2025 Cambridge IAS-Level Physics (9702) Practice Paper with Answers

The formula for density is \(\rho = \frac{m}{V} = \frac{4m}{\pi d^2 l}\). The percentage uncertainty in density is given by: \(\frac{\Delta \rho}{\rho} \times 100 = \left( \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta l}{l} \right) \times 100\). Substituting the fractional uncertainties: \(\frac{\Delta m}{m} = \frac{0.1}{25.4} \approx 0.00394\), \(\frac{\Delta d}{d} = \frac{0.02}{1.20} \approx 0.01667\), \(\frac{\Delta l}{l} = \frac{0.05}{4.50} \approx 0.01111\). Thus, total percentage uncertainty = \((0.00394 + 2(0.01667) + 0.01111) \times 100 = (0.00394 + 0.03333 + 0.01111) \times 100 = 4.838\% \approx 4.8\%\).

1 mark for correct application of error propagation for division and power rules, leading to the correct numerical percentage uncertainty of 4.8%.

Using the equation of motion \(s = ut + \frac{1}{2}at^2\) with upwards taken as positive: \(u = +12\text{ m s}^{-1}\), \(t = 4.0\text{ s}\), and \(a = -9.81\text{ m s}^{-2}\). The vertical displacement \(s\) from the edge of the cliff is: \(s = (12)(4.0) + \frac{1}{2}(-9.81)(4.0)^2 = 48 - 78.48 = -30.48\text{ m}\). The negative sign indicates the ball is below its starting point, so the height of the cliff is approximately \(30\text{ m}\) (to 2 significant figures).

1 mark for using the displacement equation with consistent signs to find 30 m.

Let the direction to the right be positive. Total momentum before collision = \((3.0)(4.0) + (2.0)(-2.0) = +8.0\text{ N s}\). For a perfectly elastic collision, the relative speed of approach equals the relative speed of separation: \(u_1 - u_2 = v_2 - v_1 \implies 4.0 - (-2.0) = v_2 - v_1 \implies v_2 = v_1 + 6.0\). Using conservation of momentum: \(3.0 v_1 + 2.0 v_2 = 8.0 \implies 3.0 v_1 + 2.0(v_1 + 6.0) = 8.0 \implies 5.0 v_1 + 12.0 = 8.0 \implies v_1 = -0.80\text{ m s}^{-1}\) (moving to the left). Thus, \(v_2 = -0.80 + 6.0 = +5.2\text{ m s}^{-1}\) (moving to the right).

1 mark for combining conservation of momentum and the elastic collision relative speed equation to solve for both velocities correctly.

Take moments about the pivot. The pivot is located at \(0.40\text{ m}\) from the left end. The force \(F\) acts downwards at the left end, giving an anticlockwise moment of \(F \times 0.40\text{ m}\). The center of gravity of the uniform beam is at its midpoint, \(1.0\text{ m}\) from the left end (distance from pivot = \(1.0 - 0.40 = 0.60\text{ m}\)). Its weight acts downwards, giving a clockwise moment of \(120\text{ N} \times 0.60\text{ m} = 72\text{ N m}\). The block of weight \(80\text{ N}\) acts downwards at the right-hand end, \(2.0\text{ m}\) from the left end (distance from pivot = \(2.0 - 0.40 = 1.6\text{ m}\)), giving a clockwise moment of \(80\text{ N} \times 1.6\text{ m} = 128\text{ N m}\). For rotational equilibrium, sum of anticlockwise moments = sum of clockwise moments: \(0.40 F = 72 + 128 = 200\text{ N m} \implies F = 500\text{ N}\).

1 mark for establishing the principle of moments about the pivot and correctly solving for F.

The Young modulus is \(E = \frac{F L}{A \Delta x}\), where the cross-sectional area \(A = \frac{\pi d^2}{4}\). Thus, extension \(\Delta x = \frac{4 F L}{\pi E d^2}\). For wire Y, the length is \(2L\) and the diameter is \(2d\). Therefore, the extension \(\Delta x_Y = \frac{4 F (2L)}{\pi E (2d)^2} = \frac{8 F L}{4 \pi E d^2} = 0.5 \left(\frac{4 F L}{\pi E d^2}\right) = 0.5 \Delta x\).

1 mark for using the Young modulus expression to deduce that extension is proportional to length/diameter^2, and correctly scaling it to find 0.5 times the original extension.

Using the Doppler effect formula for a source moving away from a stationary observer: \(f_o = f_s \left( \frac{v}{v + v_s} \right)\), where \(f_s = 800\text{ Hz}\), \(v = 340\text{ m s}^{-1}\), and \(v_s = 30\text{ m s}^{-1}\). This gives: \(f_o = 800 \times \left( \frac{340}{340 + 30} \right) = 800 \times \frac{340}{370} \approx 735\text{ Hz}\).

1 mark for choosing the correct sign in the Doppler formula (positive denominator for source moving away) and evaluating the frequency.

In bright light, the total resistance is \(4.0\text{ k}\Omega + 1.0\text{ k}\Omega = 5.0\text{ k}\Omega\). The potential difference across the fixed resistor is \(V_{\text{light}} = 12.0 \times \frac{4.0}{5.0} = 9.6\text{ V}\). In darkness, the total resistance is \(4.0\text{ k}\Omega + 20\text{ k}\Omega = 24\text{ k}\Omega\). The potential difference across the fixed resistor is \(V_{\text{dark}} = 12.0 \times \frac{4.0}{24} = 2.0\text{ V}\). The magnitude of the change in potential difference is \(|9.6\text{ V} - 2.0\text{ V}| = 7.6\text{ V}\).

1 mark for calculating the potential differences across the fixed resistor in both states and taking their difference to get 7.6 V.

A proton has a quark composition of \(uud\) and a neutron has a quark composition of \(udd\). In \(\beta^+\) decay, one up (\(u\)) quark changes into a down (\(d\)) quark (\(u \rightarrow d\)). Quark flavour changes can only occur via the weak nuclear force (weak interaction).

1 mark for identifying the correct quark flavour transition (u to d) and the weak interaction as the responsible force.

Rearranging the equation for thermal conductivity \(k\) gives:

\(k = \frac{P L}{A \Delta T}

Now, we determine the SI base units of each quantity:
- Power \)P\) has units of watts (\(W\)), where \(W = \text{J}\,\text{s}^{-1} = \text{kg}\,\text{m}^2\,\text{s}^{-3}\).
- Length \(L\) has unit \(\text{m}\).
- Area \(A\) has unit \(\text{m}^2\).
- Temperature difference \(\Delta T\) has unit \(\text{K}\).

Substituting these units into the rearranged equation:

\[ [k] = \frac{\text{kg}\,\text{m}^2\,\text{s}^{-3} \times \text{m}}{\text{m}^2 \times \text{K}} = \text{kg}\,\text{m}\,\text{s}^{-3}\,\text{K}^{-1} \]

1 mark for the correct SI base units of thermal conductivity.

When the object is projected upwards, both the weight \(mg\) (downwards) and the air resistance force \(D\) (downwards, opposing the upward velocity) act in the same direction. Therefore, the net downward force is \(mg + D\), and the magnitude of acceleration is \(g + \frac{D}{m}\), which is at its maximum at release since velocity (and therefore drag) is highest.

At the maximum height, the velocity is momentarily zero, so air resistance is zero. The only force acting on the object is its weight \(mg\), so the magnitude of acceleration is \(g\).

Thus, the magnitude of the acceleration is greatest at the moment of release and is equal to \(g\) at the maximum height.

1 mark for identifying that drag and weight act in the same direction at release (maximum acceleration) and drag is zero at maximum height (acceleration equals \(g\)).

First, find the forces acting on the box parallel to the slope:
1. Tension pulling up the slope: \(T = 120\,\text{N}\).
2. Component of weight acting down the slope: \(W_{\parallel} = mg \sin(30^\circ) = 15 \times 9.81 \times \sin(30^\circ) = 73.575\,\text{N}\).
3. Frictional force acting down the slope: \(F_{\text{frict}} = 15\,\text{N}\).

Using Newton's second law parallel to the slope:
\(F_{\text{net}} = T - W_{\parallel} - F_{\text{frict}}
\)F_{\text{net}} = 120 - 73.575 - 15 = 31.425\,\text{N}

Now, calculate the acceleration:
\(a = \frac{F_{\text{net}}}{m} = \frac{31.425\,\text{N}}{15\,\text{kg}} \approx 2.1\,\text{m}\,\text{s}^{-2}\)

1 mark for correctly resolving weight and applying Newton's second law to find the correct acceleration.

Take moments about the pivot (which is \(1.0\,\text{m}\) from \(A\)):
- The load of \(100\,\text{N}\) at \(A\) is at a distance of \(1.0\,\text{m}\) to the left of the pivot. This creates a counter-clockwise moment:
\(\text{Moment}_{\text{load}} = 100\,\text{N} \times 1.0\,\text{m} = 100\,\text{N}\,\text{m}\) counter-clockwise.
- The center of gravity of the uniform beam is at its midpoint, \(1.5\,\text{m}\) from \(A\), which is \(1.5 - 1.0 = 0.5\,\text{m}\) to the right of the pivot. The weight of the beam is \(80\,\text{N}\). This creates a clockwise moment:
\(\text{Moment}_{\text{beam}} = 80\,\text{N} \times 0.5\,\text{m} = 40\,\text{N}\,\text{m}\) clockwise.

For the beam to be in equilibrium, the net moment must be zero. Since the counter-clockwise moment (\(100\,\text{N}\,\text{m}\)) exceeds the clockwise moment (\(40\,\text{N}\,\text{m}\)), the force \(T\) at end \(B\) (which is \(2.0\,\text{m}\) to the right of the pivot) must produce a clockwise moment of:
\(100\,\text{N}\,\text{m} - 40\,\text{N}\,\text{m} = 60\,\text{N}\,\text{m}\) clockwise.

To produce a clockwise moment about the pivot, the force \(T\) at the right end \(B\) must act downwards.

Now, calculate the magnitude of \(T\):
\(T \times 2.0\,\text{m} = 60\,\text{N}\,\text{m} \implies T = 30\,\text{N}\).

Thus, \(T\) is \(30\,\text{N}\) downwards.

1 mark for correctly applying the principle of moments and determining the correct direction and magnitude of the force.

Let the load be \(F\).
- Cross-sectional area of wire \(P\) is \(A_P = \frac{\pi d^2}{4}\).
- Cross-sectional area of wire \(Q\) is \(A_Q = \frac{\pi (2d)^2}{4} = 4 A_P\).

Since stress \(\sigma = \frac{\text{Force}}{\text{Area}}\):
\(\sigma_Q = \frac{F}{A_Q} = \frac{F}{4 A_P} = \frac{\sigma_P}{4}\).

Since both wires are made of the same material, they have the same Young modulus \(E\).
Since \(\text{Strain } \epsilon = \frac{\text{Stress}}{E}\):
\(\epsilon_Q = \frac{\sigma_Q}{E} = \frac{\sigma_P}{4E} = \frac{\epsilon_P}{4}\).

Therefore, both stress and strain in wire \(Q\) are one-quarter of those in wire \(P\).

1 mark for correct calculation of both stress and strain ratios.

The phase difference \(\phi\) is related to the path difference \(\Delta x\) by the formula:

\[ \phi = \frac{2\pi}{\lambda} \times \Delta x \]

Given that the separation \(\Delta x = \frac{3}{8}\lambda\):

\[ \phi = \frac{2\pi}{\lambda} \times \frac{3}{8}\lambda = \frac{6\pi}{8} = \frac{3\pi}{4}\,\text{rad} \]

Therefore, the correct phase difference is \(\frac{3\pi}{4}\,\text{rad}\).

1 mark for applying the phase-path difference relation to find the correct phase difference.

For a tube open at both ends, the lowest frequency (fundamental mode) of a stationary wave has antinodes at both ends and a single node at the center.

The length \(L\) of the tube is related to the wavelength \(\lambda\) by:

\[ L = \frac{\lambda}{2} \implies \lambda = 2 L \]

Given \(L = 0.60\,\text{m}\):

\[ \lambda = 2 \times 0.60\,\text{m} = 1.20\,\text{m} \]

Using the wave equation \(v = f \lambda\):

\[ f = \frac{v}{\lambda} = \frac{340\,\text{m}\,\text{s}^{-1}}{1.20\,\text{m}} \approx 283\,\text{Hz} \approx 280\,\text{Hz} \]

Hence, the lowest frequency is \(280\,\text{Hz}\).

1 mark for identifying that the wavelength is twice the tube length and calculating the correct frequency.

For a negative temperature coefficient (NTC) thermistor, its resistance decreases as the temperature increases.

In a potential divider circuit, the output voltage \(V_{\text{out}}\) across the thermistor is given by:

\[ V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{thermistor}}}{R_{\text{fixed}} + R_{\text{thermistor}}} \]

As the temperature of the thermistor increases, its resistance \(R_{\text{thermistor}}\) decreases. Since the resistance of the thermistor becomes a smaller fraction of the total circuit resistance, the potential difference across it (\(V_{\text{out}}\)) must also decrease.

1 mark for correctly stating that the resistance of an NTC thermistor decreases with temperature, and explaining that this leads to a decrease in the potential difference across it.

Rearranging the equation for \( k \) gives:

\[ k = \frac{Q L}{t A \Delta \theta} \]

Now, substitute the SI base units for each quantity:
- Thermal energy \( Q \) (unit of Joule): \( \text{kg m}^2 \text{ s}^{-2} \)
- Length \( L \): \( \text{m} \)
- Time \( t \): \( \text{s} \)
- Area \( A \): \( \text{m}^2 \)
- Temperature difference \( \Delta \theta \): \( \text{K} \)

Substituting these yields:

\[ [k] = \frac{(\text{kg m}^2 \text{ s}^{-2}) \cdot \text{m}}{\text{s} \cdot \text{m}^2 \cdot \text{K}} = \text{kg m s}^{-3} \text{ K}^{-1} \]

Therefore, the correct unit is option A.

1 mark for correctly identifying the SI base units of all constituent quantities and successfully rearranging the formula to obtain the base units of thermal conductivity.

Let the ground be the reference level (height = 0) where gravitational potential energy is zero.

At the point of projection (height \( h \)), the total mechanical energy of the stone is:
\[ E = mgh + \frac{1}{2}mu^2 \]

At the maximum height (height \( 3h \)), the stone's vertical velocity is zero. The total mechanical energy is:
\[ E = 3mgh \]

By conservation of mechanical energy:
\[ mgh + \frac{1}{2}mu^2 = 3mgh \implies \frac{1}{2}mu^2 = 2mgh \implies u = \sqrt{4gh} \]

Just before the stone hits the ground (height = 0), the total mechanical energy is purely kinetic:
\[ E = \frac{1}{2}mv^2 = 3mgh \implies v = \sqrt{6gh} \]

Thus, the ratio of the final speed to the initial speed is:
\[ \frac{v}{u} = \frac{\sqrt{6gh}}{\sqrt{4gh}} = \sqrt{\frac{6}{4}} = \sqrt{1.5} \]

1 mark for using equations of motion or conservation of mechanical energy to express both the initial speed and final speed in terms of \( g \) and \( h \), and obtaining the correct ratio of \( \sqrt{1.5} \).

The relationship between Young modulus \( E \), tension \( F \), original length \( L \), cross-sectional area \( A \), and extension \( x \) is:
\[ E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{x/L} = \frac{F L}{A x} \]

Since the wire has a circular cross-section, \( A = \frac{\pi d^2}{4} \). Rearranging for extension \( x \):
\[ x = \frac{4 F L}{\pi d^2 E} \]

For wire X:
\[ x_{\text{X}} = \frac{4 F L}{\pi d^2 E} \]

For wire Y:
\[ x_{\text{Y}} = \frac{4 F (2L)}{\pi (2d)^2 (2E)} = \frac{8 F L}{8 \pi d^2 E} = \frac{F L}{\pi d^2 E} \]

Comparing the two extensions:
\[ x_{\text{X}} = 4 x_{\text{Y}} \implies \frac{x_{\text{X}}}{x_{\text{Y}}} = 4 \]

1 mark for correctly applying the Young modulus formula to express extension in terms of the given parameters and finding the ratio of 4.

First, calculate the wavelength \( \lambda \) of the wave:
\[ \lambda = \frac{v}{f} = \frac{30\text{ m s}^{-1}}{50\text{ Hz}} = 0.60\text{ m} = 60\text{ cm} \]

The phase difference \( \Delta \phi \) in radians between two points separated by a distance \( x \) is given by:
\[ \Delta \phi = \frac{x}{\lambda} \times 2\pi \]

Substitute the given values into the equation:
\[ \Delta \phi = \frac{15\text{ cm}}{60\text{ cm}} \times 2\pi = \frac{1}{4} \times 2\pi = 0.50\pi\text{ rad} \]

1 mark for calculating the wavelength of the wave and then correctly using the path difference to find the phase difference in radians.

In \( \beta^- \) decay, a neutron in the nucleus decays into a proton:
\[ \text{n} \to \text{p} + \text{e}^- + \bar{
u}_e \]

Since a neutron has quark structure \( udd \) and a proton has quark structure \( uud \), a down quark changes into an up quark (\( d \to u \)).

Let us count the total number of down quarks in the nucleus before and after the decay:
- Carbon-14 (6 protons, 8 neutrons):
- Down quarks in protons = \( 6 \times 1 = 6 \)
- Down quarks in neutrons = \( 8 \times 2 = 16 \)
- Total down quarks = 22

- Nitrogen-14 (7 protons, 7 neutrons):
- Down quarks in protons = \( 7 \times 1 = 7 \)
- Down quarks in neutrons = \( 7 \times 2 = 14 \)
- Total down quarks = 21

Therefore, the total number of down quarks in the nucleus decreases by 1.

1 mark for identifying the correct quark flavour change (down to up) and calculating the change in the total down quark count (decreases by 1).

First, relate the resistances of the wires using \( R = \rho \frac{\text{length}}{\text{area}} \), where area \( A = \pi r^2 \):
\[ R_{\text{P}} = \rho \frac{L}{\pi r^2} \]
\[ R_{\text{Q}} = \rho \frac{2L}{\pi (2r)^2} = \rho \frac{2L}{4 \pi r^2} = \frac{1}{2} R_{\text{P}} \]

Now, use Ohm's law (\( I = \frac{V}{R} \)) to determine the currents:
\[ I_{\text{P}} = \frac{V}{R_{\text{P}}} \]
\[ I_{\text{Q}} = \frac{2V}{R_{\text{Q}}} = \frac{2V}{\frac{1}{2} R_{\text{P}}} = 4 \frac{V}{R_{\text{P}}} = 4 I_{\text{P}} \]

Thus, the ratio of the currents is:
\[ \frac{I_{\text{P}}}{I_{\text{Q}}} = \frac{I_{\text{P}}}{4 I_{\text{P}}} = 0.25 \]

1 mark for calculating the correct relationship between the resistances of P and Q, and then using Ohm's law to evaluate the current ratio.

Let \( R_{\text{T}} \) be the resistance of the thermistor at room temperature. The voltage across the fixed resistor is:
\[ V_{\text{R}} = 6.0\text{ V} - 2.0\text{ V} = 4.0\text{ V} \]

Using the potential divider relationship at room temperature:
\[ \frac{V_{\text{T}}}{V_{\text{R}}} = \frac{R_{\text{T}}}{R} \implies \frac{2.0\text{ V}}{4.0\text{ V}} = \frac{R_{\text{T}}}{4.0\text{ k}\Omega} \implies R_{\text{T}} = 2.0\text{ k}\Omega \]

When the temperature is increased, the resistance of the thermistor falls to \( R'_{\text{T}} = 1.0\text{ k}\Omega \).
Using the potential divider equation to find the new voltage across the thermistor:
\[ V'_{\text{T}} = \left( \frac{R'_{\text{T}}}{R + R'_{\text{T}}} \right) \times V_{\text{supply}} = \left( \frac{1.0\text{ k}\Omega}{4.0\text{ k}\Omega + 1.0\text{ k}\Omega} \right) \times 6.0\text{ V} = \frac{1}{5} \times 6.0\text{ V} = 1.2\text{ V} \]

1 mark for calculating the resistance of the thermistor at room temperature and then using the potential divider relation with the new resistance to find the correct voltmeter reading.

Let upwards be the positive vertical direction.

First, find the speed of the ball just before impact (\( v_1 \), moving downwards):
\[ v_1 = \sqrt{2 g h_1} = \sqrt{2 \times 9.81 \times 2.0} \approx 6.264\text{ m s}^{-1} \]
Thus, the velocity before impact is \( u = -6.264\text{ m s}^{-1} \).

Next, find the speed of the ball just after impact (\( v_2 \), moving upwards):
\[ v_2 = \sqrt{2 g h_2} = \sqrt{2 \times 9.81 \times 1.2} \approx 4.852\text{ m s}^{-1} \]
Thus, the velocity after impact is \( v = +4.852\text{ m s}^{-1} \).

The change in momentum during contact is:
\[ \Delta p = m(v - u) = 0.15 \times (4.852 - (-6.264)) = 0.15 \times 11.116 = 1.667\text{ kg m s}^{-1} \]

The average net force acting on the ball during contact is:
\[ F_{\text{net}} = \frac{\Delta p}{\Delta t} = \frac{1.667\text{ kg m s}^{-1}}{0.12\text{ s}} \approx 13.90\text{ N} \]

During contact, two forces act on the ball: the upward normal force from the floor \( F_{\text{floor}} \) and the downward force of gravity \( mg \):
\[ F_{\text{net}} = F_{\text{floor}} - mg \]
\[ F_{\text{floor}} = F_{\text{net}} + mg = 13.90\text{ N} + (0.15\text{ kg} \times 9.81\text{ m s}^{-2}) = 13.90\text{ N} + 1.47\text{ N} = 15.37\text{ N} \approx 15.4\text{ N} \]

1 mark for calculating the speeds before and after impact, determining the net force using the rate of change of momentum, and adding the weight of the ball to obtain the correct contact force.

Rearranging the formula for thermal conductivity \(k\) gives: \(k = \frac{P L}{A \Delta T}\). The SI base units for the quantities are: Power \(P\) (energy per unit time) in \(\text{kg m}^2\text{ s}^{-3}\), thickness \(L\) in \(\text{m}\), area \(A\) in \(\text{m}^2\), and temperature difference \(\Delta T\) in \(\text{K}\). Substituting these base units into the expression: \([k] = \frac{(\text{kg m}^2\text{ s}^{-3}) \cdot \text{m}}{\text{m}^2 \cdot \text{K}} = \text{kg m s}^{-3}\text{ K}^{-1}\).

Correct option is A (1 mark). Incorrect options: B represents units of heat capacity, C omits the length dependence, and D represents units of pressure per kelvin.

Let \(t_1\) be the acceleration time: \(t_1 = \frac{v}{0.50} = 2v\). Let \(t_3\) be the deceleration time: \(t_3 = \frac{v}{1.0} = v\). Let \(t_2\) be the constant speed time: \(t_2 = \frac{1600}{v}\). The total time is \(T = t_1 + t_2 + t_3 = 3v + \frac{1600}{v} = 140\). Multiplying by \(v\) gives: \(3v^2 - 140v + 1600 = 0\). Solving this quadratic equation: \(v = \frac{140 \pm \sqrt{140^2 - 4 \times 3 \times 1600}}{6} = \frac{140 \pm 20}{6}\), yielding \(v = 20\text{ m s}^{-1}\) or \(v \approx 26.7\text{ m s}^{-1}\). Thus, the correct option is B.

Correct option is B (1 mark). Correct set-up of the time equation and quadratic solution leads to 20 m/s.

The sand initially has zero horizontal velocity and is accelerated to the speed of the belt \(v = 1.5\text{ m s}^{-1}\). The force \(F\) required to accelerate the sand is: \(F = v \frac{\Delta m}{\Delta t} = 1.5\text{ m s}^{-1} \times 40\text{ kg s}^{-1} = 60\text{ N}\). By Newton's third law, the force on the belt is also \(60\text{ N}\). The rate of increase of kinetic energy of the sand is: \(\frac{\Delta E_k}{\Delta t} = \frac{1}{2} \frac{\Delta m}{\Delta t} v^2 = 0.5 \times 40\text{ kg s}^{-1} \times (1.5\text{ m s}^{-1})^2 = 45\text{ W}\).

Correct option is A (1 mark). Force is calculated using momentum change rate (60 N) and KE change rate is half of power input (45 W).

Let \(x = 0\) be the left end. Support \(X\) is at \(x = 0\), support \(Y\) is at \(x = 3.0\text{ m}\), and the weight of the beam acts at \(x = 2.0\text{ m}\). When the beam begins to tilt, \(R_X = 0\). Taking moments about \(Y\): \(300\text{ N} \times (3.0 - 2.0)\text{ m} = 600\text{ N} \times (d - 3.0)\text{ m}\), where \(d\) is the position of the person. This simplifies to \(300 = 600(d - 3.0)\), so \(d - 3.0 = 0.5\text{ m}\), which gives \(d = 3.5\text{ m}\).

Correct option is B (1 mark). Taking moments about Y with normal reaction at X equal to zero yields a distance of 3.5 m.

First, find the mass flow rate of the water: \(\frac{\Delta m}{\Delta t} = \rho A v = 1000 \times (2.5 \times 10^{-3}) \times 6.0 = 15\text{ kg s}^{-1}\). The useful power is the sum of the rate of gain of potential energy and the rate of gain of kinetic energy: \(P = \left(\frac{\Delta m}{\Delta t}\right) g h + \frac{1}{2} \left(\frac{\Delta m}{\Delta t}\right) v^2 = 15 \times 9.81 \times 5.0 + 0.5 \times 15 \times 6.0^2 = 735.75 + 270 = 1005.75\text{ W} \approx 1000\text{ W}\).

Correct option is C (1 mark). Calculation must include both potential energy (736 W) and kinetic energy (270 W) contributions.

Using the formula for extension \(x = \frac{F L}{A E}\) and noting that area \(A \propto d^2\), we find that \(x \propto \frac{F L}{d^2 E}\). Since both wires experience the same tension \(F\) and have the same initial length \(L\), \(x \propto \frac{1}{d^2 E}\). Thus, \(x_X \propto \frac{1}{d^2 E}\) and \(x_Y \propto \frac{1}{(2d)^2 (2E)} = \frac{1}{8 d^2 E}\). The ratio is \(\frac{x_X}{x_Y} = 8\).

Correct option is D (1 mark). The ratio of extensions is inversely proportional to the product of area (which scales as diameter squared) and Young modulus, yielding a factor of 8.

Using the Doppler shift formula \(f_o = f_s \left( \frac{v}{v \pm v_s} \right)\): 1. When approaching: \(f_{\text{approach}} = 450 \times \left(\frac{330}{330 - 30}\right) = 495\text{ Hz}\). 2. When moving away: \(f_{\text{recede}} = 450 \times \left(\frac{330}{330 + 30}\right) = 412.5\text{ Hz}\). The change in frequency is \(\Delta f = f_{\text{approach}} - f_{\text{recede}} = 495 - 412.5 = 82.5\text{ Hz} \approx 83\text{ Hz}\).

Correct option is C (1 mark). Both source-approaching and source-receding Doppler equations must be correctly applied to find the net change.

The volume of a cylinder is \(V = A L = \frac{\pi D^2 L}{4}\). Since the volume remains constant, doubling the diameter increases the cross-sectional area \(A\) by a factor of 4. Therefore, the length \(L\) must decrease by a factor of 4 to keep volume constant: \(L_{\text{new}} = \frac{L}{4}\). Using the formula \(R = \rho \frac{L}{A}\), the new resistance is \(R_{\text{new}} = \rho \frac{L / 4}{4 A} = \frac{R}{16}\).

Correct option is A (1 mark). Conservation of volume dictates that the length decreases by 4 as area increases by 4, causing a 16-fold decrease in resistance.

The SI base units of the individual variables are:
- Frequency \(f\): \(\text{s}^{-1}\)
- Density \(\rho\): \(\text{kg}\,\text{m}^{-3}\)
- Speed \(v\): \(\text{m}\,\text{s}^{-1}\)
- Amplitude \(A\): \(\text{m}\)

Substituting these into the formula for intensity:
\([I] = [f]^2 [\rho] [v] [A]^2 = (\text{s}^{-1})^2 \times (\text{kg}\,\text{m}^{-3}) \times (\text{m}\,\text{s}^{-1}) \times (\text{m})^2\)
\([I] = \text{s}^{-2} \times \text{kg}\,\text{m}^{-3} \times \text{m}\,\text{s}^{-1} \times \text{m}^2 = \text{kg}\,\text{s}^{-3}\)

This matches option A.

1 mark for the correct SI base unit determination.

The Young modulus \(E\) of the material is given by:
\(E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{x/L} = \frac{FL}{Ax}\)

Rearranging for the extension \(x\):
\(x = \frac{FL}{AE}\)

For the second wire, the extension \(x_2\) is:
\(x_2 = \frac{F_2 L_2}{A_2 E} = \frac{(2F)(2L)}{(2A)E} = 2 \left(\frac{FL}{AE}\right) = 2x\)

Therefore, the extension of the second wire is \(2x\).

1 mark for the correct calculation of extension.

First, calculate the wavelength \(\lambda\) of the wave:
\(\lambda = \frac{v}{f} = \frac{36}{120} = 0.30\text{ m} = 30\text{ cm}\)

The phase difference \(\phi\) in radians for a path difference of \(x = 10\text{ cm}\) is:
\(\phi = \frac{x}{\lambda} \times 2\pi = \frac{10}{30} \times 2\pi = \frac{2\pi}{3}\text{ rad}\)

Therefore, the phase difference is \(\frac{2\pi}{3}\text{ rad}\).

1 mark for the correct phase difference calculation in radians.

The fringe separation is given by \(x = \frac{\lambda D}{a}\). Since the wavelength \(\lambda\), slit separation \(a\), and screen distance \(D\) remain unchanged, the fringe separation is unchanged.

When the intensity of light from one of the slits is reduced, the amplitude of light from that slit decreases. This means at the minima (dark fringes), complete destructive interference no longer occurs, making them brighter than before. At the maxima (bright fringes), the total amplitude is lower, making them less bright. As a result, the contrast between the bright and dark fringes decreases. Therefore, option B is correct.

1 mark for identifying that fringe separation remains unchanged and contrast decreases.

The current \(I\) is related to the drift velocity \(v\) by:
\(I = n A v e\)

Rearranging for drift velocity:
\(v = \frac{I}{n A e}\)

For the second wire, the diameter is twice as large, so the cross-sectional area \(A_2\) is:
\(A_2 = \pi \left(\frac{2d}{2}\right)^2 = 4 A_1 = 4A\)

Since the material is still copper, the number density \(n\) is the same. The new drift velocity \(v_2\) is:
\(v_2 = \frac{2I}{n (4A) e} = \frac{1}{2} \left(\frac{I}{n A e}\right) = 0.5v\)

This matches option B.

1 mark for the correct calculation of drift velocity.

In the dark, the potential difference across the LDR is:
\(V_{\text{dark}} = 12 \times \frac{9.0}{3.0 + 9.0} = 9.0\text{ V}\)

In the light, the potential difference across the LDR is:
\(V_{\text{light}} = 12 \times \frac{1.0}{3.0 + 1.0} = 3.0\text{ V}\)

The decrease in potential difference is:
\(\Delta V = V_{\text{dark}} - V_{\text{light}} = 9.0\text{ V} - 3.0\text{ V} = 6.0\text{ V}\)

This matches option C.

1 mark for the correct calculation of the decrease in potential difference.

The up quark has a charge of \(+\frac{2}{3}e\) and the down quark has a charge of \(-\frac{1}{3}e\).

In \(\beta^{-}\) decay, a down quark transforms into an up quark:
\(d \rightarrow u + W^{-}\)

where the \(W^{-}\) boson is the mediating exchange particle which then decays into an electron and an electron antineutrino. This corresponds to the row in Option A.

1 mark for correctly identifying quark charges and the exchange particle.

Taking the direction of initial velocity as positive:
Initial velocity \(u = +20\text{ m}\,\text{s}^{-1}\)
Final velocity \(v = -15\text{ m}\,\text{s}^{-1}\)

According to Newton's second law, the average force \(F\) is the rate of change of momentum:
\(F = \frac{\Delta p}{\Delta t} = \frac{m(v - u)}{\Delta t}\)
\(F = \frac{0.15 \times (-15 - 20)}{0.050} = \frac{0.15 \times (-35)}{0.050} = -105\text{ N}\)

The magnitude of the average force exerted by the wall on the ball is \(105\text{ N}\). This matches option C.

1 mark for calculating the correct average force magnitude.

(a) Ultimate tensile stress is the maximum tensile stress (force per unit cross-sectional area) that a material can withstand before breaking.

(b)(i)
Using the Young modulus formula:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta L / L}\)
\(\Delta L = \frac{F L}{A E}\)
\(\Delta L = \frac{120 \times 3.2}{(1.5 \times 10^{-6}) \times (2.0 \times 10^{11})}\)
\(\Delta L = \frac{384}{3.0 \times 10^5} = 1.28 \times 10^{-3}\text{ m}\) or \(1.3\text{ mm}\).

(b)(ii)
Using the formula for elastic potential energy:
\(E_p = \frac{1}{2} F \Delta L\)
\(E_p = 0.5 \times 120 \times 1.28 \times 10^{-3} = 0.0768\text{ J} \approx 0.077\text{ J}\).

(c)(i)
Plastic deformation is a type of deformation where the material does not return to its original length or shape when the applied load is removed.

(c)(ii)
The graph should show:
- A loading curve starting at the origin, showing a straight line followed by a curve of decreasing gradient.
- An unloading straight line starting from the end of the loading curve, parallel to the initial linear portion of the loading curve, meeting the horizontal axis at an extension greater than zero.

(a) [1] Maximum tensile force per unit cross-sectional area that a material can withstand before fracture.

(b)(i)
- [1] State or use formula: \(\Delta L = \frac{F L}{A E}\)
- [1] Correct substitution of values: \(\frac{120 \times 3.2}{1.5 \times 10^{-6} \times 2.0 \times 10^{11}}\)
- [1] Correct calculation to 2 or 3 s.f.: \(1.3 \times 10^{-3}\text{ m}\) (or \(1.28 \times 10^{-3}\text{ m}\))

(b)(ii)
- [1] State or use formula: \(E_p = \frac{1}{2} F \Delta L\) or \(E_p = \frac{1}{2} k x^2\)
- [1] Correct calculation: \(0.077\text{ J}\) (or \(0.0768\text{ J}\), allow ECF from b(i))

(c)(i)
- [1] Definition mentioning permanent deformation / does not return to original length when load is removed.

(c)(ii)
- [1] Loading curve with an initial straight section followed by a curve to the right.
- [1] Unloading line is straight, parallel to the initial linear loading section, and terminates at a non-zero value of extension.

(a)(i) A proton has a quark composition of \(uud\) (up, up, down).
(a)(ii) A neutron has a quark composition of \(udd\) (up, down, down).

(b)(i) A down quark (\(d\)) decays/changes into an up quark (\(u\)).
(b)(ii) The weak interaction (or weak nuclear force).

(b)(iii)
- Charge conservation:
Left-hand side (LHS) charge = \(0\) (neutron is neutral).
Right-hand side (RHS) charge = \(+1\text{ (proton)} + (-1)\text{ (electron)} + 0\text{ (antineutrino)} = 0\).
Since LHS = RHS, charge is conserved.
- Lepton number conservation:
LHS lepton number = \(0\) (neutron is a baryon).
RHS lepton number = \(0\text{ (proton)} + (+1)\text{ (electron)} + (-1)\text{ (antineutrino)} = 0\).
Since LHS = RHS, lepton number is conserved.

(c)
Protons and neutrons belong to the class of **hadrons** (or **baryons**).
Electrons and neutrinos belong to the class of **leptons**.

(a)(i) [1] \(uud\) (or up, up, down)
(a)(ii) [1] \(udd\) (or up, down, down)

(b)(i) [1] \(d \rightarrow u\) (or down to up)
(b)(ii) [1] Weak (nuclear) force / weak interaction

(b)(iii)
- [1] Charge: shows \(0 = +1 - 1 + 0\) (or equivalent statement explaining conservation of charges)
- [1] Lepton number: shows \(0 = 0 + 1 - 1\) (or equivalent statement explaining conservation of lepton numbers)

(c)
- [1] Hadrons (or baryons)
- [1] Leptons

(a) Electromotive force (e.m.f.) is the energy transferred by a source of electrical energy per unit charge (or total work done per unit charge around a complete circuit).

(b)(i) The terminal potential difference is given by:
\(V = E - I r\)

(b)(ii)
From \(E = I(R + r)\):
For \(R = 6.0\,\Omega\):
\(E = 1.5(6.0 + r)\) [Equation 1]
For \(R = 14.0\,\Omega\):
\(E = 0.75(14.0 + r)\) [Equation 2]

Equating the two expressions for \(E\):
\(1.5(6.0 + r) = 0.75(14.0 + r)\)

Divide both sides by \(0.75\):
\(2(6.0 + r) = 14.0 + r\)
\(12.0 + 2r = 14.0 + r\)
\(r = 2.0\,\Omega\).

Now substitute \(r = 2.0\,\Omega\) back into Equation 1:
\(E = 1.5(6.0 + 2.0) = 1.5 \times 8.0 = 12.0\text{ V}\).

(c) Power dissipated in the internal resistance is given by:
\(P = I^2 r\)
For \(I = 1.5\text{ A}\):
\(P = (1.5)^2 \times 2.0 = 2.25 \times 2.0 = 4.5\text{ W}\).

(a) [1] Energy transferred from other forms to electrical energy per unit charge.

(b)(i) [1] \(V = E - Ir\) (or \(E = I(R+r)\) if related to terminal p.d.)

(b)(ii)
1.
- [1] State \(E = I(R + r)\) and set up two simultaneous equations: \(E = 1.5(6.0 + r)\) and \(E = 0.75(14.0 + r)\)
- [1] Show appropriate algebraic steps to solve for \(r\)
- [1] Correct value: \(r = 2.0\,\Omega\)
2.
- [1] Substitute \(r\) back into either equation: e.g., \(E = 1.5(6.0 + 2.0)\)
- [1] Correct value: \(E = 12\text{ V}\)

(c)
- [1] State or use formula: \(P = I^2 r\)
- [1] Correct calculation: \(P = 1.5^2 \times 2.0 = 4.5\text{ W}\) (allow ECF from b(ii))

(a) Work done is the product of the force and the displacement in the direction of the force.

(b)(i)
Gain in gravitational potential energy:
\(\Delta E_p = m g h\)
\(\Delta E_p = 15 \times 9.81 \times 4.5 = 662.2\text{ J}\)
This is approximately \(660\text{ J}\).

(b)(ii)
Kinetic energy:
\(E_k = \frac{1}{2} m v^2\)
\(E_k = 0.5 \times 15 \times (1.2)^2 = 0.5 \times 15 \times 1.44 = 10.8\text{ J}\).

(b)(iii)
Total useful energy given to the box:
\(E_{\text{useful}} = \Delta E_p + E_k = 662.2 + 10.8 = 673.0\text{ J}\)

Total electrical energy input in \(15\text{ s}\):
\(E_{\text{input}} = P_{\text{input}} \times t = 350 \times 15 = 5250\text{ J}\)

Efficiency:
\(\eta = \frac{E_{\text{useful}}}{E_{\text{input}}} \times 100\% = \frac{673.0}{5250} \times 100\% = 12.8\%\) (or \(13\%\)).

(a) [1] Force \(\times\) displacement in the direction of the force.

(b)(i)
- [1] State or use formula: \(E_p = mgh\)
- [1] Correct substitution: \(15 \times 9.81 \times 4.5 = 662\text{ J}\) (showing it rounds to \(660\text{ J}\))

(b)(ii)
- [1] State or use formula: \(E_k = \frac{1}{2} m v^2\)
- [1] Correct calculation: \(10.8\text{ J}\) (or \(11\text{ J}\))

(b)(iii)
- [1] Calculate total useful work done on the box: \(662 + 10.8 = 673\text{ J}\) (or \(660 + 10.8 = 671\text{ J}\))
- [1] Calculate energy input: \(350 \times 15 = 5250\text{ J}\)
- [1] Correct efficiency calculation: \(12.8\%\) or \(13\%\) (or \(0.13\))

(a)(i) Displacement is the distance of a particle from its equilibrium (rest) position in a specified direction.
(a)(ii) Phase difference is the difference in position of two points on a wave (or different waves) expressed as an angle (in degrees or radians).

(b)(i)
Using the wave equation:
\(v = f \lambda \Rightarrow \lambda = \frac{v}{f}\)
\[\lambda = \frac{340}{850} = 0.40\text{ m}\.\]

(b)(ii)
A phase difference of \(360^\circ\) corresponds to a distance of one wavelength \(\lambda = 0.40\text{ m}\).
For a phase difference of \(60^\circ\):
\[\text{Distance } x = \frac{60^\circ}{360^\circ} \times \lambda = \frac{1}{6} \times 0.40\text{ m} = 0.0667\text{ m}\) (or \(6.7\text{ cm}\)).

(c)
Intensity is directly proportional to the square of the amplitude:
\(I \propto A^2\)
\[\frac{I_{\text{new}}}{I_0} = \left(\frac{2.5 A_0}{A_0}\right)^2 = 2.5^2 = 6.25\]
\[I_{\text{new}} = 6.25 I_0\text{ (or } 6.3 I_0\text{)}\.\]

(a)(i) [1] Distance of a point/particle from its equilibrium/mean position in a given direction.
(a)(ii) [1] Angle representing the fraction of a cycle by which one point/wave leads or lags another.

(b)(i)
- [1] State or use formula: \(v = f \lambda\)
- [1] Correct calculation: \(0.40\text{ m}\) (or \(0.4\text{ m}\))

(b)(ii)
- [1] State that \(360^\circ\) corresponds to \(\lambda\) (or \(2\pi\) corresponds to \(\lambda\))
- [1] Substitute values: \(x = \frac{60}{360} \times 0.40\)
- [1] Correct calculation: \(0.067\text{ m}\) (or \(0.0667\text{ m}\), accept ECF from b(i))

(c)
- [1] State or use relation: \(I \propto A^2\)
- [1] Correct calculation: \(6.25 I_0\) (or \(6.3 I_0\))

(a) Acceleration is the rate of change of velocity.

(b)(i)
Consider the vertical motion, taking downwards as positive:
\(s_y = u_y t + \frac{1}{2} g t^2\)
Since the ball is kicked horizontally, the initial vertical velocity is zero (\(u_y = 0\)).
\(45 = 0 + 0.5 \times 9.81 \times t^2\)
\(t^2 = \frac{45}{4.905} = 9.174\)
\(t = 3.029\text{ s}\), which is approximately \(3.0\text{ s}\).

(b)(ii)
Consider the horizontal motion:
\(s_x = v_x \times t\)
\(36 = v_x \times 3.029\)
\(v_x = \frac{36}{3.029} = 11.88\text{ m s}^{-1} \approx 12\text{ m s}^{-1}\) (using \(t = 3.0\text{ s}\) gives exactly \(12\text{ m s}^{-1}\)).

(b)(iii)
The horizontal component of velocity remains constant:
\(v_x = 11.9\text{ m s}^{-1}\)

The vertical component of velocity just before impact is:
\(v_y = u_y + g t = 0 + (9.81 \times 3.029) = 29.7\text{ m s}^{-1}\)
(or \(v_y = \sqrt{2 g s_y} = \sqrt{2 \times 9.81 \times 45} = 29.7\text{ m s}^{-1}\))

The magnitude of the resultant velocity is:
\(v = \sqrt{v_x^2 + v_y^2}\)
\(v = \sqrt{11.9^2 + 29.7^2} = \sqrt{141.6 + 882.1} = \sqrt{1023.7} = 32.0\text{ m s}^{-1}\) (or \(32\text{ m s}^{-1}\)).

(a) [1] Change in velocity per unit time (or rate of change of velocity).

(b)(i)
- [1] State or use formula: \(s_y = \frac{1}{2} g t^2\)
- [1] Correct substitution: \(45 = 0.5 \times 9.81 \times t^2\) leading to \(t = 3.03\text{ s}\) (must show at least 3 s.f. to justify the 'show' instruction)

(b)(ii)
- [1] State or use formula: \(s_x = v_x t\)
- [1] Correct calculation: \(12\text{ m s}^{-1}\) (or \(11.9\text{ m s}^{-1}\))

(b)(iii)
- [1] Calculate vertical component of velocity: \(29.7\text{ m s}^{-1}\) (or \(30\text{ m s}^{-1}\))
- [1] State or use Pythagoras' theorem: \(v = \sqrt{v_x^2 + v_y^2}\)
- [1] Correct calculation of resultant magnitude: \(32\text{ m s}^{-1}\) (accept \(31.9\text{ m s}^{-1}\) to \(32.1\text{ m s}^{-1}\))

(a) Archimedes' principle states that the upthrust acting on an object immersed in a fluid is equal to the weight of the fluid displaced by the object.

(b)(i)
The volume of water displaced by the submerged part of the cylinder is:
\(V_{\text{displaced}} = A \times h_{\text{submerged}}\)
\(V_{\text{displaced}} = (4.0 \times 10^{-3}) \times 0.22 = 8.8 \times 10^{-4}\text{ m}^3\)

Upthrust \(U\) is equal to the weight of this displaced water:
\(U = \rho_{\text{water}} V_{\text{displaced}} g\)
\(U = (1.0 \times 10^3) \times (8.8 \times 10^{-4}) \times 9.81\)
\(U = 0.88 \times 9.81 = 8.63\text{ N} \approx 8.6\text{ N}\).

(b)(ii)
Since the cylinder is floating in equilibrium, the upthrust equals its total weight:
\(W = U = 8.63\text{ N}\)

Mass of the cylinder:
\(m = \frac{W}{g} = \frac{8.63}{9.81} = 0.88\text{ kg}\).

(b)(iii)
The total volume of the cylinder is:
\(V_{\text{total}} = A \times h_{\text{total}}\)
\(V_{\text{total}} = (4.0 \times 10^{-3}) \times 0.35 = 1.4 \times 10^{-3}\text{ m}^3\)

Density of the wood:
\(\rho_{\text{wood}} = \frac{m}{V_{\text{total}}} = \frac{0.88}{1.4 \times 10^{-3}} = 628.6\text{ kg m}^{-3} \approx 630\text{ kg m}^{-3}\).

(Alternatively:
\(\rho_{\text{wood}} = \rho_{\text{water}} \times \frac{h_{\text{submerged}}}{h_{\text{total}}} = 1000 \times \frac{0.22}{0.35} = 628.6\text{ kg m}^{-3}\).)

(a) [1] Upthrust (on an object in fluid) equals weight of the fluid displaced.

(b)(i)
- [1] State or use formula: \(V_{\text{displaced}} = A \times h_{\text{submerged}}\)
- [1] Calculate volume: \(8.8 \times 10^{-4}\text{ m}^3\)
- [1] Calculate upthrust: \(8.6\text{ N}\) (or \(8.63\text{ N}\))

(b)(ii)
- [1] State that upthrust equals weight (for a floating object)
- [1] Correct calculation of mass: \(0.88\text{ kg}\)

(b)(iii)
- [1] Calculate total volume: \(1.4 \times 10^{-3}\text{ m}^3\) (or state ratio relation)
- [1] State or use formula: \(\rho = \frac{m}{V}\) (or \(\rho_{\text{wood}} = \rho_{\text{water}} \times \frac{h_{\text{sub}}}{h_{\text{total}}}\))
- [1] Correct calculation of density: \(630\text{ kg m}^{-3}\) (or \(629\text{ kg m}^{-3}\), accept ECF)

In this investigation, taking moments about the pivot (the \(5.0\text{ cm}\) mark) gives the equilibrium condition:
\[ T_{\text{pivot}} = F_{\text{spring}} L_0 - M g d - m_{\text{rule}} g d_{\text{cg}} = 0 \]
where:
- \(L_0 = 0.90\text{ m}\) is the distance from the pivot to the spring suspension point,
- \(M = 0.200\text{ kg}\) is the suspended mass,
- \(d\) is the distance of the mass from the pivot,
- \(m_{\text{rule}}\) is the mass of the metre rule.

Since \(F_{\text{spring}} = k(y_0 - h)\), where \(h\) is the height of the end of the rule and \(y_0\) is a constant base height, substituting this into the torque equation and rearranging for \(h\) yields a linear relationship:
\[ h = \left( -\frac{M g}{k L_0} \right) d + C \]
Comparing this to \(h = P d + Q\):
- The gradient \(P = -\frac{M g}{k L_0}\). With \(M = 0.200\text{ kg}\), \(g = 9.81\text{ m s}^{-2}\), \(k \approx 25\text{ N m}^{-1}\), and \(L_0 = 0.90\text{ m}\), \(P \approx -0.087\).
- The y-intercept \(Q\) is the height \(h\) when the mass is located exactly at the pivot (\(d = 0\)).

To find \(P\) and \(Q\) from the graph:
1. Choose two distant coordinates on the line of best fit: \((d_1, h_1)\) and \((d_2, h_2)\).
2. Calculate the gradient:
\[ P = \frac{h_2 - h_1}{d_2 - d_1} \]
3. Determine the y-intercept \(Q\) by reading it directly from the y-axis (if \(d = 0\) is on the scale) or calculating:
\[ Q = h_1 - P d_1 \]

**Data Collection (6 Marks)**
- [1] Successful collection of 6 sets of readings of \(d\) and \(h\) without teacher assistance.
- [1] Range of \(d\) spans at least \(0.500\text{ m}\).
- [1] Table column headings must contain the quantity and appropriate unit (e.g., \(d / \text{m}\), \(h / \text{m}\)).
- [1] Raw values of \(d\) and \(h\) are recorded to the nearest millimetre (\(0.001\text{ m}\)).
- [1] All raw values of \(h\) in the table are consistent in decimal places.
- [1] Trend of results is correct (\(h\) decreases as \(d\) increases).

**Graphical Presentation (5 Marks)**
- [1] Linear scales chosen so that the plotted points occupy more than half of the graph grid in both directions. No awkward scales (such as 3:10).
- [1] Correctly plotted points: coordinates must be accurate to within half a small square.
- [1] A best-fit straight line is drawn, showing a balanced distribution of points on either side of the line.
- [1] Plotting precision: at least 5 points plotted on the grid.
- [1] Quality mark: all data points lie within \(0.01\text{ m}\) of the best-fit line.

**Gradient and Intercept (2 Marks)**
- [1] Gradient calculation: the hypotenuse of the triangle used must be greater than half the length of the drawn line of best fit.
- [1] Y-intercept calculated correctly using a point on the line, or read directly from the y-axis if the scale starts at \(d=0\).

**Constants P and Q (3 Marks)**
- [1] Value of \(P\) set equal to the gradient, and \(Q\) set equal to the y-intercept.
- [1] Correct units provided: \(P\) is dimensionless (or \(\text{m m}^{-1}\)), and \(Q\) has unit \(\text{m}\).
- [1] Numerical values are within reasonable experimental ranges (\(P\) between \(-0.05\) and \(-0.15\); \(Q\) between \(0.30\text{ m}\) and \(0.45\text{ m}\)).

**Experimental Quality and Method (4 Marks)**
- [1] Height \(h_0\) of the unloaded rule recorded with the correct unit.
- [1] First reading of \(h\) for \(d = 0.100\text{ m}\) is within \(0.05\text{ m}\) of \(h_0\).
- [2] Repeated measurements of \(h\) performed at each position and average calculated.

Let's work through an example set of readings:

1. **For the first mass** (\(M = 100\text{ g}\)):
- Time for 10 oscillations, \(t_1 = 2.48\text{ s}\) and \(t_2 = 2.52\text{ s}\).
- Average time \(t_{\text{avg}} = 2.50\text{ s}\).
- Period \(T_1 = \frac{2.50}{10} = 0.250\text{ s}\).

2. **Percentage Uncertainty**:
- If using absolute uncertainty from human reaction time \(\Delta t = 0.2\text{ s}\):
\[ \\% \text{ uncertainty} = \frac{\Delta t}{t_{\text{avg}}} \times 100\\% = \frac{0.2}{2.50} \times 100\\% = 8.0\\% \]

3. **For the second mass** (\(M = 150\text{ g}\)):
- Time for 10 oscillations, \(t_{\text{avg}} = 3.10\text{ s}\).
- Period \(T_2 = 0.310\text{ s}\).

4. **Constant Calculation**:
- \(k_1 = \frac{T_1^2}{M_1} = \frac{0.250^2}{100} = 6.25 \times 10^{-4}\text{ s}^2\text{ g}^{-1}\]
- \)k_2 = \\frac{T_2^2}{M_2} = \\frac{0.310^2}{150} = 6.41 \\times 10^{-4}\\text{ s}^2\\text{ g}^{-1}\]

5. **Comparing values**:
- Percentage difference \(= \frac{|6.41 - 6.25|}{6.25} \times 100\\% \approx 2.56\\%\).
- Since \(2.56\\% \le 10\\%\), the results support the suggested relationship.

**Measurement of T (2 Marks)**
- [1] Raw times recorded to \(0.01\text{ s}\) with evidence of repetition.
- [1] Period \(T_1\) calculated correctly with unit (\(\text{s}\)) in the range \(0.15\text{ s} \le T_1 \le 0.40\text{ s}\).

**Percentage Uncertainty (2 Marks)**
- [1] Correct formula used: \(\frac{\text{uncertainty in } t}{t} \times 100\\%\). Absolute uncertainty in range \(0.1\text{ s} \le \Delta t \le 0.3\text{ s}\) (reaction time) or half the range of repeated values.
- [1] Calculation carried out correctly to 1 or 2 significant figures.

**Second Mass Readings (2 Marks)**
- [1] Repeats for second time \(t_2\) recorded.
- [1] Quality: \(T_2 > T_1\).

**Calculation of Constant k (2 Marks)**
- [1] Values of \(k_1\) and \(k_2\) calculated correctly with units (e.g. \(\text{s}^2\text{ g}^{-1}\)).
- [1] Number of significant figures matches the raw timing data (typically 2 or 3 sig figs).

**Analysis and Support (4 Marks)**
- [1] Correct calculation of the percentage difference between the two \(k\) values.
- [1] Clear stated criterion (e.g., "I will use 10% as the limit of experimental uncertainty").
- [2] Logical conclusion linking the percentage difference to the criterion (e.g., "Since the percentage difference of 2.6% is less than the 10% limit, the relationship is supported").

**Limitations and Improvements (8 Marks)**
- [2] **Limitation 1:** Mass slips or shifts on the plastic ruler during rapid oscillations. / **Improvement 1:** Use a secure mechanical clamp or strong double-sided tape to lock the mass to the ruler.
- [2] **Limitation 2:** The period of oscillation is too fast, making manual counting of 10 cycles difficult. / **Improvement 2:** Record the motion using a high frame-rate video camera against a grid, then analyze in slow motion.
- [2] **Limitation 3:** Oscillations decay/dampen too quickly to obtain a reliable count of 10 cycles. / **Improvement 3:** Use a stiffer rule or a motion sensor / light gate to capture the period from fewer cycles with high accuracy.
- [2] **Limitation 4:** The ruler undergoes horizontal wobbling or torsional twisting instead of pure vertical vibration. / **Improvement 4:** Implement a mechanical release mechanism (e.g. electromagnetic release) to guarantee purely vertical initial displacement.