2023 IB DP Physics 模擬試題連答案詳解

The change in momentum is equal to the impulse, which is the area under the force-time graph.

1. Calculate the impulse (area of the triangle):
\( J = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0 \text{ s} \times 6.0 \text{ N} = 12 \text{ N s} \).

2. Relate impulse to change in momentum:
\( \Delta p = m \Delta v = 12 \text{ kg m s}^{-1} \).

3. Calculate the change in velocity:
\( \Delta v = \frac{12 \text{ kg m s}^{-1}}{2.0 \text{ kg}} = 6.0 \text{ m s}^{-1} \).

4. Find the final velocity:
\( v_f = v_i + \Delta v = 3.0 \text{ m s}^{-1} + 6.0 \text{ m s}^{-1} = 9.0 \text{ m s}^{-1} \).

Award 1 mark for the correct answer C.
- Award 0 marks for other options.
- Option A incorrectly assumes no change in speed.
- Option B represents only the change in speed.
- Option D incorrectly adds the peak force directly to the velocity.

From the ideal gas equation, we have:
\( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \)

We are given:
- \( P_2 = 2P_1 \)
- \( T_2 = \frac{1}{2}T_1 \)

Rearranging for the ratio of volumes:
\( \frac{V_2}{V_1} = \left( \frac{P_1}{P_2} \right) \left( \frac{T_2}{T_1} \right) = \left( \frac{1}{2} \right) \left( \frac{1}{2} \right) = \frac{1}{4} \).

Award 1 mark for the correct answer D.
- Option A is the reciprocal of the correct ratio.
- Option B assumes no change in volume.
- Option C is obtained if only one variable's effect is considered.

1. The total resistance in the circuit is \( R_{total} = R + r \).
As \( R \) increases, \( R_{total} \) increases, which causes the current \( I = \frac{\varepsilon}{R + r} \) to decrease.

2. The terminal potential difference \( V \) is given by:
\( V = \varepsilon - I r \)
Since \( I \) decreases, the term \( I r \) decreases, which means \( V \) increases.

Award 1 mark for the correct answer A.
- Option B is the opposite of the correct behavior.
- Option C incorrectly assumes terminal potential difference is constant (true only if internal resistance is zero).
- Option D incorrectly assumes current remains constant.

The formula for the double-slit fringe spacing is:
\( s = \frac{\lambda D}{d} \)

Let the new fringe spacing be \( s' \). With \( \lambda' = 2\lambda \) and \( d' = \frac{d}{2} \):
\( s' = \frac{\lambda' D}{d'} = \frac{(2\lambda) D}{\frac{d}{2}} = 4 \left( \frac{\lambda D}{d} \right) = 4s \).

Award 1 mark for the correct answer C.
- Option A incorrectly assumes the changes cancel each other out.
- Option B only scales by one of the factors of 2.
- Option D represents an incorrect scaling factor of 8.

For nuclear fusion to occur:
1. Extremely high temperatures are needed so that nuclei have high kinetic energies to overcome the electrostatic repulsion (Coulomb barrier) between positively charged protons.
2. High density is required to ensure a sufficiently high collision rate.
3. The electrostatic repulsion must be overcome to bring the nuclei close enough for the short-range strong nuclear force to bind them.

Award 1 mark for the correct answer B.
- Option A is incorrect because low temperature would not provide enough kinetic energy, and the strong nuclear force is the attractive force that needs to be reached, not overcome.
- Option C is incorrect because high density is required to maintain collision rates.
- Option D is incorrect on all counts.

1. Find the useful power output \( P_{\text{out}} \) of the motor:
\( P_{\text{out}} = F \cdot v = m g v = 30 \text{ kg} \times 10 \text{ m s}^{-2} \times 2.0 \text{ m s}^{-1} = 600 \text{ W} \).

2. Use the efficiency formula to find the power input \( P_{\text{in}} \):
\( \text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \implies 0.60 = \frac{600 \text{ W}}{P_{\text{in}}} \)

3. Calculate \( P_{\text{in}} \):
\( P_{\text{in}} = \frac{600}{0.60} = 1000 \text{ W} = 1.0 \text{ kW} \).

Award 1 mark for the correct answer C.
- Option A is obtained by multiplying the useful power by 0.60 instead of dividing.
- Option B is the useful power output, neglecting the efficiency factor.
- Option D is a miscalculation of the efficiency ratio.

Using the equations of motion for uniform acceleration:
\( v^2 = u^2 + 2 a s \)

Let upward be the positive direction:
- Initial velocity, \( u_i = +u \)
- Final velocity, \( v = -2u \) (downward)
- Acceleration, \( a = -g \)
- Vertical displacement from launch point to ground, \( s = -h \)

Substitute these into the equation:
\( (-2u)^2 = u^2 + 2(-g)(-h) \)
\( 4u^2 = u^2 + 2gh \)
\( 3u^2 = 2gh \)
\( h = \frac{3u^2}{2g} \).

Award 1 mark for the correct answer B.
- Option A is the maximum height reached above the cliff.
- Option C is a common distraction from algebraic slips.
- Option D is obtained by neglecting the initial kinetic energy at the cliff edge.

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

Let the new power be \( P' \):
\( P' = \sigma A' (T')^4 \)

Given \( A' = \frac{A}{2} \) and \( T' = 2T \):
\( P' = \sigma \left( \frac{A}{2} \right) (2T)^4 \)
\( P' = \sigma \left( \frac{A}{2} \right) (16 T^4) = 8 \sigma A T^4 = 8P \).

Award 1 mark for the correct answer C.
- Option A incorrectly assumes power is proportional to temperature squared.
- Option B is obtained if the \( T^4 \) dependency is treated as \( T^3 \).
- Option D is obtained if the change in area is neglected.

From conservation of momentum: \(m v = (m + 3m) v_f \implies v_f = \frac{v}{4}\). The initial kinetic energy is \(E_i = \frac{1}{2} m v^2\). The final kinetic energy is \(E_f = \frac{1}{2} (4m) v_f^2 = \frac{1}{2} (4m) \left(\frac{v}{4}\right)^2 = \frac{1}{8} m v^2\). The kinetic energy lost is \(\Delta E = E_i - E_f = \frac{1}{2} m v^2 - \frac{1}{8} m v^2 = \frac{3}{8} m v^2\). The percentage lost is \(\frac{\Delta E}{E_i} = \frac{3/8}{1/2} = \frac{3}{4} = 75\%\).

Award 1 mark for the correct answer C. Award 0 marks for other options.

The pressure law states that for a constant volume of gas, pressure is directly proportional to absolute temperature: \(P \propto T\). First, convert temperatures to Kelvin: \(T_1 = 27 + 273 = 300\text{ K}\) and \(T_2 = 327 + 273 = 600\text{ K}\). Thus, the ratio of final to initial pressure is \(\frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{600}{300} = 2.0\).

Award 1 mark for the correct answer B. Option C represents the trap of failing to convert Celsius to Kelvin.

The fringe spacing formula is given by \(s = \frac{\lambda D}{d}\). The new fringe separation \(s'\) is: \(s' = \frac{\lambda' D}{d'} = \frac{(\lambda/2) D}{2d} = \frac{1}{4} \frac{\lambda D}{d} = \frac{s}{4}\).

Award 1 mark for the correct answer D. No marks are awarded for incorrect scaling factors.

The equivalent resistance of the two resistors in parallel is \(R_p = \left(\frac{1}{R} + \frac{1}{R}\right)^{-1} = \frac{R}{2}\). Since this combination is in series with the third resistor, the total equivalent resistance is \(R_{eq} = R_p + R = \frac{R}{2} + R = \frac{3}{2}R\).

Award 1 mark for the correct answer C. Standard formula execution required.

Work done is equal to the area under the force-displacement graph. For \(x = 0\) to \(x = 4\text{ m}\), the area is a triangle: \(\frac{1}{2} \times 4\text{ m} \times 10\text{ N} = 20\text{ J}\). For \(x = 4\text{ m}\) to \(x = 8\text{ m}\), the area is a rectangle: \((8 - 4)\text{ m} \times 10\text{ N} = 40\text{ J}\). The total work done is \(20\text{ J} + 40\text{ J} = 60\text{ J}\).

Award 1 mark for the correct answer B. Option C represents treating the entire shape as a rectangle (80 J).

When unpolarized light passes through the first filter, its intensity is halved: \(I_1 = \frac{I_0}{2}\). When it passes through the second filter, Malus's Law applies: \(I_2 = I_1 \cos^2(\theta) = \left(\frac{I_0}{2}\right) \cos^2(60^\circ)\). Since \(\cos(60^\circ) = \frac{1}{2}\), we have \(\cos^2(60^\circ) = \frac{1}{4}\). Therefore, \(I_2 = \frac{I_0}{2} \times \frac{1}{4} = \frac{1}{8}I_0\).

Award 1 mark for the correct answer C. Option B represents the error of omitting the halving of intensity after the first filter.

The thermal energy supplied to the liquid in time \(t\) is \(Q = P t\). This causes a temperature change \(\Delta T\), related by \(Q = m c \Delta T\). Rearranging gives \(P t = m c \Delta T \implies \frac{\Delta T}{t} = \frac{P}{m c}\). The slope of the temperature-time graph is \(k = \frac{\Delta T}{t}\). Therefore, \(k = \frac{P}{m c} \implies c = \frac{P}{m k}\).

Award 1 mark for the correct answer A.

The normal force acting on the block is \(R = m g = 2.0\text{ kg} \times 10\text{ m s}^{-2} = 20\text{ N}\). The frictional force is \(F_f = \mu_d R = 0.30 \times 20\text{ N} = 6.0\text{ N}\). The net force on the block is \(F_{net} = F - F_f = 12\text{ N} - 6.0\text{ N} = 6.0\text{ N}\). Using Newton's second law, \(a = \frac{F_{net}}{m} = \frac{6.0\text{ N}}{2.0\text{ kg}} = 3.0\text{ m s}^{-2}\).

Award 1 mark for the correct answer A.

The impulse \(J\) delivered to the body is equal to the area under the force-time graph. The graph is a triangle with base \(b = 3.0\text{ s}\) and peak height \(h = 10\text{ N}\). Area \(J = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3.0\text{ s} \times 10\text{ N} = 15\text{ N s}\). According to the impulse-momentum theorem, \(J = \Delta p = m \Delta v\). Since the body starts from rest, \(15\text{ N s} = 2.0\text{ kg} \times v\), which gives \(v = 7.5\text{ m s}^{-1}\).

[1 mark] for calculating the impulse as the area under the triangle: \(15\text{ N s}\). [1 mark] for equating the impulse to change in momentum and solving for velocity: \(v = 7.5\text{ m s}^{-1}\).

Let the initial state of the gas be described by \(p_1 V = n_1 R T_1\). After heating, the temperature is \(T_2 = 1.5 T_1\). When gas is released to return the pressure to its initial value \(p_1\) at constant volume \(V\) and constant new temperature \(T_2\), the final number of moles \(n_3\) satisfies \(p_1 V = n_3 R T_2 = n_3 R (1.5 T_1)\). Equating the two expressions for \(p_1 V\), we get \(n_1 R T_1 = n_3 R (1.5 T_1)\), which simplifies to \(n_3 = \frac{n_1}{1.5} = \frac{2}{3} n_1\). Since mass is proportional to the number of moles, the fraction of the original mass remaining is \(\frac{2}{3}\).

[1 mark] for recognizing that temperature increases by a factor of \(1.5\). [1 mark] for using the ideal gas equation to find that the number of moles must decrease by a factor of \(1.5\) (or to \(\frac{2}{3}\) of the original value) to restore the initial pressure.

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

[1 mark] for substituting the modified variables into the double-slit formula and correctly identifying the scaling factor as \(\frac{1}{4}\).

The terminal potential difference is given by \(V = E \frac{R}{R+r}\). As \(R\) increases, the fraction \(\frac{R}{R+r}\) increases continuously from near zero toward one. Thus, \(V\) increases continuously. The power dissipated in the external resistor is \(P = I^2 R = \frac{E^2 R}{(R+r)^2}\). By the maximum power transfer theorem, \(P\) is maximized when \(R = r\). Therefore, as \(R\) increases from a value less than \(r\) to a value greater than \(r\), the power \(P\) first increases, reaches a maximum at \(R = r\), and then decreases.

[1 mark] for correctly deducing that the terminal potential difference increases continuously as the external resistance increases. [1 mark] for identifying that power dissipation in the load resistor peaks when \(R = r\), so it increases then decreases.

The initial kinetic energy is \(E_k = \frac{1}{2}mv^2\). At the highest point of the trajectory, the vertical component of velocity is zero, and the horizontal component of velocity remains constant: \(v_x = v \cos(30^\circ) = v \frac{\sqrt{3}}{2}\). Therefore, the kinetic energy at the highest point is \(E_{k,\text{top}} = \frac{1}{2}mv_x^2 = \frac{1}{2}m\left(v \frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \left(\frac{1}{2}mv^2\right) = 0.75 E_k\). The ratio is \(0.75\).

[1 mark] for identifying that only horizontal velocity remains at the peak: \(v_x = v \cos(30^\circ)\). [1 mark] for squaring this factor to obtain the kinetic energy ratio: \(\cos^2(30^\circ) = 0.75\).

First, calculate the useful work output per minute: \(W = mgh = 180\text{ kg} \times 10\text{ m s}^{-2} \times 10\text{ m} = 18,000\text{ J}\). The useful power output is this work divided by the time in seconds (\(1\text{ minute} = 60\text{ s}\)): \(P_{\text{out}} = \frac{18,000\text{ J}}{60\text{ s}} = 300\text{ W}\). Since the efficiency \(\eta\) is \(60\%\) (or \(0.60\)), the electrical input power \(P_{\text{in}}\) is: \(P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{300\text{ W}}{0.60} = 500\text{ W}\).

[1 mark] for calculating the useful work output (\(18,000\text{ J}\)) and subsequent useful power output (\(300\text{ W}\)). [1 mark] for dividing by efficiency to find input power (\(500\text{ W}\)).

When unpolarized light of intensity \(I_0\) passes through the first polarizer, its intensity is reduced by half: \(I_1 = \frac{1}{2} I_0\) and it becomes vertically polarized. It then passes through the second polarizer whose transmission axis is at an angle \(\theta = 60^\circ\) relative to the vertical. By Malus's Law, the transmitted intensity is: \(I_2 = I_1 \cos^2(60^\circ) = \frac{1}{2} I_0 \times \left(\frac{1}{2}\right)^2 = \frac{1}{2} I_0 \times \frac{1}{4} = \frac{1}{8} I_0\).

[1 mark] for recognizing that the intensity after the first polarizer is \(\frac{1}{2}I_0\). [1 mark] for applying Malus's law with \(\cos(60^\circ) = 0.5\) to yield a final intensity of \(\frac{1}{8}I_0\).

First, find the number of half-lives that have elapsed: \(n = \frac{12\text{ days}}{4.0\text{ days}} = 3\). After \(3\) half-lives, the fraction of isotope \(X\) remaining is \(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\). Therefore, the number of nuclei of \(X\) is \(N_X = \frac{1}{8} N_0\). The rest of the nuclei have decayed into isotope \(Y\), so \(N_Y = N_0 - N_X = \frac{7}{8} N_0\). The ratio of the number of \(Y\) nuclei to \(X\) nuclei is: \\frac{N_Y}{N_X} = \frac{\frac{7}{8} N_0}{\frac{1}{8} N_0} = 7\).

[1 mark] for determining that 3 half-lives have elapsed, meaning \(\frac{1}{8}\) of \(X\) remains and \(\frac{7}{8}\) has decayed into \(Y\). [1 mark] for finding the ratio of \(Y\) to \(X\) is 7.

1. From conservation of linear momentum: the initial momentum is \( p_i = M v \).
2. After the explosion, the lighter fragment (\( \frac{1}{4}M \)) is at rest, so its momentum is zero. Thus, the heavier fragment (\( \frac{3}{4}M \)) must carry all the initial momentum:
\( M v = \frac{1}{4}M (0) + \frac{3}{4}M v_2 \implies v_2 = \frac{4}{3}v \).
3. The initial kinetic energy is \( E_k = \frac{1}{2} M v^2 \).
4. The kinetic energy of the heavier fragment is:
\( E'_k = \frac{1}{2} \left(\frac{3}{4}M\right) v_2^2 = \frac{1}{2} \left(\frac{3}{4}M\right) \left(\frac{4}{3}v\right)^2 = \frac{1}{2} \left(\frac{3}{4}M\right) \left(\frac{16}{9}v^2\right) = \frac{4}{3} \left(\frac{1}{2} M v^2\right) = \frac{4}{3} E_k \).

Award 1 mark for the correct option C.
- Method: Apply conservation of momentum to find the final velocity of the heavier fragment (\( \frac{4}{3}v \)), then calculate the new kinetic energy relative to \( E_k \).
- Accuracy: Correctly identify the ratio as \( \frac{4}{3} \).

1. During the isothermal compression (constant temperature \( T \)), the volume decreases from \( V \) to \( \frac{V}{2} \).
According to Boyle's law (\( P \propto \frac{1}{V} \)), the pressure doubles to \( 2P \).
2. During the constant-volume heating phase, the pressure is doubled again from \( 2P \) to \( 4P \).
3. According to Gay-Lussac's law for constant volume, pressure is directly proportional to absolute temperature (\( P \propto T \)). Since the pressure doubles from \( 2P \) to \( 4P \), the absolute temperature must also double from its current value \( T \) to \( 2T \).
Alternatively, using the ideal gas equation:
\( \frac{P_{initial} V_{initial}}{T_{initial}} = \frac{P_{final} V_{final}}{T_{final}} \implies \frac{P V}{T} = \frac{4P \cdot (V/2)}{T_{final}} \implies \frac{P V}{T} = \frac{2P V}{T_{final}} \implies T_{final} = 2T \).

Award 1 mark for the correct option B.
- Method: Apply Boyle's law to the isothermal process, then Gay-Lussac's law or the combined gas law to find the final temperature.
- Accuracy: Correctly obtain the final temperature of \( 2T \).

1. When unpolarized light passes through the first filter, its intensity is reduced to half: \( I_1 = \frac{1}{2} I_0 \).
2. According to Malus's law, the intensity after the second filter is:
\( I_2 = I_1 \cos^2(30^\circ) = \left(\frac{1}{2} I_0\right) \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{8} I_0 \).
3. The angle between the transmission axis of the second filter (\( 30^\circ \) to the vertical) and the third filter (\( 90^\circ \) to the vertical) is \( 90^\circ - 30^\circ = 60^\circ \).
4. Applying Malus's law again for the third filter:
\( I_3 = I_2 \cos^2(60^\circ) = \left(\frac{3}{8} I_0\right) \left(\frac{1}{2}\right)^2 = \frac{3}{32} I_0 \).

Award 1 mark for the correct option B.
- Method: Apply the reduction factor of \( \frac{1}{2} \) for unpolarized light, then consecutive applications of Malus's law using correct relative angles (\( 30^\circ \) and \( 60^\circ \)).
- Accuracy: Correctly perform the fractional multiplication to arrive at \( \frac{3}{32} I_0 \).

1. The terminal potential difference \( V \) across a cell is given by:
\( V = \varepsilon \frac{R}{R + r} \)
2. For the first case, \( R = 4.0\ \Omega \) and \( V = 6.0\text{ V} \):
\( 6.0 = \varepsilon \frac{4.0}{4.0 + r} \implies 6(4 + r) = 4\varepsilon \implies 24 + 6r = 4\varepsilon \implies 2\varepsilon - 3r = 12 \) (Equation 1)
3. For the second case, \( R = 9.0\ \Omega \) and \( V = 9.0\text{ V} \):
\( 9.0 = \varepsilon \frac{9.0}{9.0 + r} \implies 9 + r = \varepsilon \implies r = \varepsilon - 9 \) (Equation 2)
4. Substitute Equation 2 into Equation 1:
\( 2\varepsilon - 3(\varepsilon - 9) = 12 \implies -\varepsilon + 27 = 12 \implies \varepsilon = 15\text{ V} \).
5. Find \( r \):
\( r = 15 - 9 = 6.0\ \Omega \).

Award 1 mark for the correct option C.
- Method: Write the potential divider equation for both cases and solve the simultaneous equations for \( \varepsilon \) and \( r \).
- Accuracy: Verify that only the pair \( \varepsilon = 15\text{ V} \), \( r = 6.0\ \Omega \) satisfies both measurements.

1. The thermal energy supplied during the melting process is:
\( Q_1 = P \Delta t_1 = m L_f \implies L_f = \frac{P \Delta t_1}{m} \)
2. The thermal energy supplied to raise the temperature of the liquid is:
\( Q_2 = P \Delta t_2 = m c \Delta T \implies c = \frac{P \Delta t_2}{m \Delta T} \)
3. Taking the ratio \( \frac{L_f}{c} \):
\( \frac{L_f}{c} = \frac{\frac{P \Delta t_1}{m}}{\frac{P \Delta t_2}{m \Delta T}} = \frac{\Delta t_1}{\frac{\Delta t_2}{\Delta T}} = \Delta T \frac{\Delta t_1}{\Delta t_2} \).

Award 1 mark for the correct option B.
- Method: Formulate expressions for the thermal energy required for state change (\( Q = m L_f \)) and temperature increase (\( Q = m c \Delta T \)) in terms of power \( P \) and time, then find the ratio of \( L_f \) to \( c \).
- Accuracy: Correct algebraic simplification yielding \( \Delta T \frac{\Delta t_1}{\Delta t_2} \).

1. On a flat road, at a constant speed \( v \), the forward force from the engine balances the resistive force:
\( F_{engine} = F_r = k v^2 \).
2. The power output is \( P = F_{engine} v = k v^3 \).
3. When climbing the hill of inclination \( \theta \) at the same constant speed \( v \), the engine must overcome both the resistive force and the component of gravity parallel to the slope:
\( F'_{engine} = F_r + m g \sin\theta = k v^2 + m g \sin\theta \).
4. The new useful power required is:
\( P' = F'_{engine} v = (k v^2 + m g \sin\theta) v = k v^3 + m g v \sin\theta = P + m g v \sin\theta \).

Award 1 mark for the correct option B.
- Method: Write the force balance for both the flat road and the hill. Use \( P = F v \) to find the expression for the power required in both situations.
- Accuracy: Correctly add the power required to overcome gravity (\( m g v \sin\theta \)) to the original power \( P \).

(a) Using conservation of linear momentum: \( m_1 v = (m_1 + m_2) v_f \). Substituting values: \( 1.2 \times 5.0 = (1.2 + 1.8) v_f \implies 6.0 = 3.0 v_f \implies v_f = 2.0\text{ m s}^{-1} \).

(b) Initial kinetic energy: \( E_{ki} = \frac{1}{2} m_1 v^2 = 0.5 \times 1.2 \times (5.0)^2 = 15\text{ J} \).
Final kinetic energy: \( E_{kf} = \frac{1}{2} (m_1 + m_2) v_f^2 = 0.5 \times 3.0 \times (2.0)^2 = 6.0\text{ J} \).
Loss in kinetic energy = \( E_{ki} - E_{kf} = 15 - 6.0 = 9.0\text{ J} \).
Since there is a non-zero loss of kinetic energy, the collision is inelastic.

(c) The frictional force acting on the combined block is \( f = \mu_d N = \mu_d (m_1 + m_2) g = 0.40 \times 3.0 \times 9.81 = 11.77\text{ N} \).
Using the work-energy theorem, the work done by friction equals the loss in kinetic energy: \( f \times d = E_{kf} \implies 11.77 \times d = 6.0 \implies d = 0.51\text{ m} \) (using \( g = 9.81\text{ m s}^{-2} \)).

(a) [3 marks]
- Award [1] for using conservation of momentum.
- Award [1] for correct substitution: \( 1.2 \times 5.0 = 3.0 v_f \).
- Award [1] for final correct answer: \( 2.0\text{ m s}^{-1} \).

(b) [3 marks]
- Award [1] for correct initial kinetic energy (15 J).
- Award [1] for correct final kinetic energy (6.0 J).
- Award [1] for calculating the difference (9.0 J) and stating that kinetic energy is lost/not conserved.

(c) [4 marks]
- Award [1] for identifying the normal force as \( (m_1 + m_2)g \).
- Award [1] for calculating the friction force as \( 11.8\text{ N} \) (or stating deceleration is \( \mu_d g = 3.9\text{ m s}^{-2} \)).
- Award [1] for using work-energy theorem or kinematics equation \( v^2 = u^2 + 2as \).
- Award [1] for correct final answer of \( 0.51\text{ m} \) (accept range 0.50 - 0.52).

(a) Using the ideal gas equation: \( PV = nRT \implies n = \frac{PV}{RT} \).
Substituting the values: \( n = \frac{(1.2 \times 10^5) \times (2.5 \times 10^{-3})}{8.31 \times 300} = \frac{300}{2493} \approx 0.120\text{ mol} \).

(b) For a constant volume process: \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \).
Substituting the values: \( \frac{1.2 \times 10^5}{300} = \frac{1.8 \times 10^5}{T_2} \implies T_2 = 300 \times \frac{1.8 \times 10^5}{1.2 \times 10^5} = 450\text{ K} \).

(c) For an isothermal process: \( P_2 V_2 = P_3 V_3 \).
Since the volume is constant in step (b), \( V_2 = 2.5 \times 10^{-3}\text{ m}^3 \).
Therefore: \( (1.8 \times 10^5) \times (2.5 \times 10^{-3}) = (1.2 \times 10^5) \times V_3 \implies V_3 = 3.75 \times 10^{-3}\text{ m}^3 \) (or \( 3.8 \times 10^{-3}\text{ m}^3 \)).
On a P-V diagram:
- Process 1 to 2 is a vertical line going upwards.
- Process 2 to 3 is a downward-sloping curve (hyperbolic shape) ending at a larger volume and lower pressure.

(a) [3 marks]
- Award [1] for using \( PV = nRT \).
- Award [1] for correct substitution of all values (including \( R = 8.31\text{ J K}^{-1}\text{ mol}^{-1} \)).
- Award [1] for correct final answer: \( 0.12\text{ mol} \) (or 0.120 mol).

(b) [3 marks]
- Award [1] for using Pressure Law or \( P/T = \text{constant} \).
- Award [1] for correct numerical setup.
- Award [1] for correct final answer: \( 450\text{ K} \).

(c) [4 marks]
- Award [1] for stating Boyle's Law or \( PV = \text{constant} \) for the isothermal step.
- Award [1] for correct calculation of final volume: \( 3.8 \times 10^{-3}\text{ m}^3 \) (accept \( 3.75 \times 10^{-3}\text{ m}^3 \)).
- Award [1] for describing the first process as a vertical line pointing upwards.
- Award [1] for describing the second process as a downward curve pointing to the right.

(a) Polarized light is electromagnetic radiation whose electric field vector vibrates/oscillates in only one plane that contains the direction of wave propagation.

(b) When unpolarized light of intensity \( I_0 \) passes through the first polarizer, its intensity is reduced by exactly half because only the components of light parallel to the transmission axis pass through: \( I_1 = \frac{I_0}{2} \). When this polarized light passes through the analyzer, Malus's law applies: \( I = I_1 \cos^2\theta \). Substituting \( I_1 \) into the equation yields \( I = \frac{I_0}{2} \cos^2\theta \).

(c)(i) Using the double-slit formula: \( s = \frac{\lambda D}{d} \).
Substituting the given values: \( s = \frac{630 \times 10^{-9}\text{ m} \times 1.8\text{ m}}{0.15 \times 10^{-3}\text{ m}} = 7.56 \times 10^{-3}\text{ m} \) (or \( 7.6\text{ mm} \)).

(c)(ii) Decreasing the width of the slits causes a wider single-slit diffraction envelope (more diffraction/spreading of light). As a result, the region containing the interference fringes becomes wider (more fringes are visible), while the spacing between individual bright fringes remains unchanged because the slit separation \( d \) is constant. The overall peak intensity of the pattern also decreases.

(a) [2 marks]
- Award [1] for stating that electric field oscillations are in a single plane.
- Award [1] for stating that this plane is perpendicular to the direction of propagation/energy transfer.

(b) [3 marks]
- Award [1] for explaining that the first polarizer halves the intensity of the unpolarized light, i.e., \( I_1 = \frac{I_0}{2} \).
- Award [1] for stating Malus's Law, \( I = I_1 \cos^2\theta \).
- Award [1] for algebraically combining the steps to yield the final expression.

(c)(i) [3 marks]
- Award [1] for selecting the formula \( s = \frac{\lambda D}{d} \).
- Award [1] for correct substitution with appropriate unit conversions.
- Award [1] for the final correct value with unit: \( 7.6\text{ mm} \) or \( 7.56 \times 10^{-3}\text{ m} \).

(c)(ii) [2 marks]
- Award [1] for stating that the single-slit diffraction envelope spreads out (more fringes are visible).
- Award [1] for stating that the fringe spacing remains constant (or that the fringe intensity decreases).

(a) The relationship is: \( V = \mathcal{E} - Ir \) (or \( \mathcal{E} = I(R+r) \)).

(b) Using the relationship \( \mathcal{E} = I(R + r) \):
For the first measurement: \( \mathcal{E} = 1.5 \times (4.0 + r) = 6.0 + 1.5r \) (Equation 1)
For the second measurement: \( \mathcal{E} = 0.80 \times (9.0 + r) = 7.2 + 0.80r \) (Equation 2)
Equating the two expressions for \( \mathcal{E} \):
\( 6.0 + 1.5r = 7.2 + 0.80r \implies 0.70r = 1.2 \implies r = 1.71\\ \Omega \) (or \( 1.7\\ \Omega \)).
Substituting \( r = 1.714\\ \Omega \) into Equation 1:
\( \mathcal{E} = 6.0 + 1.5(1.714) = 8.57\text{ V} \) (or \( 8.6\text{ V} \)).

(c)(i) Power dissipated in the external resistor: \( P_{ext} = I^2 R = (1.5)^2 \times 4.0 = 2.25 \times 4.0 = 9.0\text{ W} \).

(c)(ii) Efficiency \( \eta = \frac{P_{ext}}{P_{total}} = \frac{I^2 R}{I^2(R+r)} = \frac{R}{R+r} \).
\( \eta = \frac{4.0}{4.0 + 1.71} = 0.70 \) (or \( 70\\% \)).

(a) [1 mark]
- Award [1] for writing \( V = \mathcal{E} - Ir \) or equivalent, e.g., \( \mathcal{E} = I(R + r) \).

(b) [5 marks]
- Award [1] for setting up the first equation: \( \mathcal{E} = 1.5(4.0 + r) \).
- Award [1] for setting up the second equation: \( \mathcal{E} = 0.80(9.0 + r) \).
- Award [1] for equating equations and correctly eliminating one variable.
- Award [1] for correct internal resistance \( r = 1.7\\ \Omega \) (accept 1.71 \\Omega).
- Award [1] for correct emf \( \mathcal{E} = 8.6\text{ V} \) (accept 8.57 V).

(c)(i) [2 marks]
- Award [1] for using \( P = I^2 R \) or \( P = VI \) with correct values.
- Award [1] for final correct answer of \( 9.0\text{ W} \).

(c)(ii) [2 marks]
- Award [1] for using \( \eta = \frac{R}{R+r} \) or \( \eta = \frac{P_{ext}}{P_{total}} \).
- Award [1] for correct final efficiency of \( 0.70 \) or \( 70\\% \) (accept range 0.70 - 0.71).

(a) Albedo is defined as the ratio of the total reflected power of solar radiation from the planet to the total incident solar power on the planet.

(b) The planet intercepts solar radiation over its projected cross-sectional area \( \pi R^2 \), where \( R \) is its radius. The total absorbed power is \( P_{abs} = S(1 - \alpha)\pi R^2 \). Since the planet is spherical, this absorbed energy is distributed over its entire surface area \( 4\pi R^2 \). Hence, the average intensity absorbed is: \( I_{abs} = \frac{S(1 - \alpha)\pi R^2}{4\pi R^2} = \frac{S(1 - \alpha)}{4} = \frac{1360 \times (1 - 0.30)}{4} = 238\text{ W m}^{-2} \).

(c) In radiative equilibrium, the power emitted per unit area must equal the average absorbed intensity: \( \sigma T^4 = I_{abs} \).
Substituting the Stefan-Boltzmann constant \( \sigma = 5.67 \times 10^{-8}\text{ W m}^{-2}\text{ K}^{-4} \):
\( 5.67 \times 10^{-8} T^4 = 238 \implies T^4 = 4.1975 \times 10^9 \implies T \approx 254\text{ K} \) (or \( -19^\circ\text{C} \)).

(d) The planet's surface emits long-wavelength (infrared) radiation. Greenhouse gases in the atmosphere absorb this infrared radiation and re-emit it in all directions. A portion of this re-emitted radiation is directed back to the surface (back-radiation), which provides additional heating, raising the surface temperature above the 254 K calculated in (c).

(a) [2 marks]
- Award [1] for defining albedo as the ratio of reflected power to incident power.
- Award [1] for specifying that it relates to solar radiation over the total surface/planetary level.

(b) [3 marks]
- Award [1] for recognizing that the ratio of projected area to total area is \( 1/4 \).
- Award [1] for correctly multiplying by the factor \( (1 - \alpha) \) or \( 0.70 \).
- Award [1] for completing the calculation: \( \frac{1360 \times 0.70}{4} = 238\text{ W m}^{-2} \) with clear steps.

(c) [3 marks]
- Award [1] for equating emitted intensity to absorbed intensity, e.g., \( \sigma T^4 = 238 \).
- Award [1] for correct substitution of the Stefan-Boltzmann constant.
- Award [1] for the final correct answer: \( 254\text{ K} \) (accept range 253 - 255 K or -19 C).

(d) [2 marks]
- Award [1] for stating that greenhouse gases absorb infrared/long-wavelength radiation emitted by the surface.
- Award [1] for stating that these gases re-emit this radiation back to the surface (back-radiation), resulting in a higher temperature.

(a) From the equation \(P = 0.362 \theta + 98.7\), at \(\theta = 0.0\ ^{\circ}\text{C}\), \(P = 98.7\text{ kPa}\).

(b) (i) At absolute zero, the average kinetic energy of the molecules in the gas becomes zero (or the minimum possible value), meaning all molecular translational motion ceases. Therefore, the gas exerts no pressure.
(ii) Set \(P = 0\):
\(0 = 0.362 \theta + 98.7\)
\(\theta = -\frac{98.7}{0.362} = -272.65\ ^{\circ}\text{C}\)
To two decimal places, this is \(-272.65\ ^{\circ}\text{C}\) (or rounded to \(-273\ ^{\circ}\text{C}\)).

(c) The absolute zero value is calculated as \(\theta_0 = -\frac{I}{m}\), where \(I\) is the y-intercept and \(m\) is the gradient.
The fractional uncertainty in \(\theta_0\) is the sum of the fractional uncertainties in \(I\) and \(m\):
\(\frac{\Delta \theta_0}{|\theta_0|} = 1.5\% + 2.5\% = 4.0\%\).
Thus, the absolute uncertainty in the experimental value is:
\(\Delta \theta_0 = 4.0\% \times 272.65\ ^{\circ}\text{C} = 10.9\ ^{\circ}\text{C}\) (or \(11\ ^{\circ}\text{C}\)).
The experimental range is therefore \(-272.7 \pm 10.9\ ^{\circ}\text{C}\), which spans from \(-283.6\ ^{\circ}\text{C}\) to \(-261.8\ ^{\circ}\text{C}\).
Since the accepted value of \(-273.15\ ^{\circ}\text{C}\) lies within this range, the student's result is consistent with the accepted value.

(a) [1 mark]
\bullet\ \(98.7\text{ kPa}\) [1]

(b) [3 marks]
(i) \bullet\ State that molecular kinetic energy (or molecular motion) is zero / minimum [1]
(ii) \bullet\ Setting \(P = 0\) [1]
\bullet\ \(-273\ ^{\circ}\text{C}\) or \(-272.7\ ^{\circ}\text{C}\) [1]

(c) [3 marks]
\bullet\ Calculating fractional uncertainty of \(4.0\%\) or absolute uncertainty of \(10.9\ ^{\circ}\text{C}\) (accept \(11\ ^{\circ}\text{C}\)) [1]
\bullet\ Stating range as \(-284\ ^{\circ}\text{C}\) to \(-262\ ^{\circ}\text{C}\) (or \(-283.6\ ^{\circ}\text{C}\) to \(-261.8\ ^{\circ}\text{C}\)) [1]
\bullet\ Concluding that the accepted value lies within the range with explicit comparison [1]

(a) Taking the natural logarithm of both sides of \(v = k r^2\):
\(\ln(v) = \ln(k r^2) = \ln(k) + \ln(r^2)\)
Using logarithm properties, \(\ln(r^2) = 2\ln(r)\), so:
\(\ln(v) = 2 \ln(r) + \ln(k)\)
Comparing this to the equation of a straight line, \(y = mx + c\), where \(y = \ln(v)\) and \(x = \ln(r)\), we find the gradient \(m = 2\) and the y-intercept \(c = \ln(k)\).

(b) (i) The experimental gradient is \(1.88 \pm 0.08\), representing a range of \(1.80\) to \(1.96\).
Since the theoretical gradient of \(2.00\) does not lie within this experimental range, the result is not consistent with the theoretical prediction of Stokes' Law.
(ii) The y-intercept \(c = 3.42\).
\(k = e^c = e^{3.42} \approx 30.57\text{ mm}^{-2}\text{ s}^{-1}\).
To find the absolute uncertainty in \(k\):
\(k_{max} = e^{3.42 + 0.15} = e^{3.57} \approx 35.52\text{ mm}^{-2}\text{ s}^{-1}\)
\(k_{min} = e^{3.42 - 0.15} = e^{3.27} \approx 26.31\text{ mm}^{-2}\text{ s}^{-1}\)
\(\Delta k = \frac{35.52 - 26.31}{2} = 4.6\text{ mm}^{-2}\text{ s}^{-1}\) (or using \(\Delta k \approx k \Delta c = 30.57 \times 0.15 = 4.58\)).
Rounding appropriately to the correct number of significant figures: \(k = 31 \pm 5\text{ mm}^{-2}\text{ s}^{-1}\) (or \(30.6 \pm 4.6\text{ mm}^{-2}\text{ s}^{-1}\)).

(a) [2 marks]
\bullet\ \(\ln(v) = \ln(k) + \ln(r^2)\) [1]
\bullet\ Explains comparison with \(y = mx + c\) to show \(m = 2\) and \(c = \ln(k)\) [1]

(b) [5 marks]
(i) \bullet\ Identifies range as \(1.80\) to \(1.96\) [1]
\bullet\ Concludes not consistent because 2 is outside this range [1]
(ii) \bullet\ Calculates \(k = e^{3.42} = 31\text{ mm}^{-2}\text{ s}^{-1}\) (or \(30.6\)) [1]
\bullet\ Method to find uncertainty (using max/min or calculus) [1]
\bullet\ Final value and uncertainty correct: \(31 \pm 5\) (or \(30.6 \pm 4.6\)) with units [1]

(a) Luminosity is the total power radiated by a star (or the total energy emitted per second).

(b) From the Stefan-Boltzmann law, \(L = 4\pi R^2 \sigma T^4\).
Taking the ratio of Star X to the Sun:
\(\frac{L_X}{L_{\odot}} = \left(\frac{R_X}{R_{\odot}}\right)^2 \left(\frac{T_X}{T_{\odot}}\right)^4\)
\(50 = \left(\frac{R_X}{R_{\odot}}\right)^2 \left(\frac{9600}{5800}\right)^4\)
\(50 = \left(\frac{R_X}{R_{\odot}}\right)^2 \times (1.655)^4\)
\(50 = \left(\frac{R_X}{R_{\odot}}\right)^2 \times 7.50\)
\(\left(\frac{R_X}{R_{\odot}}\right)^2 = \frac{50}{7.50} = 6.67\)
\(\frac{R_X}{R_{\odot}} = \sqrt{6.67} \approx 2.58\)
Therefore, the radius of Star X is approximately \(2.6 R_{\odot}\).

(c) Apparent brightness is given by \(b = \frac{L}{4\pi d^2}\).
Rearranging for distance \(d\):
\(d = \sqrt{\frac{L_X}{4\pi b}}\)
\(d = \sqrt{\frac{50 \times 3.85 \times 10^{26}}{4\pi \times 1.3 \times 10^{-10}}}\)
\(d = \sqrt{\frac{1.925 \times 10^{28}}{1.634 \times 10^{-9}}} = \sqrt{1.178 \times 10^{37}} \approx 3.43 \times 10^{18}\text{ m}\)
Convert distance to light years:
\(d = \frac{3.43 \times 10^{18}\text{ m}}{9.46 \times 10^{15}\text{ m ly}^{-1}} \approx 363\text{ ly}\)
To 2 significant figures, the distance is \(360\text{ ly}\).

(a) [1 mark]
\bullet\ Total power radiated / total energy emitted per second by a star [1]

(b) [3 marks]
\bullet\ Use of Stefan-Boltzmann ratio formula [1]
\bullet\ Correct substitution of temperature and luminosity ratios [1]
\bullet\ \(2.6 R_{\odot}\) (or \(2.58 R_{\odot}\)) [1]

(c) [3 marks]
\bullet\ Correct use of \(b = \frac{L}{4\pi d^2}\) to find \(d\) in meters (\(3.43 \times 10^{18}\text{ m}\)) [1]
\bullet\ Division by \(9.46 \times 10^{15}\) [1]
\bullet\ \(360\text{ ly}\) (or \(363\text{ ly}\)) [1]

(a) Proper length is the length of an object measured in the frame of reference in which the object is at rest.

(b) The Lorentz factor \(\gamma\) is:
\(\gamma = \frac{1}{\sqrt{1 - v^2/c^2}} = \frac{1}{\sqrt{1 - 0.60^2}} = 1.25\)
The contracted length \(L\) measured by the space station observer is:
\(L = \frac{L_0}{\gamma} = \frac{85.0}{1.25} = \frac{85.0}{1.25} = 68.0\text{ m}\).

(c) (i) In the spacecraft's rest frame, the speed of light is \(c\) and the distance is \(L_0 = 85.0\text{ m}\).
\(\Delta t_0 = \frac{L_0}{c} = \frac{85.0}{3.00 \times 10^8} \approx 2.833 \times 10^{-7}\text{ s}\) (or \(283\text{ ns}\)).
(ii) In the space station frame, the spacecraft moves forward at speed \(v = 0.60c\). The light pulse must cover the contracted length of the spacecraft \(L\) plus the distance the front of the spacecraft moves during the interval:
\(c \Delta t = L + v \Delta t\)
\((c - v) \Delta t = L\)
\(\Delta t = \frac{L}{c - v} = \frac{68.0}{c - 0.60c} = \frac{68.0}{0.40c} = \frac{170}{3.00 \times 10^8} \approx 5.67 \times 10^{-7}\text{ s}\) (or \(567\text{ ns}\)).
[Alternatively, using Lorentz transformation: \(t = \gamma (t' + \frac{v x'}{c^2})\) where \(t' = 2.833 \times 10^{-7}\text{ s}\) and \(x' = 85.0\text{ m}\). This yields \(t = 1.25 \times (2.833 \times 10^{-7} + 1.70 \times 10^{-7}) = 5.67 \times 10^{-7}\text{ s}\)].

(a) [1 mark]
\bullet\ The length measured in the rest frame of the object [1]

(b) [2 marks]
\bullet\ Correct calculation of \(\gamma = 1.25\) [1]
\bullet\ \(68.0\text{ m}\) [1]

(c) [4 marks]
(i) \bullet\ Expresses time as \(L_0 / c\) [1]
\bullet\ \(2.83 \times 10^{-7}\text{ s}\) (or \(2.8 \times 10^{-7}\text{ s}\)) [1]
(ii) \bullet\ Correct formulation of frame movement \((c - v) \Delta t = L\) OR Lorentz transformation equation [1]
\bullet\ \(5.67 \times 10^{-7}\text{ s}\) (or \(5.7 \times 10^{-7}\text{ s}\)) [1]

(a) A standard candle is an astronomical object with a known luminosity.
Cepheid variables are useful because they exhibit a well-defined relation between their period of pulsation and their average luminosity (the Period-Luminosity relationship).
By measuring the pulsation period of a Cepheid, astronomers can determine its luminosity. Comparing this known luminosity to its measured apparent brightness allows them to calculate the distance using the inverse-square law \(b = \frac{L}{4\pi d^2}\).

(b) First, calculate the absolute luminosity \(L\) in Watts:
\(L = 3.2 \times 10^4 \times 3.85 \times 10^{26}\text{ W} = 1.232 \times 10^{31}\text{ W}\)
Now use the apparent brightness formula:
\(b = \frac{L}{4\pi d^2}\)
\(1.6 \times 10^{-14} = \frac{1.232 \times 10^{31}}{4\pi d^2}\)
\(d^2 = \frac{1.232 \times 10^{31}}{4\pi \times 1.6 \times 10^{-14}} = 6.1275 \times 10^{43}\text{ m}^2\)
\(d = \sqrt{6.1275 \times 10^{43}} \approx 7.828 \times 10^{21}\text{ m}\)
Convert to parsecs:
\(d = \frac{7.828 \times 10^{21}}{3.09 \times 10^{16}}\text{ pc} \approx 2.533 \times 10^5\text{ pc}\)
To 2 significant figures, the distance is \(2.5 \times 10^5\text{ pc}\).

(a) [3 marks]
\bullet\ Defines standard candle as an object of known luminosity [1]
\bullet\ Identifies relationship between period and luminosity for Cepheid variables [1]
\bullet\ Explains how measured apparent brightness and calculated luminosity give distance [1]

(b) [4 marks]
\bullet\ Calculates \(L = 1.23 \times 10^{31}\text{ W}\) [1]
\bullet\ Substitutes into inverse-square equation correctly [1]
\bullet\ Obtains \(d = 7.8 \times 10^{21}\text{ m}\) [1]
\bullet\ Convert and state distance as \(2.5 \times 10^5\text{ pc}\) [1]