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The rate of energy dissipation (power) in the resistor is given by \(P = I^2 R\). Since the discharge current is given by \(I = I_0 e^{-t/RC}\), the power at time \(t\) is \(P = I_0^2 R e^{-2t/RC} = P_0 e^{-2t/RC}\), where \(P_0\) is the initial power. We are looking for the time \(t\) when \(P = \frac{1}{2} P_0\). This gives \(e^{-2t/RC} = \frac{1}{2}\), which simplifies to \(e^{2t/RC} = 2\). Taking the natural logarithm of both sides yields \(\frac{2t}{RC} = \ln 2\). Solving for \(t\) gives \(t = \frac{1}{2} RC \ln 2\).

Award 1 mark for the correct option B. Method: Identify the formula for power in a discharging capacitor and solve for the time when power is halved.

The potential difference across a discharging capacitor is given by \(V = V_0 e^{-t/RC}\). Taking the natural logarithm of both sides gives \(\ln V = \ln V_0 - \frac{t}{RC}\). A graph of \(\ln V\) against \(t\) is a straight line with gradient \(m = -\frac{1}{RC}\). Given that the gradient is \(-0.040\text{ s}^{-1}\), we have \(-\frac{1}{RC} = -0.040\), which gives the time constant \(\tau = RC = \frac{1}{0.040} = 25\text{ s}\). Since \(R = 250\text{ k}\Omega = 2.5 \times 10^5\ \Omega\), the capacitance is \(C = \frac{25}{2.5 \times 10^5} = 1.0 \times 10^{-4}\text{ F} = 100\ \mu\text{F}\).

Award 1 mark for the correct option B. Method: Relate the gradient of the log-linear discharge graph to the time constant, calculate the time constant, and use the known resistance to find capacitance.

In each discharge cycle, the charge transferred from the capacitor is \(Q = CV\). Since the discharge occurs \(f\) times per second, the average current is \(I = fQ = fCV\). Rearranging this formula in the form of a straight line through the origin, \(y = mx\), we get \(I = C(fV)\). By plotting \(I\) on the vertical axis (\(y\)) and the product \(fV\) on the horizontal axis (\(x\)), the gradient \(m\) of the line is equal to the capacitance \(C\).

Award 1 mark for the correct option A. Method: Formulate the relationship between current, frequency, capacitance, and potential difference, and match it to the equation of a straight line through the origin.

Since the mass is at its equilibrium position at \(t = 0\), its displacement \(x\) as a function of time \(t\) can be modeled as \(x(t) = x_0 \sin(\omega t)\), where the angular frequency is \(\omega = \frac{2\pi}{T}\). We set \(x(t) = \frac{\sqrt{3}}{2} x_0\), which gives \(x_0 \sin\left(\frac{2\pi t}{T}\right) = \frac{\sqrt{3}}{2} x_0\). This simplifies to \(\sin\left(\frac{2\pi t}{T}\right) = \frac{\sqrt{3}}{2}\). The smallest positive angle that satisfies this equation is \(\frac{\pi}{3}\) radians. Therefore, \(\frac{2\pi t}{T} = \frac{\pi}{3}\), which gives \(t = \frac{T}{6}\).

Award 1 mark for the correct option C. Method: Set up the sinusoidal equation of motion for simple harmonic motion starting at equilibrium, and solve for time using the inverse sine function.

The total energy \(E\) of a simple harmonic oscillator is constant and is given by \(E = E_p + E_k = \frac{1}{2} m \omega^2 A^2\). The potential energy is \(E_p = \frac{1}{2} m \omega^2 x^2\). We are given that \(E_p = 3 E_k\). Since \(E_k = E - E_p\), we can substitute this to get \(E_p = 3(E - E_p)\), which simplifies to \(4 E_p = 3 E\). Substituting the expressions for \(E_p\) and \(E\), we get \(4 \left(\frac{1}{2} m \omega^2 x^2\right) = 3 \left(\frac{1}{2} m \omega^2 A^2\right)\), which simplifies to \(4x^2 = 3A^2\). Thus, \(x = \sqrt{\frac{3}{4}} A = \frac{\sqrt{3}}{2} A\).

Award 1 mark for the correct option C. Method: Use the energy conservation equation for SHM and the given ratio of potential to kinetic energy to establish a relationship between displacement and amplitude.

The density \(\rho\) of a sphere of mass \(m\) and diameter \(d\) is given by \(\rho = \frac{m}{V} = \frac{6m}{\pi d^3}\). The fractional uncertainty in the density is given by \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3 \frac{\Delta d}{d}\). The percentage uncertainty in mass is \(\frac{0.2}{75.4} \times 100\% \approx 0.265\%\). The percentage uncertainty in diameter is \(\frac{0.02}{2.40} \times 100\% \approx 0.833\%\). Therefore, the percentage uncertainty in the density is \(\% \Delta \rho = 0.265\% + 3(0.833\%) = 0.265\% + 2.500\% = 2.765\%\), which rounds to \(2.8\%\).

Award 1 mark for the correct option C. Method: Identify the formula for density in terms of mass and diameter, combine the fractional uncertainties with appropriate weighting factors, and convert to percentage uncertainty.

The uncertainty consists of two parts: a percentage of the reading and a fixed absolute value. First, calculate the percentage-based part: \(2\% \text{ of } 4.50\text{ V} = 0.02 \times 4.50\text{ V} = 0.09\text{ V}\). Second, add the fixed absolute uncertainty to this value: \(0.09\text{ V} + 0.05\text{ V} = 0.14\text{ V}\). Therefore, the total absolute uncertainty is \(\pm 0.14\text{ V}\).

Award 1 mark for the correct option C. Method: Calculate the absolute value of the percentage uncertainty and add it to the constant offset uncertainty to find the total absolute uncertainty.

The energy released in a nuclear reaction is equal to the difference between the total binding energy of the products and the total binding energy of the reactants. First, determine the total binding energy for each nucleus by multiplying the binding energy per nucleon by the nucleon number (mass number):
ewline - For \({}_1^2\text{H}\): \(2 \times 1.11 = 2.22\text{ MeV}\)
ewline - For \({}_1^3\text{H}\): \(3 \times 2.83 = 8.49\text{ MeV}\)
ewline - For \({}_2^4\text{He}\): \(4 \times 7.07 = 28.28\text{ MeV}\)
ewline - A single neutron \({}_0^1\text{n}\) is a free nucleon and has zero binding energy.
ewline Total reactant binding energy = \(2.22 + 8.49 = 10.71\text{ MeV}\).
ewline Total product binding energy = \(28.28 + 0 = 28.28\text{ MeV}\).
ewline Energy released = \(28.28 - 10.71 = 17.57\text{ MeV}\).

Award 1 mark for the correct option C. Method: Calculate total binding energy for reactants and products by multiplying binding energy per nucleon by nucleon number, then compute the difference.

The potential difference across the discharging capacitor at time \(t\) is given by: \(V = V_0 e^{-t/RC}\). For \(t = 2RC\), the potential difference is: \(V = V_0 e^{-2}\). The energy \(E\) stored in a capacitor is proportional to the square of the potential difference (\(E = \frac{1}{2}CV^2\)). Therefore, the remaining energy \(E\) is: \(E = \frac{1}{2} C (V_0 e^{-2})^2 = \left(\frac{1}{2} C V_0^2\right) e^{-4} = E_0 e^{-4}\). The fraction of the initial energy remaining is: \(\frac{E}{E_0} = e^{-4}\).

Correct option: B. Award 1 mark for the correct mathematical derivation showing that energy is proportional to \(V^2\), giving a factor of \(e^{-4}\).

The discharge of a capacitor is described by: \(V = V_0 e^{-t/RC}\). Taking the natural logarithm of both sides: \(\ln V = \ln V_0 - \frac{t}{RC}\). A graph of \(\ln(V / \text{V})\) against \(t\) has a gradient equal to \(-\frac{1}{RC}\). Given that the gradient is \(-0.080\text{ s}^{-1}\): \(-\frac{1}{RC} = -0.080\text{ s}^{-1} \implies RC = \frac{1}{0.080} = 12.5\text{ s}\). Using \(R = 50\text{ k}\Omega = 50 \times 10^3\ \Omega\): \(C = \frac{12.5}{50 \times 10^3} = 2.5 \times 10^{-4}\text{ F} = 250\ \mu\text{F}\).

Correct option: C. Award 1 mark for using gradient = \(-1/RC\) to calculate \(C = 250\ \mu\text{F}\).

The time constant is given by \(\tau = RC\). If the time constant \(\tau\) is increased, the discharge process is slower, and the total discharge time \(t\) being measured becomes larger. Since the absolute uncertainty in the time measurement \(\Delta t\) is constant, the percentage uncertainty in the time (and hence in \(\tau\)) is given by: \(\text{Percentage uncertainty} = \frac{\Delta t}{t} \times 100\%\). By increasing \(t\) (which occurs when \(\tau = RC\) is increased), the percentage uncertainty decreases. Increasing both \(R\) and \(C\) increases \(\tau\) and therefore significantly decreases the percentage uncertainty.

Correct option: C. Award 1 mark for identifying that increasing the time constant \(\tau = RC\) increases the measured time interval, thereby decreasing the percentage uncertainty for a constant absolute uncertainty.

The displacement \(x\) of an object in simple harmonic motion starting at \(x = +A\) at \(t = 0\) can be modelled as: \(x(t) = A \cos(\omega t)\) where \( \omega = \frac{2\pi}{T} \). We require the time \(t\) at which \(x(t) = -0.5A\): \(A \cos(\omega t) = -0.5A \implies \cos(\omega t) = -0.5\). The smallest positive angle \(\theta = \omega t\) for which \(\cos\theta = -0.5\) is \(\frac{2\pi}{3}\) rad. Therefore: \(\frac{2\pi}{T} t = \frac{2\pi}{3} \implies t = \frac{T}{3}\).

Correct option: D. Award 1 mark for correctly setting up the cosine equation and solving for the angle \(\frac{2\pi}{3}\) to obtain \(t = T/3\).

The total energy \(E\) of the particle is equal to its maximum kinetic energy, which occurs at \(x = 0\): \(E = \frac{1}{2} m \omega^2 x_0^2\). Since \(\omega = 2\pi f\): \(E = \frac{1}{2} m (2\pi f)^2 x_0^2 = 2\pi^2 m f^2 x_0^2\). At any displacement \(x\), the potential energy \(E_p\) is: \(E_p = \frac{1}{2} m \omega^2 x^2\). At \(x = \frac{1}{2}x_0\): \(E_p = \frac{1}{2} m \omega^2 \left(\frac{1}{2}x_0\right)^2 = \frac{1}{4} \left(\frac{1}{2} m \omega^2 x_0^2\right) = \frac{1}{4} E\). The kinetic energy \(E_k\) is the difference between total energy and potential energy: \(E_k = E - E_p = E - \frac{1}{4} E = \frac{3}{4} E\). \(E_k = \frac{3}{4} \left(2\pi^2 m f^2 x_0^2\right) = \frac{3}{2} \pi^2 m f^2 x_0^2\).

Correct option: C. Award 1 mark for expressing \(E_k\) as \(\frac{3}{4}\) of the total energy and correctly substituting \(\omega = 2\pi f\) to find the correct algebraic expression.

For a relationship of the form \(P = \frac{V^2}{R}\), the percentage uncertainty in \(P\) is found by adding the percentage uncertainties of each term, multiplied by their power indices: \(\frac{\Delta P}{P} = 2 \left(\frac{\Delta V}{V}\right) + \frac{\Delta R}{R}\). Substituting the given values: \(\text{Percentage uncertainty in } P = 2(3.0\%) + 2.0\% = 6.0\% + 2.0\% = 8.0\%\).

Correct option: D. Award 1 mark for correctly applying the rule for combining percentage uncertainties when a term is squared and another is divided.

The percentage uncertainty in the length \(l\) is: \(\frac{\Delta l}{l} \times 100\% = \frac{0.5}{82.0} \times 100\% \approx 0.61\%\). The percentage uncertainty in the period \(T\) is equal to the percentage uncertainty in the measured total time (since the number of oscillations, 20, is an exact constant): \(\frac{\Delta T}{T} \times 100\% = \frac{0.4}{36.4} \times 100\% \approx 1.10\%\). Using the formula for \(g = \frac{4\pi^2 l}{T^2}\), the percentage uncertainty in \(g\) is: \(\frac{\Delta g}{g} \times 100\% = \frac{\Delta l}{l} \times 100\% + 2 \left(\frac{\Delta T}{T} \times 100\%\right)\). \(\frac{\Delta g}{g} \times 100\% \approx 0.61\% + 2(1.10\%) = 0.61\% + 2.20\% = 2.81\% \approx 2.8\%\).

Correct option: C. Award 1 mark for finding individual percentage uncertainties and summing them, doubling the period's percentage uncertainty, to obtain 2.8%.

A helium-4 nucleus consists of 2 protons and 2 neutrons. First, calculate the total mass of the individual constituent nucleons: \(m_{\text{nucleons}} = 2 \times m_{\text{proton}} + 2 \times m_{\text{neutron}} = 2(1.00728\text{ u}) + 2(1.00866\text{ u}) = 4.03188\text{ u}\). Next, calculate the total binding energy \(E_B\) of the nucleus: \(E_B = 4 \times 7.07\text{ MeV} = 28.28\text{ MeV}\). Convert this binding energy to mass defect \(\Delta m\) in atomic mass units (\(\text{u}\)): \(\Delta m = \frac{28.28\text{ MeV}}{931.5\text{ MeV/u}} \approx 0.03036\text{ u}\). The actual mass of the helium-4 nucleus is less than the mass of the constituent nucleons by the mass defect: \(m_{\text{nucleus}} = m_{\text{nucleons}} - \Delta m = 4.03188\text{ u} - 0.03036\text{ u} = 4.00152\text{ u} \approx 4.0015\text{ u}\).

Correct option: A. Award 1 mark for calculating the mass of constituents, finding the total binding energy of 28.28 MeV, converting it to mass defect (0.03036 u), and subtracting it to get 4.0015 u.

The initial energy stored in the capacitor is \(E_0 = \frac{1}{2} C V_0^2\). During discharge, the potential difference \(V\) across the capacitor at time \(t\) is given by \(V = V_0 e^{-t/RC}\). At \(t = 2RC\), the potential difference is \(V = V_0 e^{-2}\). The energy \(E\) remaining in the capacitor is \(E = \frac{1}{2} C V^2 = \frac{1}{2} C (V_0 e^{-2})^2 = \frac{1}{2} C V_0^2 e^{-4} = E_0 e^{-4}\). Therefore, the fraction of the initial energy remaining is \(E / E_0 = e^{-4}\).

1 mark for the correct option C.

The discharge equation is \(V = V_0 e^{-t/RC}\). Taking natural logarithms on both sides gives \(\ln(V) = \ln(V_0) - \frac{t}{RC}\). Comparing this to the equation of a straight line \(y = mx + c\) where \(y = \ln(V)\) and \(x = t\), we get the gradient \(m = -\frac{1}{RC}\). Rearranging for \(R\) gives \(R = -\frac{1}{m C}\).

1 mark for the correct option A.

First, calculate the percentage uncertainty in each quantity:
- For mass \(M\): \(\frac{0.05}{5.00} \times 100\% = 1.0\%\)
- For diameter \(d\): \(\frac{0.02}{0.80} \times 100\% = 2.5\%\)
- For length \(L\): \(\frac{1.5}{150.0} \times 100\% = 1.0\%\)

Using the rules for combining uncertainties for \(\rho \propto M d^{-2} L^{-1}\):

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

\(\frac{\Delta \rho}{\rho} = 1.0\% + 2(2.5\%) + 1.0\% = 7.0\%\)

1 mark for the correct option C.

The standard equation for simple harmonic displacement is \(x = x_0 \cos(\omega t)\). Comparing this to the given equation, we find the amplitude \(x_0 = 0.050\text{ m}\) and angular frequency \(\omega = 40\text{ rad s}^{-1}\). The acceleration is given by \(a = -\omega^2 x\). The maximum acceleration occurs when displacement is equal to the amplitude, so \(a_{\text{max}} = \omega^2 x_0 = (40)^2 \times 0.050 = 1600 \times 0.050 = 80\text{ m s}^{-2}\).

1 mark for the correct option B.

The potential energy is \(E_P = \frac{1}{2} m \omega^2 x^2\).
The kinetic energy is \(E_K = \frac{1}{2} m \omega^2 (x_0^2 - x^2)\).
We are given that \(E_K = 3 E_P\):
\(\frac{1}{2} m \omega^2 (x_0^2 - x^2) = 3 \left(\frac{1}{2} m \omega^2 x^2\right)\)
\(x_0^2 - x^2 = 3 x^2\)
\(x_0^2 = 4 x^2\)
\(x = \sqrt{\frac{x_0^2}{4}} = 0.50 x_0\).

1 mark for the correct option B.

First, exclude the anomalous reading (39.5 s). The remaining four values for 20 oscillations are:
38.2 s, 38.4 s, 38.1 s, 38.3 s

Calculate the mean time for 20 oscillations:
\(\text{Mean} = \frac{38.2 + 38.4 + 38.1 + 38.3}{4} = 38.25\text{ s}\)

The period \(T\) is:
\(T = \frac{38.25}{20} = 1.9125\text{ s} \approx 1.91\text{ s}\)

The uncertainty in the 20 oscillations is half the range of the non-anomalous readings:
\(\text{Range} = 38.4 - 38.1 = 0.3\text{ s}\)
\(\text{Uncertainty in } 20T = \frac{0.3}{2} = 0.15\text{ s}\)

The uncertainty in the period \(T\) is:
\(\Delta T = \frac{0.15}{20} = 0.0075\text{ s} \approx 0.01\text{ s}\) (to 1 significant figure).

Thus, \(T = (1.91 \pm 0.01)\text{ s}\).

1 mark for the correct option A.

Calculate the total mass before the reaction:
\(\text{Mass of reactants} = 2.0141\text{ u} + 3.0160\text{ u} = 5.0301\text{ u}\)

Calculate the total mass after the reaction:
\(\text{Mass of products} = 4.0026\text{ u} + 1.0087\text{ u} = 5.0113\text{ u}\)

Calculate the mass defect \(\Delta m\):
\(\Delta m = 5.0301\text{ u} - 5.0113\text{ u} = 0.0188\text{ u}\)

Calculate the energy released:
\(E = 0.0188 \times 931.5\text{ MeV} = 17.5122\text{ MeV} \approx 17.5\text{ MeV}\).

1 mark for the correct option B.

In any energetically favorable nuclear reaction (both fission of heavy nuclei and fusion of light nuclei), the products have higher binding energy per nucleon, which means they are more tightly bound. Consequently, the total binding energy of the products is greater than that of the reactants, releasing energy (equal to the difference in total binding energy). Statement A is incorrect because heavy nuclei undergo fission, not fusion. Statement B is incorrect because light nuclei undergo fusion, not fission. Statement D is incorrect because a high binding energy per nucleon corresponds to the most stable nucleus, not the least.

1 mark for the correct option C.

The relationship between the discharging current \(I\) and time \(t\) is given by \(I = I_0 e^{-t/RC}\). Taking natural logarithms on both sides: \(\ln(I) = \ln(I_0) - \frac{t}{RC}\). A graph of \(\ln(I)\) against \(t\) has a gradient of \(-\frac{1}{RC}\). Therefore, \(-\frac{1}{RC} = -0.25\text{ s}^{-1}\), which gives the time constant \(\tau = RC = 4.0\text{ s}\). Since \(R = 2.0\text{ k}\Omega = 2.0 \times 10^3\text{ }\Omega\), we find \(C = \frac{4.0\text{ s}}{2.0 \times 10^3\text{ }\Omega} = 2.0 \times 10^{-3}\text{ F} = 2.0\text{ mF}\).

Award 1 mark for the correct calculation of capacitance. (1 mark for accuracy of finding 2.0 mF from the gradient expression).

The energy stored in a capacitor is given by \(E = \frac{1}{2} C V^2\). Since the potential difference \(V\) decays according to \(V = V_0 e^{-t/\tau}\), the stored energy at time \(t\) is \(E = \frac{1}{2} C \left(V_0 e^{-t/\tau}\right)^2 = E_0 e^{-2t/\tau}\). We want the energy to be \(\frac{E_0}{4}\), so: \(\frac{E_0}{4} = E_0 e^{-2t/\tau} \implies e^{-2t/\tau} = 0.25\). Taking natural logarithms: \(-2t/\tau = \ln(0.25) = -\ln(4) = -2\ln(2)\). This simplifies to \(t = \tau \ln(2) \approx 0.69\tau\).

Award 1 mark for obtaining the correct time \(t \approx 0.69\tau\).

The fractional uncertainty in \(\rho\) is given by the sum of the fractional uncertainties of the independent variables, taking into account the power of \(d\): \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\). Calculating each percentage uncertainty: \(\frac{\Delta R}{R} = \frac{0.1}{2.5} \times 100\% = 4.0\%\, \)\frac{\Delta d}{d} = \frac{0.02}{1.20} \times 100\% \approx 1.67\%\, \(\frac{\Delta L}{L} = \frac{0.1}{50.0} \times 100\% = 0.20\%\. Substituting these values: \)\frac{\Delta \rho}{\rho} = 4.0\% + 2(1.67\%) + 0.20\% = 7.54\% \approx 7.5\%\.

Award 1 mark for the correct calculation of the percentage uncertainty.

Rearranging the formula for \(g\) gives \(g = \frac{2h}{t^2}\). The percentage uncertainty in \(g\) is found by adding the percentage uncertainty in \(h\) to twice the percentage uncertainty in \(t\): \(\frac{\Delta g}{g} \times 100\% = \left(\frac{\Delta h}{h} + 2\frac{\Delta t}{t}\right) \times 100\% = 2\% + 2 \times 3\% = 8\%\.

Award 1 mark for correctly determining the percentage uncertainty as 8%.

In simple harmonic motion, the velocity has its maximum magnitude when the object passes through its equilibrium position, where displacement \(x = 0\). Given \(x = x_0 \cos(\omega t)\), the displacement is zero when \(\cos(\omega t) = 0\). Since \(\omega = \frac{2\pi}{T}\), this occurs when \(\frac{2\pi}{T} t = \frac{\pi}{2}\, which simplifies to \)t = \frac{T}{4}\).

Award 1 mark for identifying the correct time corresponding to maximum velocity magnitude.

The total energy \(E\) of an object undergoing simple harmonic motion is given by \(E = \frac{1}{2} m \omega^2 A^2\). Since \(\omega = 2\pi f\), we have \(E \propto m f^2 A^2\). Let \(E'\) be the energy of the second object: \(E' \propto (2m) \left(\frac{f}{2}\right)^2 (2A)^2 = 2m \times \frac{f^2}{4} \times 4A^2 = 2 m f^2 A^2\). Comparing this to the original energy, we find \(E' = 2E\).

Award 1 mark for correctly using the proportionality relation to find that the new energy is 2E.

Binding energy is related to the mass defect \(\Delta m\) by \(\text{Binding Energy} = \Delta m \times c^2\). Using the conversion factor \(1\text{ u} = 931.5\text{ MeV}/c^2\), the mass defect in unified atomic mass units (u) is: \(\Delta m = \frac{28.3\text{ MeV}}{931.5\text{ MeV u}^{-1}} \approx 0.0304\text{ u} = 3.0 \times 10^{-2}\text{ u}\).

Award 1 mark for the correct calculation of the mass defect.

First, calculate the combined capacitance \(C_s\) of the two capacitors in series: \(\frac{1}{C_s} = \frac{1}{3.0\text{ }\mu\text{F}} + \frac{1}{6.0\text{ }\mu\text{F}} = \frac{3}{6.0\text{ }\mu\text{F}} \implies C_s = 2.0\text{ }\mu\text{F}\). The total charge \(Q\) supplied is \(Q = C_s V = 2.0\text{ }\mu\text{F} \times 12.0\text{ V} = 24.0\text{ }\mu\text{C}\). Since capacitors in series carry the same charge, the charge on the \(3.0\text{ }\mu\text{F}\) capacitor is \(24.0\text{ }\mu\text{C}\). The potential difference \(V_1\) across the \(3.0\text{ }\mu\text{F}\) capacitor is \(V_1 = \frac{Q}{C_1} = \frac{24.0\text{ }\mu\text{C}}{3.0\text{ }\mu\text{F}} = 8.0\text{ V}\).

Award 1 mark for the correct combination of charge and potential difference.

The initial energy stored in the capacitor is:
\(E_0 = \frac{1}{2} C V_0^2\)

The potential difference \(V\) across the discharging capacitor at time \(t\) is given by:
\(V = V_0 e^{-t / (RC)}\)

At time \(t = 2RC\), substituting this into the equation gives:
\(V = V_0 e^{-2RC / (RC)} = V_0 e^{-2}\)

The energy \(E\) stored at this time is:
\(E = \frac{1}{2} C V^2 = \frac{1}{2} C (V_0 e^{-2})^2 = \frac{1}{2} C V_0^2 e^{-4}\)

Therefore, the ratio of the remaining energy to the initial energy is:
\(\frac{E}{E_0} = \frac{\frac{1}{2} C V_0^2 e^{-4}}{\frac{1}{2} C V_0^2} = e^{-4}\)

1 mark for the correct answer (C).

- Correct expression for energy in a capacitor: 1 mark
- Substitution of exponential potential decay for t = 2RC: 1 mark
- Evaluation of the ratio: 1 mark

The discharge current is given by:
\(I = I_0 e^{-t / (RC)}\)

Taking the natural logarithm of both sides:
\(\ln(I) = -\frac{t}{RC} + \ln(I_0)\)

This is in the linear form \(y = mx + c\), where:
- The gradient is \(-m = -\frac{1}{RC} \implies C = -\frac{1}{mR}\)
- The vertical intercept is \(c = \ln(I_0) \implies I_0 = e^c\)

The initial charge \(Q_0\) is related to the initial current by:
\(I_0 = \frac{Q_0}{RC} \implies Q_0 = I_0 RC\)

Substituting \(RC = -\frac{1}{m}\) and \(I_0 = e^c\) gives:
\(Q_0 = e^c \left(-\frac{1}{m}\right) = -\frac{e^c}{m}\)

Thus, the correct expressions are \(C = -\frac{1}{mR}\) and \(Q_0 = -\frac{e^c}{m}\).

1 mark for the correct answer (A).

- Linking the gradient to the time constant: 1 mark
- Linking the intercept to the initial current: 1 mark
- Combining to find the correct expression for initial charge: 1 mark

For simple harmonic motion starting from maximum positive displacement at \(t = 0\), the displacement equation is:
\(x = A \cos(\omega t)\)

We want to find the minimum time \(t\) when \(x = -\frac{A}{2}\):
\(-\frac{A}{2} = A \cos(\omega t) \implies \cos(\omega t) = -\frac{1}{2}\)

The smallest positive angle that satisfies this is:
\(\omega t = \frac{2\pi}{3}\)

Since \(\omega = \frac{2\pi}{T}\), we can substitute this into the equation:
\(\left(\frac{2\pi}{T}\right) t = \frac{2\pi}{3} \implies t = \frac{T}{3}\)

1 mark for the correct answer (C).

- Correct SHM displacement equation choice: 1 mark
- Solving the phase angle for the given displacement: 1 mark
- Finding the time in terms of period T: 1 mark

The total energy \(E_T\) of the system is the sum of its kinetic energy \(E_k\) and potential energy \(E_p\):
\(E_k + E_p = E_T\)

We are given that \(E_k = 3E_p\). Substituting this into the total energy equation:
\(3E_p + E_p = E_T \implies 4E_p = E_T\)

The potential energy at displacement \(x\) is:
\(E_p = \frac{1}{2} k x^2\)

The total energy at maximum displacement \(x_0\) is:
\(E_T = \frac{1}{2} k x_0^2\)

Therefore:
\(4 \left(\frac{1}{2} k x^2\right) = \frac{1}{2} k x_0^2\)
\(4 x^2 = x_0^2 \implies x^2 = \frac{x_0^2}{4}\)
\(x = \pm \frac{x_0}{2} = \pm 0.50 x_0\)

1 mark for the correct answer (B).

- Correct formulation of total energy in terms of potential energy: 1 mark
- Relationship between displacement and energy: 1 mark
- Final calculation of displacement: 1 mark

We can rewrite the formula for \(T\) in terms of powers of variables with non-zero uncertainties:
\(T = 2\pi \cdot I^{1/2} \cdot m^{-1/2} \cdot d^{-1/2} \cdot g^{-1/2}\)

To find the overall percentage uncertainty in \(T\), we sum the fractional uncertainties multiplied by the absolute value of their powers:
\(\frac{\Delta T}{T} = \frac{1}{2}\frac{\Delta I}{I} + \frac{1}{2}\frac{\Delta m}{m} + \frac{1}{2}\frac{\Delta d}{d}\)

Substituting the given percentage uncertainties:
\(\frac{\Delta T}{T} = \frac{1}{2}(4.0\%) + \frac{1}{2}(1.5\%) + \frac{1}{2}(2.5\%)\)
\(\frac{\Delta T}{T} = 2.0\% + 0.75\% + 1.25\% = 4.0\%\)

Therefore, the percentage uncertainty in \(T\) is \(\pm 4.0\%\).

1 mark for the correct answer (B).

- Identifying the correct power rule for uncertainties: 1 mark
- Correct application of coefficients (1/2) to the variables inside the square root: 1 mark
- Calculating the correct final sum: 1 mark

To find the energy released, we calculate the difference between the total binding energy of the product and the total binding energy of the reactants.

1. **Total binding energy of reactants:**
There are two deuterium nuclei, each containing 2 nucleons:
\(\text{Total binding energy of reactants} = 2 \times (2 \times 1.11\text{ MeV}) = 4.44\text{ MeV}\)

2. **Total binding energy of products:**
There is one helium nucleus, containing 4 nucleons:
\(\text{Total binding energy of products} = 4 \times 7.07\text{ MeV} = 28.28\text{ MeV}\)

3. **Energy released:**
\(\text{Energy released} = \text{Total binding energy of products} - \text{Total binding energy of reactants}\)
\(\text{Energy released} = 28.28\text{ MeV} - 4.44\text{ MeV} = 23.84\text{ MeV}\)

1 mark for the correct answer (C).

- Correct calculation of reactant binding energy: 1 mark
- Correct calculation of product binding energy: 1 mark
- Calculating the energy difference: 1 mark

The introduction of damping to an oscillating system has two key effects on its resonance curve:
1. The resonance peak becomes lower and broader (wider), reflecting the energy losses to the viscous medium.
2. The peak frequency (resonant frequency) shifts slightly to a value below the natural undamped frequency \(f_0\).

Therefore, the resonance peak becomes wider, and the peak frequency decreases slightly below \(f_0\).

1 mark for the correct answer (B).

- Correctly identifying that damping broadens the peak: 1 mark
- Correctly identifying that damping lowers the resonant frequency: 1 mark

1. **Calculate the mean of the recorded values:**
\(\bar{d}_{\text{recorded}} = \frac{0.42 + 0.45 + 0.43 + 0.44 + 0.41}{5} = \frac{2.15}{5} = 0.43\text{ mm}\)

2. **Correct for the systematic error:**
Since the gauge reads \(0.02\text{ mm}\) too low, we must add \(0.02\text{ mm}\) to get the true reading:
\(\bar{d}_{\text{corrected}} = 0.43\text{ mm} + 0.02\text{ mm} = 0.45\text{ mm}\)

3. **Determine the absolute uncertainty:**
The absolute uncertainty is estimated as half the range of the raw readings:
\(\text{Range} = 0.45\text{ mm} - 0.41\text{ mm} = 0.04\text{ mm}\)
\(\text{Uncertainty} = \frac{0.04\text{ mm}}{2} = 0.02\text{ mm}\)

Thus, the result is expressed as \((0.45 \pm 0.02)\text{ mm}\).

1 mark for the correct answer (C).

- Calculating the average raw value: 1 mark
- Applying the systematic zero error correction correctly: 1 mark
- Estimating uncertainty using half the range: 1 mark

(a) The potential difference \(V\) at time \(t\) during discharge is given by \(V = V_0 e^{-t/\tau}\), where \(\tau = RC\).

When \(t = \tau\), the equation becomes:

\(V = V_0 e^{-1} \approx V_0 \times 0.368\)

Therefore, the potential difference decreases to \(36.8\%\) of its initial value \(V_0\), which is approximately \(37\%\).

(b)(i) Given \(V_0 = 12.0\text{ V}\), \(V = 4.4\text{ V}\), \(t = 15.0\text{ s}\), and \(R = 1.5 \times 10^5\ \Omega\).

Using \(V = V_0 e^{-t / (RC)}\):

\(4.4 = 12.0 e^{-15.0 / (1.5 \times 10^5 \times C)}\)

\(\frac{4.4}{12.0} = e^{-15.0 / (1.5 \times 10^5 \times C)}\)

Taking natural logarithms on both sides:

\(\ln(0.3667) = -\frac{15.0}{1.5 \times 10^5 \times C}\)

\(-1.0033 = -\frac{10^{-4}}{C}\)

\(C = \frac{10^{-4}}{1.0033} \approx 9.97 \times 10^{-5}\text{ F}\) (or \(1.0 \times 10^{-4}\text{ F}\)).

(b)(ii) The energy stored \(E\) in the capacitor is given by:

\(E = \frac{1}{2} C V^2\)

At \(t = 15.0\text{ s}\), \(V = 4.4\text{ V}\).

Using \(C = 9.97 \times 10^{-5}\text{ F}\):

\(E = 0.5 \times 9.97 \times 10^{-5} \times (4.4)^2 = 9.65 \times 10^{-4}\text{ J}\) (or \(9.7 \times 10^{-4}\text{ J}\)).

Using \(C = 1.0 \times 10^{-4}\text{ F}\):

\(E = 0.5 \times 1.0 \times 10^{-4} \times (4.4)^2 = 9.68 \times 10^{-4}\text{ J}\) (or \(9.7 \times 10^{-4}\text{ J}\)).

(a)
- M1: For stating or writing the discharge equation with \(t = \tau\) or \(t = RC\). [1 mark]
- A1: Showing that \(e^{-1} \approx 0.368\) and concluding it is approximately \(37\%\). [1 mark]

(b)(i)
- C1: Rearranging the formula to make \(e^{-t/RC}\) the subject: \(4.4 / 12.0\) or \(0.367\). [1 mark]
- C1: Taking natural logarithms correctly: \(\ln(0.367) = -1.00\). [1 mark]
- C1: Substituting value of \(R = 1.5 \times 10^5\ \Omega\). [1 mark]
- A1: Correct final answer for \(C\) in range \(9.9 \times 10^{-5}\text{ F}\) to \(1.0 \times 10^{-4}\text{ F}\) with correct units. [1 mark]

(b)(ii)
- C1: Recall of \(E = \frac{1}{2} C V^2\). [1 mark]
- C1: Substitution of \(V = 4.4\text{ V}\) and their value of \(C\). [1 mark]
- A0.5: Correct calculation of \(E\) to 2 or 3 sig figs: \(9.7 \times 10^{-4}\text{ J}\) (or \(9.6 \times 10^{-4}\text{ J}\)). [0.5 marks]

(a) Comparing the given equation \(x = 0.035 \cos(8.4 t)\) with the standard equation for SHM displacement \(x = x_0 \cos(\omega t)\), the amplitude \(x_0\) is the coefficient in front of the cosine term. Thus, \(x_0 = 0.035\text{ m}\).

(b)(i) The angular frequency is \(\omega = 8.4\text{ rad s}^{-1}\).

Since \(\omega = 2\pi f\):

\(f = \frac{\omega}{2\pi} = \frac{8.4}{2\pi} \approx 1.34\text{ Hz}\) (or \(1.3\text{ Hz}\)).

(b)(ii) The maximum acceleration \(a_0\) of the mass is given by:

\(a_0 = \omega^2 x_0\)

\(a_0 = (8.4)^2 \times 0.035 = 70.56 \times 0.035 = 2.47\text{ m s}^{-2}\) (or \(2.5\text{ m s}^{-2}\)).

(b)(iii) For a mass oscillating on a spring, the angular frequency satisfies \(\omega^2 = \frac{k}{m}\).

Therefore, the spring constant \(k\) is:

\(k = m \omega^2 = 0.450 \times (8.4)^2 = 0.450 \times 70.56 = 31.8\text{ N m}^{-1}\) (or \(32\text{ N m}^{-1}\)).

(a)
- A1: Correctly stating \(0.035\text{ m}\) (accept \(3.5\text{ cm}\)). [1 mark]

(b)(i)
- C1: Identifying \(\omega = 8.4\text{ rad s}^{-1}\). [1 mark]
- C1: Recall and use of \(\omega = 2\pi f\). [1 mark]
- A0.5: Correct calculation to 2 or 3 sig figs: \(1.3\text{ Hz}\) or \(1.34\text{ Hz}\). [0.5 marks]

(b)(ii)
- C1: Recall and use of \(a_0 = \omega^2 x_0\) (ignore negative sign). [1 mark]
- C1: Substitution of values: \(8.4^2 \times 0.035\). [1 mark]
- A0.5: Correct calculation to 2 or 3 sig figs: \(2.5\text{ m s}^{-2}\) or \(2.47\text{ m s}^{-2}\). [0.5 marks]

(b)(iii)
- C1: Recall and use of \(\omega^2 = k/m\) or equivalent. [1 mark]
- C1: Substitution of values: \(0.450 \times 8.4^2\). [1 mark]
- A0.5: Correct calculation to 2 or 3 sig figs with correct units: \(32\text{ N m}^{-1}\) or \(31.8\text{ N m}^{-1}\). [0.5 marks]

(a) The period \(T\) is the time for one oscillation:

\(T = \frac{t}{N} = \frac{36.4}{20} = 1.82\text{ s}\)

The absolute uncertainty in \(T\), \(\Delta T\), is given by:

\(\Delta T = \frac{\Delta t}{N} = \frac{0.2}{20} = 0.01\text{ s}\)

Therefore, \(T = 1.82 \pm 0.01\text{ s}\).

(b) Squaring both sides of the pendulum equation:

\(T^2 = 4\pi^2 \frac{L}{g} \implies g = \frac{4\pi^2 L}{T^2}\)

Substituting the measured values:

\(g = \frac{4\pi^2 \times 0.824}{1.82^2} = \frac{32.529}{3.3124} \approx 9.820\text{ m s}^{-2}\)

Since the raw data (\(L\) and \(t\)) are given to 3 significant figures, and the uncertainty is \(0.16\text{ m s}^{-2}\), \(g\) is appropriately expressed to 2 or 3 significant figures: \(9.8\text{ m s}^{-2}\) or \(9.82\text{ m s}^{-2}\).

(c) The formula for percentage uncertainty in \(g\) is:

\(\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T}\)

\(\frac{\Delta L}{L} = \frac{0.004}{0.824} \approx 0.00485\) (or \(0.49\%\))

\(\frac{\Delta T}{T} = \frac{0.01}{1.82} \approx 0.00549\) (or \(0.55\%\))

\(\frac{\Delta g}{g} = 0.00485 + 2(0.00549) = 0.00485 + 0.01098 = 0.01583\)

Therefore, the percentage uncertainty in \(g\) is \(1.58\%\) or \(1.6\%\) (accept \(2\%\)).

(a)
- C1: Show period is \(36.4 / 20 = 1.82\text{ s}\). [1 mark]
- C1: Formula for absolute uncertainty: \(\Delta T = \Delta t / 20\). [1 mark]
- A1: Correct calculation of \(\Delta T = 0.01\text{ s}\). [1 mark]

(b)
- C1: Rearranging formula correctly to make \(g\) the subject. [1 mark]
- C1: Substitution of \(L = 0.824\) and \(T = 1.82\). [1 mark]
- A1: Value of \(g = 9.8\text{ m s}^{-2}\) or \(9.82\text{ m s}^{-2}\) with correct units and 2 or 3 significant figures. [1 mark]

(c)
- C1: Correct fractional/percentage uncertainty formula: \(\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T}\). [1 mark]
- C1: Substituting values correctly: \(\frac{0.004}{0.824} + 2 \times \frac{0.01}{1.82}\). [1 mark]
- A0.5: Correct percentage uncertainty value of \(1.6\%\) (or \(1.58\%\) or \(2\%\)). [0.5 marks]

(a) The total mass of the reactants is:

\(m_{\text{initial}} = 2.014102\text{ u} + 3.016049\text{ u} = 5.030151\text{ u}\)

The total mass of the products is:

\(m_{\text{final}} = 4.002603\text{ u} + 1.008665\text{ u} = 5.011268\text{ u}\)

The mass defect \(\Delta m\) is:

\(\Delta m = m_{\text{initial}} - m_{\text{final}} = 5.030151 - 5.011268 = 0.018883\text{ u}\).

(b) To find the energy released, first convert the mass defect into kilograms:

\(\Delta m = 0.018883 \times 1.6605 \times 10^{-27}\text{ kg} = 3.1355 \times 10^{-29}\text{ kg}\)

Using Einstein's mass-energy relation \(E = \Delta m c^2\) (where \(c = 3.00 \times 10^8\text{ m s}^{-1}\)):

\(E = 3.1355 \times 10^{-29} \times (3.00 \times 10^8)^2 = 2.822 \times 10^{-12}\text{ J}\)

This is approximately \(2.8 \times 10^{-12}\text{ J}\).

(c) Let \(N\) be the number of deuterium nuclei (which is also the number of tritium nuclei).

Each reacting pair of \(^{2}_{1}\text{H}\) and \(^{3}_{1}\text{H}\) has a total mass of:

\(m_{\text{pair}} = 5.030151\text{ u} = 5.030151 \times 1.6605 \times 10^{-27}\text{ kg} = 8.3526 \times 10^{-27}\text{ kg}\)

The total mass of the mixture is \(M = 1.00\text{ g} = 1.00 \times 10^{-3}\text{ kg}\).

Thus, the number of pairs \(N\) in \(1.00\text{ g}\) of the mixture is:

\(N = \frac{M}{m_{\text{pair}}} = \frac{1.00 \times 10^{-3}}{8.3526 \times 10^{-27}} = 1.1972 \times 10^{23}\)

Since each pair undergoes one fusion reaction, the total energy released \(E_{\text{total}}\) is:

\(E_{\text{total}} = N \times E = 1.1972 \times 10^{23} \times 2.822 \times 10^{-12}\text{ J} = 3.38 \times 10^{11}\text{ J}\) (or \(3.4 \times 10^{11}\text{ J}\)).

(a)
- C1: Sum of reactant masses AND sum of product masses calculated correctly. [1 mark]
- C1: Calculation of difference in masses. [1 mark]
- A0.5: Correct answer: \(0.018883\text{ u}\) (accept \(0.0189\text{ u}\)). [0.5 marks]

(b)
- C1: Conversion of mass defect from \(\text{u}\) to \(\text{kg}\). [1 mark]
- C1: Recall and use of \(E = \Delta m c^2\). [1 mark]
- A1: Correct calculation showing \(2.82 \times 10^{-12}\text{ J}\). [1 mark]

(c)
- C1: Calculating mass of a single pair of deuterium and tritium nuclei in kilograms (\(8.35 \times 10^{-27}\text{ kg}\)). [1 mark]
- C1: Finding number of reacting pairs \(N = 1.20 \times 10^{23}\). [1 mark]
- A1: Finding total energy released: \(3.38 \times 10^{11}\text{ J}\) (accept \(3.4 \times 10^{11}\text{ J}\)). [1 mark]

(a) The diagram must include:
1. A DC power supply connected via a single-pole double-throw (SPDT) switch to a capacitor. This allows the capacitor to be connected to either the supply or the discharge circuit.
2. In the other position, the switch connects the capacitor across a resistor \(R\).
3. A voltmeter connected in parallel across the capacitor (or across the resistor).
4. Correct circuit symbols for the cell/battery, switch, capacitor, resistor, and voltmeter.

(b) Procedure:
1. Flip the switch to position 1 to connect the capacitor to the power supply. Leave it for a few seconds so that the capacitor is fully charged. The voltmeter reading will stabilize at \(V_0 = 9.0\text{ V}\).
2. Flip the switch to position 2 to start discharging. Simultaneously start the stopwatch.
3. Monitor the voltmeter. Stop the stopwatch when the potential difference across the capacitor falls to \(\frac{V_0}{e} = \frac{9.0}{2.718} \approx 3.3\text{ V}\).
4. The recorded time \(t\) on the stopwatch is equal to the time constant \(\tau\).
5. Repeat the measurement multiple times for the same resistor \(R\) and calculate the average value of \(\tau\).

(c) Graphical Analysis:
1. Since \(\tau = RC\), the relationship between \(\tau\) and \(R\) is linear.
2. Repeat the discharge procedure for several different known values of resistor \(R\), measuring the corresponding average time constant \(\tau\) for each.
3. Plot a graph of \(\tau\) (on the y-axis) against \(R\) (on the x-axis).
4. Draw a line of best fit. The graph should be a straight line that passes through the origin.
5. The gradient of this straight line is equal to the capacitance \(C\). Thus, calculating the gradient gives the value of \(C\).

(a)
- G1: Voltmeter connected in parallel across the capacitor (or resistor). [1 mark]
- G1: A double-throw switch (or equivalent arrangement) shown correctly allowing alternate charging and discharging loops. [1 mark]
- G0.5: All circuit symbols (power supply, capacitor, resistor, voltmeter, switch) correct. [0.5 marks]

(b)
- D1: Action of charging the capacitor to full voltage (stated as \(9.0\text{ V}\)). [1 mark]
- D1: Simultaneous action of starting the stopwatch and flipping the switch to discharge. [1 mark]
- D1: Explaining that the stopwatch is stopped when \(V\) reaches \(0.37 V_0 \approx 3.3\text{ V}\), which corresponds to the time constant \(\tau\). [1 mark]

(c)
- D1: Stating that \(\tau\) must be measured for several different resistors \(R\). [1 mark]
- D1: Stating that a graph of \(\tau\) against \(R\) is plotted. [1 mark]
- D1: Explaining that the gradient of this straight line is equal to the capacitance \(C\). [1 mark]

(a) The angular frequency \(\omega\) is:

\(\omega = \frac{2\pi}{T} = \frac{2\pi}{0.60} = 10.47\text{ rad s}^{-1}\)

The amplitude is \(x_0 = 5.0\text{ cm} = 0.050\text{ m}\).

The maximum speed \(v_0\) occurs at the equilibrium position and is given by:

\(v_0 = \omega x_0 = 10.47 \times 0.050 = 0.5236\text{ m s}^{-1}\)

The maximum kinetic energy \(E_{\text{k,max}}\) is:

\(E_{\text{k,max}} = \frac{1}{2} m v_0^2 = 0.5 \times 0.120 \times (0.5236)^2 = 0.060 \times 0.2741 = 0.0164\text{ J}\)

This is approximately \(0.016\text{ J}\).

(b)(i) The total energy of the oscillator is constant and is equal to \(E_{\text{total}} = E_{\text{k,max}} = 0.0164\text{ J}\).

At any displacement \(x\):

\(E_{\text{total}} = E_{\text{k}} + E_{\text{p}}\)

We are given that \(E_{\text{k}} = E_{\text{p}}\), so:

\(E_{\text{total}} = 2 E_{\text{p}} \implies E_{\text{p}} = \frac{1}{2} E_{\text{total}}\)

Since \(E_{\text{p}} = \frac{1}{2} k x^2\) and \(E_{\text{total}} = \frac{1}{2} k x_0^2\):

\(\frac{1}{2} k x^2 = \frac{1}{2} \left( \frac{1}{2} k x_0^2 \right) \implies x^2 = \frac{1}{2} x_0^2\)

\(x = \frac{x_0}{\sqrt{2}} = \frac{5.0\text{ cm}}{\sqrt{2}} \approx 3.54\text{ cm}\) (or \(0.035\text{ m}\)).

(b)(ii) The total energy is \(E_{\text{total}} = \frac{1}{2} k x_0^2 = 0.0164\text{ J}\).

The potential energy \(E_{\text{p}}\) at a displacement \(x\) is given by:

\(E_{\text{p}} = \frac{1}{2} k x^2 = E_{\text{total}} \left(\frac{x}{x_0}\right)^2\)

For \(x = 3.0\text{ cm}\) and \(x_0 = 5.0\text{ cm}\):

\(E_{\text{p}} = 0.0164 \times \left(\frac{3.0}{5.0}\right)^2 = 0.0164 \times 0.36 = 5.90 \times 10^{-3}\text{ J}\) (or \(5.9 \times 10^{-3}\text{ J}\)).

(a)
- C1: Calculation of angular frequency \(\omega = 10.5\text{ rad s}^{-1}\). [1 mark]
- C1: Recall and use of \(v_0 = \omega x_0\) to get \(0.52\text{ m s}^{-1}\). [1 mark]
- A1: Substituting into \(E_{\text{k}} = \frac{1}{2} m v^2\) to show \(0.016\text{ J}\). [1 mark]

(b)(i)
- C1: Stating that \(E_{\text{p}} = \frac{1}{2} E_{\text{total}}\) or writing \(\frac{1}{2} k x^2 = \frac{1}{4} k x_0^2\). [1 mark]
- C1: Stating or using \(x = x_0 / \sqrt{2}\). [1 mark]
- A0.5: Correctly calculating \(x = 3.5\text{ cm}\) (or \(0.035\text{ m}\)). [0.5 marks]

(b)(ii)
- C1: Recall and use of \(E_{\text{p}} = \frac{1}{2} k x^2\) or \(E_{\text{p}} = E_{\text{total}} (x/x_0)^2\). [1 mark]
- C1: Calculating spring constant \(k = 13.1\text{ N m}^{-1}\) or substituting ratio \(0.6^2 = 0.36\). [1 mark]
- A1: Finding \(E_{\text{p}} = 5.9 \times 10^{-3}\text{ J}\) (or \(0.0059\text{ J}\)). [1 mark]

(a) To identify the largest contributor, calculate the percentage uncertainty for each measured variable:

- For Resistance \(R\):
\(\text{Percentage uncertainty in } R = \frac{0.05}{4.25} \times 100\% \approx 1.18\%\)

- For Length \(L\):
\(\text{Percentage uncertainty in } L = \frac{0.002}{1.250} \times 100\% \approx 0.16\%\)

- For Diameter \(d\):
\(\text{Percentage uncertainty in } d = \frac{0.02}{0.38} \times 100\% \approx 5.26\%\)

Comparing these values, the diameter \(d\) contributes the greatest percentage uncertainty of \(5.3\%\).

(b) Using the formula \(\rho = \frac{R \pi d^2}{4 L}\), the percentage uncertainty in \(\rho\) is given by:

\(\frac{\Delta \rho}{\rho} \times 100\% = \left( \frac{\Delta R}{R} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L} \right) \times 100\%\)

Note that the exponent of \(d\) is 2, so its percentage uncertainty must be multiplied by 2.

Substituting the values:

\(\frac{\Delta \rho}{\rho} \times 100\% = 1.18\% + 2(5.26\%) + 0.16\%\)

\(\frac{\Delta \rho}{\rho} \times 100\% = 1.18\% + 10.52\% + 0.16\% = 11.86\%\)

To an appropriate number of significant figures, the percentage uncertainty is \(12\%\) (or \(11.9\%\)).

(c) To reduce the uncertainty in the diameter \(d\):
1. The student should use a digital micrometer screw gauge instead of standard calipers to improve precision.
2. The student should measure the diameter of the wire at several different positions along its length, and at mutually perpendicular orientations at each position. Stating that they take an average of these measurements reduces the impact of random errors and non-uniform cross-section.

(a)
- C1: Percentage uncertainty in \(R\) calculated as \(1.2\%\) (or \(1.18\%\)). [1 mark]
- C1: Percentage uncertainty in \(L\) calculated as \(0.16\%\). [1 mark]
- C1: Percentage uncertainty in \(d\) calculated as \(5.3\%\) (or \(5.26\%\)). [1 mark]
- A1: Correct conclusion stating that \(d\) has the largest percentage uncertainty. [1 mark]

(b)
- C1: Correct formula for total percentage uncertainty showing multiplication of the diameter's percentage uncertainty by 2: \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\). [1 mark]
- C1: Substitution of values: \(1.18 + 2(5.26) + 0.16\). [1 mark]
- A0.5: Correct final calculation: \(12\%\) or \(11.9\%\). [0.5 marks]

(c)
- B1: Stating to use a micrometer screw gauge (rather than a ruler or sliding calipers). [1 mark]
- B1: Stating to measure at multiple positions/orientations and take the average. [1 mark]

### Representative Experimental Data

| \(R / \text{k}\Omega\) | \(t_1 / \text{s}\) | \(t_2 / \text{s}\) | Mean \(t / \text{s}\) |
|---|---|---|---|
| 10.0 | 5.1 | 5.3 | 5.2 |
| 22.0 | 11.3 | 11.5 | 11.4 |
| 33.0 | 16.9 | 17.1 | 17.0 |
| 47.0 | 24.2 | 24.4 | 24.3 |
| 68.0 | 35.0 | 35.2 | 35.1 |
| 100.0 | 51.5 | 51.7 | 51.6 |

### Graph and Calculation Steps

1. **Plotting the Graph:**
* Plot \(t / \text{s}\) on the y-axis against \(R / \text{k}\Omega\) on the x-axis.
* Scale: \(x\)-axis: \(0\) to \(100\text{ k}\Omega\) (\(2\text{ cm} = 20\text{ k}\Omega\)), \(y\)-axis: \(0\) to \(60\text{ s}\) (\(2\text{ cm} = 10\text{ s}\)). This covers more than \(50\%\) of the grid.

2. **Gradient Determination:**
* Using two points on the line of best fit that are far apart:
Point 1: \((20.0, 10.3)\)
Point 2: \((90.0, 46.5)\)
\[ \text{Gradient } m = \frac{46.5 - 10.3}{90.0 - 20.0} = \frac{36.2}{70.0} = 0.517\text{ s / k}\Omega \]
* In standard SI units:
\[ m = 0.517 \times 10^{-3}\text{ s / }\Omega \]

3. **Determining Capacitance \(C\):**
* The formula gives:
\[ t = R C \ln(3) \implies \frac{t}{R} = C \ln(3) \]
* Therefore, \(\text{Gradient } m = C \ln(3)\).
\[ C = \frac{m}{\ln(3)} = \frac{0.517 \times 10^{-3}}{1.0986} \approx 4.71 \times 10^{-4}\text{ F} = 471\ \mu\text{F} \]

### Table of Results (9 Marks)
* **Successful Collection of Data (1 mark):** Six sets of values of \(R\) and \(t\) recorded with correct trend (as \(R\) increases, \(t\) increases).
* **Range of \(R\) (1 mark):** Must include \(R \le 20\text{ k}\Omega\) and \(R \ge 80\text{ k}\Omega\).
* **Column Headings (1 mark):** Each column heading must contain its quantity and unit in standard form, e.g., \(R / \text{k}\Omega\), \(t / \text{s}\).
* **Consistency of Raw Readings (1 mark):** All values of \(t\) must be recorded to the nearest 0.1 s or 0.01 s based on the precision of the digital stopwatch.
* **Significant Figures (1 mark):** Number of significant figures for \(t\) must be consistent with the timing instrument precision.
* **Data Quality (2 marks):** Points plotted on the graph should show very little scatter about the straight line of best fit. Deduct 1 mark for any clearly anomalous point.
* **Replication of Readings (2 marks):** Repeated trials for each time measurement, with the mean value computed and tabulated.

### Graph (4 Marks)
* **Axes (1 mark):** Correctly labelled with units. Scale must be chosen so that points occupy more than half of the grid in both directions. No awkward scales (e.g. 3, 7).
* **Plotting (1 mark):** Points plotted to an accuracy of better than half a small square.
* **Line of Best Fit (1 mark):** A straight line drawn with a balanced distribution of points on either side.
* **Worst Acceptable Line (1 mark):** Clearly labelled worst acceptable line (either steeper or shallower) passing through the first and last data points' extreme error bars.

### Gradient and Intercept (2 Marks)
* **Gradient calculation (1 mark):** Must use a large triangle with hypotenuse longer than half the length of the drawn line.
* **y-intercept (1 mark):** Correctly read from the axis or calculated using a point on the line of best fit.

### Calculation and Constants (5 Marks)
* **Formula identification (1 mark):** Equates the calculated gradient to \(C \ln(3)\).
* **Capacitance Value (2 marks):** Correct calculation of \(C\) in the range \(400\ \mu\text{F}\) to \(540\ \mu\text{F}\).
* **Units (1 mark):** Correct unit of capacitance (e.g. \%F, \mu\text{F}\ or \text{s}\ \Omega^{-1}\) must be shown.
* **Significant Figures (1 mark):** Capacitance expressed to 2 or 3 significant figures, matching the precision of the raw data.

### Representative Experimental Measurements

1. **First length \(L_1 = 0.600\text{ m}\):**
* Time for 10 oscillations:
\(t_1 = 9.4\text{ s}\), \(t_2 = 9.2\text{ s}\)
\(t_{\text{mean}} = 9.3\text{ s}\)
* Period \(T_1 = \frac{9.3}{10} = 0.93\text{ s}\)

2. **Uncertainty estimation:**
* Reading error/reaction time on stopwatch \(\Delta t = 0.2\text{ s}\).
* Percentage uncertainty in \(T_1\):
\[ \frac{\Delta T_1}{T_1} \times 100\% = \frac{0.2}{9.3} \times 100\% \approx 2.2\% \]

3. **Second length \(L_2 = 0.450\text{ m}\):**
* Time for 10 oscillations:
\(t_1 = 6.1\text{ s}\), \(t_2 = 5.9\text{ s}\)
\(t_{\text{mean}} = 6.0\text{ s}\)
* Period \(T_2 = \frac{6.0}{10} = 0.60\text{ s}\)

4. **Calculating \(k_1\) and \(k_2\):**
* \(k_1 = \frac{T_1^2}{L_1^3} = \frac{0.93^2}{0.600^3} = \frac{0.865}{0.216} \approx 4.00\text{ s}^2\text{ m}^{-3}\)
* \(k_2 = \frac{T_2^2}{L_2^3} = \frac{0.60^2}{0.450^3} = \frac{0.360}{0.0911} \approx 3.95\text{ s}^2\text{ m}^{-3}\)

5. **Percentage Difference and Conclusion:**
* Percentage difference between \(k_1\) and \(k_2\):
\[ \% \text{ diff} = \frac{|4.00 - 3.95|}{4.00} \times 100\% = 1.25\% \]
* Since the percentage difference of \(1.25\%\) is less than the percentage uncertainty in the period measurement (\(2.2\%\)), the experimental results support the suggested relationship.

### Measurements and Calculations (8 Marks)
* **Length Measurement (1 mark):** First value of \(L_1\) measured to the nearest mm, within \(0.580\text{ m} \le L_1 \le 0.620\text{ m}\).
* **Raw Oscillations Data (1 mark):** Raw timing data \(t\) recorded for \(N \ge 10\) complete oscillations, with repeated readings shown.
* **First Period calculation (1 mark):** \(T_1\) calculated correctly with unit (s).
* **Uncertainty analysis (1 mark):** Absolute uncertainty in raw time taken as \(0.1\text{ s}\) to \(0.3\text{ s}\), and percentage uncertainty in \(T_1\) correctly computed.
* **Second Measurements (2 marks):** Second length \(L_2 \approx 0.450\text{ m}\) and corresponding timing data recorded accurately.
* **Calculation of constants (2 marks):** \(k_1\) and \(k_2\) calculated with correct significant figures and units (\(s^2\ m^{-3}\)).

### Evaluation and Conclusion (2 Marks)
* **Conclusion (2 marks):** Correct calculation of percentage difference between \(k_1\) and \(k_2\). Clear statement of whether relationship is supported, comparing percentage difference to the experimental uncertainty (or a standard threshold such as 10%).

### Limitations & Improvements (10 Marks)
Award 1 mark for each point identified (up to 5 limitations and 5 improvements from the pairs below):

* **Limitation 1:** Only two sets of readings are insufficient to validate the proposed relationship.
* **Improvement 1:** Take multiple readings for several different lengths of the cantilever and plot a graph of \(T^2\) against \(L^3\).
* **Limitation 2:** Large uncertainty due to reaction time when timing high-frequency oscillations.
* **Improvement 2:** Increase the mass to slow down oscillations (reducing frequency), or use a light gate connected to a digital timer/data logger.
* **Limitation 3:** Oscillations die out quickly (high damping), making it hard to count 10 cycles reliably.
* **Improvement 3:** Use a heavier mass to increase the stored energy and minimize damping, or record the motion on a video with a digital timer and analyze in slow motion.
* **Limitation 4:** Ruler slips or moves inside the jaws of the clamp during oscillation.
* **Improvement 4:** Place rubber pads or sandpaper between the clamp jaws and the wooden rule to increase friction.
* **Limitation 5:** Difficult to release the cantilever without inducing side-to-side/twisting (non-vertical) oscillations.
* **Improvement 5:** Use a mechanical release mechanism, such as cutting/burning a thread holding the rule in its displaced position.

(a) Capacitance is defined as the charge stored per unit potential difference: \( C = Q/V \).

(b) Since \( C = Q/V \), the unit of capacitance is \( \text{C V}^{-1} \). Since \( R = V/I \), the unit of resistance is \( \text{V A}^{-1} \).
The unit of the time constant \( RC \) is \( \text{V A}^{-1} \times \text{C V}^{-1} = \text{C A}^{-1} \).
Since electric current \( I = Q/t \), we have \( \text{A} = \text{C s}^{-1} \), which means \( \text{C A}^{-1} = \text{s} \). Thus, the unit of \( RC \) is the second.

(c)(i) The potential difference across the discharging capacitor is given by:
\( V = V_0 e^{-t/RC} \)
Substitute the given values:
\( 3.4 = 12.0 e^{-15.0 / (R \times 470 \times 10^{-6})} \)
\( \frac{3.4}{12.0} = 0.2833 \)
\( \ln(0.2833) = -1.261 \)
\( -1.261 = -\frac{15.0}{R \times 470 \times 10^{-6}} \)
\( R \times 470 \times 10^{-6} = 11.89 \)
\( R = \frac{11.89}{470 \times 10^{-6}} = 2.53 \times 10^4\ \Omega \approx 2.5 \times 10^4\ \Omega \) (or \( 25\ \text{k}\Omega \)).

(ii) The energy stored is given by:
\( E = \frac{1}{2} C V^2 \)
At \( t = 15.0\text{ s} \), the potential difference \( V \) across the capacitor is \( 3.4\text{ V} \).
\( E = 0.5 \times (470 \times 10^{-6}) \times (3.4)^2 = 2.717 \times 10^{-3}\text{ J} \approx 2.7 \times 10^{-3}\text{ J} \) (or \( 2.7\text{ mJ} \)).

(a)
- Charge divided by potential difference (or ratio of charge to potential difference) [1]

(b)
- States unit of \( C \) is \( \text{C V}^{-1} \) and unit of \( R \) is \( \text{V A}^{-1} \) [1]
- Multiplies units and shows simplification to \( \text{s} \) [1]

(c)(i)
- Expresses relationship \( V = V_0 e^{-t/RC} \) [1]
- Correct substitution of numbers: \( 3.4 = 12.0 e^{-15.0/RC} \) [1]
- Finds time constant \( RC = 11.9\text{ s} \) [1]
- Final answer \( 2.5 \times 10^4\ \Omega \) (or \( 2.53 \times 10^4\ \Omega \)) [1]

(c)(ii)
- Recalls \( E = \frac{1}{2} C V^2 \) [1]
- Correct substitution: \( E = 0.5 \times (470 \times 10^{-6}) \times (3.4)^2 \) [1]
- Final answer \( 2.7 \times 10^{-3}\text{ J} \) (or \( 2.72 \times 10^{-3}\text{ J} \)) [1]

(a) The circuit diagram must show:
- A cell connected to one terminal of a two-way switch, which connects to the parallel combination of the capacitor and the voltmeter.
- The other terminal of the two-way switch connected to the resistor \( R \), allowing the capacitor to discharge through it.
- Voltmeter connected in parallel with the capacitor.

(b) To carry out the experiment:
1. Flip the switch to the charge position to fully charge the capacitor to \( 6.0\text{ V} \).
2. Flip the switch to the discharge position and start the stopwatch simultaneously.
3. Stop the stopwatch when the voltmeter reading falls to exactly half of the initial voltage, i.e., \( 3.0\text{ V} \).
4. Repeat the measurement for the same resistance several times and calculate an average value of \( t \).
5. Repeat the process for several other different resistors of known values.

(c)(i) The equation is in the form \( y = m x + c \) where \( y = t \), \( x = R \), and the gradient is \( m = C \ln 2 \). Since there is no y-intercept term (\( c = 0 \)), the line passes through the origin.

(ii) The gradient is given by \( \text{gradient} = C \ln 2 \), so the capacitance is determined by: \( C = \frac{\text{gradient}}{\ln 2} \).

(iii) Substitute the given gradient value into the equation:
\( C = \frac{8.3 \times 10^{-5}}{\ln 2} \)
\( C = \frac{8.3 \times 10^{-5}}{0.69315} = 1.197 \times 10^{-4}\text{ F} \approx 1.2 \times 10^{-4}\text{ F} \) (or \( 120\ \mu\text{F} \)).

(a)
- Correct symbol and placement for capacitor, voltmeter, switch and DC source [1]
- Diagram showing functional circuit with a two-way switch (or equivalent arrangement) that allows charging from the cell and discharging through the resistor [1]

(b)
- Charge capacitor to maximum, flip switch to discharge and start timer [1]
- Record time taken for voltage to reach exactly half (3.0 V) [1]
- Mentions repeating and averaging / changing resistor to get multiple data points [1]

(c)(i)
- Compares equation to \( y = mx \) showing intercept is zero [1]

(c)(ii)
- States \( C = \text{gradient} / \ln 2 \) (or equivalent rearrangement) [1]

(c)(iii)
- Shows substitution of the gradient value: \( C = 8.3 \times 10^{-5} / \ln 2 \) [1]
- Calculation step [1]
- Correct value of \( 1.2 \times 10^{-4}\text{ F} \) (or \( 120\ \mu\text{F} \)) to 2 s.f. [1]

(a) Taking the natural logarithm on both sides of the equation:
\( I = I_0 e^{-t / RC} \)
\( \ln I = \ln(I_0 e^{-t / RC}) = \ln I_0 - \frac{t}{RC} \)
If we divide by the unit \( \text{mA} \):
\( \ln(I / \text{mA}) = -\frac{1}{RC} t + \ln(I_0 / \text{mA}) \)
This equation has the linear form \( y = mx + c \), where:
- \( y = \ln(I / \text{mA}) \)
- \( x = t \)
- \( \text{gradient } m = -\frac{1}{RC} \)
- \( \text{y-intercept } c = \ln(I_0 / \text{mA}) \)

(b)(i) At \( t = 10.0\text{ s} \), \( I = 2.9\text{ mA} \).
\( \ln(I / \text{mA}) = \ln(2.9) = 1.065 \approx 1.06 \).

(ii) From the equation:
\( I = I_0 e^{-t/\tau} \)
where \( \tau = RC \) is the time constant.
Substitute \( I = 2.9\text{ mA} \), \( I_0 = 8.0\text{ mA} \), and \( t = 10.0\text{ s} \):
\( 2.9 = 8.0 e^{-10.0 / \tau} \)
\( \frac{2.9}{8.0} = 0.3625 \)
\( \ln(0.3625) = -1.0147 \)
\( -1.0147 = -\frac{10.0}{\tau} \)
\( \tau = \frac{10.0}{1.0147} = 9.855\text{ s} \approx 9.9\text{ s} \).

(iii) Using \( \tau = RC \):
\( C = \frac{\tau}{R} \)
\( C = \frac{9.855}{4.7 \times 10^3} = 2.097 \times 10^{-3}\text{ F} \approx 2.1 \times 10^{-3}\text{ F} \) (or \( 2.1\text{ mF} \) / \( 2100\ \mu\text{F} \)).

(a)
- Applies natural logarithm correctly to the equation [1]
- Identifies gradient as \( -1/RC \) [1]
- Identifies y-intercept as \( \ln(I_0 / \text{mA}) \) [1]

(b)(i)
- Correct calculation of \( \ln(2.9) = 1.06 \) (or 1.065) [1]

(b)(ii)
- Recalls \( I = I_0 e^{-t/\tau} \) [1]
- Correct substitution of values: \( 2.9 = 8.0 e^{-10.0/\tau} \) [1]
- Calculates \( \tau = 9.9\text{ s} \) (accept \( 9.8\text{ s} \) to \( 9.9\text{ s} \)) [1]

(b)(iii)
- Recalls \( C = \tau / R \) [1]
- Substitution: \( C = 9.86 / (4.7 \times 10^3) \) [1]
- Calculates \( C = 2.1 \times 10^{-3}\text{ F} \) (accept range \( 2.08 \times 10^{-3}\text{ F} \) to \( 2.11 \times 10^{-3}\text{ F} \)) [1]

(a) Simple harmonic motion is defined as motion where the acceleration is directly proportional to the displacement from a fixed position, and is always directed towards that fixed position.

(b)(i) The maximum acceleration occurs at maximum displacement:
\( a_0 = \omega^2 x_0 \)
Substitute the given values (converting displacement to meters):
\( 9.6 = \omega^2 \times 0.040 \)
\( \omega^2 = 240 \)
\( \omega = \sqrt{240} = 15.49\text{ rad s}^{-1} \approx 15.5\text{ rad s}^{-1} \).

(ii) Frequency is related to angular frequency by:
\( f = \frac{\omega}{2\pi} \)
\( f = \frac{15.49}{2\pi} = 2.465\text{ Hz} \approx 2.5\text{ Hz} \) (or \( 2.47\text{ Hz} \)).

(c)(i) Maximum kinetic energy is equal to total energy, which is given by:
\( E_k = \frac{1}{2} m \omega^2 x_0^2 \)
\( E_k = 0.5 \times 0.150 \times 240 \times (0.040)^2 = 0.0288\text{ J} \approx 0.029\text{ J} \) (or \( 2.88 \times 10^{-2}\text{ J} \)).

(ii) The kinetic energy at displacement \( x \) is:
\( E_k = \frac{1}{2} m \omega^2 (x_0^2 - x^2) \)
Substitute \( x = 0.025\text{ m} \):
\( E_k = 0.5 \times 0.150 \times 240 \times (0.040^2 - 0.025^2) \)
\( E_k = 18 \times (0.0016 - 0.000625) \)
\( E_k = 18 \times 0.000975 = 0.01755\text{ J} \approx 0.018\text{ J} \) (or \( 1.76 \times 10^{-2}\text{ J} \)).

(a)
- Acceleration proportional to displacement [1]
- Acceleration (always) directed towards a fixed point / in opposite direction to displacement [1]

(b)(i)
- Recalls \( a_0 = \omega^2 x_0 \) and substitutes: \( 9.6 = \omega^2 \times 0.040 \) [1]
- Calculation of \( \omega = 15.49\text{ rad s}^{-1} \) leading to \( 15.5\text{ rad s}^{-1} \) [1]

(b)(ii)
- Recalls \( f = \omega / 2\pi \) and substitutes [1]
- Correct calculation to 2 or 3 s.f.: \( 2.5\text{ Hz} \) (or \( 2.47\text{ Hz} \)) [1]

(c)(i)
- Recalls \( E_k = \frac{1}{2} m \omega^2 x_0^2 \) [1]
- Calculation: \( 0.029\text{ J} \) (or \( 2.88 \times 10^{-2}\text{ J} \)) [1]

(c)(ii)
- Recalls \( E_k = \frac{1}{2} m \omega^2 (x_0^2 - x^2) \) [1]
- Calculation: \( 0.018\text{ J} \) (or \( 1.76 \times 10^{-2}\text{ J} \)) [1]

(a)(i) Free oscillations occur when a system oscillates with no external periodic force acting on it. The system oscillates at its natural frequency.

(ii) Forced oscillations occur when a periodic driving force acts on a system, causing it to oscillate at the frequency of the driving force.

(b) As damping increases:
1. The maximum amplitude (peak value of the curve) decreases.
2. The peak of the curve becomes broader and flatter.
3. The resonant frequency (the frequency at which peak amplitude occurs) shifts slightly towards lower frequencies.

(c)(i) The total time \( t \) taken for 10 complete oscillations is:
\( t = 10 \times 0.80\text{ s} = 8.0\text{ s} \)
The amplitude decays exponentially:
\( A = A_0 e^{-\gamma t} \)
Substitute \( A = 2.5\text{ cm} \), \( A_0 = 5.0\text{ cm} \), and \( t = 8.0\text{ s} \):
\( 2.5 = 5.0 e^{-8.0 \gamma} \)
\( 0.5 = e^{-8.0 \gamma} \)
Take natural logarithms on both sides:
\( \ln(0.5) = -8.0 \gamma \)
\( -0.693 = -8.0 \gamma \)
\( \gamma = \frac{0.693}{8.0} = 0.0866\text{ s}^{-1} \approx 0.087\text{ s}^{-1} \).

(ii) The damping is light.
Reason: The system completes multiple oscillations (at least 10) before its amplitude decreases significantly, showing that motion continues to be periodic and decays slowly.

(a)(i)
- Oscillates with no external periodic force / oscillates at natural frequency [1]

(a)(ii)
- Driven by an external periodic force / oscillates at driver's frequency [1]

(b)
- Peak amplitude decreases [1]
- Peak becomes broader / flatter [1]
- Resonant frequency decreases / shifts left [1]

(c)(i)
- Calculates total time \( t = 8.0\text{ s} \) [1]
- Substitutes into exponential formula: \( 0.5 = e^{-8.0 \gamma} \) [1]
- Correct calculation of \( \gamma = 0.087\text{ s}^{-1} \) (or \( 0.0866\text{ s}^{-1} \)) [1]

(c)(ii)
- Light damping identified [1]
- Correct justification (system continues to oscillate many times / slow decay of amplitude) [1]

(a) Squaring both sides of \( T = 2\pi \sqrt{\frac{L}{g}} \) gives:
\( T^2 = 4\pi^2 \frac{L}{g} \)
Rearranging for \( g \):
\( g = \frac{4\pi^2 L}{T^2} \).

(b)(i) Period \( T = \frac{t}{20} = \frac{36.0}{20} = 1.80\text{ s} \).

(ii) The absolute uncertainty in time \( t \) is \( \Delta t = 0.2\text{ s} \).
Since \( T = t / 20 \), the absolute uncertainty in \( T \) is:
\( \Delta T = \frac{\Delta t}{20} = \frac{0.2}{20} = 0.01\text{ s} \).
So, \( T = 1.80 \pm 0.01\text{ s} \).

(iii) Using \( g = \frac{4\pi^2 L}{T^2} \):
\( g = \frac{4\pi^2 \times 0.800}{(1.80)^2} = 9.748\text{ m s}^{-2} \approx 9.75\text{ m s}^{-2} \).

(iv) The percentage uncertainty in \( g \) is:
\( \frac{\Delta g}{g} \times 100 = \left( \frac{\Delta L}{L} + 2 \frac{\Delta T}{T} \right) \times 100 \)
\( \frac{\Delta L}{L} = \frac{0.003}{0.800} = 0.00375 \) (or \( 0.375\% \))
\( \frac{\Delta T}{T} = \frac{0.01}{1.80} = 0.00556 \) (or \( 0.556\% \))
\( \frac{\Delta g}{g} \times 100 = (0.00375 + 2 \times 0.00556) \times 100 = 0.01487 \times 100 \approx 1.5\% \) (or \( 1.49\% \)).

(v) The absolute uncertainty in \( g \) is:
\( \Delta g = 9.748 \times 0.01487 = 0.145\text{ m s}^{-2} \approx 0.1\text{ m s}^{-2} \).
Expressing \( g \) to 1 decimal place matching the absolute uncertainty:
\( g = 9.7 \pm 0.1\text{ m s}^{-2} \) (or \( 9.75 \pm 0.15\text{ m s}^{-2} \)).

(a)
- Squares both sides and rearranges correctly to get the formula [1]

(b)(i)
- Correct calculation of \( T = 1.80\text{ s} \) [1]

(b)(ii)
- Understands \( \Delta T = \Delta t / 20 \) [1]
- Correct absolute uncertainty \( 0.01\text{ s} \) [1]

(b)(iii)
- Substitute correct values into the formula [1]
- Value \( 9.75\text{ m s}^{-2} \) (or \( 9.7\text{ m s}^{-2} \)) [1]

(b)(iv)
- Recalls fractional uncertainty equation with factor of 2 for \( T \) [1]
- Substitution of fractional uncertainties: \( 0.00375 + 2 \times 0.00556 \) [1]
- Percentage uncertainty \( 1.5\% \) (or \( 1.49\% \)) [1]

(b)(v)
- States \( 9.7 \pm 0.1\text{ m s}^{-2} \) or \( 9.75 \pm 0.15\text{ m s}^{-2} \) with consistent decimal places [1]

(a) The potential difference across a discharging capacitor is given by:
\( V = V_0 e^{-t/RC} \)
When \( t = RC \) (one time constant):
\( V = V_0 e^{-1} \)
Given \( V_0 = 10.0\text{ V} \) and \( e^{-1} \approx 0.3679 \):
\( V = 10.0 \times 0.3679 = 3.68\text{ V} \approx 3.7\text{ V} \).

(b)(i) From \( t = RC \):
\( C = \frac{t}{R} \)
\( C = \frac{47.0}{220 \times 10^3\ \Omega} = 2.136 \times 10^{-4}\text{ F} \approx 2.1 \times 10^{-4}\text{ F} \) (or \( 210\ \mu\text{F} \)).

(ii) Percentage uncertainty in \( R \):
\( \% \Delta R = \frac{10}{220} \times 100 = 4.55\% \).

(iii) Percentage uncertainty in \( t \):
\( \% \Delta t = \frac{1.5}{47.0} \times 100 = 3.19\% \).

(iv) The percentage uncertainty in \( C \) is the sum of the percentage uncertainties in \( R \) and \( t \):
\( \% \Delta C = 4.55\% + 3.19\% = 7.74\% \).
The absolute uncertainty in \( C \) is:
\( \Delta C = 2.136 \times 10^{-4} \times 0.0774 = 1.65 \times 10^{-5}\text{ F} \approx 0.2 \times 10^{-4}\text{ F} \) (or \( 20\ \mu\text{F} \)).

(v) Expressing \( C \) to match the absolute uncertainty of \( 0.2 \times 10^{-4}\text{ F} \):
\( C = (2.1 \pm 0.2) \times 10^{-4}\text{ F} \) (or \( 210 \pm 20\ \mu\text{F} \)).

(a)
- Recalls discharge equation \( V = V_0 e^{-t/RC} \) [1]
- Replaces \( t \) with \( RC \) and evaluates \( V_0 e^{-1} \approx 3.7\text{ V} \) [1]

(b)(i)
- Recalls formula \( C = t / R \) [1]
- Calculation of \( C = 2.1 \times 10^{-4}\text{ F} \) (or \( 2.14 \times 10^{-4}\text{ F} \)) [1]

(b)(ii)
- Calculates \( \% \Delta R = 4.5\% \) (or \( 4.55\% \)) [1]

(b)(iii)
- Calculates \( \% \Delta t = 3.2\% \) (or \( 3.19\% \)) [1]

(b)(iv)
- Sums percentage uncertainties (approx. 7.7%) [1]
- Multiplies percentage uncertainty by \( C \) [1]
- Correct absolute uncertainty: \( 0.2 \times 10^{-4}\text{ F} \) (or \( 20\ \mu\text{F} \)) [1]

(b)(v)
- States \( (2.1 \pm 0.2) \times 10^{-4}\text{ F} \) or \( 210 \pm 20\ \mu\text{F} \) with matching precision [1]

(a) Nuclear binding energy is the minimum energy required to completely separate the nucleons of a nucleus to infinity.

(b)(i) Total mass before reaction (reactants):
\( m_{\text{reactants}} = 2.01355 + 3.01550 = 5.02905\text{ u} \)
Total mass after reaction (products):
\( m_{\text{products}} = 4.00150 + 1.00867 = 5.01017\text{ u} \)
Mass defect \( \Delta m \):
\( \Delta m = 5.02905 - 5.01017 = 0.01888\text{ u} \).

(ii) First, convert the mass defect into kilograms:
\( \Delta m = 0.01888 \times 1.6605 \times 10^{-27}\text{ kg} = 3.13502 \times 10^{-29}\text{ kg} \)
Using Einstein's relation:
\( E = \Delta m c^2 \)
\( E = 3.13502 \times 10^{-29} \times (3.00 \times 10^8)^2 = 2.8215 \times 10^{-12}\text{ J} \)
Now convert this energy to MeV:
\( E = \frac{2.8215 \times 10^{-12}}{1.60 \times 10^{-13}\text{ J MeV}^{-1}} = 17.634\text{ MeV} \approx 17.6\text{ MeV} \).
(Note: Alternatively, using \( 1\text{ u} = 931.5\text{ MeV} \): \( 0.01888 \times 931.5 = 17.59\text{ MeV} \approx 17.6\text{ MeV} \)).

(c) The number of nucleons in a helium-4 nucleus is 4.
Total binding energy of the helium-4 nucleus:
\( E_B = 4 \times 7.07\text{ MeV} = 28.28\text{ MeV} \)
Convert this energy to Joules:
\( E_B = 28.28 \times 10^6 \times 1.60 \times 10^{-19}\text{ J} = 4.5248 \times 10^{-12}\text{ J} \approx 4.52 \times 10^{-12}\text{ J} \) (or \( 4.5 \times 10^{-12}\text{ J} \)).

(a)
- Minimum energy required to separate the nucleons of a nucleus [1]
- (separated) to infinity [1]

(b)(i)
- Sum of masses of deuterium and tritium and sum of helium and neutron [1]
- Correct subtraction to get \( 0.01888\text{ u} \) [1]

(b)(ii)
- Conversions of \( \Delta m \) to kg (or uses 931.5 MeV per u conversion factor) [1]
- Use of \( E = \Delta m c^2 \) [1]
- Correct calculation of energy: \( 17.6\text{ MeV} \) (accept range \( 17.5 - 17.7\text{ MeV} \)) [1]

(c)
- Multiplies binding energy per nucleon by 4 [1]
- Correct conversion factor (multiplies by \( 1.60 \times 10^{-13} \)) [1]
- Correct final answer: \( 4.52 \times 10^{-12}\text{ J} \) (or \( 4.5 \times 10^{-12}\text{ J} \)) [1]

(a) The time constant is the time taken for the charge (or potential difference, or current) to fall to \(1/e\) (or approximately \(37\%\)) of its initial value. (b)(i) Using \(V = V_0 e^{-t/RC}\), we substitute \(V = 4.4\text{ V}\), \(V_0 = 12.0\text{ V}\), \(t = 15.0\text{ s}\), and \(R = 47.0 \times 10^3\ \Omega\). This gives \(4.4 = 12.0 e^{-15.0 / (47.0 \times 10^3 C)}\), which leads to \(\ln(4.4/12.0) = -1.003 = -15.0 / (47.0 \times 10^3 C)\). Solving for \(C\) yields \(C = 3.18 \times 10^{-4}\text{ F}\), which is approximately \(3.2 \times 10^{-4}\text{ F}\). (ii) The charge at \(t = 25.0\text{ s}\) is \(Q = C V_0 e^{-t/RC}\). Using \(C = 3.2 \times 10^{-4}\text{ F}\), we have \(Q = 3.2 \times 10^{-4} \times 12.0 \times e^{-25.0 / (47.0 \times 10^3 \times 3.2 \times 10^{-4})} = 7.28 \times 10^{-4}\text{ C}\). If using the unrounded \(C = 3.18 \times 10^{-4}\text{ F}\), \(Q = 7.17 \times 10^{-4}\text{ C}\). (iii) The initial energy is \(E_0 = \frac{1}{2} C V_0^2 = 0.5 \times 3.18 \times 10^{-4} \times 12.0^2 = 0.0229\text{ J}\). The energy at \(t = 15.0\text{ s}\) is \(E_{15} = \frac{1}{2} C V^2 = 0.5 \times 3.18 \times 10^{-4} \times 4.4^2 = 0.0031\text{ J}\). The energy dissipated is \(E_0 - E_{15} = 0.0229 - 0.0031 = 0.0198\text{ J} \approx 2.0 \times 10^{-2}\text{ J}\) (or \(1.99 \times 10^{-2}\text{ J}\) if using \(C = 3.2 \times 10^{-4}\text{ F}\)).

(a) Time for the charge/current/potential difference to decrease to \(1/e\) (or \(37\%\)) of its initial value. [1] (b)(i) \(V = V_0 e^{-t/RC}\) stated or used. [1] Substitution of \(4.4\), \(12.0\), \(15.0\), and \(47.0 \times 10^3\). [1] Correct calculation leading to \(3.18 \times 10^{-4}\text{ F}\) with at least 3 s.f. shown. [1] (ii) \(Q = Q_0 e^{-t/RC}\) or \(Q = C V_0 e^{-t/RC}\) used. [1] Correct substitution of \(t = 25.0\text{ s}\). [1] \(Q = 7.2 \times 10^{-4}\text{ C}\) or \(7.3 \times 10^{-4}\text{ C}\). [1] (iii) \(E = \frac{1}{2} C V^2\) used to find at least one energy. [1] Change in energy \(\Delta E = E_{\text{initial}} - E_{\text{final}}\). [1] \(\Delta E = 2.0 \times 10^{-2}\text{ J}\) (or \(1.99 \times 10^{-2}\text{ J}\) or \(0.020\text{ J}\)). [1]

(a) The binding energy is the minimum energy required to completely separate a nucleus into its constituent nucleons (protons and neutrons). (b)(i) A free proton is already a single isolated nucleon, so no energy is needed to separate it from other nucleons. (b)(ii) The total binding energy of the reactants is \(2 \times 1.11\text{ MeV} + 3 \times 2.57\text{ MeV} = 2.22 + 7.71 = 9.93\text{ MeV}\). The total binding energy of the products is \(4 \times 7.07\text{ MeV} + 0 = 28.28\text{ MeV}\). The energy released is the difference in binding energy: \(28.28 - 9.93 = 18.35\text{ MeV}\). (b)(iii) First, convert the energy from MeV to Joules: \(\Delta E = 18.35 \times 10^{6} \times 1.60 \times 10^{-19}\text{ J} = 2.936 \times 10^{-12}\text{ J}\). Use Einstein's mass-energy equation \(\Delta E = \Delta m \cdot c^2\), where \(c = 3.00 \times 10^8\text{ m s}^{-1}\). This gives \(\Delta m = 2.936 \times 10^{-12} / (3.00 \times 10^8)^2 = 3.262 \times 10^{-29}\text{ kg} \approx 3.26 \times 10^{-29}\text{ kg}\).

(a) Minimum energy required to completely separate all the nucleons of a nucleus to infinity. [2] (b)(i) A proton is a single nucleon / no nucleons to separate. [1] (ii) Total binding energy of reactants = \(9.93\text{ MeV}\) calculated. [1] Total binding energy of products = \(28.28\text{ MeV}\) calculated. [1] Energy released = \(18.35\text{ MeV}\) (accept \(18.4\text{ MeV}\)). [1] (iii) Conversion of \(MeV\) to \(J\) seen (\(18.35 \times 10^6 \times 1.6 \times 10^{-19}\)). [1] Correct use of \(\Delta E = \Delta m c^2\). [1] Substitution of \(c = 3.00 \times 10^8\text{ m s}^{-1}\). [1] Mass defect = \(3.26 \times 10^{-29}\text{ kg}\) (allow \(3.27 \times 10^{-29}\text{ kg}\)). [1]

### Experimental Setup and Diagram

Draw a circuit diagram containing:
1. A DC power supply connected to a single-pole double-throw (SPDT) switch.
2. The switch connected to a capacitor of known capacitance \(C\) such that it can be charged by the supply and then flipped to discharge through the rectangular strip of conducting paper.
3. A high-resistance digital voltmeter connected in parallel across the capacitor to measure the potential difference \(V\).
4. The rectangular strip of conducting paper connected across the capacitor via low-resistance metal clamp connectors (e.g., copper foil clamps) to ensure good electrical contact.

### Procedure and Measurements

1. **Independent Variable:** The thickness \(d\) of the conducting paper strip. This can be varied by stacking multiple identical sheets of conducting paper together (so that the effective thickness becomes \(d = N \cdot d_{\text{sheet}}\) where \(N\) is the number of layers) or by obtaining sheets manufactured with different known thicknesses.
2. **Dependent Variable:** The half-discharge time \(t\) (the time taken for the voltmeter reading to fall from its initial value \(V_0\) to \(V_0 / 2\)).
3. **Control Variables:**
- The length \(L\) and width \(w\) of the strip must be kept constant throughout the experiment.
- The capacitance \(C\) must remain constant.
- The temperature of the room should be kept constant, as resistivity \(\rho\) of the paper can vary with temperature.

### Detailed Steps:
- Measure the length \(L\) and width \(w\) of the strip using a standard millimeter ruler.
- Measure the thickness \(d\) of a single sheet of conducting paper using a micrometer screw gauge. Measure at 5 to 6 different positions along the strip and calculate the mean thickness. If using a stack of \(N\) sheets, the total thickness is \(N \times d_{\text{mean}}\).
- Connect the circuit. Close the switch to the charging position until the voltmeter reads a stable initial potential difference \(V_0\).
- Flip the switch to the discharging position and simultaneously start a digital stopwatch.
- Stop the stopwatch when the voltmeter reading decreases to exactly \(\frac{V_0}{2}\).
- Repeat the procedure for the same thickness \(d\) at least three times to obtain a mean value of the half-life time \(t\).
- Change the thickness \(d\) (by altering the number of sheets in the stack or changing the sheet) and repeat the entire sequence for at least six different values of \(d\).

### Method of Analysis

The relationship is given by:
\[t = \frac{\rho L C \ln 2}{w d}\]

To test this, rewrite the equation in the linear form \(y = mx + c\):
\[t = \left( \frac{\rho L C \ln 2}{w} \right) \cdot \frac{1}{d}\]

- Plot a graph of \(t\) on the vertical axis against \(\frac{1}{d}\) on the horizontal axis.
- If the suggested relationship is valid, the plotted points should lie on a straight line passing through the origin.
- Measure the gradient \(m\) of this straight line.
- Since the gradient is \(m = \frac{\rho L C \ln 2}{w}\), we can determine the resistivity \(\rho\) using:
\[\rho = \frac{m \cdot w}{L \cdot C \ln 2}\]

### Safety Considerations

1. Use low-voltage DC power supplies (e.g., \(<12\text{ V}\)) to prevent electric shock hazards.
2. Ensure that electrolytic capacitors (if used) are connected with the correct polarity to prevent them from overheating or bursting.
3. Keep the current levels within safe limits to prevent heating of the conducting paper, which might change its resistance or present a fire hazard.

**Marking Scheme (Total: 15 Marks)**

**Defining the Problem (2 marks):**
- **DP1:** Identify \(d\) as the independent variable and \(t\) as the dependent variable. [1]
- **DP2:** State that \(L\), \(w\), and \(C\) are kept constant. [1]

**Methods of Data Collection (4 marks):**
- **MC1:** Draw a workable circuit diagram showing a capacitor, DC supply, switch, voltmeter in parallel with the capacitor, and conducting strip connected in parallel during discharge. [1]
- **MC2:** Describe the method to measure the thickness \(d\) using a micrometer screw gauge, with measurements taken at multiple points and averaged. [1]
- **MC3:** Describe the method to measure length \(L\) and width \(w\) using a ruler. [1]
- **MC4:** Describe how the timing measurement is taken: start timer when switch is flipped, stop when voltmeter reads exactly \(V_0 / 2\). [1]

**Method of Analysis (3 marks):**
- **MA1:** State that a graph of \(t\) against \(1/d\) should be plotted. [1]
- **MA2:** State that a straight line through the origin confirms the relationship. [1]
- **MA3:** State that \(\rho = \frac{m \cdot w}{L \cdot C \ln 2}\), where \(m\) is the gradient of the graph. [1]

**Safety Considerations (1 mark):**
- **SF1:** Mention a valid safety precaution, e.g., correct polarity of electrolytic capacitor to prevent explosion, or low voltage supply to avoid shock. [1]

**Additional Detail / Good Practice (5 marks max from the list below):**
- **AD1:** Use of a double-throw switch (SPDT) to switch quickly and reliably between charge and discharge modes.
- **AD2:** Discussion of how to change the thickness \(d\) by stacking multiple identical layers of paper of known thickness.
- **AD3:** Clean electrical contacts discussed, e.g., using copper foil clamps or highly conductive grease/clips to minimize contact resistance.
- **AD4:** Explanation of how the initial potential difference \(V_0\) is chosen and monitored to be constant for all trials.
- **AD5:** Repeat the timing measurements for each thickness and calculate the average time \(t\) to reduce random errors.
- **AD6:** High-resistance voltmeter (or digital oscilloscope/data logger) used to ensure that current does not bypass the conducting paper through the meter.

### Part (a)
Taking natural logarithms of both sides of the proposed equation:
\[\ln(A_n) = \ln(A_0 e^{-\lambda n}) = \ln(A_0) - \lambda n\]
Comparing this with the equation of a straight line, \(y = mx + c\):
- \(y = \ln(A_n / \text{mm})\)
- \(x = n\)
- \(m = -\lambda\) (gradient)
- \(c = \ln(A_0 / \text{mm})\) (y-intercept)
Thus, if the relationship holds, plotting \(\ln(A_n / \text{mm})\) against \(n\) yields a straight line with a constant negative gradient.

### Part (b)
The absolute uncertainty in \(y = \ln(x)\) is given by \(\Delta y \approx \frac{\Delta x}{x}\).
- For \(n = 5\), \(A_5 = 76 \pm 2\text{ mm}\):
\[\ln(76) \approx 4.331\]
\[\Delta(\ln A_n) = \frac{2}{76} \approx 0.026 \approx 0.03\]
- For \(n = 10\), \(A_{10} = 58 \pm 2\text{ mm}\):
\[\ln(58) \approx 4.060\]
\[\Delta(\ln A_n) = \frac{2}{58} \approx 0.034 \approx 0.03\]
- For \(n = 15\), \(A_{15} = 44 \pm 2\text{ mm}\):
\[\ln(44) \approx 3.784\]
\[\Delta(\ln A_n) = \frac{2}{44} \approx 0.045 \approx 0.05\]
- For \(n = 20\), \(A_{20} = 34 \pm 2\text{ mm}\):
\[\ln(34) \approx 3.526\]
\[\Delta(\ln A_n) = \frac{2}{34} \approx 0.059 \approx 0.06\]
- For \(n = 25\), \(A_{25} = 26 \pm 2\text{ mm}\):
\[\ln(26) \approx 3.258\]
\[\Delta(\ln A_n) = \frac{2}{26} \approx 0.077 \approx 0.08\]
- For \(n = 30\), \(A_{30} = 20 \pm 2\text{ mm}\):
\[\ln(20) \approx 2.996\]
\[\Delta(\ln A_n) = \frac{2}{20} \approx 0.100 \approx 0.10\]

**Processed Table:**
| \(n\) | \(A_n\) / mm | \(\ln(A_n / \text{mm})\) | Uncertainty |
|---|---|---|---|
| 5 | 76 | 4.33 | \(\pm 0.03\) |
| 10 | 58 | 4.06 | \(\pm 0.03\) |
| 15 | 44 | 3.78 | \(\pm 0.05\) |
| 20 | 34 | 3.53 | \(\pm 0.06\) |
| 25 | 26 | 3.26 | \(\pm 0.08\) |
| 30 | 20 | 3.00 | \(\pm 0.10\) |

### Part (c)
- The horizontal axis should be labelled "\(n\)" and the vertical axis should be labelled "\(\ln(A_n / \text{mm})\)".
- Error bars should be plotted vertically for each point with lengths corresponding to the calculated uncertainties.
- **Line of Best Fit:** Passes close to all points. A typical line of best fit has a gradient of \(m \approx -0.0532\) and vertical intercept of \(c \approx 4.60\).
- **Worst Acceptable Line:** The steepest or shallowest possible straight line that still passes through all the error bars. For example, passing through the top of the error bar of the first point \((5, 4.36)\) and the bottom of the error bar of the last point \((30, 2.90)\).

### Part (d)
- **Gradient of Best Fit:**
Using two points on the line of best fit, e.g., \((5, 4.33)\) and \((30, 3.00)\):
\[m = \frac{3.00 - 4.33}{30 - 5} = \frac{-1.33}{25} = -0.0532 \approx -0.053\ \text{oscillation}^{-1}\]
- **Worst Gradient (Shallowest):**
Using points \((5, 4.30)\) and \((30, 3.10)\):
\[m_{\text{worst}} = \frac{3.10 - 4.30}{30 - 5} = -0.0480\ \text{oscillation}^{-1}\]
- **Uncertainty in Gradient:**
\[\Delta m = |m_{\text{best}} - m_{\text{worst}}| = |-0.0532 - (-0.0480)| = 0.0052 \approx 0.005\ \text{oscillation}^{-1}\]
Thus, the gradient is \(-0.053 \pm 0.005\ \text{oscillation}^{-1}\).

### Part (e)
- **y-intercept of Best Fit:**
Using \(y = mx + c\) with the best-fit line:
\[c = 3.00 - (-0.0532 \times 30) = 4.60\]
- **y-intercept of Worst Line:**
Using \(y = m_{\text{worst}}x + c_{\text{worst}}\):
\[c_{\text{worst}} = 3.10 - (-0.0480 \times 30) = 4.54\]
- **Uncertainty in y-intercept:**
\[\Delta c = |c_{\text{best}} - c_{\text{worst}}| = |4.60 - 4.54| = 0.06\]
Thus, the y-intercept is \(4.60 \pm 0.06\).

### Part (f)
1. **Damping Coefficient \(\lambda\):**
\[\lambda = -m = 0.053\ \text{oscillation}^{-1}\]
The uncertainty in \(\lambda\) is \(\Delta \lambda = \Delta m = 0.005\ \text{oscillation}^{-1}\).

2. **Initial Amplitude \(A_0\):**
\[A_0 = e^c = e^{4.60} = 99.5\text{ mm} \approx 100\text{ mm}\]
To calculate the uncertainty in \(A_0\):
\[A_{0, \text{max}} = e^{4.66} = 105.6\text{ mm}\]
\[A_{0, \text{min}} = e^{4.54} = 93.7\text{ mm}\]
The uncertainty is:
\[\Delta A_0 = \frac{A_{0, \text{max}} - A_{0, \text{min}}}{2} = \frac{105.6 - 93.7}{2} \approx 6\text{ mm}\]
Thus, \(A_0 = 100 \pm 6\text{ mm}\).

**Marking Scheme (Total: 15 Marks)**

**Part (a) [2 marks]:**
- **A1:** Correct derivation showing \(\ln(A_n) = -\lambda n + \ln(A_0)\) which is linear in form. [1]
- **A2:** Identifies gradient \(m = -\lambda\) and y-intercept \(c = \ln(A_0)\). [1]

**Part (b) [3 marks]:**
- **B1:** Calculates all values of \(\ln(A_n / \text{mm})\) correctly (4.33, 4.06, 3.78, 3.53, 3.26, 3.00). [1]
- **B2:** Calculates absolute uncertainties correctly to 1 or 2 significant figures (0.03, 0.03, 0.05, 0.06, 0.08, 0.10). [1]
- **B3:** Presents all data to consistent decimal places (2 d.p. for natural logarithm values). [1]

**Part (c) [3 marks]:**
- **C1:** Plots all 6 points correctly with vertical error bars representing the calculated uncertainties. [1]
- **C2:** Draws a straight line of best fit that passes through all error bars and is balanced. [1]
- **C3:** Draws a clearly labelled worst acceptable straight line (steepest or shallowest) that passes through all error bars. [1]

**Part (d) [2 marks]:**
- **D1:** Calculates the gradient of the best-fit line using coordinates from the line (with interval \(\Delta n \ge 15\)), yielding a value in the range \(-0.051\) to \(-0.055\). [1]
- **D2:** Determines the absolute uncertainty in the gradient using the worst-fit line, yielding a value in the range \(0.003\) to \(0.007\). [1]

**Part (e) [2 marks]:**
- **E1:** Determines the y-intercept of the best-fit line correctly, yielding a value in the range \(4.55\) to \(4.65\). [1]
- **E2:** Determines the absolute uncertainty in the y-intercept using the worst-fit line, yielding a value in the range \(0.04\) to \(0.08\). [1]

**Part (f) [3 marks]:**
- **F1:** Identifies \(\lambda = 0.053 \pm 0.005\ \text{oscillation}^{-1}\) (or \(\text{cycle}^{-1}\) or no unit). [1]
- **F2:** Calculates \(A_0 = e^{y\text{-intercept}} \approx 100\text{ mm}\) (allow \(95\text{ mm}\) to \(105\text{ mm}\)) with correct unit (mm). [1]
- **F3:** Calculates absolute uncertainty in \(A_0\) correctly (yields \(6\text{ mm}\), allow range \(4\text{ mm}\) to \(8\text{ mm}\)). [1]