2024 Cambridge IAL Physics (9702) Practice Paper with Answers

The density \(\rho\) of a cylinder is given by:

\(\rho = \frac{m}{V} = \frac{m}{\pi r^2 L} = \frac{4m}{\pi d^2 L}\)

The fractional uncertainty in the density is:

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

Substituting the percentage uncertainties:

\(\frac{\Delta \rho}{\rho} \times 100\% = 1.5\% + 2(1.0\%) + 2.0\% = 5.5\%\)

1 mark for calculating the correct percentage uncertainty by summing the percentage uncertainties, doubling the uncertainty for the diameter.

By conservation of linear momentum:

\(2m \cdot v + 0 = (2m + 3m) \cdot v_f \implies v_f = \frac{2}{5}v\)

The initial kinetic energy is:

\(E_k = \frac{1}{2}(2m)v^2 = mv^2\)

The final kinetic energy is:

\(E_f = \frac{1}{2}(5m)v_f^2 = \frac{5}{2}m\left(\frac{2}{5}v\right)^2 = \frac{2}{5}mv^2\)

The kinetic energy dissipated (lost) is:

\(\Delta E = E_k - E_f = mv^2 - \frac{2}{5}mv^2 = \frac{3}{5}mv^2\)

The fraction dissipated is:

\(\frac{\Delta E}{E_k} = \frac{3}{5}\)

1 mark for conservation of momentum to find the final velocity, calculating initial and final kinetic energies, and finding the fraction of kinetic energy dissipated.

The current in the circuit is \(I = \frac{E}{R+r}\). The terminal potential difference is \(V = E - Ir = E\left(\frac{R}{R+r}\right)\). As \(R\) increases, \(I\) decreases, so \(Ir\) decreases and \(V\) increases continuously. The power dissipated in \(R\) is \(P = I^2 R = \frac{E^2 R}{(R+r)^2}\). By the maximum power transfer theorem, the power \(P\) reaches a maximum when \(R = r\). Since \(R\) starts much less than \(r\) and ends much greater than \(r\), the power \(P\) increases to a maximum (at \(R=r\)) and then decreases.

1 mark for identifying that terminal potential difference increases as external resistance increases, and power peaks when external resistance equals internal resistance.

A \(\beta^-\) emission increases the proton number \(Z\) by 1 and leaves the nucleon number \(A\) unchanged:

\(^{212}_{83}\text{Bi} \rightarrow \,^{212}_{84}\text{Po} + e^- + \bar{\nu}_e\)

An \(\alpha\) emission decreases the proton number \(Z\) by 2 and decreases the nucleon number \(A\) by 4:

\(^{212}_{84}\text{Po} \rightarrow \,^{208}_{82}\text{Pb} + \,^4_2\alpha\)

Thus, the final nucleus has proton number \(82\) and nucleon number \(208\).

1 mark for applying both conservation of charge (proton number) and conservation of mass (nucleon number) across both decays to get proton number 82 and nucleon number 208.

Let upwards be the positive direction. Using the kinematic equation:

\(s = ut + \frac{1}{2}at^2\)

Here:

\(s = -25\text{ m}\) (since the sea level is below the cliff edge)

\(u = +15\text{ m s}^{-1}\)

\(a = -g = -9.81\text{ m s}^{-2}\)

Substituting these values:

\(-25 = 15t - 4.905 t^2\)

\(4.905 t^2 - 15t - 25 = 0\)

Using the quadratic formula:

\(t = \frac{15 \pm \sqrt{(-15)^2 - 4(4.905)(-25)}}{2(4.905)} = \frac{15 \pm \sqrt{225 + 490.5}}{9.81} = \frac{15 \pm \sqrt{715.5}}{9.81} \approx 4.3\text{ s}\) (taking the positive root).

1 mark for setting up the kinematic equation with correct signs and solving the quadratic equation for the positive time value.

The period \(T\) is given by:

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

The absolute uncertainty in the period is:

\(\Delta T = \frac{\Delta t}{20} = \frac{0.3}{20} = 0.015\text{ s}\)

To express this to an appropriate number of significant figures, the absolute uncertainty is rounded to one significant figure: \(\Delta T = \pm 0.02\text{ s}\). The period must be quoted to the same number of decimal places (2 decimal places) as the absolute uncertainty:

\(T = (1.82 \pm 0.02)\text{ s}\).

1 mark for calculating the period and its uncertainty, and expressing both with correct decimal alignment and significant figures.

The potential divider equation for the voltage across the fixed resistor is:

\(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{fixed}}}{R_{\text{fixed}} + R_{\text{LDR}}}\)

In bright light (\(R_{\text{LDR}} = 1.0\text{ k}\Omega\)):

\(V_{\text{bright}} = 12 \times \frac{4.0}{4.0 + 1.0} = 9.6\text{ V}\)

In darkness (\(R_{\text{LDR}} = 20\text{ k}\Omega\)):

\(V_{\text{dark}} = 12 \times \frac{4.0}{4.0 + 20} = 2.0\text{ V}\)

The change in the reading of the voltmeter from bright light to darkness is:

\(V_{\text{dark}} - V_{\text{bright}} = 2.0\text{ V} - 9.6\text{ V} = -7.6\text{ V}\)

Thus, the voltmeter reading decreases by \(7.6\text{ V}\).

1 mark for calculating the voltmeter reading in both light conditions and finding the correct difference (a decrease of 7.6 V).

The impulse of the force is equal to the change in momentum of the particle, which is also the area under the force-time (\(F\)-\(t\)) graph.

The total impulse \(I\) from \(t = 0\) to \(t = 6.0\text{ s}\) consists of two parts:

1. From \(t = 0\) to \(4.0\text{ s}\) (triangular area):

\(I_1 = \frac{1}{2} \times 4.0\text{ s} \times 6.0\text{ N} = 12.0\text{ N s}\)

2. From \(t = 4.0\text{ s}\) to \(6.0\text{ s}\) (rectangular area):

\(I_2 = (6.0\text{ s} - 4.0\text{ s}) \times 6.0\text{ N} = 12.0\text{ N s}\)

Total impulse \(I = 12.0 + 12.0 = 24.0\text{ N s}\).

Since the particle starts from rest (\(u = 0\)):

\(I = \Delta p = m v \implies 24.0\text{ N s} = 0.50\text{ kg} \times v \implies v = 48\text{ m s}^{-1}\).

1 mark for calculating the total area under the F-t graph as 24.0 N s, and using impulse-momentum theorem to determine the final speed.

The density of a sphere is given by \(\rho = \frac{M}{V} = \frac{6M}{\pi d^3}\). The fractional uncertainty in density is \(\frac{\Delta \rho}{\rho} = \frac{\Delta M}{M} + 3 \frac{\Delta d}{d}\). Calculating the individual fractional uncertainties: \(\frac{\Delta M}{M} = \frac{3}{150} = 0.02 = 2.0\%\) and \(\frac{\Delta d}{d} = \frac{0.05}{2.50} = 0.02 = 2.0\%\). Thus, \(\frac{\Delta \rho}{\rho} = 2.0\% + 3 \times 2.0\% = 8.0\%\).

1 mark for the correct calculation of percentage uncertainty by sum of mass percentage uncertainty and 3 times diameter percentage uncertainty.

Precision is related to the spread of values (random error). The range of measured times is only 0.02 s, showing high precision. Accuracy is related to how close the average is to the true value. The systematic error of 0.05 s shifts all readings from the true value, resulting in low accuracy.

1 mark for correctly identifying high precision from small range and low accuracy due to systematic error.

Using \(s = ut + \frac{1}{2}at^2\) where upward is positive: \(u = +15\text{ m s}^{-1}\), \(a = -9.81\text{ m s}^{-2}\), and \(t = 5.0\text{ s}\). Substituting these values: \(s = (15)(5.0) + 0.5(-9.81)(5.0)^2 = 75 - 122.6 = -47.6\text{ m}\). The negative sign indicates displacement below the launching point, so the height is approximately \(48\text{ m}\).

1 mark for using the equations of motion with correct signs to find the cliff height.

By conservation of momentum, \(3m(2v) + m(-v) = (3m + m)V\), which gives \(5mv = 4mV\) and \(V = \frac{5}{4}v\). Initial kinetic energy \(E_k = \frac{1}{2}(3m)(2v)^2 + \frac{1}{2}m(-v)^2 = 6.5mv^2 = \frac{13}{2}mv^2\). Final kinetic energy \(E_f = \frac{1}{2}(4m)(\frac{5}{4}v)^2 = \frac{25}{8}mv^2\). The loss of kinetic energy is \(E_k - E_f = \frac{52}{8}mv^2 - \frac{25}{8}mv^2 = \frac{27}{8}mv^2\).

1 mark for finding the final velocity and calculating the difference between initial and final kinetic energies.

Brightly lit: \(V_{\text{out}} = 12 \times \frac{2.0}{4.0 + 2.0} = 4.0\text{ V}\). Darkened: \(V_{\text{out}} = 12 \times \frac{8.0}{4.0 + 8.0} = 8.0\text{ V}\). The reading changes from 4.0 V to 8.0 V, which is an increase of 4.0 V.

1 mark for calculating both output voltages and identifying the change as an increase of 4.0 V.

The terminal potential difference is given by \(V = E - Ir\). Rearranging to \(V = -rI + E\) shows a linear relationship between \(V\) and \(I\) with gradient \(-r\) (negative) and y-intercept \(E\) (positive).

1 mark for correctly using terminal potential difference equation to identify the line properties.

A proton consists of \(uud\) and a neutron consists of \(udd\). During beta-plus decay, a proton changes to a neutron (\(uud \rightarrow udd\)), which means an up quark (\(u\)) changes into a down quark (\(d\)).

1 mark for correctly identifying the change in quark structure from proton to neutron.

The thrust force on the rocket is \(F = v \frac{\Delta m}{\Delta t} = 120 \times 0.050 = 6.0\text{ N}\). The weight of the rocket is \(W = mg = 0.20 \times 9.81 = 1.962\text{ N}\). The net force is \(F_{\text{net}} = 6.0 - 1.962 = 4.038\text{ N}\). The acceleration is \(a = \frac{F_{\text{net}}}{m} = \frac{4.038}{0.20} = 20.19\text{ m s}^{-2}\), which rounds to \(20\text{ m s}^{-2}\).

1 mark for calculating thrust, subtracting weight to find net force, and applying Newton's second law.

The formula for the density of a cylinder is:

\(\rho = \frac{m}{V} = \frac{4m}{\pi d^2 L}\)

The fractional uncertainty in \(\rho\) is the sum of the fractional uncertainties of its constituents, accounting for exponents:

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

Calculating the percentage uncertainties of each variable:

- Percentage uncertainty in mass \(m\): \(\frac{0.1}{45.2} \times 100\% \approx 0.22\%\)
- Percentage uncertainty in diameter \(d\): \(\frac{0.02}{1.25} \times 100\% = 1.60\%\)
- Percentage uncertainty in length \(L\): \(\frac{0.05}{6.40} \times 100\% \approx 0.78\%\)

Substituting these values into the uncertainty relation:

\(\%\Delta \rho = 0.22\% + 2(1.60\%) + 0.78\% = 4.20\%\)

This rounds to \(4.2\%\).

1 mark for the correct answer C.
- Method: Recall and apply the rule for combining independent fractional/percentage uncertainties when multiplying, dividing, and raising to powers.
- Accuracy: Verify the calculations for individual percentage uncertainties and sum them up correctly.

First, rearrange Coulomb's law to make \(\epsilon_0\) the subject:

\(\epsilon_0 = \frac{Q_1 Q_2}{4 \pi F r^2}\)

Next, substitute the SI base units for each quantity:

- Charge \(Q\) has base units of \(\text{A s}\).
- Force \(F\) has base units of \(\text{kg m s}^{-2}\).
- Distance \(r\) has base units of \(\text{m}\).

The constant \(4\pi\) is dimensionless.

Now combine the base units:

\([\epsilon_0] = \frac{(\text{A s})^2}{(\text{kg m s}^{-2})(\text{m})^2} = \frac{\text{A}^2 \text{s}^2}{\text{kg m}^3 \text{s}^{-2}} = \text{kg}^{-1} \text{m}^{-3} \text{s}^4 \text{A}^2\)

1 mark for the correct answer A.
- Method: Rearrange the equation for \(\epsilon_0\), substitute correct base units for force and charge, and simplify the indices correctly.

Let's estimate the physical parameters for a person climbing a flight of stairs:

- Mass of person, \(m \approx 70\text{ kg}\)
- Weight of person, \(W = mg \approx 700\text{ N}\)
- Height of a typical flight of stairs, \(h \approx 3\text{ m}\)
- Time taken to climb, \(t \approx 7\text{ s}\)

The useful work done is equal to the increase in gravitational potential energy:

\(E_{\text{p}} = mgh \approx 70 \times 9.81 \times 3 \approx 2100\text{ J}\)

The power output is:

\(P = \frac{E_{\text{p}}}{t} \approx \frac{2100}{7} = 300\text{ W}\)

Therefore, \(300\text{ W}\) is a highly reasonable estimate for active exertion. Values of \(5\text{ W}\) or \(50\text{ W}\) are too low (less than typical metabolic rates), and \(3000\text{ W}\) is far beyond human muscle capability for more than a split second.

1 mark for the correct answer C.
- Method: Estimate mass, height, and time scale to compute power and select the closest order of magnitude.

Using the equations of constant acceleration, we define the upward direction as positive.

- Initial velocity, \(u = +15\text{ m s}^{-1}\)
- Acceleration, \(a = -g = -9.81\text{ m s}^{-2}\)
- Time of flight, \(t = 4.2\text{ s}\)
- Displacement, \(s = -h\)

Substitute these values into the kinematic equation:

\(s = ut + \frac{1}{2}at^2\)

\(-h = (15 \times 4.2) + \frac{1}{2}(-9.81)(4.2)^2\)

\(-h = 63.0 - (4.905 \times 17.64)\)

\(-h = 63.0 - 86.5 = -23.5\text{ m}\)

Therefore, the height \(h\) is approximately \(24\text{ m}\).

1 mark for the correct answer A.
- Method: Apply \(s = ut + \frac{1}{2}at^2\) with the correct signs for velocity and acceleration to find displacement.

Let the glider of mass \(2m\) have initial velocity \(u_1 = +3v\) and the glider of mass \(m\) have initial velocity \(u_2 = -v\).

Let \(v_1\) and \(v_2\) be their final velocities.

By conservation of linear momentum:

\(2m(3v) + m(-v) = 2mv_1 + mv_2\)

\(5mv = 2mv_1 + mv_2 \implies 2v_1 + v_2 = 5v\quad [1]\)

Since the collision is perfectly elastic, the relative speed of approach equals the relative speed of separation:

\(u_1 - u_2 = v_2 - v_1\)

\(3v - (-v) = v_2 - v_1 \implies v_2 - v_1 = 4v\quad [2]\)

Subtract equation [2] from [1]:

\(3v_1 = v \implies v_1 = +\frac{1}{3}v\)

Substitute \(v_1\) back into [2]:

\(v_2 = v_1 + 4v = \frac{1}{3}v + 4v = +\frac{13}{3}v\)

1 mark for the correct answer A.
- Method: Solve simultaneous equations derived from conservation of momentum and the relative speed condition for elastic collisions.

The terminal potential difference (p.d.) is the potential difference across the external resistor:

\(V = 7.2\text{ V}\)

The current \(I\) in the circuit can be calculated using Ohm's law on the external resistor:

\(I = \frac{V}{R} = \frac{7.2}{6.0} = 1.2\text{ A}\)

The relationship between e.m.f. \(E\), terminal p.d. \(V\), and internal resistance \(r\) is:

\(E = V + Ir\)

Substitute the known values:

\(9.0 = 7.2 + 1.2r\)

\(1.2r = 1.8 \implies r = \frac{1.8}{1.2} = 1.5\ \Omega\)

1 mark for the correct answer B.
- Method: Calculate current in the circuit using Ohm's Law and use \(E = V + Ir\) to find the internal resistance.

1. In the dark, \(R_{\text{LDR}} = 8.0\text{ k}\Omega\). The output voltage across the fixed resistor is:

\(V_{\text{out, dark}} = E \times \frac{R}{R + R_{\text{LDR}}} = 12.0 \times \frac{4.0}{4.0 + 8.0} = 12.0 \times \frac{4.0}{12.0} = 4.0\text{ V}\)

2. In bright light, \(R_{\text{LDR}} = 1.0\text{ k}\Omega\). The output voltage is:

\(V_{\text{out, bright}} = E \times \frac{R}{R + R_{\text{LDR}}} = 12.0 \times \frac{4.0}{4.0 + 1.0} = 12.0 \times \frac{4.0}{5.0} = 9.6\text{ V}\)

3. The increase in \(V_{\text{out}}\) is:

\(\Delta V_{\text{out}} = V_{\text{out, bright}} - V_{\text{out, dark}} = 9.6\text{ V} - 4.0\text{ V} = 5.6\text{ V}\)

1 mark for the correct answer B.
- Method: Calculate the potential divider output voltage under both conditions and subtract to find the difference.

During \(\beta^+\) decay, a proton in the nucleus turns into a neutron:

\(\text{p} \rightarrow \text{n} + \text{e}^+ + \nu_{\text{e}}\)

The quark composition of a proton is \(\text{uud}\) and that of a neutron is \(\text{udd}\). Thus, the fundamental change is that one up (\(\text{u}\)) quark changes into one down (\(\text{d}\)) quark:

\(\text{u} \rightarrow \text{d}\)

Consequently:
- The total number of up quarks in the nucleus decreases by 1.
- The total number of down quarks in the nucleus increases by 1.

1 mark for the correct answer A.
- Method: Link the nuclear transmutation (proton to neutron) to its constituent quark change (up to down).

The strain is given by the formula \( \varepsilon = \frac{e}{L} \). Note that the radius of the wire is not needed to calculate strain, which is a common distractor. For quantities related by multiplication or division, the percentage uncertainties are added together: \( \frac{\Delta \varepsilon}{\varepsilon} = \frac{\Delta e}{e} + \frac{\Delta L}{L} \). Calculating each component: \( \frac{\Delta e}{e} = \frac{0.1}{3.2} \times 100\% = 3.125\% \) and \( \frac{\Delta L}{L} = \frac{0.01}{1.50} \times 100\% \approx 0.667\% \). Adding these gives: \( 3.125\% + 0.667\% = 3.792\% \approx 3.8\% \).

1 mark for the correct calculation of individual percentage uncertainties, their summation, and rounding to two significant figures.

The SI base units are metre (m), kilogram (kg), second (s), ampere (A), kelvin (K), mole (mol), and candela (cd). Option A contains only SI base units: ampere, candela, kelvin, kilogram, second. Options B, C, and D contain derived units: coulomb, joule, newton, volt, and watt are all derived units.

1 mark for identifying the list consisting entirely of SI base units.

First, calculate the output voltage in the dark: \( V_{\text{out, dark}} = V_{\text{in}} \times \frac{R_{\text{fixed}}}{R_{\text{LDR}} + R_{\text{fixed}}} = 12 \times \frac{4.0}{16 + 4.0} = 12 \times 0.20 = 2.4 \text{ V} \). Next, calculate the output voltage in the light: \( V_{\text{out, light}} = 12 \times \frac{4.0}{1.0 + 4.0} = 12 \times 0.80 = 9.6 \text{ V} \). The change in output voltage is \( V_{\text{out, light}} - V_{\text{out, dark}} = 9.6 - 2.4 = 7.2 \text{ V} \).

1 mark for calculating the two output voltages and determining the difference.

Let the direction to the right be positive. Before collision, we have: \( m_1 = 2.0 \text{ kg} \), \( u_1 = +4.0 \text{ m s}^{-1} \), \( m_2 = 3.0 \text{ kg} \), \( u_2 = -2.0 \text{ m s}^{-1} \). For a perfectly elastic collision, the relative velocity of approach equals the relative velocity of separation: \( u_1 - u_2 = v_2 - v_1 \implies 4.0 - (-2.0) = v_2 - v_1 \implies v_2 = v_1 + 6.0 \). By conservation of linear momentum: \( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \implies 2.0(4.0) + 3.0(-2.0) = 2.0 v_1 + 3.0 v_2 \implies 2.0 = 2.0 v_1 + 3.0 v_2 \). Substituting \( v_2 \): \( 2.0 = 2.0 v_1 + 3.0(v_1 + 6.0) \implies 2.0 = 5.0 v_1 + 18.0 \implies 5.0 v_1 = -16.0 \implies v_1 = -3.2 \text{ m s}^{-1} \) (to the left). Then \( v_2 = -3.2 + 6.0 = +2.8 \text{ m s}^{-1} \) (to the right).

1 mark for using the relative speed equation and conservation of momentum to find both post-collision velocities.

Using the equations of motion with upwards as positive: \( s = ut - \frac{1}{2}gt^2 \). For \( s = 20 \text{ m} \): \( 20 = 25t - 4.905t^2 \implies 4.905t^2 - 25t + 20 = 0 \). Solving the quadratic equation for \( t \) yields: \( t = \frac{25 \pm \sqrt{(-25)^2 - 4(4.905)(20)}}{2(4.905)} = \frac{25 \pm \sqrt{625 - 392.4}}{9.81} = \frac{25 \pm \sqrt{232.6}}{9.81} \). This gives two times: \( t_1 \approx 0.99 \text{ s} \) (going up) and \( t_2 \approx 4.10 \text{ s} \) (coming down). The time spent at or above \( 20 \text{ m} \) is \( \Delta t = t_2 - t_1 = \frac{2 \times \sqrt{232.6}}{9.81} \approx 3.11 \text{ s} \approx 3.1 \text{ s} \).

1 mark for setting up the quadratic equation for displacement and finding the correct difference between the two times.

A baryon consists of three quarks. A strangeness of \( -1 \) implies the presence of exactly one strange (\( s \)) quark, which has charge \( -\frac{1}{3}e \). Let the other two quarks be \( q_1 \) and \( q_2 \). The total charge is \( q_1 + q_2 + q_s = +1e \implies q_1 + q_2 - \frac{1}{3}e = +1e \implies q_1 + q_2 = +\frac{4}{3}e \). Since up (\( u \)) quarks have charge \( +\frac{2}{3}e \) and down (\( d \)) quarks have charge \( -\frac{1}{3}e \), we must have two up quarks. This yields the quark combination \( uus \).

1 mark for evaluating the charge and strangeness of quark combinations to find the one matching the given baryon.

The volume of water ejected per second is \( \frac{V}{t} = A \times v = (2.0 \times 10^{-3} \text{ m}^2) \times (8.0 \text{ m s}^{-1}) = 1.6 \times 10^{-2} \text{ m}^3 \text{ s}^{-1} \). The mass of water lifted and ejected per second is \( \frac{m}{t} = \rho \times \frac{V}{t} = 1000 \text{ kg m}^{-3} \times 1.6 \times 10^{-2} \text{ m}^3 \text{ s}^{-1} = 16 \text{ kg s}^{-1} \). The minimum power output must supply both potential energy and kinetic energy to the water per second: \( P = P_{\text{pot}} + P_{\text{kin}} = \left(\frac{m}{t}\right) g h + \frac{1}{2} \left(\frac{m}{t}\right) v^2 = 16(9.81)(15) + \frac{1}{2}(16)(8.0)^2 = 2354.4 + 512 = 2866.4 \text{ W} \approx 2.9 \text{ kW} \).

1 mark for calculating the mass flow rate and combining both potential and kinetic energy terms per unit time to get the total power.

The wavelength \( \lambda \) is given by \( \lambda = \frac{v}{f} = \frac{340}{500} = 0.68 \text{ m} = 68 \text{ cm} \). The phase difference \( \phi \) in radians is related to the path difference \( x \) by the formula \( \phi = \frac{x}{\lambda} \times 2\pi \). Here, \( x = 51 \text{ cm} \), so \( \phi = \frac{51}{68} \times 2\pi = 0.75 \times 2\pi = 1.5\pi \text{ rad} \).

1 mark for calculating the wavelength and using it to correctly determine the phase difference in radians.

The formula for resistivity is \(\rho = \frac{\pi R d^2}{4 L}\). The percentage uncertainty in \(\rho\) is calculated by adding the individual percentage uncertainties, with the percentage uncertainty of the diameter multiplied by 2 because it is raised to the power of 2: \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L} = 2.0\% + 2 \times 1.2\% + 1.5\% = 2.0\% + 2.4\% + 1.5\% = 5.9\%\).

1 mark for the correct calculation of the combined percentage uncertainty. Award 1 mark for option B.

A zero error is a systematic error. It shifts all readings consistently in one direction by \(0.3\text{ V}\). This reduces the accuracy because the measured values are systematically further from the true values. However, it does not affect the precision, which is a measure of the spread or repeatability of the readings.

1 mark for identifying that systematic zero error decreases accuracy but leaves precision unaffected. Award 1 mark for option B.

The current in the circuit is given by \(I = \frac{E}{R + r}\). When \(R\) is increased, the total resistance of the circuit increases, so the current \(I\) decreases. The power dissipated in the internal resistance is \(P = I^2 r\); since \(I\) decreases, the power dissipated in the internal resistance decreases. The voltmeter measures the terminal potential difference, \(V = E - I r\). Since the current \(I\) decreases, the lost volts \(I r\) decrease, and thus the voltmeter reading \(V\) increases.

1 mark for analyzing that current decreases, leading to decreased power in the internal resistance and increased terminal potential difference. Award 1 mark for option B.

In bright light, the output voltage is \(V_{\text{out, bright}} = \frac{4.0\text{ k}\Omega}{4.0\text{ k}\Omega + 1.0\text{ k}\Omega} \times 12.0\text{ V} = \frac{4}{5} \times 12.0\text{ V} = 9.6\text{ V}\). In darkness, the output voltage is \(V_{\text{out, dark}} = \frac{4.0\text{ k}\Omega}{4.0\text{ k}\Omega + 20\text{ k}\Omega} \times 12.0\text{ V} = \frac{4}{24} \times 12.0\text{ V} = 2.0\text{ V}\). The decrease in output voltage is \(9.6\text{ V} - 2.0\text{ V} = 7.6\text{ V}\).

1 mark for calculating both output voltages and finding the difference. Award 1 mark for option B.

During the decay, the nucleon number remains constant at 212, and the proton number increases by 1 (from 83 to 84). This is a standard beta-minus (\(\beta^-\)) decay. In \(\beta^-\)-decay, a neutron decays into a proton, an electron (beta-minus particle), and an electron antineutrino. Therefore, particle \(X\) is a \(\beta^-\) particle and particle \(Y\) is an electron antineutrino.

1 mark for identifying that the change in proton and nucleon numbers indicates beta-minus decay, which emits an electron and an electron antineutrino. Award 1 mark for option C.

First, find the common velocity \(v\) using conservation of momentum: \(m_1 u_1 = (m_1 + m_2) v \Rightarrow 2.0 \times 6.0 = (2.0 + 4.0) v \Rightarrow 12 = 6.0 v \Rightarrow v = 2.0\text{ m s}^{-1}\). The initial kinetic energy is \(E_{\text{ki}} = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} \times 2.0 \times 6.0^2 = 36\text{ J}\). The final kinetic energy is \(E_{\text{kf}} = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} \times 6.0 \times 2.0^2 = 12\text{ J}\). The loss in kinetic energy is \(36\text{ J} - 12\text{ J} = 24\text{ J}\).

1 mark for calculating the common velocity and the difference between initial and final kinetic energies. Award 1 mark for option B.

According to Newton's second law, the force exerted is the rate of change of momentum: \(F = \frac{\Delta p}{\Delta t}\). Since the final horizontal velocity of the water is zero, the change in momentum of a mass \(m\) of water is \(\Delta p = m v\). Thus, the force is \(F = v \frac{\Delta m}{\Delta t}\), where \(\frac{\Delta m}{\Delta t} = 1.5\text{ kg s}^{-1}\) and \(v = 12\text{ m s}^{-1}\). Calculating this gives \(F = 12 \times 1.5 = 18\text{ N}\).

1 mark for using the rate of change of momentum formula and finding the force to be 18 N. Award 1 mark for option C.

Using the kinematic equation for displacement: \(s = u t + \frac{1}{2} a t^2\). Taking the upwards direction as positive: displacement \(s = -25\text{ m}\), initial velocity \(u = +15\text{ m s}^{-1}\), and acceleration \(a = -9.81\text{ m s}^{-2}\). This gives: \(-25 = 15 t - 4.905 t^2 \Rightarrow 4.905 t^2 - 15 t - 25 = 0\). Solving this quadratic equation for \(t\): \(t = \frac{15 \pm \sqrt{(-15)^2 - 4(4.905)(-25)}}{2(4.905)} = \frac{15 \pm \sqrt{225 + 490.5}}{9.81} = \frac{15 \pm 26.75}{9.81}\). Choosing the positive time solution: \(t = \frac{41.75}{9.81} \approx 4.26\text{ s}\), which rounds to \(4.3\text{ s}\).

1 mark for correctly applying the equation of motion with signs, solving the quadratic equation, and choosing the positive root. Award 1 mark for option C.

(a) Systematic error causes readings to be consistently larger or smaller than the true value (constant bias), whereas random error causes readings to be scattered around a mean value with varying magnitude and sign.

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

(c) (i) \(d = 0.50 \times 10^{-3}\text{ m}\)
\(A = \frac{\pi (0.50 \times 10^{-3})^2}{4} = 1.9635 \times 10^{-7}\text{ m}^2\)
\(\rho = \frac{5.00 \times 1.9635 \times 10^{-7}}{1.000} = 9.8175 \times 10^{-7}\ \Omega\text{ m} \approx 9.8 \times 10^{-7}\ \Omega\text{ m}\).

(ii) \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta l}{l}\)
\(\frac{\Delta R}{R} = \frac{0.15}{5.00} = 0.03 = 3\%\)
\(\frac{\Delta d}{d} = \frac{0.01}{0.50} = 0.02 = 2\%\)
\(\frac{\Delta l}{l} = \frac{0.002}{1.000} = 0.002 = 0.2\%\)
Percentage uncertainty = \(3\% + 2(2\%) + 0.2\% = 7.2\%\).

(d) Absolute uncertainty \(\Delta \rho = 9.8175 \times 10^{-7} \times 0.072 = 7.07 \times 10^{-8}\ \Omega\text{ m} \approx 0.7 \times 10^{-7}\ \Omega\text{ m}\).
Since the absolute uncertainty is expressed to 1 significant figure, \(\rho\) is written to the same decimal place:
\(\rho = (9.8 \pm 0.7) \times 10^{-7}\ \Omega\text{ m}\).

(a)
- M1: Systematic error: constant bias / readings consistently shifted in one direction. [1]
- A1: Random error: scatter of readings about a mean / variation in both directions. [1]
(b)
- B1: \(\rho = \frac{\pi R d^2}{4 l}\) or equivalent expression. [1]
(c)(i)
- C1: Calculation of cross-sectional area \(A = 1.96 \times 10^{-7}\text{ m}^2\). [1]
- A1: Correct calculation of \(\rho = 9.8 \times 10^{-7}\ \Omega\text{ m}\) (accept \(9.82 \times 10^{-7}\)). [1]
(c)(ii)
- C1: Summing fractional/percentage uncertainties: \(\frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta l}{l}\). [1]
- C1: Individual percentage values calculated: \(3\%\), \(2\%\), \(0.2\%\). [1]
- A1: Total percentage uncertainty = \(7.2\%\) (accept \(7\%\)). [1]
(d)
- C1: Absolute uncertainty calculated as \(0.7 \times 10^{-7}\ \Omega\text{ m}\). [1]
- A1: Correct final expression: \((9.8 \pm 0.7) \times 10^{-7}\ \Omega\text{ m}\). [1]

(a) Rearranging Stokes' law: \(\eta = \frac{F}{6\pi r v}\).
Units of \(F\) are \(\text{N} = \text{kg}\text{ m}\text{ s}^{-2}\).
Units of \(r\) and \(v\) are \(\text{m}\) and \(\text{m}\text{ s}^{-1}\) respectively.
Thus, units of \(\eta\) are: \(\frac{\text{kg}\text{ m}\text{ s}^{-2}}{\text{m} \times \text{m}\text{ s}^{-1}} = \frac{\text{kg}\text{ m}\text{ s}^{-2}}{\text{m}^2\text{ s}^{-1}} = \text{kg}\text{ m}^{-1}\text{ s}^{-1}\).

(b) (i) LHS unit of velocity \(v\): \(\text{m}\text{ s}^{-1}\).
RHS units: \(\frac{\text{m}^2 \times \text{kg}\text{ m}^{-3} \times \text{m}\text{ s}^{-2}}{\text{kg}\text{ m}^{-1}\text{ s}^{-1}}\).
Simplifying the numerator: \(\text{kg}\text{ s}^{-2}\).
Simplifying the whole fraction: \(\frac{\text{kg}\text{ s}^{-2}}{\text{kg}\text{ m}^{-1}\text{ s}^{-1}} = \text{m}\text{ s}^{-1}\).
Since LHS units equal RHS units, the equation is dimensionally homogeneous.

(ii) Convert \(r\) to meters: \(r = 1.5 \times 10^{-3}\text{ m}\).
Density difference: \(\rho_s - \rho_f = 7800 - 920 = 6880\text{ kg}\text{ m}^{-3}\).
Substitute values into the formula: \(v = \frac{2 \times (1.5 \times 10^{-3})^2 \times 6880 \times 9.81}{9.0 \times 0.29}\).
\(v = \frac{0.30372}{2.61} = 0.11637\text{ m}\text{ s}^{-1} \approx 0.12\text{ m}\text{ s}^{-1}\).

(a)
- C1: Identifies force \(F\) in SI base units as \(\text{kg}\text{ m}\text{ s}^{-2}\). [1]
- C1: Rearranges the formula for \(\eta\) and substitutes units for radius and velocity. [1]
- A1: Simplifies units to correctly show \(\text{kg}\text{ m}^{-1}\text{ s}^{-1}\). [1]
(b)(i)
- B1: States LHS unit is \(\text{m}\text{ s}^{-1}\). [1]
- C1: Correct substitution of base units into the RHS numerator: \(\text{m}^2 \times \text{kg}\text{ m}^{-3} \times \text{m}\text{ s}^{-2} = \text{kg}\text{ s}^{-2}\). [1]
- A1: Shows RHS simplifies to \(\text{m}\text{ s}^{-1}\), matching LHS. [1]
(b)(ii)
- C1: Calculates \((\rho_s - \rho_f) = 6880\text{ kg}\text{ m}^{-3}\) and converts \(r\) to \(1.5 \times 10^{-3}\text{ m}\). [1]
- C1: Substitutes values into formula correctly with \(g = 9.81\text{ m}\text{ s}^{-2}\). [1]
- A1: Correct numerical calculation yielding \(0.12\text{ m}\text{ s}^{-1}\) (or \(0.116\text{ m}\text{ s}^{-1}\)). [1]
- A1: Correct unit of velocity (\(\text{m}\text{ s}^{-1}\)) provided with final answer. [1]

(a) (i) At maximum height, \(v = 0\).
Using \(v^2 = u^2 + 2as\):
\(0 = 15.0^2 - 2(9.81)s\)
\(19.62s = 225 \Rightarrow s = 11.47\text{ m} \approx 11.5\text{ m}\).

(ii) Using \(v = u + at\):
\(0 = 15.0 - 9.81 t \Rightarrow t = 1.53\text{ s}\).

(b) (i) Taking the point of projection as origin and upwards as positive, the displacement \(s\) at the sea is \(-35.0\text{ m}\).
Using \(v^2 = u^2 + 2as\):
\(v^2 = 15.0^2 + 2(-9.81)(-35.0) = 225 + 686.7 = 911.7\text{ m}^2\text{ s}^{-2}\)
\(v = -30.19\text{ m}\text{ s}^{-1}\).
Speed is the magnitude: \(30.2\text{ m}\text{ s}^{-1}\).

(ii) Using \(v = u + at\):
\(-30.19 = 15.0 - 9.81 t\)
\(-45.19 = -9.81 t \Rightarrow t = 4.61\text{ s}\).

(a)(i)
- C1: Uses \(v^2 = u^2 + 2as\) with \(v = 0\) and \(a = 9.81\text{ m}\text{ s}^{-2}\). [1]
- A1: Calculates max height = \(11.5\text{ m}\). [1]
(a)(ii)
- C1: Uses \(v = u + at\) with \(v = 0\) and \(a = 9.81\text{ m}\text{ s}^{-2}\). [1]
- A1: Calculates time = \(1.53\text{ s}\). [1]
(b)(i)
- C1: Identifies total displacement \(s = -35.0\text{ m}\) (or equivalent multi-step method). [1]
- C1: Correct substitution: \(v^2 = 15.0^2 + 2(-9.81)(-35.0)\). [1]
- A1: Speed = \(30.2\text{ m}\text{ s}^{-1}\) (accept \(30\text{ m}\text{ s}^{-1}\) to 2 s.f.). [1]
(b)(ii)
- C1: Recalls or derives appropriate kinematic equation (e.g., \(v = u + at\) or quadratic \(s = ut + \frac{1}{2}at^2\)). [1]
- C1: Consistent substitutions of signs: \(-30.19 = 15.0 - 9.81 t\). [1]
- A1: Total time = \(4.61\text{ s}\) (accept \(4.6\text{ s}\)). [1]

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

(b) Let the direction to the right be positive.
Initial momentum: \(p_i = m_A u_A + m_B u_B\)
\(p_i = (0.40 \times 2.5) + (0.25 \times -1.8) = 1.0 - 0.45 = 0.55\text{ kg}\text{ m}\text{ s}^{-1}\).
After collision, glider A rebounds to the left, so \(v_A = -0.50\text{ m}\text{ s}^{-1}\).
Final momentum: \(p_f = m_A v_A + m_B v_B\)
\(p_f = (0.40 \times -0.50) + 0.25 v_B = -0.20 + 0.25 v_B\).
Using conservation of momentum: \(0.55 = -0.20 + 0.25 v_B\)
\(0.25 v_B = 0.75 \Rightarrow v_B = +3.0\text{ m}\text{ s}^{-1}\).
Since the result is positive, glider B moves to the right at \(3.0\text{ m}\text{ s}^{-1}\).

(c) Initial Kinetic Energy (KE):
\(E_{ki} = \frac{1}{2} m_A u_A^2 + \frac{1}{2} m_B u_B^2\)
\(E_{ki} = \frac{1}{2} (0.40)(2.5)^2 + \frac{1}{2} (0.25)(-1.8)^2 = 1.25 + 0.405 = 1.655\text{ J}\).
Final Kinetic Energy (KE):
\(E_{kf} = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2\)
\(E_{kf} = \frac{1}{2} (0.40)(-0.50)^2 + \frac{1}{2} (0.25)(3.0)^2 = 0.05 + 1.125 = 1.175\text{ J}\).
Since the final kinetic energy (\(1.175\text{ J}\)) is less than the initial kinetic energy (\(1.655\text{ J}\)), kinetic energy is not conserved. Hence, the collision is inelastic.

(a)
- B1: Total momentum of a system of interacting bodies is constant. [1]
- B1: Provided no external forces act (isolated/closed system). [1]
(b)
- C1: Uses conservation of momentum equation: \(m_A u_A + m_B u_B = m_A v_A + m_B v_B\). [1]
- C1: Correct signs substituted: \(u_A = +2.5, u_B = -1.8, v_A = -0.50\). [1]
- A1: Calculates value \(v_B = 3.0\text{ m}\text{ s}^{-1}\). [1]
- A1: States correct direction: to the right. [1]
(c)
- C1: Correct calculation of initial kinetic energy = \(1.66\text{ J}\) (or \(1.655\text{ J}\)). [1]
- C1: Correct calculation of final kinetic energy = \(1.18\text{ J}\) (or \(1.175\text{ J}\)). [1]
- A1: States that kinetic energy is not conserved (or \(E_{kf} < E_{ki}\)). [1]
- A1: Concludes that the collision is inelastic. [1]

(a) Electromotive force (e.m.f.) is the energy converted from other forms (e.g., chemical) to electrical energy per unit charge.

(b) (i) Total external resistance \(R_{ext} = 120 + 45 = 165\ \Omega\).
(ii) Total resistance of circuit: \(R_{tot} = 165 + 1.5 = 166.5\ \Omega\).
Current in circuit: \(I = \frac{E}{R_{tot}} = \frac{9.0}{166.5} = 0.05405\text{ A}\).
Potential difference across LDR: \(V_{LDR} = I \times R_{LDR} = 0.05405 \times 45 = 2.43\text{ V}\) (or \(2.4\text{ V}\)).

(c) (i) As the resistance of the LDR increases, the total resistance of the circuit increases. This causes the circuit current \(I\) to decrease. Since terminal potential difference is \(V = E - Ir\), a smaller current leads to a smaller "lost volts" (\(Ir\)) across the internal resistance, and therefore the terminal potential difference \(V\) increases.
(ii) Total resistance at night: \(R_{tot} = 120 + 1500 + 1.5 = 1621.5\ \Omega\).
Current: \(I = \frac{9.0}{1621.5} = 0.00555\text{ A}\).
P.d. across LDR: \(V_{LDR} = I \times R_{LDR} = 0.00555 \times 1500 = 8.33\text{ V}\) (or \(8.3\text{ V}\)).

(a)
- B1: Energy converted from other forms to electrical energy. [1]
- B1: Per unit charge. [1]
(b)(i)
- B1: Correct total external resistance = \(165\ \Omega\). [1]
(b)(ii)
- C1: Calculates total circuit resistance including internal resistance = \(166.5\ \Omega\). [1]
- C1: Calculates current: \(I = \frac{9.0}{166.5} = 0.054\text{ A}\). [1]
- A1: Calculates potential difference across LDR = \(2.43\text{ V}\) (accept \(2.4\text{ V}\)). [1]
(c)(i)
- B1: Explains that higher LDR resistance reduces circuit current. [1]
- B1: Explains that reduced current reduces 'lost volts' (\(Ir\)) across internal resistance, so terminal potential difference (\(E - Ir\)) increases. [1]
(c)(ii)
- C1: Calculates new current: \(I = 5.55 \times 10^{-3}\text{ A}\). [1]
- A1: Calculates new p.d. across LDR = \(8.3\text{ V}\) (accept \(8.33\text{ V}\)). [1]

(a) The beta-minus particle is an electron \(_{-1}^{0}\text{e}\) (or \(_{-1}^{0}\beta\)). It is accompanied by an electron antineutrino \(\bar{\nu}_e\) (or simply \(\bar{\nu}\)):
\[_{6}^{14}\text{C} \rightarrow _{7}^{14}\text{N} + _{-1}^{0}\text{e} + \bar{\nu}_e\]

(b) (i) During \(\beta^-\) decay, a neutron decays into a proton. A neutron has a quark composition of \(udd\) and a proton has \(uud\). Therefore, a down quark changes to an up quark (\(d \rightarrow u\)).
(ii) The exchange particle is the \(W^-\) boson.
(iii) The electron and antineutrino belong to the class of fundamental particles called leptons.

(c) (i) Proton quark composition is up-up-down (\(uud\)) and neutron quark composition is up-down-down (\(udd\)).
(ii) The quark flavour is the down quark (\(d\)).
Reasoning: A down quark has a charge of \(-\frac{1}{3}e\). Three down quarks have a total charge of \(3 \times \left(-\frac{1}{3}e\right) = -1e\), which matches the charge of the baryon. (An up quark has a charge of \(+\frac{2}{3}e\), so three up quarks would give \(+2e\)).

(a)
- B1: Correct symbols and proton/nucleon numbers for Carbon and Nitrogen: \(_{6}^{14}\text{C}\) and \(_{7}^{14}\text{N}\). [1]
- B1: Correct symbol for the beta-minus particle: \(_{-1}^{0}\text{e}\) (or \(_{-1}^{0}\beta\)). [1]
- B1: Correct symbol for the electron antineutrino: \(\bar{\nu}_e\) or \(\bar{\nu}\) with nucleon number 0 and proton number 0. [1]
(b)(i)
- B1: Identifies that a neutron turns into a proton. [1]
- B1: States that a down quark changes to an up quark (\(d \rightarrow u\)). [1]
(b)(ii)
- B1: \(W^-\) boson (accept \(W\) boson). [1]
(b)(iii)
- B1: Leptons. [1]
(c)(i)
- B1: Proton is \(uud\) and neutron is \(udd\) (both required). [1]
(c)(ii)
- B1: Down quark (\(d\)). [1]
- B1: Explains that \(3 \times \left(-\frac{1}{3}e\right) = -e\). [1]

(a) Gravitational potential at a point is defined as the work done per unit mass in bringing a small test mass from infinity to that point.

(b)(i) To find the escape velocity, we equate the kinetic energy to the gravitational potential energy needed to escape to infinity:
\( \frac{1}{2} m v_e^2 = \frac{G M m}{R} \)
\( v_e = \sqrt{\frac{2 G M}{R}} \)
\( v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 4.80 \times 10^{24}}{6.00 \times 10^6}} = \sqrt{1.0672 \times 10^8} \approx 1.033 \times 10^4 \text{ m s}^{-1} \approx 10.3 \text{ km s}^{-1} \).

(ii) Using conservation of energy:
\( \frac{1}{2} m v^2 - \frac{G M m}{R} = -\frac{G M m}{r} \)
\( \frac{1}{2} v^2 - \frac{G M}{R} = -\frac{G M}{r} \)
\( \frac{1}{2} (5.00 \times 10^3)^2 - \frac{6.67 \times 10^{-11} \times 4.80 \times 10^{24}}{6.00 \times 10^6} = -\frac{6.67 \times 10^{-11} \times 4.80 \times 10^{24}}{r} \)
\( 1.25 \times 10^7 - 5.336 \times 10^7 = -\frac{3.2016 \times 10^{14}}{r} \)
\( -4.086 \times 10^7 = -\frac{3.2016 \times 10^{14}}{r} \)
\( r = \frac{3.2016 \times 10^{14}}{4.086 \times 10^7} \approx 7.84 \times 10^6 \text{ m} \).

(iii) Gravitational potential energy is given by:
\( E_p = -\frac{G M m}{r} \)
\( E_p = -\frac{6.67 \times 10^{-11} \times 4.80 \times 10^{24} \times 150}{7.836 \times 10^6} \approx -6.13 \times 10^9 \text{ J} \).

(a)
- Work done per unit mass [1 mark]
- in bringing a small test mass from infinity to the point [1 mark]

(b)(i)
- Formula \( v_e = \sqrt{2GM/R} \) used or derived [1 mark]
- Correct substitution shown to yield \( 1.03 \times 10^4 \text{ m s}^{-1} \) or \( 10.3 \text{ km s}^{-1} \) [1 mark]

(b)(ii)
- Conservation of energy equation stated: \( \frac{1}{2}mv^2 - \frac{GMm}{R} = -\frac{GMm}{r} \) [1 mark]
- Calculation of initial potential energy per unit mass \( -5.34 \times 10^7 \text{ J kg}^{-1} \) [1 mark]
- Equation simplified to find \( \frac{GM}{r} = 4.09 \times 10^7 \text{ J kg}^{-1} \) [1 mark]
- Correct calculation of \( r = 7.84 \times 10^6 \text{ m} \) [1 mark]

(b)(iii)
- Use of \( E_p = -\frac{GMm}{r} \) with calculated \( r \) [1 mark]
- Correct answer: \( -6.13 \times 10^9 \text{ J} \) (negative sign required) [1 mark]

(a) Any two assumptions: 1. The volume of the molecules is negligible compared with the volume of the container. 2. There are no intermolecular forces between molecules except during collisions. 3. All collisions are perfectly elastic. 4. The time of collisions is negligible compared to the time between collisions.

(b)(i) Using the ideal gas equation:
\( p V = n R T \)
\( T = 27 + 273.15 = 300.15 \text{ K} \) (or \( 300 \text{ K} \))
\( p = \frac{1.80 \times 8.31 \times 300}{0.040} = 1.12185 \times 10^5 \text{ Pa} \approx 1.12 \times 10^5 \text{ Pa} \).

(ii) Total kinetic energy of an ideal gas is given by:
\( E_k = \frac{3}{2} n R T \)
\( E_k = 1.5 \times 1.80 \times 8.31 \times 300 = 6731.1 \text{ J} \approx 6.73 \times 10^3 \text{ J} \).

(iii) The total mass of helium is:
\( m = n \times M_{mol} = 1.80 \times 4.00 \times 10^{-3} \text{ kg} = 7.20 \times 10^{-3} \text{ kg} \).
Using \( E_k = \frac{1}{2} m v_{rms}^2 \):
\( v_{rms} = \sqrt{\frac{2 E_k}{m}} = \sqrt{\frac{2 \times 6731.1}{7.20 \times 10^{-3}}} = \sqrt{1.86975 \times 10^6} \approx 1.37 \times 10^3 \text{ m s}^{-1} \).
Alternatively, using \( v_{rms} = \sqrt{\frac{3 R T}{M_{mol}}} = \sqrt{\frac{3 \times 8.31 \times 300}{4.00 \times 10^{-3}}} = 1.37 \times 10^3 \text{ m s}^{-1} \).

(a)
- Any correct assumption (e.g., volume of molecules is negligible, perfectly elastic collisions, no intermolecular forces except during collisions, continuous random motion) [1 mark]
- A second correct assumption from the list [1 mark]

(b)(i)
- Conversion of temperature to Kelvin: \( 300 \text{ K} \) [1 mark]
- Correct calculation of pressure: \( 1.12 \times 10^5 \text{ Pa} \) (allow \( 1.1 \times 10^5 \text{ Pa} \)) [1 mark]

(b)(ii)
- Use of formula \( E_k = \frac{3}{2}nRT \) or \( E_k = \frac{3}{2}NkT \) [1 mark]
- Correct calculation: \( 6.73 \times 10^3 \text{ J} \) (accept \( 6.7 \times 10^3 \text{ J} \)) [1 mark]

(b)(iii)
- Correct calculation of the total mass of gas: \( 7.20 \times 10^{-3} \text{ kg} \) (or mass of single atom \( 6.64 \times 10^{-27} \text{ kg} \)) [1 mark]
- Use of formula \( v_{rms} = \sqrt{\frac{3RT}{M_{mol}}} \) or \( \frac{1}{2}mv_{rms}^2 = \text{energy} \) [1 mark]
- Correct substitution of values [1 mark]
- Correct calculation: \( 1.37 \times 10^3 \text{ m s}^{-1} \) (accept \( 1.4 \times 10^3 \text{ m s}^{-1} \)) [1 mark]

(a) Simple harmonic motion is defined as motion where the acceleration of the object is directly proportional to its displacement from its equilibrium position and is always directed towards that equilibrium position (i.e., in the opposite direction to the displacement, \( a = -\omega^2 x \)).

(b)(i) Angular frequency is calculated by:
\( \omega = 2 \pi f = 2 \times \pi \times 2.50 = 5.00 \pi \approx 15.71 \text{ rad s}^{-1} \approx 15.7 \text{ rad s}^{-1} \).

(ii) Maximum acceleration is given by:
\( a_{max} = \omega^2 x_0 \)
\( a_{max} = (15.71)^2 \times 0.080 = 246.7 \times 0.080 \approx 19.7 \text{ m s}^{-2} \).

(iii) Kinetic energy at a given displacement \( x \) is:
\( E_k = \frac{1}{2} m \omega^2 (x_0^2 - x^2) \)
\( E_k = 0.5 \times 0.350 \times (15.71)^2 \times (0.080^2 - 0.040^2) \)
\( E_k = 43.18 \times (0.0064 - 0.0016) = 43.18 \times 0.0048 \approx 0.207 \text{ J} \).

(iv) Kinetic energy is maximum when the displacement is zero (at the equilibrium position, \( x = 0 \)).
\( E_{k,max} = \frac{1}{2} m \omega^2 x_0^2 = 0.5 \times 0.350 \times (15.71)^2 \times (0.080)^2 = 43.18 \times 0.0064 \approx 0.276 \text{ J} \).

(a)
- Acceleration is directly proportional to displacement [1 mark]
- Acceleration is directed towards the equilibrium position (or opposite to displacement) [1 mark]

(b)(i)
- Correct calculation showing \( 2\pi \times 2.50 = 15.7 \text{ rad s}^{-1} \) [1 mark]

(b)(ii)
- Use of \( a_{max} = \omega^2 x_0 \) [1 mark]
- Correct calculation of \( 19.7 \text{ m s}^{-2} \) (accept \( 19.6 \text{ m s}^{-2} \) if using 15.7) [1 mark]

(b)(iii)
- Use of formula \( E_k = \frac{1}{2} m \omega^2 (x_0^2 - x^2) \) [1 mark]
- Correct substitution: \( 0.5 \times 0.350 \times 15.71^2 \times (0.080^2 - 0.040^2) \) [1 mark]
- Correct calculation: \( 0.207 \text{ J} \) (accept \( 0.21 \text{ J} \)) [1 mark]

(b)(iv)
- Identifies displacement as zero (equilibrium position) [1 mark]
- Correct calculation of maximum kinetic energy as \( 0.276 \text{ J} \) (accept \( 0.28 \text{ J} \)) [1 mark]

(a) The time constant is the time taken for the charge, potential difference, or current in a discharging capacitor circuit to fall to \( 1/e \) (approximately 37.8%) of its initial value.

(b)(i) Initial charge stored is:
\( Q_0 = C V_0 = 220 \times 10^{-6} \text{ F} \times 12.0 \text{ V} = 2.64 \times 10^{-3} \text{ C} \) (or \( 2.64 \text{ mC} \)).

(ii) Time constant is:
\( \tau = R C = 15.0 \times 10^3 \text{ }\Omega \times 220 \times 10^{-6} \text{ F} = 3.30 \text{ s} \).

(iii) Potential difference is given by:
\( V = V_0 e^{-t/RC} \)
\( V = 12.0 \times e^{-5.00 / 3.30} = 12.0 \times e^{-1.515} = 12.0 \times 0.2198 \approx 2.64 \text{ V} \).

(iv) Energy stored initially is:
\( E_{initial} = \frac{1}{2} C V_0^2 = 0.5 \times 220 \times 10^{-6} \times 12.0^2 = 0.01584 \text{ J} \)
Energy stored at \( t = 5.00 \text{ s} \) is:
\( E_{final} = \frac{1}{2} C V^2 = 0.5 \times 220 \times 10^{-6} \times 2.637^2 = 0.000765 \text{ J} \)
Energy lost:
\( \Delta E = E_{initial} - E_{final} = 0.01584 - 0.000765 \approx 0.0151 \text{ J} \) (or \( 15.1 \text{ mJ} \)).

(a)
- Time for charge/p.d./current to fall to \( 1/e \) (or 37%) of its initial value [1 mark]

(b)(i)
- Formula \( Q = CV \) used [1 mark]
- Correct calculation of charge: \( 2.64 \times 10^{-3} \text{ C} \) [1 mark]

(b)(ii)
- Correct calculation of time constant: \( 3.30 \text{ s} \) [1 mark]

(b)(iii)
- Recall of discharge equation \( V = V_0 e^{-t/RC} \) [1 mark]
- Correct substitution with calculated time constant [1 mark]
- Correct calculation of \( V = 2.64 \text{ V} \) [1 mark]

(b)(iv)
- Formula for energy stored \( E = \frac{1}{2}CV^2 \) used [1 mark]
- Calculated energy change \( \Delta E = \frac{1}{2}C(V_0^2 - V_2^2) \) [1 mark]
- Correct calculation of energy lost: \( 0.0151 \text{ J} \) (accept \( 1.5 \times 10^{-2} \text{ J} \)) [1 mark]

(a) 1. The magnetic force \( F = Bqv \) is always perpendicular to the direction of motion (velocity) of the charged particle (by Fleming's Left-Hand Rule).
2. Since the force is perpendicular to the displacement, no work is done on the particle, so its speed remains constant.
3. A constant magnitude force acting perpendicular to the velocity provides a centripetal acceleration, which results in circular motion.

(b)(i) Centripetal force is provided by the magnetic force:
\( B q v = \frac{m v^2}{r} \)
\( r = \frac{m v}{B q} \)
\( r = \frac{9.11 \times 10^{-31} \text{ kg} \times 3.50 \times 10^6 \text{ m s}^{-1}}{2.40 \times 10^{-3} \text{ T} \times 1.60 \times 10^{-19} \text{ C}} \)
\( r = \frac{3.1885 \times 10^{-24}}{3.840 \times 10^{-22}} = 8.303 \times 10^{-3} \text{ m} \approx 8.3 \text{ mm} \).

(ii) Frequency \( f \) of the circular motion is:
\( v = ̉\omega r = 2 \pi f r \Rightarrow f = \frac{v}{2 \pi r} \)
\( f = \frac{3.50 \times 10^6}{2 \pi \times 8.303 \times 10^{-3}} \approx 6.71 \times 10^7 \text{ Hz} \) (or \( 67.1 \text{ MHz} \)).

(iii) The proton has a much larger mass than the electron (approx. 1836 times greater). Because kinetic energy is given by \( E_k = \frac{p^2}{2m} \), the momentum is \( p = \sqrt{2m E_k} \).
Since the radius of the path is \( r = \frac{p}{B q} = \frac{\sqrt{2 m E_k}}{B q} \) and both have the same kinetic energy and charge magnitude, the proton will have a much larger momentum and therefore a much larger radius than the electron.

(a)
- Magnetic force is perpendicular to velocity / direction of motion [1 mark]
- Force has constant magnitude and does not change speed / acts perpendicular to motion [1 mark]
- This force acts as the centripetal force resulting in circular motion [1 mark]

(b)(i)
- Formula \( B q v = \frac{m v^2}{r} \) used [1 mark]
- Algebraic rearrangement to \( r = \frac{m v}{B q} \) [1 mark]
- Correct substitution showing \( 8.30 \times 10^{-3} \text{ m} \) [1 mark]

(b)(ii)
- Use of \( v = 2\pi f r \) or equivalent [1 mark]
- Correct calculation of frequency: \( 6.71 \times 10^7 \text{ Hz} \) (or \( 6.7 \times 10^7 \text{ Hz} \)) [1 mark]

(b)(iii)
- State that the radius increases [1 mark]
- Reason: Proton has a much larger mass, which results in larger momentum at the same kinetic energy (or using \( r \propto \sqrt{m} \)) [1 mark]

(a) Work function energy is the minimum energy required to remove a single electron from the surface of a metal.

(b)(i) Energy of a photon is:
\( E = \frac{h c}{\lambda} \)
\( E = \frac{6.63 \times 10^{-34} \text{ J s} \times 3.00 \times 10^8 \text{ m s}^{-1}}{240 \times 10^{-9} \text{ m}} \)
\( E = 8.2875 \times 10^{-19} \text{ J} \approx 8.29 \times 10^{-19} \text{ J} \).

(ii) Convert work function to joules:
\( \Phi = 3.20 \text{ eV} \times 1.60 \times 10^{-19} \text{ J eV}^{-1} = 5.12 \times 10^{-19} \text{ J} \)
Using Einstein's photoelectric equation:
\( E_{k,max} = E - \Phi \)
\( E_{k,max} = 8.2875 \times 10^{-19} - 5.12 \times 10^{-19} = 3.1675 \times 10^{-19} \text{ J} \approx 3.17 \times 10^{-19} \text{ J} \).

(iii) The de Broglie wavelength \( \lambda_{db} \) is:
\( \lambda_{db} = \frac{h}{p} = \frac{h}{\sqrt{2 m_e E_k}} \)
\( p = \sqrt{2 \times 9.11 \times 10^{-31} \text{ kg} \times 3.1675 \times 10^{-19} \text{ J}} = \sqrt{5.771 \times 10^{-49}} = 7.597 \times 10^{-25} \text{ kg m s}^{-1} \)
\( \lambda_{db} = \frac{6.63 \times 10^{-34}}{7.597 \times 10^{-25}} \approx 8.73 \times 10^{-10} \text{ m} \) (or \( 0.873 \text{ nm} \)).

(a)
- Minimum energy [1 mark]
- to remove/liberate an electron from the metal surface [1 mark]

(b)(i)
- Formula \( E = hc/\lambda \) stated or used [1 mark]
- Correct substitution of values [1 mark]
- Correct calculation of energy: \( 8.29 \times 10^{-19} \text{ J} \) [1 mark]

(b)(ii)
- Conversion of eV to Joules: \( 3.20 \times 1.60 \times 10^{-19} = 5.12 \times 10^{-19} \text{ J} \) [1 mark]
- Recall of \( E_{k,max} = E - \Phi \) [1 mark]
- Correct subtraction to get \( 3.17 \times 10^{-19} \text{ J} \) [1 mark]

(b)(iii)
- Use of de Broglie relation \( \lambda = h/p \) and \( p = \sqrt{2mE_k} \) [1 mark]
- Correct calculation of wavelength: \( 8.73 \times 10^{-10} \text{ m} \) [1 mark]

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

(b)(i) Using the relation for activity:
\( A = \lambda N \)
\( \lambda = \frac{A}{N} = \frac{1.80 \times 10^{11} \text{ Bq}}{4.20 \times 10^{16}} = 4.286 \times 10^{-6} \text{ s}^{-1} \approx 4.29 \times 10^{-6} \text{ s}^{-1} \).

(ii) Half-life \( t_{1/2} \) is given by:
\( t_{1/2} = \frac{\ln 2}{\lambda} \)
\( t_{1/2} = \frac{0.69315}{4.286 \times 10^{-6}} = 1.617 \times 10^5 \text{ s} \approx 1.6 \times 10^5 \text{ s} \).

(iii) The decay equation for the number of nuclei remaining is:
\( N = N_0 e^{-\lambda t} \)
\( N = 4.20 \times 10^{16} \times e^{-(4.286 \times 10^{-6}) \times (5.00 \times 10^5)} \)
\( N = 4.20 \times 10^{16} \times e^{-2.143} \)
\( N = 4.20 \times 10^{16} \times 0.1173 = 4.927 \times 10^{15} \approx 4.93 \times 10^{15} \).

(a)
- Probability of decay [1 mark]
- per unit time [1 mark]

(b)(i)
- Formula \( A = \lambda N \) used [1 mark]
- Correct calculation of \( \lambda = 4.29 \times 10^{-6} \text{ s}^{-1} \) [1 mark]

(b)(ii)
- Formula \( t_{1/2} = \frac{\ln 2}{\lambda} \) stated [1 mark]
- Substitution showing calculation to approx \( 1.6 \times 10^5 \text{ s} \) [1 mark]

(b)(iii)
- Decay equation \( N = N_0 e^{-\lambda t} \) (or equivalent ratio method) used [1 mark]
- Calculation of exponent value: \( \lambda t = 2.14 \) [1 mark]
- Substitution of correct values [1 mark]
- Correct final calculation of active nuclei: \( 4.93 \times 10^{15} \) (accept range \( 4.90 \times 10^{15} \) to \( 5.00 \times 10^{15} \)) [1 mark]

(a) 1. A radiotracer is injected into the body and decays by emitting positrons (beta-plus decay).
2. The emitted positrons travel a short distance in tissue before colliding with electrons.
3. This collision results in annihilation, which converts mass to energy, producing two gamma-ray photons travelling in opposite directions (back-to-back).
4. Detectors surrounding the patient register these photons. The difference in the arrival times of the two photons is processed by a computer to locate the site of annihilation along the line of response.

(b)(i) Two photons must be produced to conserve linear momentum (since the initial momentum of the electron-positron system is approximately zero, the two photons must travel in opposite directions with equal and opposite momenta).

(ii) The mass of an electron (or positron) is \( m_e = 9.11 \times 10^{-31} \text{ kg} \).
Total mass annihilated is \( \Delta m = 2 m_e = 1.822 \times 10^{-30} \text{ kg} \).
Total energy released is:
\( E_{total} = \Delta m c^2 = 1.822 \times 10^{-30} \times (3.00 \times 10^8)^2 = 1.6398 \times 10^{-13} \text{ J} \).
Since two identical photons are produced, the energy of each photon is:
\( E_{photon} = \frac{1}{2} E_{total} = m_e c^2 = 9.11 \times 10^{-31} \times 9.00 \times 10^{16} = 8.199 \times 10^{-14} \text{ J} \).
Converting this to MeV:
\( E_{photon} = \frac{8.199 \times 10^{-14} \text{ J}}{1.60 \times 10^{-13} \text{ J MeV}^{-1}} \approx 0.512 \text{ MeV} \).

(iii) Using \( E = \frac{h c}{\lambda} \):
\( \lambda = \frac{h c}{E_{photon}} = \frac{6.63 \times 10^{-34} \text{ J s} \times 3.00 \times 10^8 \text{ m s}^{-1}}{8.199 \times 10^{-14} \text{ J}} \approx 2.43 \times 10^{-12} \text{ m} \) (or \( 2.43 \text{ pm} \)).

(a)
- Radioactive tracer decays to emit a positron [1 mark]
- Positron annihilates with an electron in the tissue [1 mark]
- Annihilation produces two gamma-ray photons travelling in opposite directions [1 mark]
- Detectors measure arrival times to calculate difference and locate position of tracer [1 mark]

(b)(i)
- To conserve linear momentum [1 mark]

(b)(ii)
- Recall of \( E = m c^2 \) [1 mark]
- Correct substitution of values using mass of electron: \( 9.11 \times 10^{-31} \times (3.00 \times 10^8)^2 \) [1 mark]
- Conversion from J to MeV to get \( 0.512 \text{ MeV} \) (accept range \( 0.51 \) to \( 0.513 \)) [1 mark]

(b)(iii)
- Use of \( E = hc/\lambda \) [1 mark]
- Correct calculation of wavelength: \( 2.43 \times 10^{-12} \text{ m} \) [1 mark]

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

(b) (i) The number of parent nuclei \(N_X\) remaining at time \(t\) is given by the radioactive decay law:
\[N_X = N_0 e^{-\lambda t}\]
Since each decay of a nucleus of \(X\) produces one nucleus of \(Y\), the number of daughter nuclei is:
\[N_Y = N_0 - N_X = N_0 - N_0 e^{-\lambda t} = N_0(1 - e^{-\lambda t})\]

(ii) At time \(t_1\), we are given:
\[\frac{N_Y}{N_X} = 3.0\]
Substituting the expressions for \(N_Y\) and \(N_X\):
\[\frac{N_0(1 - e^{-\lambda t_1})}{N_0 e^{-\lambda t_1}} = 3.0\]
\[\frac{1 - e^{-\lambda t_1}}{e^{-\lambda t_1}} = 3.0 \implies e^{\lambda t_1} - 1 = 3.0\]
\[e^{\lambda t_1} = 4.0\]
Taking the natural logarithm of both sides:
\[\lambda t_1 = \ln(4.0) = 2 \ln(2)\]
Since \(\lambda = \frac{\ln(2)}{T_{1/2}}\):
\[\frac{\ln(2)}{T_{1/2}} t_1 = 2 \ln(2) \implies t_1 = 2.0 T_{1/2}\]

(c) The time \(t_1 = 2.0 T_{1/2}\). Since \(T_{1/2} = 6.0\text{ hours}\), \(t_1 = 12.0\text{ hours}\).
Because the activity \(A\) is directly proportional to the number of parent nuclei \(N_X\), the activity decreases by the same factor:
\[A = A_0 e^{-\lambda t_1} = A_0 e^{-2\ln(2)} = A_0 / 4\]
Alternatively, after two half-lives, the activity is reduced to \((1/2)^2 = 1/4\) of its initial value:
\[A = \frac{1.2 \times 10^8\text{ Bq}}{4} = 3.0 \times 10^7\text{ Bq}\]

(a)
- probability of decay of a nucleus [1]
- per unit time [1]

(b)(i)
- \(N_X = N_0 e^{-\lambda t}\) stated or used [1]
- \(N_Y = N_0 - N_X\) leading to \(N_Y = N_0(1 - e^{-\lambda t})\) [1]

(b)(ii)
- \(\frac{1 - e^{-\lambda t_1}}{e^{-\lambda t_1}} = 3.0\) or \(e^{\lambda t_1} = 4.0\) [1]
- \(\lambda t_1 = \ln(4.0)\) or \(2 \ln(2)\) [1]
- substitution of \(\lambda = \frac{\ln(2)}{T_{1/2}}\) leading to \(t_1 = 2.0 T_{1/2}\) [1]

(c)
- recognition that \(t_1\) corresponds to exactly 2 half-lives (or \(12.0\text{ hours}\)) [1]
- formula for decay of activity or relation \(A = A_0 / 4\) [1]
- \(A = 3.0 \times 10^7\text{ Bq}\) [1]

(a) The gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point.

(b) (i) The formula for gravitational potential \(\phi\) at a distance \(R\) from the centre of a point mass \(M\) is:
\[\phi = - \frac{G M}{R}\]
Substituting the values:
\[\phi = - \frac{(6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}) \times (6.0 \times 10^{24}\text{ kg})}{6.4 \times 10^6\text{ m}}\]
\[\phi = -6.253 \times 10^7\text{ J kg}^{-1} \approx -6.3 \times 10^7\text{ J kg}^{-1}\]

(ii) By conservation of energy, the loss of kinetic energy is equal to the gain in gravitational potential energy:
\[E_{k,\text{initial}} = E_{p,\text{final}} - E_{p,\text{initial}}\]
Where:
\[E_{p} = m \phi = - \frac{G M m}{r}\]
So:
\[E_{k,\text{initial}} = - \frac{G M m}{r_{\text{max}}} - \left(- \frac{G M m}{R}\right)\]
\[4.5 \times 10^{10} = - \frac{G M m}{r_{\text{max}}} + \frac{G M m}{R}\]
Let us first calculate \(\frac{G M m}{R}\):
\[\frac{G M m}{R} = 1200\text{ kg} \times 6.2531 \times 10^7\text{ J kg}^{-1} = 7.504 \times 10^{10}\text{ J}\]
Thus:
\[4.5 \times 10^{10} = - \frac{G M m}{r_{\text{max}}} + 7.504 \times 10^{10}\]
\[\frac{G M m}{r_{\text{max}}} = 7.504 \times 10^{10} - 4.5 \times 10^{10} = 3.004 \times 10^{10}\text{ J}\]
Solving for \(r_{\text{max}}\):
\[r_{\text{max}} = \frac{G M m}{3.004 \times 10^{10}}\]
\[r_{\text{max}} = \frac{(6.67 \times 10^{-11}) \times (6.0 \times 10^{24}) \times 1200}{3.004 \times 10^{10}} = 1.5987 \times 10^7\text{ m} \approx 1.6 \times 10^7\text{ m}\]

(iii) The gravitational potential at infinity is defined as zero.

(iv) The gravitational force is attractive. Since potential is defined as zero at infinity, and work is done *by* the gravitational field as a mass moves from infinity towards the planet, the potential energy decreases, making it negative. Alternatively, work must be done *on* the mass to move it to infinity where potential is zero.

(a)
- work done per unit mass [1]
- in bringing a small test mass from infinity to that point [1]

(b)(i)
- formula \(\phi = - \frac{G M}{R}\) stated or shown [1]
- substitution shown leading to \(-6.25 \times 10^7\text{ J kg}^{-1}\) or \(-6.3 \times 10^7\text{ J kg}^{-1}\) [1]

(b)(ii)
- conservation of energy stated or implied by equation, e.g., \(\Delta E_p = \Delta E_k\) or \(E_{k} = - \frac{G M m}{r_{\text{max}}} - (- \frac{G M m}{R})\) [1]
- calculation of initial potential energy \(E_{p,\text{initial}} = -7.50 \times 10^{10}\text{ J}\) (or change in potential \(\Delta \phi = 3.75 \times 10^7\text{ J kg}^{-1}\)) [1]
- correct equation for \(r_{\text{max}}\): \(\frac{G M m}{r_{\text{max}}} = 3.00 \times 10^{10}\text{ J}\) (or final potential \(\phi_{\text{final}} = -2.50 \times 10^7\text{ J kg}^{-1}\)) [1]
- final answer \(r_{\text{max}} = 1.6 \times 10^7\text{ m}\) (accept \(1.60 \times 10^7\text{ m}\)) [1]

(b)(iii)
- Zero [1]

(b)(iv)
- Gravitational force is attractive (or work is done by the field / work must be done to move mass to infinity) [1]

### Experimental Design

#### 1. Diagram and Experimental Setup
- The loudspeaker is connected to the output of the signal generator.
- The microphone is positioned coaxially with the loudspeaker at a fixed distance \(x\).
- The sheets of paper are supported (e.g., using a clamp and stand) between the loudspeaker and the microphone.
- The microphone is connected to the Y-input of an oscilloscope.

#### 2. Procedure and Measurements
- Switch on the signal generator and set it to a constant audible frequency (e.g., \(1000\text{ Hz}\)) and constant amplitude.
- Measure the peak-to-peak voltage \(V_0\) on the oscilloscope screen with no sheets of paper (\(N = 0\)). The intensity \(I_0\) is proportional to \(V_0^2\).
- Place \(N = 1\) sheet of paper between the speaker and the microphone, ensuring it fully covers the path of the sound.
- Measure the new peak-to-peak voltage \(V\) on the oscilloscope.
- Repeat this process by increasing the number of sheets of paper \(N\) one by one up to at least \(N = 6\), recording \(V\) for each value of \(N\).

#### 3. Control of Variables
- Keep the distance between the loudspeaker and the microphone constant throughout the experiment.
- Keep the frequency and amplitude of the signal generator output constant.
- Ensure the orientation of the microphone and loudspeaker remains unchanged.
- Keep background noise to a minimum; perform the experiment in a quiet room or place the apparatus inside a sound-absorbing enclosure.

#### 4. Analysis of Data
- Since \(I \propto V^2\), we can write \(V^2 = V_0^2 e^{-\alpha N}\), which simplifies to:

\(V = V_0 e^{-\frac{\alpha}{2} N}\)

- Taking the natural logarithm of both sides:

\(\ln V = -\frac{\alpha}{2} N + \ln V_0\)

- Plot a graph of \(\ln V\) against \(N\).
- If the relationship is valid, the graph should be a straight line with a negative gradient.
- The gradient of the line is equal to \(-\frac{\alpha}{2}\), so \(\alpha = -2 \times \text{gradient}\).

#### 5. Safety Precautions
- Loud sounds can cause damage to hearing. Wear ear defenders or keep the volume of the signal generator at a safe level.

#### Marking Scheme (15 marks total)

**Defining the Problem (4 marks):**
- **MC1:** Identify \(N\) as the independent variable.
- **MC2:** Identify intensity \(I\) (or peak-to-peak voltage \(V\)) as the dependent variable.
- **MC3:** State that the frequency/amplitude of the sound source must be kept constant.
- **MC4:** State that the distance between the loudspeaker and the microphone must be kept constant.

**Methods of Data Collection (4 marks):**
- **MD1:** Draw a labeled diagram showing the loudspeaker, microphone, and sheets of paper positioned between them.
- **MD2:** Show connections of the signal generator to the loudspeaker and the microphone to the oscilloscope (or sound level meter).
- **MD3:** Describe how the sheets of paper are held or clamped in position.
- **MD4:** Explain how the sound amplitude/intensity is measured (e.g., measuring peak-to-peak voltage \(V\) on the oscilloscope).

**Method of Analysis (3 marks):**
- **MA1:** Plot a graph of \(\ln V\) against \(N\) (or \(\ln I\) against \(N\)).
- **MA2:** State that a straight-line graph with a negative gradient confirms the relationship.
- **MA3:** Relate the absorption coefficient to the gradient: \(\alpha = -2 \times \text{gradient}\) (if using \(\ln V\)) or \(\alpha = -\text{gradient}\) (if using \(\ln I\)).

**Safety Precautions (1 mark):**
- **MS1:** Mention wearing ear protection or keeping sound intensity levels within safe limits to avoid hearing damage.

**Additional Details / Reliability (3 marks):**
- **AD1:** Describe a method to minimize background noise (e.g., sound-insulated chamber, quiet room, or measuring background noise with speaker off to subtract it).
- **AD2:** Explain how to ensure the paper sheets are parallel to the speaker/microphone face to prevent reflections/scattering at angles.
- **AD3:** Repeat measurements of \(V\) for each \(N\) and take the average to reduce random errors.

### Detailed Solution

#### Part (a)
Comparing \(\frac{1}{C} = -\left(\frac{\varepsilon_r - 1}{\varepsilon_r \varepsilon_0 A}\right) d + \frac{D}{\varepsilon_0 A}\) to \(y = mx + c\):
- **Gradient \(m\)**: \(-\frac{\varepsilon_r - 1}{\varepsilon_r \varepsilon_0 A}\)
- **y-intercept \(c\)**: \(\frac{D}{\varepsilon_0 A}\)

#### Part (b)
We calculate \(\frac{1}{C}\) and its absolute uncertainty using \(\Delta\left(\frac{1}{C}\right) \approx \frac{\Delta C}{C^2}\):
- For \(d = 1.0\text{ mm}\): \(1/C = 1/56 = 0.0179\text{ pF}^{-1}\). Uncertainty \(\Delta(1/C) = 2 / 56^2 \approx 0.0006\text{ pF}^{-1}\).
- For \(d = 2.0\text{ mm}\): \(1/C = 1/64 = 0.0156\text{ pF}^{-1}\). Uncertainty \(\Delta(1/C) = 2 / 64^2 \approx 0.0005\text{ pF}^{-1}\).
- For \(d = 3.0\text{ mm}\): \(1/C = 1/76 = 0.0132\text{ pF}^{-1}\). Uncertainty \(\Delta(1/C) = 2 / 76^2 \approx 0.0003\text{ pF}^{-1}\).
- For \(d = 4.0\text{ mm}\): \(1/C = 1/88 = 0.0114\text{ pF}^{-1}\). Uncertainty \(\Delta(1/C) = 3 / 88^2 \approx 0.0004\text{ pF}^{-1}\).
- For \(d = 5.0\text{ mm}\): \(1/C = 1/114 = 0.0088\text{ pF}^{-1}\). Uncertainty \(\Delta(1/C) = 4 / 114^2 \approx 0.0003\text{ pF}^{-1}\).

#### Part (c), (d) & (e)
- **Gradient of best-fit line \(m\)**:
Using \((1.0, 0.0179)\) and \((5.0, 0.0088)\):
\(m = \frac{0.0088 - 0.0179}{5.0 - 1.0} = -0.002275 \approx -0.00228\text{ pF}^{-1}\text{ mm}^{-1}\)
- **Worst acceptable line**:
Passes through the upper error limit of the first point and the lower error limit of the last point:
\((1.0, 0.0179 + 0.0007) = (1.0, 0.0186)\)
\((5.0, 0.0088 - 0.0003) = (5.0, 0.0085)\)
\(m_{\text{worst}} = \frac{0.0085 - 0.0186}{5.0 - 1.0} = -0.002525\text{ pF}^{-1}\text{ mm}^{-1}\)
- **Uncertainty in gradient \(\Delta m\)**: \(|-0.00228 - (-0.00253)| = 0.00025\text{ pF}^{-1}\text{ mm}^{-1}\)

- **y-intercept of best-fit line \(c\)**:
\(c = 0.0179 - (-0.00228 \times 1.0) = 0.0202\text{ pF}^{-1}\)
- **y-intercept of worst acceptable line**:
\(c_{\text{worst}} = 0.0085 - (-0.002525 \times 5.0) = 0.0211\text{ pF}^{-1}\)
- **Uncertainty in y-intercept \(\Delta c\)**: \(|0.0202 - 0.0211| = 0.0009\text{ pF}^{-1}\)

#### Part (f)
1. **To find \(D\)**:
\(c = \frac{D}{\varepsilon_0 A} \implies D = c \times 10^{12}\text{ F}^{-1} \times 8.85 \times 10^{-12}\text{ F m}^{-1} \times 3.40 \times 10^{-2}\text{ m}^2\)
\(D = 0.0202 \times 8.85 \times 3.40 \times 10^{-2} = 6.08 \times 10^{-3}\text{ m} = 6.1\text{ mm}\)

Uncertainty in \(D\):
\(\frac{\Delta D}{D} = \frac{\Delta c}{c} + \frac{\Delta A}{A} = \frac{0.0009}{0.0202} + \frac{0.08}{3.40} = 0.0446 + 0.0235 = 0.0681\)
\
\(\Delta D = 0.0681 \times 6.08\text{ mm} \approx 0.4\text{ mm}\)
So, \(D = (6.1 \pm 0.4)\text{ mm}\).

2. **To find \(\varepsilon_r\)**:
Convert \(m\) to SI units: \(m_{\text{SI}} = m \times 10^{15}\text{ F}^{-1}\text{ m}^{-1}\)
\(G = -m_{\text{SI}} \varepsilon_0 A = -(-0.00228 \times 10^{15}) \times 8.85 \times 10^{-12} \times 3.40 \times 10^{-2} \approx 0.686\)
Since \(\frac{\varepsilon_r - 1}{\varepsilon_r} = G \implies \varepsilon_r = \frac{1}{1 - G}\):
\(\varepsilon_r = \frac{1}{1 - 0.686} = 3.18 \approx 3.2\)

Uncertainty in \(\varepsilon_r\):
Using \(G_{\text{max}} = 2.53 \times 8.85 \times 3.48 \times 10^{-2} = 0.779 \implies \varepsilon_{r,\text{max}} = \frac{1}{1 - 0.779} = 4.52\)
Using \(G_{\text{min}} = 2.03 \times 8.85 \times 3.32 \times 10^{-2} = 0.596 \implies \varepsilon_{r,\text{min}} = \frac{1}{1 - 0.596} = 2.48\)
\(\Delta \varepsilon_r = \frac{4.52 - 2.48}{2} = 1.02 \approx 1.0\)
So, \(\varepsilon_r = 3.2 \pm 1.0\).

#### Marking Scheme (15 marks total)

**Part (a) [2 marks]:**
- **M1:** Correct expression for the gradient: \(m = -\frac{\varepsilon_r - 1}{\varepsilon_r \varepsilon_0 A}\)
- **M2:** Correct expression for the y-intercept: \(c = \frac{D}{\varepsilon_0 A}\)

**Part (b) [3 marks]:**
- **M3:** All 5 values of \(\frac{1}{C}\) calculated correctly to 3 significant figures.
- **M4:** Column headed with appropriate unit: \(\frac{1}{C}\ /\text{ pF}^{-1}\).
- **M5:** Absolute uncertainties calculated correctly and matching the precision of \(\frac{1}{C}\).

**Part (c) [3 marks]:**
- **M6:** Five points plotted correctly with error bars on \(\frac{1}{C}\).
- **M7:** Line of best fit drawn cleanly through the points.
- **M8:** Worst acceptable line drawn and clearly labeled.

**Part (d) [2 marks]:**
- **M9:** Gradient of best-fit line calculated correctly using points on the line separated by at least half the length of the line.
- **M10:** Uncertainty in gradient calculated correctly using \(|m_{\text{best}} - m_{\text{worst}}|\).

**Part (e) [2 marks]:**
- **M11:** y-intercept of best-fit line determined correctly.
- **M12:** Uncertainty in y-intercept calculated correctly using \(|c_{\text{best}} - c_{\text{worst}}|\).

**Part (f) [3 marks]:**
- **M13:** Value of \(D\) calculated correctly with units (e.g., \(6.1\text{ mm}\) or \(6.1 \times 10^{-3}\text{ m}\)).
- **M14:** Value of \(\varepsilon_r\) calculated correctly as a dimensionless quantity (value around \(3.2\)).
- **M15:** Absolute uncertainties for both \(D\) and \(\varepsilon_r\) calculated and stated correctly with appropriate SF.