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The percentage uncertainty in \(g\) is given by:
\[\frac{\Delta g}{g} \times 100\% = \left(\frac{\Delta L}{L} + 2 \frac{\Delta T}{T}\right) \times 100\%\]

Calculate each individual percentage uncertainty:
\[\frac{\Delta L}{L} \times 100\% = \frac{0.004}{0.800} \times 100\% = 0.5\%\]
\[\frac{\Delta T}{T} \times 100\% = \frac{0.03}{1.80} \times 100\% \approx 1.67\%\]

Therefore, the percentage uncertainty in \(g\) is:
\[\frac{\Delta g}{g} \times 100\% = 0.5\% + 2(1.67\%) = 0.5\% + 3.33\% = 3.83\% \approx 3.8\%\]

1 mark for calculating the fractional uncertainties of L and T correctly, doubling the fractional uncertainty of T, and summing them to obtain 3.8%.

First, calculate the volume \(V\) of the sphere:
\[V = \frac{\pi d^3}{6} = \frac{\pi (1.62)^3}{6} \approx 2.226 \text{ cm}^3\]

The density \(\rho\) is:
\[\rho = \frac{m}{V} = \frac{12.4}{2.226} \approx 5.57 \text{ g cm}^{-3}\]

The relative uncertainty in \(\rho\) is:
\[\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3 \frac{\Delta d}{d}\]
\[\frac{\Delta \rho}{\rho} = \frac{0.1}{12.4} + 3 \times \frac{0.02}{1.62} \approx 0.00806 + 0.03704 = 0.0451\]

Now, calculate the absolute uncertainty \(\Delta \rho\):
\[\Delta \rho = 0.0451 \times 5.57 \approx 0.25 \text{ g cm}^{-3}\]

Expressing the absolute uncertainty to 1 significant figure gives \(0.3 \text{ g cm}^{-3}\). The calculated value must be quoted to the same decimal place as its uncertainty:
\[\rho = (5.6 \pm 0.3) \text{ g cm}^{-3}\]

1 mark for calculating the density to be 5.57 and the absolute uncertainty to be 0.25, and correctly rounding the uncertainty to 1 s.f. (0.3) and matching the density decimal places (5.6).

Using the equation of motion with upward as the positive direction:
\[s = ut + \frac{1}{2}at^2\]

Here, the final displacement \(s = -h\), the initial velocity \(u = +15.0 \text{ m s}^{-1}\), the acceleration \(a = -g = -9.81 \text{ m s}^{-2}\), and the time interval \(t = 4.50 \text{ s}\):
\[-h = (15.0 \times 4.50) + \frac{1}{2}(-9.81)(4.50)^2\]
\[-h = 67.5 - 4.905(20.25)\]
\[-h = 67.5 - 99.33 = -31.83 \text{ m}\]

Thus, \(h \approx 31.8 \text{ m}\).

1 mark for using the kinematic equation with consistent signs for velocity and acceleration to solve for the displacement, yielding a cliff height of 31.8 m.

Before the collision, the total kinetic energy is:
\[E_i = \frac{1}{2} (3M) (2v)^2 = 6Mv^2\]

The total initial momentum is:
\[p_i = (3M)(2v) = 6Mv\]

After the collision, the combined mass of the trolleys is \(4M\) and they move with speed \(V_f\). Using the conservation of linear momentum:
\[6Mv = 4M V_f \implies V_f = 1.5v\]

The total kinetic energy after the collision is:
\[E_f = \frac{1}{2} (4M) (1.5v)^2 = 2M (2.25v^2) = 4.5Mv^2\]

The ratio of kinetic energy is:
\[\frac{E_f}{E_i} = \frac{4.5Mv^2}{6Mv^2} = 0.75\]

1 mark for determining the final speed using conservation of momentum and calculating the ratio of kinetic energies to be 0.75.

In bright light:
\[V_{\text{LDR, light}} = \frac{R_{\text{LDR, light}}}{R_1 + R_{\text{LDR, light}}} \times V_{\text{supply}} = \frac{2.0 \text{ k}\Omega}{4.0 \text{ k}\Omega + 2.0 \text{ k}\Omega} \times 12.0 \text{ V} = 4.0 \text{ V}\]

In darkness:
\[V_{\text{LDR, dark}} = \frac{R_{\text{LDR, dark}}}{R_1 + R_{\text{LDR, dark}}} \times V_{\text{supply}} = \frac{10.0 \text{ k}\Omega}{4.0 \text{ k}\Omega + 10.0 \text{ k}\Omega} \times 12.0 \text{ V} \approx 8.57 \text{ V}\]

Calculate the change in potential difference:
\[\Delta V = V_{\text{LDR, dark}} - V_{\text{LDR, light}} = 8.57 \text{ V} - 4.0 \text{ V} = 4.57 \text{ V} \approx 4.6 \text{ V}\]

1 mark for calculating the potential differences across the LDR in both states and finding the difference to be 4.6 V.

The upthrust force is equal to the weight of the fluid displaced by the cylinder. Since the bottom of the cylinder is at a depth of \(0.15 \text{ m}\), the cylinder is submerged to a depth of \(d = 0.15 \text{ m}\).

The submerged volume is:
\[V_{\text{sub}} = A \times d = 2.5 \times 10^{-3} \text{ m}^2 \times 0.15 \text{ m} = 3.75 \times 10^{-4} \text{ m}^3\]

The upthrust force \(U\) is:
\[U = \rho V_{\text{sub}} g = 1200 \text{ kg m}^{-3} \times 3.75 \times 10^{-4} \text{ m}^3 \times 9.81 \text{ m s}^{-2} = 4.41 \text{ N} \approx 4.4 \text{ N}\]

1 mark for determining the correct volume of displaced liquid using the submerged depth of 0.15 m, and calculating the corresponding weight of the displaced liquid to be 4.4 N.

The quark structure of a neutron is \(\text{udd}\) and that of a proton is \(\text{uud}\). Thus, during the decay, one down quark changes into an up quark:
- Change in up quarks: \(+1\)
- Change in down quarks: \(-1\)

Both the produced electron (a lepton) and the electron antineutrino (an antilepton) are members of the lepton family. Because there were no leptons initially, the total number of leptons increases:
- Change in number of leptons: \(+2\)

1 mark for identifying the correct changes: +1 for up quarks, -1 for down quarks, and +2 for leptons.

An air column closed at one end has a node at the closed end and an antinode at the open end. The possible wavelengths of stationary waves in this pipe are given by:
\[L = n \frac{\lambda}{4} \quad \text{where } n = 1, 3, 5, \dots\]

For the fundamental mode (first harmonic), \(n = 1\):
\[L = \frac{\lambda_1}{4} \implies \lambda_1 = 4L \implies f_1 = \frac{v}{4L}\]

The first overtone (the next resonant mode) occurs at \(n = 3\) (third harmonic):
\[L = \frac{3\lambda_3}{4} \implies \lambda_3 = \frac{4L}{3}\]

The frequency of this overtone is:
\[f_3 = \frac{v}{\lambda_3} = \frac{3v}{4L}\]

1 mark for recognizing that the first overtone for a pipe closed at one end is the third harmonic, and correctly expressing its frequency as 3v / 4L.

The relation for fractional uncertainty is given by \(\frac{\Delta g}{g} = \frac{\Delta h}{h} + 2\frac{\Delta t}{t}\). Substituting the given percentage uncertainties: \(\text{Percentage uncertainty in } g = 1.5\% + 2 \times 2.0\% = 5.5\%\).

Award 1 mark for the correct answer B. Method: Identify the rule for compounding independent fractional/percentage uncertainties when powers are involved. Calculate: 1.5% + 2 * 2.0% = 5.5%.

Using the equation of motion \(s = ut + \frac{1}{2}at^2\) with upwards as the positive direction: \(u = +15\text{ m s}^{-1}\), \(a = -9.81\text{ m s}^{-2}\), and \(t = 4.5\text{ s}\). Substituting these values: \(s = (15)(4.5) + \frac{1}{2}(-9.81)(4.5)^2 = 67.5 - 99.3 = -31.8\text{ m}\). The negative sign indicates displacement is downwards from the release point. The height of the cliff is therefore \(32\text{ m}\) (to 2 s.f.).

Award 1 mark for the correct answer A. Method: Use displacement equation with appropriate signs for direction. Resolve to get height of cliff.

Using conservation of linear momentum (taking right as positive): \(p_i = m_1u_1 + m_2u_2 = (0.40)(3.0) + (0.60)(-2.0) = 1.2 - 1.2 = 0\text{ kg m s}^{-1}\). Since the final momentum is zero, the final velocity of the stuck-together cars must be zero, meaning final kinetic energy is zero. The initial kinetic energy is \(E_{ki} = \frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}(0.40)(3.0)^2 + \frac{1}{2}(0.60)(-2.0)^2 = 1.8 + 1.2 = 3.0\text{ J}\). Thus, the lost kinetic energy is \(3.0\text{ J} - 0\text{ J} = 3.0\text{ J}\).

Award 1 mark for the correct answer D. Method: Apply conservation of momentum to find the final velocity, calculate initial and final kinetic energies, and find the difference.

Initially, the voltage across the fixed resistor \(R = 3.0\text{ k}\Omega\) is \(4.0\text{ V}\). Therefore, the voltage across the thermistor \(R_T\) is \(12.0\text{ V} - 4.0\text{ V} = 8.0\text{ V}\). Since potential difference is directly proportional to resistance in a series circuit: \(\frac{R_T}{R} = \frac{8.0\text{ V}}{4.0\text{ V}} = 2\), which means \(R_T = 2 \times 3.0\text{ k}\Omega = 6.0\text{ k}\Omega\). At \(T_2\), the thermistor resistance decreases by a factor of 4: \(R_T' = \frac{6.0\text{ k}\Omega}{4} = 1.5\text{ k}\Omega\). The new potential difference across the fixed resistor is \(V_R' = 12\text{ V} \times \frac{3.0\text{ k}\Omega}{3.0\text{ k}\Omega + 1.5\text{ k}\Omega} = 12\text{ V} \times \frac{2}{3} = 8.0\text{ V}\).

Award 1 mark for the correct answer B. Method: Determine initial resistance of the thermistor using the potential divider ratio. Calculate the new thermistor resistance, then recalculate the voltage share of the fixed resistor.

The pressure \(p\) on the base is given by \(p = \frac{F}{A}\), where \(F\) is the force exerted by the weight of the liquid. Thus, \(F = p \times A = (6.0 \times 10^3\text{ Pa}) \times 0.050\text{ m}^2 = 300\text{ N}\). The weight of the liquid \(W = mg = 300\text{ N}\). The mass of the liquid \(m = \frac{300}{9.81} \approx 30.6\text{ kg}\), which rounds to \(31\text{ kg}\) to two significant figures. (Note that the density is redundant details).

Award 1 mark for the correct answer A. Method: Relate base pressure to force and area, equate force to weight, and find mass.

The charge of an up (u) quark is \(+\frac{2}{3}e\) and a down (d) quark is \(-\frac{1}{3}e\). Let us calculate the charges for each combination: A: \(+\frac{2}{3} - \frac{1}{3} - \frac{1}{3} = 0\). B: \(+\frac{2}{3} + \frac{2}{3} - \frac{1}{3} = +1e\). C: \(+\frac{2}{3} + \frac{2}{3} + \frac{2}{3} = +2e\). D: The strange quark (s) has a charge of \(-\frac{1}{3}e\), so \(+\frac{2}{3} - \frac{1}{3} - \frac{1}{3} = 0\). Therefore, combination B represents a baryon with a net charge of \(+1e\).

Award 1 mark for the correct answer B. Method: Identify individual quark charges (u = +2/3, d = -1/3, s = -1/3) and sum them up for each option to find the one that yields +1.

Young modulus \(E = \frac{\text{stress}}{\text{strain}}\), so \(\text{strain} = \frac{\text{stress}}{E}\). First, calculate the tensile stress: \(\text{stress} = \frac{F}{A} = \frac{150\text{ N}}{1.5 \times 10^{-6}\text{ m}^2} = 1.0 \times 10^8\text{ Pa}\). Next, calculate the strain: \(\text{strain} = \frac{1.0 \times 10^8\text{ Pa}}{2.0 \times 10^{11}\text{ Pa}} = 5.0 \times 10^{-4}\).

Award 1 mark for the correct answer A. Method: Compute stress as force over area, then divide stress by Young modulus to get strain.

A precise set of readings has very little spread (high repeatability). An inaccurate set has a mean value that differs significantly from the true/accepted value (due to systematic error). Looking at the sets: Set A: Mean \(\approx 589\text{ nm}\) (accurate) and small spread (precise). Set B: Mean \(\approx 582.5\text{ nm}\) (inaccurate, far from \(589\text{ nm}\)) and small spread of only \(1\text{ nm}\) (precise). Set C: Large spread (imprecise). Set D: Large spread (imprecise). Therefore, Set B is precise but inaccurate.

Award 1 mark for the correct answer B. Method: Distinguish between the definitions of precision (closeness of values) and accuracy (closeness to target value).

The formula for the calculated density is \(\rho = \frac{6 m}{\pi d^3}\). The fractional uncertainty in density \(\rho\) is given by \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3 \frac{\Delta d}{d}\). Substituting the percentage uncertainties: percentage uncertainty in \(\rho = 1.2\% + 3 \times 1.5\% = 1.2\% + 4.5\% = 5.7\%\).

1 mark for the correct option C. Method: State that the fractional uncertainty in density is the sum of the fractional uncertainty in mass and three times the fractional uncertainty in diameter. Calculate \(1.2\% + 3 \times 1.5\% = 5.7\%\).

Taking the vertically upward direction as positive, the initial velocity is \(u = +15\text{ m s}^{-1}\), the acceleration is \(a = -9.81\text{ m s}^{-2}\), and the time is \(t = 4.0\text{ s}\). Using the equations of motion, the displacement \(s\) is given by \(s = ut + \frac{1}{2}at^2\). Substituting the values: \(s = (15 \times 4.0) + \frac{1}{2}(-9.81)(4.0)^2 = 60 - 78.48 = -18.48\text{ m}\). The negative sign indicates that the ground is \(18.48\text{ m}\) below the starting point, so the height of the cliff is approximately \(18\text{ m}\).

1 mark for the correct option B. Method: Set up the kinematic equation for displacement using correct signs for velocity and acceleration. Calculate the displacement and round to two significant figures to get \(18\text{ m}\).

By conservation of linear momentum: \(2m(v) + m(-v) = 2m v_X + m v_Y\), which simplifies to \(v = 2 v_X + v_Y\). For a perfectly elastic collision, the relative speed of approach equals the relative speed of separation: \(u_X - u_Y = v_Y - v_X \implies v - (-v) = v_Y - v_X \implies 2v = v_Y - v_X\). Solving these two simultaneous equations: subtracting the second equation from the first gives \(3v_X = -v \implies v_X = -\frac{1}{3}v\). Substituting back into the second equation gives \(v_Y = 2v + v_X = 2v - \frac{1}{3}v = \frac{5}{3}v\).

1 mark for the correct option A. Method: Formulate simultaneous equations for conservation of momentum and conservation of kinetic energy (via relative speed of approach and separation) and solve for the final velocities of both objects.

The current \(I\) in the circuit is given by \(I = \frac{E}{R + r}\). As \(R\) decreases, the total resistance \(R+r\) decreases, causing the current \(I\) to increase. The terminal potential difference \(V\) is given by \(V = E - Ir\). Since \(I\) increases and \(r\) is constant, the lost volts \(Ir\) increases, so \(V\) decreases. The power \(P\) dissipated in the internal resistance is \(P = I^2 r\). Since \(I\) increases, \(P\) must increase. Thus, \(V\) decreases and \(P\) increases.

1 mark for the correct option B. Method: Determine the change in current from Ohm's law. Use \(V = E - Ir\) to deduce the change in terminal potential difference and \(P = I^2 r\) to deduce the change in internal power dissipation.

The total pressure \(p\) at the bottom of the beaker is the sum of the atmospheric pressure \(p_0\), the hydrostatic pressure exerted by the column of liquid X of depth \(h\), and the hydrostatic pressure exerted by the column of liquid Y of depth \(2h\). This is expressed as: \(p = p_0 + \rho_X g h + \rho_Y g (2h) = p_0 + g h (\rho_X + 2\rho_Y)\).

1 mark for the correct option B. Method: Identify and sum all three pressure contributions: atmospheric pressure \(p_0\), the pressure from liquid X of depth \(h\), and the pressure from liquid Y of depth \(2h\).

A neutron has the quark structure \(udd\), while a proton has the quark structure \(uud\). In beta-minus decay, one down quark changes into an up quark, so the quark flavour change is \(d \rightarrow u\). Lepton number must be conserved. Before the decay, the neutron has a lepton number of 0. After the decay, the proton has lepton number 0, the electron has lepton number +1, and the electron antineutrino has lepton number -1. The total lepton number remains 0, meaning the change in total lepton number is 0.

1 mark for the correct option A. Method: Identify the change in quark configuration from neutron to proton and use the conservation of lepton number to determine that the net change in lepton number is zero.

For a pipe closed at one end and open at the other, the third harmonic (first overtone) consists of a node at the closed end, followed by an antinode, a node, and finally an antinode at the open end. The length of the pipe \(L\) is equal to \(\frac{3}{4}\lambda\), so the wavelength is \(\lambda = \frac{4}{3}L\). The distance from the closed end (\(x = 0\), which is a node) to the next displacement node is half a wavelength, \(\frac{\lambda}{2}\). Substituting the expression for \(\lambda\): distance = \(\frac{1}{2} \left(\frac{4}{3}L\right) = \frac{2}{3}L\).

1 mark for the correct option C. Method: Determine the wavelength of the third harmonic in terms of \(L\) as \(\lambda = \frac{4}{3}L\). Identify that the next node occurs at a distance of \(\frac{\lambda}{2}\) from the closed end and calculate this value.

The elastic potential energy stored in a wire obeying Hooke's law is \(U = \frac{1}{2} F \Delta L\). Young modulus \(E\) is given by \(E = \frac{F L}{A \Delta L}\), which can be rearranged to express the force as \(F = \frac{E A \Delta L}{L}\). Substituting this expression for \(F\) into the energy formula gives: \(U = \frac{1}{2} \left(\frac{E A \Delta L}{L}\right) \Delta L = \frac{E A (\Delta L)^2}{2 L}\).

1 mark for the correct option B. Method: Recall the formula for work done / strain energy and the definition of Young modulus. Substitute the force from the Young modulus equation into the strain energy equation to find the correct algebraic expression.

The formula for the density of a cylindrical rod is:

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

To find the percentage uncertainty, we sum the percentage uncertainties of each variable multiplied by their respective powers:

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

Substituting the given percentage uncertainties:

\(\frac{\Delta \rho}{\rho} \times 100\% = 2.0\% + 2(2.0\%) + 1.0\% = 7.0\%\)

Therefore, the percentage uncertainty in the calculated density is \(7.0\%\).

1 mark for identifying the correct fractional uncertainty combination where the uncertainty in diameter is multiplied by 2, and adding the values to obtain 7.0%.

We use the equations of uniformly accelerated motion. Let the upward direction be positive.

The initial velocity is \(u = +15.0\text{ m s}^{-1}\).
The acceleration is \(a = -g = -9.81\text{ m s}^{-2}\).
The total time of flight is \(t = 4.50\text{ s}\).

The displacement \(s\) from the top of the cliff to the bottom of the cliff is given by:

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

\(s = (15.0)(4.50) + \frac{1}{2}(-9.81)(4.50)^2\)

\(s = 67.5 - 99.33 = -31.83\text{ m}\)

Since the displacement is negative, the bottom of the cliff is \(31.8\text{ m}\) below the starting point. Thus, the height of the cliff is \(31.8\text{ m}\).

1 mark for using the correct kinematic equation with consistent positive/negative signs to find the cliff height of 31.8 m.

1. **Initial Kinetic Energy (\(E_{ki}\)):**

\(E_{ki} = \frac{1}{2}(3M)v^2 = \frac{3}{2}Mv^2\)

2. **Conservation of Momentum:**

\(3Mv + M(0) = (3M + M)v_f \implies 3Mv = 4M v_f \implies v_f = \frac{3}{4}v\)

3. **Final Kinetic Energy (\(E_{kf}\)):**

\(E_{kf} = \frac{1}{2}(4M)v_f^2 = \frac{1}{2}(4M)\left(\frac{3}{4}v\right)^2 = 2M \left(\frac{9}{16}v^2\right) = \frac{9}{8}Mv^2\)

4. **Kinetic Energy Lost (\(\Delta E_k\)):**

\(\Delta E_k = E_{ki} - E_{kf} = \frac{12}{8}Mv^2 - \frac{9}{8}Mv^2 = \frac{3}{8}Mv^2\)

5. **Fraction of Initial Kinetic Energy Lost:**

\(\text{Fraction} = \frac{\Delta E_k}{E_{ki}} = \frac{\frac{3}{8}Mv^2}{\frac{3}{2}Mv^2} = \frac{3}{8} \times \frac{2}{3} = \frac{1}{4} = 0.25\)

1 mark for calculating both initial and final kinetic energies correctly using conservation of momentum, and finding the ratio of lost energy to initial energy to be 0.25.

Initially, at \(20\text{ }^\circ\text{C}\):

\(V_{\text{out, initial}} = V_{\text{supply}} \times \frac{R}{R + R_{\text{th}}} = 12.0 \times \frac{4.0}{4.0 + 8.0} = 4.0\text{ V}\)

The output potential difference across the fixed resistor increases by \(2.0\text{ V}\), so at \(60\text{ }^\circ\text{C}\):

\(V_{\text{out, final}} = 4.0 + 2.0 = 6.0\text{ V}\)

Using the potential divider equation for the final state:

\(V_{\text{out, final}} = V_{\text{supply}} \times \frac{R}{R + R_{\text{th, final}}}\)

\(6.0 = 12.0 \times \frac{4.0}{4.0 + R_{\text{th, final}}}\)

\(\frac{4.0}{4.0 + R_{\text{th, final}}} = 0.5\)

\(4.0 + R_{\text{th, final}} = 8.0\text{ k}\Omega \implies R_{\text{th, final}} = 4.0\text{ k}\Omega\)

1 mark for calculating the new output voltage as 6.0 V and setting up the potential divider equation correctly to find the final thermistor resistance.

The total hydrostatic pressure \(p\) at the bottom of the tank is the sum of the pressures exerted by both liquid layers:

\(p = \rho_A g h_A + \rho_B g h_B\)

Substituting the given values (using \(g = 9.81\text{ m s}^{-2}\)):

\(p = (800 \times 9.81 \times 0.60) + (1200 \times 9.81 \times 0.40)\)

\(p = 4708.8 + 4708.8 = 9417.6\text{ Pa} \approx 9.4\text{ kPa}\)

Note that the cross-sectional area of the tank is not needed because hydrostatic pressure depends only on fluid density, depth, and gravitational field strength.

1 mark for using the formula \(p = \rho g h\) for both liquids and correctly summing the values to yield 9.4 kPa.

A neutron has the quark composition \(udd\) and a proton has the quark composition \(uud\). Therefore, in \(\beta^-\)\ decay, one of the down (\(d\)) quarks changes into an up (\(u\)) quark.

This process is mediated by the weak force, and specifically, the mediating particle is the \(W^-\)\ boson.

1 mark for identifying the quark transition as down to up and the boson as the W- boson.

For a tube closed at one end and open at the other, the boundaries require a node at the closed end and an antinode at the open end.

The fundamental mode (first harmonic) has a wavelength \(\lambda_1\) such that:

\(L = \frac{\lambda_1}{4} \implies \lambda_1 = 4L = 4 \times 0.68 = 2.72\text{ m}\)

The next resonant frequency above the fundamental corresponds to the third harmonic (since only odd harmonics exist for a closed-open tube), where:

\(L = \frac{3\lambda_3}{4} \implies \lambda_3 = \frac{4L}{3} = \frac{4 \times 0.68}{3} \approx 0.907\text{ m}\)

The frequency of this mode is:

\(f_3 = \frac{v}{\lambda_3} = \frac{340}{\frac{4 \times 0.68}{3}} = \frac{340 \times 3}{2.72} = 375\text{ Hz}\)

1 mark for identifying the correct boundary conditions for the second resonant mode of a closed-open tube and calculating the correct frequency.

1. **Elastic Potential Energy (\(E_p\)):**

Since the wire obeys Hooke's law, the force is proportional to extension, and the stored elastic potential energy is given by:

\(E_p = \frac{1}{2} F x = \frac{1}{2} \times 60\text{ N} \times (2.0 \times 10^{-3}\text{ m}) = 0.060\text{ J}\)

2. **Young Modulus (\(E\)):**

\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{x / L} = \frac{F L}{A x}\)

\(E = \frac{60 \times 2.0}{(1.5 \times 10^{-7}) \times (2.0 \times 10^{-3})} = \frac{120}{3.0 \times 10^{-10}} = 4.0 \times 10^{11}\text{ Pa}\)

1 mark for calculating the potential energy as 0.060 J and Young Modulus as 4.0 x 10^11 Pa.

The percentage uncertainty in a quantity calculated by multiplying and dividing other quantities is found by adding their individual percentage uncertainties, multiplying each by the power to which it is raised in the equation. For \(\rho = \frac{R \pi d^2}{4 L}\), the percentage uncertainty in \(\rho\) is given by: \(\frac{\Delta \rho}{\rho} \times 100\% = (\frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}) \times 100\%\). Substituting the values: \(2.0\% + 2(2.0\%) + 0.5\% = 2.0\% + 4.0\% + 0.5\% = 6.5\%\).

1 mark for correctly applying the power rule to the uncertainty of diameter \(d\) (doubling it to 4.0%) and summing all percentage uncertainties to obtain 6.5%.

Taking the upward direction as positive, the initial velocity \(u = +v\), the acceleration \(a = -g\), and the vertical displacement \(s = -h\) where \(h\) is the height of the cliff. Using the equation of motion \(s = ut + \frac{1}{2}at^2\): \(-h = vt - \frac{1}{2}gt^2\). Multiplying both sides by \(-1\) gives \(h = \frac{1}{2}gt^2 - vt\).

1 mark for using the kinematic equation \(s = ut + \frac{1}{2}at^2\) with correct signs according to the chosen direction convention and rearranging it for \(h\).

Let the direction of the initial velocity of A be positive. So, \(u_A = +u\) and \(u_B = -2u\). By conservation of momentum: \(3m(u) + m(-2u) = 3m(v_A) + m(v_B)\), which simplifies to \(u = 3v_A + v_B\) (Equation 1). For a perfectly elastic collision, the relative speed of approach equals the relative speed of separation: \(u_A - u_B = v_B - v_A \implies u - (-2u) = v_B - v_A \implies 3u = v_B - v_A \implies v_B = v_A + 3u\) (Equation 2). Substituting Equation 2 into Equation 1: \(u = 3v_A + v_A + 3u \implies -2u = 4v_A \implies v_A = -0.5u\). Substituting this back into Equation 2 gives \(v_B = -0.5u + 3u = +2.5u\).

1 mark for writing the correct simultaneous equations representing conservation of momentum and conservation of kinetic energy (relative speed equation) and solving them to obtain \(v_A = -0.5u\) and \(v_B = +2.5u\).

For a floating cylinder in equilibrium, the upward upthrust equals the downward weight of the cylinder. Weight of cylinder \(W = m g = A L \rho_c g\). Upthrust \(U = \) weight of displaced liquid \(= A h \rho_l g\). Equating the two: \(A h \rho_l g = A L \rho_c g\). Dividing both sides by \(A g\) gives \(h = L \left(\frac{\rho_c}{\rho_l}\right)\).

1 mark for equating the expression for upthrust \(A h \rho_l g\) to the cylinder's weight \(A L \rho_c g\) and solving for \(h\).

The Young modulus is defined as \(E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{x/L} = \frac{FL}{Ax}\), which gives tension force \(F = \frac{EAx}{L}\). The elastic potential energy stored is \(U = \frac{1}{2} F x\). Substituting \(F\) into this energy equation yields \(U = \frac{1}{2} \left(\frac{EAx}{L}\right) x = \frac{EAx^2}{2L}\).

1 mark for deriving the tension force \(F = \frac{EAx}{L}\) and correctly substituting it into the strain energy formula \(U = \frac{1}{2}Fx\).

For a tube closed at one end, there is a node at the closed end and an antinode at the open end. The possible boundary conditions require the column length \(L\) to be an odd multiple of quarter-wavelengths: \(L = n \frac{\lambda}{4}\) where \(n = 1, 3, 5, \dots\). This gives \(
\lambda = \frac{4L}{n}\). Using \(v = f\lambda\), the possible frequencies are \(f = \frac{nv}{4L}\) where \(n\) is an odd integer. For \(n=3\), the frequency is \(f = \frac{3v}{4L}\).

1 mark for identifying the general equation for the frequency of a closed-open tube as \(f = \frac{nv}{4L}\) for odd \(n\), and matching it to the given options.

In \(\beta^-\) decay, a neutron (quark composition \(udd\)) decays into a proton (quark composition \(uud\)), emitting an electron (\(e^-\)) and an electron antineutrino (\(\bar{\nu}_e\)). This represents a quark flavour change of down to up (\(d \to u\)). The electron, being a lepton, has a lepton number of \(+1\). The antineutrino, being an antilepton, has a lepton number of \(-1\).

1 mark for identifying the correct quark transformation (\(d \to u\)) and the correct lepton numbers (\(+1\) for the electron, \(-1\) for the antineutrino).

The relationship between terminal potential difference \(V\), electromotive force \(E\), current \(I\), and internal resistance \(r\) is \(V = E - Ir\). In a graph of \(V\) (y-axis) against \(I\) (x-axis), comparing this to the straight-line equation \(y = mx + c\) shows that the y-intercept \(c\) is equal to \(E\) (e.m.f.) and the gradient \(m\) is \(-r\). Thus, the magnitude of the gradient is the internal resistance \(r\).

1 mark for identifying the y-intercept as the electromotive force (e.m.f.) and the magnitude of the gradient as the internal resistance using the equation \(V = E - Ir\).

(a) Calculate cross-sectional area \(A = \frac{\pi d^2}{4} = \frac{\pi \times (0.38 \times 10^{-3})^2}{4} = 1.134 \times 10^{-7}\text{ m}^2\). Resistivity \(\rho = \frac{R A}{L} = \frac{8.6 \times 1.134 \times 10^{-7}}{1.240} = 7.865 \times 10^{-7}\ \Omega\text{ m}\). (b) Calculate the fractional/percentage uncertainty of each component: \(\frac{\Delta R}{R} = \frac{0.2}{8.6} \approx 2.33\%\), \(\frac{\Delta L}{L} = \frac{0.002}{1.240} \approx 0.16\%\), \(\frac{\Delta d}{d} = \frac{0.02}{0.38} \approx 5.26\%\). Total percentage uncertainty \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2\left(\frac{\Delta d}{d}\right) + \frac{\Delta L}{L} = 2.33\% + 2(5.26\%) + 0.16\% = 13.01\% \approx 13\%\). (c) The absolute uncertainty \(\Delta \rho = 0.1301 \times 7.865 \times 10^{-7} = 1.02 \times 10^{-7}\ \Omega\text{ m}\). Rounding the absolute uncertainty to one significant figure gives \(1 \times 10^{-7}\ \Omega\text{ m}\). The final value is therefore expressed as \(\rho = (7.9 \pm 1.0) \times 10^{-7}\ \Omega\text{ m}\).

(a) C1: Recall of \(A = \frac{\pi d^2}{4}\) or substitution of diameter. C1: Substitution into formula \(\rho = \frac{R A}{L}\). A1: Correct calculation of \(7.9 \times 10^{-7}\ \Omega\text{ m}\) (accept \(7.87 \times 10^{-7}\ \Omega\text{ m}\)). (b) C1: Identifies percentage uncertainty in \(d\) is doubled in the error equation. C1: Sum of the percentage uncertainties of \(R\), \(L\), and \(2 \times d\). A1: Correct calculation to obtain \(13\%\) (accept answers in range 13% to 13.1%). (c) C1: Correct calculation of absolute uncertainty as \(1 \times 10^{-7}\ \Omega\text{ m}\) (or \(1.0 \times 10^{-7}\ \Omega\text{ m}\)). A1: Expresses final value to same decimal place precision as uncertainty: \((7.9 \pm 1.0) \times 10^{-7}\ \Omega\text{ m}\).

(a) For vertical motion: \(s = ut + \frac{1}{2}gt^2\). Since initial vertical velocity \(u_y = 0\), we have \(35 = 0 + \frac{1}{2}(9.81)t^2\), giving \(t = \sqrt{\frac{70}{9.81}} = 2.67\text{ s}\), which is approximately \(2.7\text{ s}\). (b) Horizontal distance \(x = v_x \times t = 12 \times 2.67 = 32.0\text{ m}\). (c) Just before hitting the ground, the horizontal component of velocity is still \(v_x = 12\text{ m s}^{-1}\). The vertical component is \(v_y = u_y + gt = 0 + 9.81 \times 2.67 = 26.2\text{ m s}^{-1}\). The magnitude of the velocity is \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{12^2 + 26.2^2} = 28.8\text{ m s}^{-1} \approx 29\text{ m s}^{-1}\). (d) Air resistance acts vertically upwards opposing the downward motion. This reduces the downward acceleration to less than \(g\). Consequently, the average vertical speed is lower, causing the ball to take a longer time to reach the ground.

(a) M1: Recalls or uses \(s = \frac{1}{2}gt^2\). A1: Shows substitution clearly to obtain \(t = 2.67\text{ s}\) (concluding to \(2.7\text{ s}\)). (b) C1: Recalls \(s_x = v_x t\). A1: Calculates horizontal distance as \(32\text{ m}\) (or \(32.0\text{ m}\)). (c) C1: Calculates final vertical speed \(v_y = 26.2\text{ m s}^{-1}\). C1: Uses Pythagoras theorem to combine components. A1: Obtains final speed \(29\text{ m s}^{-1}\) (or \(28.8\text{ m s}^{-1}\)). (d) B1: States that there is an upward force/resistance which reduces the downward acceleration. B1: States that the time to reach the ground increases.

(a) Taking to the right as positive. Initial total momentum \(P_i = m_A u_A + m_B u_B = 0.40(3.0) + 0 = 1.20\text{ kg m s}^{-1}\). Final total momentum \(P_f = m_A v_A + m_B v_B = 0.40(-0.60) + 0.60 v_B = -0.24 + 0.60 v_B\). By conservation of momentum: \(1.20 = -0.24 + 0.60 v_B \Rightarrow 0.60 v_B = 1.44 \Rightarrow v_B = 2.4\text{ m s}^{-1}\). Since \(v_B\) is positive, block B moves to the right. (b) Calculate initial kinetic energy: \(E_k(initial) = \frac{1}{2} m_A u_A^2 = 0.5 \times 0.40 \times 3.0^2 = 1.80\text{ J}\). Calculate final kinetic energy: \(E_k(final) = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 = 0.5 \times 0.40 \times (-0.60)^2 + 0.5 \times 0.60 \times 2.4^2 = 0.072 + 1.728 = 1.80\text{ J}\). Since the initial kinetic energy equals the final kinetic energy, the collision is perfectly elastic. (c) During the collision, block A exerts a force on block B, and block B exerts an equal and opposite force on block A (Newton's third law). Since force is the rate of change of momentum (\(F = \frac{\Delta p}{\Delta t}\)) and the contact time \(\Delta t\) is identical for both blocks, the change in momentum of block A is equal and opposite to the change in momentum of block B. Hence, the total change in momentum of the system is zero.

(a) C1: Recalls total initial momentum is \(1.20\text{ kg m s}^{-1}\). C1: Correct expression for final momentum taking signs into account (\(v_A = -0.60\text{ m s}^{-1}\)). A1: Calculates magnitude of velocity of B as \(2.4\text{ m s}^{-1}\). A1: States direction is to the right. (b) C1: Correctly calculates initial kinetic energy as \(1.80\text{ J}\). C1: Correctly calculates final kinetic energy as \(1.80\text{ J}\). A1: Concludes collision is elastic because kinetic energy is conserved. (c) B1: States that forces between blocks during collision are equal in magnitude and opposite in direction. B1: Connects equal and opposite forces to equal and opposite momentum changes over the same time interval.

(a) (i) Total resistance of circuit \(R_T = r + R + R_{th} = 1.5 + 6.0 + 4.5 = 12.0\ \Omega\). Current \(I = \frac{E}{R_T} = \frac{9.0}{12.0} = 0.75\text{ A}\). (ii) Terminal p.d. \(V = E - Ir = 9.0 - (0.75 \times 1.5) = 7.875\text{ V} \approx 7.9\text{ V}\). (iii) Power \(P = I^2 R = (0.75)^2 \times 6.0 = 3.375\text{ W} \approx 3.4\text{ W}\). (b) As temperature of NTC thermistor increases, its resistance decreases. Consequently, the total resistance of the circuit decreases, which causes the current \(I\) in the circuit to increase. The 'lost volts' within the cell, given by \(I r\), increases. Since terminal p.d. \(V = E - I r\), the terminal potential difference of the battery decreases.

(a) (i) C1: Finds total circuit resistance \(12.0\ \Omega\). A1: Correctly calculates current \(I = 0.75\text{ A}\). (ii) C1: Recalls or uses formula \(V = E - Ir\) or \(V = I(R + R_{th})\). A1: Calculates terminal p.d. as \(7.9\text{ V}\) (or \(7.88\text{ V}\)). (iii) A1: Calculates power dissipated as \(3.4\text{ W}\) (or \(3.38\text{ W}\)). (b) B1: States resistance of NTC thermistor decreases when temperature increases. B1: States that this causes the circuit current to increase. B1: Explains that higher current increases the internal potential drop (lost volts) causing terminal p.d. to decrease.

(a) Upthrust is the net upward force exerted on an object fully or partially submerged in a fluid, caused by the difference in hydrostatic pressure between the bottom and the top of the object. (b) Volume of cylinder \(V = A h = 4.0 \times 10^{-3} \times 0.25 = 1.0 \times 10^{-3}\text{ m}^3\). Upthrust \(U = \rho_{oil} V g = 850 \times 1.0 \times 10^{-3} \times 9.81 = 8.3385\text{ N} \approx 8.3\text{ N}\). (c) When submerged, the forces acting on the cylinder are: Upward force from spring balance (reading \(T = 12.4\text{ N}\)), Upward upthrust (\(U = 8.34\text{ N}\)), Downward weight \(W\). Thus, \(W = T + U = 12.4 + 8.34 = 20.74\text{ N}\). Mass of cylinder \(m = \frac{W}{g} = \frac{20.74}{9.81} = 2.114\text{ kg}\). Density of the cylinder metal \(\rho_{metal} = \frac{m}{V} = \frac{2.114}{1.0 \times 10^{-3}} = 2114\text{ kg m}^{-3} \approx 2.1 \times 10^3\text{ kg m}^{-3}\).

(a) B1: Upward force exerted by fluid on an object. B1: Mentions this is due to difference in pressure between top and bottom surfaces. (b) C1: Calculates volume of cylinder \(V = 1.0 \times 10^{-3}\text{ m}^3\). A1: Calculates upthrust \(8.3\text{ N}\) (or \(8.34\text{ N}\)). (c) C1: Recognizes total weight of cylinder is the sum of reading and upthrust: \(W = 20.7\text{ N}\) (or \(20.74\text{ N}\)). C1: Calculates mass of cylinder as \(2.11\text{ kg}\). A1: Calculates density as \(2.1 \times 10^3\text{ kg m}^{-3}\) (accept \(2100\text{ kg m}^{-3}\) or \(2110\text{ kg m}^{-3}\)).

(a) The complete decay equation is: \(^{32}_{15}\text{P} \rightarrow \ ^{32}_{16}\text{S} + \ ^{0}_{-1}\beta + \bar{\nu}_e\) where \(\bar{\nu}_e\) is an electron antineutrino. (b) A neutron consists of one up (u) quark and two down (d) quarks (\(udd\)). A proton consists of two up (u) quarks and one down (d) quark (\(uud\)). Therefore, one down (d) quark changes into an up (u) quark. (c) Quantities conserved include: charge, baryon number, lepton number, and mass-energy. (d) Beta decay is caused by the weak interaction (or weak nuclear force).

(a) B1: Correct symbols and proton/nucleon numbers for \(^{32}_{15}\text{P}\), \(^{32}_{16}\text{S}\), and \(^{0}_{-1}\beta\). B1: Includes electron antineutrino symbol (\(\bar{\nu}\) or \(\bar{\nu}_e\)). (b) B1: States composition of neutron (\(udd\)) and proton (\(uud\)). B1: Clearly states that a down quark changes to an up quark (\(d \rightarrow u\)). (c) B3: 1 mark for each correct conserved quantity listed (e.g. charge, baryon number, lepton number, momentum, mass-energy) up to 3 marks. (d) B1: Correctly names the 'weak interaction' or 'weak nuclear force'.

(a) Sound waves from the loudspeaker travel down the tube and reflect at the closed end. The incident and reflected waves, which have the same frequency and amplitude, travel in opposite directions and superpose/interfere, creating a stationary wave. (b) For the fundamental frequency, there is a displacement node (N) at the closed end of the tube and a displacement antinode (A) at the open end. The diagram should show the displacement envelope opening up from the closed end (zero amplitude) to the open end (maximum amplitude). (c) For a tube closed at one end, only odd harmonics exist. The fundamental has \(\lambda_1 = 4L\). The first overtone is the third harmonic, where \(L = \frac{3}{4}\lambda_3\). Hence, \(\lambda_3 = \frac{4}{3}L = \frac{4}{3} \times 0.85 = 1.133\text{ m}\). The frequency \(f_3 = \frac{v}{\lambda_3} = \frac{340}{1.133} = 300\text{ Hz}\).

(a) B1: Sound wave reflects at the closed end of the tube. B1: Incident and reflected waves travel in opposite directions and overlap/superpose. B1: Mentions that the waves must have the same frequency / speed. (b) B1: Shows zero displacement (node) at the closed end and maximum displacement (antinode) at the open end. B1: Nodes and antinodes are correctly labelled with N and A. (c) C1: Recalls that the first overtone (third harmonic) has a wavelength pattern where \(L = \frac{3}{4}\lambda\). C1: Calculates \(\lambda = 1.13\text{ m}\). A1: Obtains frequency of \(300\text{ Hz}\).

### Theoretical Context
The period of a simple pendulum of length \(l\) is given by:
\[ T_p = 2\pi \sqrt{\frac{l}{g}} \]

In this constrained configuration:
- For one half-cycle, the pendulum oscillates freely with a length \(L\). The duration of this half-cycle is:
\[ t_1 = \pi \sqrt{\frac{L}{g}} \]

- For the other half-cycle, the string hits the rod, so the pendulum oscillates with an effective length \(L - h\). The duration of this half-cycle is:
\[ t_2 = \pi \sqrt{\frac{L-h}{g}} \]

Thus, the total period \(T\) of one complete oscillation is:
\[ T = t_1 + t_2 = \pi \sqrt{\frac{L}{g}} + \pi \sqrt{\frac{L-h}{g}} \]

Comparing this to the proposed equation:
\[ T = m (L - h)^{1/2} + C \]

We find:
- \(m = \frac{\pi}{\sqrt{g}}\)
- \(C = \pi \sqrt{\frac{L}{g}}\)

### Sample Calculations and Expected Graph Parameters
Using \(g = 9.81\text{ m s}^{-2}\) and \(L = 0.600\text{ m}\):
- \(m = \frac{\pi}{\sqrt{9.81}} \approx 1.002\text{ s m}^{-1/2}\)
- \(C = \pi \sqrt{\frac{0.600}{9.81}} \approx 0.777\text{ s}\)

**Example Data Points:**
- For \(h = 0.150\text{ m}\), \(L-h = 0.450\text{ m}\):
\[ T = 1.002 \times \sqrt{0.450} + 0.777 \approx 1.45\text{ s} \]
For 10 oscillations, \(t \approx 14.5\text{ s}\).

- For \(h = 0.400\text{ m}\), \(L-h = 0.200\text{ m}\):
\[ T = 1.002 \times \sqrt{0.200} + 0.777 \approx 1.23\text{ s} \]
For 10 oscillations, \(t \approx 12.3\text{ s}\).

Plotting \(T\) vs \((L - h)^{1/2}\) yields a straight line with a gradient of approximately \(1.00\text{ s m}^{-1/2}\) and a vertical intercept of \(0.78\text{ s}\).

#### 1. Data Collection (6 Marks)
- **Table Structure:** Table of results must include raw data of at least 6 different values of \(h\) with appropriate headings and units. (1 mark)
- **Range:** The range of \(h\) values must be wide (e.g. at least \(h \le 0.150\text{ m}\) and \(h \ge 0.400\text{ m}\)). (1 mark)
- **Unit Formatting:** Column headings must include appropriate units separated by solidus: \(h / \text{m}\), \(t / \text{s}\), \(T / \text{s}\), \((L-h)^{1/2} / \text{m}^{1/2}\). (1 mark)
- **Raw Data Precision:** Raw time \(t\) must be measured to \(0.01\text{ s}\) and repeated. (1 mark)
- **Derived Quantities:** Values of \(T\) and \((L-h)^{1/2}\) calculated correctly. (1 mark)
- **Significant Figures:** The number of significant figures in \((L-h)^{1/2}\) must match the significant figures of the raw value \((L-h)\) (typically 3 SF). (1 mark)

#### 2. Graph Plotting (4 Marks)
- **Axes:** Scale must be linear, with no awkward fractions. Axes must be labeled with quantity and unit. (1 mark)
- **Plotting:** All points must be plotted within half a small square on the grid. (1 mark)
- **Line of Best Fit:** Straight line of best fit drawn with a balanced scatter of points above and below the line. (1 mark)
- **Quality of Data:** Low scatter of data points relative to the best-fit line. (1 mark)

#### 3. Analysis of Graph (4 Marks)
- **Gradient Calculation:** Must use a large triangle with hypotenuse length at least half of the drawn line. Coordinates of vertices must be read correctly. (1 mark)
- **Intercept:** Vertical intercept determined from the line or calculated using a point on the line. (1 mark)

#### 4. Constants & Units (2 Marks)
- **Value of m:** Value of \(m\) is equal to the gradient, within a range of \(0.95\text{ to }1.05\text{ s m}^{-1/2}\). (1 mark)
- **Value of C:** Value of \(C\) is equal to the vertical intercept, with unit \(\text{s}\), within a range of \(0.72\text{ to }0.82\text{ s}\). (1 mark)

#### 5. Experimental Quality (4 Marks)
- **Timing repetition:** Measurements of \(t\) are repeated and averaged for each setup. (1 mark)
- **Oscillations count:** At least 10 oscillations are timed to reduce relative uncertainty in manual timing. (1 mark)
- **Precision of L:** Ruler held vertically, and parallax error minimized when setting up \(L = 0.600\text{ m}\). (2 marks)

### Typical Results and Calculations

1. **For \(x_1 = 2.0\text{ cm} = 0.020\text{ m}\):**
- Readings of \(d\): \(19.5\text{ cm}\), \(20.5\text{ cm}\), \(20.0\text{ cm}\).
- Average distance \(d_1 = 20.0\text{ cm} = 0.200\text{ m}\).
- Absolute uncertainty in \(d\) (half-range) \(= \frac{20.5 - 19.5}{2} = 0.5\text{ cm}\).
- If the half-range is smaller than the precision of reading, use the resolution of the scale (e.g., \(0.1\text{ cm}\) or \(1.0\text{ cm}\) depending on how easily stopping position is determined).
- Let us use an absolute uncertainty of \(\Delta d = 1.0\text{ cm}\) due to sliding deviation.
- Percentage uncertainty:
\[ \frac{1.0}{20.0} \times 100\% = 5\% \]

2. **For \(x_2 = 5.0\text{ cm} = 0.050\text{ m}\):**
- Readings of \(d\): \(123\text{ cm}\), \(127\text{ cm}\), \(125\text{ cm}\).
- Average distance \(d_2 = 125\text{ cm} = 1.25\text{ m}\).

3. **Calculation of constant \(k\):**
- \(k_1 = \frac{d_1}{x_1^2} = \frac{0.200}{0.020^2} = 500\text{ m}^{-1}\)
- \(k_2 = \frac{d_2}{x_2^2} = \frac{1.25}{0.050^2} = 500\text{ m}^{-1}\)

4. **Comparison:**
- Percentage difference between \(k_1\) and \(k_2\):
\[ \text{Difference} = \frac{|500 - 500|}{500} \times 100\% = 0\% \]
- Since the percentage difference is less than the percentage uncertainty in \(d\) (which is \(5\%\)), the relationship \(d = k x^2\) is strongly supported.

#### 1. Measurements & Uncertainty (4 Marks)
- **Raw Readings:** Raw measurements of \(x\) and \(d\) must be recorded with appropriate SI or metric units. (1 mark)
- **Repetition:** Multiple trials of \(d\) are taken to get an average. (1 mark)
- **Absolute Uncertainty:** Correct calculation of absolute uncertainty in \(d\) (either half-range or a reasonable value of \(0.5\text{ to }2.0\text{ cm}\) to account for stopping position variance). (1 mark)
- **Percentage Uncertainty:** Calculated correctly with formula: \(\frac{\Delta d}{d} \times 100\%\). (1 mark)

#### 2. Test of Proportionality (3 Marks)
- **Second Reading:** Accurate measurement of \(d\) for the larger extension \(x = 5.0\text{ cm}\). (1 mark)
- **Calculation of k:** Two values of \(k = d/x^2\) calculated with correct decimal places matching raw data. (1 mark)
- **Comparison Criterion:** Clear comparison between the percentage difference in \(k\) values and the percentage uncertainty in \(d\) to state whether the hypothesis is supported. (1 mark)

#### 3. Limitations & Improvements (8 Marks)
- **Limitations (any 4):**
1. Two sets of readings are not enough to confirm a relationship.
2. Friction of the desk surface is non-uniform along the path of the block.
3. Difficult to release the block consistently without imparting spin or an extra force.
4. Difficult to determine exactly when/where the block stops because it may rotate or slide sideways.
5. Aligning the ruler with the center of the elastic band to measure the extension \(x\) has a high parallax error.
- **Improvements (any 4 matching the limitations):**
1. Take multiple readings for different values of \(x\) and plot a graph of \(d\) against \(x^2\).
2. Use a smooth guide track or clean the bench surface thoroughly to ensure uniform friction.
3. Use a mechanical release mechanism (e.g. a simple latch, pin, or electromagnet trigger).
4. Position a high-speed camera or video camera with a scale next to the path to accurately determine the final resting position frame-by-frame.
5. Attach a horizontal scale beneath the band to directly read the extension without alignment error.

#### 4. Safety and Setup (5 Marks)
- **Clamping:** Securely clamps stands to bench to avoid tipping. (1 mark)
- **Clean path:** Ensures path is free of obstacles. (1 mark)
- **Sufficient range:** Uses appropriate elastic band stiffness to ensure the block does not slide off the bench. (1 mark)
- **Significant Figures:** Keeps \(k\) to 2 or 3 significant figures. (1 mark)
- **Unit Consistency:** Consistent use of units (e.g., all in meters or all in centimeters). (1 mark)