = 5.064 \times 10^{-3}\text{ m} \approx 5.06\text{ mm}\).

(d) Decreasing the slit width increases the angle of diffraction for each single slit. Since the double-slit interference pattern is modulated by the single-slit diffraction envelope, the central maximum of the envelope becomes broader. Consequently, more interference fringes are seen with greater brightness further from the center, while the fringe spacing remains unchanged.

(a)
- [1 mark] for calculating \(d = 2.00 \times 10^{-6}\text{ m}\)
- [1 mark] for using \(d \sin\theta = n\lambda\)
- [1 mark] for final answer of \(39.3^{\circ}\)

(b)
- [1 mark] for setting up inequality \(\sin\theta \le 1\)
- [1 mark] for calculating the ratio \(d/\lambda \approx 3.16\)
- [1.25 marks] for rounding down to the integer \(n = 3\)

(c)
- [1 mark] for using the fringe spacing formula \(s = \lambda D/d\)
- [1 mark] for correct substitution of SI units
- [1 mark] for final answer of \(5.06\text{ mm}\) (or \(5.06 \times 10^{-3}\text{ m}\))

(d)
- [1 mark] for stating that the single-slit envelope widens
- [1 mark] for concluding that more fringes become visible / more uniform intensity (with no change in spacing)

(a) The free-body diagram should show the following four forces acting on the crate:
1. Weight \(W = mg\) acting vertically downwards.
2. Normal reaction force \(R\) acting perpendicular to the ramp surface, pointing away from the ramp.
3. Frictional force \(F_f\) acting down the ramp (parallel to the ramp surface, opposing motion).
4. Tension force \(T\) acting up the ramp (parallel to the ramp surface).

(b) Because the crate moves at constant velocity, the forces are in equilibrium:
Resolving forces perpendicular to the incline:
\(R = m g \cos\theta = 80.0 \times 9.81 \times \cos(15.0^{\circ}) = 784.8 \times 0.9659 = 758.1\text{ N}\).

Frictional force is:
\(F_f = \mu R = 0.20 \times 758.1 = 151.6\text{ N}\).

Resolving forces parallel to the incline:
\(T = m g \sin\theta + F_f = (80.0 \times 9.81 \times \sin(15.0^{\circ})) + 151.6 = 203.1 + 151.6 = 354.7\text{ N}\).
This is indeed approximately \(350\text{ N}\).

(c) The useful power output of the motor is:
\(P_{\text{out}} = T v = 354.7 \times 1.20 = 425.6\text{ W}\) (or \(426\text{ W}\)).

(d) The efficiency \(\eta\) of the motor is:
\(\eta = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{425.6}{650} \approx 0.655\) or \(65.5\%\).

(a)
- [2.25 marks] Award 0.5 marks for each correctly drawn and labelled vector (Weight downwards, Normal reaction perpendicular to ramp, Friction downwards along ramp, Tension upwards along ramp). Deduct 0.25 marks if vectors do not touch the body.

(b)
- [1 mark] for resolving normal force: \(R = mg\cos\theta\)
- [1 mark] for calculating \(F_f = \mu R\)
- [1 mark] for equation \(T = mg\sin\theta + F_f\)
- [1 mark] for correct substitution to yield \(355\text{ N}\)

(c)
- [1 mark] for using \(P = Tv\)
- [1 mark] for final answer of \(426\text{ W}\) (accept \(420\text{ W}\) if using rounded \(350\text{ N}\))

(d)
- [1 mark] for formula for efficiency \(\eta = P_{out}/P_{in}\)
- [1 mark] for correct substitution
- [1 mark] for final answer of \(65.5\%\) (or \(65\%\))

(a) The electric field strength is:
\(E = \frac{V}{d} = \frac{120\text{ V}}{0.0200\text{ m}} = 6000\text{ V m}^{-1}\) (or \(6.00 \times 10^3\text{ N C}^{-1}\)).

(b) The horizontal motion has a constant speed. The time taken to cross the plates of length \(L\) is:
\(t = \frac{L}{u} = \frac{0.100}{2.00 \times 10^7} = 5.00 \times 10^{-9}\text{ s}\) (or \(5.00\text{ ns}\)).

(c) The electrostatic force on the electron is:
\(F_e = |q| E = 1.60 \times 10^{-19}\text{ C} \times 6000\text{ V m}^{-1} = 9.60 \times 10^{-16}\text{ N}\).

The acceleration is:
\(a = \frac{F_e}{m_e} = \frac{9.60 \times 10^{-16}\text{ N}}{9.11 \times 10^{-31}\text{ kg}} \approx 1.054 \times 10^{15}\text{ m s}^{-2}\).

Since the electron is negatively charged, it is accelerated upwards (towards the positive plate).

(d) The vertical deflection \(y\) is given by kinematics (with initial vertical velocity of zero):
\(y = \frac{1}{2} a t^2\)
\(y = \frac{1}{2} \times (1.054 \times 10^{15}\text{ m s}^{-2}) \times (5.00 \times 10^{-9}\text{ s})^2\)
\(y = 0.527 \times 10^{15} \times 2.50 \times 10^{-16} = 0.13175\text{ m} \approx 1.32 \times 10^{-2}\text{ m}\) (or \(1.32\text{ cm}\)).

(a)
- [1 mark] for using \(E = V/d\)
- [1 mark] for final answer of \(6.00 \times 10^3\text{ V m}^{-1}\) (or \(\text{N C}^{-1}\))

(b)
- [1 mark] for using \(t = L/u\)
- [1.25 marks] for final answer of \(5.00 \times 10^{-9}\text{ s}\) (or \(5.00\text{ ns}\))

(c)
- [1 mark] for calculating force \(F = q E\)
- [1 mark] for using \(a = F/m\) to get \(1.05 \times 10^{15}\text{ m s}^{-2}\)
- [1 mark] for specifying direction as "upwards" or "towards positive plate"

(d)
- [1 mark] for using \(y = \frac{1}{2}at^2\)
- [1 mark] for substituting \(a\) and \(t\) correctly
- [2 marks] for final answer of \(1.32 \times 10^{-2}\text{ m}\) (or \(1.32\text{ cm}\))