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

The equation must be homogeneous, so the units on both sides of the equation must be identical.

The SI base units of force \( F_D \) are \( \text{kg m s}^{-2} \).

Therefore, the units of the term \( \alpha v^3 \) must also be \( \text{kg m s}^{-2} \):

\( [\alpha v^3] = [\alpha] \times (\text{m s}^{-1})^3 = \text{kg m s}^{-2} \)

\( [\alpha] \times \text{m}^3 \text{s}^{-3} = \text{kg m s}^{-2} \)

\( [\alpha] = \frac{\text{kg m s}^{-2}}{\text{m}^3 \text{s}^{-3}} = \text{kg m}^{-2} \text{s} \).

1 mark for the correct option.
- Award 1 mark for calculating the correct SI base units of force and equating it to the units of the term involving \( \alpha \).

Let the ground be the reference point \( y = 0 \).

For ball \( P \) dropped from height \( H \), its height after time \( t \) is:
\( y_P = H - \frac{1}{2}gt^2 \)

For ball \( Q \) projected upwards from the ground, its height after time \( t \) is:
\( y_Q = ut - \frac{1}{2}gt^2 \)

They collide at height \( y = \frac{H}{3} \). Therefore:
\( y_P = \frac{H}{3} \implies H - \frac{1}{2}gt^2 = \frac{H}{3} \implies \frac{1}{2}gt^2 = \frac{2}{3}H \implies t = \sqrt{\frac{4H}{3g}} \)

At the same instant, for ball \( Q \):
\( y_Q = \frac{H}{3} \implies ut - \frac{1}{2}gt^2 = \frac{H}{3} \)

Substituting \( \frac{1}{2}gt^2 = \frac{2}{3}H \) into this equation:
\( ut - \frac{2}{3}H = \frac{H}{3} \implies ut = H \implies u = \frac{H}{t} \)

Now, substitute \( t = \sqrt{\frac{4H}{3g}} \):
\( u = \frac{H}{\sqrt{\frac{4H}{3g}}} = \sqrt{\frac{3gH}{4}} \).

1 mark for the correct option.
- Award 1 mark for setting up equations of motion for both balls, solving for the time of collision, and using it to express the initial velocity of ball Q.

Let the left end of the plank be at position \( z = 0 \).
- Rope A is at \( z = 0 \).
- Rope B is at \( z = 3.0\text{ m} \) (since it is \( 1.0\text{ m} \) from the right end of the \( 4.0\text{ m} \) plank).
- The weight of the uniform plank (\( 150\text{ N} \)) acts at its center of gravity, \( z = 2.0\text{ m} \).
- The person (\( 600\text{ N} \)) is at \( z = x \).

Rope A is on the verge of going slack when the tension \( T_A = 0 \). At this limit, we can take moments about the contact point of Rope B (\( z = 3.0\text{ m} \)):

Taking moments about \( z = 3.0\text{ m} \):
- The plank's weight acts at \( z = 2.0\text{ m} \), which is \( 1.0\text{ m} \) to the left of B. This creates an anticlockwise moment of:
\( 150\text{ N} \times 1.0\text{ m} = 150\text{ N m} \).
- The person at \( z = x \) (where \( x > 3.0\text{ m} \)) is at a distance of \( (x - 3.0)\text{ m} \) to the right of B, creating a clockwise moment of:
\( 600\text{ N} \times (x - 3.0)\text{ m} \).

Equating the clockwise and anticlockwise moments about B:
\( 600(x - 3.0) = 150 \times 1.0 \)
\( x - 3.0 = \frac{150}{600} = 0.25 \)
\( x = 3.25\text{ m} \).

If the person moves beyond \( 3.25\text{ m} \), the clockwise moment exceeds the anticlockwise moment, causing Rope A to go slack (the plank would tip).

1 mark for the correct option.
- Award 1 mark for taking moments about the position of Rope B at the limiting condition where the tension in Rope A is zero.

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

Rearranging this for extension \( \Delta L \):
\( \Delta L = \frac{FL}{EA} \)

Since \( A = \frac{\pi d^2}{4} \), we have \( A \propto d^2 \), which gives:
\( \Delta L \propto \frac{L}{E d^2} \)

For wire X:
\( \Delta L_X \propto \frac{L}{E d^2} \)

For wire Y:
\( \Delta L_Y \propto \frac{2L}{(2E)(2d)^2} = \frac{2L}{8 E d^2} = \frac{1}{4}\left(\frac{L}{E d^2}\right) \)

Taking the ratio:
\( \frac{\Delta L_X}{\Delta L_Y} = \frac{1}{1/4} = 4 \).

1 mark for the correct option.
- Award 1 mark for expressing the extension in terms of length, Young modulus, and diameter, and correctly determining the ratio.

Let the resistance of the original wire of length \( 1.5\text{ m} \) be \( R \).

The power dissipated initially is:
\( P = \frac{V^2}{R} = 36\text{ W} \)

When the wire is cut into three equal pieces, the resistance of each piece is proportional to its length. Therefore, each piece has resistance:
\( R_{\text{piece}} = \frac{R}{3} \)

These three pieces are connected in parallel across the same potential difference \( V = 12\text{ V} \).

The power dissipated by a single piece is:
\( P_{\text{piece}} = \frac{V^2}{R_{\text{piece}}} = \frac{V^2}{R/3} = 3 \left(\frac{V^2}{R}\right) = 3 \times 36\text{ W} = 108\text{ W} \)

Since there are three pieces in parallel, the total power dissipated is:
\( P_{\text{total}} = 3 \times P_{\text{piece}} = 3 \times 108\text{ W} = 324\text{ W} \).

1 mark for the correct option.
- Award 1 mark for showing that cutting the wire divides resistance by 3, and parallel connection leads to 9 times the original power.

The initial baryon has the quark composition \( \text{uud} \) (a proton).
One \( \text{u} \) quark changes into a \( \text{d} \) quark, changing the baryon to \( \text{udd} \) (a neutron).

This quark change is:
\( \text{u} \rightarrow \text{d} + \text{e}^+ + \nu_{\text{e}} \)

This represents \( \beta^+ \) decay, which emits a positron (\( \beta^+ \) particle) and an electron neutrino.

1 mark for the correct option.
- Award 1 mark for identifying the decay as beta-plus and listing the positron and electron neutrino as the emitted leptons.

Let the initial amplitude of the light wave from each slit be \( A \).
Since intensity is proportional to amplitude squared (\( I \propto A^2 \)), the initial maximum amplitude from constructive interference is:
\( A_{\text{max}} = A + A = 2A \)
Therefore, the initial maximum intensity is:
\( I_0 \propto (2A)^2 = 4A^2 \)

When the intensity of one slit is reduced to 25%, its new intensity is \( 0.25 I \). Its new amplitude is:
\( A_1 = \sqrt{0.25} A = 0.5A \)

The amplitude of the other slit is unchanged: \( A_2 = A \).

For the new bright fringes (maximum intensity), the amplitudes interfere constructively:
\( A_{\text{max}}' = A_1 + A_2 = 0.5A + A = 1.5A \)

The new maximum intensity is:
\( I_{\text{max}}' \propto (1.5A)^2 = 2.25A^2 \)

Comparing this to \( I_0 \):
\( \frac{I_{\text{max}}'}{I_0} = \frac{2.25A^2}{4A^2} = 0.5625 \approx 0.56 \).

1 mark for the correct option.
- Award 1 mark for relating amplitude to the square root of intensity, summing the amplitudes for constructive interference, and squaring the result to find the new intensity relative to \( I_0 \).

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

Given \( \phi = \frac{3\pi}{5}\text{ rad} \) and \( \Delta x = 0.18\text{ m} \):
\( \frac{3\pi}{5} = \frac{2\pi \times 0.18}{\lambda} \)

Simplifying this equation:
\( \frac{3}{5} = \frac{0.36}{\lambda} \implies \lambda = \frac{5 \times 0.36}{3} = 0.60\text{ m} \)

The wave speed \( v \) is:
\( v = f \lambda = 25\text{ Hz} \times 0.60\text{ m} = 15\text{ m s}^{-1} \).

1 mark for the correct option.
- Award 1 mark for calculating the wavelength using the phase difference formula and then calculating the wave speed using the wave equation.

Power \(P\) is related to resistance \(R\) and current \(I\) by \(P = I^2 R\), hence \(R = P / I^2\). The SI base unit of power is the watt (\(\text{W}\)), which is equivalent to \(\text{kg}\ \text{m}^2\ \text{s}^{-3}\). Dividing this by amperes squared (\(\text{A}^2\)) gives \(\text{kg}\ \text{m}^2\ \text{s}^{-3}\ \text{A}^{-2}\).

1 mark for identifying the correct combination of base units by relating electrical resistance to mechanical base units.

Using the equations of motion for constant acceleration, let upwards be positive. Thus, initial velocity \(u = +15\ \text{m}\ \text{s}^{-1}\), acceleration \(a = -9.81\ \text{m}\ \text{s}^{-2}\), and time \(t = 4.5\ \text{s}\). The displacement \(s\) is given by \(s = ut + \frac{1}{2}at^2\). Substituting the values: \(s = (15)(4.5) + 0.5(-9.81)(4.5)^2 = 67.5 - 99.33 = -31.83\ \text{m}\). The negative sign indicates a displacement below the starting level, so the height of the cliff is \(32\ \text{m}\) to two significant figures.

1 mark for using the correct equations of motion and computing the displacement value.

By conservation of linear momentum: \(m v = m (-0.2 v) + 3 m v_2\). This simplifies to \(1.2 v = 3 v_2\), which yields \(v_2 = 0.4 v\). The initial kinetic energy is \(E_k = 0.5 m v^2\). The final kinetic energy is the sum of the kinetic energies of both gliders: \(E_f = 0.5 m (-0.2 v)^2 + 0.5 (3m) (0.4 v)^2 = 0.02 m v^2 + 0.24 m v^2 = 0.26 m v^2\). The ratio is \(0.26 / 0.5 = 0.52\).

1 mark for applying conservation of momentum to find the post-collision speed of the second glider and calculating the ratio of kinetic energies.

The center of gravity of the uniform plank acts at its midpoint, \(2.0\ \text{m}\) from the left end, which is \(1.0\ \text{m}\) to the right of the pivot. The force of \(50\ \text{N}\) is at the right end, which is \(3.0\ \text{m}\) to the right of the pivot. Taking moments about the pivot: clockwise moments must equal anticlockwise moments. Therefore, \(W \times 1.0 = (100 \times 1.0) + (50 \times 3.0)\), giving \(W = 100 + 150 = 250\ \text{N}\).

1 mark for establishing the correct balance of anticlockwise and clockwise moments about the pivot.

The volume flow rate is \(V/t = A v = (2.5 \times 10^{-3}\ \text{m}^2)(6.0\ \text{m}\ \text{s}^{-1}) = 0.015\ \text{m}^3\ \text{s}^{-1}\). The mass flow rate is \(m/t = \rho (V/t) = 1000 \times 0.015 = 15\ \text{kg}\ \text{s}^{-1}\). The rate of increase of potential energy is \((m/t) g h = 15 \times 9.81 \times 15 = 2207.25\ \text{W}\). The rate of increase of kinetic energy is \(0.5 (m/t) v^2 = 0.5 \times 15 \times 6.0^2 = 270\ \text{W}\). The total useful power output is \(P_{\text{out}} = 2207.25 + 270 = 2477.25\ \text{W}\). The input electrical power is \(P_{\text{in}} = P_{\text{out}} / 0.60 = 2477.25 / 0.60 = 4128.75\ \text{W} \approx 4.1\ \text{kW}\).

1 mark for combining mass flow rate calculations, mechanical power components, and efficiency to find the total electrical power input.

The elastic potential energy stored is given by \(E = \frac{1}{2} F x\). Using Young's modulus \(E_Y = \frac{F L}{A x}\), the extension is \(x = \frac{F L}{A E_Y}\), so \(E = \frac{F^2 L}{2 A E_Y}\). Since \(F\), \(L\), and \(E_Y\) are identical for both wires, \(E\) is inversely proportional to the cross-sectional area \(A\). Because the diameter of P is twice that of Q, the cross-sectional area of P is four times that of Q (\(A_P = 4 A_Q\)). Thus, the energy ratio is \(E_P / E_Q = A_Q / A_P = 0.25\).

1 mark for establishing the relationship between energy and cross-sectional area, and applying the ratio.

First, find the wavelength \(\lambda = v / f = 340 / 850 = 0.40\ \text{m} = 40\ \text{cm}\). The phase difference \(\Delta \phi\) for a path difference \(x\) is given by \(\Delta \phi = \frac{2\pi x}{\lambda} = \frac{2\pi \times 15}{40} = 0.75\pi\ \text{rad}\).

1 mark for calculating wavelength and finding the correct phase difference in radians.

Let \(I\) be the total current leaving the battery. This current passes through the single series resistor and then splits equally between the two parallel resistors. Thus, the current through each of the parallel resistors is \(0.5 I\). The power dissipated in one parallel resistor is \(P_{\text{one}} = (0.5 I)^2 R = 0.25 I^2 R\). The total resistance of the circuit is \(R + 0.5 R = 1.5 R\), so the total power dissipated in the circuit is \(P_{\text{total}} = I^2 (1.5 R) = 1.5 I^2 R\). The ratio is \(P_{\text{one}} / P_{\text{total}} = 0.25 / 1.5 = 1/6\).

1 mark for expressing individual and total power dissipation in terms of the total current and determining the ratio.

Resistivity is given by \(\rho = \frac{R A}{L}\). Since resistance is \(R = \frac{V}{I}\) and potential difference is \(V = \frac{W}{Q} = \frac{W}{I t}\), the SI base units of resistance are \(\frac{\text{kg m}^2 \text{s}^{-2}}{\text{A}^2 \text{s}} = \text{kg m}^2 \text{s}^{-3} \text{A}^{-2}\). Substituting this into the resistivity formula gives the SI base units of resistivity as \(\text{kg m}^2 \text{s}^{-3} \text{A}^{-2} \times \frac{\text{m}^2}{\text{m}} = \text{kg m}^3 \text{s}^{-3} \text{A}^{-2}\).

1 mark for the correct derivation of SI base units of resistivity.

Using the equation \(v^2 = u^2 + 2as\) for the downward motion from the cliff edge to the sea: the initial downward velocity is \(u\), the final downward velocity is \(3u\), the acceleration is \(g\), and the displacement is \(h\). Thus, \((3u)^2 = u^2 + 2gh\), which simplifies to \(9u^2 = u^2 + 2gh\). This gives \(8u^2 = 2gh\), so \(h = \frac{4u^2}{g}\).

1 mark for applying the equations of motion to find the height in terms of u and g.

Let the left-hand end be at position \(x = 0\) and the right-hand end at \(x = 4.0\text{ m}\). The center of gravity of the beam is at \(x = 2.0\text{ m}\). Support \(B\) is at \(x = 3.0\text{ m}\). At the threshold of tipping about support \(B\), the force at support \(A\) is zero. Taking moments about support \(B\): the anticlockwise moment is due to the weight of the beam, which is \(240\text{ N} \times (3.0\text{ m} - 2.0\text{ m}) = 240\text{ N m}\). The clockwise moment is due to the person of weight \(600\text{ N}\) standing at a distance \(d\) to the right of support \(B\), which is \(600\text{ N} \times d\). Setting these equal, we find \(600 d = 240\), so \(d = 0.40\text{ m}\). The distance from the right-hand end is \(1.0\text{ m} - 0.40\text{ m} = 0.60\text{ m}\).

1 mark for equating the moments about support B to find the distance.

The rate of useful energy output consists of the rate of gain of gravitational potential energy and kinetic energy. Rate of gain of GPE = \(\frac{\Delta m}{\Delta t} g h = 8.0 \times 9.81 \times 12 = 941.8\text{ W}\). Rate of gain of KE = \(\frac{1}{2} \frac{\Delta m}{\Delta t} v^2 = \frac{1}{2} \times 8.0 \times (5.0)^2 = 100\text{ W}\). Total useful power output = \(941.8 + 100 = 1041.8\text{ W}\). Efficiency = \(\frac{\text{useful power output}}{\text{power input}} = \frac{1041.8}{1500} \approx 0.69\) or \(69\%\).

1 mark for calculating both GPE and KE rates and finding the total efficiency.

The strain energy stored in a wire is \(U = \frac{1}{2} F x\), where \(x\) is the extension. Since \(x = \frac{F L}{A E}\), the strain energy can be written as \(U = \frac{F^2 L}{2 A E}\). Since both wires are made of the same material (same \(E\)) and carry the same load \(F\), the strain energy is proportional to \(\frac{L}{A}\), which is proportional to \(\frac{L}{d^2}\). For wire \(X\), \(U_X \propto \frac{L}{d^2}\). For wire \(Y\), \(U_Y \propto \frac{2L}{(2d)^2} = \frac{L}{2d^2}\). Therefore, the ratio \(\frac{U_X}{U_Y} = 2\).

1 mark for establishing the relationship between strain energy, length, and diameter, and finding the correct ratio.

The initial resistance of the wire is \(R\), and the initial power is \(P = \frac{V^2}{R}\). When cut in half, each piece has a resistance of \(\frac{R}{2}\). Connecting them in parallel across the same potential difference \(V\) means each piece experiences the full potential difference \(V\). The power dissipated by each half is \(\frac{V^2}{R/2} = 2P\). The total power dissipated by both halves in parallel is \(2P + 2P = 4P\).

1 mark for using resistance and parallel connections to determine the new total power.

In beta-plus decay, a proton decays into a neutron, a positron, and an electron neutrino. A proton has quark composition \(\text{uud}\) and a neutron has composition \(\text{udd}\). Thus, an up quark changes to a down quark (\(\text{u} \rightarrow \text{d}\)). The leptons produced in this process are the positron (the antiparticle of the electron) and the electron neutrino.

1 mark for identifying both the quark change and the leptons produced.

The phase difference \(\Delta \phi\) is related to the path difference \(\Delta x\) by \(\Delta \phi = \frac{2\pi \Delta x}{\lambda}\). Substituting the given values: \(\frac{3\pi}{5} = \frac{2\pi \times 0.150}{\lambda}\). This simplifies to \(\lambda = \frac{5 \times 0.300}{3} = 0.500\text{ m}\). The speed of the wave is \(v = f \lambda = 50.0 \times 0.500 = 25.0\text{ m s}^{-1}\).

1 mark for calculating the wavelength and then using the wave equation to find the speed.

We determine the SI base units of each quantity on the right-hand side:
- Pressure \( P \): Since \( P = \frac{\text{Force}}{\text{Area}} \), the units are \( \text{N}\,\text{m}^{-2} = (\text{kg}\,\text{m}\,\text{s}^{-2})\,\text{m}^{-2} = \text{kg}\,\text{m}^{-1}\,\text{s}^{-2} \).
- Wavelength \( \lambda \): \( \text{m} \).
- Speed \( v \): \( \text{m}\,\text{s}^{-1} \).
- Density \( \rho \): \( \text{kg}\,\text{m}^{-3} \).

Substituting these into the expression for \( X \):
\( [X] = \frac{[P][\lambda]}{[v][\rho]} = \frac{(\text{kg}\,\text{m}^{-1}\,\text{s}^{-2}) \times \text{m}}{(\text{m}\,\text{s}^{-1}) \times (\text{kg}\,\text{m}^{-3})} \)

Simplify the numerator:
\( (\text{kg}\,\text{m}^{-1}\,\text{s}^{-2}) \times \text{m} = \text{kg}\,\text{s}^{-2} \)

Simplify the denominator:
\( (\text{m}\,\text{s}^{-1}) \times (\text{kg}\,\text{m}^{-3}) = \text{kg}\,\text{m}^{-2}\,\text{s}^{-1} \)

Divide the numerator by the denominator:
\( [X] = \frac{\text{kg}\,\text{s}^{-2}}{\text{kg}\,\text{m}^{-2}\,\text{s}^{-1}} = \text{m}^2\,\text{s}^{-1} \)

1 mark for the correct option A.
Method: [1 mark] for converting all variables into their respective SI base units, and [1 mark] for correctly algebraically simplifying the ratio to find the base units of X.

The horizontal component of the velocity is constant:
\( v_x = u \)

The time of flight \( t \) is related to the horizontal distance \( d \) by:
\( t = \frac{d}{u} \)

The vertical component of the velocity when hitting the ground is:
\( v_y = gt = \frac{gd}{u} \)

The final speed \( v \) is the magnitude of the resultant velocity vector:
\( v = \sqrt{v_x^2 + v_y^2} = \sqrt{u^2 + \left(\frac{gd}{u}\right)^2} = \sqrt{u^2 + \frac{g^2 d^2}{u^2}} \)

1 mark for the correct option B.
Method: [1 mark] for finding the time of flight in terms of d and u, finding the vertical velocity component, and applying Pythagoras' theorem to determine the speed.

The total weight of the plank is:
\( W = m g = 25 \times 9.81 = 245.25\text{ N} \)

Since the plank is in vertical equilibrium, the sum of the upward tensions equals the downward weight:
\( T_A + T_B = W \)
\( 100 + T_B = 245.25 \implies T_B = 145.25\text{ N} \)

Let \( x \) be the distance of the centre of gravity from end A. Taking moments about end A:
\( W \times x = T_B \times L \)
\( 245.25 \times x = 145.25 \times 4.0 \)
\( x = \frac{581}{245.25} \approx 2.37\text{ m} \)

To two significant figures, this is 2.4 m.

1 mark for the correct option C.
Method: [1 mark] for finding the tension in rope B using translational equilibrium, then setting up a correct moment equation about point A to solve for the distance of the centre of gravity.

The extension \( x \) of a wire is given by:
\( x = \frac{F L}{A E} = \frac{4 F L}{\pi d^2 E} \)

Since the wires are connected in series, the tensile force \( F \) in both wires is identical. Thus, the extension is proportional to:
\( x \propto \frac{L}{d^2 E} \)

We compare the steel and brass wires:
\( \frac{x_{\text{steel}}}{x_{\text{brass}}} = \left(\frac{L_{\text{steel}}}{L_{\text{brass}}}\right) \times \left(\frac{d_{\text{brass}}}{d_{\text{steel}}}\right)^2 \times \left(\frac{E_{\text{brass}}}{E_{\text{steel}}}\right) \)

Given:
- \( \frac{L_{\text{steel}}}{L_{\text{brass}}} = 2 \)
- \( \frac{d_{\text{brass}}}{d_{\text{steel}}} = 2 \)
- \( \frac{E_{\text{brass}}}{E_{\text{steel}}} = \frac{1.0 \times 10^{11}}{2.0 \times 10^{11}} = 0.5 \)

Substitute these values:
\( \frac{x_{\text{steel}}}{x_{\text{brass}}} = 2 \times 2^2 \times 0.5 = 2 \times 4 \times 0.5 = 4.0 \)

1 mark for the correct option C.
Method: [1/2 mark] for recognizing that tension is equal in series, and [1/2 mark] for substituting the ratios of length, diameter (squared), and Young modulus to obtain the correct ratio of 4.0.

First, calculate the wavelength \( \lambda \) of the wave:
\( \lambda = \frac{v}{f} = \frac{300}{500} = 0.60\text{ m} = 60\text{ cm} \)

The path difference \( \Delta x \) between the two points is 15 cm.

The phase difference \( \Delta \phi \) is given by:
\( \Delta \phi = \frac{2\pi \Delta x}{\lambda} = \frac{2\pi \times 15}{60} = \frac{30\pi}{60} = \frac{\pi}{2}\text{ rad} \)

1 mark for the correct option B.
Method: [1 mark] for computing the wavelength, and using the formula relating path difference and phase difference to find the correct value in radians.

The power \( P \) dissipated in the variable resistor is:
\( P = V I \)

Using the relationship for a closed circuit with internal resistance:
\( V = E - I r \implies I = \frac{E - V}{r} \)

Substituting \( I \) into the power equation:
\( P = V \left(\frac{E - V}{r}\right) = \frac{E V - V^2}{r} \)

This is a quadratic equation representing an inverted parabola:
- When \( R \to 0 \), we have \( V \to 0 \), so \( P \to 0 \).
- When \( R \to \infty \), we have \( V \to E \), so \( P \to 0 \).
- A maximum power is reached when \( R = r \) (which occurs at \( V = E/2 \)).

Thus, the graph of \( P \) against \( V \) starts at the origin, rises to a maximum, and then decreases back to zero when \( V = E \).

1 mark for the correct option B.
Method: [1 mark] for linking P, V, and current I using internal resistance equations, and analyzing the limiting behavior for R tending to zero and infinity to determine the graphical shape.

Let's analyze the properties before and after the decay:

1. **Change in up quarks**:
- A neutron (\(\text{n}\)) has the quark composition \( \text{udd} \) (one up quark, two down quarks).
- A proton (\(\text{p}\)) has the quark composition \( \text{uud} \) (two up quarks, one down quark).
- The number of up quarks increases from 1 to 2.
- Change in up quarks = \( +1 \).

2. **Change in total lepton number**:
- Before decay: The neutron is a baryon, so lepton number \( L = 0 \).
- After decay:
- Proton: \( L = 0 \)
- Electron (\(\text{e}^-\)): \( L = +1 \)
- Electron antineutrino (\(\bar{\nu}_{\text{e}}\)): \( L = -1 \)
- Total lepton number after decay = \( 0 + 1 + (-1) = 0 \).
- Lepton number is conserved, so the change in total lepton number is \( 0 \).

1 mark for the correct option A.
Method: [1/2 mark] for showing that the change in the up quark count is +1, and [1/2 mark] for showing that the total lepton number is conserved (change = 0).

1. **Calculate the final velocity of the crate**:
Since acceleration is constant:
\( s = \frac{1}{2}(u + v)t \)
\( 9.0 = \frac{1}{2}(0 + v)(3.0) \implies v = \frac{18.0}{3.0} = 6.0\text{ m s}^{-1} \)

2. **Calculate the gain in kinetic energy**:
\( E_{\text{k}} = \frac{1}{2} m v^2 = \frac{1}{2} \times 20 \times (6.0)^2 = 360\text{ J} \)

3. **Calculate the gain in gravitational potential energy**:
\( \Delta E_{\text{p}} = m g h = 20 \times 9.81 \times 9.0 = 1765.8\text{ J} \)

4. **Calculate total useful work output**:
\( E_{\text{out}} = E_{\text{k}} + \Delta E_{\text{p}} = 360 + 1765.8 = 2125.8\text{ J} \)

5. **Calculate the electrical energy input**:
\( \text{Efficiency} = \frac{E_{\text{out}}}{E_{\text{in}}} \implies 0.60 = \frac{2125.8}{E_{\text{in}}} \)
\( E_{\text{in}} = \frac{2125.8}{0.60} \approx 3543\text{ J} \approx 3.5\text{ kJ} \)

1 mark for the correct option C.
Method: [1 mark] for calculating both kinetic and gravitational potential energy changes of the crate, summing them to find the useful mechanical work output, and dividing by efficiency to find the total electrical energy input.

Rearranging the equation for thermal conductivity gives:

\( k = \frac{q x}{\Delta T} \)

The unit of \( q \) is \( \text{W m}^{-2} \). Since \( 1 \text{ W} = 1 \text{ J s}^{-1} = 1 \text{ kg m}^2 \text{s}^{-3} \), the SI base units of \( q \) are:

\( \text{kg m}^2 \text{s}^{-3} \text{m}^{-2} = \text{kg s}^{-3} \)

Substituting the base units of \( q \), \( x \) (\( \text{m} \)), and \( \Delta T \) (\( \text{K} \)) into the rearranged equation:

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

1 mark for the correct choice. Method: Convert power to base units, determine base units of \( q \), rearrange for \( k \), and combine units.

Using the equation of motion \( v^2 = u^2 + 2as \):

Taking the downward direction as positive:
- Initial velocity, \( v_0 = -u \) (upwards is negative)
- Final velocity, \( v = 3u \) (downwards is positive)
- Acceleration, \( a = g \)
- Displacement, \( s = h \)

Substituting these values:

\( (3u)^2 = (-u)^2 + 2gh \)

\( 9u^2 = u^2 + 2gh \)

\( 8u^2 = 2gh \)

\( h = \frac{4u^2}{g} \)

1 mark for correct option. Award for correct application of \( v^2 = u^2 + 2as \) with consistent signs.

For the shelf to remain in rotational equilibrium, the sum of clockwise moments about the hinge must equal the sum of anticlockwise moments about the hinge.

Clockwise moments:
- The weight of the uniform shelf \( W \) acts at its center of gravity (distance \( \frac{L}{2} \) from the hinge): \( \text{Moment}_1 = W \times \frac{L}{2} \)
- The weight of the object \( 2W \) acts at a distance of \( \frac{L}{4} \): \( \text{Moment}_2 = 2W \times \frac{L}{4} = \frac{W L}{2} \)

Total clockwise moment = \( \frac{W L}{2} + \frac{W L}{2} = W L \)

Anticlockwise moment:
- The vertical component of the tension is \( T \sin\theta \), acting at distance \( L \): \( \text{Moment}_{\text{anticlockwise}} = T \sin\theta \times L \)

Equating the moments:

\( T L \sin\theta = W L \)

\( T = \frac{W}{\sin\theta} \)

1 mark for correct option. Award for setting up the sum of moments about the hinge correctly and solving for \( T \).

The Young modulus \( E \) is defined as:

\( E = \frac{\text{stress}}{\text{strain}} = \frac{F / A}{e / L} = \frac{F L}{A e} \)

Rearranging for extension \( e \):

\( e = \frac{F L}{A E} \)

Since the cross-sectional area \( A = \frac{\pi d^2}{4} \), the extension is:

\( e \propto \frac{L}{d^2 E} \)

For wire X:

\( e_X \propto \frac{L}{d^2 E} \)

For wire Y:

\( e_Y \propto \frac{2L}{(2d)^2 (2E)} = \frac{2L}{8 d^2 E} = \frac{1}{4} \frac{L}{d^2 E} \)

Therefore, the ratio is:

\( \frac{e_X}{e_Y} = \frac{1}{1/4} = 4 \)

1 mark for correct option. Award for deriving the relation of extension in terms of length, diameter, and Young modulus, and taking the ratio.

The useful power output is the power dissipated in the external resistor \( R \):

\( P_{\text{out}} = I^2 R \)

The total power supplied by the cell is:

\( P_{\text{total}} = E I \)

Since \( I = \frac{E}{R+r} \), the e.m.f. can be expressed as \( E = I(R + r) \). Therefore, the total power is:

\( P_{\text{total}} = I^2(R + r) \)

The efficiency \( \eta \) of the power transfer is:

\( \eta = \frac{P_{\text{out}}}{P_{\text{total}}} = \frac{I^2 R}{I^2(R + r)} = \frac{R}{R+r} \)

1 mark for correct option. Method: Identify useful power output and total power input, use Ohm's law to substitute for e.m.f. or current, and simplify the ratio.

A neutron has a quark composition of \( \text{udd} \), while a proton has a quark composition of \( \text{uud} \). During beta-minus decay, one of the down (\( \text{d} \)) quarks in the neutron changes into an up (\( \text{u} \)) quark, converting the neutron into a proton.

All beta decays are governed by the weak nuclear force (or weak interaction).

1 mark for correct option. Identify quark change from \( \text{udd} \) to \( \text{uud} \) and recall that weak interaction is responsible for beta decay.

The fringe separation \( x \) is given by:

\( x = \frac{\lambda D}{a} \)

Initially, for red light:

\( x = \frac{6.4 \times 10^{-7} \times D}{a} \)

Finally, for blue-green light with the slit separation halved (\( a' = 0.5a \)):

\( x' = \frac{4.8 \times 10^{-7} \times D}{0.5a} = 2 \times \frac{4.8 \times 10^{-7} \times D}{a} \)

Taking the ratio of the new and original fringe separations:

\( \frac{x'}{x} = \frac{2 \times 4.8 \times 10^{-7}}{6.4 \times 10^{-7}} = 2 \times 0.75 = 1.5 \)

Thus, the new fringe separation is \( 1.5x \).

1 mark for correct option. Method: Use the double-slit equation, substitute the new wavelength and halved slit separation, and compute the ratio to obtain \( 1.5x \).

First, calculate the wavelength \( \lambda \) of the wave using the wave equation:

\( v = f \lambda \implies \lambda = \frac{v}{f} = \frac{120}{250} = 0.48\text{ m} = 48\text{ cm} \)

The separation between the two points is \( x = 16\text{ cm} \).

The phase difference \( \phi \) in radians is:

\( \phi = \frac{x}{\lambda} \times 2\pi = \frac{16}{48} \times 2\pi = \frac{1}{3} \times 2\pi = \frac{2\pi}{3}\text{ rad} \)

1 mark for correct option. Method: Calculate wavelength \( \lambda \) first, then apply the formula for phase difference \( \phi = \frac{x}{\lambda} \times 2\pi \) with consistent units.

a) A base quantity is a fundamental physical quantity that cannot be defined in terms of other physical quantities. Two other base units are the ampere (A) and the kelvin (K). b) Rearranging the formula gives \(\eta = \frac{F}{6\pi r v}\). The constant \(6\pi\) has no units. The SI base units of force \(F\) are \(\text{kg}\ \text{m}\ \text{s}^{-2}\). The SI unit of radius \(r\) is \(\text{m}\), and of velocity \(v\) is \(\text{m}\ \text{s}^{-1}\). Substituting these into the formula gives base units of \(\eta = \frac{\text{kg}\ \text{m}\ \text{s}^{-2}}{\text{m} \times \text{m}\ \text{s}^{-1}} = \text{kg}\ \text{m}^{-1}\ \text{s}^{-1}\). c) The percentage uncertainty in \(\eta\) is the sum of the percentage uncertainties of \(F\), \(r\), and \(v\): \(\% \text{ uncertainty} = 2.5\% + 1.2\% + 1.8\% = 5.5\%\). d) An empty 500 mL plastic bottle has a mass of about 15 g (or 0.015 kg). The weight is \(W = m g = 0.015 \times 9.81 \approx 0.15\text{ N}\). Any estimate between 0.08 N and 0.3 N is acceptable.

a) [1] mark for definition of base quantity. [1] mark for two correct base units (e.g. A, K, mol). b) [1] mark for expressing \(\eta\) as the subject. [1] mark for substituting base units of force. [1] mark for correct simplification to show the final unit. c) [1] mark for identifying that percentage uncertainties must be added. [1] mark for correct calculation of 5.5%. d) [1] mark for a reasonable estimate of the weight (0.08 N to 0.3 N).

a) Acceleration is defined as the rate of change of velocity. b) i) Using \(v = u + a t\): \(v = 15.0 + (3.20 \times 4.00) = 27.8\ \text{m}\ \text{s}^{-1}\). ii) Distance traveled during powered phase: \(s_1 = u t + \frac{1}{2} a t^2 = 15.0 \times 4.00 + 0.5 \times 3.20 \times 4.00^2 = 60.0 + 25.6 = 85.6\ \text{m}\). Height above the base of the cliff at this point = \(45.0 + 85.6 = 130.6\ \text{m}\). Distance traveled during unpowered phase until highest point: \(v^2 = u^2 + 2as \implies 0 = 27.8^2 - 2 \times 9.81 \times s_2 \implies s_2 = 39.4\ \text{m}\). Total height = \(130.6 + 39.4 = 170\ \text{m}\). iii) From the moment the motor shuts off, the displacement to the ground is \(-130.6\ \text{m}\). Using \(s = u t + \frac{1}{2} a t^2\): \(-130.6 = 27.8 t - 4.905 t^2\). Rearranging gives \(4.905 t^2 - 27.8 t - 130.6 = 0\). Solving the quadratic equation gives \(t = \frac{27.8 + \sqrt{(-27.8)^2 - 4 \times 4.905 \times (-130.6)}}{9.81} = 8.72\ \text{s}\).

a) [1] mark for rate of change of velocity. b) i) [1] mark for recall of \(v=u+at\). [1] mark for correct calculation of 27.8 m/s. ii) [1] mark for powered height 85.6 m. [1] mark for unpowered height 39.4 m. [1] mark for adding to cliff height to get 170 m. iii) [1] mark for setting up kinematic equation with correct signs. [1] mark for correct substitution. [1] mark for final answer of 8.72 s.

a) i) The center of gravity is the point at which the entire weight of the body may be considered to act. ii) The torque of a couple is the product of one of the forces and the perpendicular distance between the lines of action of the forces. b) i) There is an upward force \(T_A\) at A (x = 0), a downward force of 120 N at the center of gravity (x = 1.0 m from A), and an upward force \(T_C\) at C (x = 1.8 m from A). ii) Taking moments about A: \(T_C \times 1.8 = 120 \times 1.0 \implies T_C = 66.7\ \text{N} \approx 67\ \text{N}\). Since the beam is in vertical equilibrium, \(T_A + T_C = 120 \implies T_A = 120 - 66.7 = 53.3\ \text{N} \approx 53\ \text{N}\).

a) i) [1] mark for weight acting through this point. ii) [1] mark for force multiplied by perpendicular distance. [1] mark for specifying distance between the lines of action of the forces. b) i) [1] mark for recognizing the upward tensions at A and C. [1] mark for identifying the downward weight at 1.0 m from A. ii) [1] mark for setting up a correct moment equation. [1] mark for calculating \(T_C = 67\ \text{N}\). [1] mark for using vertical equilibrium. [1] mark for calculating \(T_A = 53\ \text{N}\).

a) Gravitational potential energy is energy stored due to an object's position in a gravitational field, whereas elastic potential energy is energy stored due to the deformation or stretching/compression of an object. b) i) Useful work done is the change in gravitational potential energy: \(W = m g h = 35 \times 9.81 \times 12 = 4120\ \text{J} \approx 4100\ \text{J}\). ii) \(\text{Efficiency} = \frac{\text{Useful work}}{\text{Total energy input}} = \frac{4120}{5400} = 0.763\) or \(76\%\). iii) At constant speed, useful output power \(P = F v = m g v = 35 \times 9.81 \times 1.8 = 618\ \text{W} \approx 620\ \text{W}\).

a) [1] mark for explaining gravitational potential energy (position in field). [1] mark for explaining elastic potential energy (deformation). b) i) [1] mark for recalling \(E_p = mgh\). [1] mark for calculating 4100 J (or 4120 J). ii) [1] mark for efficiency formula. [1] mark for calculating 76% (or 0.76). iii) [1] mark for recalling \(P = Fv\). [1] mark for calculating 620 W (or 618 W).

a) i) Stress is force per unit cross-sectional area. ii) Strain is extension per unit original length. b) i) Using Young modulus formula \(E = \frac{F L}{A x}\): \(x_{\text{steel}} = \frac{F L_{\text{steel}}}{A_{\text{steel}} E_{\text{steel}}} = \frac{180 \times 1.5}{(2.0 \times 10^{-6}) \times (2.0 \times 10^{11})} = 6.75 \times 10^{-4}\ \text{m}\) (or 0.68 mm). ii) Extension of brass wire: \(x_{\text{brass}} = \frac{F L_{\text{brass}}}{A_{\text{brass}} E_{\text{brass}}} = \frac{180 \times 2.0}{(4.0 \times 10^{-6}) \times (1.0 \times 10^{11})} = 9.0 \times 10^{-4}\ \text{m}\) (or 0.90 mm). Total extension = \(6.75 \times 10^{-4} + 9.0 \times 10^{-4} = 1.58 \times 10^{-3}\ \text{m}\) (or 1.6 mm).

a) i) [1] mark for force / area. ii) [1] mark for extension / original length. b) i) [1] mark for rearranging Young modulus formula. [1] mark for correct substitution. [1] mark for answer 0.68 mm. ii) [1] mark for calculating extension of brass wire (0.90 mm). [1] mark for adding the two extensions. [1] mark for total extension 1.6 mm (or 1.58 mm).

a) Electromotive force (e.m.f.) is the energy converted from other forms to electrical energy per unit charge, while potential difference (p.d.) is the energy converted from electrical energy to other forms per unit charge. b) i) Using \(V = E - I r\): \(5.4 = E - 0.50 r\) (Equation 1), and \(4.2 = E - 1.50 r\) (Equation 2). Subtracting Equation 2 from Equation 1: \(1.2 = 1.00 r \implies r = 1.2\ \Omega\). ii) Substituting \(r = 1.2\ \Omega\) into Equation 1: \(E = 5.4 + (0.50 \times 1.2) = 6.0\ \text{V}\). iii) Power dissipated in the internal resistance: \(P = I^2 r = 1.50^2 \times 1.2 = 2.25 \times 1.2 = 2.7\ \text{W}\).

a) [1] mark for e.m.f. as energy per unit charge converted into electrical form. [1] mark for p.d. as electrical energy per unit charge converted into other forms. b) i) [1] mark for setting up two equations using \(V=E-Ir\). [1] mark for method of solving equations. [1] mark for \(r = 1.2\ \Omega\). ii) [1] mark for substitution method. [1] mark for \(E = 6.0\ \text{V}\). iii) [1] mark for formula \(P=I^2 r\). [1] mark for calculation of 2.7 W.

a) The two main groups of hadrons are: 1. Baryons, composed of three quarks (or three antiquarks). 2. Mesons, composed of a quark and an antiquark. b) i) The weak interaction (or weak nuclear force). ii) A neutron consists of 'udd' quarks and a proton consists of 'uud' quarks. Therefore, the quark change is: \(\text{d} \rightarrow \text{u} + \text{e}^- + \bar{\nu}_\text{e}\). The charges of the particles are: \(Q_\text{d} = -\frac{1}{3}e\), \(Q_\text{u} = +\frac{2}{3}e\), \(Q_{\text{e}^-} = -e\), and \(Q_{\bar{\nu}_\text{e}} = 0\). Comparing charges: LHS: \(-\frac{1}{3}e\); RHS: \(+\frac{2}{3}e - e + 0 = -\frac{1}{3}e\). Since LHS = RHS, charge is conserved. iii) The electron (\(\text{e}^-\)).

a) [1] mark for naming Baryons and Mesons. [1] mark for Baryons as 3 quarks. [1] mark for Mesons as quark-antiquark pair. b) i) [1] mark for identifying the weak interaction. ii) [1] mark for writing the quark equation: \(\text{d} \rightarrow \text{u} + \text{e}^- + \bar{\nu}_\text{e}\). [1] mark for stating correct quark charges. [1] mark for showing that the sum of charges on both sides is equal. iii) [1] mark for stating the electron.

### Sample Results & Detailed Worked Solution:

**(a) Measurement of thread length:**
\(L = 45.2\text{ cm} = 0.452\text{ m}\)

**(b) First reading for \(d = 20.0\text{ cm}\):**
\(d = 0.200\text{ m}\)
\(t_1 = 14.8\text{ s}\), \(t_2 = 14.6\text{ s}\)
\(\text{Mean } t = 14.7\text{ s}\)
\(T = 1.47\text{ s}\)

**(c) Table of Results (6 sets):**
| \(d / \text{m}\) | \(t_1 / \text{s}\) | \(t_2 / \text{s}\) | \(t_{\text{mean}} / \text{s}\) | \(T / \text{s}\) | \((1/d) / \text{m}^{-1}\) |
|---|---|---|---|---|---|
| 0.150 | 19.5 | 19.3 | 19.4 | 1.94 | 6.67 |
| 0.200 | 14.8 | 14.6 | 14.7 | 1.47 | 5.00 |
| 0.300 | 10.1 | 10.3 | 10.2 | 1.02 | 3.33 |
| 0.400 | 7.9 | 7.7 | 7.8 | 0.78 | 2.50 |
| 0.500 | 6.5 | 6.3 | 6.4 | 0.64 | 2.00 |
| 0.600 | 5.4 | 5.6 | 5.5 | 0.55 | 1.67 |

**(d) Graph of \(T/\text{s}\) against \((1/d)/\text{m}^{-1}\):**
- Points are plotted on a linear grid.
- Trend is a straight line passing through the points.
- Best-fit line drawn.

**(e) Gradient and y-intercept:**
- Choose two points on the best-fit line: \((1.67, 0.55)\) and \((6.67, 1.94)\).
- \(\text{Gradient} = \frac{1.94 - 0.55}{6.67 - 1.67} = \frac{1.39}{5.00} = 0.278\text{ s m}\).
- \(\text{y-intercept} = T - \text{gradient} \times (1/d) = 0.55 - 0.278 \times 1.67 = 0.55 - 0.46 = 0.09\text{ s}\).

**(f) Determine \(p\) and \(q\):**
Comparing \(T = \frac{p}{d} + q\) with \(y = m x + c\):
- \(p = \text{gradient} = 0.278\text{ s m}\) (or \(27.8\text{ s cm}\))
- \(q = \text{y-intercept} = 0.09\text{ s}\)

**Marking Scheme (Total: 20 marks)**

**(a) Length measurement [1 mark]**
- [1] Value of \(L\) recorded to nearest millimetre, with appropriate unit (e.g., \(45.2\text{ cm}\) or \(0.452\text{ m}\)), within the range \(40.0\text{ cm}\) to \(50.0\text{ cm}\).

**(b) First set of readings [2 marks]**
- [1] Value of \(t\) recorded to \(0.1\text{ s}\) or \(0.01\text{ s}\).
- [1] Correct calculation of \(T = t / 10\).

**(c) Table of results [6 marks]**
- [1] **Data collection**: Six sets of readings of \(d\) and \(t\) recorded, demonstrating the correct trend (as \(d\) increases, \(t\) decreases).
- [1] **Range**: Separations must cover a range of at least \(40.0\text{ cm}\) (e.g., \(15.0\text{ cm}\) to \(55.0\text{ cm}\)).
- [1] **Column headings**: Headings in the table must include units in accepted form, e.g., \(d/\text{m}\), \(t/\text{s}\), \(T/\text{s}\), \((1/d)/\text{m}^{-1}\).
- [1] **Consistency**: All raw values of \(d\) recorded to the nearest millimetre (e.g., \(0.150\text{ m}\) or \(15.0\text{ cm}\)).
- [1] **Significant figures**: Significant figures for \((1/d)\) must be the same as, or one more than, the number of significant figures in the raw \(d\) values.
- [1] **Calculation**: Values of \(1/d\) and \(T\) calculated correctly.

**(d) Graph [5 marks]**
- [1] **Axes**: Clearly labelled with quantity and unit. Scales must be linear and chosen such that the plotted points occupy more than half the grid in both directions.
- [1] **Plotting**: All points plotted to within half a small square.
- [1] **Line of best fit**: Straight line drawn, with a balanced distribution of points on either side.
- [1] **Quality**: Scatter of points about the line of best fit is small, showing high precision.
- [1] **Plotted points** must not be excessively large (diameter < 1 mm).

**(e) Gradient and y-intercept [2 marks]**
- [1] **Gradient**: Determined using a triangle with a hypotenuse of length greater than half the length of the drawn line. Calculation must be correct.
- [1] **y-intercept**: Calculated using a point on the line and the gradient, or read directly from the y-axis (if \(x = 0\) is on the axis).

**(f) Constants \(p\) and \(q\) [4 marks]**
- [1] Value of \(p\) equated to gradient, recorded to 2 or 3 significant figures.
- [1] Value of \(q\) equated to y-intercept, recorded to 2 or 3 significant figures.
- [1] Correct unit for \(p\) (e.g., \\text{s m}\).
- [1] Correct unit for \(q\) (e.g., \\text{s}\).

### Sample Results & Detailed Worked Solution:

**(a) Unstretched length:**
\(l_0 = 2.5\text{ cm} = 0.025\text{ m}\)

**(b) For \(y = 40.0\text{ cm} = 0.400\text{ m}\):**
- Stretched length \(l = 7.3\text{ cm}\)
- Extension \(x = 7.3 - 2.5 = 4.8\text{ cm}\)
- Absolute uncertainty in \(x\):
Each length reading \(l\) and \(l_0\) has an uncertainty of \\pm 0.1\\text{ cm}\) due to parallax and ruler alignment.
Combining these, the absolute uncertainty in \(x = l - l_0\) is \\Delta x \\approx 0.2\\text{ cm}\).
- Percentage uncertainty:
\[ \\% \text{ uncertainty} = \frac{\Delta x}{x} \times 100\\% = \frac{0.2}{4.8} \times 100\\% \approx 4.2\\% \]

**(c) For \(y = 70.0\text{ cm} = 0.700\text{ m}\):**
- Stretched length \(l_2 = 10.9\text{ cm}\)
- New extension \(x_2 = 10.9 - 2.5 = 8.4\text{ cm}\)

**(d) Calculating constant \(k\):**
- First value: \(k_1 = \frac{x_1}{y_1} = \frac{4.8\text{ cm}}{40.0\text{ cm}} = 0.120\)
- Second value: \(k_2 = \frac{x_2}{y_2} = \frac{8.4\text{ cm}}{70.0\text{ cm}} = 0.120\)

Comparison:
- Percentage difference between \(k_1\) and \(k_2\):
\[ \text{Difference} = \frac{|0.120 - 0.120|}{0.120} \times 100\\% = 0\\% \]
- Since the percentage difference (\(0\\%\)) is much less than the estimated experimental percentage uncertainty (e.g., \(10\\%\)), the suggested relationship \(x = k y\) is strongly supported.

**(e) Evaluation (Limitations & Improvements):**
1. **Limitation:** Hard to judge exactly when the rule is horizontal by eye.
**Improvement:** Use a spirit level placed on the wooden rule.
2. **Limitation:** Ruler may slip or rotate at the pivot point during loading or adjusting.
**Improvement:** Use a fixed axle / G-clamp to secure the pivot pin to the stand.
3. **Limitation:** Parallax error when measuring the length of the spring against the vertical metre rule.
**Improvement:** Use a fiducial marker attached to the spring, or clamp the metre rule close and parallel to the spring.
4. **Limitation:** Two readings of \(y\) are not sufficient to draw a reliable conclusion about proportionality.
**Improvement:** Take multiple readings of \(y\) and plot a graph of \(x\) against \(y\) to see if the line is straight and passes through the origin.

**Marking Scheme (Total: 20 marks)**

**(a) Unstretched length [1 mark]**
- [1] Value of \(l_0\) recorded to the nearest millimetre with correct unit.

**(b) First measurement and uncertainty [4 marks]**
- [1] Value of \(l\) recorded to the nearest millimetre and \(l > l_0\).
- [1] Value of \(x\) calculated correctly.
- [1] **Absolute uncertainty**: Justifies \\Delta x\\ in the range \(0.1\text{ cm}\) to \(0.3\text{ cm}\) based on taking two independent readings (\(l\) and \(l_0\)).
- [1] **Percentage uncertainty**: Correct calculation of percentage uncertainty to 2 significant figures.

**(c) Second measurement [2 marks]**
- [1] Value of \(l_2\) recorded to the nearest millimetre with correct unit.
- [1] Value of \(x_2\) calculated correctly, showing \(x_2 > x\).

**(d) Proportionality analysis [5 marks]**
- [1] Calculation of first constant \(k_1\) with correct units (or dimensionless if both \(x\) and \(y\) are in cm).
- [1] Calculation of second constant \(k_2\) with correct units.
- [1] **Percentage difference calculation**: Shows clear working to find the percentage difference between \(k_1\) and \(k_2\).
- [1] **Criterion statement**: States a reasonable criterion for agreement (e.g., \"I will accept the relationship if the percentage difference is less than \(10\\%\)\").
- [1] **Conclusion**: Makes a correct, consistent conclusion based on comparing the percentage difference to the criterion.

**(e) Evaluation [8 marks - 4 marks for limitations, 4 marks for improvements]**
- [1] **Limitation 1**: Difficulty in leveling the rule / ensuring horizontality.
- [1] **Improvement 1**: Use a spirit level on the rule / alignment grid behind.
- [1] **Limitation 2**: Friction/slipping at the pivot.
- [1] **Improvement 2**: Use a low-friction pin/axle assembly or clamp the pivot securely.
- [1] **Limitation 3**: Parallax error in measuring spring length due to separation between ruler and spring.
- [1] **Improvement 3**: Mount ruler close to spring / use a fiducial indicator.
- [1] **Limitation 4**: Only two sets of readings are insufficient to validate the relationship.
- [1] **Improvement 4**: Take a wider range of readings and plot a graph of \(x\) against \(y\).