2023 AQA IAS-Level Physics (9630) 模擬試題連答案詳解

(a) The time period of a mass-spring system is given by T = 2\pi \sqrt{m/k}. Substituting the values: T = 2\pi \sqrt{0.350/14.0} = 2\pi \sqrt{0.025} \approx 0.9935 s. The frequency f = 1/T = 1/0.9935 \approx 1.01 Hz.
(b) The maximum speed v_{max} = \omega A = 2\pi f A. Using the angular frequency \omega = \sqrt{k/m} = \sqrt{14.0/0.350} = \sqrt{40} \approx 6.325 rad s^{-1}. Therefore, v_{max} = 6.325 \times 0.085 \approx 0.538 m s^{-1}.
(c) The total energy of the system is E_{total} = 0.5 k A^2. When kinetic energy is equal to potential energy, the total energy is twice the potential energy: E_{total} = 2 E_p \Rightarrow 0.5 k A^2 = 2 (0.5 k x^2) \Rightarrow A^2 = 2 x^2. Thus, x = A / \sqrt{2} = 0.085 / \sqrt{2} \approx 0.060 m.

(a) [2 marks]
- Formula for T or f used correctly (1 mark)
- Final calculated frequency of 1.01 Hz (1 mark)
(b) [2 marks]
- Formula for maximum speed used correctly (1 mark)
- Correct value of 0.538 m s^{-1} or 0.54 m s^{-1} (1 mark)
(c) [2.6 marks]
- Identifies that total energy is twice the potential energy (1 mark)
- Derives or states x = A / \sqrt{2} (1 mark)
- Correct value of 0.060 m (0.6 marks)

(a) The period T is the time taken for one complete oscillation: T = 42.4 / 20 = 2.12 s.
(b) The period of a simple pendulum is T = 2\pi \sqrt{L/g}. Rearranging for g gives g = 4\pi^2 L / T^2. Substituting the values: g = 4\pi^2 \times 1.25 / 2.12^2 \approx 49.348 / 4.4944 \approx 10.98 m s^{-2} \approx 11.0 m s^{-2}.
(c) Since the formula for the period T does not depend on the mass of the bob, doubling the mass has no effect on the period of oscillation. It remains 2.12 s.

(a) [1.6 marks]
- Correct calculation of the period: 2.12 s (1.6 marks)
(b) [3 marks]
- Rearranging formula to make g the subject (1 mark)
- Correct substitution of L and T (1 mark)
- Correct value of g as 11.0 m s^{-2} (accept range 10.9 to 11.0) (1 mark)
(c) [2 marks]
- States there is no change to the period (1 mark)
- Explanation referencing that period is independent of mass (1 mark)

(a) The gravitational force acts as the centripetal force: G M m / r^2 = m v^2 / r, which simplifies to v = \sqrt{G M / r}. Substituting the values: v = \sqrt{6.67 \times 10^{-11} \times 4.80 \times 10^{24} / 1.20 \times 10^7} = \sqrt{3.2016 \times 10^{14} / 1.20 \times 10^7} = \sqrt{2.668 \times 10^7} \approx 5.165 \times 10^3 m s^{-1} \approx 5.17 \times 10^3 m s^{-1}, which is approximately 5.2 \times 10^3 m s^{-1}.
(b) The orbital period T is T = 2\pi r / v = 2\pi \times 1.20 \times 10^7 / 5.165 \times 10^3 \approx 1.4598 \times 10^4 s. Converting to hours: 1.4598 \times 10^4 / 3600 \approx 4.055 hours.
(c) The formula for orbital speed, v = \sqrt{G M / r}, does not contain the mass of the satellite. Therefore, doubling the mass of the satellite has no effect on its orbital speed.

(a) [2.6 marks]
- Setting GMm/r^2 = mv^2/r (1 mark)
- Correct substitution of G, M, and r (1 mark)
- Correctly showing value of 5.17 \times 10^3 m s^{-1} (0.6 marks)
(b) [2 marks]
- Correct formula for period T = 2\pi r / v (1 mark)
- Correct final value in hours: 4.05 hours (accept 4.0 to 4.1 hours) (1 mark)
(c) [2 marks]
- Correctly stating that there is no effect (1 mark)
- Explanation referencing that satellite mass cancels out (1 mark)

(a) For an orbit to be geostationary: 1. The orbit must be directly above the Earth's equator (equatorial orbit). 2. The satellite must rotate in the same direction as Earth's rotation (from west to east) with a period of exactly 24 hours.
(b) According to Kepler's Third Law: (T_1 / T_2)^2 = (r_1 / r_2)^3. Substituting the given values: (24.0 / 12.0)^2 = (4.22 \times 10^7 / r_2)^3 \Rightarrow 4 = (4.22 \times 10^7 / r_2)^3. Taking the cube root: 4^{1/3} \approx 1.5874 = 4.22 \times 10^7 / r_2. Rearranging gives r_2 = 4.22 \times 10^7 / 1.5874 \approx 2.658 \times 10^7 m \approx 2.66 \times 10^7 m.

(a) [2.6 marks]
- Stating the orbit must be equatorial (1.3 marks)
- Stating the period must be 24 hours / matching Earth's rotation (1.3 marks)
(b) [4 marks]
- Recalling Kepler's Third Law ratio (1 mark)
- Substituting values correctly (1 mark)
- Performing the cube root step (1 mark)
- Final correct orbital radius: 2.66 \times 10^7 m (accept 2.65 to 2.67 \times 10^7 m) (1 mark)

(a) Using s = u t + 0.5 a t^2 with u = 0: s = 0.5 a t^2 \Rightarrow 110 = 0.5 \times a \times 5.5^2 \Rightarrow 110 = 15.125 a \Rightarrow a = 110 / 15.125 \approx 7.27 m s^{-2}, which is approximately 7.3 m s^{-2}.
(b) Using v = u + a t: v = 0 + 7.2727 \times 5.5 = 40.0 m s^{-1}.
(c) Using v^2 = u^2 + 2 a_d s for the deceleration phase, where u = 40.0 m s^{-1}, v = 0, and s = 45 m: 0 = 40.0^2 + 2 \times a_d \times 45 \Rightarrow 0 = 1600 + 90 a_d \Rightarrow a_d = -1600 / 90 \approx -17.78 m s^{-2}. The magnitude of the deceleration is 17.8 m s^{-2}.

(a) [2 marks]
- Correct formula s = 0.5 a t^2 used (1 mark)
- Showing acceleration is 7.27 or 7.3 m s^{-2} (1 mark)
(b) [2 marks]
- Correct kinematic formula used (1 mark)
- Final velocity of 40 m s^{-1} (1 mark)
(c) [2.6 marks]
- Correct kinematic formula v^2 = u^2 + 2as used (1 mark)
- Substituting u = 40, v = 0, and s = 45 correctly (1 mark)
- Correct magnitude of 17.8 m s^{-2} (accept 17.7 to 18 m s^{-2}) (0.6 marks)

(a) At maximum height, v = 0. Using v^2 = u^2 + 2 a s with u = 15.0 m s^{-1} and a = -9.81 m s^{-2}: 0 = 15.0^2 + 2 \times (-9.81) \times s \Rightarrow 19.62 s = 225 \Rightarrow s = 225 / 19.62 \approx 11.47 m \approx 11.5 m.
(b) Using s = u t + 0.5 a t^2 with u = +15.0 m s^{-1}, a = -9.81 m s^{-2}, and t = 4.20 s: s = (15.0 \times 4.20) + 0.5 \times (-9.81) \times (4.20)^2 = 63.0 - 0.5 \times 9.81 \times 17.64 = 63.0 - 86.5242 = -23.52 m. This displacement means the stone landed 23.52 m below the cliff top. Thus, the height of the cliff is 23.5 m.

(a) [3 marks]
- Identifying v = 0 at maximum height (1 mark)
- Correct formula and substitution (1 mark)
- Answer of 11.5 m (accept 11.4 to 11.5 m) (1 mark)
(b) [3.6 marks]
- Correct equation of motion s = ut + 0.5at^2 selected (1 mark)
- Consistent use of signs for u and a (1 mark)
- Calculated displacement of -23.5 m (1 mark)
- Interpreted as a cliff height of 23.5 m (accept 23.5 to 24 m) (0.6 marks)

(a) The relationship between activity and decay constant is A = \lambda N. Therefore, \lambda = A / N = 8.50 \times 10^{12} / 3.40 \times 10^{18} = 2.50 \times 10^{-6} s^{-1}.
(b) The half-life is related to the decay constant by T_{1/2} = \ln(2) / \lambda = 0.69315 / 2.50 \times 10^{-6} = 2.7726 \times 10^5 s. Converting to hours: 2.7726 \times 10^5 / 3600 \approx 77.0 hours.
(c) Radioactive decay is random because: 1. We cannot predict when a specific nucleus will decay. 2. Each nucleus in the sample has an equal probability of decaying in any given time interval.

(a) [2 marks]
- Formula A = \lambda N used (1 mark)
- Correct value of 2.50 \times 10^{-6} s^{-1} (1 mark)
(b) [2.6 marks]
- Formula T_{1/2} = \ln(2) / \lambda used (1 mark)
- Correct value in seconds: 2.77 \times 10^5 s (0.6 marks)
- Correct conversion to 77.0 hours (accept 77 hours) (1 mark)
(c) [2 marks]
- Stating that we cannot predict which nucleus or when a nucleus decays (1 mark)
- Stating that the decay probability per unit time is constant (1 mark)

(a) Mean diameter d = (0.42 + 0.44 + 0.43 + 0.43) / 4 = 0.43 mm. The range is 0.44 - 0.42 = 0.02 mm. The absolute uncertainty is range / 2 = 0.02 / 2 = 0.01 mm. Thus, d = 0.43 \pm 0.01 mm.
(b) The cross-sectional area A = \pi d^2 / 4. The percentage uncertainty in A is 2 \times (percentage uncertainty in d). Percentage uncertainty in d = (0.01 / 0.43) \times 100\% \approx 2.326\%. Therefore, percentage uncertainty in A = 2 \times 2.326\% \approx 4.65\% \approx 4.7\%.
(c) Resistivity \rho = R A / L. The percentage uncertainty in \rho is the sum of the percentage uncertainties of R, A, and L. Percentage uncertainty in R = (0.2 / 12.4) \times 100\% \approx 1.61\%. Percentage uncertainty in L = (0.02 / 1.50) \times 100\% \approx 1.33\%. Total percentage uncertainty in \rho = 1.61\% + 4.65\% + 1.33\% = 7.59\% \approx 7.6\%.

(a) [2 marks]
- Calculates mean diameter as 0.43 mm (1 mark)
- Calculates absolute uncertainty as 0.01 mm (1 mark)
(b) [2 marks]
- Calculates percentage uncertainty of diameter as 2.3% (1 mark)
- Obtains percentage uncertainty of area as 4.7% (accept 4.6% to 4.7%) (1 mark)
(c) [2.6 marks]
- Calculates percentage uncertainty in R as 1.6% (0.6 marks)
- Calculates percentage uncertainty in L as 1.3% (0.6 marks)
- Correctly adds the percentage uncertainties of R, A, and L (0.7 marks)
- Obtains final percentage uncertainty in resistivity as 7.6% (accept 7.5% to 7.7%) (0.7 marks)

**(a) Calculate the maximum velocity:**
Using the equation of motion for the first stage:
\(v_{\text{max}} = u + a t_1\)
Since the van starts from rest, \(u = 0\):
\(v_{\text{max}} = 0 + 1.5\text{ m s}^{-2} \times 12.0\text{ s} = 18.0\text{ m s}^{-1}\)

**(b) Calculate the total time taken:**
- Stage 1 (Acceleration): \(t_1 = 12.0\text{ s}\)
- Stage 2 (Constant velocity): \(t_2 = \frac{s_2}{v_{\text{max}}} = \frac{300\text{ m}}{18.0\text{ m s}^{-1}} = 16.67\text{ s}\)
- Stage 3 (Deceleration): \(t_3 = 8.0\text{ s}\)

Total time \(T = t_1 + t_2 + t_3 = 12.0 + 16.67 + 8.0 = 36.67\text{ s}\)
Rounding to 3 significant figures: \(T = 36.7\text{ s}\).

**(c) Calculate the total distance travelled:**
- Distance during acceleration stage (Stage 1):
\(s_1 = \frac{u + v_{\text{max}}}{2} \times t_1 = \frac{0 + 18.0}{2} \times 12.0 = 108\text{ m}\)
- Distance during constant velocity stage (Stage 2):
\(s_2 = 300\text{ m}\)
- Distance during deceleration stage (Stage 3):
\(s_3 = \frac{v_{\text{max}} + 0}{2} \times t_3 = \frac{18.0}{2} \times 8.0 = 72\text{ m}\)

Total distance \(D = s_1 + s_2 + s_3 = 108 + 300 + 72 = 480\text{ m}\).

**(a) [2.0 Marks]**
- 1 Mark: Correct formula used (\(v = u + at\)) or substitution of values.
- 1 Mark: Correct calculation of \(18.0\text{ m s}^{-1}\) (or \(18\text{ m s}^{-1}\)) with appropriate unit.

**(b) [2.3 Marks]**
- 1.0 Mark: Calculates the time for the second stage correctly: \(t_2 = 16.7\text{ s}\) (or \(16.67\text{ s}\)).
- 1.3 Marks: Sums the three time intervals to get \(36.7\text{ s}\) (accept range \(36.6\text{ s}\) to \(37\text{ s}\) depending on intermediate rounding; 1 mark if process is correct but arithmetic error occurs).

**(c) [2.3 Marks]**
- 1.0 Mark: Correctly calculates \(s_1 = 108\text{ m}\) and \(s_3 = 72\text{ m}\) using equations of motion.
- 1.3 Marks: Correctly sums the three distances to give \(480\text{ m}\) (accept answer based on ECF from previous parts).

**(a) Calculate the value of \(g\):**
Rearranging the formula for \(g\):
\(g = \frac{2h}{t^2}\)
Substituting the measured values:
\(g = \frac{2 \times 1.450}{(0.54)^2} = \frac{2.90}{0.2916} \approx 9.945\text{ m s}^{-2}\)
Since \(t\) is given to 2 significant figures, \(g\) should be stated to 2 significant figures:
\(g \approx 9.9\text{ m s}^{-2}\).

**(b) Calculate the percentage uncertainties:**
(i) Percentage uncertainty in \(h\):
\(\% \text{ uncertainty in } h = \frac{0.002}{1.450} \times 100 \approx 0.1379\% \approx 0.14\%\)
(ii) Percentage uncertainty in \(t\):
\(\% \text{ uncertainty in } t = \frac{0.02}{0.54} \times 100 \approx 3.704\% \approx 3.7\%\)

**(c) Calculate the absolute uncertainty in \(g\):**
From the formula \(g = \frac{2h}{t^2}\), the percentage uncertainty in \(g\) is given by:
\(\% \text{ uncertainty in } g = (\% \text{ uncertainty in } h) + 2 \times (\% \text{ uncertainty in } t)\)
\(\% \text{ uncertainty in } g = 0.138\% + 2 \times 3.704\% = 0.138\% + 7.408\% = 7.546\%\)

Absolute uncertainty in \(g\):
\(\Delta g = g \times \frac{\% \text{ uncertainty in } g}{100}\)
\(\Delta g = 9.945 \times \frac{7.546}{100} \approx 0.750\text{ m s}^{-2}\)

Thus, the absolute uncertainty rounded to 1 significant figure is \(0.8\text{ m s}^{-2}\) (or \(0.75\text{ m s}^{-2}\) if 2 significant figures are retained).

**(a) [2.0 Marks]**
- 1.0 Mark: Rearranging equation to \(g = \frac{2h}{t^2}\) and correct substitution.
- 1.0 Mark: Calculation of \(9.9\text{ m s}^{-2}\) (accept \(9.95\text{ m s}^{-2}\)) with correct units.

**(b) [2.0 Marks]**
- 1.0 Mark: Percentage uncertainty in \(h = 0.14\%\) (accept \(0.138\%\)).
- 1.0 Mark: Percentage uncertainty in \(t = 3.7\%\) (accept \(3.70\%\)).

**(c) [2.6 Marks]**
- 1.0 Mark: Identification of relation for combining uncertainties: \(\% \text{ uncertainty in } g = \% \text{ in } h + 2 \times (\% \text{ in } t)\).
- 0.6 Marks: Correct calculation of total percentage uncertainty as \(7.5\%\) or \(7.55\%\) (accept \(7.6\%\) if rounded).
- 1.0 Mark: Correct absolute uncertainty of \(0.8\text{ m s}^{-2}\) (accept \(0.75\text{ m s}^{-2}\) or value consistent with ECF from (a) and (b)).

The total energy \(E\) of a simple harmonic oscillator is given by the formula:

\[E = \frac{1}{2} k A^2\]

where \(k\) is the spring constant and \(A\) is the amplitude of the oscillation.

Let the new spring constant be \(k' = 0.5k\) and the new amplitude be \(A' = 2A\).

The new total energy \(E'\) is:

\[E' = \frac{1}{2} k' (A')^2 = \frac{1}{2} (0.5k) (2A)^2\]

\[E' = \frac{1}{2} \times 0.5k \times 4A^2 = 2 \left( \frac{1}{2} k A^2 \right) = 2E\]

Therefore, the new energy is \(2E\).

1 mark for identifying the correct relationship \(E \propto k A^2\) and deducing that the new energy is \(2E\), corresponding to option C.

Let the displacement \(x\) of the particle as a function of time \(t\) be represented by:

\[x = A \cos(\omega t)\]

where \(A\) is the maximum displacement (amplitude) and \(x(0) = A\) at \(t = 0\).

We want to find the time \(t\) when the displacement is half the amplitude:

\[x = 0.5A \implies A \cos(\omega t) = 0.5A \implies \cos(\omega t) = 0.5\]

The minimum positive solution for \(\omega t\) is:

\[\omega t = \frac{\pi}{3}\]

Since \(\omega = \frac{2\pi}{T}\), we substitute this into the equation:

\[\left(\frac{2\pi}{T}\right) t = \frac{\pi}{3} \implies t = \frac{T}{6}\]

Therefore, the minimum time taken is \(\frac{T}{6}\).

1 mark for using the displacement equation and solving for the time \(t = \frac{T}{6}\), corresponding to option C.

According to Kepler's Third Law for circular orbits, the square of the orbital period \(T\) is directly proportional to the cube of the orbital radius \(r\):

\[T^2 \propto r^3\]

The mass of the satellite does not affect the orbital period.

For the first satellite: \(T_1^2 \propto r^3\).

For the second satellite: \(T_2^2 \propto (4r)^3 = 64r^3\).

Taking the ratio of the two relations:

\[\frac{T_2^2}{T_1^2} = \frac{64r^3}{r^3} = 64 \implies T_2 = \sqrt{64} T_1 = 8T\]

Therefore, the orbital period of the second satellite is \(8T\).

1 mark for applying Kepler's Third Law \(T^2 \propto r^3\) and showing that the mass of the satellite has no effect, resulting in \(8T\), corresponding to option C.

For a satellite in a stable circular orbit, the centripetal force is provided by the gravitational attraction:

\[\frac{m v^2}{R} = \frac{G M m}{R^2} \implies m v^2 = \frac{G M m}{R}\]

The kinetic energy \(E_k\) is:

\[E_k = \frac{1}{2} m v^2 = \frac{G M m}{2 R}\]

The gravitational potential energy \(E_p\) (with zero at infinity) is:

\[E_p = -\frac{G M m}{R}\]

Therefore, the ratio of kinetic energy to gravitational potential energy is:

\[\frac{E_k}{E_p} = \frac{\frac{G M m}{2 R}}{-\frac{G M m}{R}} = -0.5\]

1 mark for determining expressions for \(E_k\) and \(E_p\) and taking their ratio to find \(-0.5\), corresponding to option B.

Using the equation of motion for uniform acceleration from rest (\(u = 0\)):

\[s = \frac{1}{2} a t^2\]

For the first \(2.0\text{ s}\):

\[d_1 = \frac{1}{2} a (2.0)^2 = 2.0a\]

For the total time of the first \(4.0\text{ s}\) (from the start):

\[s_{\text{total}} = \frac{1}{2} a (4.0)^2 = 8.0a\]

The distance travelled in the second \(2.0\text{ s}\) interval is:

\[d_2 = s_{\text{total}} - d_1 = 8.0a - 2.0a = 6.0a\]

Now, calculate the ratio:

\[\frac{d_2}{d_1} = \frac{6.0a}{2.0a} = 3.0\]

1 mark for calculating the distances as functions of acceleration \(a\) and showing the ratio is \(3.0\), corresponding to option C.

We can use the kinematic equation for displacement \(s\) in a uniform gravitational field. Taking the upwards direction as positive:

- Initial velocity, \(u = +15\text{ m s}^{-1}\)
- Acceleration, \(a = -g = -9.81\text{ m s}^{-2}\)
- Time of flight, \(t = 5.0\text{ s}\)

Using:

\[s = u t + \frac{1}{2} a t^2\]

\[s = (15)(5.0) + \frac{1}{2}(-9.81)(5.0)^2\]

\[s = 75 - 4.905 \times 25 = 75 - 122.625 = -47.625\text{ m}\]

The negative sign indicates that the final position of the ball (the sea) is \(47.6\text{ m}\) below the starting position (the cliff edge). Therefore, the height of the cliff is approximately \(48\text{ m}\) (to 2 significant figures).

1 mark for using the equations of motion with consistent signs to calculate the displacement and rounding to 2 s.f., corresponding to option B.

1. **Penetrating power:**
- Alpha particles (\(\alpha\)) are stopped by a thin sheet of paper or a few centimetres of air, so they cannot penetrate the \(5\text{ mm}\) aluminium sheet.
- Beta-minus particles (\(\beta^-\)) can penetrate air but are typically stopped by a few millimetres of aluminium (usually up to \(3\text{ mm}\) to \(5\text{ mm}\)). A \(5\text{ mm}\) sheet of aluminium will stop virtually all beta-minus radiation.
- Gamma radiation (\(\gamma\)) is highly penetrating and easily passes through \(5\text{ mm}\) of aluminium with minimal absorption.

Thus, only gamma radiation emerges from the aluminium sheet.

2. **Behaviour in a magnetic field:**
- Gamma rays are uncharged electromagnetic photons, so they do not experience any magnetic force and remain undeflected.

Therefore, only gamma emerges, and it is undeflected.

1 mark for identifying that only gamma radiation penetrates \(5\text{ mm}\) of aluminium and that neutral gamma photons are not deflected in a magnetic field, corresponding to option B.

The volume \(V\) of a uniform cylinder (wire) is given by:

\[V = \pi \left(\frac{d}{2}\right)^2 L = \frac{\pi}{4} d^2 L\]

The percentage uncertainty in the volume \(V\) is:

\[\frac{\Delta V}{V} \times 100\% = \left( 2 \frac{\Delta d}{d} + \frac{\Delta L}{L} \right) \times 100\%\]

Calculate the fractional uncertainty in the diameter \(d\):

\[\frac{\Delta d}{d} = \frac{0.01}{0.40} = 0.025 \quad (2.5\%)\]

Calculate the fractional uncertainty in the length \(L\) (ensuring units are consistent, here both values are in cm, so they cancel out):

\[\frac{\Delta L}{L} = \frac{0.2}{80.0} = 0.0025 \quad (0.25\%)\]

Now, sum the percentage uncertainties, doubling the contribution from \(d\):

\[\% \text{ uncertainty in } V = 2(2.5\%) + 0.25\% = 5.0\% + 0.25\% = 5.25\%\]

Rounding to two significant figures gives \(5.3\%\).

1 mark for correctly applying the rule for combining uncertainties in products/powers and calculating the value of \(5.3\%\), corresponding to option D.

The relationship between speed \(v\) and displacement \(x\) in simple harmonic motion is given by \(v = \omega \sqrt{A^2 - x^2}\). The maximum speed is reached when \(x = 0\), which gives \(v_{\text{max}} = \omega A\). When the displacement is \(x = \frac{1}{2}A\), we substitute this into the equation: \(v = \omega \sqrt{A^2 - \left(\frac{1}{2}A\right)^2} = \omega \sqrt{A^2 - \frac{1}{4}A^2} = \omega \sqrt{\frac{3}{4}A^2} = \frac{\sqrt{3}}{2}\omega A\). Since \(v_{\text{max}} = \omega A\), this simplifies to \(v = \frac{\sqrt{3}}{2}v_{\text{max}}\).

B is the correct option. Award 1 mark for the correct selection. No partial marks.

By Kepler's Third Law, the square of the orbital period \(T\) is proportional to the cube of the orbital radius \(R\): \(T^2 \propto R^3\). Therefore, \(\left(\frac{T_{\text{new}}}{T_{\text{old}}}\right)^2 = \left(\frac{4R}{R}\right)^3 = 4^3 = 64\). Taking the square root of both sides gives \(\frac{T_{\text{new}}}{T_{\text{old}}} = \sqrt{64} = 8\). Thus, the ratio of the new orbital period to the original orbital period is 8.

C is the correct option. Award 1 mark for the correct selection. No partial marks.

Using the equation of motion for constant acceleration, \(s = ut + \frac{1}{2}at^2\). Taking the upward direction as positive: displacement \(s = -48\text{ m}\), initial velocity \(u = +15\text{ m s}^{-1}\), and acceleration \(a = -g = -9.81\text{ m s}^{-2}\). This gives \(-48 = 15t - 4.905t^2\), which simplifies to the quadratic equation \(4.905t^2 - 15t - 48 = 0\). Solving for \(t\) using the quadratic formula: \(t = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(4.905)(-48)}}{2(4.905)}\), which simplifies to \(t = \frac{15 \pm \sqrt{225 + 941.76}}{9.81} = \frac{15 \pm 34.16}{9.81}\). Since time must be positive, \(t = \frac{49.16}{9.81} \approx 5.0\text{ s}\).

C is the correct option. Award 1 mark for the correct selection. No partial marks.

Let the initial activity of Y be \(A_0\). The initial activity of X is therefore \(4A_0\). After 16 hours: For isotope X (half-life 4.0 hours), 16 hours represents exactly 4 half-lives. The activity becomes \(4A_0 \times \left(\frac{1}{2}\right)^4 = 4A_0 \times \frac{1}{16} = \frac{1}{4}A_0\). For isotope Y (half-life 8.0 hours), 16 hours represents exactly 2 half-lives. The activity becomes \(A_0 \times \left(\frac{1}{2}\right)^2 = \frac{1}{4}A_0\). The ratio of the activity of X to the activity of Y after 16 hours is \(\frac{\frac{1}{4}A_0}{\frac{1}{4}A_0} = 1\), which is a ratio of 1:1.

A is the correct option. Award 1 mark for the correct selection. No partial marks.

The formula for resistivity is \(\rho = \frac{RA}{L} = \frac{R \pi d^2}{4 L}\). The fractional uncertainty in \(\rho\) is given by: \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\). Calculating the individual percentage uncertainties: \(\frac{\Delta R}{R} \times 100\% = \frac{0.1}{4.0} \times 100\% = 2.5\%\), \(\frac{\Delta d}{d} \times 100\% = \frac{0.01}{0.50} \times 100\% = 2.0\%\) (so \(2\frac{\Delta d}{d} = 4.0\%\)), \(\frac{\Delta L}{L} \times 100\% = \frac{0.02}{1.20} \times 100\% \approx 1.67\%\). Adding these together: \(2.5\% + 4.0\% + 1.67\% = 8.17\%\), which rounds to \(8.2\%\).

C is the correct option. Award 1 mark for the correct selection. No partial marks.

From Einstein's photoelectric equation, the maximum kinetic energy is \(E_{\text{k,max}} = hf - \phi = \frac{hc}{\lambda} - \phi\), where \(\phi\) is the work function of the metal. If the wavelength is halved to \(\frac{\lambda}{2}\), the new maximum kinetic energy is \(E_{\text{k,max}}' = \frac{hc}{\lambda/2} - \phi = 2\frac{hc}{\lambda} - \phi\). We can express this as: \(E_{\text{k,max}}' = 2\left(E_{\text{k,max}} + \phi\right) - \phi = 2E_{\text{k,max}} + \phi\). Since the work function \(\phi\) is a positive quantity, \(E_{\text{k,max}}' > 2E_{\text{k,max}}\).

B is the correct option. Award 1 mark for the correct selection. No partial marks.

(a) The angular frequency is given by \(\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{42}{0.35}} = \sqrt{120} \approx 10.95\text{ rad s}^{-1}\). The maximum acceleration is \(a_{\max} = \omega^2 A = 120 \times 0.080 = 9.6\text{ m s}^{-2}\).

(b) Using the displacement equation for a system starting at maximum displacement: \(x = A \cos(\omega t)\). Substituting \(x = 0.040\text{ m}\) and \(A = 0.080\text{ m}\) gives \(0.040 = 0.080 \cos(10.95 t)\), so \cos(10.95 t) = 0.5\). This yields \(10.95 t = \frac{\pi}{3}\), leading to \(t = \frac{\pi}{3 \times 10.95} \approx 0.0956\text{ s}\) (or \(0.096\text{ s}\)).

(a) [2 Marks] - Award 1 mark for correct calculation of \(\omega^2 = 120\) or \(\omega = 10.95\text{ rad s}^{-1}\). Award 1 mark for calculating \(a_{\max} = 120 \times 0.080 = 9.6\text{ m s}^{-2}\) with clear working.

(b) [3.5 Marks] - Award 1 mark for identifying the correct SHM equation \(x = A \cos(\omega t)\). Award 1 mark for substituting the values correctly: \(0.04 = 0.08 \cos(10.95 t)\). Award 1 mark for calculating \(t = 0.096\text{ s}\) (accept range 0.095 to 0.096). Award 0.5 marks for correct unit (s) and sensible significant figures (2 or 3 sf).

(a) The time period of a simple pendulum is given by \(T = 2\pi \sqrt{\frac{L}{g}}\). For \(L = 1.40\text{ m}\) and \(g = 9.81\text{ m s}^{-2}\): \(T = 2\pi \sqrt{\frac{1.40}{9.81}} \approx 2.374\text{ s}\). The frequency \(f = \frac{1}{T} = \frac{1}{2.374} \approx 0.421\text{ Hz}\).

(b) When accelerating upwards at \(a = 1.25\text{ m s}^{-2}\), the effective acceleration of free fall inside the elevator increases to \(g' = g + a = 9.81 + 1.25 = 11.06\text{ m s}^{-2}\). The new period of oscillation is \(T' = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{1.40}{11.06}} \approx 2.236\text{ s}\) (or \(2.24\text{ s}\)).

(a) [2 Marks] - Award 1 mark for using \(T = 2\pi \sqrt{L/g}\) to find \(T = 2.37\text{ s}\). Award 1 mark for \(f = 0.42\text{ Hz}\) (or \(0.421\text{ Hz}\)).

(b) [3.5 Marks] - Award 1.5 marks for calculating the effective acceleration \(g' = 11.06\text{ m s}^{-2}\) (award 0.5 marks for attempt to add, 1 mark for correct calculation). Award 1 mark for substituting \(g'\) into the period equation. Award 1 mark for final value of \(2.24\text{ s}\) (accept 2.23 - 2.24).

(a) The orbital radius \(r\) is the sum of Mars' radius and the altitude: \(r = R + h = 3.39 \times 10^6 + 2.50 \times 10^5 = 3.64 \times 10^6\text{ m}\). Equating centripetal force and gravitational force: \(\frac{m v^2}{r} = \frac{G M m}{r^2} \Rightarrow v = \sqrt{\frac{G M}{r}}\). Substituting values: \(v = \sqrt{\frac{6.67 \times 10^{-11} \times 6.42 \times 10^{23}}{3.64 \times 10^6}} \approx 3430\text{ m s}^{-1} \approx 3.4 \times 10^3\text{ m s}^{-1}\).

(b) The orbital period is \(T = \frac{2\pi r}{v} = \frac{2\pi \times 3.64 \times 10^6}{3430} \approx 6668\text{ s}\). Converting to hours: \(T = \frac{6668}{3600} \approx 1.85\text{ hours}\).

(a) [2 Marks] - Award 1 mark for calculating the correct orbit radius \(r = 3.64 \times 10^6\text{ m}\). Award 1 mark for substituting into \(v = \sqrt{GM/r}\) to show \(3.43 \times 10^3\text{ m s}^{-1}\).

(b) [3.5 Marks] - Award 1 mark for using \(T = 2\pi r / v\) (or equivalent Kepler's 3rd law calculation). Award 1 mark for calculating \(T \approx 6670\text{ s}\). Award 1 mark for dividing by 3600 to convert to hours. Award 0.5 marks for correct final answer of \(1.85\text{ hours}\) (accept 1.8 to 1.9).

(a) Kepler's Third Law states that \(T^2 \propto r^3\) (or \(\frac{T^2}{r^3} = \text{constant}\)).

(b) Using the relationship: \(\frac{T_B^2}{T_A^2} = \frac{r_B^3}{r_A^3}\). Therefore, \(r_B = r_A \left( \frac{T_B}{T_A} \right)^{2/3}\). Substituting the given values: \(r_B = 1.20 \times 10^7 \times \left( \frac{12.0}{4.50} \right)^{2/3} = 1.20 \times 10^7 \times (2.667)^{2/3} \approx 2.31 \times 10^7\text{ m}\).

(c) The orbital speed of a satellite is given by \(v = \sqrt{\frac{GM}{r}}\). Since satellite B has a larger orbital radius than satellite A (\(r_B > r_A\)), its orbital speed \(v_B\) must be lower than the orbital speed \(v_A\).

(a) [1 Mark] - State \(T^2 \propto r^3\) or equivalent equation.

(b) [2.5 Marks] - Award 1 mark for setting up the ratio equation \(r_B^3 / r_A^3 = T_B^2 / T_A^2\). Award 1 mark for correct substitution and algebra. Award 0.5 marks for the final correct answer of \(2.31 \times 10^7\text{ m}\) (accept \(2.3 \times 10^7\text{ m}\)).

(c) [2 Marks] - Award 1 mark for stating that speed \(v\) is inversely proportional to \(\sqrt{r}\) (or \(v = \sqrt{GM/r}\)). Award 1 mark for stating that satellite B has a lower speed because it has a larger radius.

(a) The distance traveled during the reaction time (thinking distance) is \(d_1 = v \times t = 24.0 \times 0.45 = 10.8\text{ m}\). The total stopping distance is the sum of thinking distance and braking distance: \(d_{\text{total}} = 10.8 + 48.0 = 58.8\text{ m}\).

(b) Using the equation of motion \(v^2 = u^2 + 2as\) for the braking phase, where final velocity \(v = 0\), initial velocity \(u = 24.0\text{ m s}^{-1}\), and distance \(s = 48.0\text{ m}\): \(0 = 24.0^2 + 2(a)(48.0)\), which gives \(96.0 a = -576\). Thus, \(a = -6.0\text{ m s}^{-2}\). The magnitude of the deceleration is \(6.0\text{ m s}^{-2}\).

(a) [2.5 Marks] - Award 1 mark for calculating thinking distance: \(10.8\text{ m}\). Award 1 mark for adding thinking and braking distances. Award 0.5 marks for correct final answer of \(58.8\text{ m}\).

(b) [3 Marks] - Award 1 mark for identifying the formula \(v^2 = u^2 + 2as\). Award 1 mark for correct substitution: \(0 = 576 + 96a\). Award 1 mark for correct final magnitude of deceleration: \(6.0\text{ m s}^{-2}\) (accept \(-6.0\text{ m s}^{-2}\) if clear they mean magnitude is 6.0).

(a) At maximum height, the vertical velocity is \(0\). Using \(v^2 = u^2 + 2as\) with \(u = 15.0\text{ m s}^{-1}\) and \(a = -9.81\text{ m s}^{-2}\): \(0 = 15.0^2 - 2(9.81)h\), giving \(19.62 h = 225\), hence \(h \approx 11.47\text{ m}\) (or \(11.5\text{ m}\)).

(b) Taking upwards as positive, we can find the overall displacement \(s\) at \(t = 4.20\text{ s}\) using \(s = ut + \frac{1}{2}at^2\): \(s = (15.0 \times 4.20) + 0.5 \times (-9.81) \times (4.20)^2 = 63.0 - 86.52 = -23.52\text{ m}\). The negative sign indicates the ball is below the starting point, so the height of the cliff is \(23.5\text{ m}\).

(a) [2.5 Marks] - Award 1 mark for identifying \(v = 0\) at maximum height and selecting \(v^2 = u^2 + 2as\). Award 1 mark for substituting values correctly: \(15^2 = 2 \times 9.81 \times h\). Award 0.5 marks for final answer of \(11.5\text{ m}\) (or \(11.47\text{ m}\)).

(b) [3 Marks] - Award 1.5 marks for using \(s = ut + \frac{1}{2}at^2\) with signs handled correctly (upwards positive, acceleration negative). Award 1 mark for calculating displacement \(s = -23.5\text{ m}\). Award 0.5 marks for expressing cliff height as a positive value \(23.5\text{ m}\) (accept range 23.5 to 23.6).

(a) Using the activity equation \(A = A_0 e^{-\lambda t}\), we get \(4.0 \times 10^4 = 3.2 \times 10^5 e^{-\lambda \times 12.0}\). This simplifies to \(e^{-12\lambda} = 0.125\). Taking the natural log on both sides: \(-12\lambda = \ln(0.125) = -2.079\). Thus, \(\lambda = \frac{2.079}{12.0} \approx 0.173\text{ hour}^{-1}\), which is approximately \(0.17\text{ hour}^{-1}\).

(b) The half-life \(T_{1/2}\) is given by \(T_{1/2} = \frac{\ln(2)}{\lambda} = \frac{0.693}{0.1733} = 4.0\text{ hours}\). Converting to minutes: \(4.0 \times 60 = 240\text{ minutes}\).

(c) Physical differences include: 1. An alpha particle is a helium nucleus (mass \(\approx 4\text{ u}\), charge \(+2e\)), whereas a beta-minus particle is an electron (mass \(\approx 1/1840\text{ u}\), charge \(-1e\)). 2. Alpha particles have much higher ionizing power but lower penetration than beta-minus particles.

(a) [2 Marks] - Award 1 mark for using \(A = A_0 e^{-\lambda t}\) and substituting values correctly. Award 1 mark for showing \(\lambda = 0.173\text{ hour}^{-1}\) clearly via logarithms.

(b) [2 Marks] - Award 1 mark for calculating \(T_{1/2} = 4.0\text{ hours}\) (using either \(\ln(2)/\lambda\) or noticing that the activity drops to exactly 1/8th, which represents 3 half-lives). Award 1 mark for converting hours to minutes (\(240\text{ minutes}\)).

(c) [1.5 Marks] - Award 0.5 marks for each distinct valid physical difference (up to 1.5 marks). Acceptable differences: mass, charge, ionizing power, penetrating power, or constituent composition.

(a) The cross-sectional area \(A = \frac{\pi d^2}{4}\). The percentage uncertainty in \(A\) is twice the percentage uncertainty in diameter \(d\):
\%\text{ uncertainty in } d = \frac{0.02}{0.40} \times 100\% = 5\%\).
Therefore, \(\%\text{ uncertainty in } A = 2 \times 5\% = 10\%\).

(b) First calculate the resistivity \(\rho = \frac{R A}{L}\):
\(A = \frac{\pi \times (0.40 \times 10^{-3})^2}{4} = 1.257 \times 10^{-7}\text{ m}^2\).
\(\rho = \frac{4.8 \times 1.257 \times 10^{-7}}{1.25} = 4.83 \times 10^{-7}\ \Omega\text{ m}\).
Next, sum the percentage uncertainties:
\(\%\text{ uncertainty in } R = \frac{0.2}{4.8} \times 100\% \approx 4.17\%\).
\(\%\text{ uncertainty in } L = \frac{0.01}{1.25} \times 100\% = 0.8\%\).
\(\%\text{ uncertainty in } \rho = \%\text{ in } R + \%\text{ in } A + \%\text{ in } L = 4.17\% + 10\% + 0.8\% = 14.97\% \approx 15\%\).
Absolute uncertainty in \(\rho = 14.97\% \times 4.83 \times 10^{-7} \approx 0.72 \times 10^{-7}\ \Omega\text{ m} \approx 0.7 \times 10^{-7}\ \Omega\text{ m}\).
Thus, \(\rho = (4.8 \pm 0.7) \times 10^{-7}\ \Omega\text{ m}\).

(a) [2 Marks] - Award 1 mark for finding percentage uncertainty in \(d\) is 5%. Award 1 mark for doubling to find 10% for \(A\).

(b) [3.5 Marks] - Award 1 mark for calculating central value of resistivity: \(4.8 \times 10^{-7}\ \Omega\text{ m}\). Award 1 mark for summing the percentage uncertainties (\(4.17\% + 10\% + 0.8\% = 15\%\)). Award 1 mark for calculating absolute uncertainty: \(0.7 \times 10^{-7}\ \Omega\text{ m}\). Award 0.5 marks for stating the final values with correct units (\(\Omega\text{ m}\)) and appropriate precision.

(a) Using the double-slit interference formula:
\(w = \frac{\lambda D}{s}\)

Substituting the given values into the equation:
\(w = \frac{6.33 \times 10^{-7}\text{ m} \times 1.80\text{ m}}{0.250 \times 10^{-3}\text{ m}}\)
\(w = 4.5576 \times 10^{-3}\text{ m} = 4.56\text{ mm}\)

This is approximately \(4.6\text{ mm}\), as required.

(b) The distance between the central maximum and the fourth-order bright fringe is equal to four fringe spacings (\(4w'\)).
\(4w' = 14.4\text{ mm} = 1.44 \times 10^{-2}\text{ m}\)
\(w' = \frac{14.4\text{ mm}}{4} = 3.60\text{ mm} = 3.60 \times 10^{-3}\text{ m}\)

Using the formula for the new wavelength \(\lambda'\):
\(\lambda' = \frac{w' s}{D}\)
\(\lambda' = \frac{3.60 \times 10^{-3}\text{ m} \times 0.250 \times 10^{-3}\text{ m}}{1.80\text{ m}}\)
\(\lambda' = 5.00 \times 10^{-7}\text{ m}\) (or \(500\text{ nm}\))

(c) If the width of each slit is increased slightly:
- The fringes become brighter (as more light intensity passes through the wider slits).
- The overall width of the diffraction envelope decreases, so fewer fringes are visible within the central maximum.

(a)
- [1 mark] Correct substitution into the fringe spacing formula with appropriate unit conversions (e.g., \(0.250 \times 10^{-3}\text{ m}\)): \(w = \frac{6.33 \times 10^{-7} \times 1.80}{0.250 \times 10^{-3}}\).
- [1 mark] Obtains a calculated value of \(4.56\text{ mm}\) (or \(4.558\text{ mm}\)) and links it to the target value of \(4.6\text{ mm}\).

(b)
- [1 mark] Recognises that the distance to the fourth-order bright fringe is \(4w'\), obtaining \(w' = 3.60\text{ mm}\) (or writes a correct algebraic expression for \(\lambda'\) such as \(\lambda' = \frac{y_4 s}{4 D}\)).
- [1 mark] Obtains the correct final answer: \(\lambda' = 5.00 \times 10^{-7}\text{ m}\) (or \(5.0 \times 10^{-7}\text{ m}\) or \(500\text{ nm}\)).

(c)
- [1 mark] States that the fringes become brighter OR that fewer fringes are visible / the central diffraction envelope becomes narrower. Do not accept answers saying that the fringe spacing changes.

(a) The student should open the switch/disconnect the circuit between taking readings. This prevents a continuous current which would cause the cell to warm up, and internal resistance of cells varies significantly with temperature.

(b) By comparing the equation \(V = -r I + \varepsilon\) to the equation of a straight line \(y = mx + c\):
- The vertical intercept (y-intercept) is equal to the electromotive force \(\varepsilon\).
- The gradient of the line is equal to \(-r\), meaning the internal resistance \(r\) is the magnitude of the gradient.

(c)(i) The y-intercept represents \(\varepsilon\).
\(\varepsilon = 1.48 \pm 0.02\text{ V}\)

(c)(ii) The internal resistance is represented by the magnitude of the gradient:
\(r = 0.68\ \Omega\)
The absolute uncertainty in \(r\) is \(0.04\ \Omega\).
Percentage uncertainty in \(r\) = \(\frac{0.04}{0.68} \times 100\% \approx 5.88\%\) (accept \(5.9\%\) or \(6\%\)).

(a)
- Award 1 mark for stating that the circuit should be switched off/disconnected between taking readings.
- Award 1 mark for stating this prevents the cell from heating up / keeps the temperature constant (since resistance changes with temperature).

(b)
- Award 1 mark for identifying that the y-intercept corresponds to the emf \(\varepsilon\).
- Award 1 mark for identifying that the internal resistance \(r\) is the magnitude of the gradient (or gradient = \(-r\)).

(c)(i)
- Award 1 mark for the value \(\varepsilon = 1.48\text{ V}\).
- Award 1 mark for the correct absolute uncertainty of \(\pm 0.02\text{ V}\).

(c)(ii)
- Award 1 mark for \(r = 0.68\ \Omega\) (ignore sign if negative is given, but unit must be correct).
- Award 1 mark for percentage uncertainty \(= 5.9\%\) (accept \(5.88\%\) or \(6\%\)).

(a) Mean diameter \(d\):
\[d = \frac{0.41 + 0.43 + 0.42 + 0.42 + 0.42}{5} = 0.42\text{ mm}\]
- Taking measurements at different positions is necessary to account for any variation in the wire's cross-sectional area along its length.
- Taking measurements at different orientations is necessary because the cross-section of the wire may not be perfectly circular (it might be elliptical).

(b)
- A metre ruler is suitable for \(L_0\) because the length is large (around 2 m), so the percentage uncertainty is extremely small (\(\approx 0.1\%\)).
- However, the extension \(\Delta L\) is very small (around 1.8 mm). A metre ruler's resolution (typically 1 mm) would lead to an unacceptably large percentage uncertainty (\(\approx 50\%\)).
- A suitable instrument to measure \(\Delta L\) is a travelling microscope, vernier scale, or extensometer.

(c) First, calculate the cross-sectional area \(A\):
\[A = \frac{\pi d^2}{4} = \frac{\pi (0.42 \times 10^{-3}\text{ m})^2}{4} = 1.3854 \times 10^{-7}\text{ m}^2\]

Now, calculate the Young modulus \(E\):
\[E = \frac{F L_0}{A \Delta L} = \frac{25.0 \times 2.050}{1.3854 \times 10^{-7} \times 1.8 \times 10^{-3}} = 2.057 \times 10^{11}\text{ Pa}\]

Since the extension (1.8 mm) and diameter (0.42 mm) are given to 2 significant figures, the final answer should be rounded to 2 significant figures:
\[E = 2.1 \times 10^{11}\text{ Pa} \quad (\text{or } \text{N m}^{-2})\]

(a)
- Award 1 mark for correct mean diameter of \(0.42\text{ mm}\).
- Award 1 mark for explaining that multiple positions account for non-uniformity of thickness/diameter along the length.
- Award 1 mark for explaining that multiple orientations account for non-circularity / oval shape of the cross-section.

(b)
- Award 1 mark for explaining that a metre ruler has an insignificantly low percentage uncertainty for large \(L_0\) but would give an unacceptably large percentage uncertainty for small \(\Delta L\).
- Award 1 mark for naming a travelling microscope, vernier scale, micrometer, or extensometer.

(c)
- Award 1 mark for calculating correct area \(A = 1.39 \times 10^{-7}\text{ m}^2\) (allow rounding in steps).
- Award 1 mark for substitution of values into the correct formula: \(E = \frac{F L_0}{A \Delta L}\).
- Award 1 mark for the final answer of \(2.1 \times 10^{11}\text{ Pa}\) (or \(\text{N m}^{-2}\)) quoted to 2 s.f. (accept \(2.06 \times 10^{11}\text{ Pa}\) if 3 s.f. is used, but penalise incorrect powers of 10 or missing/incorrect units).

The total energy \( E_T \) in simple harmonic motion is proportional to the square of the amplitude: \( E_T = \frac{1}{2} k A^2 \). At displacement \( x = 0.60 A \), the potential energy \( E_P \) is: \( E_P = \frac{1}{2} k x^2 = \frac{1}{2} k (0.60 A)^2 = 0.36 \left(\frac{1}{2} k A^2\right) = 0.36 E_T \). The kinetic energy \( E_K \) is: \( E_K = E_T - E_P = E_T - 0.36 E_T = 0.64 E_T \). Therefore, the ratio of kinetic energy to potential energy is: \( \frac{E_K}{E_P} = \frac{0.64 E_T}{0.36 E_T} = \frac{16}{9} \approx 1.78 \).

1 mark for the correct answer (D).

The time period of a mass-spring system is given by \( T = 2\pi \sqrt{\frac{m}{k}} \). When the mass becomes \( 2m \) and the spring constant becomes \( \frac{k}{2} \), the new period \( T' \) is: \( T' = 2\pi \sqrt{\frac{2m}{k/2}} = 2\pi \sqrt{\frac{4m}{k}} = 2 \left( 2\pi \sqrt{\frac{m}{k}} \right) = 2T \).

1 mark for the correct answer (D).

For a satellite in a circular orbit of radius \( r \), the gravitational force provides the centripetal force: \( \frac{G M m}{r^2} = \frac{m v^2}{r} \implies v = \sqrt{\frac{GM}{r}} \). Thus, the orbital speed \( v \) is inversely proportional to the square root of the orbital radius \( r \): \( v \propto \frac{1}{\sqrt{r}} \). Therefore, the ratio is: \( \frac{v_P}{v_Q} = \sqrt{\frac{r_Q}{r_P}} = \sqrt{\frac{4r_P}{r_P}} = \sqrt{4} = 2.0 \).

1 mark for the correct answer (C).

Using the equation of motion \( s = ut + \frac{1}{2}at^2 \) with \( u = 0 \): For the first 2.0 seconds: \( d_1 = \frac{1}{2} a (2.0)^2 = 2.0 a \). For the first 4.0 seconds, the total distance traveled is: \( D = \frac{1}{2} a (4.0)^2 = 8.0 a \). The distance traveled in the next 2.0 seconds is: \( d_2 = D - d_1 = 8.0 a - 2.0 a = 6.0 a \). Therefore, the ratio is: \( \frac{d_2}{d_1} = \frac{6.0 a}{2.0 a} = 3 \).

1 mark for the correct answer (C).

After a time interval of \( 3 T_{1/2} \) (three half-lives), the fraction of the original active nuclei remaining is: \( \left(\frac{1}{2}\right)^3 = \frac{1}{8} \). The fraction of active nuclei that has decayed is: \( 1 - \frac{1}{8} = \frac{7}{8} \).

1 mark for the correct answer (D).

The volume of the wire is given by \( V = \pi r^2 L = \frac{\pi d^2 L}{4} \). The percentage uncertainty in the diameter \( d \) is: \( \frac{\Delta d}{d} \times 100\% = \frac{0.02}{0.40} \times 100\% = 5.0\% \). The percentage uncertainty in the length \( L \) is: \( \frac{\Delta L}{L} \times 100\% = \frac{0.1}{100.0} \times 100\% = 0.1\% \). Since \( V \propto d^2 L \), the percentage uncertainty in \( V \) is: \( \frac{\Delta V}{V} \times 100\% = 2 \left(\frac{\Delta d}{d} \times 100\%\right) + \left(\frac{\Delta L}{L} \times 100\%\right) = 2(5.0\%) + 0.1\% = 10.1\% \).

1 mark for the correct answer (B).

First, calculate the mean of the measurements: \( \text{Mean} = \frac{1.42 + 1.46 + 1.40 + 1.44}{4} = 1.43 \text{ s} \). The uncertainty in the mean can be estimated as half the range of the measurements: \( \text{Range} = 1.46 - 1.40 = 0.06 \text{ s} \). \( \text{Uncertainty} = \frac{\text{Range}}{2} = \frac{0.06}{2} = 0.03 \text{ s} \). Thus, the period of oscillation is \( (1.43 \pm 0.03) \text{ s} \).

1 mark for the correct answer (B).

According to Kepler's Third Law, the square of the orbital period \( T \) is proportional to the cube of the orbital radius \( r \): \( T^2 \propto r^3 \). Let \( T_E \) and \( r_E \) be the period and radius of Earth's orbit, and \( T_p \) and \( r_p \) be those of the planet. We have: \( \left(\frac{T_p}{T_E}\right)^2 = \left(\frac{r_p}{r_E}\right)^3 \). Since \( \frac{r_p}{r_E} = 9.0 \) and \( T_E = 1.0 \text{ Earth year} \): \( T_p^2 = (9.0)^3 = 729 \). Taking the square root: \( T_p = \sqrt{729} = 27 \text{ Earth years} \).

1 mark for the correct answer (C).

The total energy of the oscillating system is proportional to the square of the amplitude: \(E_{\text{total}} = \frac{1}{2} k A^2\).

The potential energy at a displacement \(x\) is given by: \(E_{\text{p}} = \frac{1}{2} k x^2\).

When \(x = \frac{1}{3}A\):
\(E_{\text{p}} = \frac{1}{2} k \left(\frac{1}{3}A\right)^2 = \frac{1}{18} k A^2 = \frac{1}{9} E_{\text{total}}\).

The kinetic energy is the difference between total energy and potential energy:
\(E_{\text{k}} = E_{\text{total}} - E_{\text{p}} = E_{\text{total}} - \frac{1}{9} E_{\text{total}} = \frac{8}{9} E_{\text{total}}\).

The ratio of kinetic energy to potential energy is:
\(\frac{E_{\text{k}}}{E_{\text{p}}} = \frac{\frac{8}{9} E_{\text{total}}}{\frac{1}{9} E_{\text{total}}} = 8\).

1 mark for the correct answer A.

- Correct deduction of potential energy proportion (\(1/9\) of total).
- Correct deduction of kinetic energy proportion (\(8/9\) of total).
- Ratio of \(8\) determined.

The centripetal force is provided by the gravitational attraction:
\(\frac{m v^2}{r} = \frac{G M m}{r^2}\)

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

Therefore, orbital speed is inversely proportional to the square root of the orbital radius:
\(v \propto \frac{1}{\sqrt{r}}\)

When the radius increases from \(R\) to \(3R\), the speed changes by a factor of:
\(\frac{v_{\text{new}}}{v_{\text{old}}} = \sqrt{\frac{R}{3R}} = \frac{1}{\sqrt{3}}\).

1 mark for the correct answer C.

- Uses relation \(v \propto \frac{1}{\sqrt{r}}\).
- Calculates ratio when radius is multiplied by 3 as \(\frac{1}{\sqrt{3}}\).

The velocity \(v\) is the first derivative of displacement \(s\) with respect to time \(t\):
\(v = \frac{ds}{dt} = \frac{d}{dt} (\beta t^3) = 3\beta t^2\)

The acceleration \(a\) is the derivative of velocity \(v\) with respect to time \(t\):
\(a = \frac{dv}{dt} = \frac{d}{dt} (3\beta t^2) = 6\beta t\).

1 mark for the correct answer B.

- Differentiates the displacement function once to find velocity.
- Differentiates again to find acceleration.

Let the initial activity of both isotopes at \(t = 0\) be \(A_0\).

For isotope \(X\):
The number of half-lives elapsed in \(24.0\text{ hours}\) is:
\(n_X = \frac{24.0}{4.0} = 6\)
So the activity of \(X\) is:
\(A_X = A_0 \left(\frac{1}{2}\right)^6 = \frac{A_0}{64}\)

For isotope \(Y\):
The number of half-lives elapsed in \(24.0\text{ hours}\) is:
\(n_Y = \frac{24.0}{12.0} = 2\)
So the activity of \(Y\) is:
\(A_Y = A_0 \left(\frac{1}{2}\right)^2 = \frac{A_0}{4}\)

The ratio is:
\(\frac{A_X}{A_Y} = \frac{A_0 / 64}{A_0 / 4} = \frac{4}{64} = \frac{1}{16}\).

1 mark for the correct answer A.

- Determines remaining activity of X (\(1/64\) of initial).
- Determines remaining activity of Y (\(1/4\) of initial).
- Divides the activities to get \(1/16\).

The density \(\rho\) of a sphere is given by:
\(\rho = \frac{m}{V} = \frac{m}{\frac{4}{3}\pi (d/2)^3} = \frac{6m}{\pi d^3}\)

Therefore, density depends on mass and the third power of diameter:
\(\rho \propto \frac{m}{d^3}\)

The percentage uncertainty in density is calculated by adding the percentage uncertainty in mass to three times the percentage uncertainty in diameter:
\(\%\Delta \rho = \%\Delta m + 3\%\Delta d\)

Calculate each percentage uncertainty:
\(\%\Delta m = \frac{0.1}{15.4} \times 100\% \approx 0.65\%\)
\(\%\Delta d = \frac{0.2}{12.6} \times 100\% \approx 1.59\%\)

Now, calculate the total percentage uncertainty:
\(\%\Delta \rho \approx 0.65\% + 3(1.59\%) = 0.65\% + 4.76\% = 5.41\% \approx 5.4\%\).

1 mark for the correct answer C.

- Correctly identifies that \(\%\Delta \rho = \%\Delta m + 3\%\Delta d\).
- Computes individual percentage uncertainties.
- Correctly rounds the final sum to \(5.4\%\).

The total mass of the two-block system is:
\(M = m_1 + m_2 = 2.0\text{ kg} + 3.0\text{ kg} = 5.0\text{ kg}\)

Applying Newton's second law to the system, the acceleration \(a\) is:
\(a = \frac{F}{M} = \frac{15\text{ N}}{5.0\text{ kg}} = 3.0\text{ m s}^{-2}\)

The only horizontal force acting on the block \(m_2\) is the contact force \(C\) from block \(m_1\):
\(C = m_2 a = 3.0\text{ kg} \times 3.0\text{ m s}^{-2} = 9.0\text{ N}\).

1 mark for the correct answer B.

- Finds total system acceleration of \(3.0\text{ m s}^{-2}\).
- Calculates the contact force required to accelerate \(m_2\) at this rate, giving \(9.0\text{ N}\).

(a)
Using the equation for the frequency of a mass-spring system:
\(f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\)
Substituting the values:
\(f = \frac{1}{2\pi} \sqrt{\frac{14}{0.35}} = \frac{1}{2\pi} \sqrt{40} \approx 1.007\text{ Hz}\)
Rounding to 2 significant figures gives:
\(f = 1.0\text{ Hz}\)

(b)
The angular frequency is:
\(\omega = \sqrt{\frac{k}{m}} = \sqrt{40} \approx 6.32\text{ rad s}^{-1}\)
The maximum velocity is achieved when the block passes through the equilibrium position and is given by:
\(v_{\text{max}} = \omega A\)
where \(A = 0.080\text{ m}\) is the amplitude of oscillation.
\(v_{\text{max}} = 6.32 \times 0.080 \approx 0.506\text{ m s}^{-1}\)
Rounding to 2 significant figures gives:
\(v_{\text{max}} = 0.51\text{ m s}^{-1}\)

(c)
The magnitude of the acceleration for a body executing simple harmonic motion is given by:
\(|a| = \omega^2 x\)
Substituting \(\omega^2 = \frac{k}{m} = 40\text{ s}^{-2}\) and \(x = 0.050\text{ m}\):
\(|a| = 40 \times 0.050 = 2.0\text{ m s}^{-2}\)

(a)
- C1: Correct formula for the frequency of a mass-spring system seen or implied.
- C1: Correct substitution of values: \(\sqrt{14 / 0.35}\) or \(\sqrt{40}\).
- A1: Correct answer to 2 or more significant figures: \(1.0\text{ Hz}\) or \(1.01\text{ Hz}\).

(b)
- C1: Recalls or derives \(v_{\text{max}} = \omega A\).
- C1: Correctly calculates \(\omega = 6.32\text{ rad s}^{-1}\) or substitutes \(\sqrt{40} \times 0.080\).
- A1: Correct final value for velocity: \(0.51\text{ m s}^{-1}\) (accept \(0.50\text{ m s}^{-1}\) to \(0.51\text{ m s}^{-1}\)).

(c)
- C1: Recalls or uses the definition of acceleration magnitude \(|a| = \omega^2 x\).
- C1.3: Correct substitution of values: \(40 \times 0.050\).
- A1: Correct answer with appropriate units: \(2.0\text{ m s}^{-2}\).

(a)
For a simple pendulum executing small-amplitude oscillations, the period \(T\) is:
\(T = 2\pi \sqrt{\frac{L}{g}}\)
Substitute the given values:
\(T = 2\pi \sqrt{\frac{1.25}{9.81}} = 2\pi \sqrt{0.1274} \approx 2.244\text{ s}\)
Rounding to 3 significant figures:
\(T = 2.24\text{ s}\)

(b)
The maximum linear displacement \(A\) along the arc is:
\(A = L \theta_{\text{rad}}\)
First, convert the angle to radians:
\(\theta_{\text{rad}} = 6.0^\circ \times \frac{\pi}{180^\circ} = 0.1047\text{ rad}\)
Then calculate \(A\):
\(A = 1.25 \times 0.1047 = 0.131\text{ m}\) (or using \(L \sin\theta = 1.25 \sin(6.0^\circ) = 0.131\text{ m}\)).

(c)
The maximum speed is given by:
\(v_{\text{max}} = \omega A\)
where \(\omega = \frac{2\pi}{T} = \sqrt{\frac{g}{L}} = \sqrt{\frac{9.81}{1.25}} \approx 2.80\text{ rad s}^{-1}\).
\(v_{\text{max}} = 2.80 \times 0.131 = 0.367\text{ m s}^{-1}\)
Alternatively, using conservation of energy:
\(h = L(1 - \cos\theta) = 1.25(1 - \cos 6^\circ) = 0.00685\text{ m}\)
\(v_{\text{max}} = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.00685} = 0.367\text{ m s}^{-1} \approx 0.37\text{ m s}^{-1}\).

(a)
- C1: Recalls \(T = 2\pi \sqrt{L/g}\).
- C1: Correct substitution of \(1.25\) and \(9.81\).
- A1: Correct answer: \(2.24\text{ s}\) (accept \(2.2\text{ s}\) to \(2.25\text{ s}\)).

(b)
- C1: Uses either \(A = L \theta\) with angle converted to radians, or \(A = L \sin\theta\).
- C1: Correct angle conversion: \(6.0^\circ = 0.105\text{ rad}\).
- A1: Correct linear displacement: \(0.13\text{ m}\) or \(0.131\text{ m}\).

(c)
- C1: Recalls \(v_{\text{max}} = \omega A\) or \(E_k = E_p\) method.
- C1.3: Correct calculation of \(\omega \approx 2.8\text{ rad s}^{-1}\) or height \(h \approx 0.0068\text{ m}\).
- A1: Correct speed: \(0.37\text{ m s}^{-1}\) (accept \(0.36\text{ m s}^{-1}\) to \(0.37\text{ m s}^{-1}\)).

(a)
Using Newton's law of gravitation:
\(F = \frac{G M m}{r^2}\)
Substituting the values:
\(F = \frac{(6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}) \times (4.8 \times 10^{24}\text{ kg}) \times 1200\text{ kg}}{(8.4 \times 10^6\text{ m})^2}\)
\(F = \frac{3.8419 \times 10^{17}}{7.056 \times 10^{13}} \approx 5445\text{ N}\)
Rounding to 2 significant figures gives:
\(F = 5.4 \times 10^3\text{ N}\)

(b)
For a circular orbit, the centripetal force is provided by the gravitational force:
\(\frac{m v^2}{r} = \frac{G M m}{r^2}\)
Solving for the orbital speed \(v\):
\(v = \sqrt{\frac{G M}{r}}\)
\(v = \sqrt{\frac{6.67 \times 10^{-11} \times 4.8 \times 10^{24}}{8.4 \times 10^6}} = \sqrt{\frac{3.2016 \times 10^{14}}{8.4 \times 10^6}} = \sqrt{3.811 \times 10^7} \approx 6174\text{ m s}^{-1}\)
This is approximately \(6.2 \times 10^3\text{ m s}^{-1}\), as required.

(c)
The period \(T\) of the orbit is given by:
\(T = \frac{2\pi r}{v}\)
Substituting the values:
\(T = \frac{2\pi \times 8.4 \times 10^6}{6174} \approx 8545\text{ s}\)
Converting seconds to hours:
\(T_{\text{hours}} = \frac{8545}{3600} \approx 2.37\text{ hours}\)
Rounding to 2 significant figures gives:
\(T = 2.4\text{ hours}\)

(a)
- C1: Recalls Newton's law of gravitation formula: \(F = G M m / r^2\).
- C1: Correctly substitutes values including \(G = 6.67 \times 10^{-11}\).
- A1: Evaluates to \(5400\text{ N}\) or \(5.4 \times 10^3\text{ N}\).

(b)
- C1: Equates centripetal and gravitational forces to obtain \(v^2 = G M / r\).
- C1.3: Shows intermediate step \(v = \sqrt{3.8 \times 10^7}\).
- A1: Obtains \(6.17 \times 10^3\text{ m s}^{-1}\) and states that this rounds to \(6.2 \times 10^3\text{ m s}^{-1}\).

(c)
- C1: Uses \(T = 2\pi r / v\).
- C1: Obtains period in seconds: \(8.5 \times 10^3\text{ s}\) (accept \(8.5 \times 10^3\) to \(8.6 \times 10^3\)).
- A1: Correct division by 3600 to get \(2.4\text{ hours}\) (accept \(2.37\text{ hours}\) to \(2.40\text{ hours}\)).

(a)
Using the equation of motion:
\(v = u + at\)
Substituting \(u = 15\text{ m s}^{-1}\), \(a = 2.4\text{ m s}^{-2}\), and \(t = 5.0\text{ s}\):
\(v = 15 + (2.4 \times 5.0) = 15 + 12 = 27\text{ m s}^{-1}\)

(b)
The first \(15.0\text{ s}\) consists of two phases:
1. Acceleration phase (0 to 5.0 s):
\(s_1 = ut + \frac{1}{2}at^2\)
\(s_1 = (15 \times 5.0) + 0.5 \times 2.4 \times (5.0)^2 = 75 + 30 = 105\text{ m}\)

2. Constant speed phase (5.0 s to 15.0 s, duration of 10.0 s):
\(s_2 = v \times t = 27 \times 10.0 = 270\text{ m}\)

Total distance in the first 15.0 s:
\(s_{\text{total}} = s_1 + s_2 = 105 + 270 = 375\text{ m}\)

(c)
For the braking phase, the initial speed is \(u_b = 27\text{ m s}^{-1}\), the final speed is \(v_b = 0\text{ m s}^{-1}\), and the distance is \(s_b = 45\text{ m}\).
Using the equation:
\(v_b^2 = u_b^2 + 2as_b\)
\(0 = 27^2 + 2(a)(45)\)
\(0 = 729 + 90a\)
\(90a = -729\)
\(a = -8.1\text{ m s}^{-2}\)
Thus, the magnitude of the deceleration is \(8.1\text{ m s}^{-2}\).

(a)
- C1: Recalls or selects \(v = u + at\).
- C1: Substitutes the correct values: \(15 + 2.4 \times 5\).
- A1: Obtains \(27\text{ m s}^{-1}\).

(b)
- C1: Calculates distance during acceleration phase: \(105\text{ m}\).
- C1: Calculates distance during constant velocity phase: \(270\text{ m}\).
- A1: Sums both parts correctly to get \(375\text{ m}\).

(c)
- C1: Selects suitable SUVAT equation: \(v^2 = u^2 + 2as\).
- C1.3: Substitutes correctly: \(0 = 27^2 + 2a(45)\).
- A1: Arrives at correct magnitude: \(8.1\text{ m s}^{-2}\) (ignore negative sign for magnitude).

(a)
First, convert the half-life from days to seconds:
\(T_{1/2} = 8.0\text{ days} = 8.0 \times 24 \times 3600\text{ s} = 6.912 \times 10^5\text{ s}\)
Now use the relation between half-life and decay constant:
\(\lambda = \frac{\ln 2}{T_{1/2}}\)
\(\lambda = \frac{0.69315}{6.912 \times 10^5} \approx 1.0028 \times 10^{-6}\text{ s}^{-1}\)
This is approximately \(1.0 \times 10^{-6}\text{ s}^{-1}\), as required.

(b)
The relationship between activity \(A\) and the number of active nuclei \(N\) is:
\(A = \lambda N\)
Substituting the values to find the initial number of nuclei \(N_0\):
\(N_0 = \frac{A_0}{\lambda} = \frac{3.5 \times 10^7\text{ Bq}}{1.0028 \times 10^{-6}\text{ s}^{-1}} \approx 3.49 \times 10^{13}\)
Rounding to 2 significant figures gives:
\(N_0 = 3.5 \times 10^{13}\)

(c)
The number of half-lives that have elapsed in 24 days is:
\(n = \frac{24\text{ days}}{8.0\text{ days}} = 3.0\)
The remaining activity \(A\) after 3 half-lives is:
\(A = A_0 \left(\frac{1}{2}\right)^3 = 3.5 \times 10^7 \times 0.125 = 4.375 \times 10^6\text{ Bq}\)
Rounding to 2 significant figures:
\(A = 4.4 \times 10^6\text{ Bq}\)

(a)
- C1: Converts days to seconds correctly: \(6.91 \times 10^5\text{ s}\).
- C1: Recalls and uses \(\lambda = \ln(2)/T_{1/2}\).
- A1: Shows explicit steps leading to \(1.00 \times 10^{-6}\text{ s}^{-1}\).

(b)
- C1: Recalls or uses \(A = \lambda N\).
- C1: Correct substitution of initial activity and decay constant.
- A1: Correct calculation to 2 significant figures: \(3.5 \times 10^{13}\).

(c)
- C1: Identifies that 24 days corresponds to exactly 3 half-lives, or uses \(A = A_0 e^{-\lambda t}\).
- C1.3: Shows correct ratio multiplication: \(A_0 / 8\) or substitutes into exponential equation.
- A1: Correct answer: \(4.4 \times 10^6\text{ Bq}\) (accept range \(4.38 \times 10^6\) to \(4.40 \times 10^6\text{ Bq}\)).

(a)
Since \(A = \frac{\pi d^2}{4}\), the percentage uncertainty in \(A\) is twice the percentage uncertainty in \(d\):
\(\% \Delta A = 2 \times \% \Delta d\)
\(\% \Delta d = \frac{0.02}{0.46} \times 100\% \approx 4.348\%\)
\(\% \Delta A = 2 \times 4.348\% = 8.696\%\)
Rounding to 2 significant figures gives \(8.7\%\).

(b)
First, calculate the cross-sectional area \(A\):
\(A = \frac{\pi \times (0.46 \times 10^{-3}\text{ m})^2}{4} = 1.6619 \times 10^{-7}\text{ m}^2\)
Now, calculate the resistivity \(\rho\):
\(\rho = \frac{R A}{L} = \frac{4.2 \times 1.6619 \times 10^{-7}}{0.850} \approx 8.212 \times 10^{-7}\ \Omega\text{ m}\)
Rounding to 2 significant figures:
\(\rho = 8.2 \times 10^{-7}\ \Omega\text{ m}\)

(c)
The percentage uncertainty in resistivity \(\rho\) is the sum of the percentage uncertainties of \(R\), \(A\), and \(L\):
\(\% \Delta \rho = \% \Delta R + \% \Delta A + \% \Delta L\)
\(\% \Delta R = \frac{0.1}{4.2} \times 100\% \approx 2.381\%\)
\(\% \Delta L = \frac{0.005}{0.850} \times 100\% \approx 0.588\%\)
\(\% \Delta \rho = 2.381\% + 8.696\% + 0.588\% = 11.665\%\)
Now, calculate the absolute uncertainty in \(\rho\):
\(\Delta \rho = 11.665\% \times 8.212 \times 10^{-7}\ \Omega\text{ m} = 0.958 \times 10^{-7}\ \Omega\text{ m}\)
Typically, absolute uncertainties are stated to 1 or 2 significant figures. Stating it as:
\(\Delta \rho = 1.0 \times 10^{-7}\ \Omega\text{ m}\) (or \(0.96 \times 10^{-7}\ \Omega\text{ m}\)).

(a)
- C1: Identifies that percentage uncertainty in Area is twice that of diameter: \(\%\Delta A = 2 \times \%\Delta d\).
- C1: Correct calculation of \(\%\Delta d = 4.35\%\).
- A1: Correct answer: \(8.7\%\) (accept range \(8.6\%\) to \(8.7\%\)).

(b)
- C1: Correctly calculates Area: \(1.66 \times 10^{-7}\text{ m}^2\).
- C1.3: Correct substitution into the resistivity equation.
- A1: Correct final resistivity: \(8.2 \times 10^{-7}\ \Omega\text{ m}\) (accept \(8.21 \times 10^{-7}\ \Omega\text{ m}\)).

(c)
- C1: Adds fractional or percentage uncertainties of \(R\), \(A\), and \(L\) together.
- C1: Obtains overall percentage uncertainty: \(11.7\%\) (accept \(11.6\%\) to \(12\%\)).
- A1: Obtains absolute uncertainty: \(0.96 \times 10^{-7}\ \Omega\text{ m}\) or \(1.0 \times 10^{-7}\ \Omega\text{ m}\).

(a)
Convert the orbital period from days to seconds:
\(T = 365.25\text{ days} \times 24\text{ hours/day} \times 3600\text{ seconds/hour}\)
\(T = 365.25 \times 86400 = 31,557,600\text{ s}\)
In scientific notation, this is:
\(T \approx 3.16 \times 10^7\text{ s}\)

(b)
For a body in a circular gravitational orbit, the gravitational force provides the centripetal force:
\(F_g = F_c\)
\(\frac{G M m}{r^2} = \frac{m v^2}{r}\)
Simplifying by dividing by \(m\) and multiplying by \(r\):
\(\frac{G M}{r} = v^2\)
The orbital speed \(v\) is related to the period \(T\) by:
\(v = \frac{2\pi r}{T}\)
Substituting this into the speed equation:
\(\frac{G M}{r} = \left(\frac{2\pi r}{T}\right)^2 = \frac{4\pi^2 r^2}{T^2}\)
Rearranging to make \(T^2\) the subject:
\(T^2 = \left(\frac{4\pi^2}{G M}\right) r^3\)

(c)
Rearranging the derived formula for the mass of the Sun \(M\):
\(M = \frac{4\pi^2 r^3}{G T^2}\)
Substituting \(r = 1.50 \times 10^{11}\text{ m}\), \(T = 3.1558 \times 10^7\text{ s}\), and \(G = 6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}\):
\(M = \frac{4\pi^2 \times (1.50 \times 10^{11})^3}{6.67 \times 10^{-11} \times (3.1558 \times 10^7)^2}\)
\(M = \frac{39.4784 \times 3.375 \times 10^{33}}{6.67 \times 10^{-11} \times 9.959 \times 10^{14}} = \frac{1.3324 \times 10^{35}}{6.6427 \times 10^4} \approx 2.006 \times 10^{30}\text{ kg}\)
Rounding to 2 significant figures gives:
\(M = 2.0 \times 10^{30}\text{ kg}\)

(a)
- C1: Uses factor of 24 and 3600 to convert days to seconds.
- C1: Correct calculation setup: \(365.25 \times 86400\).
- A1: Correct final answer: \(3.16 \times 10^7\text{ s}\) (accept \(3.15 \times 10^7\text{ s}\) if 365 days are used).

(b)
- C1: Equates gravitational force expression and centripetal force expression.
- C1.3: Introduces \(v = 2\pi r / T\) into the resulting expression.
- A1: Clearly rearranges steps with algebraic accuracy to arrive at the final given relationship.

(c)
- C1: Rearranges formula correctly to make \(M\) the subject.
- C1: Substitutes the correct numerical values including \(G\) and the calculated \(T\).
- A1: Correctly calculates \(2.0 \times 10^{30}\text{ kg}\) (accept \(1.9 \times 10^{30}\text{ kg}\) to \(2.0 \times 10^{30}\text{ kg}\)).

The total energy \(E\) of a simple harmonic oscillator is given by \(E = \frac{1}{2} k A^2\), where \(k\) is the spring constant and \(A\) is the amplitude.

At a displacement \(x = \frac{A}{2}\), the potential energy \(E_p\) is:
\(E_p = \frac{1}{2} k x^2 = \frac{1}{2} k \left(\frac{A}{2}\right)^2 = \frac{1}{4} \left(\frac{1}{2} k A^2\right) = \frac{1}{4} E\)

The kinetic energy \(E_k\) is the difference between the total energy and the potential energy:
\(E_k = E - E_p = E - \frac{1}{4} E = \frac{3}{4} E\)

The ratio of the kinetic energy to the potential energy is:
\(\frac{E_k}{E_p} = \frac{\frac{3}{4} E}{\frac{1}{4} E} = 3\)

Therefore, the ratio is \(3 : 1\).

Award 1 mark for the correct answer C.

The orbital speed \(v\) of a satellite of mass \(m\) orbiting a planet of mass \(M\) at radius \(r\) is given by equating the gravitational force to the centripetal force:
\(\frac{G M m}{r^2} = \frac{m v^2}{r} \implies v = \sqrt{\frac{G M}{r}}\)

Therefore, the orbital speed is inversely proportional to the square root of the orbital radius:
\(v \propto \frac{1}{\sqrt{r}}\)

We can find the ratio of their speeds as:
\(\frac{v_{\text{X}}}{v_{\text{Y}}} = \sqrt{\frac{r_{\text{Y}}}{r_{\text{X}}}} = \sqrt{\frac{2r_{\text{X}}}{r_{\text{X}}}} = \sqrt{2} \approx 1.41\)

Award 1 mark for the correct answer C.

Using the equation of motion \(s = u t + \frac{1}{2} a t^2\), where \(u = 0\) (starts from rest):

For the first \(t\) seconds, the distance \(d_1\) is:
\(d_1 = \frac{1}{2} a t^2\)

For the total time of \(2t\) seconds, the total distance \(s_{\text{total}}\) is:
\(s_{\text{total}} = \frac{1}{2} a (2t)^2 = 2 a t^2 = 4 d_1\)

The distance \(d_2\) travelled in the next \(t\) seconds is:
\(d_2 = s_{\text{total}} - d_1 = 4 d_1 - d_1 = 3 d_1\)

Thus, the ratio \(\frac{d_2}{d_1}\) is \(3\).

Award 1 mark for the correct answer C.

1. A thin sheet of paper completely absorbs all \(\alpha\) radiation. Therefore, the reduction in count rate when paper is added is due to the absorption of \(\alpha\) particles:
\(\text{Count rate due to } \alpha = 800 - 500 = 300\text{ s}^{-1}\).

2. A few millimetres of aluminium absorbs all \(\beta^-\) radiation (and also \(\alpha\) radiation). The remaining count rate of \(50\text{ s}^{-1}\) is due entirely to \(\gamma\) radiation:
\(\text{Count rate due to } \gamma = 50\text{ s}^{-1}\).

3. The count rate due to \(\beta^-\) radiation is the difference between the count rate with paper (where only \(\alpha\) has been stopped) and the count rate with aluminium (where both \(\alpha\) and \(\beta^-\) have been stopped):
\(\text{Count rate due to } \beta^- = 500 - 50 = 450\text{ s}^{-1}\).

4. The ratio of the count rates is:
\(\alpha : \beta^- : \gamma = 300 : 450 : 50 = 6 : 9 : 1\).

Award 1 mark for the correct answer B.

To find the percentage uncertainty in the calculated resistivity \(\rho\), we add the percentage uncertainties of each component, noting that the uncertainty of the diameter \(d\) must be multiplied by 2 because it is squared in the formula:

\(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\)

Calculate the percentage uncertainties for each measured quantity:
- \(\frac{\Delta R}{R} = \frac{0.6}{24.0} \times 100 = 2.5\%\)
- \(\frac{\Delta d}{d} = \frac{0.01}{0.50} \times 100 = 2.0\%\)
- \(\frac{\Delta L}{L} = \frac{0.01}{1.00} \times 100 = 1.0\%\)

Now, combine these:
\(\frac{\Delta \rho}{\rho} = 2.5\% + 2(2.0\%) + 1.0\% = 2.5\% + 4.0\% + 1.0\% = 7.5\%\)

Therefore, the percentage uncertainty is \(7.5\%\).

Award 1 mark for the correct answer D.

The time period \(T\) of a simple pendulum is given by:
\(T = 2\pi \sqrt{\frac{l}{g}}\)

This means that the period is inversely proportional to the square root of the gravitational field strength:
\(T \propto \frac{1}{\sqrt{g}}\)

On the new planet, the gravitational field strength is \(g' = 0.40g\). Thus, the new period \(T'\) is:
\(T' = T \times \sqrt{\frac{g}{0.40g}} = T \times \frac{1}{\sqrt{0.40}} \approx 1.58 T\)

Award 1 mark for the correct answer C.

According to Kepler's third law, the square of the orbital period \(T\) of a satellite is directly proportional to the cube of its orbital radius \(R\):
\(T^2 \propto R^3 \implies T \propto R^{3/2}\)

For the second satellite with orbital radius \(4R\), the new period \(T'\) is:
\(T' \propto (4R)^{3/2} = 4^{3/2} \times R^{3/2} = 8 \times T\)

Therefore, the period is \(8T\).

Award 1 mark for the correct answer C.

The equation \(s = k t^2\) has the same form as \(s = u t + \frac{1}{2} a t^2\) with \(u = 0\). This indicates that the car accelerates from rest with a constant acceleration \(a = 2k\).

For constant acceleration, we can relate velocity \(v\) to displacement \(s\) using the equation:
\(v^2 = u^2 + 2 a s\)

Since \(u = 0\), we have:
\(v^2 = 2 a s\)

The kinetic energy \(E_{\text{k}}\) of the car is:
\(E_{\text{k}} = \frac{1}{2} m v^2 = \frac{1}{2} m (2 a s) = (m a) s\)

Since the mass \(m\) and acceleration \(a\) are constant, \(E_{\text{k}}\) is directly proportional to the displacement \(s\) (\(E_{\text{k}} \propto s\)).

A graph of \(E_{\text{k}}\) against \(s\) is therefore a straight line of positive gradient passing through the origin.

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The maximum velocity of a simple harmonic oscillator is given by \(v_0 = \omega A\) and its maximum acceleration is \(a_0 = \omega^2 A\). Dividing the maximum acceleration by the maximum velocity yields \(\frac{a_0}{v_0} = \omega\). Substituting this into the time period formula \(T = \frac{2\pi}{\omega}\) gives \(T = \frac{2\pi v_0}{a_0}\).

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For a circular orbit, the gravitational force provides the centripetal force: \(\frac{GMm}{R^2} = \frac{mv^2}{R}\), which gives \(mv^2 = \frac{GMm}{R}\). The kinetic energy is \(E_k = \frac{1}{2}mv^2 = \frac{GMm}{2R}\). Since the kinetic energy is inversely proportional to the radius \(R\), tripling the radius decreases the kinetic energy by a factor of 3 to \(\frac{E_k}{3}\).

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Using the equation of motion \(s = ut + \frac{1}{2}at^2\) with \(u = 0\), the distance travelled in the first interval \(t\) is \(d = \frac{1}{2}at^2\). In the total time of \(2t\), the total distance travelled from the start is \(s = \frac{1}{2}a(2t)^2 = 4 \times \left(\frac{1}{2}at^2\right) = 4d\). Thus, the distance travelled specifically during the second time interval is the total distance minus the first interval distance: \(4d - d = 3d\).

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Using the magnetic force equation \(\vec{F} = q(\vec{v} \times \vec{B})\): 1) \(\alpha\) particles are positively charged, so they deflect to the right. 2) \(\beta^-\) particles are negatively charged, so they deflect to the left. 3) \(\gamma\) photons carry no charge and therefore travel undeviated.

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The relative uncertainty equation for density is \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2 \frac{\Delta r}{r} + \frac{\Delta L}{L}\). Substituting the given percentage values gives \(1.5\% + 2 \times 1.2\% + 2.0\% = 1.5\% + 2.4\% + 2.0\% = 5.9\%\).

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The total energy is \(E_{\text{total}} = E_k + E_p = 4E_p\) because \(E_k = 3E_p\). Using the formulas \(E_{\text{total}} = \frac{1}{2}kA^2\) and \(E_p = \frac{1}{2}kx^2\), we have \(\frac{1}{2}kA^2 = 4 \times \frac{1}{2}kx^2\), which simplifies to \(A^2 = 4x^2\) and thus \(x = \pm \frac{A}{2}\).

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For a stable circular orbit, \(\frac{GMm}{R^2} = \frac{mv^2}{R}\), which yields \(v = \sqrt{\frac{GM}{R}}\). Thus, orbital speed is inversely proportional to the square root of orbital radius: \(v \propto \frac{1}{\sqrt{R}}\). Therefore, \(\frac{v_Y}{v_X} = \sqrt{\frac{R_X}{R_Y}} = \sqrt{\frac{1}{4}} = \frac{1}{2}\), which means \(v_Y = \frac{v_X}{2}\).

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