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The density \(\rho\) of a sphere of mass \(m\) and diameter \(d\) is given by: \(\rho = \frac{m}{V} = \frac{m}{\frac{4}{3}\pi (d/2)^3} = \frac{6m}{\pi d^3}\). Using the rules for combining uncertainties, the fractional uncertainty in density is: \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3 \frac{\Delta d}{d}\). Calculate the fractional uncertainties: \(\frac{\Delta m}{m} = \frac{0.05}{4.50} \approx 0.0111\) (or \(1.11\%\)) and \(\frac{\Delta d}{d} = \frac{0.02}{1.20} \approx 0.0167\) (or \(1.67\%\)). Substituting these values: \(\frac{\Delta \rho}{\rho} = 0.0111 + 3(0.0167) = 0.0111 + 0.0500 = 0.0611\). Converting this to a percentage gives: \(\text{Percentage uncertainty} = 0.0611 \times 100\% \approx 6.1\%\).

1 mark for calculating the correct percentage uncertainty of 6.1% (Option D).

A zero error is a systematic error, which consistently shifts all readings away from the true value in one direction, thereby decreasing the accuracy of the result. Taking multiple measurements at different points and calculating their mean reduces the random error associated with local variations and measurement fluctuations, which increases the precision of the result. Therefore, statement A is correct.

1 mark for identifying that zero error decreases accuracy and taking multiple measurements increases precision (Option A).

The Young modulus \(E\) of the material is given by \(E = \frac{F L}{A \Delta L}\), where \(F\) is the load, \(L\) is the length, \(A\) is the cross-sectional area, and \(\Delta L\) is the extension. Rearranging for extension gives: \(\Delta L = \frac{F L}{A E}\). Since both wires are made of the same material and carry the same load, \(E\) and \(F\) are constant, meaning \(\Delta L \propto \frac{L}{A}\). Since cross-sectional area \(A = \frac{\pi d^2}{4}\), we have \(A \propto d^2\), so \(\Delta L \propto \frac{L}{d^2}\). For wire X, \(L_X = 2 L_Y\) and \(d_X = 0.5 d_Y\). Thus, the ratio of extensions is: \(\frac{\Delta L_X}{\Delta L_Y} = \frac{L_X}{L_Y} \times \left(\frac{d_Y}{d_X}\right)^2 = 2 \times \left(\frac{1}{0.5}\right)^2 = 2 \times 4 = 8\).

1 mark for deriving the relation between extension, length, and diameter, and correctly calculating the ratio as 8 (Option D).

For two identical springs connected in parallel, the equivalent spring constant is \(k_{\text{eq}} = 2k = 2 \times 200\text{ N m}^{-1} = 400\text{ N m}^{-1}\). The extension \(x\) under a load of \(F = 40\text{ N}\) is given by \(x = \frac{F}{k_{\text{eq}}} = \frac{40\text{ N}}{400\text{ N m}^{-1}} = 0.10\text{ m}\). The total elastic potential energy stored is \(E_p = \frac{1}{2} F x = \frac{1}{2} \times 40\text{ N} \times 0.10\text{ m} = 2.0\text{ J}\). Alternatively, \(E_p = \frac{1}{2} k_{\text{eq}} x^2 = \frac{1}{2} \times 400\text{ N m}^{-1} \times (0.10\text{ m})^2 = 2.0\text{ J}\).

1 mark for finding the combined parallel spring constant and using it to find the correct total energy of 2.0 J (Option B).

All three statements are correct: Statement 1 describes the main proof that the atom's mass and charge are concentrated in a tiny volume, leaving most of the atom as empty space. Statement 2 describes how large-angle deflections can only be caused by a massive, concentrated positive charge (the nucleus). Statement 3 describes a necessary experimental condition: if the foil were thick, multiple collisions would occur, obscuring the single-collision scattering pattern that reveals nuclear structure.

1 mark for identifying that statements 1, 2, and 3 are all correct (Option D).

The initial kinetic energy is \(E_{k,\text{initial}} = \frac{1}{2} m u^2\). At the highest point, the vertical component of the velocity is zero (\(v_y = 0\)), and because there is no air resistance, the horizontal component of velocity remains constant: \(v_x = u \cos(30.0^\circ) = u \frac{\sqrt{3}}{2}\). Thus, the velocity at the highest point is \(v = u \frac{\sqrt{3}}{2}\). The kinetic energy at the highest point is \(E_{k,\text{highest}} = \frac{1}{2} m v^2 = \frac{1}{2} m \left(u \frac{\sqrt{3}}{2}\right)^2 = \frac{1}{2} m u^2 \times \frac{3}{4} = 0.75 E_{k,\text{initial}}\). This gives a ratio of \(0.75\).

1 mark for identifying the horizontal velocity component at the highest point and calculating the ratio of the kinetic energies to be 0.75 (Option D).

The gradient of a distance-time (\(s-t\)) graph represents speed. At the start (\(t = 0\)), the stone is released from rest, so its speed is zero, meaning the initial gradient is zero. As it falls, its speed increases, so the gradient of the curve increases. Once it reaches its terminal velocity, its speed becomes constant, which means the gradient becomes constant and non-zero (represented by a straight line of constant positive slope). Thus, the gradient is initially zero and increases to a constant non-zero value.

1 mark for correctly matching the physical features of the falling motion to the gradient behavior of the distance-time graph (Option A).

The current in the circuit is \(I = \frac{E}{R + r}\). When \(R\) is decreased, the total circuit resistance decreases, so the current \(I\) increases. The potential difference across the internal resistance is \(V_r = I r\); since \(I\) increases, \(V_r\) must increase. The terminal potential difference (the voltmeter reading) is \(V = E - I r\); since \(I r\) increases and \(E\) is constant, \(V\) must decrease.

1 mark for correctly analyzing the changes in current and determining that terminal potential difference decreases while internal potential difference increases (Option B).

The formula for density is \(\rho = \frac{m}{V} = \frac{4m}{\pi d^2 L}\). The fractional uncertainty equation is \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\). First, compute the individual percentage uncertainties: \%\Delta m = (0.5 / 25.0) \times 100\% = 2\%\), \%\Delta d = (0.02 / 2.00) \times 100\% = 1\%\), and \%\Delta L = (0.1 / 5.0) \times 100\% = 2\%\). Summing these up with the power factor for \(d\): \%\Delta \rho = 2\% + 2(1\%) + 2\% = 6\%.

Award 1 mark for the correct answer C. No partial credit is awarded for multiple-choice questions.

A zero error is a systematic error because it affects all measurements consistently in the same direction. Since the instrument reads \(-0.03\text{ mm}\) when empty (which is below zero), all measured readings are too low by \(0.03\text{ mm}\). To obtain the correct actual diameter, she must add \(0.03\text{ mm}\) to the mean of the five readings.

Award 1 mark for the correct answer D. No partial credit is awarded for multiple-choice questions.

The strain energy \(W\) stored in an elastically deformed wire is given by \(W = \frac{1}{2} F x\). The Young modulus is defined as \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{x / L} = \frac{F L}{A x}\). Rearranging this for force gives \(F = \frac{E A x}{L}\). Substituting this expression for \(F\) into the strain energy formula gives \(W = \frac{1}{2} \left( \frac{E A x}{L} \right) x = \frac{E A x^2}{2 L}\).

Award 1 mark for the correct answer B. No partial credit is awarded for multiple-choice questions.

A baryon is made of three quarks. Let the quark composition be \(q_1 q_2 \text{s}\). The sum of the charges of the quarks must equal the total charge of the baryon, which is \(+1e\). Thus, \(q_1 + q_2 + (-\frac{1}{3}e) = +1e\), which simplifies to \(q_1 + q_2 = +\frac{4}{3}e\). The only combination of up and down quarks that can sum to \(+\frac{4}{3}e\) is two up quarks (since \(\frac{2}{3}e + \frac{2}{3}e = \frac{4}{3}e\)). Therefore, the quark composition is \(\text{u u s}\).

Award 1 mark for the correct answer A. No partial credit is awarded for multiple-choice questions.

The object is projected upwards with initial velocity \(v_0 = 24\text{ m s}^{-1}\) and reaches its highest point where \(v = 0\). Setting \(24 - 4.0t = 0\) yields \(t = 6.0\text{ s}\). The upward distance travelled is the area under the \(v\)-\(t\) graph from \(t = 0\) to \(t = 6\text{ s}\), which is \(\frac{1}{2} \times 6.0\text{ s} \times 24\text{ m s}^{-1} = 72\text{ m}\). Because there is no air resistance, the return trip downwards covers the same distance of \(72\text{ m}\). Thus, the total distance travelled is \(72\text{ m} + 72\text{ m} = 144\text{ m}\).

Award 1 mark for the correct answer B. No partial credit is awarded for multiple-choice questions.

For an NTC thermistor, a decrease in temperature causes its resistance to increase. Since the thermistor is in series with the fixed resistor, the increase in thermistor resistance increases the total resistance of the circuit. The total current \(I\) from the power supply decreases. The potential difference across the fixed resistor, given by \(V_{\text{fixed}} = I R\), must decrease because the resistance \(R\) is constant and the current \(I\) decreases.

Award 1 mark for the correct answer C. No partial credit is awarded for multiple-choice questions.

Rearranging the formula for \(g\) gives: \(T^2 = 4\pi^2 \frac{L}{g} \implies g = 4\pi^2 \frac{L}{T^2}\). The percentage uncertainty in \(g\) is therefore given by the combination of the percentage uncertainties of \(L\) and \(T\): \(\%\Delta g = \%\Delta L + 2(\%\Delta T)\). Substituting the given percentage uncertainties: \(\%\Delta g = 2\% + 2(3\%) = 8\%\).

Award 1 mark for the correct answer C. No partial credit is awarded for multiple-choice questions.

The Young modulus \(E\) is given by \(E = \frac{\text{stress}}{\text{strain}} = \frac{F L}{A x}\), where \(x\) is the extension and \(A\) is the cross-sectional area. Rearranging for extension yields \(x = \frac{F L}{A E}\). Since \(A = \pi d^2 / 4\), the extension can be written as \(x \propto \frac{L}{d^2}\) because \(F\) and \(E\) are identical for both wires. For wire X, \(x_{\text{X}} \propto \frac{L}{d^2}\). For wire Y, \(x_{\text{Y}} \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{L}{2d^2}\). The ratio of extensions is \(\frac{x_{\text{X}}}{x_{\text{Y}}} = \frac{L/d^2}{L/(2d^2)} = 2\).

Award 1 mark for the correct answer B. No partial credit is awarded for multiple-choice questions.

The density \( \rho \) of a cylinder is given by the formula:
\[ \rho = \frac{m}{V} = \frac{4m}{\pi d^2 L} \]

Using the rules for combining uncertainties:
\[ \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\left(\frac{\Delta d}{d}\right) + \frac{\Delta L}{L} \]

Substituting the percentage uncertainties:
\[ \frac{\Delta \rho}{\rho} = 2.0\% + 2(2.0\%) + 1.0\% = 7.0\% \]

1 mark for the correct calculation of the percentage uncertainty, accounting for the factor of 2 in the diameter term due to its square in the volume formula.

The relation between true diameter \( d_{\text{true}} \) and recorded reading \( d_{\text{read}} \) is:
\[ d_{\text{true}} = d_{\text{read}} - (\text{zero error}) \]
\[ d_{\text{true}} = d_{\text{read}} - (-0.03\text{ mm}) = d_{\text{read}} + 0.03\text{ mm} \]

Since \( d_{\text{read}} < d_{\text{true}} \), the uncorrected calculated cross-sectional area will be smaller than the true area. A zero error is an instrumental error that remains constant under the same conditions, making it a systematic error.

1 mark for identifying that the calculated area is smaller than the true area and that this is due to a systematic error.

The Young modulus \( E \) is defined as:
\[ E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{x/L} = \frac{FL}{A x} \]

Rearranging for extension \( x \):
\[ x = \frac{FL}{AE} = \frac{4FL}{\pi d^2 E} \]

Since \( F \) and \( E \) are identical for both wires:
\[ x \propto \frac{L}{d^2} \]

Using the ratios given:
\[ \frac{x_{\text{X}}}{x_{\text{Y}}} = \frac{L_{\text{X}}}{L_{\text{Y}}} \times \left( \frac{d_{\text{Y}}}{d_{\text{X}}} \right)^2 = 2 \times (2)^2 = 8 \]

1 mark for correctly applying the Young modulus formula to derive the relationship between extension, length, and diameter, and obtaining the ratio of 8.

The work done to stretch the rubber band during loading is \( 2.5\text{ J} \).
The work recovered from the rubber band as it unloads is equal to the area under the unloading curve, which is \( 1.8\text{ J} \).
The difference between the work done during loading and unloading is the thermal energy dissipated (hysteresis loss):
\[ \text{Thermal energy} = 2.5\text{ J} - 1.8\text{ J} = 0.7\text{ J} \]

1 mark for correctly identifying the thermal energy as the difference between the areas and the work recovered as the area under the unloading curve.

A neutron has a quark composition of \( udd \) and a proton has \( uud \). During \( \beta^- \) decay, one down quark (\( d \)) changes into an up quark (\( u \)):
\[ d \rightarrow u + e^- + \bar{\nu}_e \]

At the fundamental vertex, a down quark of charge \( -\frac{1}{3}e \) emits a negative \( W^- \) boson to become an up quark of charge \( +\frac{2}{3}e \). The \( W^- \) boson then decays into the electron and electron antineutrino.

1 mark for selecting the option with the correct quark transition (d to u) and the correct charge-conserving weak exchange boson (W-).

Let upwards be the positive direction.
- Initial velocity \( u_i = +u \)
- Acceleration \( a = -g \)
- Total time of flight \( t = T \)
- Vertical displacement \( s = -h \) (since the ground is a distance \( h \) below the launch point)

Using the equations of motion:
\[ s = u_i t + \frac{1}{2} a t^2 \]
\[ -h = uT - \frac{1}{2} g T^2 \]

Multiplying both sides by \( -1 \) yields:
\[ h = \frac{1}{2} g T^2 - uT \]

1 mark for using the equations of motion with consistent vector signs to obtain the correct expression.

Let \( R \) be the resistance of the fixed resistor (\( 4.0\text{ k}\Omega \)) and \( R_{\text{th}} \) be the resistance of the thermistor.

At room temperature:
\[ V_{\text{out}} = V_{\text{supply}} \left( \frac{R}{R + R_{\text{th}}} \right) \]
\[ 3.0 = 9.0 \left( \frac{4.0}{4.0 + R_{\text{th}}} \right) \]
\[ \frac{4.0}{4.0 + R_{\text{th}}} = \frac{1}{3} \implies R_{\text{th}} = 8.0\text{ k}\Omega \]

At the high temperature, \( R_{\text{th}}' = 1.0\text{ k}\Omega \):
\[ V_{\text{out}}' = 9.0 \left( \frac{4.0}{4.0 + 1.0} \right) = 9.0 \times 0.8 = 7.2\text{ V} \]

The change in the voltmeter reading is:
\[ \Delta V = 7.2\text{ V} - 3.0\text{ V} = 4.2\text{ V}\text{ (increase)} \]

1 mark for determining the initial thermistor resistance, calculating the new potential difference across the fixed resistor, and finding the correct difference of 4.2 V.

Precision is indicated by the range of the measured values. A smaller range means higher precision. Student R has the smallest range (\( 0.10\text{ m s}^{-2} \)), so R's results are the most precise.

Accuracy is indicated by how close the mean value is to the accepted true value (\( 9.81\text{ m s}^{-2} \)). Students P and Q have means very close to \( 9.81\text{ m s}^{-2} \). Students R and S have means far from the true value. Comparing R and S, R's mean of \( 10.55\text{ m s}^{-2} \) is highly inaccurate, and R is significantly more precise than S. Thus, R represents high precision combined with low accuracy.

1 mark for identifying that the smallest range corresponds to the highest precision, and a mean far from the true value corresponds to the lowest accuracy.

To find the percentage uncertainty in the calculated resistivity \(\rho\), we sum the fractional (or percentage) uncertainties of each term in the product/quotient formula, taking into account powers:

\[ \frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L} \]

Calculate the percentage uncertainty for each quantity:
- For resistance \(R\): \(\frac{0.05}{2.50} \times 100\% = 2.0\%\)
- For diameter \(d\): \(\frac{0.01}{0.40} \times 100\% = 2.5\%\)
- For length \(L\): \(\frac{0.003}{1.500} \times 100\% = 0.2\%\)

Now substitute these values into the uncertainty relation:

\[ \text{Percentage uncertainty} = 2.0\% + 2(2.5\%) + 0.2\% = 7.2\% \]

Therefore, the percentage uncertainty is \(7.2\%\).

1 mark for calculating the percentage uncertainties of individual terms and correctly multiplying the diameter uncertainty by 2 before summing, yielding 7.2%.

Her reaction time causes a systematic delay at the start of every measurement, meaning the stopwatch is consistently started late. This introduces a systematic error, which cannot be reduced or eliminated by taking multiple readings and calculating an average.

However, human timing variations from trial to trial around this systematic offset are random. Calculating the average of multiple independent measurements reduces the impact of these random variations, thereby reducing the random error.

1 mark for identifying that systematic error is unaffected by averaging, whereas random error is reduced.

The Young modulus \(E\) of the material (steel) is constant. Using the formula for Young modulus:

\[ E = \frac{F L}{A x} \implies x = \frac{F L}{A E} \]

The extension of the first wire is \(x_1 = \frac{F L}{A E}\).

The elastic potential energy stored in the first wire is:

\[ E_1 = \frac{1}{2} F x_1 = \frac{F^2 L}{2 A E} \]

For the second wire, length is \(2L\) and area is \(2A\). Its extension \(x_2\) under the same force \(F\) is:

\[ x_2 = \frac{F (2L)}{(2A) E} = \frac{F L}{A E} = x_1 \]

Since the extension and the force are identical to those of the first wire, the elastic potential energy stored in the second wire is:

\[ E_2 = \frac{1}{2} F x_2 = \frac{1}{2} F x_1 = E_1 \]

Therefore, the ratio \(\frac{E_2}{E_1} = 1.0\).

1 mark for establishing that the extension is unchanged and thus the elastic potential energy stored remains the same (ratio 1.0).

First, calculate the cross-sectional area \(A\) of the wire from its diameter \(d\):

\[ A = \frac{\pi d^2}{4} = \frac{\pi (0.80 \times 10^{-3}\text{ m})^2}{4} \approx 5.03 \times 10^{-7}\text{ m}^2 \]

Next, calculate the extension \(x\) using Young modulus \(E = \frac{F L}{A x}\):

\[ x = \frac{F L}{A E} = \frac{120\text{ N} \times 2.0\text{ m}}{(5.03 \times 10^{-7}\text{ m}^2) \times (1.5 \times 10^{11}\text{ Pa})} \approx 3.18 \times 10^{-3}\text{ m} \]

Finally, calculate the strain energy (elastic potential energy) stored:

\[ E_s = \frac{1}{2} F x = \frac{1}{2} \times 120\text{ N} \times 3.18 \times 10^{-3}\text{ m} \approx 0.19\text{ J} \]

1 mark for correctly finding the cross-sectional area and extension, then applying the strain energy formula to arrive at 0.19 J.

The observation that most \(\alpha\)-particles pass through the foil undeflected means they did not encounter any obstacles or strong electrostatic fields. This directly implies that the vast majority of the volume of the gold atom is empty space. (The small volume and large mass of the nucleus are deduced from the rare, large-angle deflections, not the undeflected ones).

1 mark for identifying that the high transmission rate of alpha particles indicates that the atom consists mostly of empty space.

In projectile motion, the horizontal component of velocity remains constant, so the horizontal velocity at impact is \(v_x = v\).

The stone hits the sea at an angle of \(45^\circ\) to the horizontal, which means:

\[ \tan(45^\circ) = \frac{v_y}{v_x} \implies 1 = \frac{v_y}{v} \implies v_y = v \]

For the vertical motion, using the equation of motion \(v_y^2 = u_y^2 + 2 g h\) where the initial vertical velocity is \(u_y = 0\):

\[ v^2 = 2 g h \implies h = \frac{v^2}{2g} \]

1 mark for establishing that the vertical velocity equals the horizontal velocity at impact, and applying equations of motion to find the correct expression for height.

We can use the kinematic equation \(s = u t + \frac{1}{2} a t^2\).

Let the upward direction be positive. The parameters are:
- Displacement \(s = -2.0\text{ m}\) (since the ball ends up 2.0 m below its starting position)
- Initial velocity \(u = +15\text{ m s}^{-1}\)
- Acceleration \(a = -g = -9.81\text{ m s}^{-2}\)

Substitute these values into the equation:

\[ -2.0 = 15 t - 4.905 t^2 \]

Rearrange into a quadratic equation:

\[ 4.905 t^2 - 15 t - 2.0 = 0 \]

Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\[ t = \frac{15 \pm \sqrt{(-15)^2 - 4(4.905)(-2.0)}}{2 \times 4.905} \]
\[ t = \frac{15 \pm \sqrt{225 + 39.24}}{9.81} = \frac{15 \pm 16.26}{9.81} \]

Since time must be positive:

\[ t = \frac{15 + 16.26}{9.81} \approx 3.19\text{ s} \approx 3.2\text{ s} \]

1 mark for setting up the displacement equation with correct signs and resolving the quadratic equation to get 3.2 s.

The terminal potential difference \(V\) is given by \(V = I R = \frac{E R}{R + r}\). As \(R\) increases, the total resistance of the circuit increases, which reduces the current. This reduces the internal potential drop \(I r\), thereby causing \(V = E - I r\) to increase continuously toward \(E\).

The power dissipated in the external resistor is \(P = I^2 R = \frac{E^2 R}{(R+r)^2}\).
- When \(R \ll r\), \(P \approx \frac{E^2 R}{r^2}\) which is very small.
- When \(R \gg r\), \(P \approx \frac{E^2}{R}\) which is also very small.
- The power \(P\) reaches a maximum when \(R = r\) (maximum power transfer theorem).

Thus, as \(R\) increases, \(V\) increases continuously, and \(P\) increases to a maximum and then decreases.

1 mark for identifying that the terminal potential difference increases continuously with external resistance, while power reaches a peak when external resistance matches internal resistance.

The formula for density is \( \rho = \frac{4m}{\pi d^2 h} \). The fractional uncertainty in density is given by \( \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta h}{h} \). First, calculate the individual percentage uncertainties: \( \% \text{ uncertainty in } m = \frac{0.1}{25.0} \times 100 = 0.40\% \). \( \% \text{ uncertainty in } d = \frac{0.05}{12.00} \times 100 \approx 0.417\% \). \( \% \text{ uncertainty in } h = \frac{0.2}{40.0} \times 100 = 0.50\% \). Now, sum them up with the appropriate multipliers: \( \% \text{ uncertainty in } \rho = 0.40\% + 2(0.417\%) + 0.50\% = 0.40\% + 0.834\% + 0.50\% = 1.734\% \approx 1.7\% \).

1 mark for identifying the correct relationship for combining uncertainties and calculating the total percentage uncertainty to be 1.7%.

Systematic errors, such as a calibration error where the stopwatch runs too fast, affect the accuracy of the measurements because they shift all readings in one direction away from the true value. Random variations, such as the human reaction time in starting and stopping the timer, cause a spread in repeated measurements and thus reduce the precision. Repeating the measurements and taking an average reduces the impact of random error (improving precision) but does not remove the systematic error (accuracy remains affected).

1 mark for correctly matching systematic error with accuracy and random variation (reaction time) with precision.

Young modulus is defined as \( E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{x/L} \), which gives the tension force as \( F = \frac{E A x}{L} \). For the steel wire: \( F_{\text{steel}} = \frac{(2E) A x}{L} = \frac{2 E A x}{L} \). For the brass wire: \( F_{\text{brass}} = \frac{E (2A) x}{2L} = \frac{E A x}{L} \). The ratio of the tension in the steel wire to the brass wire is therefore \( \frac{F_{\text{steel}}}{F_{\text{brass}}} = \frac{2EAx/L}{EAx/L} = 2.0 \).

1 mark for using the formula for tension in terms of Young modulus, substituting the correct parameters for both wires, and calculating the ratio of 2.0.

The total work done on the wire during loading is represented by the entire area under the loading curve \( \text{OP} \) down to the extension axis. When the wire is unloaded, some of this energy is recovered as elastic strain energy, which is represented by the area under the unloading line \( \text{PQ} \) down to the extension axis. The difference between these two areas is the area enclosed by the loop \( \text{OPQ} \, which represents the work done in permanently deforming the wire (energy dissipated as thermal energy).

1 mark for identifying that the net work done in permanent deformation (energy dissipated) is represented by the enclosed area OPQ.

A proton has a quark structure of up-up-down (\( uud \)), and a neutron has a quark structure of up-down-down (\( udd \)). During beta-plus decay, a proton converts into a neutron: \( p \rightarrow n + e^+ + \nu_e \). Comparing the quark compositions, we see that one up quark (\( u \)) in the proton changes into a down quark (\( d \)) to form the neutron.

1 mark for identifying the quark compositions of a proton and neutron and correctly concluding that one up quark changes to a down quark.

For horizontal projectile motion, the vertical motion is independent of the horizontal motion. The time of flight \( t \) is determined by the height: \( h = \frac{1}{2}gt^2 \implies t = \sqrt{\frac{2h}{g}} \). The horizontal distance is \( d = v t = v \sqrt{\frac{2h}{g}} \). For the second ball, the height is \( 2h \), so the time of flight is \( t' = \sqrt{\frac{2(2h)}{g}} = \sqrt{2} t \). The horizontal speed is \( 2v \), so the new horizontal distance is \( d' = (2v) t' = 2v (\sqrt{2} t) = 2\sqrt{2} vt = 2\sqrt{2} d \).

1 mark for setting up equations for time of flight and horizontal distance, finding the ratio of times of flight, and calculating the final horizontal distance as \( 2\sqrt{2}d \).

An increase in light intensity causes the resistance of a light-dependent resistor (LDR) to decrease. In a series potential divider circuit, the output voltage across the LDR is given by \( V_{\text{out}} = V \times \frac{R_{\text{LDR}}}{R + R_{\text{LDR}}} \). As the resistance of the LDR decreases, the proportion of the total resistance of the circuit contributed by the LDR decreases. Consequently, the potential difference across it, which is \( V_{\text{out}} \), also decreases.

1 mark for correctly identifying that increasing light intensity decreases the LDR resistance, and that this leads to a decrease in the output voltage.

Let \( N_0 \) be the initial number of atoms of \( X \). At time \( t \), the number of atoms of \( X \) is \( N_X \) and the number of atoms of \( Y \) is \( N_Y = N_0 - N_X \). We are given \( \frac{N_Y}{N_X} = 15 \implies \frac{N_0 - N_X}{N_X} = 15 \implies \frac{N_0}{N_X} - 1 = 15 \implies \frac{N_0}{N_X} = 16 \). This means \( N_X = \frac{1}{16}N_0 = \left(\frac{1}{2}\right)^4 N_0 \). Thus, 4 half-lives have elapsed. The time \( t \) is \( 4 \times 2.0 \text{ hours} = 8.0 \text{ hours} \).

1 mark for setting up the ratio equation to find that the remaining fraction of X is 1/16, identifying that this represents 4 half-lives, and calculating the time to be 8.0 hours.

(a) \(g = \frac{2h}{t^2} = \frac{2 \times 1.25}{(0.51)^2} = 9.61\text{ m s}^{-2}\) (or \(9.6\text{ m s}^{-2}\)).
(b) (i) Percentage uncertainty in \(g\) is given by:
\(\frac{\Delta g}{g} \times 100 = \left( \frac{\Delta h}{h} + 2 \frac{\Delta t}{t} \right) \times 100\)
\(\frac{\Delta h}{h} \times 100 = \frac{0.02}{1.25} \times 100 = 1.6\%\)
\(\frac{\Delta t}{t} \times 100 = \frac{0.01}{0.51} \times 100 \approx 1.96\%\)
Total percentage uncertainty = \(1.6\% + 2 \times 1.96\% = 5.52\% \approx 5.5\%\).
(ii) Absolute uncertainty \(\Delta g = 9.61 \times 0.0552 = 0.53\text{ m s}^{-2}\). Since absolute uncertainties are typically quoted to 1 significant figure, \(\Delta g = 0.5\text{ m s}^{-2}\).
Therefore, \(g = 9.6 \pm 0.5\text{ m s}^{-2}\).
(c) Two possible improvements:
1. Use light gates connected to a digital timer to measure \(t\) directly, which eliminates human reaction time.
2. Increase the height \(h\) from which the ball is dropped so that the time of fall \(t\) is larger, thereby reducing the fractional/percentage uncertainty in \(t\).

(a) [2 marks]
C1 for substitution: \(2 \times 1.25 / 0.51^2\)
A1 for final answer: \(9.6\text{ m s}^{-2}\) (or \(9.61\text{ m s}^{-2}\))

(b) (i) [3 marks]
C1 for fractional uncertainty formula: \(\frac{\Delta h}{h} + 2\frac{\Delta t}{t}\)
C1 for calculating percentage uncertainties: \(1.6\%\) and \(2 \times 1.96\% = 3.92\%\)
A1 for showing addition to get \(5.52\%\) or \(5.5\%\)
(ii) [2 marks]
C1 for calculating absolute uncertainty: \(9.6 \times 0.055 = 0.53\) or \(9.61 \times 0.0552 = 0.53\)
A1 for rounding to 1 s.f. and matching the decimal place of \(g\): \(0.5\text{ m s}^{-2}\) (accept \(0.53\) if stated clearly)

(c) [3 marks]
B1 for stating a valid method (e.g. use of light gates / electronic timer OR increasing the height \(h\))
B1 for explaining how this method works (e.g. eliminates human reaction time OR increases the time \(t\))
B1 for explaining the effect on uncertainty (e.g. reduces the systematic/random error in time OR decreases the percentage uncertainty in \(t\))

(a) Ohm law: \(R = V / I\).
Potential difference: \(V = W / Q = \text{work done} / \text{charge}\).
SI base units: \(W\) is \(\text{kg m}^2\text{ s}^{-2}\), \(Q\) is \(\text{A s}\).
So \(V\) is \(\text{kg m}^2\text{ s}^{-3}\text{ A}^{-1}\).
Hence, \(R\) is \(\text{kg m}^2\text{ s}^{-3}\text{ A}^{-2}\).
Resistivity: \(\rho = R \times A / L\).
SI base unit of \(A\) is \(\text{m}^2\) and \(L\) is \(\text{m}\).
Thus, unit of \(\rho\) is \(\text{kg m}^2\text{ s}^{-3}\text{ A}^{-2} \times \text{m}^2 / \text{m} = \text{kg m}^3\text{ s}^{-3}\text{ A}^{-2}\).

(b) \(d = 0.38 \times 10^{-3}\text{ m}\).
\(\rho = \frac{\pi \times (0.38 \times 10^{-3})^2 \times 12.4}{4 \times 1.500} = 9.38 \times 10^{-7}\ \Omega\text{ m}\).

(c) Percentage uncertainty in \(\rho\):
\(\frac{\Delta \rho}{\rho} \times 100 = \left( 2 \frac{\Delta d}{d} + \frac{\Delta R}{R} + \frac{\Delta L}{L} \right) \times 100\)
\(2 \frac{\Delta d}{d} \times 100 = 2 \times \frac{0.02}{0.38} \times 100 = 10.53\%\)
\(\frac{\Delta R}{R} \times 100 = \frac{0.2}{12.4} \times 100 = 1.61\%\)
\(\frac{\Delta L}{L} \times 100 = \frac{0.002}{1.500} \times 100 = 0.13\%\)
Total percentage uncertainty = \(10.53\% + 1.61\% + 0.13\% = 12.27\% \approx 12\%\).

(d) Absolute uncertainty: \(\Delta \rho = 9.38 \times 10^{-7} \times 0.1227 = 1.15 \times 10^{-7}\ \Omega\text{ m}\).
If using 1 s.f.: \(\Delta \rho = 1 \times 10^{-7}\ \Omega\text{ m}\) and \(\rho = (9 \pm 1) \times 10^{-7}\ \Omega\text{ m}\).
If using 2 s.f.: \(\Delta \rho = 1.2 \times 10^{-7}\ \Omega\text{ m}\) and \(\rho = (9.4 \pm 1.2) \times 10^{-7}\ \Omega\text{ m}\).

(a) [3 marks]
B1 for base units of work/energy (\(\text{kg m}^2\text{ s}^{-2}\)) and charge (\(\text{A s}\)) to get unit of \(V\) as \(\text{kg m}^2\text{ s}^{-3}\text{ A}^{-1}\)
B1 for unit of resistance \(R\) as \(\text{kg m}^2\text{ s}^{-3}\text{ A}^{-2}\)
B1 for correct substitution of units into the resistivity equation to show final result \(\text{kg m}^3\text{ s}^{-3}\text{ A}^{-2}\)

(b) [2 marks]
C1 for substitution of all values in consistent units: \(\pi \times (0.38 \times 10^{-3})^2 \times 12.4 / (4 \times 1.500)\)
A1 for value: \(9.4 \times 10^{-7}\ \Omega\text{ m}\) (accept \(9.38 \times 10^{-7}\ \Omega\text{ m}\))

(c) [3 marks]
C1 for identifying that the uncertainty in \(d\) is doubled: \(2 \times \frac{\Delta d}{d}\)
C1 for summing fractional uncertainties: \(2 \left(\frac{0.02}{0.38}\right) + \frac{0.2}{12.4} + \frac{0.002}{1.500}\)
A1 for final percentage uncertainty: \(12\%\) (accept \(12.3\%\))

(d) [2 marks]
C1 for calculating absolute uncertainty: \(1.15 \times 10^{-7}\) or \(1.2 \times 10^{-7}\) or \(1 \times 10^{-7}\)
A1 for final value and uncertainty with unit, with matching decimal places: e.g. \((9.4 \pm 1.2) \times 10^{-7}\ \Omega\text{ m}\) or \((9 \pm 1) \times 10^{-7}\ \Omega\text{ m}\)

(a) (i) Tensile stress is defined as the force per unit cross-sectional area.
(ii) Tensile strain is defined as the extension per unit original length.

(b) (i) \(\text{Stress} = \frac{\text{Force}}{\text{Area}} = \frac{30}{1.5 \times 10^{-7}} = 2.0 \times 10^8\text{ Pa}\) (or \(\text{N m}^{-2}\)).
(ii) Young modulus \(E = \frac{\text{stress}}{\text{strain}}\).
\(\text{Strain} = \frac{\text{stress}}{E} = \frac{2.0 \times 10^8}{1.1 \times 10^{11}} = 1.82 \times 10^{-3}\).
\(\text{Extension } x = \text{strain} \times \text{length} = 1.82 \times 10^{-3} \times 2.4 = 4.36 \times 10^{-3}\text{ m}\) (or \(4.4\text{ mm}\)).

(c) In elastic deformation, the wire returns to its original length when the deforming force/load is removed. In plastic deformation, the wire does not return to its original length when the load is removed; there is permanent deformation.

(a) (i) [1 mark]
B1 for force per unit cross-sectional area.
(ii) [1 mark]
B1 for extension per unit original length.

(b) (i) [2 marks]
C1 for formula or substitution: \(30 / (1.5 \times 10^{-7})\)
A1 for \(2.0 \times 10^8\text{ Pa}\) (or \(\text{N m}^{-2}\))
(ii) [3 marks]
C1 for formula for Young modulus: \(E = \text{stress} / \text{strain}\) or \(E = \frac{FL}{A x}\)
C1 for substitution: \(x = \frac{30 \times 2.4}{1.5 \times 10^{-7} \times 1.1 \times 10^{11}}\)
A1 for \(4.4 \times 10^{-3}\text{ m}\) (or \(4.36 \times 10^{-3}\text{ m}\))

(c) [3 marks]
B1 for elastic deformation description (returns to original shape/length when load is removed).
B1 for plastic deformation description (does not return to original shape/length OR permanent extension).
B1 for mentioning the elastic limit (beyond the elastic limit, the deformation becomes plastic).

(a) The nuclear equation is:
\(^{24}_{11}\text{Na} \rightarrow ^{24}_{12}\text{Mg} + \,^0_{-1}\text{e} + \bar{\nu}_e\)

(b) According to Einstein mass-energy equivalence, mass and energy are equivalent: \(\Delta E = \Delta m c^2\). The decrease in mass (mass defect) between the parent nucleus and the decay products is converted into the kinetic energy of the emitted beta-minus particle, the antineutrino, and the magnesium nucleus.

(c) (i) Proton composition: \(uud\) (up, up, down).
Neutron composition: \(udd\) (up, down, down).
(ii) During beta-minus decay, a neutron decays into a proton. This means one of the down quarks in the neutron changes into an up quark: \(d \rightarrow u\). This change is accompanied by the emission of an electron (beta-minus particle) and an electron antineutrino.

(a) [3 marks]
B1 for magnesium-24 represented as \(^{24}_{12}\text{Mg}\)
B1 for beta-minus particle represented as \(\beta^-\), \(e^-\), or \(^0_{-1}\text{e}\)
B1 for antineutrino represented as \(\bar{\nu}\) or \(\bar{\nu}_e\)

(b) [2 marks]
B1 for referencing mass-energy equivalence: \(E = m c^2\) or mass defect \(\Delta m\) is converted into energy \(\Delta E\).
B1 for stating that this energy becomes the kinetic energy of the decay products.

(c) (i) [2 marks]
B1 for proton: \(uud\) (up, up, down).
B1 for neutron: \(udd\) (up, down, down).
(ii) [3 marks]
B1 for stating that a neutron changes to a proton.
B1 for stating that a down quark changes to an up quark (or \(d \rightarrow u\)).
B1 for mentioning emission of W-minus boson as the mediator of this weak interaction (or writing a complete quark equation: \(udd \rightarrow uud + e^- + \bar{\nu}_e\)).

(a) The package is moving with the helicopter before release, so its initial velocity is \(4.5\text{ m s}^{-1}\) in the upward direction.

(b) (i) Taking upwards as the positive direction, \(u = +4.5\text{ m s}^{-1}\), \(a = -9.81\text{ m s}^{-2}\), and \(v = 0\) at the maximum height.
Using \(v^2 = u^2 + 2as\):
\(0 = 4.5^2 + 2(-9.81)s\)
\(19.62s = 20.25 \Rightarrow s = 1.03\text{ m}\)
The maximum height above the ground = \(120 + 1.03 = 121\text{ m}\) (to 3 s.f.).

(ii) Using \(s = ut + \frac{1}{2}at^2\), where the total displacement is \(s = -120\text{ m}\):
\(-120 = 4.5t - 4.905t^2\)
\(4.905t^2 - 4.5t - 120 = 0\)
Using the quadratic formula:
\(t = \frac{-(-4.5) \pm \sqrt{(-4.5)^2 - 4(4.905)(-120)}}{2 \times 4.905}\)
\(t = \frac{4.5 \pm \sqrt{20.25 + 2354.4}}{9.81} = \frac{4.5 \pm 48.73}{9.81}\)
For a positive time, \(t = \frac{4.5 + 48.73}{9.81} = 5.43\text{ s}\) (or \(5.4\text{ s}\) to 2 s.f.).

(c) The velocity-time graph is a straight line with a constant negative gradient equal to \(-9.81\text{ m s}^{-2}\). It starts at \(v = +4.5\text{ m s}^{-1}\) at \(t = 0\), crosses the time axis (where \(v = 0\)) at \(t \approx 0.46\text{ s}\), and ends at \(t \approx 5.4\text{ s}\) with a final velocity of \(-48.8\text{ m s}^{-1}\).

(a) [1 mark]
B1 for \(4.5\text{ m s}^{-1}\) directed upwards.

(b) (i) [3 marks]
C1 for choosing the correct equation: \(v^2 = u^2 + 2as\)
C1 for calculating the upward displacement: \(s = 1.03\text{ m}\)
A1 for adding to the initial height to get \(121\text{ m}\) (or \(121.0\text{ m}\))
(ii) [4 marks]
C1 for identifying correct displacement: \(s = -120\text{ m}\)
C1 for substitution into quadratic equation: \(-120 = 4.5t - 4.905t^2\) (or equivalent multi-step method)
C1 for solving quadratic equation correctly for \(t\)
A1 for final answer \(5.4\text{ s}\) (or \(5.43\text{ s}\))

(c) [2 marks]
B1 for a straight line with a negative gradient (representing constant acceleration \(-g\)).
B1 for identifying at least two key values (e.g. y-intercept at \(+4.5\text{ m s}^{-1}\), x-intercept at \(0.46\text{ s}\), or final time \(5.4\text{ s}\)).

(a) The electromotive force (e.m.f.) of the battery is the energy converted from other forms (such as chemical energy) to electrical energy per unit charge that passes through the battery.

(b) (i) From the loop rule or Ohm law, the total e.m.f. in the circuit equals the sum of the potential differences:
\(E = I(R + r)\)
Divide both sides by \(I \cdot E\):
\(\frac{E}{I E} = \frac{I(R + r)}{I E}\)
\(\frac{1}{I} = \frac{R + r}{E} = \frac{1}{E} R + \frac{r}{E}\). This is a linear equation of the form \(y = mx + c\).
(ii) Comparing \(\frac{1}{I} = \frac{1}{E} R + \frac{r}{E}\) to \(y = mx + c\):
- The gradient \(m\) is \(\frac{1}{E}\). Therefore, \(E = \frac{1}{m} = \frac{1}{\text{gradient}}\).
- The y-intercept \(c\) is \(\frac{r}{E}\). Therefore, \(r = c \times E = \frac{\text{y-intercept}}{\text{gradient}}\).

(c) (i) \(E = \frac{1}{0.16} = 6.25\text{ V}\).
(ii) \(r = 0.24 \times 6.25 = 1.5\ \Omega\).

(a) [2 marks]
B1 for stating energy converted/work done per unit charge.
B1 for specifying conversion from chemical (or other) energy to electrical energy (or work done in driving charge around a complete circuit).

(b) (i) [2 marks]
B1 for starting from \(E = I(R + r)\) or \(E = I R + I r\).
B1 for algebraic manipulation to reach the final form: \(\frac{1}{I} = \frac{1}{E} R + \frac{r}{E}\).
(ii) [3 marks]
B1 for stating that gradient \(m = 1/E\) and hence \(E = 1/\text{gradient}\).
B1 for stating that y-intercept \(c = r/E\).
B1 for stating that \(r = \text{y-intercept} \times E\) or \(r = \text{y-intercept} / \text{gradient}\).

(c) (i) [1 mark]
A1 for \(E = 6.3\text{ V}\) or \(6.25\text{ V}\).
(ii) [2 marks]
C1 for substitution: \(r = 0.24 \times 6.25\) (or \(r = 0.24 / 0.16\))
A1 for \(r = 1.5\ \Omega\).

### Example Results and Data Analysis

**1. Measurements:**
- Unstretched length of spring, \(L_0 = 3.5\text{ cm} = 0.035\text{ m}\)

**2. Table of Results:**

| \(x / \text{cm}\) | \(x / \text{m}\) | \(l / \text{cm}\) | \(l / \text{m}\) | \(e / \text{m}\) |
|---|---|---|---|---|
| 10.0 | 0.100 | 22.0 | 0.220 | 0.185 |
| 20.0 | 0.200 | 20.1 | 0.201 | 0.166 |
| 30.0 | 0.300 | 18.1 | 0.181 | 0.146 |
| 40.0 | 0.400 | 16.2 | 0.162 | 0.127 |
| 50.0 | 0.500 | 14.2 | 0.142 | 0.107 |
| 60.0 | 0.600 | 12.2 | 0.122 | 0.087 |
| 70.0 | 0.700 | 10.3 | 0.103 | 0.068 |

**3. Graph of \(e\) against \(x\):**
- A graph of \(e\text{ / m}\) (y-axis) against \(x\text{ / m}\) (x-axis) is plotted.
- Points are plotted accurately as a straight line with a negative gradient.

**4. Gradient and Intercept Calculation:**
- Choosing two points on the line of best fit: \((0.150, 0.175)\) and \((0.650, 0.077)\).
- \(\text{Gradient } A = \frac{0.077 - 0.175}{0.650 - 0.150} = \frac{-0.098}{0.500} = -0.196\)
- \(\text{y-intercept } B = y - A x = 0.175 - (-0.196 \times 0.150) = 0.175 + 0.0294 = 0.204\text{ m}\)

**5. Determination of Constants:**
- Given: \(\text{Gradient } A = -\frac{m g}{k L}\)
\[-0.196 = -\frac{0.200 \times 9.81}{k \times 0.800}\]
\[0.196 = \frac{1.962}{0.800 k} \implies k = 12.5\text{ N m}^{-1}\]
- Given: \(\text{y-intercept } B = \frac{m g}{k} + \frac{W}{2 k}\)
\[0.204 = \frac{0.200 \times 9.81}{12.5} + \frac{W}{2 \times 12.5}\]
\[0.204 = 0.157 + \frac{W}{25.0} \implies 0.047 = \frac{W}{25.0} \implies W = 1.18\text{ N}\]

**6. Final Values:**
- \(k = 12.5\text{ N m}^{-1}\)
- \(W = 1.18\text{ N}\)

**[1 mark]** Measurement of unstretched length \(L_0\) recorded with correct unit (to nearest mm).
**[1 mark]** Successful collection of at least 6 sets of readings of \(x\) and \(l\).
**[1 mark]** Range of \(x\) is at least 50.0 cm.
**[1 mark]** Table headings are correct with units: \(x / \text{cm}\) (or \(x / \text{m}\)), \(l / \text{cm}\) (or \(l / \text{m}\)), \(e / \text{cm}\) (or \(e / \text{m}\)).
**[1 mark]** Raw readings of \(x\) and \(l\) are consistent to the nearest millimetre.
**[1 mark]** Extension \(e\) is calculated correctly and is consistent in significant figures with the raw measurements.
**[1 mark]** Quality of data: points on the graph lie close to a straight line showing a clear negative correlation.
**[1 mark]** Linear graph axes: appropriate scales covering at least half of the grid in both dimensions, clearly labelled.
**[1 mark]** Plotting of points: accurate to within half a small grid square.
**[1 mark]** Best-fit straight line: a single thin, sharp line showing balanced distribution of points.
**[1 mark]** Gradient calculation: uses a large triangle with hypotenuse length at least half the length of the line.
**[1 mark]** y-intercept calculation: correctly read from the axis at \(x=0\) or calculated by substituting coordinates of a point on the line into \(y = mx + c\).
**[2 marks]** Determination of \(k\): equating gradient to \(-\frac{m g}{k L}\) with substitution of \(m = 0.200\text{ kg}\), \(g = 9.81\text{ m s}^{-2}\), and \(L = 0.800\text{ m}\). (1 mark for working, 1 mark for correct value in range \(8\text{ to }20\text{ N m}^{-1}\)).
**[2 marks]** Determination of \(W\): equating intercept to \(\frac{m g}{k} + \frac{W}{2 k}\) with correct substitution. (1 mark for working, 1 mark for correct value in range \(0.5\text{ to }2.0\text{ N}\)).
**[1 mark]** Correct units for both \(k\) (\(\text{N m}^{-1}\)) and \(W\) (\(\text{N}\)).
**[2 marks]** Quality mark: experimental values of \(k\) and \(W\) are realistic, and all calculations are shown clearly with consistent significant figures.

### Example Results and Data Analysis

**1. First Height (\(h_1 = 15.0\text{ cm}\)):**
- Raw values of sliding distance: \(d_{1,a} = 24.5\text{ cm}\), \(d_{1,b} = 25.5\text{ cm}\), \(d_{1,c} = 25.0\text{ cm}\).
- Average distance: \(d_1 = 25.0\text{ cm}\).

**2. Percentage Uncertainty in \(d_1\):**
- Due to variations in collision alignment and stopping position, the absolute uncertainty \(\Delta d_1\) is estimated as \(1.0\text{ cm}\).
- Percentage uncertainty: \(\frac{\Delta d_1}{d_1} \times 100\% = \frac{1.0}{25.0} \times 100\% = 4.0\%\).

**3. Second Height (\(h_2 = 30.0\text{ cm}\)):**
- Raw values of sliding distance: \(d_{2,a} = 49.0\text{ cm}\), \(d_{2,b} = 51.0\text{ cm}\), \(d_{2,c} = 50.0\text{ cm}\).
- Average distance: \(d_2 = 50.0\text{ cm}\).

**4. Calculation of Constants:**
- For \(h_1 = 15.0\text{ cm} = 0.150\text{ m}\), \(d_1 = 0.250\text{ m}\):
\[C_1 = \frac{d_1}{h_1} = \frac{0.250}{0.150} = 1.67\]
- For \(h_2 = 30.0\text{ cm} = 0.300\text{ m}\), \(d_2 = 0.500\text{ m}\):
\[C_2 = \frac{d_2}{h_2} = \frac{0.500}{0.300} = 1.67\]

**5. Evaluation:**
- Percentage difference between \(C_1\) and \(C_2\):
\[\frac{|C_1 - C_2|}{\text{Average of } C} \times 100\% = \frac{|1.67 - 1.67|}{1.67} \times 100\% = 0\%\]
- Since \(0\%\) is well within the typical experimental tolerance of 10%, the proposed relationship \(d = C h\) is strongly supported.

**6. Limitations and Improvements Table:**

| Source of Limitation / Uncertainty | Corresponding Improvement |
|---|---|
| 1. Two sets of readings are insufficient to draw a reliable conclusion. | Take several readings for a wide range of heights \(h\) and plot a graph of \(d\) against \(h\) to check for linearity. |
| 2. The collision is not always head-on, causing the block to slide at an angle or rotate. | Place parallel guide rails on the bench surface to constrain the sliding motion of the block. |
| 3. Difficult to release the sphere from the exact height without imparting an initial velocity. | Use an electromagnet or mechanical release gate to release the sphere consistently from rest. |
| 4. Difficulty in identifying the exact final position of the sliding block due to friction variations. | Use a video camera with a slow-motion playback aligned with a scale alongside the path to record the exact stopping point. |

**[1 mark]** Value of \(d_1\) recorded to the nearest millimetre with correct unit.
**[1 mark]** Repeated readings of \(d_1\) recorded to show consistency.
**[2 marks]** Percentage uncertainty in \(d_1\) calculated correctly: 1 mark for using a reasonable absolute uncertainty \(\Delta d_1\) (between 0.5 cm and 1.5 cm) and showing working; 1 mark for correct percentage calculation.
**[1 mark]** Value of \(d_2\) recorded with correct unit, showing that \(d_2 > d_1\).
**[2 marks]** Calculation of \(C_1\) and \(C_2\) with working shown and correct arithmetic.
**[1 mark]** Significant figures of \(C_1\) and \(C_2\) are consistent with the raw measurements of \(d\) and \(h\) (usually 2 or 3 s.f.).
**[2 marks]** Comparison and Conclusion: 1 mark for calculating the percentage difference between \(C_1\) and \(C_2\); 1 mark for making a valid conclusion stating whether the relationship is supported (comparing the difference to 10%).

**[10 marks]** Detailed Evaluation (1 mark for each valid limitation up to 4, and 1 mark for each corresponding improvement up to 4):
- **Limitation 1:** Only two values of \(h\) are used, which is insufficient to validate the relationship.
- **Improvement 1:** Take multiple readings of \(h\) and plot a graph of \(d\) against \(h\).
- **Limitation 2:** Non-ideal/non-head-on collision causes the block to rotate or slide at an angle.
- **Improvement 2:** Design/use a guiding track or parallel rails on the bench to keep the block sliding straight.
- **Limitation 3:** Difficult to ensure a consistent starting position for the block.
- **Improvement 3:** Draw a reference baseline on the bench or use a physical stop to place the block identically each time.
- **Limitation 4:** Variations in the friction of the bench surface cause inconsistencies in \(d\).
- **Improvement 4:** Use a highly uniform, low-friction surface (such as a smooth acrylic sheet) to ensure consistent deceleration.

\(g = \frac{2s}{t^2}\)

(a) Systematic error is a constant deviation of the measured values from the true value in one direction. Random error is a scatter of measured values about a mean value.

(b)(i) \(g = \frac{2 \times 1.50}{(0.55)^2} \approx 9.917\text{ m s}^{-2}\). Since \(t\) has 2 significant figures, \(g\) should be expressed to 2 significant figures, which is \(9.9\text{ m s}^{-2}\).

(b)(ii) \(\frac{\Delta g}{g} = \frac{\Delta s}{s} + 2\frac{\Delta t}{t} = \frac{0.02}{1.50} + 2 \times \frac{0.01}{0.55} = 0.0133 + 0.0364 = 0.0497\).

\(\Delta g = 9.917 \times 0.0497 \approx 0.493\text{ m s}^{-2} \approx 0.5\text{ m s}^{-2}\).

So, \(g = (9.9 \pm 0.5)\text{ m s}^{-2}\).

(c) The percentage uncertainty in \(t\) contributes more because \(\frac{2\Delta t}{t} = 3.6\%\) whereas \(\frac{\Delta s}{s} = 1.3\%\). This is because the uncertainty in \(t\) is doubled in the calculation of the uncertainty in \(g\).

(a) Systematic error: constant difference between measured values and true value [1]
Random error: scatter of readings about a mean / unpredictable fluctuations [1]

(b)(i) Formula \(g = \frac{2s}{t^2}\) used [1]
Correct calculation to 9.917 [1]
Answer to 2 s.f. i.e., 9.9 m s\(^{-2}\) [1]

(b)(ii) Fractional uncertainty equation \(\frac{\Delta g}{g} = \frac{\Delta s}{s} + 2\frac{\Delta t}{t}\) [1]
Correct substitution to find \(\Delta g / g \approx 0.050\) [1]
Absolute uncertainty \(\Delta g \approx 0.5\) (or 0.49) m s\(^{-2}\) [1]

(c) Uncertainty in \(t\) contributes more because its percentage uncertainty is doubled [1]
Calculation shown: \(\% \text{ uncertainty of } t = 1.8\% \times 2 = 3.6\%\), whereas \(\% \text{ uncertainty of } s = 1.3\%\) [1]

(a) \(\rho = R \frac{A}{L}\) and \(R = \frac{V}{I}\).
Unit of \(V\) is \(\text{W A}^{-1} = \text{J s}^{-1}\text{ A}^{-1} = \text{kg m}^2\text{ s}^{-3}\text{ A}^{-1}\).
Unit of \(R\) is \(\text{kg m}^2\text{ s}^{-3}\text{ A}^{-2}\).
Unit of \(\rho\) is \(\text{kg m}^2\text{ s}^{-3}\text{ A}^{-2} \times \frac{\text{m}^2}{\text{m}} = \text{kg m}^3\text{ A}^{-2}\text{ s}^{-3}\).

(b)(i) \(\rho = \frac{2.80 \times \pi \times (0.45 \times 10^{-3})^2}{4 \times 1.20 \times 1.250} = \frac{1.7813 \times 10^{-6}}{6.00} = 2.97 \times 10^{-7}\ \Omega\text{ m}\).

(b)(ii) \(\frac{\Delta \rho}{\rho} = \frac{\Delta V}{V} + \frac{\Delta I}{I} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L} = \frac{0.05}{2.80} + \frac{0.02}{1.20} + 2 \times \frac{0.01}{0.45} + \frac{0.002}{1.250} = 0.0179 + 0.0167 + 0.0444 + 0.0016 = 0.0806\).
Percentage uncertainty = \(8.06\%\) (or \(8.1\%\)).

(c) The diameter \(d\). Its percentage uncertainty is the largest and is doubled in the calculation of the uncertainty of \(\rho\).

(a) Expresses \(V\) or \(R\) in base units [1]
Recognises \(\rho = R A / L\) and multiplies by correct units for area and length [1]
Arrives at final correct combination: \(\text{kg m}^3\text{ A}^{-2}\text{ s}^{-3}\) [1]

(b)(i) Correct conversion of diameter \(d\) to meters [1]
Substitution of all values into the formula [1]
Correct calculation to \(2.97 \times 10^{-7}\ \Omega\text{ m}\) [1]

(b)(ii) Expression for fractional uncertainty [1]
Correct calculation of individual percentage uncertainties [1]
Correct final sum: \(8.1\%\) (accept \(8\%\) or \(8.06\%\)) [1]

(c) Diameter \(d\) named with explanation referencing the doubling of its percentage uncertainty [1]

(a) By definition, Young modulus \(E = \frac{\text{stress}}{\text{strain}} = \frac{T / A}{x / L} = \frac{T L}{A x}\). Rearranging this formula directly gives \(T = \frac{E A x}{L}\).

(b) Since both wires have the same length \(L\) and extension \(x\), the term \(\frac{x}{L}\) is constant. Therefore, \(\frac{T_P}{T_Q} = \frac{E_P A_P}{E_Q A_Q}\).
Substituting the values:
\(\frac{T_P}{T_Q} = \frac{(2.0 \times 10^{11}) \times (2.0 \times 10^{-6})}{(1.0 \times 10^{11}) \times (4.0 \times 10^{-6})} = \frac{4.0 \times 10^5}{4.0 \times 10^5} = 1.0\).

(c)(i) The total downward force is the weight of the load: \(W = mg = 120 \times 9.81 = 1177.2\text{ N}\).
Since the support bar remains horizontal, the sum of the tensions must equal the weight: \(T_P + T_Q = 1177.2\text{ N}\).
Since \(\frac{T_P}{T_Q} = 1.0\), we have \(T_P = T_Q\).
Therefore, \(T_P = T_Q = \frac{1177.2}{2} = 588.6\text{ N} \approx 590\text{ N}\).

(c)(ii) Using wire \(P\):
\(T_P = \frac{E_P A_P x}{L} \implies 588.6 = \frac{2.0 \times 10^{11} \times 2.0 \times 10^{-6} \times x}{2.5}\)
\(588.6 = 1.6 \times 10^5 \times x \implies x = \frac{588.6}{1.6 \times 10^5} = 3.68 \times 10^{-3}\text{ m} \approx 3.7\text{ mm}\).

(a) Definition of Young modulus as stress/strain [1]
Correct algebraic rearrangement showing step-by-step derivation to \(T = \frac{E A x}{L}\) [1]

(b) Identifies that \(x\) and \(L\) are equal for both wires, so \(T \propto E A\) [1]
Correct substitution of values [1]
Correct ratio of 1.0 [1]

(c)(i) Calculates weight of load = 1177 N (or 1180 N) [1]
States \(T_P + T_Q = W\) [1]
Calculates correct tension for each wire = 590 N (or 589 N) [1]

(c)(ii) Correct substitution of either wire's parameters [1]
Calculates correct extension = \(3.7 \times 10^{-3}\text{ m}\) (or \(3.68 \times 10^{-3}\text{ m}\)) [1]

(a)(i) Limit of proportionality (or elastic limit).
(ii) Plastic deformation.

(b) In the linear region (from \(0\) to \(P\)):
\(E = \frac{\text{stress}}{\text{strain}} = \frac{F L}{A x}\)
\(F = 60\text{ N}\), \(x = 2.0 \times 10^{-3}\text{ m}\), \(L = 1.8\text{ m}\), \(A = 1.2 \times 10^{-7}\text{ m}^2\).
\(E = \frac{60 \times 1.8}{1.2 \times 10^{-7} \times 2.0 \times 10^{-3}} = \frac{108}{2.4 \times 10^{-10}} = 4.5 \times 10^{11}\text{ Pa}\).

(c) The work done is the area under the force-extension curve from \(0\) to \(4.0\text{ mm}\).
Area under linear region \(0\) to \(2.0\text{ mm}\):
\(W_{\text{linear}} = \frac{1}{2} \times 60 \times 2.0 \times 10^{-3} = 0.060\text{ J}\).
Area under curved region \(2.0\text{ mm}\) to \(4.0\text{ mm}\) can be approximated by a trapezium:
\(W_{\text{curved}} \approx \frac{60 + 80}{2} \times (4.0 - 2.0) \times 10^{-3} = 70 \times 2.0 \times 10^{-3} = 0.140\text{ J}\).
Total work done \(\approx 0.060 + 0.140 = 0.200\text{ J}\). (Allow range \(0.18\text{ J}\) to \(0.22\text{ J}\)).

(d) The work recovered during unloading is the area under the unloading line from \(4.0\text{ mm}\) to \(1.5\text{ mm}\):
\(W_{\text{recovered}} = \frac{1}{2} \times 80\text{ N} \times (4.0 - 1.5) \times 10^{-3}\text{ m} = 0.100\text{ J}\).
Thermal energy dissipated = Total work done - Work recovered
\(E_{\text{thermal}} = 0.200\text{ J} - 0.100\text{ J} = 0.100\text{ J}\).

(a)(i) Limit of proportionality (accept elastic limit) [1]
(ii) Plastic deformation [1]

(b) Identifies elastic region up to \(60\text{ N}\) and uses \(E = \frac{F L}{A x}\) [1]
Correct values substituted [1]
Correct calculation of Young modulus: \(4.5 \times 10^{11}\text{ Pa}\) [1]

(c) States that work done is the area under the curve [1]
Calculates area of linear part = 0.060 J or estimates curved part [1]
Sums areas correctly to get total work done = \(0.20\text{ J}\) (accept range \(0.18 - 0.22\text{ J}\)) [1]

(d) Calculates work recovered during unloading = 0.10 J [1]
Subtracts recovered work from total work to find thermal energy = \(0.10\text{ J}\) (accept range \(0.08 - 0.12\text{ J}\)) [1]

(a) The decay constant is the probability of decay per unit time of a nucleus.

(b) Using the relation between activity \(A\) and number of nuclei \(N\):
\(A = \lambda N \implies \lambda = \frac{A}{N}\)
\(\lambda = \frac{4.2 \times 10^{11}}{3.5 \times 10^{18}} = 1.2 \times 10^{-7}\text{ s}^{-1}\).

(c) The half-life \(t_{1/2}\) is given by:
\(t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.69315}{1.2 \times 10^{-7}\text{ s}^{-1}} = 5.776 \times 10^6\text{ s}\).
Converting to days:
\(t_{1/2} = \frac{5.776 \times 10^6}{24 \times 3600} = 66.85\text{ days} \approx 67\text{ days}\).

(d) The activity after \(t = 120\text{ days}\) is:
\(t = 120 \times 86400 = 1.0368 \times 10^7\text{ s}\).
\(A = A_0 e^{-\lambda t} = (4.2 \times 10^{11}) \times e^{-(1.2 \times 10^{-7} \times 1.0368 \times 10^7)}\)
\(A = 4.2 \times 10^{11} \times e^{-1.244} \approx 4.2 \times 10^{11} \times 0.288 = 1.21 \times 10^{11}\text{ Bq}\).

(a) Probability of decay of a nucleus [1]
Per unit time [1]

(b) Formula \(A = \lambda N\) stated [1]
Correct calculation showing \(\lambda = 1.2 \times 10^{-7}\text{ s}^{-1}\) [1]

(c) Formula \(t_{1/2} = \frac{\ln 2}{\lambda}\) used [1]
Time in seconds calculated = \(5.78 \times 10^6\text{ s}\) [1]
Correct conversion to days: 67 days (accept 66.8 - 67.0 days) [1]

(d) Converts 120 days to seconds (\(1.04 \times 10^7\text{ s}\)) [1]
Uses \(A = A_0 e^{-\lambda t}\) with correct substitution [1]
Correct activity value: \(1.2 \times 10^{11}\text{ Bq}\) (accept \(1.2 \times 10^{11}\text{ Bq}\) or \(1.21 \times 10^{11}\text{ Bq}\)) [1]

(a) The mass defect is the difference between the total mass of the individual nucleons (protons and neutrons) that make up a nucleus and the actual mass of the nucleus.

(b) Mass of reactants = \(2.01355\text{ u} + 3.01550\text{ u} = 5.02905\text{ u}\).
Mass of products = \(4.00150\text{ u} + 1.00867\text{ u} = 5.01017\text{ u}\).
Mass defect \(\Delta m\) = Mass of reactants - Mass of products
\(\Delta m = 5.02905 - 5.01017 = 0.01888\text{ u}\).
In \(\text{kg}\):
\(\Delta m = 0.01888 \times 1.6605 \times 10^{-27}\text{ kg} = 3.1350 \times 10^{-29}\text{ kg} \approx 3.14 \times 10^{-29}\text{ kg}\).

(c)(i) Energy \(E = \Delta m c^2\)
\(E = (3.1350 \times 10^{-29}\text{ kg}) \times (3.00 \times 10^8\text{ m s}^{-1})^2 = 2.8215 \times 10^{-12}\text{ J} \approx 2.82 \times 10^{-12}\text{ J}\).

(c)(ii) Since \(1\text{ eV} = 1.60 \times 10^{-19}\text{ J}\):
\(E = \frac{2.8215 \times 10^{-12}\text{ J}}{1.60 \times 10^{-19}\text{ J eV}^{-1}} = 1.763 \times 10^7\text{ eV} \approx 17.6\text{ MeV}\).

(a) Difference between constituent nucleons' individual total mass and the mass of the nucleus [1]
Identifies constituents as protons and neutrons [1]

(b) Calculates mass of reactants and mass of products correctly [1]
Calculates \(\Delta m\) in \(\text{u}\) = \(0.01888\text{ u}\) [1]
Converts to \(\text{kg}\) correctly: \(3.14 \times 10^{-29}\text{ kg}\) (accept \(3.13 \times 10^{-29}\) to \(3.14 \times 10^{-29}\)) [1]

(c)(i) Formula \(E = \Delta m c^2\) used [1]
Substitution of mass in \(\text{kg}\) [1]
Correct calculation: \(2.82 \times 10^{-12}\text{ J}\) [1]

(c)(ii) Correct conversion factor from \(\text{J}\) to \(\text{eV}\) [1]
Correct calculation to \(17.6\text{ MeV}\) (accept \(17.5\) to \(17.7\)) [1]

(a) The horizontal component of velocity remains constant because there is no horizontal force acting on the package, since air resistance is negligible. Gravity acts only vertically.

(b)(i) Considering vertical motion with constant acceleration \(g = 9.81\text{ m s}^{-2}\) and initial vertical velocity \(u_y = 0\):
\(s_y = u_y t + \frac{1}{2} g t^2\)
\(120 = 0 + \frac{1}{2} \times 9.81 \times t^2\)
\(t^2 = \frac{240}{9.81} \approx 24.465 \implies t \approx 4.95\text{ s}\).

(b)(ii) Considering horizontal motion with constant velocity \(v_x = 45\text{ m s}^{-1}\):
\(s_x = v_x t = 45 \times 4.946 \approx 222.6\text{ m} \approx 223\text{ m}\) (or \(220\text{ m}\) using 2 s.f.).

(c) Vertical velocity just before hitting the water:
\(v_y = u_y + g t = 0 + 9.81 \times 4.946 \approx 48.52\text{ m s}^{-1}\).
Horizontal velocity: \(v_x = 45\text{ m s}^{-1}\).
Magnitude of velocity:
\(v = \sqrt{v_x^2 + v_y^2} = \sqrt{45^2 + 48.52^2} \approx 66.2\text{ m s}^{-1}\).
Direction of velocity \(\theta\) below the horizontal:
\(\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{48.52}{45}\right) \approx 47.1^\circ\).

(a) No horizontal force acting / air resistance is negligible [1]
Gravity only acts in the vertical direction [1]

(b)(i) Uses vertical motion equation \(s_y = \frac{1}{2} g t^2\) [1]
Substitute values \(120\text{ m}\) and \(9.81\text{ m s}^{-2}\) [1]
Correct time = 4.95 s (accept 4.9 s) [1]

(b)(ii) Uses \(s_x = v_x t\) [1]
Correct distance = 223 m (or 220 m) [1]

(c) Calculates vertical velocity component \(v_y \approx 48.5\text{ m s}^{-1}\) [1]
Calculates magnitude using Pythagoras = \(66\text{ m s}^{-1}\) (accept 66.1 - 66.3) [1]
Calculates angle using trigonometry = \(47^\circ\) below horizontal (accept 47.1 - 47.2) [1]

(a) The circuit diagram should show a series loop containing: a DC power source (battery), a fixed resistor \(R\), and a thermistor \(T\). The voltmeter is connected in parallel across the thermistor \(T\).

(b) Using the potential divider formula:
\(V_{\text{out}} = V \times \frac{R_T}{R + R_T}\)
\(V_{\text{out}} = 9.0\text{ V} \times \frac{2.4\text{ k}\Omega}{1.5\text{ k}\Omega + 2.4\text{ k}\Omega}\)
\(V_{\text{out}} = 9.0 \times \frac{2400}{3900} = 5.54\text{ V} \approx 5.5\text{ V}\).

(c)(i) At \(35^\circ\text{C}\), \(V_{\text{out}} = 3.6\text{ V}\).
\(3.6 = 9.0 \times \frac{R_T}{1500 + R_T}\)
\(0.40 \times (1500 + R_T) = R_T\)
\(600 + 0.40 R_T = R_T \implies 0.60 R_T = 600\)
\(R_T = 1000\ \Omega = 1.0\text{ k}\Omega\).

(c)(ii) Power dissipated in the thermistor:
\(P = \frac{V_{\text{out}}^2}{R_T}\)
\(P = \frac{3.6^2}{1000} = 1.296 \times 10^{-2}\text{ W} = 1.30 \times 10^{-2}\text{ W} = 13\text{ mW}\).

(a) Correct potential divider circuit diagram with correct symbols for battery, resistor, and thermistor [1]
Voltmeter connected in parallel across the thermistor [1]

(b) Potential divider equation quoted or implied [1]
Correct values substituted [1]
Correct voltmeter reading = 5.5 V (accept 5.54 V) [1]

(c)(i) Uses \(V_{\text{out}} = 3.6\text{ V}\) with potential divider formula or finds current in circuit first [1]
Shows correct calculation steps [1]
Correct thermistor resistance = \(1.0\text{ k}\Omega\) (or \(1000\ \Omega\)) [1]

(c)(ii) Correct power formula used (e.g., \(P = V^2 / R\) or \(P = I^2 R\) or \(P = V I\)) [1]
Correct calculation to \(13\text{ mW}\) (or \(0.013\text{ W}\)) [1]

(a) The decay constant is defined as the probability of decay of a nucleus per unit time.

(b) (i) Convert the half-life from years to seconds:
\(T_{1/2} = 5.27\text{ years} = 5.27 \times 365 \times 24 \times 3600\text{ s} = 1.662 \times 10^8\text{ s}\) (or \(1.663 \times 10^8\text{ s}\) using 365.25 days).
Using the relationship between decay constant and half-life:
\(\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.69315}{1.662 \times 10^8} = 4.17 \times 10^{-9}\text{ s}^{-1}\).

(ii) Use the relationship between activity and number of nuclei:
\(A_0 = \lambda N_0\)
\(N_0 = \frac{A_0}{\lambda} = \frac{4.50 \times 10^5}{4.17 \times 10^{-9}} = 1.079 \times 10^{14} \approx 1.08 \times 10^{14}\) nuclei.

(iii) The time interval is \(t = 12.0\text{ years} = 12.0 \times 365 \times 24 \times 3600 = 3.784 \times 10^8\text{ s}\).
Use the exponential decay law:
\(A = A_0 e^{-\lambda t} = (4.50 \times 10^5) \times e^{-(4.17 \times 10^{-9} \times 3.784 \times 10^8)}\)
\(A = (4.50 \times 10^5) \times e^{-1.578} = (4.50 \times 10^5) \times 0.2064 = 9.29 \times 10^4\text{ Bq}\).
(Alternatively, using \(A = A_0 (0.5)^{t/T_{1/2}} = 4.50 \times 10^5 \times (0.5)^{12.0/5.27} = 9.28 \times 10^4\text{ Bq}\)).

(iv) First calculate the remaining number of nuclei \(N\) at 12.0 years:
\(N = \frac{A}{\lambda} = \frac{9.29 \times 10^4}{4.17 \times 10^{-9}} = 2.228 \times 10^{13}\) nuclei.
The number of nuclei decayed is:
\(\Delta N = N_0 - N = 1.079 \times 10^{14} - 2.228 \times 10^{13} = 8.562 \times 10^{13}\) nuclei.
The molar mass of Cobalt-60 is \(60\text{ g mol}^{-1}\).
The mass \(m\) decayed is given by:
\(m = \frac{\Delta N}{N_A} \times 60\text{ g}\)
\(m = \frac{8.562 \times 10^{13}}{6.02 \times 10^{23}} \times 60\text{ g} = 8.537 \times 10^{-9}\text{ g} = 8.54 \times 10^{-12}\text{ kg}\).

(a)
- C1: 'probability of decay of a nucleus' or 'fraction of remaining nuclei that decay'
- A1: 'per unit time'

(b)(i)
- C1: converts half-life to seconds correctly: \(T_{1/2} \approx 1.66 \times 10^8\text{ s}\)
- A1: calculates \(\lambda = \ln 2 / T_{1/2}\) to show \(4.17 \times 10^{-9}\text{ s}^{-1}\)

(b)(ii)
- C1: recalls and rearranges \(A = \lambda N\)
- A1: key substitution and final answer: \(1.08 \times 10^{14}\) (accept \(1.07 \times 10^{14}\) to \(1.08 \times 10^{14}\))

(b)(iii)
- C1: recalls and substitutes into \(A = A_0 e^{-\lambda t}\) or \(A = A_0 (0.5)^{t/T_{1/2}}\)
- A1: final answer in range \(9.28 \times 10^4\text{ Bq}\) to \(9.30 \times 10^4\text{ Bq}\)

(b)(iv)
- C1: calculates number of decayed nuclei \(\Delta N = 8.56 \times 10^{13}\) (or \(8.57 \times 10^{13}\))
- A1: calculates mass correctly in kg: \(8.54 \times 10^{-12}\text{ kg}\) (accept range \(8.50 \times 10^{-12}\) to \(8.60 \times 10^{-12}\text{ kg}\))

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

(ii) As charge \(Q\) builds up on the plates of the capacitor, the electric field strength \(E\) between the plates increases in direct proportion (\(E \propto Q\)). Because the potential difference \(V\) between the plates is given by the product of the uniform electric field strength and the plate separation (\(V = Ed\)), \(V\) must also be directly proportional to \(E\). Consequently, \(V\) is directly proportional to \(Q\).

(b) (i) Use the capacitor discharge formula for potential difference:
\(V = V_0 e^{-t / RC}\)
Divide both sides by \(V_0\):
\(\frac{V}{V_0} = e^{-t / RC}\)
Take the natural logarithm of both sides:
\(\ln\left(\frac{3.3}{9.0}\right) = -\frac{15}{120 \times 10^3 \times C}\)
\(-1.0033 = -\frac{15}{1.20 \times 10^5 \times C}\)
Solve for \(C\):
\(C = \frac{15}{1.0033 \times 1.20 \times 10^5} = 1.246 \times 10^{-4}\text{ F} \approx 125\ \mu\text{F}\).

(ii) The energy stored in a capacitor is given by \(E = \frac{1}{2} C V^2\).
The energy lost during the first 15 seconds is the difference between the initial energy stored and the remaining energy stored:
\(\Delta E = \frac{1}{2} C V_0^2 - \frac{1}{2} C V^2 = \frac{1}{2} C (V_0^2 - V^2)\)
Using \(C = 125\ \mu\text{F}\):
\(\Delta E = 0.5 \times (125 \times 10^{-6}\text{ F}) \times (9.0^2 - 3.3^2)\)
\(\Delta E = 62.5 \times 10^{-6} \times (81 - 10.89) = 62.5 \times 10^{-6} \times 70.11\)
\(\Delta E = 4.38 \times 10^{-3}\text{ J}\) (or \(4.38\text{ mJ}\)).

(iii) The current at any time \(t\) is given by \(I = \frac{V}{R}\).
The initial current \(I_0\) at \(t = 0\) is:
\(I_0 = \frac{V_0}{R} = \frac{9.0\text{ V}}{120 \times 10^3\ \Omega} = 7.5 \times 10^{-5}\text{ A} = 75\ \mu\text{A}\).
The graph of current \(I\) against time \(t\) is an exponential decay curve starting from \(75\ \mu\text{A}\) at \(t = 0\). At \(t = 15\text{ s}\) (one time constant), the current drops to \(I_0 / e \approx 28\ \mu\text{A}\), and at \(t = 30\text{ s}\) (two time constants) it drops to \(I_0 / e^2 \approx 10\ \mu\text{A}\).

(a)(i)
- B1: 'charge (on one plate) per unit potential difference' or \(Q/V\) with symbols defined.

(a)(ii)
- B1: electric field strength \(E\) is directly proportional to charge \(Q\) on plates.
- B1: potential difference \(V\) is directly proportional to \(E\) (since \(V = Ed\)), leading to \(V \propto Q\).

(b)(i)
- C1: recalls \(V = V_0 e^{-t/RC}\)
- C1: substitutes values correctly: \(3.3 = 9.0 e^{-15 / (1.20 \times 10^5 \times C)}\)
- A1: carries out correct logarithmic manipulation to show \(C \approx 1.25 \times 10^{-4}\text{ F}\) or \(125\ \mu\text{F}\)

(b)(ii)
- C1: recalls \(E = \frac{1}{2}CV^2\) and attempts to find change in energy.
- A1: calculates \(4.38 \times 10^{-3}\text{ J}\) (or \(4.4 \times 10^{-3}\text{ J}\)).

(b)(iii)
- B1: exponential decay curve shape starting from a non-zero current on y-axis and asymptoting to the t-axis (not touching or crossing it).
- B1: y-intercept clearly labeled as \(75\) (or \(7.5 \times 10^{-5}\text{ A}\)) with the curve passing through approximately \(28\ \mu\text{A}\) at \(15\text{ s}\) and \(10\ \mu\text{A}\) at \(30\text{ s}\).

The relationship between amplitude and time is given by \(A = A_0 e^{-\lambda t}\). Taking the natural logarithm of both sides yields: \(\ln A = -\lambda t + \ln A_0\). Therefore, a graph of \(\ln A\) against \(t\) is linear with a gradient of \(-\lambda\). To test the relationship \(\lambda = C I^n\), we take logarithms of both sides: \(\lg \lambda = n \lg I + \lg C\). Plotting \(\lg \lambda\) against \(\lg I\) will result in a straight line if the relationship is valid. The gradient of this line is \(n\), and the y-intercept is \(\lg C\), from which \(C = 10^{\text{intercept}}\) can be calculated. This procedure is repeated for at least six different currents, ensuring that the initial displacement of the plate and the position of the electromagnet poles remain constant for each trial.

Defining the variables (2 marks): 1. Independent variable: current \(I\) in the electromagnet. 2. Dependent variable: damping constant \(\lambda\). (Accept control variables: spring constant/mass of plate, starting amplitude, position of electromagnet). Methods of data collection (4 marks): 1. Diagram showing a spring supporting an aluminum plate situated between the poles of an electromagnet. 2. Circuit diagram showing a DC power supply, ammeter, rheostat, and electromagnet connected in series. 3. Use of a motion sensor/data logger, or video camera with a vertical scale to measure amplitude as a function of time. 4. Describe how the plate is displaced vertically and released from rest to begin oscillations. Method of analysis (3 marks): 1. Plot a graph of \(\ln A\) against \(t\) to obtain \(\lambda\) from the gradient (gradient = \(-\lambda\)). 2. Plot a graph of \(\lg \lambda\) (or \(\ln \lambda\)) against \(\lg I\) (or \(\ln I\)). 3. Identify that \(n = \text{gradient}\) and \(\lg C = \text{y-intercept}\) (or \(\ln C = \text{y-intercept}\)). Safety considerations (1 mark): 1. Use of G-clamp to secure retort stand; or safety screen; or turn off electromagnet to prevent overheating. Additional details (5 marks): 1. Use a rheostat / variable power supply to keep current constant during a single decay measurement. 2. Ensure aluminum plate does not touch the pole pieces of the electromagnet during oscillations. 3. Avoid placing ferromagnetic materials near the apparatus. 4. Repeat measurements of decay for each current and average the values of \(\lambda\). 5. Describe how to calibrate the motion sensor or ensure high frames-per-second on the video camera to accurately capture peak amplitudes.

Taking the natural logarithm of both sides of the equation yields: \(\ln(R/\Omega) = \ln(R_0/\Omega) + \frac{\beta}{T}\). Comparing this with \(y = mx + c\), a plot of \(\ln(R/\Omega)\) on the y-axis against \(1/T\) on the x-axis gives a straight line where the gradient \(m = \beta\) and the y-intercept \(c = \ln R_0\).\
\
(a) Calculated values for \(\ln(R/\Omega)\) using \(\ln R\) and absolute uncertainty \(\Delta(\ln R) \approx \Delta R / R\):\
1. \(8.07 \pm 0.03\)\
2. \(7.69 \pm 0.04\)\
3. \(7.33 \pm 0.03\)\
4. \(6.99 \pm 0.04\)\
5. \(6.66 \pm 0.04\)\
6. \(6.35 \pm 0.04\)\
\
(b) Plotted points with vertical error bars of height equal to \(2 \times \Delta(\ln R)\). The best-fit line is drawn through the center of the points. The worst acceptable straight line is drawn passing from the top of the error bar of the first point to the bottom of the error bar of the last point.\
\
(c) Using points on the best-fit line: \((3.411 \times 10^{-3}, 8.07)\) and \((2.914 \times 10^{-3}, 6.35)\):\
\(\text{Gradient } m = \frac{8.07 - 6.35}{(3.411 - 2.914) \times 10^{-3}} = 3.46 \times 10^3\)\
Using the worst acceptable line: \((3.411 \times 10^{-3}, 8.10)\) and \((2.914 \times 10^{-3}, 6.31)\):\
\(\text{Worst gradient } m_{\text{worst}} = \frac{8.10 - 6.31}{(3.411 - 2.914) \times 10^{-3}} = 3.60 \times 10^3\)\
\(\text{Uncertainty in gradient} = |3.46 - 3.60| \times 10^3 = 0.14 \times 10^3\)\
Best-fit y-intercept: \(c = 8.07 - (3.461 \times 10^3 \times 3.411 \times 10^{-3}) = -3.73\)\
Worst y-intercept: \(c_{\text{worst}} = 8.10 - (3.602 \times 10^3 \times 3.411 \times 10^{-3}) = -4.18\)\
\(\text{Uncertainty in y-intercept} = |-3.73 - (-4.18)| = 0.45\)\
\
(d) \(\beta = m = (3.46 \pm 0.14) \times 10^3\text{ K}\)\
\(R_0 = e^{c} = e^{-3.73} = 0.024\ \Omega\)\
\(\text{Maximum } R_0 = e^{-3.73 + 0.45} = 0.038\ \Omega\)\
\(\text{Minimum } R_0 = e^{-3.73 - 0.45} = 0.015\ \Omega\)\
\(\text{Uncertainty in } R_0 = \frac{0.038 - 0.015}{2} \approx 0.011\ \Omega\) (or using \(R_0 \Delta c = 0.024 \times 0.45 = 0.010\ \Omega\)). Thus, \(R_0 = 0.024 \pm 0.010\ \Omega\).

Part (a) [3 marks]:\
- 1 mark for all values of \(\ln R\) calculated correctly to 2 decimal places.\
- 2 marks for all absolute uncertainties in \(\ln R\) calculated correctly (accept values rounded to 1 s.f., e.g., 0.03, 0.04, 0.03, 0.04, 0.04, 0.04). Subtract 1 mark for any arithmetic slip.\
\
Part (b) [4 marks]:\
- 1 mark for axes labeled correctly with quantities and units (y-axis: \(\ln(R/\Omega)\), x-axis: \((1/T) / 10^{-3}\text{ K}^{-1}\)).\
- 1 mark for plotting all 6 points correctly with their corresponding vertical error bars.\
- 1 mark for drawing a single, straight line of best fit.\
- 1 mark for drawing a clearly identified worst acceptable straight line that passes through all error bars.\
\
Part (c) [4 marks]:\
- 1 mark for finding the gradient of the best-fit line using coordinates that span at least half the length of the line.\
- 1 mark for finding the gradient of the worst acceptable line, and calculating the absolute uncertainty as the difference between the two gradients.\
- 1 mark for determining the y-intercept of the best-fit line by substitution of a point on the line into \(y = mx + c\).\
- 1 mark for determining the y-intercept of the worst acceptable line, and calculating the absolute uncertainty in the y-intercept.\
\
Part (d) [4 marks]:\
- 1 mark for equating \(\beta\) to the best-fit gradient and assigning the unit Kelvin (K).\
- 1 mark for equating the absolute uncertainty in \(\beta\) to the uncertainty in the gradient.\
- 1 mark for calculating \(R_0 = e^c\) with the correct unit (\(\Omega\)).\
- 1 mark for calculating the absolute uncertainty in \(R_0\) using the extreme values (max/min) method or \(R_0 \Delta c\).