2025 Cambridge IAL Physics (9702) Practice Paper with Answers

To find the overall uncertainty of the calculated density, we sum the individual fractional uncertainties. Since \(d\) is squared, its fractional uncertainty contribution is multiplied by 2:

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

Substituting the given values:

\(\frac{\Delta \rho}{\rho} = 1.5\% + 2(1.0\%) + 2.0\% = 5.5\%\)

1 mark for the correct answer B.
- Correctly identifies that the power of 2 for diameter multiplies its percentage uncertainty by 2.
- Correctly adds the percentage uncertainties to obtain 5.5%.

Using the potential divider formula, we calculate the potential difference across the fixed resistor \(V_{\text{fixed}}\):

1. In the dark:
\(V_{\text{fixed}} = E \times \frac{R}{R + R_{\text{LDR}}} = 12.0\text{ V} \times \frac{400\ \Omega}{400\ \Omega + 800\ \Omega} = 4.0\text{ V}\)

2. In the light:
\(V_{\text{fixed}} = 12.0\text{ V} \times \frac{400\ \Omega}{400\ \Omega + 200\ \Omega} = 8.0\text{ V}\)

3. The change in the potential difference:
\(\Delta V = 8.0\text{ V} - 4.0\text{ V} = +4.0\text{ V}\)

Hence, the potential difference across the fixed resistor increases by \(4.0\text{ V}\).

1 mark for the correct answer C.
- Method: Correct potential divider ratio calculations for both situations.
- Accuracy: Correctly computes the change of +4.0 V.

1. Initial charge on the first capacitor:
\(Q = C_1 V_1 = (4.0 \times 10^{-6}\text{ F}) \times 50\text{ V} = 2.0 \times 10^{-4}\text{ C}\)

2. Initial electrostatic energy:
\(E_i = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} \times (4.0 \times 10^{-6}\text{ F}) \times (50\text{ V})^2 = 5.0 \times 10^{-3}\text{ J} = 5.0\text{ mJ}\)

3. Total capacitance after connecting in parallel:
\(C_{\text{total}} = C_1 + C_2 = 4.0\ \mu\text{F} + 6.0\ \mu\text{F} = 10.0\ \mu\text{F}\)

4. Charge is conserved, so total charge \(Q_{\text{total}} = 2.0 \times 10^{-4}\text{ C}\). Final electrostatic energy:
\(E_f = \frac{Q_{\text{total}}^2}{2 C_{\text{total}}} = \frac{(2.0 \times 10^{-4}\text{ C})^2}{2 \times (10.0 \times 10^{-6}\text{ F})} = 2.0 \times 10^{-3}\text{ J} = 2.0\text{ mJ}\)

5. Energy lost:
\(\Delta E = E_i - E_f = 5.0\text{ mJ} - 2.0\text{ mJ} = 3.0\text{ mJ}\)

1 mark for the correct answer B.
- Calculates initial energy correctly (5.0 mJ).
- Recognizes conservation of charge and calculates the new shared voltage or directly calculates final energy (2.0 mJ).
- Calculates the difference accurately.

1. Initial kinetic energy:
\(E_{\text{ki}} = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} \times (3.0\text{ kg}) \times (4.0\text{ m s}^{-1})^2 = 24.0\text{ J}\)

2. Conservation of linear momentum during the collision:
\(m_1 u_1 = (m_1 + m_2) v\)
\((3.0\text{ kg}) \times (4.0\text{ m s}^{-1}) = (3.0\text{ kg} + 1.0\text{ kg}) \times v\)
\(12.0 = 4.0 v \implies v = 3.0\text{ m s}^{-1}\)

3. Final kinetic energy of the combined system:
\(E_{\text{kf}} = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} \times (4.0\text{ kg}) \times (3.0\text{ m s}^{-1})^2 = 18.0\text{ J}\)

4. Loss in kinetic energy:
\(\Delta E_{\text{k}} = E_{\text{ki}} - E_{\text{kf}} = 24.0\text{ J} - 18.0\text{ J} = 6.0\text{ J}\)

1 mark for the correct answer A.
- Method: Momentum conservation equation used correctly to find the common final speed.
- Accuracy: Correct energy difference calculated.

Using the first law of thermodynamics: \(\Delta U = q + w\), where:
- \(\Delta U\) is the increase in internal energy.
- \(q\) is the heat energy supplied to the system. Since heat is transferred to the gas, \(q = +450\text{ J}\).
- \(w\) is the work done on the system. Since the gas performs work on the surroundings, the work done on the gas is negative, so \(w = -150\text{ J}\).

Substituting these values:
\(\Delta U = +450\text{ J} + (-150\text{ J}) = +300\text{ J}\)

1 mark for the correct answer C.
- Method: Correct application of the First Law of Thermodynamics, with correct sign conventions.
- Accuracy: Obtains +300 J.

The relationship for maximum acceleration in simple harmonic motion is:
\(a_{\text{max}} = \omega^2 x_0\)

where:
- \(x_0 = 5.0\text{ cm} = 0.050\text{ m}\) (amplitude)
- \(\omega = \frac{2\pi}{T} = \frac{2\pi}{2.0\text{ s}} = \pi\text{ rad s}^{-1}\)

Calculating the maximum acceleration:
\(a_{\text{max}} = (\pi)^2 \times 0.050\text{ m} \approx 9.87 \times 0.050 \approx 0.49\text{ m s}^{-2}\)

1 mark for the correct answer B.
- Correct equation used for maximum acceleration.
- Correct conversion of amplitude from cm to m, leading to the correct value.

1. The difference in energy levels \(\Delta E\) is:
\(\Delta E = E_{\text{initial}} - E_{\text{final}} = -1.5\text{ eV} - (-3.4\text{ eV}) = 1.9\text{ eV}\)

2. Convert the energy change to Joules:
\(\Delta E = 1.9 \times (1.60 \times 10^{-19}\text{ J}) = 3.04 \times 10^{-19}\text{ J}\)

3. Use the photon energy formula \(\Delta E = h f\) to solve for frequency \(f\):
\(f = \frac{\Delta E}{h} = \frac{3.04 \times 10^{-19}\text{ J}}{6.63 \times 10^{-34}\text{ J s}} \approx 4.6 \times 10^{14}\text{ Hz}\)

1 mark for the correct answer A.
- Method: Correctly converts energy difference in eV to Joules, and uses the relation f = E/h.
- Accuracy: Obtains approximately 4.6 * 10^14 Hz.

For an ideal gas at constant volume, the absolute thermodynamic temperature \(T\) is directly proportional to its pressure \(P\):

\(T = T_{\text{tr}} \left(\frac{P}{P_{\text{tr}}}\right)\)

where:
- \(T_{\text{tr}} = 273.16\text{ K}\) is the temperature of the triple point of water.
- \(P_{\text{tr}} = 4.00 \times 10^4\text{ Pa}\)
- \(P = 5.50 \times 10^4\text{ Pa}\)

Substituting the values:
\(T = 273.16 \times \frac{5.50 \times 10^4\text{ Pa}}{4.00 \times 10^4\text{ Pa}} = 273.16 \times 1.375 = 375.6\text{ K} \approx 376\text{ K}\)

1 mark for the correct answer B.
- Relates absolute temperature directly to pressure for a constant volume ideal gas.
- Calculates the value correctly to 3 significant figures.

To find the percentage uncertainty in \(\rho\), we sum the fractional uncertainties of each component (multiplied by their power):

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

Calculate each term:
- Diameter component: \(2 \times \frac{0.01}{0.50} = 2 \times 2.0\% = 4.0\%\)
- Resistance component: \rac{0.3}{12.0} = 2.5\%\)
- Length component: \(\frac{0.002}{1.000} = 0.2\%\)

Sum the percentage uncertainties:
\(4.0\% + 2.5\% + 0.2\% = 6.7\%\)

1 mark for calculating individual percentage uncertainties, doubling the diameter uncertainty, and summing them to obtain 6.7%.

Let right be the positive direction.
Initial total momentum \(p_i\):
\(p_i = m_P u_P + m_Q u_Q = (0.20 \times 3.0) + (0.30 \times (-2.0)) = 0.60 - 0.60 = 0\text{ kg m s}^{-1}\).

Since momentum is conserved, the final total momentum \(p_f = 0\text{ kg m s}^{-1}\).
After the collision, the velocity of \(P\) is \(v_P = -1.5\text{ m s}^{-1}\).
\(p_f = m_P v_P + m_Q v_Q = 0\)
\((0.20 \times (-1.5)) + 0.30 v_Q = 0\)
\(-0.30 + 0.30 v_Q = 0 \implies v_Q = 1.0\text{ m s}^{-1}\) (to the right).

To determine if the collision is elastic, we compare kinetic energy:
Initial kinetic energy \(E_{k,i} = \frac{1}{2}(0.20)(3.0)^2 + \frac{1}{2}(0.30)(2.0)^2 = 0.90 + 0.60 = 1.50\text{ J}\).
Final kinetic energy \(E_{k,f} = \frac{1}{2}(0.20)(-1.5)^2 + \frac{1}{2}(0.30)(1.0)^2 = 0.225 + 0.15 = 0.375\text{ J}\).
Since \(E_{k,f} < E_{k,i}\), kinetic energy is not conserved, meaning the collision is inelastic.

1 mark for using conservation of linear momentum to find the velocity of glider Q as 1.0 m s⁻¹ to the right, and comparing initial and final kinetic energies to deduce that the collision is inelastic.

Using the first law of thermodynamics, \(\Delta U = q + w\), where \(q\) is the thermal energy supplied to the gas and \(w\) is the work done on the gas. For a complete cycle \(X \rightarrow Y \rightarrow Z \rightarrow X\), the net change in internal energy is zero (\(\Delta U_{\text{cycle}} = 0\)).

For process \(X \rightarrow Y\):
\(q = +400\text{ J}\)
\(w = -150\text{ J}\) (work done by the gas)
\(\Delta U_{XY} = +400 - 150 = +250\text{ J}\)

For process \(Y \rightarrow Z\):
\(q = -280\text{ J}\) (heat released)
\(w = 0\) (constant volume)
\(\Delta U_{YZ} = -280\text{ J}\)

For the cycle:
\(\Delta U_{XY} + \Delta U_{YZ} + \Delta U_{ZX} = 0\)
\(250 - 280 + \Delta U_{ZX} = 0 \implies \Delta U_{ZX} = +30\text{ J}\)

For process \(Z \rightarrow X\):
Since the process is adiabatic, \(q = 0\).
Therefore, \(\Delta U_{ZX} = w_{ZX} = +30\text{ J}\).
The work done on the gas is \(+30\text{ J}\).

1 mark for applying the first law of thermodynamics to each stage of the cycle and using the fact that the total internal energy change over a closed cycle is zero to calculate work done on the gas as +30 J.

First, find the angular frequency \(\omega\):
\(\omega = 2\pi f = 2\pi \times 20 = 40\pi\text{ rad s}^{-1} \approx 125.7\text{ rad s}^{-1}\)

The maximum acceleration magnitude \(a_{\text{max}}\) is:
\(a_{\text{max}} = \omega^2 x_0 = (40\pi)^2 \times (5.0 \times 10^{-3}\text{ m}) \approx 79\text{ m s}^{-2}\)

The speed \(v\) at displacement \(x = 3.0\text{ mm}\) is:
\(v = \omega \sqrt{x_0^2 - x^2} = (40\pi) \sqrt{(5.0 \times 10^{-3})^2 - (3.0 \times 10^{-3})^2} = 40\pi \times 4.0 \times 10^{-3}\text{ m} \approx 0.50\text{ m s}^{-1}\)

1 mark for calculating angular frequency \(\omega\), and using it to obtain \(a_{\text{max}} = 79\text{ m s}^{-2}\) and \(v = 0.50\text{ m s}^{-1}\).

The energy of the emitted photon is equal to the change in energy levels:
\(\Delta E = E_{\text{initial}} - E_{\text{final}} = -1.5\text{ eV} - (-3.4\text{ eV}) = 1.9\text{ eV}\)

Convert this energy into joules:
\(\Delta E = 1.9 \times 1.60 \times 10^{-19}\text{ J} = 3.04 \times 10^{-19}\text{ J}\)

Now, find the frequency \(f\) using the photon energy equation \(\Delta E = h f\):
\(f = \frac{\Delta E}{h} = \frac{3.04 \times 10^{-19}\text{ J}}{6.63 \times 10^{-34}\text{ J s}} \approx 4.6 \times 10^{14}\text{ Hz}\)

1 mark for calculating the energy difference in joules and applying Planck's relation to find the frequency.

At the lowest point of the circular path, both the tension \(T\) (acting upwards towards the center) and the weight \(mg\) (acting downwards) affect the centripetal acceleration. The net centripetal force is:

\[F_{\text{net}} = T - mg = \frac{m v^2}{r}\]

Rearranging to solve for the tension \(T\):

\[T = mg + \frac{m v^2}{r} = m \left( g + \frac{v^2}{r} \right)\]

Substitute the given values:

\[T = 0.15 \times \left( 9.81 + \frac{6.0^2}{0.80} \right)\]
\[T = 0.15 \times (9.81 + 45.0) = 0.15 \times 54.81 \approx 8.2\text{ N}\]

1 mark for identifying the correct equation for the net force at the lowest point of circular motion and calculating the correct tension value.

For an ideal gas at constant volume, pressure is directly proportional to thermodynamic temperature \(T\) (in Kelvin):

\[\frac{P_1}{T_1} = \frac{P_2}{T_2}\]

Substitute the values:

\[T_2 = T_1 \times \frac{P_2}{P_1} = 273.16 \times \frac{1.50 \times 10^5}{1.20 \times 10^5} = 273.16 \times 1.25 = 341.45\text{ K}\]

To find the temperature \(\theta\) in degrees Celsius:

\[\theta = T_2 - 273.15 = 341.45 - 273.15 = 68.30^\circ\text{C} \approx 68^\circ\text{C}\]

1 mark for using the pressure-temperature relationship for a gas thermometer on the Kelvin scale and converting the resulting temperature correctly to degrees Celsius.

The energy delivered to the resistor is equal to the decrease in energy stored in the capacitor.

Initial energy stored, \(E_i = \frac{1}{2} C V_i^2\):
\[E_i = \frac{1}{2} \times (470 \times 10^{-6}\text{ F}) \times (12.0\text{ V})^2 = 0.03384\text{ J} = 33.84\text{ mJ}\]

Final energy stored, \(E_f = \frac{1}{2} C V_f^2\):
\[E_f = \frac{1}{2} \times (470 \times 10^{-6}\text{ F}) \times (6.0\text{ V})^2 = 0.00846\text{ J} = 8.46\text{ mJ}\]

Energy delivered, \(\Delta E = E_i - E_f\):
\[\Delta E = 33.84\text{ mJ} - 8.46\text{ mJ} = 25.38\text{ mJ} \approx 25\text{ mJ}\]

1 mark for computing initial and final stored energy correctly using E = 0.5 C V² and finding the difference to get 25 mJ.

The density of a uniform wire is given by the formula \(\rho = \frac{4m}{\pi d^2 L}\). The fractional uncertainty in \(\rho\) is: \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\). Calculating each percentage uncertainty: \(\frac{\Delta m}{m} = \frac{0.05}{5.00} \times 100\% = 1.0\%\), \(\frac{\Delta L}{L} = \frac{0.5}{50.0} \times 100\% = 1.0\%\), and \(\frac{\Delta d}{d} = \frac{0.02}{1.00} \times 100\% = 2.0\%\). Substituting these values: \(\frac{\Delta \rho}{\rho} = 1.0\% + 2(2.0\%) + 1.0\% = 6.0\%\).

1 mark for the correct calculation of the percentage uncertainty of density, accounting for the power of 2 in diameter.

Initially, the potential difference across the thermistor is \(8.0\text{ V}\). Since the supply voltage is \(12.0\text{ V}\), the potential difference across the fixed resistor is \(12.0\text{ V} - 8.0\text{ V} = 4.0\text{ V}\). Because the current is the same through both, the ratio of resistance is equal to the ratio of voltage: \(\frac{R_T}{4.0\text{ k}\Omega} = \frac{8.0\text{ V}}{4.0\text{ V}} = 2.0\), giving the initial thermistor resistance \(R_T = 8.0\text{ k}\Omega\). When the temperature increases, the resistance of the thermistor decreases by \(75\%\) to: \(R_T' = (1 - 0.75) \times 8.0\text{ k}\Omega = 2.0\text{ k}\Omega\). The new voltmeter reading across the thermistor is: \(V_T' = 12.0\text{ V} \times \frac{2.0\text{ k}\Omega}{2.0\text{ k}\Omega + 4.0\text{ k}\Omega} = 4.0\text{ V}\).

1 mark for the correct calculation of the final voltage across the thermistor after its resistance changes.

Initially, the capacitor has capacitance \(C_0\), potential difference \(V\), and energy stored \(E_0 = \frac{1}{2} C_0 V^2\). When the dielectric is inserted while connected, capacitance becomes \(C_1 = 3 C_0\), and potential difference remains \(V\). The charge on the capacitor becomes \(Q_1 = C_1 V = 3 C_0 V = 3 Q_0\). When disconnected, this charge is trapped: \(Q_{\text{trapped}} = 3 Q_0\). Removing the dielectric returns the capacitance to \(C_0\). The final energy stored is \(E_{\text{final}} = \frac{Q_{\text{trapped}}^2}{2 C_0} = \frac{(3 Q_0)^2}{2 C_0} = 9 \left(\frac{Q_0^2}{2 C_0}\right) = 9 E_0\).

1 mark for the correct derivation of energy scaling based on charge conservation and capacitance change.

By conservation of momentum: \(3M \cdot u = (3M + 2M) \cdot v \implies v = \frac{3}{5} u\). The initial kinetic energy is \(E_{k,i} = \frac{1}{2} (3M) u^2 = 1.5 M u^2\). The final kinetic energy is \(E_{k,f} = \frac{1}{2} (5M) \left(\frac{3}{5} u\right)^2 = 0.9 M u^2\). The loss in kinetic energy is \(\Delta E_k = 1.5 M u^2 - 0.9 M u^2 = 0.6 M u^2\). The fraction of kinetic energy lost is \(\frac{\Delta E_k}{E_{k,i}} = \frac{0.6}{1.5} = 0.40\).

1 mark for calculating the final velocity and correctly evaluating the kinetic energy loss fraction.

For process \(X \rightarrow Y\): \(\Delta U_{XY} = Q + W = +600\text{ J} - 400\text{ J} = +200\text{ J}\). For process \(Y \rightarrow Z\): isothermal compression means \(\Delta U_{YZ} = 0\). For a complete cycle: \(\Delta U_{\text{total}} = \Delta U_{XY} + \Delta U_{YZ} + \Delta U_{ZX} = 0 \implies +200\text{ J} + 0 + \Delta U_{ZX} = 0 \implies \Delta U_{ZX} = -200\text{ J}\). For process \(Z \rightarrow X\), volume is constant, so \(W_{ZX} = 0\). Thus, \(\Delta U_{ZX} = Q_{ZX} + 0 \implies Q_{ZX} = -200\text{ J}\).

1 mark for applying the first law of thermodynamics to each process and finding both correct values.

The total mechanical energy is \(E = \frac{1}{2} m \omega^2 A^2\). The potential energy at displacement \(x\) is \(E_p = \frac{1}{2} m \omega^2 x^2\), and the kinetic energy is \(E_k = E - E_p = \frac{1}{2} m \omega^2 (A^2 - x^2)\). We require \(E_k = 3 E_p \implies \frac{1}{2} m \omega^2 (A^2 - x^2) = 3 \left(\frac{1}{2} m \omega^2 x^2\right) \implies A^2 - x^2 = 3 x^2 \implies A^2 = 4 x^2 \implies x = \pm 0.50 A\).

1 mark for setting up the equation relating kinetic and potential energy in terms of amplitude and displacement, and solving for x.

The energy required to excite an ion from \(n=1\) to \(n=2\) is \(-6.0 - (-24.0) = 18.0\text{ eV}\). To excite from \(n=1\) to \(n=3\) requires \(-1.5 - (-24.0) = 22.5\text{ eV}\). The bombarding electrons have \(20.0\text{ eV}\), which is sufficient to cause the \(n=1 \rightarrow n=2\) transition, but not the \(n=1 \rightarrow n=3\) transition. Thus, the ions can only be excited to a maximum state of \(n=2\). Upon transition back to the ground state, they can emit a photon of maximum energy \(18.0\text{ eV}\).

1 mark for identifying the correct threshold energy level reached and determining the resulting maximum transition energy.

Let \(N\) be the normal force. Since the walls of the cone are inclined at \(\theta\) to the vertical, the normal force \(N\) is directed at angle \(\theta\) above the horizontal. Resolving forces: vertically, \(N \sin\theta = mg\); horizontally (providing the centripetal force), \(N \cos\theta = \frac{mv^2}{r}\). Dividing these two equations gives \(\frac{N \sin\theta}{N \cos\theta} = \tan\theta = \frac{mg}{mv^2/r} = \frac{gr}{v^2}\). Rearranging yields \(v = \sqrt{\frac{gr}{\tan\theta}}\).

1 mark for setting up the correct force component equations and solving for speed v.

The formula for resistivity is \(\rho = \frac{R \pi d^2}{4 L}\). The relative 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 fractional uncertainties: For \(R\), \(\frac{\Delta R}{R} = \frac{0.05}{4.50} \approx 1.11\%\). For \(d\), \(\frac{\Delta d}{d} = \frac{0.01}{0.38} \approx 2.63\%\). For \(L\), \(\frac{\Delta L}{L} = \frac{0.003}{1.200} = 0.25\%\). Summing these according to the rule gives \(\frac{\Delta \rho}{\rho} = 1.11\% + 2(2.63\%) + 0.25\% = 6.62\%\), which rounds to \(6.6\%\).

1 mark for calculating the correct percentage uncertainty by summing the individual percentage uncertainties, remembering to multiply the diameter uncertainty by 2.

The formula for volume is \(V = \frac{\pi d^2 l}{4}\). The percentage uncertainty in a product or quotient is the sum of the percentage uncertainties of the individual quantities, multiplied by their powers. Therefore, the percentage uncertainty in \(V\) is: \(\frac{\Delta V}{V} \times 100 = 2 \times \left(\frac{\Delta d}{d} \times 100\right) + \left(\frac{\Delta l}{l} \times 100\right)\). Substituting the values: \(\frac{\Delta d}{d} \times 100 = \frac{0.02}{1.20} \times 100 = 1.67\%\) and \(\frac{\Delta l}{l} \times 100 = \frac{0.5}{50.0} \times 100 = 1.0\%\). Thus: \(\frac{\Delta V}{V} \times 100 = 2 \times 1.67\% + 1.0\% = 4.33\% \approx 4.3\%\).

1 mark for the correct calculation of percentage uncertainties and propagation.

The initial output voltage across the fixed resistor is given by: \(V_{\text{out, initial}} = V_{\text{supply}} \times \frac{R_{\text{fixed}}}{R_{\text{thermistor, initial}} + R_{\text{fixed}}} = 12.0 \times \frac{4.0}{8.0 + 4.0} = 4.0\text{ V}\). After heating, the thermistor resistance decreases to \(2.0\text{ k}\Omega\). The final output voltage is: \(V_{\text{out, final}} = 12.0 \times \frac{4.0}{2.0 + 4.0} = 8.0\text{ V}\). The change in the output voltage is: \(V_{\text{out, final}} - V_{\text{out, initial}} = 8.0 - 4.0 = 4.0\text{ V}\) (an increase of \(4.0\text{ V}\)).

1 mark for calculating the initial and final output voltages correctly and finding the correct difference.

The discharging of a capacitor is described by the equation \(Q = Q_0 e^{-\frac{t}{\tau}}\), where \(\tau = RC\) is the time constant. Given that at \(t = 2.0\text{ s}\), \(Q = 0.25 Q_0\): \(0.25 Q_0 = Q_0 e^{-\frac{2.0}{\tau}} \implies 0.25 = e^{-\frac{2.0}{\tau}}\). Taking the natural logarithm of both sides: \(\ln(0.25) = -\frac{2.0}{\tau} \implies -1.386 = -\frac{2.0}{\tau} \implies \tau = \frac{2.0}{1.386} \approx 1.4\text{ s}\). Alternatively, since falling to \(25\%\) requires two half-lives, one half-life is \(t_{1/2} = 1.0\text{ s}\). Since \(t_{1/2} = \tau \ln(2)\), we have \(\tau = \frac{1.0}{\ln(2)} \approx 1.4\text{ s}\).

1 mark for using the decay equation or half-life relationship to calculate the time constant.

First, define the direction of block X's motion as positive. By conservation of linear momentum: \(m_X u_X + m_Y u_Y = (m_X + m_Y) v\), where \(u_X = +6.0\text{ m s}^{-1}\) and \(u_Y = -2.0\text{ m s}^{-1}\). \((2.0)(6.0) + (3.0)(-2.0) = (2.0 + 3.0) v \implies 12.0 - 6.0 = 5.0 v \implies v = 1.2\text{ m s}^{-1}\). The initial kinetic energy is: \(E_{\text{k, initial}} = \frac{1}{2} m_X u_X^2 + \frac{1}{2} m_Y u_Y^2 = \frac{1}{2}(2.0)(6.0)^2 + \frac{1}{2}(3.0)(-2.0)^2 = 36.0 + 6.0 = 42.0\text{ J}\). The final kinetic energy is: \(E_{\text{k, final}} = \frac{1}{2}(m_X + m_Y) v^2 = \frac{1}{2}(5.0)(1.2)^2 = 3.6\text{ J}\). The loss in kinetic energy is: \(\Delta E_{\text{k}} = E_{\text{k, initial}} - E_{\text{k, final}} = 42.0 - 3.6 = 38.4\text{ J}\).

1 mark for applying conservation of momentum to find the final velocity, calculating initial and final kinetic energies, and finding the difference.

According to the first law of thermodynamics, \(\Delta U = q + w\), where \(\Delta U\) is the increase in internal energy, \(q\) is the thermal energy supplied to the system, and \(w\) is the work done on the system. Here, \(q = +400\text{ J}\) (since thermal energy is supplied to the gas) and \(w = -150\text{ J}\) (since work is done by the gas on the surroundings). Therefore: \(\Delta U = +400 + (-150) = +250\text{ J}\). Thus, the internal energy of the gas increases by \(250\text{ J}\).

1 mark for applying the first law of thermodynamics with correct signs for work and heat.

Since \(x = 0\) at \(t = 0\), the displacement can be modeled by the equation \(x = x_0 \sin(\omega t)\), where the angular frequency \(\omega = \frac{2\pi}{T}\). At time \(t = \frac{T}{12}\): \(\omega t = \frac{2\pi}{T} \times \frac{T}{12} = \frac{\pi}{6}\text{ radians}\). Substituting this into the displacement equation: \(x = x_0 \sin\left(\frac{\pi}{6}\right) = 0.50 x_0\).

1 mark for expressing displacement in terms of a sine function and substituting the correct phase angle.

The energy of the emitted photon is equal to the difference in energy between the two states: \(\Delta E = E_{\text{initial}} - E_{\text{final}} = -1.5\text{ eV} - (-3.4\text{ eV}) = 1.9\text{ eV}\). Converting this energy to Joules: \(\Delta E = 1.9 \times 1.60 \times 10^{-19}\text{ J} = 3.04 \times 10^{-19}\text{ J}\). The energy of a photon is related to its wavelength \(\lambda\) by \(E = \frac{hc}{\lambda}\), where \(h = 6.63 \times 10^{-34}\text{ J s}\) and \(c = 3.00 \times 10^8\text{ m s}^{-1}\). Rearranging for \(\lambda\): \(\lambda = \frac{hc}{E} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{3.04 \times 10^{-19}} \approx 6.54 \times 10^{-7}\text{ m}\).

1 mark for finding the energy difference, converting to Joules, and applying the photon energy equation.

Let \(T\) be the tension in the string. Resolving forces on the mass: Vertically, \(T \cos\theta = mg\). Horizontally, the centripetal force is provided by the horizontal component of tension, \(T \sin\theta = \frac{mv^2}{r}\). Dividing the horizontal equation by the vertical equation: \(\frac{T \sin\theta}{T \cos\theta} = \frac{mv^2 / r}{mg} \implies \tan\theta = \frac{v^2}{rg}\). Rearranging for \(v\): \(v = \sqrt{rg \tan\theta}\).

1 mark for setting up the vertical and horizontal force equations and correctly deriving the expression for \(v\).

(a) A systematic error causes measurements to be consistently shifted from the true value in one direction (e.g., zero error on a micrometer). A random error causes measurements to vary unpredictably about the mean value (due to human reaction time or fluctuating conditions).

(b) The formula for resistivity is:

\(\rho = \frac{R A}{L} = \frac{R \pi d^2}{4 L}\)

Substitute the central values:

\(d = 0.38 \times 10^{-3}\text{ m}\)
\(A = \frac{\pi \times (0.38 \times 10^{-3})^2}{4} = 1.134 \times 10^{-7}\text{ m}^2\)
\(\rho = \frac{4.8 \times 1.134 \times 10^{-7}}{1.25} = 4.355 \times 10^{-7}\\ \Omega\text{ m} \approx 4.4 \times 10^{-7}\\ \Omega\text{ m}\)

(c) The fractional uncertainty equation is:

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

Substitute the fractional uncertainties:

\(\frac{\Delta R}{R} = \frac{0.2}{4.8} = 0.0417\)
\(\frac{\Delta d}{d} = \frac{0.01}{0.38} = 0.0263 \implies 2\frac{\Delta d}{d} = 0.0526\)
\(\frac{\Delta L}{L} = \frac{0.02}{1.25} = 0.0160\)

Sum of fractional uncertainties:

\(\frac{\Delta \rho}{\rho} = 0.0417 + 0.0526 + 0.0160 = 0.1103\)

Calculate absolute uncertainty:

\(\Delta \rho = 0.1103 \times 4.355 \times 10^{-7} = 0.48 \times 10^{-7}\\ \Omega\text{ m} \approx 0.5 \times 10^{-7}\\ \Omega\text{ m}\)

Thus, \(\rho = (4.4 \pm 0.5) \times 10^{-7}\\ \Omega\text{ m}\).

(d) The student can take multiple readings of the diameter at different positions and orientations along the wire and calculate a mean value.

(a)
- Systematic error: constant bias/shift in one direction from true value [1]
- Random error: scatter/spread of readings about the mean value [1]

(b)
- Correct formula used for cross-sectional area or direct substitution: \(A = 1.13 \times 10^{-7}\text{ m}^2\) [1]
- Calculation of central value: \(4.355 \times 10^{-7}\\ \Omega\text{ m}\) [1]
- Value rounded to 2 significant figures: \(4.4 \times 10^{-7}\\ \Omega\text{ m}\) [1]

(c)
- Correct fractional uncertainty formula with factor of 2 for \(d\) [1]
- Individual fractional terms calculated correctly (at least two of: 0.0417, 0.0526, 0.0160) [1]
- Correct sum of fractional uncertainties (approx. 0.11 or 11%) [1]
- Correct absolute uncertainty calculated and expressed to 1 significant figure: \(0.5 \times 10^{-7}\\ \Omega\text{ m}\) [1]

(d)
- Take multiple measurements at different orientations/positions along the wire and find the average [1]

(a) The total momentum of a closed system remains constant, provided there are no external forces acting on the system.

(b) Choose the direction to the right as positive.

Initial momentum of glider A: \(p_A = m_A u_A = 0.350 \times (+2.40) = +0.840\text{ kg m s}^{-1}\)
Initial momentum of glider B: \(p_B = m_B u_B = 0.150 \times (-1.20) = -0.180\text{ kg m s}^{-1}\)
Total initial momentum: \(P_{\text{initial}} = 0.840 - 0.180 = +0.660\text{ kg m s}^{-1}\)

Total mass after collision: \(M = m_A + m_B = 0.350 + 0.150 = 0.500\text{ kg}\)
By conservation of momentum:
\(M v = P_{\text{initial}}\)
\(0.500 \times v = 0.660\)
\(v = 1.32\text{ m s}^{-1}\) (directed to the right)

(c) Initial Kinetic Energy:
\(E_{k,\text{initial}} = \frac{1}{2} m_A u_A^2 + \frac{1}{2} m_B u_B^2\)
\(E_{k,\text{initial}} = \frac{1}{2}(0.350)(2.40)^2 + \frac{1}{2}(0.150)(-1.20)^2 = 1.008 + 0.108 = 1.116\text{ J}\)

Final Kinetic Energy:
\(E_{k,\text{final}} = \frac{1}{2} (m_A + m_B) v^2 = \frac{1}{2}(0.500)(1.32)^2 = 0.4356\text{ J}\)

Loss in Kinetic Energy:
\(\Delta E_k = 1.116 - 0.4356 = 0.6804\text{ J} \approx 0.680\text{ J}\)

(d) The collision is inelastic because there is a loss of kinetic energy (kinetic energy is not conserved). It is converted into thermal energy and sound.

(a)
- Statement: Total momentum of a system of interacting bodies is constant [1]
- Condition: Provided no external forces act / closed system [1]

(b)
- Uses conservation of momentum with appropriate signs: \(m_A u_A + m_B u_B = (m_A + m_B) v\) [1]
- Correct substitution: \(0.350 \times 2.40 - 0.150 \times 1.20 = 0.500 v\) [1]
- Correct final answer: \(1.32\text{ m s}^{-1}\) (allow 3 sig figs) [1]

(c)
- Correct formula and calculation for initial K.E. (\(1.116\text{ J}\)) [1]
- Correct formula and calculation for final K.E. (\(0.436\text{ J}\)) [1]
- Correct difference: \(0.680\text{ J}\) (accept \(0.68\text{ J}\)) [1]

(d)
- Identifies the collision as inelastic [1]
- Mentions that kinetic energy is not conserved or is lost to other forms [1]

(a) The first law of thermodynamics can be written as:

\(\Delta U = q + w\)

where \(\Delta U\) is the increase in internal energy of the system, \(q\) is the heat energy supplied to the system, and \(w\) is the work done ON the system (or \(\Delta U = q - w\) where \(w\) is work done BY the system).

(b) Work done by the gas during expansion at constant pressure is given by:

\(W = p \Delta V\)
\(W = 2.4 \times 10^5 \times (3.5 \times 10^{-3} - 1.5 \times 10^{-3})\)
\(W = 2.4 \times 10^5 \times 2.0 \times 10^{-3} = 480\text{ J}\)

(c) Using the first law of thermodynamics:

\(\Delta U = q - W_{\text{by}}\)

Where thermal energy supplied to the gas is \(q = +820\text{ J}\), and work done by the gas is \(W_{\text{by}} = 480\text{ J}\).

\(\Delta U = 820 - 480 = +340\text{ J}\)

This represents an increase of \(340\text{ J}\) in internal energy.

(d) The total change in internal energy for the complete process is \(0\text{ J}\). Internal energy of an ideal gas is a function of its temperature only. Since the gas is cooled back to its initial temperature, the final internal energy equals the initial internal energy, resulting in zero net change.

(a)
- State: \(\Delta U = q + w\) (or \(\Delta U = q - w\)) with terms correctly defined [1]
- Correct sign conventions explained (e.g., \(q\) is heat added to system, \(w\) is work done on system) [1]

(b)
- Uses formula: \(W = p \Delta V\) [1]
- Substitute values: \(2.4 \times 10^5 \times (3.5 \times 10^{-3} - 1.5 \times 10^{-3})\) [1]
- Correct result: \(480\text{ J}\) [1]

(c)
- Applies first law of thermodynamics with correct signs: \(\Delta U = 820 - 480\) [1]
- Correct calculation: \(340\text{ J}\) [1]
- Clearly identifies this as an increase in internal energy (or positive sign shown) [1]

(d)
- States total change is zero [1]
- Explains that internal energy of an ideal gas depends only on temperature, and initial and final temperatures are the same [1]

(a) For a mass-spring system, the angular frequency \(\omega\) is given by:

\(\omega = \sqrt{\frac{k}{m}}\)

Substitute \(k = 80\text{ N m}^{-1}\) and \(m = 0.45\text{ kg}\):

\(\omega = \sqrt{\frac{80}{0.45}} = \sqrt{177.78} = 13.33\text{ rad s}^{-1} \approx 13\text{ rad s}^{-1}\)

(b) The maximum velocity \(v_{\text{max}}\) of the block is given by:

\(v_{\text{max}} = \omega x_0\)

where \(x_0 = 0.060\text{ m}\) is the amplitude of oscillation.

\(v_{\text{max}} = 13.33 \times 0.060 = 0.80\text{ m s}^{-1}\)

(c) The kinetic energy \(E_k\) at a displacement \(x\) is given by:

\(E_k = \frac{1}{2} m \omega^2 (x_0^2 - x^2) = \frac{1}{2} k (x_0^2 - x^2)\)

Substitute \(k = 80\text{ N m}^{-1}\), \(x_0 = 0.060\text{ m}\), and \(x = 0.030\text{ m}\):

\(E_k = \frac{1}{2} \times 80 \times (0.060^2 - 0.030^2)\)
\(E_k = 40 \times (0.0036 - 0.0009) = 40 \times 0.0027 = 0.108\text{ J} \approx 0.11\text{ J}\)

(d) The magnitude of the maximum acceleration \(a_{\text{max}}\) is:

\(a_{\text{max}} = \omega^2 x_0 = (13.33)^2 \times 0.060 = 10.7\text{ m s}^{-2} \approx 11\text{ m s}^{-2}\)

Simple harmonic motion is defined by the relation:

\(a = -\omega^2 x\)

which means acceleration is directly proportional to displacement and is always directed towards the equilibrium position (opposite direction to displacement).

(a)
- Formula: \(\omega = \sqrt{k/m}\) used [1]
- Correct calculation shown with at least 3 sig figs: \(13.3\text{ rad s}^{-1}\) [1]

(b)
- Formula: \(v_{\text{max}} = \omega x_0\) used [1]
- Correct calculation: \(0.80\text{ m s}^{-1}\) [1]

(c)
- Formula: \(E_k = \frac{1}{2} k(x_0^2 - x^2)\) or \(\frac{1}{2} m \omega^2(x_0^2 - x^2)\) [1]
- Correct substitution of values [1]
- Correct calculation: \(0.11\text{ J}\) (or \(0.108\text{ J}\)) [1]

(d)
- Correct maximum acceleration calculation: \(11\text{ m s}^{-2}\) (or \(10.7\text{ m s}^{-2}\)) [1]
- Defines SHM relation: acceleration is directly proportional to displacement [1]
- Acceleration is in the opposite direction to displacement / directed towards equilibrium [1]

(a) Coherent light sources have a constant phase difference (and therefore the same frequency or wavelength).

(b) The double-slit interference equation is:

\(x = \frac{\lambda D}{a}\)

Where:
\(\lambda = 632.8 \times 10^{-9}\text{ m}\)
\(D = 1.80\text{ m}\)
\(a = 0.250 \times 10^{-3}\text{ m}\)

Substitute values:

\(x = \frac{632.8 \times 10^{-9} \times 1.80}{0.250 \times 10^{-3}}\)
\(x = 4.556 \times 10^{-3}\text{ m} \approx 4.56\text{ mm}\)

(c) (i) Halving the slit separation \(a\): Since fringe width \(x \propto 1/a\), halving \(a\) will double the fringe separation. The fringes become wider apart (separation becomes \(9.12\text{ mm}\)).
(ii) Replacing with \(450\text{ nm}\) light: Since \(x \propto \lambda\) and the new wavelength is shorter, the fringe separation decreases. The fringes are closer together.

(d) If one slit is covered, the double-slit interference pattern disappears. Instead, a broader single-slit diffraction pattern is observed (with a bright central maximum and much weaker outer maxima).

(a)
- Constant phase difference [1]
- Same frequency / wavelength [1]

(b)
- Formula: \(x = \frac{\lambda D}{a}\) [1]
- Correct substitution with standard SI units [1]
- Correct calculation: \(4.56 \times 10^{-3}\text{ m}\) (or \(4.56\text{ mm}\) or \(4.6\text{ mm}\)) [1]

(c) (i)
- Statement: Fringe spacing / separation doubles [1]
- Reason: Fringe width is inversely proportional to slit separation (\(x \propto 1/a\)) [1]

(c) (ii)
- Statement: Fringe spacing / separation decreases [1]
- Reason: Fringe width is directly proportional to wavelength (\(x \propto \lambda\)) [1]

(d)
- Interference pattern disappears / single-slit diffraction pattern seen [1]

(a) The fixed resistor and the thermistor must be connected in series across the \(9.0\text{ V}\) supply. The output voltage \(V_{\text{out}}\) must be measured across the thermistor. As the temperature increases, the resistance of the NTC thermistor decreases, so it takes a smaller fraction of the total voltage, decreasing \(V_{\text{out}}\).

(b) Using the potential divider equation:

\(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{th}}}{R + R_{\text{th}}}\)

Substitute \(V_{\text{in}} = 9.0\text{ V}\), \(R_{\text{th}} = 2400\\ \Omega\), and \(R = 1200\\ \Omega\):

\(V_{\text{out}} = 9.0 \times \frac{2400}{1200 + 2400} = 9.0 \times \frac{2400}{3600} = 6.0\text{ V}\)

(c) (i) With \(V_{\text{out}} = 2.5\text{ V}\):

\(2.5 = 9.0 \times \frac{R_{\text{th}}}{1200 + R_{\text{th}}}\)
\(2.5(1200 + R_{\text{th}}) = 9.0 R_{\text{th}}\)
\(3000 + 2.5 R_{\text{th}} = 9.0 R_{\text{th}}\)
\(6.5 R_{\text{th}} = 3000\)
\(R_{\text{th}} = \frac{3000}{6.5} = 461.5\\ \Omega \approx 460\\ \Omega\)

(ii) The calculated resistance at this temperature is \(460\\ \Omega\), which is significantly less than the initial resistance of \(2400\\ \Omega\). Since this is an NTC thermistor, its resistance decreases as temperature increases. Hence, the temperature must have increased.

(a)
- Fixed resistor and thermistor connected in series [1]
- \(V_{\text{out}}\) taken across the thermistor [1]

(b)
- Correct formula: \(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{th}}}{R + R_{\text{th}}}\) [1]
- Substitution: \(9.0 \times \frac{2400}{3600}\) [1]
- Correct calculation: \(6.0\text{ V}\) [1]

(c) (i)
- Potential divider formula rearranged or current calculation: \(I = \frac{9.0 - 2.5}{1200} = 5.42 \times 10^{-3}\text{ A}\) [1]
- Correct calculation of resistance: \(R_{\text{th}} = \frac{2.5}{5.42 \times 10^{-3}}\) [1]
- Final value: \(460\\ \Omega\) (allow \(461\\ \Omega\) or \(462\\ \Omega\)) [1]

(c) (ii)
- Observes that resistance has decreased (from \(2400\\ \Omega\) to \(460\\ \Omega\)) [1]
- Connects decrease in resistance of NTC thermistor to an increase in temperature [1]

(a) Internal energy is the sum of a random distribution of kinetic and potential energies associated with the molecules of a system.

(b) (i) Since volume is constant, work done \(W = p \Delta V = 0\).
(ii) From the first law of thermodynamics, \(\Delta U = q + W\). Since \(W = 0\), \(q = \Delta U = 600 \text{ J}\).

(c) (i) For an ideal gas, there are no intermolecular forces, so the potential energy of the molecules is zero. The internal energy depends entirely on the kinetic energy of the molecules, which is directly proportional to thermodynamic temperature. Since the temperature is constant, there is no change in internal energy.
(ii) Here \(\Delta U = 0\). Work done by the gas is \(659 \text{ J}\), so work done on the gas is \(W = -659 \text{ J}\). Thus, \(0 = q - 659 \implies q = +659 \text{ J}\).

(d) (i) The volume at C is found using \(p_B V_B = p_C V_C \implies (3.0 \times 10^5)(2.0 \times 10^{-3}) = (1.0 \times 10^5) V_C \implies V_C = 6.0 \times 10^{-3} \text{ m}^3\).
Work done on the gas: \(W = -p \Delta V = -1.0 \times 10^5 \times (2.0 \times 10^{-3} - 6.0 \times 10^{-3}) = +400 \text{ J}\).
(ii) For a complete cycle, net \(\Delta U = 0\), so \(\Delta U_{CA} = -600 \text{ J}\). For CA: \(\Delta U = q + W \implies -600 = q + 400 \implies q = -1000 \text{ J}\). Hence, \(1000 \text{ J}\) of thermal energy is transferred from the gas.

(a) Sum of random distribution of kinetic and potential energies [1] of particles/molecules of system [1].
(b)(i) Correct value of 0 J [1].
(b)(ii) Correct value of 600 J [1].
(c)(i) Ideal gas has no intermolecular forces / potential energy is zero [1]; internal energy depends only on temperature, which is constant [1].
(c)(ii) Correct value of 659 J (or +659 J) [1].
(d)(i) Volume at C = \(6.0 \times 10^{-3} \text{ m}^3\) [1], Work done = \(400 \text{ J}\) [1].
(d)(ii) Correct value of 1000 J (or -1000 J) [1].

(a) Capacitance is the charge stored per unit potential difference across a conductor.

(b) For capacitors in parallel, the total capacitance \(C_p = C_1 + C_2 = 47\ \mu\text{F} + 22\ \mu\text{F} = 69\ \mu\text{F}\).

(c) (i) \(Q = C_p V = 69 \times 10^{-6} \times 12 = 8.28 \times 10^{-4} \text{ C}\) (or \(8.3 \times 10^{-4} \text{ C}\)).
(ii) \(E = \frac{1}{2} C_p V^2 = 0.5 \times 69 \times 10^{-6} \times 12^2 = 4.97 \times 10^{-3} \text{ J}\) (or \(5.0 \times 10^{-3} \text{ J}\)).

(d) (i) Time constant \(\tau = R C = 150 \times 10^3 \times 69 \times 10^{-6} = 10.35 \text{ s} \approx 10 \text{ s}\).
(ii) Using \(V = V_0 e^{-t / \tau}\):
\(3.0 = 12 e^{-t / 10.35} \implies 0.25 = e^{-t / 10.35} \implies \ln(0.25) = -t / 10.35\)
\(t = -10.35 \times (-1.386) = 14.3 \text{ s}\) (or \(14 \text{ s}\) using \(\tau = 10 \text{ s}\)).

(a) Charge per unit potential difference [1].
(b) Direct addition shown: \(47 + 22 = 69\ \mu\text{F}\) [1].
(c)(i) Formula \(Q = CV\) used [1], value \(8.3 \times 10^{-4} \text{ C}\) (or \(8.28 \times 10^{-4} \text{ C}\)) [1].
(c)(ii) Formula \(E = \frac{1}{2} CV^2\) used [1], value \(5.0 \times 10^{-3} \text{ J}\) (or \(4.97 \times 10^{-3} \text{ J}\)) [1].
(d)(i) Multiplication of \(150 \times 10^3\) by \(69 \times 10^{-6}\) to get \(10.35 \text{ s}\) [1].
(d)(ii) Exponential formula \(V = V_0 e^{-t / \tau}\) used [1], substitution \(3.0 / 12 = 0.25\) and taking logs [1], final value \(14 \text{ s}\) (or \(14.3 \text{ s}\)) [1].

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

(b) (i) Parallel: \(0.40 \times 2.5 = 0.40 \times 1.1 \cos(60^\circ) + 0.60 v \cos(\theta) \implies 1.0 = 0.22 + 0.60 v \cos(\theta)\).
(ii) Perpendicular: \(0 = 0.40 \times 1.1 \sin(60^\circ) - 0.60 v \sin(\theta) \implies 0.60 v \sin(\theta) = 0.381\).

(c) (i) & (ii):
From (b)(i): \(0.60 v \cos(\theta) = 0.78 \implies v \cos(\theta) = 1.30\).
From (b)(ii): \(v \sin(\theta) = 0.635\).
Dividing the equations: \(\tan(\theta) = 0.635 / 1.30 = 0.4885 \implies \theta = 26.0^\circ\).
Then: \(v = 1.30 / \cos(26.0^\circ) = 1.45 \text{ m s}^{-1} \approx 1.5 \text{ m s}^{-1}\).

(d) Initial kinetic energy: \(E_{k,i} = \frac{1}{2} m_A u_A^2 = 0.5 \times 0.40 \times 2.5^2 = 1.25 \text{ J}\).
Final kinetic energy: \(E_{k,f} = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 = 0.5 \times 0.40 \times 1.1^2 + 0.5 \times 0.60 \times 1.45^2 = 0.242 + 0.631 = 0.873 \text{ J}\).
Since final KE \(0.87 \text{ J}\) is less than initial KE \(1.25 \text{ J}\), the collision is inelastic.

(a) Total momentum remains constant [1] for an isolated/closed system (or no external forces) [1].
(b)(i) \(1.0 = 0.22 + 0.60 v \cos(\theta)\) or equivalent [1].
(b)(ii) \(0 = 0.381 - 0.60 v \sin(\theta)\) or equivalent [1].
(c)(i) Dividing equations to find \(\tan(\theta)\) or squaring and adding [1], resolving components [1], obtaining speed \(v = 1.45 \text{ m s}^{-1}\) (allow \(1.4\) or \(1.5\)) [1].
(c)(ii) Correct angle \(\theta = 26^\circ\) (or \(26.0^\circ\)) [2] (1 mark for method, 1 mark for correct value).
(d) Correct calculation of initial KE and final KE showing final KE is less, concluding inelastic [1].

(a) Simple harmonic motion is defined as motion where the acceleration of the system is directly proportional to its displacement from equilibrium [1] and is always directed towards the equilibrium position [1].

(b) (i) Comparing with \(x = x_0 \cos(\omega t)\), the amplitude \(x_0 = 0.080 \text{ m}\).
(ii) The angular frequency \(\omega = 15 \text{ rad s}^{-1}\). Since \(\omega = 2\pi f\), the frequency \(f = 15 / (2\pi) = 2.39 \text{ Hz} \approx 2.4 \text{ Hz}\).
(iii) Maximum speed \(v_{\max} = \omega x_0 = 15 \times 0.080 = 1.2 \text{ m s}^{-1}\).

(c) Total energy \(E = \frac{1}{2} m \omega^2 x_0^2\) or \(\frac{1}{2} m v_{\max}^2\).
\(E = 0.5 \times 0.25 \times (1.2)^2 = 0.125 \times 1.44 = 0.18 \text{ J}\).

(a) Acceleration proportional to displacement [1], directed towards fixed point/equilibrium [1].
(b)(i) \(0.080 \text{ m}\) [1].
(b)(ii) Use of \(\omega = 2\pi f\) with \(\omega = 15\) [1], final value \(2.4 \text{ Hz}\) (or \(2.39 \text{ Hz}\)) [1].
(b)(iii) Use of \(v = \omega x_0\) [1], final value \(1.2 \text{ m s}^{-1}\) [1].
(c) Use of \(E = \frac{1}{2} m \omega^2 x_0^2\) or \(E = \frac{1}{2} m v_{\max}^2\) [1], correct substitution [1], final value \(0.18 \text{ J}\) [1].

(a) A photon is a discrete packet or quantum of electromagnetic energy.

(b) Electrons in the ion can only occupy discrete energy levels. When an electron falls from a higher energy level to a lower level, it emits a single photon. The energy of this photon is exactly equal to the difference in energy between the two levels (\(\Delta E = hf\)). Since the energy levels are discrete, only photons of specific, discrete frequencies (and hence wavelengths) are emitted, forming distinct lines on the spectrum.

(c) (i) \(\Delta E = E_3 - E_1 = -3.4 - (-13.6) = 10.2 \text{ eV}\).
In Joules: \(\Delta E = 10.2 \times 1.60 \times 10^{-19} = 1.63 \times 10^{-18} \text{ J}\).
(ii) Using \(\Delta E = \frac{hc}{\lambda}\):
\(\lambda = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{1.63 \times 10^{-18}} = 1.22 \times 10^{-7} \text{ m}\).

(a) Packet / quantum [1] of electromagnetic energy [1].
(b) Discrete/specific energy levels [1], photon emitted when electron falls to lower level [1], energy difference corresponds to a single frequency/wavelength \(E = hf\) [1].
(c)(i) Energy difference calculated as \(10.2 \text{ eV}\) [1], conversion to Joules to give \(1.63 \times 10^{-18} \text{ J}\) [1].
(c)(ii) Formula \(\lambda = hc/E\) stated or used [1], substitution of Planck constant and speed of light [1], final value \(1.22 \times 10^{-7} \text{ m}\) (or \(1.2 \times 10^{-7} \text{ m}\)) [1].

(a) At the top of the path, both forces act vertically downwards towards the center of the circle. The diagram must show two arrows pointing downwards from the sphere: one for the tension \(T\) and one for the weight \(W\).

(b) (i) Centripetal acceleration \(a_c = \frac{v_T^2}{r} = \frac{3.5^2}{0.80} = 15.3 \text{ m s}^{-2}\) (or \(15 \text{ m s}^{-2}\)).
(ii) The net downward force provides the centripetal force: \(T + mg = m a_c \implies T = m(a_c - g)\).
\(T = 0.15 \times (15.31 - 9.81) = 0.15 \times 5.5 = 0.825 \text{ N} \approx 0.83 \text{ N}\).

(c) At the bottom of the path, tension acts upwards and weight acts downwards: \(T - mg = \frac{m v_B^2}{r}\).
\(T = mg + \frac{m v_B^2}{r} = 0.15 \times 9.81 + \frac{0.15 \times 6.1^2}{0.80} = 1.47 + 6.98 = 8.45 \text{ N} \approx 8.5 \text{ N}\).

(a) Arrow for Weight (W / mg) pointing vertically downwards [1], Arrow for Tension (T) pointing vertically downwards [1].
(b)(i) Formula \(a = v^2/r\) used [1], value \(15 \text{ m s}^{-2}\) (or \(15.3 \text{ m s}^{-2}\)) [1].
(b)(ii) Equation \(T + mg = ma_c\) or \(T + mg = mv^2/r\) [1], correct substitution [1], value \(0.83 \text{ N}\) (or \(0.825 \text{ N}\)) [1].
(c) Equation \(T - mg = mv^2/r\) [1], substitution of values with \(v = 6.1\) [1], value \(8.5 \text{ N}\) (or \(8.45 \text{ N}\)) [1].

(a) The thermodynamic scale of temperature is a temperature scale that is independent of the properties of any particular substance [1] and has its absolute zero at 0 K [1].

(b) (i) \(R_{100} = 100.0 \times (1 + 3.90 \times 10^{-3} \times 100) = 100.0 \times 1.39 = 139.0\ \Omega\).
(ii) \(R_{-50} = 100.0 \times (1 + 3.90 \times 10^{-3} \times (-50)) = 100.0 \times (1 - 0.195) = 80.5\ \Omega\).

(c) (i) \(E = P \times t = 120 \times (3.0 \times 60) = 21600 \text{ J}\) (or \(2.16 \times 10^4 \text{ J}\)).
(ii) Using \(\Delta Q = m c \Delta\theta\):
\(21600 = 0.45 \times 4200 \times \Delta\theta \implies 21600 = 1890 \Delta\theta\)
\(\Delta\theta = 11.43^\circ\text{C} \approx 11.4^\circ\text{C}\).
(iii) Using the final resistance \(154.6\ \Omega\) to find final temperature \(\theta_f\):
\(154.6 = 100.0 (1 + 3.90 \times 10^{-3} \theta_f) \implies 1.546 = 1 + 3.90 \times 10^{-3} \theta_f\)
\(0.546 = 3.90 \times 10^{-3} \theta_f \implies \theta_f = 140.0^\circ\text{C}\).
\(\theta_i = \theta_f - \Delta\theta = 140.0 - 11.43 = 128.6^\circ\text{C} \approx 129^\circ\text{C}\).

(a) Independent of properties of a physical substance [1], absolute zero is the minimum possible temperature (or 0 K) [1].
(b)(i) \(139.0\ \Omega\) [1].
(b)(ii) \(80.5\ \Omega\) [1].
(c)(i) \(21600 \text{ J}\) (or \(2.16 \times 10^4 \text{ J}\)) [1].
(c)(ii) Use of \(Q = mc\Delta\theta\) [1], final value \(11.4^\circ\text{C}\) [1].
(c)(iii) Use of thermometer equation to find final temperature \(\theta_f = 140.0^\circ\text{C}\) [1], subtraction of \(\Delta\theta\) [1], final value \(129^\circ\text{C}\) (or \(128.6^\circ\text{C}\)) [1].

(a) When ultrasound waves hit the piezoelectric crystal, they cause it to compress and expand (vibrate) [1]. This alternating stress/pressure creates an alternating potential difference across the faces of the crystal [1]. This electrical signal is then detected and processed [1].

(b) Intensity reflection coefficient \(\alpha = \frac{(Z_2 - Z_1)^2}{(Z_2 + Z_1)^2}\)
\(\alpha = \frac{(1.70 \times 10^6 - 1.38 \times 10^6)^2}{(1.70 \times 10^6 + 1.38 \times 10^6)^2} = \frac{(0.32 \times 10^6)^2}{(3.08 \times 10^6)^2} = \frac{0.1024}{9.4864} = 0.0108 \approx 0.011\).

(c) The acoustic impedance of air is very different from that of skin/tissue [1]. If there is an air gap, almost all of the ultrasound intensity is reflected at the air-skin boundary [1]. The coupling gel has an acoustic impedance similar to skin, which matches the impedances and ensures high transmission of ultrasound into the body [1].

(d) Using \(I = I_0 e^{-\mu x}\):
\(0.45 I_0 = I_0 e^{-\mu (3.5)} \implies 0.45 = e^{-3.5 \mu}\)
\(\ln(0.45) = -3.5 \mu \implies -0.7985 = -3.5 \mu \implies \mu = 0.228 \text{ cm}^{-1} \approx 0.23 \text{ cm}^{-1}\) (or \(23 \text{ m}^{-1}\)).

(a) Ultrasound waves cause crystal to vibrate/compress [1], which creates an alternating potential difference across the crystal [1], giving an electrical signal [1].
(b) Formula stated or used [1], correct calculation to give \(0.011\) (or \(1.1\%\)) [1].
(c) Impedance of air is very different from skin/tissue [1], leading to almost 100% reflection at the boundary [1], gel has similar impedance to skin to reduce reflection / maximize transmission [1].
(d) Formula \(I = I_0 e^{-\mu x}\) used with correct substitution [1], final value of \(0.23 \text{ cm}^{-1}\) (or \(23 \text{ m}^{-1}\)) [1].

(a) Capacitance is defined as the charge stored per unit potential difference (or ratio of charge on one plate of the capacitor to the potential difference across the plates).

(b)(i) The discharging equation is given by \( V = V_0 e^{-t/\tau} \) where \( \tau = RC \).
When \( t = \tau \):
\( V = V_0 e^{-1} \)
\( V / V_0 = e^{-1} \approx 0.368 \approx 37\% \) of the initial potential difference.

(ii) 1. Taking the natural logarithm of both sides of the discharging equation:
\( \ln(V) = \ln(V_0) - \frac{1}{\tau} t \)
This is in the form \( y = mx + c \) where \( x = t \). Therefore, the gradient \( m \) of the graph of \( \ln(V) \) against \( t \) is equal to \( -\frac{1}{\tau} \).
Thus, the time constant \( \tau = -\frac{1}{\text{gradient}} \).

2. From the gradient:
\( \tau = -\frac{1}{-0.0454\text{ s}^{-1}} \approx 22.03\text{ s} \)
Since \( \tau = RC \), we have:
\( R = \frac{\tau}{C} = \frac{22.026\text{ s}}{220 \times 10^{-6}\text{ F}} \approx 1.00 \times 10^5\ \Omega \) (or \( 100\text{ k}\Omega \)).

(iii) The initial energy stored in the capacitor is:
\( E_0 = \frac{1}{2} C V_0^2 = \frac{1}{2} \times (220 \times 10^{-6}\text{ F}) \times (12.0\text{ V})^2 = 0.01584\text{ J} \)
At time \( t = 15.0\text{ s} \), the energy stored is:
\( E = E_0 e^{-2t/\tau} \) (since energy is proportional to \( V^2 \))
\( E = 0.01584 \times e^{-2 \times 15.0 / 22.026} \)
\( E = 0.01584 \times e^{-1.362} \approx 0.01584 \times 0.2561 \approx 4.06 \times 10^{-3}\text{ J} \) (or \( 4.1 \times 10^{-3}\text{ J} \)).

Alternatively, calculate the potential difference at \( t = 15.0\text{ s} \):
\( V = 12.0 \times e^{-15.0 / 22.03} \approx 6.07\text{ V} \)
Then energy is:
\( E = \frac{1}{2} C V^2 = \frac{1}{2} \times (220 \times 10^{-6}) \times (6.07)^2 \approx 4.05 \times 10^{-3}\text{ J} \).

(a) C1: charge per unit potential difference [1]
(b)(i) M1: State \( V = V_0 e^{-t/RC} \) and substitute \( t = RC \) (or \( t = \tau \)) to get \( V = V_0 e^{-1} \) [1]
A1: Calculate \( e^{-1} \approx 0.368 \) and state this is approx. 37% [1]
(b)(ii) 1. C1: Use \( \ln(V) = \ln(V_0) - t/\tau \) to show linear relationship [1]
A1: State that gradient is \( -1/\tau \) (or \( \tau = -1/\text{gradient} \)) [1]
2. C1: Calculate \( \tau = 22.03\text{ s} \) (or use algebraic expression \( R = -1/(\text{gradient} \times C) \)) [1]
A1: \( R = 1.0 \times 10^5\ \Omega \) (or \( 100\text{ k}\Omega \)) [1]
(b)(iii) C1: Calculate initial energy \( E_0 = 1.58 \times 10^{-2}\text{ J} \) OR find \( V = 6.07\text{ V} \) at \( t = 15.0\text{ s} \) [1]
C1: Use of \( E = E_0 e^{-2t/\tau} \) or \( E = \frac{1}{2} C V^2 \) [1]
A1: \( E = 4.1 \times 10^{-3}\text{ J} \) (allow \( 4.0 \times 10^{-3}\text{ J} \) to \( 4.1 \times 10^{-3}\text{ J} \)) [1]

(a) The first law of thermodynamics is written as \( \Delta U = q + w \), where \( \Delta U \) is the increase in internal energy, \( q \) is the heating (thermal energy transferred to the system), and \( w \) is the work done on the system.
Sign convention: \( q \) is positive for thermal energy added to the system, and \( w \) is positive for work done on the system.

(b)(i) 1. For a constant pressure process, work done by the gas is:
\( W_{\text{by}} = P \Delta V = (3.0 \times 10^5\text{ Pa}) \times (4.0 \times 10^{-3}\text{ m}^3 - 1.5 \times 10^{-3}\text{ m}^3) \)
\( W_{\text{by}} = 3.0 \times 10^5 \times 2.5 \times 10^{-3} = 750\text{ J} \).

2. The work done on the gas is \( w = -750\text{ J} \).
Given that thermal energy supplied to the gas is \( q = +1150\text{ J} \).
By the first law of thermodynamics:
\( \Delta U_{A \to B} = q + w = 1150\text{ J} - 750\text{ J} = +400\text{ J} \).

(ii) For stage \( B \to C \), the volume is constant, so no work is done on the gas (\( w = 0 \)).
Thermal energy is removed, so \( q = -980\text{ J} \).
Thus, \( \Delta U_{B \to C} = q + w = -980\text{ J} \).
Over a complete closed cycle, the net change in internal energy is zero:
\( \Delta U_{\text{net}} = \Delta U_{A \to B} + \Delta U_{B \to C} + \Delta U_{C \to A} = 0 \)
\( +400\text{ J} - 980\text{ J} + \Delta U_{C \to A} = 0 \implies \Delta U_{C \to A} = +580\text{ J} \).
For stage \( C \to A \):
\( \Delta U_{C \to A} = q + w \)
\( +580\text{ J} = q + 620\text{ J} \implies q = 580 - 620 = -40\text{ J} \).
This represents \( 40\text{ J} \) of thermal energy transferred out of (removed from) the gas.

(iii) The net change in internal energy over the complete cycle is \( 0\text{ J} \).
Reason: Internal energy is a state function that depends only on the state of the system (such as temperature). Since the gas returns to its initial state \( A \), its temperature and internal energy return to their initial values, making the net change zero.

(a) B1: State \( \Delta U = q + w \) with correct identification of all terms [1]
B1: State that \( q \) is positive when heat is added and \( w \) is positive when work is done ON the gas [1]
(b)(i) 1. C1: Use of \( W = P \Delta V \) [1]
A1: Work done by gas = \( 750\text{ J} \) [1]
2. C1: State \( w = -750\text{ J} \) (work done on gas is negative of work done by gas) [1]
A1: \( \Delta U_{A \to B} = +400\text{ J} \) [1]
(b)(ii) C1: Calculate or state \( \Delta U_{B \to C} = -980\text{ J} \) or state total \( \Delta U \) is zero [1]
A1: \( q = -40\text{ J} \) (or 40 J transferred from the gas) [1]
(b)(iii) B1: State net change is zero [1]
B1: Explain that internal energy depends on state / temperature, and system returns to its initial state [1]

### Detailed Design Plan:

1. **Diagram and Experimental Setup:**
- Set up a rigid retort stand clamped to the laboratory bench.
- Suspend the vertical steel wire holding the center of the horizontal brass rod.
- Position the small magnets at the ends of the brass rod directly above two thick copper plates supported on adjustable laboratory jacks.
- Place a protractor centered on the axis of the torsion wire directly beneath the rod to measure the angular displacement \(\theta\).

2. **Variables:**
- Independent variable: separation \(d\) between the bottom face of the magnets and the top surface of the copper plates.
- Dependent variable: damping coefficient \(\gamma\).
- Controlled variables: length and tension of the steel wire, mass and moment of inertia of the brass rod, strength of the neodymium magnets, material and thickness of the copper plates, and initial angle of twist \(\theta_0\).

3. **Procedure & Measurements:**
- Measure the separation \(d\) using a vernier caliper or a micrometer screw gauge. Keep the copper plates perfectly horizontal.
- Twist the rod by a small, fixed angle (e.g., \(\theta_0 \approx 15^\circ\)) and release it carefully to ensure the motion is purely torsional without any lateral wobble.
- Record the oscillations using a video camera with an on-screen digital timer. Frame the video to clearly capture both the protractor scale and the timer.
- Play back the video frame-by-frame to record the maximum angular displacement \(\theta\) of each successive swing and the corresponding elapsed time \(t\).

4. **Analysis of Damping Coefficient \(\gamma\):**
- From the equation \(\theta = \theta_0 e^{-\gamma t}\), taking natural logarithms yields:
\(\ln(\theta) = -\gamma t + \ln(\theta_0)\)
- For a fixed separation \(d\), plot a graph of \(\ln(\theta)\) on the y-axis against \(t\) on the x-axis.
- A straight line of negative gradient will confirm exponential damping. The gradient of this line is equal to \(-\gamma\), from which \(\gamma = -\text{gradient}\).

5. **Analysis of the relationship between \(\gamma\) and \(d\):**
- Repeat the entire procedure for at least six different separations \(d\) (ranging from \(2\text{ mm}\) to \(15\text{ mm}\)).
- Since \(\gamma = p d^{-q}\), taking natural logarithms gives:
\(\ln(\gamma) = -q \ln(d) + \ln(p)\)
- Plot a graph of \(\ln(\gamma)\) on the y-axis against \(\ln(d)\) on the x-axis.
- A straight line confirms the proposed relationship.
- The gradient of this line is equal to \(-q\), so \(q = -\text{gradient}\).
- The y-intercept of this line is equal to \(\ln(p)\), so \(p = e^{\text{y-intercept}}\).

6. **Safety and Experimental Details:**
- Use a G-clamp to secure the retort stand to the bench to prevent tipping.
- Use a transparent draft shield or box around the setup to eliminate air drafts which would introduce stray damping forces.
- Keep magnetic objects and iron filings away from the strong neodymium magnets.

**Defining the Problem (3 marks):**
- [1] Identify the independent variable as the separation \(d\) and the dependent variable as the damping coefficient \(\gamma\).
- [1] State that the initial angular displacement \(\theta_0\) (or wire properties, rod inertia, magnet strength) must be kept constant.
- [1] Specify a suitable range of separations \(d\) (e.g., \(2\text{ mm}\) to \(15\text{ mm}\)) to ensure significant eddy-current damping without physical contact.

**Methods of Data Collection (5 marks):**
- [1] Draw a labeled diagram showing the torsion wire, horizontal rod with magnets, copper plates directly underneath, and laboratory jacks or spacers.
- [1] Detail on measuring \(d\) using a vernier caliper or a micrometer screw gauge.
- [1] Detail on measuring angular displacement \(\theta\) using a protractor placed under the rod.
- [1] Use of a video camera with a timer (or data logger with an angle sensor) to record \(\theta\) at specified times \(t\).
- [1] Detail on ensuring oscillations are torsional only, by twisting the rod from the center and releasing it carefully.

**Analysis of Data (3 marks):**
- [1] Plot a graph of \(\ln(\theta)\) against \(t\) to find the damping coefficient \(\gamma\), where \(\gamma = -\text{gradient}\).
- [1] Plot a graph of \(\ln(\gamma)\) against \(\ln(d)\).
- [1] Identify that a straight line confirms the proposed relationship, with \(q = -\text{gradient}\) and \(p = e^{\text{y-intercept}}\).

**Additional Detail / Safety (4 marks):**
- [1] Use a G-clamp to secure the retort stand to the bench.
- [1] Use a transparent draft box or shield to prevent air currents from affecting the damping.
- [1] Repeat measurements of \(d\) at different orientations to ensure the copper plate is horizontal and parallel to the magnet's path.
- [1] Keep the initial amplitude small (e.g., \(< 15^\circ\)) so that simple harmonic approximations hold and motion remains stable.

### (a) Expressions for Gradient and Intercept
The equation is:
\(\tau = C R + C R_0\)
Comparing this to the linear equation \(y = mx + c\):
- **Gradient** \(m = C\) (if \(R\) is in \(\Omega\)).
- **y-intercept** \(c = C R_0\).

### (b) Graph Plotting
- **X-axis:** \(R / \text{k}\Omega\) scaled from \(0\) to \(110\) with a major grid division of \(10\text{ k}\Omega\).
- **Y-axis:** \(\tau / \text{s}\) scaled from \(0\) to \(26\) with a major grid division of \(2\text{ s}\).
- **Points plotted:** \((15.0, 5.5)\), \((27.0, 8.1)\), \((39.0, 10.8)\), \((56.0, 14.5)\), \((82.0, 20.2)\), \((100.0, 24.2)\).
- **Error bars:** Plotted vertically for each point extending \(\pm 0.2\text{ s}\) above and below the plotted coordinate.

### (c) Best-fit and Worst Acceptable Lines
- **Best-fit line** is drawn straight through the points, yielding a gradient of \(0.220\text{ s / k}\Omega\) and a y-intercept of \(2.20\text{ s}\).
- **Worst acceptable line (WAL)** is drawn as the shallowest or steepest possible straight line passing through all error bars. For example, passing through the bottom of the first error bar \((15.0, 5.3)\) and the top of the last error bar \((100.0, 24.4)\):
\(m_{\text{worst}} = \frac{24.4 - 5.3}{100.0 - 15.0} = 0.2247\text{ s / k}\Omega\)
\(c_{\text{worst}} = 24.4 - (0.2247 \times 100.0) = 1.93\text{ s}\).

### (d) Gradient and Intercept calculations
- **Gradient of best-fit line:**
\(m = \frac{24.2 - 5.5}{100.0 - 15.0} = 0.220\text{ s / k}\Omega\)
In SI units, \(m = 2.20 \times 10^{-4}\text{ s }\Omega^{-1}\).
- **Uncertainty in gradient:**
\(\Delta m = |0.2247 - 0.220| = 0.005\text{ s / k}\Omega\) (or \(5 \times 10^{-6}\text{ s }\Omega^{-1}\)).
- **y-intercept of best-fit line:**
\(c = 2.20\text{ s}\).
- **Uncertainty in y-intercept:**
\(\Delta c = |1.93 - 2.20| = 0.27\text{ s} \approx 0.25\text{ s}\).

### (e) Determining \(C\) and \(R_0\)
- **Capacitance \(C\):**
\(C = m = 2.20 \times 10^{-4}\text{ F} = 220\text{ }\mu\text{F}\)
\(\Delta C = \Delta m = 5\text{ }\mu\text{F}\)
Therefore, \(C = 220 \pm 5\text{ }\mu\text{F}\).
- **Resistance \(R_0\):**
\(R_0 = \frac{c}{C} = \frac{2.20\text{ s}}{2.20 \times 10^{-4}\text{ F}} = 10000\text{ }\Omega = 10.0\text{ k}\Omega\)
Using fractional uncertainties:
\(\frac{\Delta R_0}{R_0} = \frac{\Delta c}{c} + \frac{\Delta C}{C} = \frac{0.25}{2.20} + \frac{5}{220} = 0.1136 + 0.0227 = 0.1363\)
\(\Delta R_0 = 10.0 \times 0.1363 \approx 1.4\text{ k}\Omega\)
Therefore, \(R_0 = 10.0 \pm 1.4\text{ k}\Omega\).

**(a) Expressions (1 mark):**
- [1] State that the gradient is equal to \(C\) and the vertical intercept is equal to \(C R_0\) (where \(R\) is in \(\Omega\)).

**(b) Graph and Error Bars (3 marks):**
- [1] All six points plotted correctly to within half a small square.
- [1] Suitable scales used on axes (not awkward, covering at least half of the grid).
- [1] Vertically oriented error bars plotted correctly at length \(\pm 0.2\text{ s}\).

**(c) Best-fit and Worst Acceptable Lines (2 marks):**
- [1] Single clear best-fit line drawn through the centers of the points.
- [1] Worst acceptable line drawn that passes through the top of the first/last error bar and the bottom of the last/first error bar.

**(d) Gradient and Intercept calculations (4 marks):**
- [1] Correct determination of the gradient of the best-fit line using points on the line separated by at least half the length of the line.
- [1] Correct determination of the intercept of the best-fit line.
- [1] Uncertainty in gradient calculated as \(|\text{gradient of best-fit} - \text{gradient of worst WAL}|\).
- [1] Uncertainty in intercept calculated as \(|\text{intercept of best-fit} - \text{intercept of worst WAL}|\).

**(e) Value and Uncertainty of \(C\) and \(R_0\) (5 marks):**
- [1] Value of \(C\) in range \(215\text{ }\mu\text{F}\) to \(225\text{ }\mu\text{F}\) derived from gradient.
- [1] Absolute uncertainty in \(C\) calculated correctly and expressed to 1 or 2 significant figures.
- [1] Value of \(R_0\) in range \(9.0\text{ k}\Omega\) to \(11.0\text{ k}\Omega\) computed from \(\frac{c}{C}\).
- [1] Absolute uncertainty in \(R_0\) computed correctly using fractional uncertainties or extreme values: \(\frac{\Delta R_0}{R_0} = \frac{\Delta c}{c} + \frac{\Delta C}{C}\).
- [1] Correct SI units given for both final values (\(\mu\text{F}\) or \(\text{F}\) for \(C\); \(\text{k}\Omega\) or \(\Omega\) for \(R_0\)).