2023 Cambridge IAL Physics (9702) Practice Paper with Answers

The percentage uncertainty in resistivity \(\rho\) is determined by summing the percentage uncertainties of each component, accounting for powers in the formula:

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

1. Calculate the percentage uncertainty for each measured value:
- For resistance \(R\): \(\frac{0.5}{25.0} \times 100\% = 2.0\%\)
- For diameter \(d\): \(\frac{0.01}{0.40} \times 100\% = 2.5\%\)
- For length \(L\): \(\frac{0.003}{1.500} \times 100\% = 0.2\%\)

2. Combine these values:

\frac{\Delta \rho}{\rho} = 2.0\% + 2(2.5\%) + 0.2\% = 7.2\%

Award 1 mark for the correct answer B. Allowances: 0.5 marks for calculating the individual percentage uncertainties, and 0.5 marks for correctly applying the power rule to double the diameter's percentage uncertainty, resulting in 7.2%.

1. Calculate the total initial kinetic energy \(E_{ki}\):

E_{ki} = \frac{1}{2}(2m)u^2 + \frac{1}{2}(3m)\left(-\frac{1}{2}u\right)^2 = mu^2 + \frac{3}{8}mu^2 = \frac{11}{8}mu^2 = 1.375 mu^2

2. Use the conservation of linear momentum to find the final velocity \(v\) of the combined mass (taking right as positive):

p_{initial} = p_{final}
2m(u) + 3m\left(-\frac{1}{2}u\right) = (2m + 3m)v
2mu - 1.5mu = 5mv
0.5mu = 5mv \implies v = 0.1u

3. Calculate the final kinetic energy \(E_{kf}\):

E_{kf} = \frac{1}{2}(5m)(0.1u)^2 = 2.5m(0.01u^2) = 0.025mu^2

4. Calculate the loss in kinetic energy:

\text{Loss} = E_{ki} - E_{kf} = 1.375mu^2 - 0.025mu^2 = 1.35mu^2 = \frac{27}{20}mu^2

Award 1 mark for the correct answer B. Deduce marks for sign errors in momentum or failing to account for kinetic energy of both blocks initially.

The relationship between phase difference \(\Delta \phi\) and path difference \(\Delta x\) is given by:

\Delta \phi = \frac{2\pi \Delta x}{\lambda}

Substitute the given values to solve for wavelength \(\lambda\):

\frac{\pi}{3} = \frac{2\pi \times 0.12}{\lambda} \implies \frac{1}{3} = \frac{0.24}{\lambda} \implies \lambda = 0.72\text{ m}

Using the wave equation to find the speed \(v\):

v = f \lambda = 50\text{ Hz} \times 0.72\text{ m} = 36\text{ m s}^{-1}

Award 1 mark for the correct answer C. Correct application of the phase difference formula to find wavelength, followed by wave speed formula.

Initially, the power dissipated is:

P = \frac{E^2}{R}

Cutting the wire into three equal lengths divides the resistance of each piece proportionally to its length, so each piece has resistance:

R' = \frac{R}{3}

When connected in parallel, the reciprocal of the equivalent resistance \(R_{eq}\) is:

\frac{1}{R_{eq}} = \frac{1}{R'} + \frac{1}{R'} + \frac{1}{R'} = \frac{3}{R'} = \frac{3}{R/3} = \frac{9}{R} \implies R_{eq} = \frac{R}{9}

Now, calculate the new total power dissipated \(P'\) across the same e.m.f. \(E\):

P' = \frac{E^2}{R_{eq}} = \frac{E^2}{R/9} = 9\frac{E^2}{R} = 9P

Award 1 mark for the correct answer D. Standard reasoning should show that reducing resistance to 1/9 of the original value causes power to increase by a factor of 9 under constant potential difference.

1. Calculate the cross-sectional area \(A\) of the wire:

A = \frac{\pi d^2}{4} = \frac{\pi (1.2 \times 10^{-3})^2}{4} \approx 1.131 \times 10^{-6}\text{ m}^2

2. Calculate the tension force \(F\) on the wire:

F = mg = 5.0\text{ kg} \times 9.81\text{ m s}^{-2} = 49.05\text{ N}

3. Calculate the extension \(\Delta L\) from the Young modulus equation:

\Delta L = \frac{FL}{AE} = \frac{49.05 \times 2.0}{(1.131 \times 10^{-6}) \times (1.1 \times 10^{11})} \approx 7.885 \times 10^{-4}\text{ m}

4. Calculate the elastic potential energy stored:

E_p = \frac{1}{2} F \Delta L = \frac{1}{2} \times 49.05 \times 7.885 \times 10^{-4} \approx 1.93 \times 10^{-2}\text{ J}

Award 1 mark for the correct answer B. Common errors include forgetting the 1/2 factor in the potential energy equation (which leads to choice C) or using diameter instead of radius for the area.

The gravitational potential energy \(E_p\) of the satellite is given by:

E_p = -\frac{GMm}{r}

For a stable circular orbit, the centripetal force is provided by the gravitational attraction:

\frac{mv^2}{r} = \frac{GMm}{r^2} \implies mv^2 = \frac{GMm}{r}

The kinetic energy \(E_k\) of the satellite is:

E_k = \frac{1}{2}mv^2 = \frac{GMm}{2r}

By comparing the two expressions, we find:

E_p = -2 \left(\frac{GMm}{2r}\right) = -2E_k

Award 1 mark for the correct answer A. Note that gravitational potential energy is negative because potential is defined as zero at infinity, leading to the negative sign in the relationship.

Since the liquid is already at its boiling point and heat losses are negligible, the total electrical energy supplied equals the energy used for vaporization:

Q = P t = 60\text{ W} \times (10 \times 60\text{ s}) = 36000\text{ J}

The mass of liquid vaporized is:

m = 15\text{ g} = 0.015\text{ kg}

Using the formula for latent heat:

Q = mL \implies L = \frac{Q}{m} = \frac{36000\text{ J}}{0.015\text{ kg}} = 2.4 \times 10^6\text{ J kg}^{-1} = 2.4\text{ MJ kg}^{-1}

Award 1 mark for the correct answer B. Common errors include using the total mass of the liquid (0.20 kg) rather than the vaporized mass, or failing to convert minutes to seconds.

Substitute the SI base units of density and speed into the defining equation for acoustic impedance:

- Density \(\rho\) is mass per unit volume: \([\rho] = \text{kg m}^{-3}\)
- Speed \(c\) is distance per unit time: \([c] = \text{m s}^{-1}\)

Now combine them:

[Z] = [\rho] \times [c] = \text{kg m}^{-3} \times \text{m s}^{-1} = \text{kg m}^{-2} \text{s}^{-1}

Award 1 mark for the correct answer B. Ensure that the units of density and speed are correctly combined to simplify the overall expression.

The formula for density is \(\rho = \frac{m}{\pi r^2 h}\). The fractional uncertainty in \(\rho\) is given by \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta h}{h}\). Substituting the values: \(\frac{\Delta m}{m} = \frac{0.5}{50.0} = 0.010\) (or 1.0%), \(\frac{\Delta r}{r} = \frac{0.04}{2.00} = 0.020\) (or 2.0%), and \(\frac{\Delta h}{h} = \frac{0.1}{10.0} = 0.010\) (or 1.0%). Therefore, the percentage uncertainty in \(\rho\) is \(1.0\% + 2 \times 2.0\% + 1.0\% = 6.0\%\).

1 mark for the correct calculation of percentage uncertainty.

Let the upward direction be positive. The initial velocity is \(u = +v\). The final velocity just before hitting the water is \(v_f = -3v\). The acceleration is \(a = -g\). Using the equation of motion \(v_f^2 = u^2 + 2as\), we get \((-3v)^2 = v^2 + 2(-g)(-h)\) where \(h\) is the displacement downwards (the height of the cliff). This simplifies to \(9v^2 = v^2 + 2gh\), which gives \(8v^2 = 2gh\). Solving for \(h\) yields \(h = \frac{4v^2}{g}\).

1 mark for the correct derivation of the cliff height.

The total hydrostatic pressure \(P\) is the sum of the pressures exerted by each liquid layer: \(P = \rho_{\text{oil}} g h_{\text{oil}} + \rho_{\text{water}} g h_{\text{water}}\). Substituting the values: \(P = (800 \times 9.81 \times 0.150) + (1000 \times 9.81 \times 0.250) = 1177.2 + 2452.5 = 3629.7\text{ Pa}\), which rounds to \(3630\text{ Pa}\).

1 mark for the correct calculation of total pressure.

Taking the direction to the right as positive: Initial momentum \(p_{\text{initial}} = m_x u_x + m_y u_y = (2.0 \times 6.0) + (3.0 \times -2.0) = 12.0 - 6.0 = 6.0\text{ kg m s}^{-1}\). The total mass is \(M = 2.0 + 3.0 = 5.0\text{ kg}\). The final velocity is \(v = \frac{p_{\text{initial}}}{M} = \frac{6.0}{5.0} = 1.2\text{ m s}^{-1}\). The initial kinetic energy is \(E_{\text{k, initial}} = \frac{1}{2} m_x u_x^2 + \frac{1}{2} m_y u_y^2 = \frac{1}{2}(2.0)(6.0)^2 + \frac{1}{2}(3.0)(-2.0)^2 = 36.0 + 6.0 = 42.0\text{ J}\). The final kinetic energy is \(E_{\text{k, final}} = \frac{1}{2} M v^2 = \frac{1}{2}(5.0)(1.2)^2 = 3.6\text{ J}\). Thus, the loss in kinetic energy is \(42.0 - 3.6 = 38.4\text{ J}\).

1 mark for the correct determination of the kinetic energy lost.

The wavelength \(\lambda\) of the wave is given by \(\lambda = \frac{v}{f} = \frac{20}{50} = 0.40\text{ m} = 40\text{ cm}\). The phase difference \(\Delta \phi\) is related to the path difference \(\Delta x\) by \(\Delta \phi = \frac{\Delta x}{\lambda} \times 2\pi\). For \(\Delta x = 15\text{ cm}\), \(\Delta \phi = \frac{15}{40} \times 2\pi = \frac{3}{8} \times 2\pi = \frac{3\pi}{4}\text{ rad}\).

1 mark for the correct phase difference calculation.

Diffraction (the spreading of waves) becomes more significant when the wavelength \(\lambda\) is comparable to or larger than the gap width \(w\). Therefore, decreasing the gap width \(w\) relative to the wavelength increases the angle of diffraction, causing the waves to spread out more. Decreasing the wavelength or increasing the frequency reduces diffraction, and changing the amplitude has no effect on the angle of diffraction.

1 mark for identifying that decreasing the gap width increases diffraction.

Since mass and density remain constant, the volume of the wire remains constant. When the length is doubled to \(2L\), the cross-sectional area must halve to \(A/2\). The initial resistance is \(R = \rho \frac{L}{A}\). The new resistance is \(R' = \rho \frac{2L}{A/2} = 4R\). Since the potential difference \(V\) remains constant, the power dissipated is inversely proportional to resistance (\(P = \frac{V^2}{R}\)). Therefore, the new power is \(P' = \frac{V^2}{4R} = \frac{P}{4}\).

1 mark for finding the correct relation for new power.

The total resistance in the circuit is \(R + r\). As \(R\) increases, the total resistance increases, which causes the current \(I = \frac{E}{R + r}\) to decrease. The terminal potential difference \(V\) across the cell is given by \(V = E - Ir\). Since \(I\) decreases, the internal potential drop \(Ir\) decreases, and therefore \(V\) increases (approaching the e.m.f. \(E\) as \(R\) approaches infinity).

1 mark for the correct combination of changes for current and potential difference.

First, calculate the percentage uncertainty in each measured value:

Percentage uncertainty in \(R = \frac{0.1}{4.0} \times 100\% = 2.5\%\)
Percentage uncertainty in \(L = \frac{0.02}{2.00} \times 100\% = 1.0\%\)
Percentage uncertainty in \(d = \frac{0.01}{0.50} \times 100\% = 2.0\%\)

According to the formula \(\rho = \frac{R \pi d^2}{4L}\), the fractional uncertainty in \(\rho\) is given by:

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

Thus, the percentage uncertainty is:

\(\% \Delta \rho = 2.5\% + 2(2.0\%) + 1.0\% = 2.5\% + 4.0\% + 1.0\% = 7.5\%\)

1 mark for the correct calculation of percentage uncertainty. Reject other values due to failing to double the uncertainty of the diameter, or incorrectly combining uncertainties.

The initial momentum of the ball is \(p_{\text{initial}} = mu\) (taking the direction towards the wall as positive).

Since the collision is elastic, the ball rebounds with the same speed \(u\) in the opposite direction, so the final momentum is \(p_{\text{final}} = -mu\).

The change in momentum of the ball is:
\(\Delta p = p_{\text{final}} - p_{\text{initial}} = -mu - mu = -2mu\).

The magnitude of the change in momentum is \(2mu\).

By Newton's second law, the average force is the rate of change of momentum:
\(F = \frac{\Delta p}{\Delta t} = \frac{2mu}{\Delta t}\).

1 mark for the correct expression of average force. Reject options that do not account for the change in direction of momentum (e.g., resulting in \(mu/\Delta t\)).

The phase difference \(\phi\) is related to the path difference \(\Delta x\) by the equation:
\(\phi = \frac{2\pi \Delta x}{\lambda}\)

Substitute the given values to find the wavelength \(\lambda\):
\(\frac{\pi}{3} = \frac{2\pi (0.15)}{\lambda}\)

Solving for \(\lambda\):
\(\frac{1}{3} = \frac{0.30}{\lambda} \implies \lambda = 0.90\ \text{m}\)

Now, use the wave equation to find the speed \(v\):
\(v = f \lambda = 120\ \text{Hz} \times 0.90\ \text{m} = 108\ \text{m s}^{-1}\)

1 mark for the correct wave speed. Correctly determines the wavelength and then applies \(v = f\lambda\).

Let \(w\) be the fringe spacing (the distance between adjacent bright fringes).

The distance between the central bright fringe and the third-order bright fringe corresponds to 3 fringe spacings:
\(3w = 1.2\ \text{cm} = 0.012\ \text{m} \implies w = 0.0040\ \text{m} = 4.0\ \text{mm}\)

The formula for double-slit fringe spacing is:
\(w = \frac{\lambda D}{a}\)

Rearranging to solve for the screen distance \(D\):
\(D = \frac{w a}{\lambda}\)

Substitute the values in SI units:
\(D = \frac{(4.0 \times 10^{-3}\ \text{m}) \times (0.30 \times 10^{-3}\ \text{m})}{600 \times 10^{-9}\ \text{m}} = \frac{1.2 \times 10^{-6}}{6.0 \times 10^{-7}} = 2.0\ \text{m}\)

1 mark for the correct calculation of screen distance \(D\). Correctly identifies that 3 fringe spaces make up \(1.2\ \text{cm}\).

Let the resistance of the original wire be \(R\). The power dissipated is:
\(P = \frac{V^2}{R}\)

When cut into two equal halves, each half has a resistance of \(R/2\).

When connected in parallel, the equivalent resistance \(R_p\) is:
\(\frac{1}{R_p} = \frac{1}{R/2} + \frac{1}{R/2} = \frac{2}{R} + \frac{2}{R} = \frac{4}{R} \implies R_p = \frac{R}{4}\)

The total power dissipated in the parallel combination with the same potential difference \(V\) is:
\(P' = \frac{V^2}{R_p} = \frac{V^2}{R/4} = 4 \left( \frac{V^2}{R} \right) = 4P\)

1 mark for the correct ratio of power. Reject options that incorrectly relate resistance to length or fail to apply parallel circuit rules.

In the dark:
- The resistance of the LDR is \(6.0\ \text{k}\Omega\).
- Total circuit resistance \(R_{\text{dark}} = 3.0\ \text{k}\Omega + 6.0\ \text{k}\Omega = 9.0\ \text{k}\Omega\).
- Voltmeter reading across the LDR: \(V_{\text{dark}} = 9.0\ \text{V} \times \frac{6.0\ \text{k}\Omega}{9.0\ \text{k}\Omega} = 6.0\ \text{V}\).

In bright light:
- The resistance of the LDR is \(1.5\ \text{k}\Omega\).
- Total circuit resistance \(R_{\text{bright}} = 3.0\ \text{k}\Omega + 1.5\ \text{k}\Omega = 4.5\ \text{k}\Omega\).
- Voltmeter reading across the LDR: \(V_{\text{bright}} = 9.0\ \text{V} \times \frac{1.5\ \text{k}\Omega}{4.5\ \text{k}\Omega} = 3.0\ \text{V}\).

The change in the voltmeter reading is \(V_{\text{bright}} - V_{\text{dark}} = 3.0\ \text{V} - 6.0\ \text{V} = -3.0\ \text{V}\) (a decrease of \(3.0\ \text{V}\)).

1 mark for identifying the correct potential divider calculations for both states and finding the correct difference of 3.0 V (decrease).

The relationship for extension \(\Delta L\) in terms of Young Modulus \(E\) is given by:
\(E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L} \implies \Delta L = \frac{FL}{AE}\)

Since the cross-sectional area of a wire with diameter \(D\) is \(A = \frac{\pi D^2}{4}\), we have:
\(\Delta L = \frac{4FL}{\pi D^2 E}\)

Since the load \(F\) and material (Young Modulus \(E\)) are the same for both wires:
\(\Delta L \propto \frac{L}{D^2}\)

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

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

Therefore, the ratio is:
\(\frac{\Delta L_X}{\Delta L_Y} = \frac{\frac{L}{d^2}}{\frac{1}{2} \frac{L}{d^2}} = 2\)

1 mark for the correct ratio of 2. Correctly applies the dependence of area on the square of the diameter.

A neutron consists of one up (u) quark and two down (d) quarks, so its quark structure is \(udd\).

A proton consists of two up (u) quarks and one down (d) quark, so its quark structure is \(uud\).

During beta-minus decay, a neutron converts to a proton. This means one of the down quarks changes flavor into an up quark: \(d \to u\).

1 mark for selecting the correct quark transition (down to up).

The formula for resistivity is \(\rho = \frac{\pi R d^2}{4 L}\).

The fractional uncertainty in \(\rho\) is given by:

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

First, calculate the percentage uncertainty for each component:

1. For \(R\): \(\frac{0.1}{4.0} \times 100\% = 2.5\%\)

2. For \(d\): \(\frac{0.01}{0.50} \times 100\% = 2.0\%\)

3. For \(L\): \(\frac{0.02}{2.00} \times 100\% = 1.0\%\)

Substituting these values into the uncertainty equation:

Percentage uncertainty in \(\rho = 2.5\% + 2 \times (2.0\%) + 1.0\% = 2.5\% + 4.0\% + 1.0\% = 7.5\%\).

1 mark for the correct calculation of the combined percentage uncertainty (7.5%).

Let the initial direction of the \(0.20\text{ kg}\) sphere be positive.

Using the law of conservation of linear momentum:

\(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\)

\((0.20 \times 15) + (0.30 \times (-5.0)) = (0.20 \times (-3.0)) + (0.30 \times v_2)\)

\(3.0 - 1.5 = -0.60 + 0.30 v_2\)

\(1.5 = -0.60 + 0.30 v_2\)

\(2.1 = 0.30 v_2 \implies v_2 = 7.0\text{ m s}^{-1}\)

To determine whether the collision is elastic or inelastic, we compare the relative speed of approach to the relative speed of separation:

- Relative speed of approach = \(u_1 - u_2 = 15 - (-5.0) = 20\text{ m s}^{-1}\)

- Relative speed of separation = \(v_2 - v_1 = 7.0 - (-3.0) = 10\text{ m s}^{-1}\)

Since the relative speed of approach does not equal the relative speed of separation, kinetic energy is not conserved, which means the collision is inelastic.

1 mark for correctly determining the final speed of 7.0 m/s and identifying the collision as inelastic.

The total pressure at the bottom is the sum of the atmospheric pressure and the hydrostatic pressures exerted by the two liquid columns:

\(p_{\text{total}} = p_0 + \rho_1 g h_1 + \rho_2 g h_2\)

- The height of the upper liquid column, \(h_1 = 0.60\text{ m}\)

- The height of the lower liquid column, \(h_2 = 1.80\text{ m} - 0.60\text{ m} = 1.20\text{ m}\)

Calculating the hydrostatic pressures:

\(\rho_1 g h_1 = 800\text{ kg m}^{-3} \times 9.81\text{ m s}^{-2} \times 0.60\text{ m} = 4709\text{ Pa}\)

\(\rho_2 g h_2 = 1200\text{ kg m}^{-3} \times 9.81\text{ m s}^{-2} \times 1.20\text{ m} = 14126\text{ Pa}\)

Adding these to the atmospheric pressure:

\(p_{\text{total}} = 1.01 \times 10^5\text{ Pa} + 4709\text{ Pa} + 14126\text{ Pa} = 1.198 \times 10^5\text{ Pa} \approx 1.20 \times 10^5\text{ Pa}\).

1 mark for the correct application of the hydrostatic pressure formula including atmospheric pressure to obtain 1.20 x 10^5 Pa.

A phase difference of \(60^\circ\) is equivalent to a fraction of a full cycle:

\(\frac{60^\circ}{360^\circ} = \frac{1}{6}\ \text{cycle}\)

Therefore, the distance \(\Delta x = 0.18\text{ m}\) represents \(\frac{1}{6}\) of a wavelength \(\lambda\):

\(\lambda = 6 \times 0.18\text{ m} = 1.08\text{ m}\)

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

\(v = 125\text{ Hz} \times 1.08\text{ m} = 135\text{ m s}^{-1}\).

1 mark for calculating the wavelength as 1.08 m and subsequently the correct wave speed of 135 m s^-1.

The fringe spacing is given by \(x = \frac{\lambda D}{a}\).

Let the new fringe spacing be \(x'\):

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

Given:

- \(\lambda' = \frac{475}{633} \lambda \approx 0.75 \lambda\)

- \(D' = 2D\)

- \(a' = \frac{a}{2}\)

Substituting these new parameters into the expression for \(x'\):

\(x' = \frac{(0.75 \lambda) (2D)}{\frac{a}{2}} = 4 \times 0.75 \times \frac{\lambda D}{a} = 3.0 x\).

1 mark for establishing the correct ratio of fringe spacings, showing the new spacing is 3.0x.

The current in the circuit is given by \(I = \frac{E}{R+r}\).

The power dissipated in the resistor is \(P = I^2 R = \frac{E^2 R}{(R+r)^2}\).

1. When \(R = r\):

\(P_1 = \frac{E^2 r}{(r+r)^2} = \frac{E^2 r}{4r^2} = \frac{E^2}{4r}\)

2. When \(R = 3r\):

\(P_2 = \frac{E^2 (3r)}{(3r+r)^2} = \frac{3 E^2 r}{16r^2} = \frac{3 E^2}{16r}\)

Comparing the two values:

\(\frac{P_1}{P_2} = \frac{E^2 / (4r)}{3 E^2 / (16r)} = \frac{1}{4} \times \frac{16}{3} = \frac{4}{3}\).

1 mark for deriving correct power expressions in terms of internal resistance and calculating their ratio.

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

The strain energy is \(W = \frac{1}{2} F x = \frac{F^2 L}{2 A E} = \frac{2 F^2 L}{\pi d^2 E}\).

This shows that strain energy is directly proportional to length \(L\) and inversely proportional to the square of diameter \(d\):

\(W \propto \frac{L}{d^2}\)

For the second wire:

- Length is \(2L\)

- Diameter is \(\frac{1}{2}d\)

Hence, the strain energy in the second wire \(W_2\) is:

\(W_2 \propto \frac{2L}{\left(\frac{1}{2}d\right)^2} = \frac{2L}{\frac{1}{4}d^2} = 8 \left(\frac{L}{d^2}\right)\)

Therefore, \(W_2 = 8W\).

1 mark for using the scaling relationship of strain energy with length and diameter to obtain 8W.

In beta-minus decay, a neutron decays to a proton:

- Neutron composition: \(\text{udd}\)

- Proton composition: \(\text{uud}\)

Therefore, one down quark turns into an up quark, resulting in the number of up quarks increasing by 1, and the number of down quarks decreasing by 1.

Before the decay, the lepton number is 0. After the decay, the system contains an electron (lepton number = +1) and an electron antineutrino (lepton number = -1). The total lepton number is \(+1 + (-1) = 0\), meaning there is no change in the total lepton number of the system.

1 mark for correctly identifying the change in quarks and checking the conservation of total lepton number.

The percentage uncertainty in each quantity is:
- For resistance \(R\):
\(\frac{0.05}{2.50} \times 100\% = 2.0\%\)
- For diameter \(d\):
\(\frac{0.01}{0.40} \times 100\% = 2.5\%\)
- For length \(L\):
\(\frac{0.03}{1.50} \times 100\% = 2.0\%\)

The formula for resistivity is \(\rho = \frac{R \pi d^2}{4 L}\). The percentage uncertainty in \(\rho\) is given by:
\(\frac{\Delta \rho}{\rho} \times 100\% = \left( \frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L} \right) \times 100\%\)

Substituting the values:
\(\text{Percentage uncertainty} = 2.0\% + 2(2.5\%) + 2.0\% = 9.0\%\)

1 mark for calculating individual percentage uncertainties and correctly summing them according to the rules of combining uncertainties, resulting in 9.0%.

Let the initial direction of motion of the ball of mass \(m\) be positive.
By conservation of linear momentum:
\(m u + 0 = m\left(-\frac{u}{5}\right) + 3m v\)
where \(v\) is the velocity of the second ball after collision.

Simplifying:
\(u = -\frac{u}{5} + 3v \implies \frac{6}{5}u = 3v \implies v = \frac{2}{5}u\)

The initial kinetic energy of the system is:
\(E_{ki} = \frac{1}{2} m u^2\)

The final kinetic energy of the system is:
\(E_{kf} = \frac{1}{2} m \left(-\frac{u}{5}\right)^2 + \frac{1}{2}(3m)\left(\frac{2}{5}u\right)^2\)
\(E_{kf} = \frac{1}{2} m \left(\frac{u^2}{25}\right) + \frac{3}{2} m \left(\frac{4u^2}{25}\right) = \frac{1}{2} m u^2 \left(\frac{1}{25} + \frac{12}{25}\right) = \frac{13}{25} \left(\frac{1}{2} m u^2\right)\)

The kinetic energy lost is:
\(\Delta E_k = E_{ki} - E_{kf} = \left(1 - \frac{13}{25}\right) E_{ki} = \frac{12}{25} E_{ki}\)

Therefore, the fraction of initial kinetic energy lost is \(\frac{12}{25}\).

1 mark for using conservation of momentum to find the post-collision velocity of the second ball, calculating initial and final kinetic energies, and finding the fraction of energy lost.

The relationship between phase difference \(\Delta \phi\) and path difference \(d\) is given by:
\(\Delta \phi = \frac{2\pi d}{\lambda}\)

Given \(\Delta \phi = \frac{2\pi}{3}\,\text{rad}\) and \(d = 0.25\,\text{m}\):
\(\frac{2\pi}{3} = \frac{2\pi (0.25)}{\lambda}\)
\(\frac{1}{3} = \frac{0.25}{\lambda} \implies \lambda = 0.75\,\text{m}\)

The speed of the wave \(v\) is given by:
\(v = f \lambda = 120\,\text{Hz} \times 0.75\,\text{m} = 90\,\text{m}\,\text{s}^{-1}\)

1 mark for calculating the wavelength from the phase difference relation, then correctly applying \(v = f \lambda\) to find the speed of the wave.

The formula for the double-slit fringe width is:
\(x = \frac{\lambda D}{a}\)

Let the new parameters be \(\lambda' = 1.2 \lambda\), \(a' = 0.5 a\), and \(D' = 2 D\).
The new fringe width \(x'\) is:
\(x' = \frac{\lambda' D'}{a'} = \frac{(1.2 \lambda)(2 D)}{0.5 a} = \frac{2.4}{0.5} \frac{\lambda D}{a} = 4.8 x\)

1 mark for using the double-slit formula and scaling the variables correctly to determine the factor of 4.8.

The resistance of a wire is given by \(R = \frac{\rho L}{A} = \frac{\rho L}{\pi d^2 / 4}\).

For the first wire, the resistance is \(R_1 = \frac{4 \rho L}{\pi d^2}\).
For the second wire, the resistance is:
\(R_2 = \frac{\rho (2L)}{\pi (2d)^2 / 4} = \frac{2 \rho L}{4 \pi d^2 / 4} = \frac{1}{2} R_1\).

The power dissipated under constant potential difference \(V\) is:
\(P = \frac{V^2}{R}\)

Therefore, the power dissipated in the second wire is:
\(P_2 = \frac{V^2}{R_2} = \frac{V^2}{\frac{1}{2} R_1} = 2 \left(\frac{V^2}{R_1}\right) = 2P\)

1 mark for determining that the second wire has half the resistance of the first, and thus twice the power dissipation when connected across the same potential difference.

The elastic strain energy \(E\) stored in a wire under tension is given by:
\(E = \frac{1}{2} F x\)
where \(x\) is the extension.

From Young modulus \(E_Y = \frac{F/A}{x/L}\), the extension is:
\(x = \frac{F L}{A E_Y}\)

Therefore, the strain energy is:
\(E = \frac{F^2 L}{2 A E_Y}\)

Since \(F\) and \(E_Y\) are the same for both wires, and area \(A \propto d^2\):
\(E \propto \frac{L}{d^2}\)

For wire X:
\(E_X \propto \frac{L}{d^2}\)

For wire Y:
\(E_Y \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{L}{2d^2}\)

Thus, the ratio is:
\(\frac{E_X}{E_Y} = \frac{L / d^2}{L / (2d^2)} = 2\)

1 mark for identifying the proportional relationship between elastic strain energy and the dimensions of the wire under a constant force, and calculating the ratio to be 2.

Let \(r\) be the distance from the center of the planet where the net gravitational field strength is zero.
The distance from this point to the center of the moon is \(R - r\).

Equating the magnitude of the gravitational field strength of the planet and the moon at this point:
\(\frac{G M}{r^2} = \frac{G (0.04M)}{(R - r)^2}\)

Cancel out \(G M\):
\(\frac{1}{r^2} = \frac{0.04}{(R - r)^2}\)

Taking the square root of both sides:
\(\frac{1}{r} = \frac{0.2}{R - r}\)

Cross-multiplying:
\(R - r = 0.2 r\)
\(R = 1.2 r \implies r = \frac{R}{1.2} \approx 0.83 R\)

1 mark for setting up the equation for equal and opposite gravitational fields, solving for distance \(r\), and obtaining approximately 0.83 R.

According to the first law of thermodynamics:
\(\Delta U = q + w\)
where:
- \(\Delta U\) is the change in internal energy,
- \(q\) is the thermal energy transferred to the system (\(q = +800\,\text{J}\) since it is absorbed),
- \(w\) is the work done on the system.

Since the gas does \(300\,\text{J}\) of work on the surroundings, the work done on the gas is \(w = -300\,\text{J}\).

Thus:
\(\Delta U = 800\,\text{J} + (-300\,\text{J}) = +500\,\text{J}\)

The internal energy of the gas increases by \(500\,\text{J}\).

1 mark for correctly identifying the sign conventions for heat supplied to and work done by the system, and applying the first law of thermodynamics to find the increase of 500 J.

(a) Systematic errors cause all measurements to deviate from the true value in the same direction / by a constant amount each time. In contrast, random errors cause measurements to vary unpredictably on either side of the true value. Systematic errors cannot be eliminated by repeating and averaging, whereas random errors can be reduced through this process.

(b) The formula for resistivity is:
\(\rho = \frac{R A}{L} = \frac{R \pi d^2}{4 L}\)

Substitute the given values:
\(\rho = \frac{4.25 \times \pi \times (0.38 \times 10^{-3})^2}{4 \times 1.240}\)
\(\rho = \frac{4.25 \times 3.1416 \times 1.444
\times 10^{-7}}{4.960}\)
\(\rho \approx 3.89 \times 10^{-7}\ \Omega\text{ m}\)

Rounding to 2 significant figures gives:
\(\rho = 3.9 \times 10^{-7}\ \Omega\text{ m}\)

(c) The fractional uncertainty in resistivity is given by:
\(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\)

Calculate individual terms:
\(\frac{\Delta R}{R} = \frac{0.05}{4.25} \approx 0.0118\)
\(2 \frac{\Delta d}{d} = 2 \times \frac{0.02}{0.38} \approx 0.1053\)
\(\frac{\Delta L}{L} = \frac{0.002}{1.240} \approx 0.0016\)

Total fractional uncertainty:
\(\frac{\Delta \rho}{\rho} = 0.0118 + 0.1053 + 0.0016 = 0.1187\)

Absolute uncertainty:
\
\Delta \rho = 0.1187 \times 3.89 \times 10^{-7} \approx 0.46 \times 10^{-7}\ \Omega\text{ m}\)

Rounding the absolute uncertainty to 1 significant figure:
\(\Delta \rho = \pm 0.5 \times 10^{-7}\ \Omega\text{ m}\)
Thus, \(\rho = (3.9 \pm 0.5) \times 10^{-7}\ \Omega\text{ m}\).

(a)
- Systematic errors cause measurements to deviate consistently in one direction / have a constant offset. [1]
- Random errors cause measurements to vary unpredictably about the mean. [1]
- Averaging reduces random errors but does not reduce systematic errors. [0.5]

(b)
- Recalls and uses \(\rho = \frac{R \pi d^2}{4 L}\). [1]
- Correct substitution of values with SI units. [1]
- Evaluates answer to \(3.9 \times 10^{-7}\ \Omega\text{ m}\) (or \(3.89 \times 10^{-7}\ \Omega\text{ m}\)). [1]

(c)
- Correct algebraic sum of fractional uncertainties, doubling the fractional uncertainty of diameter. [1]
- Correctly calculates total fractional uncertainty to be \(0.12\) (or \(0.119\)). [1]
- Multiplies fractional uncertainty by value of \(\rho\) and rounds absolute uncertainty to 1 s.f. (accepts \(\pm 0.5 \times 10^{-7}\) or \(\pm 0.46 \times 10^{-7}\)). [1]

(a) Vertical component:
\(u_y = u \sin \theta = 25.0 \sin(30.0^\circ) = 12.5\text{ m s}^{-1}\)

Horizontal component:
\(u_x = u \cos \theta = 25.0 \cos(30.0^\circ) = 21.65\text{ m s}^{-1} \approx 21.7\text{ m s}^{-1}\)

(b) Using the equations of motion for the vertical direction, taking upwards as the positive direction:
\(s_y = u_y t + \frac{1}{2} a_y t^2\)
Here, \(s_y = -45.0\text{ m}\), \(u_y = 12.5\text{ m s}^{-1}\), and \(a_y = -9.81\text{ m s}^{-2}\).

\(-45.0 = 12.5 t - 4.905 t^2\)

Rearranging into a standard quadratic equation:
\(4.905 t^2 - 12.5 t - 45.0 = 0\)

Solve using the quadratic formula:
\(t = \frac{-(-12.5) \pm \sqrt{(-12.5)^2 - 4(4.905)(-45.0)}}{2(4.905)}\)
\(t = \frac{12.5 \pm \sqrt{156.25 + 882.9}}{9.81}\)
\(t = \frac{12.5 \pm \sqrt{1039.15}}{9.81}\)
\(t = \frac{12.5 \pm 32.24}{9.81}\)

This yields two values:
\(t_1 = 4.56\text{ s}\) and \(t_2 = -2.01\text{ s}\).

Since time must be positive, the time taken is \(t = 4.56\text{ s}\).

(c) The horizontal motion has constant velocity because air resistance is neglected:
\(x = u_x \times t\)
\(x = 21.65 \times 4.56 \approx 98.7\text{ m}\)
(Allow \(99.0\text{ m}\) if rounding to 2 significant figures or using intermediate rounded values).

(a)
- Shows vertical component using sine function. [0.5]
- Correct calculation of horizontal component as \(21.7\text{ m s}^{-1}\) (or \(21.65\text{ m s}^{-1}\)). [1]

(b)
- Identifies vertical displacement as \(-45.0\text{ m}\) (with correct sign convention). [1]
- Formulates the quadratic equation with correct signs. [1]
- Demonstrates correct substitution into the quadratic formula. [1]
- Obtains final positive time of \(4.56\text{ s}\). [1]

(c)
- Recalls that horizontal velocity remains constant. [1]
- Multiplies horizontal component by the time calculated in part (b). [1]
- Obtains correct final distance of \(98.7\text{ m}\) (accept range \(98.5\text{ m}\) to \(99.0\text{ m}\) depending on rounding). [1]

(a) The principle of conservation of momentum states that the total momentum of a closed system remains constant, provided no external resultant force acts on it.

(b) Let glider A have mass \(m_A = 0.45\text{ kg}\) and initial velocity \(u_A = 1.20\text{ m s}^{-1}\).
Let glider B have mass \(m_B = 0.25\text{ kg}\) and initial velocity \(u_B = 0\text{ m s}^{-1}\).

After the collision:
Final velocity of B is \(v_B = 1.50\text{ m s}^{-1}\).
Final velocity of A is \(v_A\).

By conservation of momentum:
\(m_A u_A + m_B u_B = m_A v_A + m_B v_B\)
\((0.45 \times 1.20) + (0.25 \times 0) = (0.45 \times v_A) + (0.25 \times 1.50)\)
\(0.54 = 0.45 v_A + 0.375\)
\(0.45 v_A = 0.54 - 0.375 = 0.165\)
\(v_A = \frac{0.165}{0.45} \approx 0.367\text{ m s}^{-1}\)

Thus, the velocity of glider A after the collision is \(0.37\text{ m s}^{-1}\) (or \(0.367\text{ m s}^{-1}\)) in the original direction.

(c) A collision is perfectly elastic if the total kinetic energy is conserved (or if the relative speed of approach equals the relative speed of separation).

Using relative speeds:
Relative speed of approach = \(u_A - u_B = 1.20 - 0 = 1.20\text{ m s}^{-1}\)
Relative speed of separation = \(v_B - v_A = 1.50 - 0.367 = 1.133\text{ m s}^{-1}\)

Since the relative speed of approach \((1.20\text{ m s}^{-1})\) is not equal to the relative speed of separation \((1.13\text{ m s}^{-1})\), the collision is inelastic.

Alternatively, calculate kinetic energy:
Initial KE = \(\frac{1}{2} m_A u_A^2 = 0.5 \times 0.45 \times 1.20^2 = 0.324\text{ J}\)
Final KE = \(\frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 = 0.5 \times 0.45 \times 0.367^2 + 0.5 \times 0.25 \times 1.50^2 \approx 0.0303 + 0.2813 = 0.3116\text{ J}\)
Since Initial KE \(\neq\) Final KE, the collision is inelastic.

(a)
- Total momentum of a closed/isolated system remains constant. [1]
- In the absence of an external resultant force. [1]

(b)
- Recalls and uses conservation of momentum equation. [1]
- Correctly substitutes values into the equation. [1]
- Obtains intermediate value \(0.45 v_A = 0.165\). [1]
- Final answer: \(0.37\text{ m s}^{-1}\) (or \(0.367\text{ m s}^{-1}\)). [0.5]

(c)
- States that for an elastic collision, kinetic energy is conserved or relative speed of approach equals relative speed of separation. [1]
- Calculates both relative speeds (approach = \(1.20\text{ m s}^{-1}\), separation = \(1.13\text{ m s}^{-1}\)) OR calculates both initial and final kinetic energies (initial = \(0.32\text{ J}\), final = \(0.31\text{ J}\)). [1]
- Draws a correct conclusion based on the calculations (i.e. collision is inelastic because they are not equal). [1]

(a) A progressive wave is a wave that transfers energy from one point to another through oscillations of the medium, without any permanent transfer of the medium itself.

(b) Comparing the given equation \(y = 4.5 \times 10^{-6} \sin(2500 t - 8.2 x)\) with the general wave equation \(y = A \sin(\omega t - k x)\):

(i) Angular frequency \(\omega = 2500\text{ rad s}^{-1}\).
Using \(\omega = 2 \pi f\):
\(f = \frac{\omega}{2 \pi} = \frac{2500}{2 \pi} \approx 397.89\text{ Hz}\)
Rounding to 3 significant figures: \(f = 398\text{ Hz}\).

(ii) Wave number \(k = 8.2\text{ m}^{-1}\).
Using \(k = \frac{2 \pi}{\lambda}\):
\(\lambda = \frac{2 \pi}{k} = \frac{2 \pi}{8.2} \approx 0.7662\text{ m}\)
Rounding to 2 significant figures: \(\lambda = 0.77\text{ m}\).

(iii) The speed of the wave is given by:
\(v = f \lambda\)
Using the unrounded values:
\(v = 397.89 \times 0.7662 \approx 304.9\text{ m s}^{-1}\)
Alternatively, \(v = \frac{\omega}{k} = \frac{2500}{8.2} \approx 305\text{ m s}^{-1}\).

(a)
- Transfer of energy. [1]
- Without transferring matter/medium. [0.5]

(b)(i)
- Identifies \(\omega = 2500\text{ rad s}^{-1}\). [1]
- Correctly recalls and uses \(\omega = 2\pi f\). [1]
- Obtains \(398\text{ Hz}\) (or \(4.0 \times 10^2\text{ Hz}\)). [0.5]

(b)(ii)
- Identifies \(k = 8.2\text{ m}^{-1}\) as the coefficient of \(x\). [1]
- Recalls and uses \(k = \frac{2\pi}{\lambda}\). [1]
- Obtains \(0.77\text{ m}\) (or \(0.766\text{ m}\)). [0.5]

(b)(iii)
- Recalls and uses \(v = f \lambda\) or \(v = \frac{\omega}{k}\). [1]
- Obtains \(305\text{ m s}^{-1}\) (allow range \(300\text{ m s}^{-1}\) to \(310\text{ m s}^{-1}\) depending on rounding of frequency and wavelength). [1]

(a) For stable and visible interference fringes to be observed:
1. The two sources of light must be coherent (they must have a constant phase relationship/difference and the same frequency).
2. The waves must overlap/superpose.
3. The sources should have similar amplitudes to produce high contrast fringes (clear bright and dark regions).

(b) The double-slit equation is:
\(x = \frac{\lambda D}{a}\)

Identify the given parameters and convert to SI units:
\(\lambda = 633\text{ nm} = 633 \times 10^{-9}\text{ m}\)
\(a = 0.450\text{ mm} = 0.450 \times 10^{-3}\text{ m}\)
\(D = 2.20\text{ m}\)

Substitute values into the equation:
\(x = \frac{633 \times 10^{-9} \times 2.20}{0.450 \times 10^{-3}}\)
\(x = 3.095 \times 10^{-3}\text{ m}\)

So the fringe spacing is \(3.10\text{ mm}\) (or \(3.1 \times 10^{-3}\text{ m}\)).

(c) The fringe spacing formula is \(x = \frac{\lambda D}{a}\). Since \(D\) and \(a\) remain constant, the fringe spacing is directly proportional to the wavelength (\(x \propto \lambda\)). Blue light has a shorter wavelength (\(450\text{ nm}\)) than the red light (\(633\text{ nm}\)). Therefore, replacing the red laser with a blue laser causes the fringe spacing to decrease (the fringes will be closer together).

(a)
- Sources must be coherent (or constant phase difference). [1]
- Light waves must overlap. [1]
- Similar amplitudes to ensure high contrast / dark fringes are completely dark. [0.5]

(b)
- Recalls and uses \(x = \frac{\lambda D}{a}\). [1]
- Correctly converts standard SI prefixes (\(\text{nm}\) to \(10^{-9}\text{ m}\) and \(\text{mm}\) to \(10^{-3}\text{ m}\)). [1]
- Obtains \(3.1 \times 10^{-3}\text{ m}\) (or \(3.1\text{ mm}\) / \(3.10 \times 10^{-3}\text{ m}\)). [1]

(c)
- States that fringe spacing is directly proportional to wavelength (\(x \propto \lambda\)). [1]
- Recognises that blue light has a shorter wavelength than red light. [1]
- Concludes that the fringe spacing decreases (fringes are closer together). [1]

(a) Power is related to voltage and resistance by the formula:
\(P = \frac{V^2}{R}\)

Rearranging for resistance \(R\):
\(R = \frac{V^2}{P}\)

Given \(V = 230\text{ V}\) and \(P = 1.80\text{ kW} = 1800\text{ W}\):
\(R = \frac{230^2}{1800} = \frac{52900}{1800} \approx 29.39\ \Omega\)
This is approximately \(29.3\ \Omega\).

(b) To find the charge \(Q\) passing through the heating element, we use:
\(Q = I \times t\)

First, find the current \(I\) flowing through the element:
\(I = \frac{P}{V} = \frac{1800}{230} \approx 7.826\text{ A}\)

Convert time to seconds:
\(t = 45.0\text{ minutes} = 45.0 \times 60 = 2700\text{ s}\)

Calculate the charge:
\(Q = 7.826 \times 2700 \approx 21130\text{ C}\)

In standard scientific notation, \(Q = 2.11 \times 10^4\text{ C}\).

(c) Assuming the resistance \(R = 29.39\ \Omega\) remains constant, the power output at \(V_{\text{new}} = 210\text{ V}\) is:
\(P_{\text{new}} = \frac{V_{\text{new}}^2}{R} = \frac{210^2}{29.39} = \frac{44100}{29.39} \approx 1500\text{ W} = 1.50\text{ kW}\)

(a)
- Recalls and uses \(P = \frac{V^2}{R}\). [1]
- Substitutes values into the formula to show \(29.3\ \Omega\) or \(29.4\ \Omega\). [1]

(b)
- Recalls and uses \(P = I V\) or equivalent to find current \(I = 7.83\text{ A}\). [1]
- Converts minutes to seconds correctly (\(45.0 \times 60 = 2700\text{ s}\)). [1]
- Recalls and uses \(Q = I t\). [1]
- Obtains \(2.11 \times 10^4\text{ C}\) (accept range \(2.10 \times 10^4\) to \(2.12 \times 10^4\)). [0.5]

(c)
- Recalls and uses \(P = \frac{V^2}{R}\) with new voltage. [1]
- Substitutes new voltage of \(210\text{ V}\) and resistance from (a). [1]
- Obtains correct power output of \(1.50\text{ kW}\) (or \(1500\text{ W}\)). [1]

(a) Hooke's law states that the force applied to a body is directly proportional to its extension, provided that the limit of proportionality is not exceeded.

(b) First, calculate the tension force \(F\) acting on the wire due to the hanging mass:
\(F = m g = 6.0 \times 9.81 = 58.86\text{ N}\)

Young modulus \(E\) is given by:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{e / L} = \frac{F L}{A e}\)

Rearranging to make extension \(e\) the subject:
\(e = \frac{F L}{A E}\)

Substitute the given values:
\(e = \frac{58.86 \times 2.50}{(1.20 \times 10^{-6}) \times (2.0 \times 10^{11})}\)
\(e = \frac{147.15}{240000} = 6.13 \times 10^{-4}\text{ m} = 0.613\text{ mm}\)

So, the extension is \(0.61\text{ mm}\) (or \(6.1 \times 10^{-4}\text{ m}\)).

(c) Assuming Hooke's law is obeyed, the elastic potential energy \(E_p\) stored in the wire is given by:
\(E_p = \frac{1}{2} F e\)

Substitute the values for force and extension:
\(E_p = 0.5 \times 58.86 \times (6.13 \times 10^{-4})\)
\(E_p \approx 0.0180\text{ J}\)

Thus, the stored elastic potential energy is \(0.018\text{ J}\) (or \(1.8 \times 10^{-2}\text{ J}\)).

(a)
- Force is proportional to extension. [1]
- Provided the limit of proportionality is not exceeded. [1]

(b)
- Calculates the tension force \(F = 58.9\text{ N}\) (or \(58.86\text{ N}\)) using \(g = 9.81\text{ m s}^{-2}\). [1]
- Recalls and uses Young modulus formula \(E = \frac{F L}{A e}\). [1]
- Correctly rearranges for extension \(e = \frac{F L}{A E}\). [0.5]
- Substitutes values into equation and obtains \(6.1 \times 10^{-4}\text{ m}\) (or \(0.61\text{ mm}\)). [1]

(c)
- Recalls and uses \(E_p = \frac{1}{2} F e\) or \(E_p = \frac{1}{2} k e^2\). [1]
- Correctly substitutes values for force and extension. [1]
- Calculates stored elastic potential energy to be \(0.018\text{ J}\) (or \(1.8 \times 10^{-2}\text{ J}\)). [1]

### (a) Table of Results

The values of \( d^3 / 10^3 \text{ cm}^3 \) are calculated to 3 significant figures, matching the precision of \( d \):

| \( d / \text{cm} \) | \( y / \text{cm} \) | \( d^3 / 10^3 \text{ cm}^3 \) |
| :---: | :---: | :---: |
| 15.0 | 0.7 | 3.38 |
| 20.0 | 1.6 | 8.00 |
| 25.0 | 3.1 | 15.6 |
| 30.0 | 5.2 | 27.0 |
| 35.0 | 8.3 | 42.9 |
| 40.0 | 12.3 | 64.0 |

### (b) Graph Plotting
- **Scale**: x-axis from 0 to 70 with 10 units per 2 cm; y-axis from 0 to 14 with 2 units per 2 cm. Points occupy more than 50% of the grid.
- **Plotted points**: Points are plotted correctly at (3.38, 0.7), (8.00, 1.6), (15.6, 3.1), (27.0, 5.2), (42.9, 8.3), and (64.0, 12.3).
- **Line of Best Fit**: A straight line passing symmetrically through the points.
- **Worst Acceptable Line**: A line passing through the centroid but with a different gradient, passing through the first point's upper error limit and the last point's lower error limit.

### (c) Gradient and y-intercept
- **Gradient calculation**:
Using coordinates on the line of best fit, e.g., \((5.0, 1.0)\) and \((60.0, 11.5)\):
\[ \text{Gradient} = \frac{11.5 - 1.0}{60.0 - 5.0} = \frac{10.5}{55.0} \approx 0.191 \text{ cm} / (10^3 \text{ cm}^3) \]
- **y-intercept calculation**:
Using \( y = mx + C \):
\[ 1.0 = (0.191 \times 5.0) + C \implies C = 1.0 - 0.955 = 0.045 \text{ cm} \approx 0.05 \text{ cm} \]
(Acceptable intercept range from graph reading or calculation: \( 0.0 \) to \( 0.2 \text{ cm} \)).

### (d) Determination of constants \( P \) and \( Q \)
- Since \( y = P d^3 + Q \) and the x-axis is \( X = d^3 / 10^3 \), the equation plotted is:
\[ y = (10^3 P) X + Q \]
- Therefore, the gradient \( m = 10^3 P \):
\[ P = \frac{m}{10^3} = \frac{0.191}{10^3} = 1.91 \times 10^{-4} \text{ cm}^{-2} \]
- The y-intercept represents \( Q \):
\[ Q = 0.05 \text{ cm} \]

### (e) Analysis of a thinner strip
- (i) Since \( P \) represents the constant of proportionality and the gradient is directly proportional to \( P \) (\( \text{gradient} = 10^3 P \)), an increase in \( P \) by a factor of 1.50 means the **gradient will increase by a factor of 1.50** (the line becomes steeper).
- (ii) New gradient:
\[ \text{New Gradient} = 1.50 \times 0.191 = 0.287 \text{ cm} / (10^3 \text{ cm}^3) \]

**(a) Table of Results (4 marks):**
- **[1]** Column headers include correct quantities and units: \( d / \text{cm} \), \( y / \text{cm} \), and \( d^3 / 10^3 \text{ cm}^3 \) (or equivalent).
- **[1]** All raw data of \( d \) and \( y \) are recorded to the same absolute precision (1 decimal place for cm, representing 1 mm scale division).
- **[1]** Significant figures for \( d^3 / 10^3 \) are calculated to 3 s.f. (matching the s.f. of \( d \)).
- **[1]** Correctly calculated values of \( d^3 / 10^3 \) (allow \( \pm 1 \) on last decimal place).

**(b) Graph Plotting (4 marks):**
- **[1]** Axes are labeled with quantities and units. Linear scales chosen so that points occupy more than half of the grid in both horizontal and vertical directions.
- **[1]** All points plotted correctly within half a small grid square.
- **[1]** Line of best fit is straight with balanced distribution of points about the line.
- **[1]** Worst acceptable line is clearly identified, passing through the centroid and the extremes of the dataset.

**(c) Gradient and y-intercept (4 marks):**
- **[1]** Large triangle used for gradient calculation (hypotenuse length must be greater than half the length of the drawn line).
- **[1]** Correct gradient value obtained in the range \( 0.185 \) to \( 0.200 \text{ cm} / (10^3 \text{ cm}^3) \).
- **[1]** y-intercept calculated correctly using coordinates of a point on the line of best fit, or read directly from y-axis if x=0 is included.
- **[1]** y-intercept value in the range \( 0.0 \) to \( 0.2 \text{ cm} \).

**(d) Determining constants \( P \) and \( Q \) (4 marks):**
- **[1]** Value of \( P \) correctly calculated by dividing gradient by \( 10^3 \) (value in range \( 1.8 \times 10^{-4} \) to \( 2.0 \times 10^{-4} \)).
- **[1]** Value of \( Q \) equated directly to y-intercept.
- **[1]** Unit for \( P \) is \( \text{cm}^{-2} \) (or \( \text{m}^{-2} \)) and unit for \( Q \) is \( \text{cm} \) (or \( \text{m} \)).
- **[1]** Values of \( P \) and \( Q \) written to 2 or 3 significant figures.

**(e) Second Student's Experiment (4 marks):**
- **[1]** (i) States that the gradient will increase.
- **[1]** (i) Explains that because the gradient is equal to \( 10^3 P \) and \( P \) increases, the gradient must increase proportionally.
- **[1]** (ii) Correct substitution of candidate's gradient from (c) multiplied by 1.50.
- **[1]** (ii) Correct final value of new gradient in range \( 0.275 \) to \( 0.300 \text{ cm} / (10^3 \text{ cm}^3) \).

### Specimen Results

**(a)(i) Measurement of \(d_1\)**
\(d_1 = 25.0\text{ cm}\) (or \(0.250\text{ m}\))

**(a)(ii) Percentage Uncertainty in \(d_1\)**
An absolute uncertainty \(\Delta d\) of \(0.2\text{ cm}\) (or \(2\text{ mm}\)) is appropriate because the alignment of both ends of two separate threads must be judged.
\[ \text{Percentage Uncertainty} = \frac{\Delta d}{d_1} \times 100\% = \frac{0.2\text{ cm}}{25.0\text{ cm}} \times 100\% = 0.8\% \]

**(b)(i) Measurement of \(t_1\)**
Time for 10 oscillations, \(t_1 = 14.24\text{ s}\)

**(b)(ii) Calculation of Period \(T_1\)**
\(T_1 = \frac{t_1}{10} = 1.424\text{ s}\)

**(c) Measurement of \(d_2\)**
Adjusted distance, \(d_2 = 15.0\text{ cm}\) (or \(0.150\text{ m}\))

**(d) Measurement of \(t_2\) and Period \(T_2\)**
Time for 10 oscillations, \(t_2 = 18.46\text{ s}\)
\(T_2 = \frac{18.46}{10} = 1.846\text{ s}\)

**(e)(i) Calculation of Constants \(C_1\) and \(C_2\)**
\(C_1 = T_1^2 \times d_1 = (1.424)^2 \times 0.250 = 2.028 \times 0.250 = 0.507\text{ s}^2\text{ m}\)
\(C_2 = T_2^2 \times d_2 = (1.846)^2 \times 0.150 = 3.408 \times 0.150 = 0.511\text{ s}^2\text{ m}\)

**(e)(ii) Evaluation of Constant Relationship**
Calculate the percentage difference between \(C_1\) and \(C_2\):
\[ \% \text{ difference} = \frac{|C_2 - C_1|}{C_1} \times 100\% = \frac{|0.511 - 0.507|}{0.507} \times 100\% = 0.79\% \]

*Criterion:* A standard criterion for a simple school laboratory experiment is that if the percentage difference is less than \(10\%\), the relationship is supported.
*Conclusion:* Since the percentage difference of \(0.79\%\) is significantly less than the specified limit of \(10\%\), the results support the suggestion that \(C\) is constant.

**(f) Limitations and Improvements Table**

| # | Difficulty / Source of Uncertainty (Limitation) | Corresponding Improvement |
|---|---|---|
| 1 | Two sets of readings are not enough to draw a valid conclusion. | Take more readings for several different distances \(d\) and plot a graph of \(T^2\) against \(1/d\). |
| 2 | Difficult to keep the two threads vertical and parallel / difficult to measure \(d\) accurately because threads may skew. | Use a set-square/spirit level to ensure threads are perpendicular to the support, and measure \(d\) at both top and bottom to ensure they are equal. |
| 3 | Difficult to identify the exact end of an oscillation / human reaction time error in starting and stopping the stopwatch. | Place a vertical reference marker (fiducial mark) at the centre of the oscillation, or use a motion sensor/light gate connected to a data logger. |
| 4 | The wooden strip does not just rotate; it also oscillates laterally like a standard pendulum when released. | Use a mechanical release mechanism (e.g., burning a thread) to release the strip symmetrically, or use a guide pin at the centre of rotation to prevent lateral movement. |

### Marking Scheme

* **(a)(i)** [1 mark]
* Value of \(d_1\) recorded to the nearest millimetre with correct unit (e.g. \(25.0\text{ cm}\) or \(0.250\text{ m}\)).

* **(a)(ii)** [2 marks]
* Absolute uncertainty \(\Delta d\) in the range \(1\text{ mm}\) to \(4\text{ mm}\) chosen and justified based on aligning two separate threads [1 mark].
* Correct calculation of percentage uncertainty with working shown [1 mark].

* **(b)(i)** [1 mark]
* Raw time \(t_1\) recorded with unit (seconds) to at least \(0.1\text{ s}\) or \(0.01\text{ s}\) for at least 5 oscillations (10 recommended).

* **(b)(ii)** [1 mark]
* Period \(T_1\) calculated correctly as \(t_1 / 10\).

* **(c)** [1 mark]
* Second value of \(d_2\) recorded, showing \(d_2 < d_1\).

* **(d)** [2 marks]
* Raw time \(t_2\) recorded to consistent precision as \(t_1\) [1 mark].
* Correct calculation of \(T_2\) [1 mark].

* **(e)(i)** [2 marks]
* Calculation of both \(C_1\) and \(C_2\) is correct [1 mark].
* Unit of \(C\) is correct (e.g., \(\text{s}^2\text{ m}\) or \(\text{s}^2\text{ cm}\)) [1 mark].

* **(e)(ii)** [2 marks]
* Correct calculation of percentage difference between \(C_1\) and \(C_2\) [1 mark].
* Explicit comparison to a stated criterion (e.g., \(10\%\) or the percentage uncertainty calculated in (a)(ii)) leading to a logical conclusion on whether the relationship is supported [1 mark].

* **(f) Table of Limitations and Improvements** [8 marks, 1 mark per point up to 4 pairs]:
* **Limitation 1:** Two readings not enough to draw a valid conclusion.
* **Improvement 1:** Take more readings and plot a graph of \(T^2\) against \(1/d\).
* **Limitation 2:** Non-parallel threads / difficulty keeping them vertical.
* **Improvement 2:** Use a set-square/spirit level / measure \(d\) at both top and bottom.
* **Limitation 3:** Human reaction time / difficulty in identifying end of oscillation.
* **Improvement 3:** Use a fiducial marker / light gate with data logger.
* **Limitation 4:** Lateral swinging (pendulum mode) instead of pure rotation.
* **Improvement 4:** Mechanical release / central guiding pin.

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

(b)(i) Let \(r\) be the distance from the centre of the Earth. At the point of zero net gravitational field strength:
\(\frac{G M_E}{r^2} = \frac{G M_M}{(d-r)^2}\)
Taking the square root on both sides:
\(\frac{\sqrt{M_E}}{r} = \frac{\sqrt{M_M}}{d-r}\)
\(\frac{\sqrt{5.97 \times 10^{24}}}{r} = \frac{\sqrt{7.35 \times 10^{22}}}{3.84 \times 10^8 - r}\)
\(\frac{2.443 \times 10^{12}}{r} = \frac{2.711 \times 10^{11}}{3.84 \times 10^8 - r}\)
\(9.012 \times (3.84 \times 10^8 - r) = r\)
\(3.46 \times 10^9 - 9.012 r = r\)
\(10.012 r = 3.46 \times 10^9 \implies r \approx 3.46 \times 10^8\text{ m}\).

(ii) The gravitational potential \(\phi\) at this point is the sum of the potentials due to Earth and Moon:
\(\phi = -\frac{G M_E}{r} - \frac{G M_M}{d-r}\)
Here, \(r = 3.456 \times 10^8\text{ m}\) and \(d-r = 3.84 \times 10^7\text{ m}\).
\(\phi = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{3.456 \times 10^8} - \frac{6.67 \times 10^{-11} \times 7.35 \times 10^{22}}{3.84 \times 10^7}\)
\(\phi = -1.152 \times 10^6 - 1.277 \times 10^6 = -2.43 \times 10^6\text{ J kg}^{-1}\).

(iii) Potential at Earth's surface \(\phi_i \approx -\frac{G M_E}{R_E}\) (neglecting Moon's small contribution):
\(\phi_i = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.37 \times 10^6} = -6.251 \times 10^7\text{ J kg}^{-1}\).
\(\Delta \phi = \phi_f - \phi_i = -2.43 \times 10^6 - (-6.251 \times 10^7) = 6.01 \times 10^7\text{ J kg}^{-1}\).
Work done \(W = m \Delta \phi = 1200 \times 6.01 \times 10^7 = 7.21 \times 10^{10}\text{ J}\).

(a)
- Work done per unit mass: M1
- In moving a mass from infinity to the point: A1

(b)(i)
- Equating field strengths \(G M_E / r^2 = G M_M / (d-r)^2\): C1
- Correct substitution of values: C1
- Clear algebraic step showing \(r \approx 3.46 \times 10^8\text{ m}\): A1

(b)(ii)
- Use of formula \(\phi = -GM/r\) and adding potentials: C1
- Correct substitution for both potentials: C1
- Correct final answer: \(-2.43 \times 10^6\text{ J kg}^{-1}\) (allow \(-2.4 \times 10^6\)): A1

(b)(iii)
- Calculation of potential at Earth surface: C1
- Work done = \(m \Delta \phi\) leading to \(7.2 \times 10^{10}\text{ J}\): A1

(a) Assumptions of kinetic theory:
1. Large number of molecules/atoms moving in rapid, random motion.
2. Volume of gas molecules is negligible compared to the volume of the container.
3. All collisions (between molecules and with walls) are perfectly elastic.
4. No intermolecular forces except during collisions.
5. Duration of collisions is negligible compared to the time between collisions.
(Any three of the above)

(b)(i) Use \(P V = N k T\) where \(T = 27 + 273.15 = 300\text{ K}\):
\(P = \frac{N k T}{V} = \frac{3.2 \times 10^{23} \times 1.38 \times 10^{-23} \times 300}{0.025}\)
\(P = \frac{1324.8}{0.025} = 5.30 \times 10^4\text{ Pa}\).

(ii) Mean translational kinetic energy:
\(\langle E_k \rangle = \frac{3}{2} k T = 1.5 \times 1.38 \times 10^{-23} \times 300 = 6.21 \times 10^{-21}\text{ J}\).

(iii) r.m.s. speed:
\(v_{\text{rms}} = \sqrt{\frac{3 k T}{m}} = \sqrt{\frac{2 \langle E_k \rangle}{m}}\)
\(v_{\text{rms}} = \sqrt{\frac{2 \times 6.21 \times 10^{-21}}{6.64 \times 10^{-27}}} = \sqrt{1.8705 \times 10^6} = 1.37 \times 10^3\text{ m s}^{-1}\).

(a)
- Any three valid assumptions: B3 (1 mark each)

(b)(i)
- Temperature converted to \(300\text{ K}\) and formula \(P V = N k T\) used: C1
- Correct answer: \(5.30 \times 10^4\text{ Pa}\) (or \(5.3 \times 10^4\)): A1

(b)(ii)
- Use of \(\langle E_k \rangle = \frac{3}{2} k T\): C1
- Correct answer: \(6.21 \times 10^{-21}\text{ J}\): A1

(b)(iii)
- Use of \(v_{\text{rms}} = \sqrt{3 k T / m}\) or \(v_{\text{rms}} = \sqrt{2 E_k / m}\): C1
- Correct substitution of helium atom mass: C1
- Correct answer: \(1.37 \times 10^3\text{ m s}^{-1}\) (allow \(1.4 \times 10^3\)): A1

(a) The first law of thermodynamics is written as \(\Delta U = Q + W\) (or \(Q = \Delta U + W\) with appropriate signs):
- \(\Delta U\) is the change in internal energy.
- \(Q\) is the thermal energy transferred to the system (positive if added, negative if lost).
- \(W\) is the work done on the system (positive if work is done on the gas, negative if work is done by the gas during expansion).

(b)(i) Since \(A \to B\) is isothermal (constant temperature), the change in internal energy \(\Delta U = 0\).
From \(\Delta U = Q + W\):
\(0 = -540 + W \implies W = +540\text{ J}\).

(b)(ii)
1. Work done by the gas \(W_{\text{by}} = P \Delta V = 3.6 \times 10^5 \times (3.0 \times 10^{-3} - 1.5 \times 10^{-3}) = 3.6 \times 10^5 \times 1.5 \times 10^{-3} = 540\text{ J}\).
2. \(\Delta U = Q + W_{\text{on}}\). Here \(W_{\text{on}} = -540\text{ J}\) and \(Q = +780\text{ J}\).
\(\Delta U = 780 - 540 = +240\text{ J}\).

(b)(iii) For a complete cycle, the net change in internal energy is zero:
\(\Delta U_{AB} + \Delta U_{BC} + \Delta U_{CA} = 0\)
\(0 + 240 + \Delta U_{CA} = 0 \implies \Delta U_{CA} = -240\text{ J}\).

(a)
- Correct equation: \(\Delta U = Q + W\) (or equivalent): B1
- Correct identification of all three terms: B1
- Correct sign conventions (e.g. \(Q\) added, \(W\) work done ON system): B1

(b)(i)
- \(\Delta U = 0\) stated (since isothermal): B1
- Work done \(W = +540\text{ J}\): B1

(b)(ii)
1.
- Use of \(W = P \Delta V\): C1
- Correct value: \(540\text{ J}\): A1
2.
- \(\Delta U = 780 - 540 = +240\text{ J}\): B1

(b)(iii)
- Idea that sum of internal energy changes around cycle is zero: C1
- Correct calculation: \(-240\text{ J}\): A1

(a) Simple harmonic motion is motion where the acceleration of the body is directly proportional to its displacement from a fixed reference point, and is always directed towards that fixed point.

(b)(i) Maximum kinetic energy occurs at displacement \(x = 0\):
\(E_{k,\text{max}} = \frac{1}{2} m \omega^2 x_0^2\)
\(0.180 = 0.5 \times 0.350 \times \omega^2 \times (0.060)^2\)
\(0.180 = 0.175 \times 0.0036 \times \omega^2\)
\(0.180 = 0.00063 \omega^2 \implies \omega^2 = 285.7 \implies \omega = 16.9\text{ rad s}^{-1}\).

(ii) \(f = \frac{\omega}{2 \pi} = \frac{16.9}{2 \pi} = 2.69\text{ Hz}\).

(iii) Total energy \(E_T = E_k + E_p = 0.180\text{ J}\).
If \(E_k = E_p\), then \(E_p = 0.090\text{ J}\).
\(E_p = \frac{1}{2} m \omega^2 x^2 = E_T \left(\frac{x}{x_0}\right)^2\)
\(\frac{1}{2} = \left(\frac{x}{x_0}\right)^2 \implies x = \frac{x_0}{\sqrt{2}} = \frac{6.0\text{ cm}}{\sqrt{2}} = 4.24\text{ cm}\) (or \(0.0424\text{ m}\)).

(c)(i) For light damping, the period of oscillation remains constant (or increases by an extremely negligible amount).
(ii) The total energy of the oscillation decreases exponentially with time because mechanical energy is dissipated as thermal energy due to resistive forces.

(a)
- Acceleration proportional to displacement: B1
- Acceleration and displacement in opposite directions (or directed towards a fixed point): B1

(b)(i)
- Use of \(E_{k,\text{max}} = \frac{1}{2} m \omega^2 x_0^2\): C1
- Substitution of values in correct units (converting cm to m): C1
- Answer: \(16.9\text{ rad s}^{-1}\) (allow \(17\text{ rad s}^{-1}\)): A1

(b)(ii)
- Use of \(f = \omega / 2\pi\) yielding \(2.69\text{ Hz}\) (or \(2.7\text{ Hz}\)): B1

(b)(iii)
- Equating \(E_k = E_p\) to give \(E_p = 0.5 E_T\): C1
- Correct calculation showing \(x = x_0 / \sqrt{2} = 4.24\text{ cm}\) (or \(0.0424\text{ m}\)): A1

(c)(i)
- Constant / no significant change: B1

(c)(ii)
- Energy decreases (exponentially) because work is done against resistive forces: B1

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

(b)(i) \(C = \frac{\varepsilon_0 A}{d}\)
\(C = \frac{8.85 \times 10^{-12} \times 0.12}{1.5 \times 10^{-3}} = 7.08 \times 10^{-10}\text{ F}\), which is approximately \(7.1 \times 10^{-10}\text{ F}\).

(ii) \(E = \frac{1}{2} C V^2 = 0.5 \times 7.08 \times 10^{-10} \times 120^2 = 5.10 \times 10^{-6}\text{ J}\).

(c)(i) Initial charge stored on first capacitor: \(Q = C_1 V_1 = 7.08 \times 10^{-10} \times 120 = 8.50 \times 10^{-8}\text{ C}\).
When connected in parallel, the total capacitance is \(C_T = C_1 + C_2 = 7.08 \times 10^{-10} + 3.5 \times 10^{-10} = 1.058 \times 10^{-9}\text{ F}\).
Since charge is conserved, the final potential difference \(V_f\) is:
\(V_f = \frac{Q}{C_T} = \frac{8.50 \times 10^{-8}}{1.058 \times 10^{-9}} = 80.3\text{ V}\).

(ii) During the redistribution of charge, current flows through the connecting wires which have some electrical resistance. This causes energy to be dissipated as thermal energy (heat).

(a)
- Charge per unit potential difference: B1

(b)(i)
- Formula \(C = \varepsilon_0 A / d\) used with correct substitution: C1
- Correct calculated value \(7.08 \times 10^{-10}\text{ F}\) shown: A1

(b)(ii)
- Use of \(E = \frac{1}{2} C V^2\): C1
- Correct answer: \(5.10 \times 10^{-6}\text{ J}\) (or \(5.1 \times 10^{-6}\)): A1

(c)(i)
- Calculation of initial charge \(Q = 8.50 \times 10^{-8}\text{ C}\): C1
- Total capacitance \(C_T = 1.06 \times 10^{-9}\text{ F}\): C1
- Final potential difference \(80.3\text{ V}\) (or \(80\text{ V}\)): A1

(c)(ii)
- Sparking / resistance in the wires leads to: M1
- Dissipation of energy as thermal energy / heat: A1

(a) Faraday's law states that the magnitude of the induced electromotive force (e.m.f.) is directly proportional to the rate of change of magnetic flux linkage.

(b)(i) Area \(A = \pi r^2 = \pi \times (0.045)^2 = 6.36 \times 10^{-3}\text{ m}^2\).
Flux linkage \(\Phi = N B A = 150 \times 0.12 \times 6.36 \times 10^{-3} = 0.114\text{ Wb-turns}\) (or \(0.115\text{ Wb-turns}\)).

(ii) Since the coil is rotated by \(90^\circ\), the final magnetic flux through the coil is zero.
Change in flux linkage \(\Delta \Phi = 0.115\text{ Wb-turns}\).
Induced e.m.f. \(E = \frac{\Delta \Phi}{\Delta t} = \frac{0.115}{0.18} = 0.639\text{ V}\) (or \(0.64\text{ V}\)).

(c) Lenz's law states that the direction of the induced e.m.f. is such that it opposes the change in magnetic flux that produces it.
Explanation:
- As the coil is rotated, the magnetic flux through it decreases.
- The induced current produces its own magnetic field.
- To oppose the decrease in flux, the magnetic field produced by the induced current must be in the same direction as the external magnetic field to try to maintain the initial magnetic flux.

(a)
- Induced e.m.f. is proportional to: M1
- Rate of change of magnetic flux linkage: A1

(b)(i)
- Area calculation \(A = 6.36 \times 10^{-3}\text{ m}^2\): C1
- Flux linkage \(\Phi = 0.115\text{ Wb-turns}\): A1

(b)(ii)
- Change in flux linkage identified as \(0.115\text{ Wb-turns}\): C1
- Induced e.m.f. = \(0.639\text{ V}\): A1

(c)
- Statement of Lenz's law: B1
- Flux through the coil decreases as it rotates: B1
- Induced current produces its own magnetic field: B1
- Induced field is in the direction of the external field to oppose the reduction in flux: B1

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

(b)(i) \(E = \frac{h c}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{310 \times 10^{-9}} = 6.42 \times 10^{-19}\text{ J}\).

(ii) Convert work function to joules:
\(\Phi = 2.28 \times 1.60 \times 10^{-19} = 3.65 \times 10^{-19}\text{ J}\).
Using Einstein's photoelectric equation:
\(E = \Phi + E_{k,\text{max}}\)
\(E_{k,\text{max}} = 6.416 \times 10^{-19} - 3.648 \times 10^{-19} = 2.77 \times 10^{-19}\text{ J}\), which is approximately \(2.8 \times 10^{-19}\text{ J}\).

(iii) Stopping potential \(V_s = \frac{E_{k,\text{max}}}{e} = \frac{2.77 \times 10^{-19}}{1.60 \times 10^{-19}} = 1.73\text{ V}\).

(c)(i) No effect: The maximum kinetic energy depends only on the photon energy (frequency/wavelength) and the work function, which are both unchanged.
(ii) Increases: Increasing intensity increases the rate of photons incident on the metal surface. Since one photon interacts with one electron, the rate of photoelectron emission increases proportionally.

(a)
- Minimum energy: B1
- Required to release an electron from the metal surface: B1

(b)(i)
- Use of \(E = hc/\lambda\): C1
- Correct calculation: \(6.42 \times 10^{-19}\text{ J}\): A1

(b)(ii)
- Conversion of eV to Joules (\(3.65 \times 10^{-19}\text{ J}\)): C1
- Use of \(E = \Phi + E_{k,\text{max}}\): C1
- Answer calculated to at least 3 s.f. (\(2.77 \times 10^{-19}\text{ J}\)): A1

(b)(iii)
- Correct calculation of \(V_s = 1.73\text{ V}\): B1

(c)(i)
- No change AND explanation (depends only on photon energy / wavelength): B1

(c)(ii)
- Increases AND explanation (intensity increases rate of photons, 1-to-1 interaction): B1

(a)(i) Wien's displacement law states that the wavelength of maximum emission intensity \(\lambda_{\text{max}}\) is inversely proportional to the thermodynamic temperature \(T\) of the black body: \(\lambda_{\text{max}} T = \text{constant}\).
(ii) Hubble's law states that the recessional speed \(v\) of a distant galaxy is directly proportional to its distance \(d\) from the observer: \(v = H_0 d\).

(b)(i) Using Wien's law:
\(\lambda_{\text{max}} = \frac{2.90 \times 10^{-3}\text{ m K}}{T} = \frac{2.90 \times 10^{-3}}{3500} = 8.29 \times 10^{-7}\text{ m}\) (or \(830\text{ nm}\)).

(ii) Stefan-Boltzmann law:
\(L = 4 \pi R^2 \sigma T^4\)
\(1.2 \times 10^{31} = 4 \pi R^2 \times (5.67 \times 10^{-8}) \times (3500)^4\)
\(1.2 \times 10^{31} = 4 \pi R^2 \times 5.67 \times 10^{-8} \times 1.5006 \times 10^{14}\)
\(1.2 \times 10^{31} = 1.070 \times 10^8 R^2\)
\(R^2 = 1.1215 \times 10^{23} \implies R = 3.35 \times 10^{11}\text{ m}\).

(c)(i) Redshift \(z = \frac{v}{c}\):
\(v = z c = 0.045 \times 3.00 \times 10^8 = 1.35 \times 10^7\text{ m s}^{-1}\).

(ii) Using Hubble's law \(v = H_0 d\):
\(d = \frac{v}{H_0} = \frac{1.35 \times 10^7}{2.3 \times 10^{-18}} = 5.87 \times 10^{24}\text{ m}\).

(a)(i)
- Statement of Wien's law: \(\lambda_{\text{max}} \propto 1/T\) or \(\lambda_{\text{max}} T = \text{constant}\) (defining terms): B1
(a)(ii)
- Statement of Hubble's law: recessional speed is proportional to distance: B1

(b)(i)
- Use of \(\lambda_{\text{max}} T = 2.90 \times 10^{-3}\): C1
- Correct answer: \(8.29 \times 10^{-7}\text{ m}\): A1

(b)(ii)
- Use of \(L = 4 \pi R^2 \sigma T^4\): C1
- Correct substitution: C1
- Correct answer: \(3.35 \times 10^{11}\text{ m}\) (allow \(3.4 \times 10^{11}\)): A1

(c)(i)
- Correct speed: \(1.35 \times 10^7\text{ m s}^{-1}\): B1

(c)(ii)
- Use of \(v = H_0 d\): C1
- Correct distance: \(5.87 \times 10^{24}\text{ m}\) (allow \(5.9 \times 10^{24}\)): A1

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

(b)(i) Using the gravitational potential formula:
\(\phi = -\frac{GM}{R}\)
\(\phi = -\frac{6.67 \times 10^{-11} \times 6.4 \times 10^{23}}{3.4 \times 10^6}\)
\(\phi = -1.256 \times 10^7\text{ J kg}^{-1} \approx -1.3 \times 10^7\text{ J kg}^{-1}\).

(ii) Initial total energy is the sum of kinetic and gravitational potential energy:
\(E_{\text{total}} = E_k + E_p = E_k + m \phi\)
\(E_p = 1200 \times (-1.2555 \times 10^7) = -1.507 \times 10^{10}\text{ J}\)
\(E_{\text{total}} = 9.0 \times 10^9 - 1.507 \times 10^{10} = -6.07 \times 10^9\text{ J}\)
At the maximum distance \(d\), the probe's radial speed is zero, meaning its kinetic energy is zero (since it is launched vertically):
\(E_{pf} = -\frac{GMm}{d} = -6.07 \times 10^9\text{ J}\)
\(-\frac{6.67 \times 10^{-11} \times 6.4 \times 10^{23} \times 1200}{d} = -6.07 \times 10^9\)
\(d = \frac{5.123 \times 10^{16}}{6.07 \times 10^9} = 8.44 \times 10^6\text{ m}\)
So, \(d = 8.4 \times 10^6\text{ m}\) (or \(8.44 \times 10^6\text{ m}\)).

(iii) The maximum distance would be less. When launched at an angle, the probe has angular momentum about the planet's centre. To conserve angular momentum, the probe must have a non-zero tangential velocity at its highest point, meaning its final kinetic energy is not zero. Consequently, less kinetic energy is converted into gravitational potential energy.

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

(b)(i)
- State/use \(\phi = -GM/R\) [1]
- Substitute values: \(\phi = -\frac{6.67 \times 10^{-11} \times 6.4 \times 10^{23}}{3.4 \times 10^6} = -1.26 \times 10^7\text{ J kg}^{-1}\) showing approximation to \(-1.3 \times 10^7\text{ J kg}^{-1}\) [1]

(ii)
- Clear statement of conservation of energy: \(E_k + E_{pi} = E_{pf}\) [1]
- Calculate initial potential energy: \(-1.51 \times 10^{10}\text{ J}\) [1]
- Calculate final potential energy: \(-6.07 \times 10^9\text{ J}\) [1]
- Final value of \(d = 8.4 \times 10^6\text{ m}\) (accept range \(8.4 \times 10^6\text{ m}\) to \(8.5 \times 10^6\text{ m}\)) [1]

(iii)
- State that max distance is less [1]
- Explain that probe must retain tangential velocity / kinetic energy at maximum distance to conserve angular momentum [1]

(a) The internal energy of a system is defined as the sum of a random distribution of kinetic and potential energies associated with the atoms or molecules of the system.

(b)(i) \(+q\) represents the thermal (heat) energy supplied to or absorbed by the system.
(ii) \(-w\) represents the work done by the system (such as during expansion).

(c)(i) For process \(A \to B\), the volume is constant, so \(p/T = \text{constant}\):
\(\frac{p_A}{T_A} = \frac{p_B}{T_B}\)
\(T_B = T_A \times \frac{p_B}{p_A} = 300 \times \frac{3.0 \times 10^5}{1.0 \times 10^5} = 900\text{ K}\).

(ii) During \(C \to A\), the pressure is constant at \(p_A = 1.0 \times 10^5\text{ Pa}\). The gas is compressed from \(V_C = 6.0 \times 10^{-3}\text{ m}^3\) to \(V_A = 2.0 \times 10^{-3}\text{ m}^3\):
\(w = p \Delta V = 1.0 \times 10^5 \times (6.0 \times 10^{-3} - 2.0 \times 10^{-3}) = 400\text{ J}\).
Since it is a compression, work is done on the gas, so the work done on the gas is \(+400\text{ J}\).

(iii) For a monoatomic ideal gas, the internal energy is given by \(U = \frac{3}{2} nRT = \frac{3}{2} pV\).
For the process \(A \to B\):
\(\Delta U_{AB} = \frac{3}{2} (p_B V_B - p_A V_A) = \frac{3}{2} (3.0 \times 10^5 \times 2.0 \times 10^{-3} - 1.0 \times 10^5 \times 2.0 \times 10^{-3})\)
\(\Delta U_{AB} = \frac{3}{2} (600 - 200) = 600\text{ J}\)
Alternatively, using \(\Delta U = \frac{3}{2} n R \Delta T\):
\(nR = \frac{p_A V_A}{T_A} = \frac{200}{300} = \frac{2}{3}\text{ J K}^{-1}\)
\(\Delta U_{AB} = \frac{3}{2} \times \frac{2}{3} \times (900 - 300) = 600\text{ J}\).

(a)
- Sum of random distribution [1]
- of kinetic and potential energies of molecules [1]

(b)
(i) Heat supplied to/absorbed by system [1]
(ii) Work done by system [1]

(c)
(i)
- Recall and use \(p_A / T_A = p_B / T_B\) (or \(pV=nRT\)) [1]
- \(T_B = 900\text{ K}\) [1]

(ii)
- Use of \(w = p \Delta V\) [1]
- Correct calculation: \(w = 1.0 \times 10^5 \times 4.0 \times 10^{-3} = 400\text{ J}\) [1]

(iii)
- Recall \(\Delta U = \frac{3}{2} n R \Delta T\) or \(\Delta U = \frac{3}{2} \Delta (pV)\) for monoatomic gas [1]
- Calculate \(\Delta U_{AB} = 600\text{ J}\) [1]

### 1. Diagram and Experimental Setup
* **Setup**: Clamp a copper tube vertically. Ensure it is perfectly vertical using a plumb line or spirit level.
* **Drop Mechanism**: Position a non-magnetic guide tube (e.g., plastic) at the top of the copper tube to ensure the magnet is released centrally along the tube axis without tumbling.
* **Timing System**: Since copper is opaque, standard light gates cannot view the magnet inside. Wrap two small search coils of copper wire around the outside of the tube near the bottom (with a known separation \(d\), measured using a meter rule).
* **Detection**: Connect these coils to the inputs of a dual-channel storage oscilloscope or a fast data logger. As the magnet falls through each coil, the changing magnetic flux induces a brief electromotive force (e.g., a voltage pulse). The time interval \(\Delta t\) between the peaks of these two pulses is measured using the oscilloscope's timebase.

### 2. Variables and Measurements
* **Independent Variable**: Wall thickness \(t\) of the copper tube.
* **Dependent Variable**: Terminal velocity \(v\) of the magnet.
* **Control Variables**:
* Inner diameter of the copper tubes (must be identical so the air gap and magnetic coupling remain constant).
* Material of the tubes (constant electrical resistivity).
* The same neodymium magnet (constant magnetic field strength and mass).
* Temperature of the laboratory (resistivity of copper varies with temperature).

### 3. Measuring Wall Thickness \(t\)
* Use a digital micrometer screw gauge or Vernier calipers to measure the outer diameter \(D_{outer}\) and the inner diameter \(D_{inner}\) at several orientations around the rim of each tube.
* Calculate the average wall thickness using:
\[ t = \frac{D_{outer} - D_{inner}}{2} \]

### 4. Determination of Terminal Velocity \(v\)
* The magnet must reach terminal velocity before passing the first search coil. Place the first coil a significant distance down the tube (e.g., more than halfway down a 1.0 m tube) and verify that terminal velocity is reached by showing that further drops over a longer distance do not change the measured speed.
* For each tube, release the magnet and record \(\Delta t\).
* Calculate terminal velocity using:
\[ v = \frac{d}{\Delta t} \]
* Repeat the drop several times for each tube to obtain an average value of \(v\).

### 5. Analysis of Data
* The proposed relationship is \(v = \beta t^{-1}\).
* Rearranging gives: \(v = \beta \left(\frac{1}{t}\right)\).
* Plot a graph of \(v\) on the y-axis against \(1/t\) on the x-axis.
* If the relationship is valid, the plotted points should lie on a straight line that passes through the origin.
* The gradient of this line is equal to the constant \(\beta\):
\[ \beta = \text{gradient} \]

**Marking Scheme (Total: 15 Marks)**

**Defining Variables [Maximum 3 marks]**
* **[1 mark]** Identify \(t\) as the independent variable and \(v\) as the dependent variable.
* **[1 mark]** Identify that the internal diameter of the tube must be kept constant.
* **[1 mark]** Identify that the magnet mass/strength must be kept constant (or keep the tube temperature/material constant).

**Methods of Data Collection [Maximum 5 marks]**
* **[1 mark]** Draw a clear, labeled diagram showing a vertically clamped copper tube with a system to drop the magnet, and a soft cushion/box at the bottom to catch the magnet safely.
* **[1 mark]** Describe the use of Vernier calipers or micrometer to measure both the outer and inner diameters, and show the correct formula \(t = (D_{outer} - D_{inner})/2\).
* **[1 mark]** Describe a viable timing method to find the velocity inside the opaque tube, e.g., using search coils wrapped around the tube connected to a dual-channel oscilloscope/data-logger (or using a motion sensor at the top looking down).
* **[1 mark]** Explain how the velocity is calculated: \(v = d/\Delta t\) where \(d\) is the distance between the search coils / sensors.
* **[1 mark]** Describe how terminal velocity is ensured (e.g., placing the sensors near the bottom of a long tube, or showing that \(v\) is constant by measuring speeds over different consecutive segments).

**Method of Analysis [Maximum 2 marks]**
* **[1 mark]** State that a graph of \(v\) against \(1/t\) should be plotted.
* **[1 mark]** State that the relationship is valid if the graph is a straight line through the origin, and that \(\beta\) is calculated as the gradient of this graph.

**Safety Considerations [Maximum 1 mark]**
* **[1 mark]** Identify a valid safety precaution, e.g., keep fingers clear of the tube when dropping very strong magnets to avoid pinching, or keep magnetic storage devices/credit cards away from the strong neodymium magnet.

**Additional Details [Maximum 4 marks]**
* **[1 mark]** Measure outer/inner diameters at several positions and orientations along the tube and average to account for non-uniformity.
* **[1 mark]** Use a plumb line or spirit level to ensure the tube is perfectly vertical.
* **[1 mark]** Use a non-magnetic guide tube or release mechanism at the top to ensure the magnet falls centrally without tumbling or hitting the sides.
* **[1 mark]** Clean the inside of the tube before trials to remove any dirt/oxidation that might add mechanical friction.
* **[1 mark]** Repeat the drop multiple times for each tube and take an average of \(\Delta t\) to reduce random error.

**Part (a)**
Squaring both sides of the equation:
\(f^2 = \frac{T}{4 L^2 \mu}\)
Comparing with the linear equation \(y = mx + c\), where \(y = f^2\) and \(x = T\):
\(\text{gradient } m = \frac{1}{4 L^2 \mu}\)

**Part (b)**
The absolute uncertainty in \(f^2\), denoted as \(\Delta(f^2)\), is given by \(\Delta(f^2) \approx 2 f \Delta f\), where \(\Delta f = 3\text{ Hz}\):
- For \(T = 15.0\text{ N}\): \(f^2 = 117^2 = 13.689 \times 10^3\text{ Hz}^2 \approx 13.7 \times 10^3\text{ Hz}^2\); \(\Delta(f^2) \approx 2 \times 117 \times 3 = 702\text{ Hz}^2 \approx 0.7 \times 10^3\text{ Hz}^2\)
- For \(T = 25.0\text{ N}\): \(f^2 = 151^2 = 22.801 \times 10^3\text{ Hz}^2 \approx 22.8 \times 10^3\text{ Hz}^2\); \(\Delta(f^2) \approx 2 \times 151 \times 3 = 906\text{ Hz}^2 \approx 0.9 \times 10^3\text{ Hz}^2\)
- For \(T = 35.0\text{ N}\): \(f^2 = 179^2 = 32.041 \times 10^3\text{ Hz}^2 \approx 32.0 \times 10^3\text{ Hz}^2\); \(\Delta(f^2) \approx 2 \times 179 \times 3 = 1074\text{ Hz}^2 \approx 1.1 \times 10^3\text{ Hz}^2\)
- For \(T = 45.0\text{ N}\): \(f^2 = 203^2 = 41.209 \times 10^3\text{ Hz}^2 \approx 41.2 \times 10^3\text{ Hz}^2\); \(\Delta(f^2) \approx 2 \times 203 \times 3 = 1218\text{ Hz}^2 \approx 1.2 \times 10^3\text{ Hz}^2\)
- For \(T = 55.0\text{ N}\): \(f^2 = 224^2 = 50.176 \times 10^3\text{ Hz}^2 \approx 50.2 \times 10^3\text{ Hz}^2\); \(\Delta(f^2) \approx 2 \times 224 \times 3 = 1344\text{ Hz}^2 \approx 1.3 \times 10^3\text{ Hz}^2\)
- For \(T = 65.0\text{ N}\): \(f^2 = 244^2 = 59.536 \times 10^3\text{ Hz}^2 \approx 59.5 \times 10^3\text{ Hz}^2\); \(\Delta(f^2) \approx 2 \times 244 \times 3 = 1464\text{ Hz}^2 \approx 1.5 \times 10^3\text{ Hz}^2\)

**Part (c)**
- Plot \(f^2 / 10^3\text{ Hz}^2\) against \(T / \text{N}\). Error bars are vertical and range from \(\pm 0.7\) (at \(T=15.0\text{ N}\)) to \(\pm 1.5\) (at \(T=65.0\text{ N}\)).
- Draw the line of best fit through the points.
- Draw the worst acceptable line (the steepest or shallowest possible line that passes through all error bars, usually from top of first error bar to bottom of last error bar).

**Part (d)**
- Best-fit gradient \(m_{\text{best}} \approx 0.916 \times 10^3\text{ Hz}^2\text{ N}^{-1} = 916\text{ Hz}^2\text{ N}^{-1}\).
- Worst acceptable gradient \(m_{\text{worst}} \approx 0.960 \times 10^3\text{ Hz}^2\text{ N}^{-1} = 960\text{ Hz}^2\text{ N}^{-1}\).
- Uncertainty in gradient \(\Delta m = |m_{\text{best}} - m_{\text{worst}}| \approx 44\text{ Hz}^2\text{ N}^{-1}\).
- Best-fit y-intercept \(c_{\text{best}} \approx 0.0 \times 10^3\text{ Hz}^2\).
- Uncertainty in y-intercept \(\Delta c = |c_{\text{best}} - c_{\text{worst}}| \approx 1.4 \times 10^3\text{ Hz}^2\).

**Part (e)**
- Mass per unit length: \(\mu = \frac{1}{4 L^2 m}\)
Using \(L = 0.450\text{ m}\) and \(m = 916\text{ Hz}^2\text{ N}^{-1}\):
\(\mu = \frac{1}{4 \times (0.450)^2 \times 916} = 1.35 \times 10^{-3}\text{ kg m}^{-1}\)
- Uncertainty in \(\mu\):
\(\frac{\Delta \mu}{\mu} = 2 \frac{\Delta L}{L} + \frac{\Delta m}{m} = 2\left(\frac{0.005}{0.450}\right) + \frac{44}{916} = 0.0222 + 0.0480 = 0.0702\)
\(\Delta \mu = 0.0702 \times 1.35 \times 10^{-3} = 0.095 \times 10^{-3} \approx 0.10 \times 10^{-3}\text{ kg m}^{-1}\)
Therefore: \(\mu = (1.4 \pm 0.1) \times 10^{-3}\text{ kg m}^{-1}\)

**Part (a)** [1 Mark]
- Correct expression for gradient: \(\frac{1}{4 L^2 \mu}\).

**Part (b)** [3 Marks]
- 1 Mark: Correctly calculated values of \(f^2 / 10^3\text{ Hz}^2\) to 3 significant figures (13.7, 22.8, 32.0, 41.2, 50.2, 59.5).
- 1 Mark: Correct calculations of absolute uncertainties in \(f^2 / 10^3\text{ Hz}^2\) (0.7, 0.9, 1.1, 1.2, 1.3, 1.5).
- 1 Mark: Values are recorded consistently to 1 decimal place matching the precision of the uncertainties.

**Part (c)** [4 Marks]
- 1 Mark: Horizontal and vertical axes chosen with linear scales and correctly labelled with units. Plotted points occupy more than half of the grid.
- 1 Mark: All 6 points plotted correctly to within half a small square.
- 1 Mark: Vertical error bars plotted correctly with lengths matching the calculated absolute uncertainties.
- 1 Mark: Best-fit and worst acceptable lines drawn, clearly labelled.

**Part (d)** [4 Marks]
- 1 Mark: Gradient of best fit determined using a triangle with hypotenuse of at least half the length of the line.
- 1 Mark: Uncertainty in gradient determined as \(|m_{\text{best}} - m_{\text{worst}}|\).
- 1 Mark: y-intercept of best fit determined by direct reading from \(x=0\) or calculation using \(y=mx+c\).
- 1 Mark: Uncertainty in y-intercept determined as \(|c_{\text{best}} - c_{\text{worst}}|\).

**Part (e)** [3 Marks]
- 1 Mark: Value of \(\mu\) calculated in range \(1.30 \times 10^{-3}\) to \(1.40 \times 10^{-3}\) with correct unit \(\text{kg m}^{-1}\).
- 1 Mark: Uncertainty in \(\mu\) calculated using \(\frac{\Delta \mu}{\mu} = 2 \frac{\Delta L}{L} + \frac{\Delta m}{m}\).
- 1 Mark: Final value of \(\mu\) and its uncertainty expressed to appropriate significant figures (usually 1 or 2 s.f. for uncertainty and consistent decimal places for the value).