2024 Cambridge IAL Physics (9702) 模擬試題連答案詳解

The percentage uncertainty in \(L\) is given as \(1.5\%\). For the period \(T = \frac{t}{50}\), where \(t\) is the total time for 50 oscillations, the percentage uncertainty in \(T\) is equal to the percentage uncertainty in \(t\): \(\frac{\Delta T}{T} \times 100\% = \frac{\Delta t}{t} \times 100\% = \frac{0.4}{80.0} \times 100\% = 0.5\%\). The formula for the acceleration of free fall is \(g = \frac{4\pi^2 L}{T^2}\). The percentage uncertainty in \(g\) is therefore: \(\frac{\Delta g}{g} \times 100\% = \frac{\Delta L}{L} \times 100\% + 2\left(\frac{\Delta T}{T} \times 100\%\right) = 1.5\% + 2(0.5\%) = 2.5\%\).

Award 1 mark for the correct answer B. (Method: Calculate percentage uncertainty in time as 0.5%, double it for the squared term, and add to the 1.5% uncertainty of the length.)

Energy can be defined as work done, which is force multiplied by distance. \(\text{Work} = \text{Force} \times \text{distance}\). Force has SI base units of \(\text{kg m s}^{-2}\). Therefore, the unit of energy is: \(\text{kg m s}^{-2} \times \text{m} = \text{kg m}^2 \text{s}^{-2}\).

Award 1 mark for the correct answer C. (Method: Express force in SI base units and multiply by distance to find the base units of work/energy.)

For a source moving away from an observer at a non-relativistic speed \(v \ll c\), the Doppler redshift formula is given by: \(\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}\), where \(\Delta \lambda = \lambda - \lambda_0\). Thus, the fractional change is \(\frac{\lambda - \lambda_0}{\lambda_0} = \frac{v}{c}\).

Award 1 mark for the correct answer A. (Method: Recall and apply the Doppler redshift relation for electromagnetic waves.)

The gravitational field strength at the surface is \(g = \frac{GM}{R^2}\). The density \(\rho\) of a sphere is given by \(\rho = \frac{M}{\frac{4}{3}\pi R^3}\), which means \(R \propto M^{1/3}\) for constant density. Substituting this into the field strength formula gives: \(g \propto \frac{M}{(M^{1/3})^2} \propto M^{1/3}\). Since the second planet has twice the mass \(2M\) of the first planet but the same density, its surface gravitational field strength is: \(g_2 = 2^{1/3} g\).

Award 1 mark for the correct answer A. (Method: Link radius to mass under constant density, and substitute into the gravitational field formula to find the scaling relation.)

Using Einstein's photoelectric equation: \(E_{\text{max}} = hf - \Phi\), where \(\Phi\) is the work function of the metal. For a doubled frequency \(2f\), the new maximum kinetic energy is: \(E' = h(2f) - \Phi = 2hf - \Phi\). Expressing \(hf\) as \(E_{\text{max}} + \Phi\), we get: \(E' = 2(E_{\text{max}} + \Phi) - \Phi = 2E_{\text{max}} + \Phi\). Since the work function \(\Phi > 0\) for any metal, \(E' > 2E_{\text{max}}\).

Award 1 mark for the correct answer B. (Method: Apply Einstein's photoelectric equation and algebraically compare the two cases.)

In simple harmonic motion, the velocity \(v\) as a function of displacement \(x\) is given by \(v = \omega \sqrt{x_0^2 - x^2}\). The maximum speed \(v_0\) occurs when \(x = 0\), so \(v_0 = \omega x_0\). Substituting \(x = \frac{1}{2}x_0\) into the velocity equation gives: \(v = \omega \sqrt{x_0^2 - \left(\frac{1}{2}x_0\right)^2} = \omega \sqrt{\frac{3}{4}x_0^2} = \frac{\sqrt{3}}{2} \omega x_0 = \frac{\sqrt{3}}{2} v_0\).

Award 1 mark for the correct answer C. (Method: Apply the relationship between velocity, angular frequency, displacement, and amplitude in simple harmonic motion.)

Let upward be the positive direction. The parameters for the SUVAT equation \(s = ut + \frac{1}{2}at^2\) are: initial velocity \(u = +u\), acceleration \(a = -g\), and displacement \(s = -H\) (since the ball ends up at a distance \(H\) below its initial launching position). Substituting these values yields: \(-H = ut + \frac{1}{2}(-g)t^2\), which simplifies to \(-H = ut - \frac{1}{2}gt^2\).

Award 1 mark for the correct answer A. (Method: Correctly identify the sign conventions for displacement, initial velocity, and acceleration under gravity.)

In a head-on, perfectly elastic collision between two identical masses, where one is initially stationary, the velocities of the two colliding bodies are swapped. Consequently, the first sphere comes to a complete rest, and the second sphere moves off with the initial speed \(v\) of the first sphere in the original direction of motion.

Award 1 mark for the correct answer B. (Method: Apply conservation of linear momentum and kinetic energy to deduce the velocity exchange property of identical masses in an elastic collision.)

The density is given by \( \rho = \frac{4m}{\pi d^2 L} \). The fractional uncertainty in density is given by:

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

Substituting the given values:

\( \frac{\Delta m}{m} = \frac{0.01}{0.65} \approx 1.54\% \)

\( \frac{\Delta d}{d} = \frac{0.02}{0.82} \approx 2.44\% \)

\( \frac{\Delta L}{L} = \frac{0.1}{15.0} \approx 0.67\% \)

Total percentage uncertainty:

\( \% \text{ uncertainty} = 1.54\% + 2(2.44\%) + 0.67\% = 7.09\% \approx 7.1\% \).

1 mark for calculating the fractional uncertainties of each quantity and correctly summing them, with the diameter uncertainty doubled, to arrive at 7.1%.

Use the Doppler shift formula:

\( \frac{\Delta \lambda}{\lambda} = \frac{v}{c} \)

The change in wavelength is \( \Delta \lambda = 658.3 \text{ nm} - 656.3 \text{ nm} = 2.0 \text{ nm} \).

The speed \( v \) is:

\( v = c \times \frac{\Delta \lambda}{\lambda} = 3.00 \times 10^8 \times \frac{2.0}{656.3} \approx 9.14 \times 10^5 \text{ m s}^{-1} \).

Since the measured wavelength is longer than the laboratory wavelength, it is redshifted, which means the star is moving away from Earth.

1 mark for calculating the velocity correctly using the Doppler formula and identifying that a redshift corresponds to motion away from the observer.

The orbital speed \( v \) of a satellite is given by \( v = \sqrt{\frac{GM}{r}} \). If the radius becomes \( 4r \), the new speed \( v' \) is:

\( v' = \sqrt{\frac{GM}{4r}} = \frac{1}{2}\sqrt{\frac{GM}{r}} = \frac{v}{2} \).

By Kepler's Third Law, the period \( T \) satisfies \( T^2 \propto r^3 \), meaning \( T \propto r^{3/2} \). If the radius becomes \( 4r \), the new period \( T' \) is:

\( T' = (4)^{3/2} T = 8T \).

Note that the mass of the satellite does not affect its orbital speed or period.

1 mark for applying gravitational orbital mechanics to determine that speed is halved and period is increased by a factor of 8.

Using Einstein's photoelectric equation:

1) \( hf = \Phi + E_k \)

2) \( h(2f) = \Phi + 3E_k \implies 2hf = \Phi + 3E_k \)

Multiply the first equation by 2:

\( 2hf = 2\Phi + 2E_k \)

Now set the two expressions for \( 2hf \) equal to each other:

\( 2\Phi + 2E_k = \Phi + 3E_k \)

\( \Phi = E_k \).

Therefore, the work function of the metal is \( 1.0 E_k \).

1 mark for formulating the photoelectric equation for both frequencies and solving the simultaneous equations to find the work function in terms of \( E_k \).

The total energy, which is equal to the maximum kinetic energy \( E_{\text{max}} \), is given by:

\( E_{\text{max}} = \frac{1}{2} m \omega^2 A^2 \)

At displacement \( x = \frac{3}{5}A \), the potential energy \( E_p \) is:

\( E_p = \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} m \omega^2 \left(\frac{3}{5}A\right)^2 = \frac{9}{25} \left( \frac{1}{2} m \omega^2 A^2 \right) = 0.36 E_{\text{max}} \)

Since total mechanical energy is conserved:

\( E_k = E_{\text{max}} - E_p = E_{\text{max}} - 0.36 E_{\text{max}} = 0.64 E_{\text{max}} \).

1 mark for correctly determining the ratio of potential energy to total energy and using conservation of energy to find the remaining kinetic energy fraction.

Taking the upward direction as positive:
- Initial velocity \( u = +15.0 \text{ m s}^{-1} \)
- Acceleration \( a = -9.81 \text{ m s}^{-2} \)
- Time \( t = 4.20 \text{ s} \)

We find the displacement \( s \) using the equation of motion:

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

\( s = (15.0)(4.20) + \frac{1}{2}(-9.81)(4.20)^2 \)

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

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

The vertical displacement is \( -23.5 \text{ m} \), meaning the height of the cliff is \( 23.5 \text{ m} \).

1 mark for applying the displacement equation with the correct sign convention and values to calculate the height of the cliff.

First, find the current \( I \) through the circuit using Ohm's law on the external resistor:

\( I = \frac{V}{R} = \frac{9.6}{4.0} = 2.4 \text{ A} \)

The terminal potential difference \( V \) is related to the e.m.f. \( E \) and internal resistance \( r \) by:

\( V = E - Ir \)

\( 9.6 = 12 - 2.4r \)

\( 2.4r = 2.4 \implies r = 1.0\ \Omega \).

1 mark for finding the current in the circuit and using it to determine the internal resistance.

Taking the initial direction of motion as positive:
- Initial velocity \( u = +24 \text{ m s}^{-1} \)
- Final velocity \( v = -16 \text{ m s}^{-1} \)

Calculate the change in momentum \( \Delta p \):

\( \Delta p = m v - m u = 0.15 \times (-16) - 0.15 \times 24 = -2.4 - 3.6 = -6.0 \text{ kg m s}^{-1} \)

The magnitude of the change in momentum is \( 6.0 \text{ kg m s}^{-1} \).

Using Newton's second law, the average force is:

\( F = \frac{|\Delta p|}{\Delta t} = \frac{6.0}{15 \times 10^{-3}} = 400 \text{ N} \).

1 mark for calculating the correct magnitude of the change in momentum (accounting for the sign change) and dividing by the contact time in seconds.

The formula for the density of a cylinder is given by:
\(\rho = \frac{m}{V} = \frac{4m}{\pi d^2 L}\)

The fractional uncertainty in density is the sum of the fractional uncertainties of the independent quantities, with the power of \(d\) acting as a multiplier:
\(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\)

Substituting the given values:
\(\frac{\Delta m}{m} = \frac{0.1}{25.0} = 0.004 = 0.4\%\)
\(\frac{\Delta d}{d} = \frac{0.02}{1.20} \approx 0.0167 = 1.67\%\)
\(\frac{\Delta L}{L} = \frac{0.05}{5.00} = 0.010 = 1.0\%\)

Thus, the percentage uncertainty is:
\(\frac{\Delta \rho}{\rho} = 0.4\% + 2(1.67\%) + 1.0\% = 4.74\% \approx 4.7\%\)

1 mark for calculating the correct percentage uncertainty using the formula.

- Precision refers to how close the measured values are to one another (small random error).
- Accuracy refers to how close the average (mean) of the measurements is to the true value (small systematic error).

Let us analyze Student R:
- Range of measurements = \(10.10 - 9.40 = 0.70\text{ m s}^{-2}\), which indicates a wide spread, hence low precision.
- Mean value = \(\frac{9.40 + 9.85 + 10.10 + 9.90}{4} = 9.8125\text{ m s}^{-2}\), which is extremely close to the true value of \(9.81\text{ m s}^{-2}\), hence high accuracy.

Therefore, Student R has low precision but high accuracy.

1 mark for identifying the student with low precision (large spread) and high accuracy (mean closest to the true value).

Using the equations of motion for an object falling from rest:
\(s = \frac{1}{2}gt^2\)

Comparing this with \(y = mx\), where \(y = s\) and \(x = t^2\), the gradient \(G\) is:
\(G = \frac{1}{2}g \implies g = 2G\)

Therefore, the best value for \(g\) is:
\(g_{\text{best}} = 2 \times 4.85 = 9.70\text{ m s}^{-2}\)

The uncertainty in the gradient \(G\) is:
\(\Delta G = |G_{\text{best}} - G_{\text{worst}}| = |4.85 - 4.60| = 0.25\text{ m s}^{-2}\)

The absolute uncertainty in \(g\) is:
\(\Delta g = 2 \Delta G = 2 \times 0.25 = 0.50\text{ m s}^{-2}\)

Thus, the experimental value is \((9.7 \pm 0.5)\text{ m s}^{-2}\). Note that the value and its uncertainty must be stated to the same number of decimal places (1 decimal place here because the absolute uncertainty is quoted to 1 significant figure as \(0.5\)).

1 mark for calculating the correct experimental value and its uncertainty to the correct decimal formatting.

First, calculate the redshift \(z\):
\(z = \frac{\Delta \lambda}{\lambda} = \frac{678.5 - 656.3}{656.3} = \frac{22.2}{656.3} \approx 0.033826\)

The recession velocity \(v\) is given by:
\(v = zc = 0.033826 \times 3.00 \times 10^8\text{ m s}^{-1} \approx 1.015 \times 10^7\text{ m s}^{-1}\)

According to Hubble's law:
\(v = H_0 d \implies d = \frac{v}{H_0}\)

Substitute the values:
\(d = \frac{1.015 \times 10^7}{2.3 \times 10^{-18}} \approx 4.41 \times 10^{24}\text{ m}\)

1 mark for calculating the correct distance using redshift and Hubble's law.

The gravitational potential energy \(E_p\) of the satellite is given by:
\(E_p = -\frac{GMm}{r}\)

For a circular orbit, the centripetal force is provided by the gravitational force:
\(\frac{GMm}{r^2} = \frac{mv^2}{r} \implies m v^2 = \frac{GMm}{r}\)

The kinetic energy \(E_k\) is:
\(E_k = \frac{1}{2}mv^2 = \frac{GMm}{2r}\)

The total energy \(E_{\text{total}}\) is the sum of \(E_k\) and \(E_p\):
\(E_{\text{total}} = E_k + E_p = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}\)

1 mark for combining kinetic and gravitational potential energy correctly.

First, find the energy of the incident photons:
\(E = hf = 6.63 \times 10^{-34} \times 1.2 \times 10^{15} = 7.956 \times 10^{-19}\text{ J}\)

Convert the work function \(\Phi\) from electronvolts to Joules:
\(\Phi = 3.2\text{ eV} \times 1.60 \times 10^{-19}\text{ J/eV} = 5.12 \times 10^{-19}\text{ J}\)

Using the photoelectric equation:
\(E_{k,\max} = hf - \Phi = 7.956 \times 10^{-19} - 5.12 \times 10^{-19} = 2.836 \times 10^{-19}\text{ J}\)

Using \(E_{k,\max} = \frac{1}{2}m_e v_{\max}^2\):
\(v_{\max} = \sqrt{\frac{2 E_{k,\max}}{m_e}} = \sqrt{\frac{2 \times 2.836 \times 10^{-19}}{9.11 \times 10^{-31}}} \approx 7.89 \times 10^5\text{ m s}^{-1}\)

This rounds to \(7.9 \times 10^5\text{ m s}^{-1}\).

1 mark for the correct calculation of maximum speed of photoelectrons using photoelectric equation.

The total energy \(E\) of a simple harmonic oscillator is given by:
\(E = \frac{1}{2} m \omega^2 x_0^2\)

where:
\(\omega = 2\pi f = 2\pi \times 4.0 = 8\pi\text{ rad s}^{-1}\)
\(x_0 = 5.0\text{ cm} = 0.050\text{ m}\)

Substituting the values:
\(E = \frac{1}{2} \times 0.20 \times (8\pi)^2 \times (0.050)^2\)
\(E = 0.10 \times 64\pi^2 \times 0.0025 = 0.016\pi^2 \approx 0.158\text{ J}\)

This is approximately \(0.16\text{ J}\).

1 mark for calculating the total energy correctly using the SHM energy formula.

The equations for maximum height \(H\) and horizontal range \(R\) are:
\(H = \frac{u^2 \sin^2 \theta}{2g}\)
\(R = \frac{u^2 \sin(2\theta)}{g} = \frac{2u^2 \sin\theta \cos\theta}{g}\)

Given that \(R = 3H\), we can write:
\(\frac{2u^2 \sin\theta \cos\theta}{g} = 3 \left(\frac{u^2 \sin^2 \theta}{2g}\right)\)

Simplifying this expression:
\(2 \sin\theta \cos\theta = \frac{3}{2} \sin^2 \theta\)

Dividing both sides by \(\cos\theta\) and \(\sin\theta\) (since \(\theta \neq 0\)):
\(2 = \frac{3}{2} \tan\theta\)
\(\tan\theta = \frac{4}{3}\)
\(\theta = \arctan(1.333) \approx 53.1^\circ\)

Thus, \(\theta \approx 53^\circ\).

1 mark for correctly equating range and height to find \(\tan\theta\) and the angle.

The percentage uncertainty in length L is \frac{0.004}{0.800} \times 100\% = 0.5\% . The percentage uncertainty in period T is \frac{0.03}{1.80} \times 100\% = 1.67\% . The percentage uncertainty in g is given by adding the percentage uncertainty in L to twice the percentage uncertainty in T : \frac{\Delta g}{g} \times 100\% = \frac{\Delta L}{L} \times 100\% + 2 \left( \frac{\Delta T}{T} \times 100\% \right) = 0.5\% + 2(1.67\%) = 3.84\% , which is approximately 3.8\% .

1 mark for the correct option C.

The redshift equation is \frac{\Delta \lambda}{\lambda} = \frac{v}{c} , where \lambda is the source (laboratory) wavelength of 656 nm. The change in wavelength is \Delta \lambda = 686 - 656 = 30 \text{ nm} . The relative speed v is calculated as: v = c \frac{\Delta \lambda}{\lambda} = 3.00 \times 10^8 \times \frac{30}{656} = 1.37 \times 10^7 \text{ m s}^{-1} \approx 1.4 \times 10^7 \text{ m s}^{-1} . Since the measured wavelength is greater than the laboratory wavelength, the light is redshifted, meaning the galaxy is moving away from the Earth.

1 mark for the correct option D.

The gravitational potential at the surface of the planet is \phi_{1} = -\frac{GM}{R} and at a distance of 3R from the center is \phi_{2} = -\frac{GM}{3R} . The work done against the gravitational field is equal to the increase in gravitational potential energy: W = m(\phi_{2} - \phi_{1}) = m\left(-\frac{GM}{3R} - \left(-\frac{GM}{R}\right)\right) = m\left(\frac{GM}{R} - \frac{GM}{3R}\right) = \frac{2GMm}{3R} .

1 mark for the correct option B.

Using Einstein's photoelectric equation: for frequency f , hf = \Phi + E_k . For frequency 2f , h(2f) = \Phi + 3E_k . Substituting hf = \Phi + E_k into the second equation gives 2(\Phi + E_k) = \Phi + 3E_k , which simplifies to 2\Phi + 2E_k = \Phi + 3E_k , hence \Phi = E_k = 1.0 E_k .

1 mark for the correct option B.

The speed v in simple harmonic motion is given by v = \omega \sqrt{A^2 - x^2} , where \omega = \frac{2\pi}{T} . Substituting x = \frac{1}{2} A gives v = \frac{2\pi}{T} \sqrt{A^2 - \left(\frac{1}{2}A\right)^2} = \frac{2\pi}{T} \sqrt{A^2 - \frac{1}{4}A^2} = \frac{2\pi}{T} \sqrt{\frac{3}{4}A^2} = \frac{\pi A \sqrt{3}}{T} .

1 mark for the correct option C.

Using the equation of motion s = ut + \frac{1}{2} a t^2 with the upward direction defined as positive: initial velocity u = +15.0 \text{ m s}^{-1} , acceleration a = -9.81 \text{ m s}^{-2} , and time t = 5.00 \text{ s} . The displacement is s = (15.0 \times 5.00) + \frac{1}{2} (-9.81) (5.00)^2 = 75.0 - 122.625 = -47.625 \text{ m} . Thus, the cliff height is 47.6 \text{ m} below the release point.

1 mark for the correct option A.

The initial resistance is R = \rho \frac{L_0}{A_0} . When stretched, the new length is L_1 = 1.10 L_0 . Since the volume V = A L remains constant, the new cross-sectional area is A_1 = \frac{A_0}{1.10} . The new resistance is R_1 = \rho \frac{L_1}{A_1} = \rho \frac{1.10 L_0}{A_0 / 1.10} = 1.10^2 \times \rho \frac{L_0}{A_0} = 1.21 R .

1 mark for the correct option B.

By conservation of momentum with the first glider's direction as positive: (3m)(u) + (m)(-2u) = (3m + m) v_f , which gives v_f = \frac{1}{4}u . The initial kinetic energy of the system is E_{k,i} = \frac{1}{2}(3m)u^2 + \frac{1}{2}m(-2u)^2 = 1.5 m u^2 + 2 m u^2 = 3.5 m u^2 . The final kinetic energy of the system is E_{k,f} = \frac{1}{2}(4m)v_f^2 = 2m \left(\frac{u}{4}\right)^2 = \frac{1}{8} m u^2 = 0.125 m u^2 . The loss in kinetic energy is E_{k,i} - E_{k,f} = 3.5 m u^2 - 0.125 m u^2 = 3.375 m u^2 = \frac{27}{8} m u^2 .

1 mark for the correct option D.

The formula for resistivity is \(\rho = \frac{R A}{L} = \frac{R \pi d^2}{4 L}\). The fractional uncertainty in resistivity is given by: \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\). Percentage uncertainty in \(R\) is \(\frac{0.5}{25.0} \times 100\% = 2.0\%\). Percentage uncertainty in \(d\) is \(\frac{0.01}{0.40} \times 100\% = 2.5\%\). Percentage uncertainty in \(L\) is \(\frac{0.003}{1.500} \times 100\% = 0.2\%\). Therefore, the percentage uncertainty in \(\rho\) is \(2.0\% + 2 \times 2.5\% + 0.2\% = 7.2\%\).

1 mark for calculating correct percentage uncertainties for each term and adding them correctly with the factor of 2 for diameter.

A systematic error is a constant bias in measurements that affects all readings in the same way (shifting them in one direction). A zero offset on the ruler means every distance measured will have the exact same constant error, shifting the entire set of measurements, which results in a systematic error. The other options describe random fluctuations which cause random errors.

1 mark for identifying the zero offset as the only source of systematic error.

First, calculate the redshift \(z\): \(z = \frac{\Delta \lambda}{\lambda_0} = \frac{672.7 - 656.3}{656.3} = 0.02499\). Next, determine the recession velocity \(v\): \(v = z c = 0.02499 \times 3.00 \times 10^8 = 7.50 \times 10^6 \text{ m s}^{-1}\). Using Hubble's Law \(v = H_0 d\), find the distance \(d\): \(d = \frac{v}{H_0} = \frac{7.50 \times 10^6}{2.3 \times 10^{-18}} = 3.26 \times 10^{24} \text{ m} \approx 3.3 \times 10^{24} \text{ m}\).

1 mark for calculating redshift, recession velocity, and applying Hubble's Law to obtain the correct distance.

The gravitational potential energy of a probe of mass \(m\) at the surface of the planet is \(E_p = m \phi_0\). Since \(\phi_0\) is negative, this represents a potential well. To escape, the change in potential energy is \(\Delta E_p = E_{\text{final}} - E_{\text{initial}} = 0 - m \phi_0 = -m \phi_0\). By conservation of energy, the minimum kinetic energy required at launch must equal this increase in potential energy, which is \(-m \phi_0\).

1 mark for recognizing that escape requires total energy to be at least zero, leading to \(E_k = -m \phi_0\).

The Einstein photoelectric equation is \(E_{\text{max}} = hf - \Phi\), where \(\Phi\) is the work function. When the frequency is doubled, the new maximum kinetic energy \(E_{\text{max}}'\) is: \(E_{\text{max}}' = h(2f) - \Phi = 2hf - \Phi\). From the first equation, \(hf = E_{\text{max}} + \Phi\). Substituting this into the new equation gives: \(E_{\text{max}}' = 2(E_{\text{max}} + \Phi) - \Phi = 2E_{\text{max}} + \Phi\). Since the work function \(\Phi\) of any metal is a positive constant, the new kinetic energy must be strictly more than \(2E_{\text{max}}\).

1 mark for using the photoelectric equation to demonstrate that doubling the frequency results in a maximum kinetic energy greater than twice the original.

The equation for simple harmonic motion displacement is \(x = x_0 \cos(\omega t)\). Comparing this with \(x = 0.050 \cos(10 \pi t)\) gives: \(x_0 = 0.050 \text{ m}\) and \(\omega = 10 \pi \text{ rad s}^{-1}\). The maximum velocity of the object is \(v_0 = \omega x_0 = 10 \pi \times 0.050 = 0.50 \pi \text{ m s}^{-1}\). The maximum kinetic energy is \(E_{\text{max}} = \frac{1}{2} m v_0^2 = \frac{1}{2} \times 0.20 \times (0.50 \pi)^2 = 0.025 \pi^2 \approx 0.247 \text{ J} \approx 0.25 \text{ J}\).

1 mark for identifying the angular frequency and amplitude, and calculating the maximum kinetic energy correctly to 2 significant figures.

Using the kinematic equation \(s = ut + \frac{1}{2} a t^2\) with the upwards direction taken as positive: initial velocity \(u = +15.0 \text{ m s}^{-1}\), acceleration \(a = -9.81 \text{ m s}^{-2}\), and time \(t = 5.00 \text{ s}\). Substituting these gives: \(s = (15.0 \times 5.00) + \frac{1}{2} \times (-9.81) \times (5.00)^2 = 75.0 - 122.625 = -47.6 \text{ m}\). The displacement is negative, indicating the final position is \(47.6 \text{ m}\) below the release point. Thus, the cliff height is \(47.6 \text{ m}\).

1 mark for applying the displacement-time equation with consistent signs to find the cliff height.

Using Newton's second law in terms of momentum, the average force is \(F = \frac{\Delta p}{\Delta t}\). Taking the initial direction of the ball's motion as positive: initial momentum \(p_i = m u = 0.15 \times 20 = 3.0 \text{ kg m s}^{-1}\). Final momentum \(p_f = m v = 0.15 \times (-12) = -1.8 \text{ kg m s}^{-1}\). The change in momentum is \(\Delta p = p_f - p_i = -1.8 - 3.0 = -4.8 \text{ kg m s}^{-1}\). The magnitude of the average force is \(F = \left| \frac{\Delta p}{\Delta t} \right| = \frac{4.8}{0.040} = 120 \text{ N}\).

1 mark for calculating the change in momentum (taking direction into account) and dividing by the contact time to get the average force.

(a) \(\text{Density} = \frac{\text{mass}}{\text{volume}}\).
Unit of mass = \(\text{kg}\).
Unit of volume = \(\text{m}^3\).
Therefore, unit of density = \(\frac{\text{kg}}{\text{m}^3} = \text{kg m}^{-3}\).

(b) \(V = \frac{\pi d^2}{4} L = \frac{\pi \times 1.24^2}{4} \times 8.50 = 10.27\text{ cm}^3\).
\(\rho = \frac{m}{V} = \frac{124.3}{10.27} = 12.108\text{ g cm}^{-3} \approx 12.1\text{ g cm}^{-3}\).

(c) (i) Using \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\):
\(\frac{\Delta m}{m} = \frac{0.1}{124.3} = 0.000804\) (or \(0.080\%\))
\(\frac{\Delta d}{d} = \frac{0.02}{1.24} = 0.0161\) (or \(1.61\%\))
\(\frac{\Delta L}{L} = \frac{0.05}{8.50} = 0.00588\) (or \(0.59\%\))
Percentage uncertainty = \(0.080\% + 2(1.61\%) + 0.59\% = 3.89\% \approx 3.9\%\) (accept \(4\%\)).

(ii) Absolute uncertainty \(\Delta \rho = 12.108 \times 0.0389 = 0.47\text{ g cm}^{-3} \approx 0.5\text{ g cm}^{-3}\).
Therefore, \(\rho = (12.1 \pm 0.5)\text{ g cm}^{-3}\).

(a) [1] State: Density = mass / volume, and substitute base units: kg / m^3.

(b) [3]
- 1 mark for calculating Volume = 10.27 cm^3
- 1 mark for the formula: rho = m / V
- 1 mark for correct value: 12.1 g cm^-3 (or 1.21 x 10^4 kg m^-3)

(c)(i) [3]
- 1 mark for correct fractional/percentage uncertainty formula: dm/m + 2*dd/d + dL/L
- 1 mark for substituting values: 0.0008 + 2(0.0161) + 0.0059
- 1 mark for calculating: 3.9% (or 4.0%)

(c)(ii) [2]
- 1 mark for calculating absolute uncertainty: 0.5 g cm^-3
- 1 mark for writing the final density to 1 decimal place matching the uncertainty: (12.1 +/- 0.5) g cm^-3

(a) (i) \(u_y = u \sin(30^\circ) = 18 \sin(30^\circ) = 9.0\text{ m s}^{-1}\).

(ii) Taking downwards as positive:
\(s = -45\text{ m}\) (relative to launch point), \(u_y = 9.0\text{ m s}^{-1}\), and \(a = -9.81\text{ m s}^{-2}\).
\(s = u_y t + \frac{1}{2} a t^2 \implies -45 = 9.0 t - 4.905 t^2\)
\(4.905 t^2 - 9.0 t - 45 = 0\)
Solving the quadratic equation:
\(t = \frac{9.0 \pm \sqrt{(-9.0)^2 - 4(4.905)(-45)}}{2(4.905)}\)
\(t = \frac{9.0 \pm \sqrt{81 + 882.9}}{9.81} = \frac{9.0 + 31.05}{9.81} = 4.08\text{ s} \approx 4.1\text{ s}\).

(b) \(u_x = u \cos(30^\circ) = 18 \cos(30^\circ) = 15.6\text{ m s}^{-1}\).
Horizontal distance \(x = u_x t = 15.6 \times 4.08 = 63.6\text{ m} \approx 64\text{ m}\).

(c) If air resistance were present, there would be a horizontal drag force that causes horizontal deceleration. Thus, the average horizontal speed would be lower, and the ball would travel a smaller horizontal distance before hitting the ground.

(a)(i) [1] 9.0 m s^-1
(a)(ii) [3]
- 1 mark for using s = ut + 0.5 a t^2 with correct signs
- 1 mark for setting up the quadratic equation correctly
- 1 mark for calculating t = 4.1 s (or 4.08 s)

(b) [2]
- 1 mark for calculating horizontal velocity = 15.6 m s^-1
- 1 mark for distance = 64 m (or 63.6 m)

(c) [2]
- 1 mark for stating that air resistance opposes horizontal motion / causes horizontal deceleration
- 1 mark for stating that this reduces horizontal velocity and therefore decreases the horizontal distance.

(a) The total momentum of an isolated/closed system (where no external forces act) remains constant.

(b) (i) By conservation of momentum:
\(m_A u_A + m_B u_B = (m_A + m_B) v\)
\(0.80 \times 3.5 + 0 = (0.80 + 1.2) v\)
\(2.8 = 2.0 v \implies v = 1.4\text{ m s}^{-1}\).

(ii) Initial kinetic energy:
\(E_k = \frac{1}{2} m_A u_A^2 = 0.5 \times 0.80 \times 3.5^2 = 4.90\text{ J}\).
Final kinetic energy:
\(E_k = \frac{1}{2} (m_A + m_B) v^2 = 0.5 \times 2.0 \times 1.4^2 = 1.96\text{ J}\).
Loss in kinetic energy = \(4.90 - 1.96 = 2.94\text{ J} \approx 2.9\text{ J}\).

(c) By Newton's third law, the force that trolley A exerts on trolley B is equal in magnitude and opposite in direction to the force that trolley B exerts on trolley A. Since the contact time \(\Delta t\) is identical for both, the impulse (\(F \Delta t\)) experienced by each trolley is equal and opposite. Since impulse equals change in momentum, their changes in momentum are equal and opposite.

(a) [2]
- 1 mark for total momentum remains constant
- 1 mark for specifying in a closed system / no external force acting

(b)(i) [2]
- 1 mark for correct momentum conservation equation
- 1 mark for v = 1.4 m s^-1

(b)(ii) [3]
- 1 mark for initial KE = 4.9 J
- 1 mark for final KE = 1.96 J
- 1 mark for calculating the difference: 2.9 J (or 2.94 J)

(c) [2]
- 1 mark for stating that the forces between trolleys are equal and opposite (Newton's third law)
- 1 mark for linking this to impulse (F x t) being equal and opposite, and hence equal and opposite momentum changes.

(a) (i) Stress is force per unit cross-sectional area.
(ii) Young modulus is the ratio of stress to strain.

(b) (i) \(E = \frac{\text{stress}}{\text{strain}} = \frac{F / A}{x / L} = \frac{FL}{A x}\)
Rearranging for extension \(x\):
\(x = \frac{FL}{A E} = \frac{35 \times 2.4}{1.5 \times 10^{-7} \times 1.2 \times 10^{11}}\)
\(x = \frac{84}{1.8 \times 10^4} = 4.67 \times 10^{-3}\text{ m} = 4.7\text{ mm}\).

(ii) Strain energy \(E_s = \frac{1}{2} F x = 0.5 \times 35 \times 4.67 \times 10^{-3} = 0.0817\text{ J} \approx 0.082\text{ J}\).

(c) Under elastic deformation, the material returns to its original shape and dimensions when the load is removed. Under plastic deformation, permanent (irreversible) deformation remains after the load is removed.

(a)(i) [1] Force / Area
(a)(ii) [1] Stress / Strain

(b)(i) [3]
- 1 mark for formula: E = FL / Ax
- 1 mark for correct substitution of values
- 1 mark for extension = 4.7 mm (or 4.67 mm)

(b)(ii) [2]
- 1 mark for formula: E_s = 0.5 * F * x
- 1 mark for strain energy = 0.082 J (or 0.0817 J)

(c) [1] Elastic deformation is fully recovered when load is removed whereas plastic deformation is permanent.

(a) Using \(R = \frac{\rho L}{A}\):
\(R = \frac{1.1 \times 10^{-6} \times 3.0}{2.5 \times 10^{-7}} = 13.2\ \Omega \approx 13\ \Omega\).

(b) (i) \(I = \frac{V}{R} = \frac{12}{13.2} = 0.909\text{ A} \approx 0.91\text{ A}\) (or \(0.92\text{ A}\) using \(13\ \Omega\)).
(ii) \(P = V I = 12 \times 0.909 = 10.9\text{ W} \approx 11\text{ W}\) (or \(P = \frac{V^2}{R} = \frac{12^2}{13.2} = 10.9\text{ W}\)).

(c) (i) The total resistance of the circuit decreases. Because the two identical resistors are connected in parallel, the total resistance is halved (to \(6.6\ \Omega\)).
(ii) The total power dissipated increases/doubles. Since \(P = \frac{V^2}{R}\) and total resistance \(R\) is halved while potential difference remains constant, the power dissipated is doubled to \(22\text{ W}\).

(a) [2]
- 1 mark for formula: R = rho * L / A
- 1 mark for correct calculation showing 13.2 ohms (approx 13 ohms)

(b)(i) [1] I = V/R = 0.91 A (or 0.92 A)
(b)(ii) [2]
- 1 mark for formula: P = VI or P = V^2 / R
- 1 mark for power = 11 W (or 10.9 W)

(c)(i) [2]
- 1 mark for stating that total resistance is halved / decreases
- 1 mark for explanation (e.g. 1/R_total = 1/R + 1/R, or R_total = 6.6 ohms)

(c)(ii) [2]
- 1 mark for stating that total power increases / doubles
- 1 mark for explaining that both wires have the same potential difference of 12 V and thus each dissipates 11 W.

(a) Simple harmonic motion is a motion where the acceleration of the system is directly proportional to its displacement from its equilibrium position, and is always directed towards that equilibrium position.

(b) (i) Amplitude \(x_0 = 0.045\text{ m}\).
(ii) Angular frequency \(\omega = 12\text{ rad s}^{-1}\).

(c) (i) \(\omega = 2 \pi f \implies f = \frac{\omega}{2 \pi} = \frac{12}{2 \pi} = 1.91\text{ Hz} \approx 1.9\text{ Hz}\).

(ii) \(a_{\text{max}} = \omega^2 x_0 = 12^2 \times 0.045 = 144 \times 0.045 = 6.48\text{ m s}^{-2} \approx 6.5\text{ m s}^{-2}\).

(a) [2]
- 1 mark for: acceleration is directly proportional to displacement
- 1 mark for: acceleration is always directed towards a fixed/equilibrium position

(b)(i) [1] Amplitude = 0.045 m
(b)(ii) [1] Angular frequency = 12 rad s^-1 (or s^-1)

(c)(i) [2]
- 1 mark for using f = omega / 2*pi
- 1 mark for frequency = 1.9 Hz (or 1.91 Hz)

(c)(ii) [2]
- 1 mark for using a_max = omega^2 * x_0
- 1 mark for maximum acceleration = 6.5 m s^-2 (or 6.48 m s^-2)

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

(b) (i) Photon energy \(E = \frac{h c}{\lambda}\).
\(E = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{2.60 \times 10^{-7}} = 7.65 \times 10^{-19}\text{ J}\).
In \(\text{eV}\): \(E = \frac{7.65 \times 10^{-19}}{1.60 \times 10^{-19}} = 4.781\text{ eV} \approx 4.78\text{ eV}\).

(ii) Using the photoelectric equation:
\(E_k = hf - \Phi = 4.78\text{ eV} - 2.28\text{ eV} = 2.50\text{ eV}\).

(iii) Maximum kinetic energy in Joules:
\(E_k = 2.50 \times 1.60 \times 10^{-19} = 4.00 \times 10^{-19}\text{ J}\).
Since \(E_k = \frac{1}{2} m_e v^2\):
\(v = \sqrt{\frac{2 \times E_k}{m_e}} = \sqrt{\frac{2 \times 4.00 \times 10^{-19}}{9.11 \times 10^{-31}}} = 9.38 \times 10^5\text{ m s}^{-1} \approx 9.4 \times 10^5\text{ m s}^{-1}\).

(a) [2]
- 1 mark for: minimum energy required
- 1 mark for: to release/liberate an electron from the surface of a metal

(b)(i) [3]
- 1 mark for formula: E = hc / lambda
- 1 mark for energy in Joules = 7.65 x 10^-19 J
- 1 mark for converting to eV = 4.78 eV

(b)(ii) [2]
- 1 mark for the photoelectric equation: KE_max = E - Work Function
- 1 mark for KE_max = 2.50 eV

(b)(iii) [2]
- 1 mark for converting KE to Joules (4.0 x 10^-19 J) and using 1/2 m v^2
- 1 mark for maximum speed = 9.4 x 10^5 m s^-1

**Step-by-step physical analysis:**

1. **Theoretical Relationship:**
The period of a physical pendulum consisting of a uniform rod of mass \( M \) and length \( L_0 \) with an added point mass \( m \) at distance \( d \) is given by:
\[ T = 2\pi \sqrt{\frac{I}{(M + m) g y_{\text{com}}}} \]
where \( I \) is the moment of inertia about the pivot, and \( y_{\text{com}} \) is the distance from the pivot to the center of mass.

This relationship simplifies to an empirical linear form when we plot variables chosen to linearize the system:
\[ T^2 d = a d^2 + b \]
Here, plotting \( T^2 d \) on the y-axis against \( d^2 \) on the x-axis yields a straight line where:
- \( \text{Gradient} = a \)
- \( \text{y-intercept} = b \)

2. **Worked Example Calculations:**
Let the raw measurements be:
- \( d = 40.0 \text{ cm} = 0.400 \text{ m} \)
- \( 10 \text{ oscillations} = 13.52 \text{ s} \implies T = 1.352 \text{ s} \)
- \( T^2 d = (1.352)^2 \times 0.400 = 0.731 \text{ s}^2\text{ m} \)
- \( d^2 = 0.160 \text{ m}^2 \)

By performing this for multiple \( d \) values:
- For \( d = 0.200 \text{ m} \): \( T = 1.140 \text{ s} \implies T^2 d = 0.260 \text{ s}^2\text{ m} \), \( d^2 = 0.040 \text{ m}^2 \)
- For \( d = 0.800 \text{ m} \): \( T = 1.831 \text{ s} \implies T^2 d = 2.682 \text{ s}^2\text{ m} \), \( d^2 = 0.640 \text{ m}^2 \)

Calculating the gradient \( a \):
\[ a = \frac{2.682 - 0.260}{0.640 - 0.040} = \frac{2.422}{0.600} \approx 4.04 \text{ s}^2\text{ m}^{-1} \]

Calculating the intercept \( b \):
\[ b = T^2 d - a d^2 = 0.260 - (4.04 \times 0.040) = 0.260 - 0.1616 = 0.0984 \text{ s}^2\text{ m} \]

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

* **Initial Measurements & Routine (3 marks):**
- **[1]** Value of raw \( d \) recorded to the nearest millimetre with correct unit (e.g. \( 40.0 \pm 0.1 \text{ cm} \)).
- **[1]** Value of \( t \) recorded for at least 10 oscillations, with repeating measurements shown.
- **[1]** Correct calculation of \( T \) with appropriate unit (seconds).

* **Table of Results (6 marks):**
- **[1]** Table includes headings with correct units: \( d/\text{cm} \), \( t/\text{s} \), \( T/\text{s} \), \( d^2/\text{cm}^2 \), \( T^2 d/\text{s}^2\text{ cm} \) (or SI units).
- **[1]** Raw readings of \( d \) are consistent to the nearest millimetre.
- **[1]** Minimum of 6 sets of different readings taken.
- **[1]** Selected range of \( d \) is wide (difference between maximum and minimum values of \( d \ge 45.0 \text{ cm} \)).
- **[1]** Column \( T^2 d \) calculated to the correct number of significant figures (matching the lease SF of \( d \) and \( T \), typically 3 SF).
- **[1]** Quality of data: points lie close to a straight line (trend shows \( T^2 d \) increases as \( d^2 \) increases).

* **Graph & Linear Fit (5 marks):**
- **[1]** Axes: scale chosen so that plotted points occupy more than half of the grid in both directions. Axis labeled with quantities and units.
- **[1]** Plotting: points plotted accurately to within half a small square. No thick blobs.
- **[1]** Line of best fit: straight line drawn with a balanced distribution of points on either side.
- **[1]** Precision of reading values: no anomalous points ignored without explanation.

* **Gradient & Intercept (2 marks):**
- **[1]** Gradient: uses a triangle with a hypotenuse greater than half the length of the drawn line.
- **[1]** Intercept: read directly from the y-axis if the scale starts at zero, or calculated using a point on the line \( y - mx \).

* **Analysis of Constants (4 marks):**
- **[1]** Constant \( a \) set equal to gradient with appropriate working.
- **[1]** Constant \( b \) set equal to intercept with appropriate working.
- **[1]** Correct units for both: \( a \) in \( \text{s}^2\text{ m}^{-1} \) (or \( \text{s}^2\text{ cm}^{-1} \)) and \( b \) in \( \text{s}^2\text{ m} \) (or \( \text{s}^2\text{ cm} \)).
- **[1]** Values of constants are physically reasonable.

**Step-by-step physical analysis:**

1. **Theoretical deflection of a cantilever:**
The deflection \( d \) of a cantilever loaded with a point force \( F = mg \) at its free end is given by:
\[ d = \frac{F L^3}{3 E I} \]
where \( E \) is Young's Modulus and \( I \) is the area moment of inertia. Rearranging this yields:
\[ \frac{L^3}{d} = \frac{3 E I}{F} = k \]
If \( E \), \( I \), and \( F \) are constant, \( k \) should remain constant.

2. **Example experimental values:**
- **First measurement:**
\( L_1 = 40.0 \text{ cm} = 0.400 \text{ m} \)
\( y_0 = 50.2 \text{ cm} \), \( y_1 = 44.1 \text{ cm} \implies d_1 = 6.1 \text{ cm} = 0.061 \text{ m} \)
\( k_1 = \frac{(0.400)^3}{0.061} = 1.05 \text{ m}^2 \)
- **Second measurement:**
\( L_2 = 30.0 \text{ cm} = 0.300 \text{ m} \)
\( y_0 = 50.2 \text{ cm} \), \( y_1 = 47.6 \text{ cm} \implies d_2 = 2.6 \text{ cm} = 0.026 \text{ m} \)
\( k_2 = \frac{(0.300)^3}{0.026} = 1.04 \text{ m}^2 \)

3. **Uncertainty Analysis:**
- Since \( d \) is determined by subtracting two ruler readings, and the uncertainty in each reading is \( \pm 1 \text{ mm} \), the absolute uncertainty in \( d \) is:
\[ \Delta d \approx 2 \text{ mm} = 0.2 \text{ cm} \]
- For the first reading: \( \% \text{ uncertainty} = \frac{0.2}{6.1} \times 100 \% \approx 3.3 \% \).

4. **Comparison:**
- Percentage difference between \( k_1 \) and \( k_2 \):
\[ \% \text{ diff} = \frac{|1.05 - 1.04|}{1.045} \times 100 \% \approx 0.96 \% \]
- Since \( 0.96 \% < 10 \% \) (or less than the experimental percentage uncertainty of \( 3.3 \% \)), the suggestion that \( k \) is constant is strongly supported.

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

* **Measurements & Deflection (3 marks):**
- **[1]** Measured \( L \) to nearest millimetre with unit (e.g. \( L = 40.0 \pm 0.1 \text{ cm} \)).
- **[1]** Measured \( y_0 \) and \( y_1 \) to nearest millimetre, showing raw readings.
- **[1]** Deflection \( d \) calculated correctly with correct unit.

* **Percentage Uncertainty (2 marks):**
- **[1]** Standard absolute uncertainty \( \Delta d \) estimated realistically in range \( 1 \text{ mm} \text{ to } 2 \text{ mm} \).
- **[1]** Correct percentage uncertainty calculation: \( \frac{\Delta d}{d} \times 100\% \).

* **First Constant \( k \) (2 marks):**
- **[1]** Calculated \( k_1 \) correctly from \( L_1^3 / d_1 \).
- **[1]** Correct unit for \( k \) stated as \( \text{cm}^2 \) or \( \text{m}^2 \).

* **Second Constant \( k \) (3 marks):**
- **[1]** Measured second projecting length \( L_2 \approx 30.0 \text{ cm} \).
- **[1]** Measured second deflection \( d_2 \) showing all intermediate steps.
- **[1]** Correctly calculated second \( k \) value.

* **Comparison and Conclusion (2 marks):**
- **[1]** Calculated percentage difference between the two values of \( k \).
- **[1]** Explicitly stated criterion (e.g. comparison with \( 10\% \) limit or the calculated percentage uncertainty in \( d \)) to make a valid concluding statement.

* **Limitations & Improvements (8 marks - 1 mark for each point, max 4 pairs):**
- **Limitation 1:** Deflection \( d \) is relatively small, which leads to a large percentage uncertainty.
- **Improvement 1:** Use a larger mass \( M \) or a thinner/more flexible ruler to increase deflection.
- **Limitation 2:** The flat rule slips or shifts at the clamp when loaded, altering the effective \( L \).
- **Improvement 2:** Place high-friction rubber pads between the rule and the bench/clamping blocks.
- **Limitation 3:** Difficult to read the vertical scale precisely because of parallax error.
- **Improvement 3:** Use a small mirror aligned behind the vertical scale (align reflection) or mount a digital travel sensor.
- **Limitation 4:** The suspended mass oscillates or swings, making it hard to measure a stable vertical height.
- **Improvement 4:** Place the suspended mass inside a container of oil/water to damp the swing, or use a video camera to record and take the average of the extrema of oscillation.

(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) For a satellite of mass \(m\) in a circular orbit of radius \(r\) around a planet of mass \(M\), the centripetal force is provided by the gravitational force: \(G M m / r^2 = m v^2 / r\). Simplifying this gives \(v^2 = G M / r\). The gravitational potential is defined as \(\phi = -G M / r\). Substituting this into the speed equation yields \(v^2 = -\phi\). (c)(i) The altitude is equal to the radius \(R\), so the orbital radius is \(r = 2R = 2 \times 5.50 \times 10^6\text{ m} = 1.10 \times 10^7\text{ m}\). The gravitational potential is \(\phi = -G M / r = -(6.67 \times 10^{-11} \times 4.80 \times 10^{24}) / (1.10 \times 10^7) = -2.91 \times 10^7\text{ J kg}^{-1}\). (ii) The kinetic energy is \(E_k = \frac{1}{2} m v^2 = \frac{1}{2} m (-\phi) = 0.5 \times 650 \times 2.91 \times 10^7 = 9.46 \times 10^9\text{ J}\).

(a) Work done per unit mass [1] in bringing a test mass from infinity to the point [1]. (b) Equating gravitational force to centripetal force: \(G M m / r^2 = m v^2 / r\) [1]. Rearranging to find \(v^2 = G M / r\) [1]. Using \(\phi = -G M / r\) to show \(v^2 = -\phi\) [1]. (c)(i) Recalling \(r = 2R = 1.10 \times 10^7\text{ m}\) [1]. Correct calculation of potential to give \(-2.91 \times 10^7\text{ J kg}^{-1}\) (including negative sign) [1]. (ii) Recalling \(E_k = \frac{1}{2} m v^2\) or \(E_k = -\frac{1}{2} m \phi\) [1]. Correct substitution of values [1]. Final answer of \(9.46 \times 10^9\text{ J}\) [1].

(a) Redshift is the increase in the observed wavelength (or decrease in frequency) of electromagnetic radiation due to the relative motion of the source away from the observer. (b)(i) The change in wavelength is \(\Delta \lambda = 494.3\text{ nm} - 486.1\text{ nm} = 8.2\text{ nm}\). The redshift is \(z = \Delta \lambda / \lambda = 8.2 / 486.1 = 0.01687 \approx 0.0169\). (ii) Using \(v = z c\), the recession speed is \(v = 0.01687 \times 3.00 \times 10^8\text{ m s}^{-1} = 5.06 \times 10^6\text{ m s}^{-1}\), which is approximately \(5.1 \times 10^6\text{ m s}^{-1}\). (c)(i) Using Hubble's law \(v = H_0 d\), the distance is \(d = v / H_0 = 5.06 \times 10^6 / (2.2 \times 10^{-18}) = 2.3 \times 10^{24}\text{ m}\). (ii) The age of the Universe is estimated as \(T \approx 1 / H_0 = 1 / (2.2 \times 10^{-18}) = 4.55 \times 10^{17}\text{ s}\). In years, this is \(4.55 \times 10^{17} / (365.25 \times 24 \times 3600) = 1.44 \times 10^{10}\text{ years}\) (or 14 billion years).

(a) Apparent increase in wavelength [1] due to the source moving away from the observer [1]. (b)(i) Correct calculation of \(\Delta \lambda = 8.2\text{ nm}\) [1]. Calculation of \(z = 0.0169\) [1]. (ii) Recalling \(v = z c\) [1]. Showing calculation leading to \(5.06 \times 10^6\text{ m s}^{-1}\) (must show at least 3 sig figs before rounding) [1]. (c)(i) Recalling Hubble's law \(v = H_0 d\) and rearranging [1]. Correct answer of \(2.3 \times 10^{24}\text{ m}\) (accept range \(2.3 \times 10^{24}\) to \(2.32 \times 10^{24}\)) [1]. (ii) Recalling \(T = 1 / H_0\) [1]. Conversion from seconds to years to obtain \(1.4 \times 10^{10}\text{ years}\) [1].

(a) The two conditions are: (1) The acceleration of the system is directly proportional to its displacement from the equilibrium position. (2) The acceleration is always directed towards the equilibrium position (opposite direction to displacement). (b)(i) Comparing the equation to \(x = x_0 \cos(\omega t)\), the angular frequency is \(\omega = 15.0\text{ rad s}^{-1}\). The frequency is \(f = \omega / (2\pi) = 15.0 / (2\pi) = 2.39\text{ Hz}\). (ii) The maximum kinetic energy is equal to the total energy of the oscillator, which is given by \(E_{k,\text{max}} = \frac{1}{2} m \omega^2 x_0^2\). Substituting the values: \(E_{k,\text{max}} = 0.5 \times 0.350 \times (15.0)^2 \times (0.080)^2 = 0.252\text{ J}\). (iii) The total energy is the sum of kinetic and potential energies: \(E_{\text{total}} = E_k + E_p\). When \(E_k = E_p\), then \(E_{\text{total}} = 2 E_p\). This means \(\frac{1}{2} m \omega^2 x_0^2 = 2 \left(\frac{1}{2} m \omega^2 x^2\right)\), which simplifies to \(x_0^2 = 2 x^2\). Thus, \(x = x_0 / \sqrt{2} = 0.080 / \sqrt{2} = 0.057\text{ m}\) (or \(5.7\text{ cm}\)).

(a) Acceleration proportional to displacement [1]. Acceleration and displacement in opposite directions / acceleration directed to equilibrium point [1]. (b)(i) Identifying \(\omega = 15.0\text{ rad s}^{-1}\) [1]. Correct value of \(f = 2.39\text{ Hz}\) [1]. (ii) Recalling formula \(E_k = \frac{1}{2} m \omega^2 x_0^2\) or finding maximum speed first [1]. Substitution of values [1]. Correct answer of \(0.252\text{ J}\) [1]. (iii) Stating relation \(E_{\text{total}} = 2 E_p\) or \(x = x_0 / \sqrt{2}\) [1]. Substitution of \(x_0 = 0.080\text{ m}\) [1]. Correct answer of \(0.057\text{ m}\) (or \(5.7\text{ cm}\)) [1].

(a) The work function energy of a metal is defined as the minimum energy of a photon required to release an electron from the surface of the metal. (b)(i) The energy of a photon is given by \(E = h c / \lambda\). Substituting the values: \(E = (6.63 \times 10^{-34} \times 3.00 \times 10^8) / (380 \times 10^{-9}) = 5.234 \times 10^{-19}\text{ J}\). In electron-volts: \(E = 5.234 \times 10^{-19} / (1.60 \times 10^{-19}) = 3.27\text{ eV}\). (ii) No photoelectric emission occurs because the photon energy (3.27 eV) is less than the work function energy (4.3 eV). Since the interaction is a one-to-one collision between a photon and an electron, individual photons do not have sufficient energy to liberate an electron. (c)(i) For \(\lambda = 240\text{ nm}\), the photon energy is \(E = h c / \lambda = (6.63 \times 10^{-34} \times 3.00 \times 10^8) / (240 \times 10^{-9}) = 8.288 \times 10^{-19}\text{ J}\). The work function in joules is \(\Phi = 4.3 \times 1.60 \times 10^{-19} = 6.88 \times 10^{-19}\text{ J}\). The maximum kinetic energy is \(E_{k,\text{max}} = E_{\text{photon}} - \Phi = 8.288 \times 10^{-19} - 6.88 \times 10^{-19} = 1.41 \times 10^{-19}\text{ J}\). (ii) The stopping potential \(V_s\) is given by \(e V_s = E_{k,\text{max}}\), so \(V_s = 1.41 \times 10^{-19} / 1.60 \times 10^{-19} = 0.88\text{ V}\).

(a) Minimum photon energy [1] required to remove an electron from the metal surface [1]. (b)(i) Recalling \(E = h c / \lambda\) [1]. Converting joules to eV to obtain \(3.27\text{ eV}\) (must show calculation clearly) [1]. (ii) Stating photon energy is less than work function [1]. Explaining the one-to-one nature of photon-electron interaction [1]. (c)(i) Recalling Einstein's photoelectric equation [1]. Calculating the new photon energy in joules as \(8.29 \times 10^{-19}\text{ J}\) [1]. Subtracting the work function in joules to obtain \(1.41 \times 10^{-19}\text{ J}\) [1]. (ii) Dividing the kinetic energy by \(1.60 \times 10^{-19}\text{ C}\) to obtain \(0.88\text{ V}\) [1].

(a) Any two of: (1) Total volume of gas molecules is negligible compared to the volume of the container. (2) There are no intermolecular forces except during collisions. (3) Collisions of molecules with each other and the walls are perfectly elastic. (4) Time of collision is negligible compared to time between collisions. (5) Molecules are in continuous random motion. (b)(i) Using the equation of state \(p V = n R T\) where \(T = 27 + 273.15 = 300\text{ K}\): \(n = p V / (R T) = (1.6 \times 10^5 \times 0.045) / (8.31 \times 300) = 7200 / 2493 = 2.89\text{ mol}\). (ii) The total kinetic energy is \(E_{\text{total}} = \frac{3}{2} n R T\) (or \rac{3}{2} p V\). \(E_{\text{total}} = 1.5 \times 1.6 \times 10^5 \times 0.045 = 10800\text{ J} = 1.08 \times 10^4\text{ J}\). (iii) The r.m.s. speed can be calculated from \(E_{\text{total}} = \frac{1}{2} M_{\text{total}} c_{\text{rms}}^2\). Total mass of helium \(M_{\text{total}} = n \times M = 2.89 \times 4.0 \times 10^{-3}\text{ kg} = 0.01156\text{ kg}\). Alternatively, using \(c_{\text{rms}} = \sqrt{3 R T / M}\) where \(M = 4.0 \times 10^{-3}\text{ kg mol}^{-1}\): \(c_{\text{rms}} = \sqrt{(3 \times 8.31 \times 300) / (4.0 \times 10^{-3})} = \sqrt{1.8698 \times 10^6} = 1370\text{ m s}^{-1}\).

(a) Any two assumptions stated correctly [1 mark each, max 2 marks]. (b)(i) Conversion of temperature to \(300\text{ K}\) [1]. Recalling \(p V = n R T\) [1]. Correct answer of \(2.89\text{ mol}\) [1]. (ii) Recalling total kinetic energy formula \(\frac{3}{2} p V\) or \(\frac{3}{2} n R T\) [1]. Correct calculation of \(1.08 \times 10^4\text{ J}\) [1]. (iii) Recalling formula \(c_{\text{rms}} = \sqrt{3 R T / M}\) or equivalent [1]. Correct value of molar mass in kg, \(4.0 \times 10^{-3}\text{ kg mol}^{-1}\) [1]. Correct final answer of \(1370\text{ m s}^{-1}\) (allow 1360 to 1380) [1].

(a) Capacitance is defined as the charge stored per unit potential difference. (b)(i) Initial energy stored: \(E_0 = \frac{1}{2} C V^2 = 0.5 \times 470 \times 10^{-6} \times (12.0)^2 = 0.03384\text{ J} \approx 3.38 \times 10^{-2}\text{ J}\). (ii) Time constant \(\tau = R C = 15 \times 10^3 \times 470 \times 10^{-6} = 7.05\text{ s}\). (iii) During discharge, the potential difference across the capacitor changes according to \(V = V_0 e^{-t/RC}\). Since energy is proportional to \(V^2\), the energy remaining is given by \(E = E_0 e^{-2t/RC}\). At \(t = 5.0\text{ s}\): \(E = 0.03384 \times e^{-2 \times 5.0 / 7.05} = 0.03384 \times e^{-1.4184} = 0.03384 \times 0.2421 = 8.18 \times 10^{-3}\text{ J}\). (iv) The energy dissipated in the resistor is the change in the energy stored in the capacitor: \(\Delta E = E_0 - E = 0.03384 - 0.00818 = 0.02566\text{ J}\). Average power dissipated is \(P = \Delta E / t = 0.02566 / 5.0 = 5.13 \times 10^{-3}\text{ W}\) (or 5.1 mW).

(a) Charge divided by potential difference [1]. (b)(i) Recalling energy formula \(E = \frac{1}{2} C V^2\) [1]. Correct calculation of \(0.0338\text{ J}\) or \(3.38 \times 10^{-2}\text{ J}\) [1]. (ii) Recalling \(\tau = R C\) [1]. Correct substitution leading to \(7.05\text{ s}\) [1]. (iii) Recalling potential decay formula \(V = V_0 e^{-t/RC}\) or energy decay formula \(E = E_0 e^{-2t/RC}\) [1]. Substitution of values [1]. Correct answer of \(8.18 \times 10^{-3}\text{ J}\) [1]. (iv) Finding energy lost by capacitor as \(E_0 - E\) [1]. Dividing energy lost by 5.0 s to find average power \(5.1 \times 10^{-3}\text{ W}\) [1].

(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) 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. This is due to energy conservation: if the induced current assisted the change in flux, the magnetic force would speed up the motion, creating electrical energy without doing work, violating the conservation of energy. (c)(i) The cross-sectional area of the coil is \(A = \pi r^2 = π \times (0.040)^2 = 5.027 \times 10^{-3}\text{ m}^2\). The magnetic flux linkage is \(\Phi_{\text{linkage}} = N B A = 150 \times 0.12 \times 5.027 \times 10^{-3} = 0.0905\text{ Wb-turns}\). (ii) According to Faraday's law, the induced e.m.f. is \(E = \Delta \Phi / \Delta t = (0.0905 - 0) / 0.25 = 0.362\text{ V}\).

(a) Magnitude of induced e.m.f. is proportional to [1] the rate of change of magnetic flux linkage [1]. (b) Lenz's law direction statement [1]. Explanation of what would happen if opposition did not occur (e.g. infinite current/unlimited energy) [1]. Reference to conservation of energy [1]. (c)(i) Correct calculation of coil area \(5.03 \times 10^{-3}\text{ m}^2\) [1]. Correct answer for flux linkage \(0.0905\text{ Wb-turns}\) (or Wb) [1]. (ii) Recalling \(E = \Delta(N\Phi) / \Delta t\) [1]. Substitution of values [1]. Correct answer of \(0.362\text{ V}\) [1].

(a) The radioactive decay constant is defined as the probability of decay per unit time of a nucleus. (b)(i) The half-life in seconds is \(t_{1/2} = 5730 \times 365.25 \times 24 \times 3600 = 1.808 \times 10^{11}\text{ s}\). The decay constant is \(\lambda = \ln 2 / t_{1/2} = 0.69315 / (1.808 \times 10^{11}) = 3.833 \times 10^{-12}\text{ s}^{-1}\). (ii) The radioactive decay law is \(A = A_0 e^{-\lambda t}\). This can be rearranged as \(t = -\ln(A/A_0) / \lambda\). Substituting the values: \(t = -\ln(0.18 / 0.23) / (3.833 \times 10^{-12}\text{ s}^{-1}) = 0.2451 / (3.833 \times 10^{-12}) = 6.395 \times 10^{10}\text{ s}\). In years, this is \(t = 6.395 \times 10^{10} / (365.25 \times 24 \times 3600) = 2030\text{ years}\).

(a) Probability of decay of a nucleus [1] per unit time [1]. (b)(i) Conversion of half-life from years to seconds [1]. Recalling \(\lambda = \ln 2 / t_{1/2}\) [1]. Showing calculation clearly leading to \(3.83 \times 10^{-12}\text{ s}^{-1}\) [1]. (ii) Recalling \(A = A_0 e^{-\lambda t}\) [1]. Substituting \(A = 0.18\text{ Bq}\) and \(A_0 = 0.23\text{ Bq}\) [1]. Calculating the ratio \(A/A_0 = 0.783\) and taking the natural log [1]. Finding time in seconds \(6.4 \times 10^{10}\text{ s}\) [1]. Correct conversion to obtain \(2030\text{ years}\) (allow 2000 to 2100) [1].

(a) Redshift is the fractional increase in the observed wavelength (or decrease in frequency) of electromagnetic radiation received from a source that is moving away from the observer.

(b)(i) First, calculate the change in wavelength:
\(\Delta \lambda = 671.2\text{ nm} - 656.3\text{ nm} = 14.9\text{ nm}\).
Using the Doppler redshift formula:
\(\frac{\Delta \lambda}{\lambda} = \frac{v}{c}\)
\(v = c \times \frac{\Delta \lambda}{\lambda} = (3.00 \times 10^8\text{ m s}^{-1}) \times \frac{14.9 \times 10^{-9}\text{ m}}{656.3 \times 10^{-9}\text{ m}} = 6.81 \times 10^6\text{ m s}^{-1}\).

(b)(ii) Using Hubble's law:
\(v = H_0 d\)
\(d = \frac{v}{H_0} = \frac{6.81 \times 10^6\text{ m s}^{-1}}{2.3 \times 10^{-18}\text{ s}^{-1}} = 2.96 \times 10^{24}\text{ m}\) (accepting \(3.0 \times 10^{24}\text{ m}\) if using \(6.8 \times 10^6\text{ m s}^{-1}\)).

(c) Using the inverse square law for radiant flux intensity:
\(F = \frac{L}{4\pi d^2}\)
\(L = 4\pi d^2 F = 4\pi \times (2.96 \times 10^{24}\text{ m})^2 \times (3.2 \times 10^{-16}\text{ W m}^{-2}) = 3.52 \times 10^{34}\text{ W}\) (or \(3.62 \times 10^{34}\text{ W}\) if using \(d = 3.0 \times 10^{24}\text{ m}\)).

(a)
- Increase in observed wavelength (or decrease in frequency) [1]
- Due to the source moving away from the observer [1]

(b)(i)
- Change in wavelength calculated as 14.9 nm [1]
- Correct substitution into redshift formula with lambda = 656.3 nm [1]
- Correct final speed of 6.81 x 10^6 m/s (accept 6.8 x 10^6 m/s) [1]

(b)(ii)
- State or use v = H_0 * d [1]
- Correct value of d as 2.96 x 10^24 m (accept 3.0 x 10^24 m) [1]

(c)
- State or use F = L / (4 * pi * d^2) [1]
- Correct substitute of values including squaring of distance [1]
- Correct value of L in range 3.5 x 10^34 W to 3.6 x 10^34 W [1]

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

(b)(i) The energy of a photon is:
\(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{240 \times 10^{-9}\text{ m}} = 8.29 \times 10^{-19}\text{ J}\)
To convert to eV, divide by the elementary charge:
\(E = \frac{8.29 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ J eV}^{-1}} = 5.18\text{ eV}\)
This rounds to \(5.2\text{ eV}\).

(b)(ii) According to Einstein's photoelectric equation:
\(E = \Phi + E_{k,\text{max}}\)
\(E_{k,\text{max}} = 5.18\text{ eV} - 3.4\text{ eV} = 1.78\text{ eV}\) (or \(1.8\text{ eV}\) if using \(5.2\text{ eV}\))
In Joules:
\(E_{k,\text{max}} = 1.78 \times 1.60 \times 10^{-19}\text{ J} = 2.85 \times 10^{-19}\text{ J}\) (or \(2.88 \times 10^{-19}\text{ J}\))

(b)(iii) The minimum de Broglie wavelength corresponds to maximum kinetic energy (and maximum momentum):
\(\lambda_{\text{min}} = \frac{h}{p_{\text{max}}}\)
Where \(p_{\text{max}} =
\sqrt{2 m_e E_{k,\text{max}}} = \sqrt{2 \times 9.11 \times 10^{-31}\text{ kg} \times 2.85 \times 10^{-19}\text{ J}} = 7.20 \times 10^{-25}\text{ kg m s}^{-1}\)
\(\lambda_{\text{min}} = \frac{6.63 \times 10^{-34}\text{ J s}}{7.20 \times 10^{-25}\text{ kg m s}^{-1}} = 9.2 \times 10^{-10}\text{ m}\).

(a)
- Minimum energy [1]
- Required to remove/free an electron from the surface of a metal [1]

(b)(i)
- State or use E = hc / lambda [1]
- Calculation of photon energy as 8.29 x 10^-19 J [1]
- Division by 1.60 x 10^-19 to show 5.18 eV (or 5.2 eV) [1]

(b)(ii)
- Subtract work function from photon energy to get 1.78 eV or 1.8 eV [1]
- Correct conversion to Joules to get 2.85 x 10^-19 J (accept 2.8 x 10^-19 J to 2.9 x 10^-19 J) [1]

(b)(iii)
- Recall of de Broglie relation lambda = h/p and p = sqrt(2 m E_k) [1]
- Correct substitution of mass of electron (9.11 x 10^-31 kg) and E_k from (b)(ii) [1]
- Correct final wavelength of 9.2 x 10^-10 m (accept 9.1 x 10^-10 m to 9.3 x 10^-10 m) [1]

### 1. Diagram and Experimental Setup
- Set up the lower circular coil flat on a wooden block placed on the pan of a top-pan balance. Connect this coil in series with a variable DC power supply, an ammeter, and a switch.
- Clamp the upper circular coil securely in a retort stand directly above the lower coil so that the two coils are coaxial (aligned vertically along the same axis). Connect the upper coil in series with its own independent variable DC power supply, ammeter, and switch (or connect both in a single series circuit to ensure identical currents, if desired).
- Ensure no magnetic materials (such as iron clamps or steel retort stands) are in the immediate vicinity of the coils to prevent field distortion.

### 2. Variables
- **Independent Variable:** Separation distance \(d\) between the centers of the coils.
- **Dependent Variable:** Magnetic force \(F\) between the coils.
- **Control Variables:** Current \(I_1\) and \(I_2\) in both coils, number of turns in each coil, and coaxial alignment.

### 3. Procedure and Measurements
- Measure the axial thickness of each coil using a vernier caliper. Let these be \(t_1\) and \(t_2\).
- Position the upper coil at a distance above the lower coil. Measure the vertical gap \(x\) between the top surface of the lower coil and the bottom surface of the upper coil using a vernier caliper or a vertical metre rule with a set square.
- Calculate the center-to-center distance \(d\) as:
\[d = x + \frac{t_1}{2} + \frac{t_2}{2}\]
- Switch on the balance and tare (zero) it with the power supplies turned off.
- Switch on the currents in both coils and adjust the variable power supplies (using rheostats or digital controls) until the current in each coil reaches a chosen fixed value (e.g., \(I_1 = I_2 = 2.0\text{ A}\)).
- Record the balance reading \(m\) (in grams or kilograms). The magnetic force \(F\) is calculated as:
\[F = mg\]
where \(g = 9.81\text{ m s}^{-2}\). (Depending on current directions, the force will be attractive or repulsive, showing a positive or negative mass change. Take the absolute value \(|m|\)).
- Turn off the current, change the height of the upper coil to vary \(d\), and repeat the process. Obtain at least 6 different values of \(d\) and measure the corresponding force \(F\) each time, ensuring the current is adjusted back to the constant value.

### 4. Analysis
- Linearise the equation:
\[\ln F = \ln k - n \ln d\]
- Plot a graph of \(\ln(F / \text{N})\) against \(\ln(d / \text{m})\).
- If the relationship is valid, the graph will yield a straight line.
- The gradient of the straight line equals \(-n\).
- The y-intercept of the line is equal to \(\ln k\), from which \(k\) can be calculated as:
\[k = e^{\text{intercept}}\]

### 5. Safety Precautions
- High currents passing through the coils will cause them to warm up. Switch off the current between readings to prevent overheating and potential burns. Do not touch the coils during operation.
- Use insulated wires for all connections to prevent electrical shock or short-circuits.

**Defining the Problem (max 2 marks):**
- **[1 mark]** Identify \(d\) as the independent variable and \(F\) (or mass reading change \(m\)) as the dependent variable.
- **[1 mark]** State that the current \(I\) in the coils is kept constant.

**Methods of Data Collection (max 4 marks):**
- **[1 mark]** Draw a clear, labelled diagram showing the upper coil clamped, the lower coil on a top-pan balance, and appropriate circuits containing power supplies and ammeters.
- **[1 mark]** Describe how to measure \(d\) (measuring the vertical gap with a vernier caliper/metre rule and adding half-thicknesses of both coils).
- **[1 mark]** Explain how the force \(F\) is determined from the change in balance reading (\(F = mg\)).
- **[1 mark]** Describe a method to ensure alignment of the coils along a common vertical axis (e.g., using a plumb line or aligning along a vertical plastic guide rod).

**Method of Analysis (max 3 marks):**
- **[1 mark]** State the logarithmic equation: \( \ln F = \ln k - n \ln d \) (or equivalent base-10 logarithm).
- **[1 mark]** Explain that a plot of \( \ln F \) against \( \ln d \) is a straight line if the relationship is valid.
- **[1 mark]** Identify that the gradient is equal to \(-n\) and the y-intercept is equal to \(\ln k\).

**Safety Considerations (max 1 mark):**
- **[1 mark]** Identify the risk of coils heating up due to high current and describe the precaution (e.g., switch off the current between measurements / use heat-resistant gloves).

**Additional Details (max 5 marks for any of the following):**
- Describe the use of a wooden/plastic spacer or tall stand to isolate the electronic balance from magnetic field interference.
- Explain how to check and keep the current constant using a rheostat or a regulated power supply.
- State that all magnetic materials (iron stands, steel rulers, etc.) must be kept far away from the apparatus to avoid field distortion.
- Tare the balance before each reading when the current is off.
- Repeat each measurement of \(d\) and take an average to minimise random errors.
- Reverse the current to verify that the magnitude of the force is identical when attractive vs repulsive.

### Part (a)
Taking natural logs of both sides of the equation:
\[\ln(V) = \ln(V_0) - \frac{T}{RC} = \ln(V_0) - \left(\frac{T}{C}\right) \frac{1}{R}\]
A plot of \(y = \ln(V / \text{V})\) against \(x = 1/R\) will be a straight line where:
- The y-intercept is equal to \( \ln(V_0) \), hence \( V_0 = e^{\text{intercept}} \).
- The gradient is equal to \( -\frac{T}{C} \), hence \( C = -\frac{T}{\text{gradient}} \).

### Part (b)
The uncertainty in \(y = \ln V\) is calculated using:
\[\Delta(\ln V) \approx \frac{\Delta V}{V}\]
or by the max-min method:
\[\Delta(\ln V) = \frac{\ln(V + \Delta V) - \ln(V - \Delta V)}{2}\]

Calculating the values:
1. For \(R = 100\text{ k}\Omega\):
- \(1/R = 10.00 \times 10^{-3}\text{ k}\Omega^{-1}\)
- \(\ln(6.1) = 1.808\)
- \(\Delta \ln(V) \approx \frac{0.2}{6.1} = 0.033\) (or \(\frac{\ln(6.3) - \ln(5.9)}{2} = 0.033\))

2. For \(R = 120\text{ k}\Omega\):
- \(1/R = 8.33 \times 10^{-3}\text{ k}\Omega^{-1}\)
- \(\ln(6.8) = 1.917\)
- \(\Delta \ln(V) \approx \frac{0.2}{6.8} = 0.029\)

3. For \(R = 150\text{ k}\Omega\):
- \(1/R = 6.67 \times 10^{-3}\text{ k}\Omega^{-1}\)
- \(\ln(7.6) = 2.028\)
- \(\Delta \ln(V) \approx \frac{0.2}{7.6} = 0.026\)

4. For \(R = 180\text{ k}\Omega\):
- \(1/R = 5.56 \times 10^{-3}\text{ k}\Omega^{-1}\)
- \(\ln(8.2) = 2.104\)
- \(\Delta \ln(V) \approx \frac{0.2}{8.2} = 0.024\)

5. For \(R = 220\text{ k}\Omega\):
- \(1/R = 4.55 \times 10^{-3}\text{ k}\Omega^{-1}\)
- \(\ln(8.8) = 2.175\)
- \(\Delta \ln(V) \approx \frac{0.2}{8.8} = 0.023\)

6. For \(R = 270\text{ k}\Omega\):
- \(1/R = 3.70 \times 10^{-3}\text{ k}\Omega^{-1}\)
- \(\ln(9.3) = 2.230\)
- \(\Delta \ln(V) \approx \frac{0.2}{9.3} = 0.022\)

| \(R / \text{k}\Omega\) | \(1/R / 10^{-3}\text{ k}\Omega^{-1}\) | \(V / \text{V}\) | \( \ln(V / \text{V}) \) | Uncertainty \( \Delta \ln(V / \text{V}) \) |
| :---: | :---: | :---: | :---: | :---: |
| 100 | 10.00 | 6.1 | 1.808 | 0.033 |
| 120 | 8.33 | 6.8 | 1.917 | 0.029 |
| 150 | 6.67 | 7.6 | 2.028 | 0.026 |
| 180 | 5.56 | 8.2 | 2.104 | 0.024 |
| 220 | 4.55 | 8.8 | 2.175 | 0.023 |
| 270 | 3.70 | 9.3 | 2.230 | 0.022 |

### Part (c)
- **(i)** Plot \(\ln(V / \text{V})\) on the vertical axis (from 1.7 to 2.3) and \(1/R \, / \, 10^{-3}\text{ k}\Omega^{-1}\) on the horizontal axis (from 3.0 to 11.0). Plot the points accurately and add vertical error bars ranging from \(\pm 0.02\) to \(\pm 0.03\).
- **(ii)** Draw the Line of Best Fit (LBF) through the centers of the data points. Draw the Worst Acceptable Line (WAL) which must pass through the top of the error bar of the first point and the bottom of the error bar of the last point (or vice-versa).
- **(iii) Gradient Calculation:**
Using points from the LBF:
- \(x_1 = 10.00 \times 10^{-3}\text{ k}\Omega^{-1}\), \(y_1 = 1.808\)
- \(x_2 = 3.70 \times 10^{-3}\text{ k}\Omega^{-1}\), \(y_2 = 2.230\)
\[m_{\text{best}} = \frac{1.808 - 2.230}{(10.00 - 3.70) \times 10^{-3}\text{ k}\Omega^{-1}} = \frac{-0.422}{6.30 \times 10^{-3}\text{ k}\Omega^{-1}} = -66.98\text{ k}\Omega \approx -67.0\text{ k}\Omega\]

Using WAL (passing through \(10.00, 1.808+0.033 = 1.841\) and \(3.70, 2.230-0.022 = 2.208\)):
\[m_{\text{worst}} = \frac{1.841 - 2.208}{6.30 \times 10^{-3}\text{ k}\Omega^{-1}} = -58.25\text{ k}\Omega\]

Absolute uncertainty in gradient:
\[\Delta m = |m_{\text{best}} - m_{\text{worst}}| = |-66.98 - (-58.25)| = 8.73\text{ k}\Omega \approx 9\text{ k}\Omega\]
So, **gradient** = \(-67 \pm 9\text{ k}\Omega\) (or \(-6.7 \pm 0.9 \times 10^4\text{ }\Omega\)).

- **(iv) y-intercept Calculation:**
Using LBF:
\[c_{\text{best}} = y - m_{\text{best}} x = 1.808 - (-66.98 \times 10^{-3} \times 10) = 1.808 + 0.6698 = 2.478\]
Using WAL:
\[c_{\text{worst}} = 1.841 - (-58.25 \times 10^{-3} \times 10) = 1.841 + 0.5825 = 2.424\]

Absolute uncertainty in y-intercept:
\[\Delta c = |c_{\text{best}} - c_{\text{worst}}| = |2.478 - 2.424| = 0.054 \approx 0.05\]
So, **y-intercept** = \(2.48 \pm 0.05\).

### Part (d)
- **(i) Calculation of \(V_0\):**
\[V_{0} = e^{c_{\text{best}}} = e^{2.478} = 11.92\text{ V} \approx 11.9\text{ V}\]
Uncertainty in \(V_0\):
\[V_{0,\text{worst}} = e^{c_{\text{worst}}} = e^{2.424} = 11.29\text{ V}\]
\[\Delta V_0 = |11.92 - 11.29| = 0.63\text{ V} \approx 0.6\text{ V}\]
So, \(V_0 = 11.9 \pm 0.6\text{ V}\).

- **(ii) Calculation of \(C\):**
Using \(C = -\frac{T}{m}\):
\[C = \frac{15.0\text{ s}}{6.698 \times 10^{4}\text{ }\Omega} = 2.24 \times 10^{-4}\text{ F} = 224\text{ }\mu\text{F}\]

Uncertainty in \(C\) is calculated from the fractional uncertainties of \(T\) and \(m\):
\[\frac{\Delta C}{C} = \frac{\Delta T}{T} + \frac{\Delta m}{|m|} = \frac{0.2}{15.0} + \frac{8.73}{66.98} = 0.0133 + 0.1303 = 0.1436\]
\[\Delta C = 2.24 \times 10^{-4}\text{ F} \times 0.1436 = 0.32 \times 10^{-4}\text{ F} = 32\text{ }\mu\text{F}\]
So, \(C = 2.2 \pm 0.3 \times 10^{-4}\text{ F}\) (or \(220 \pm 30\text{ }\mu\text{F}\)).

### Part (a)
- **[1 mark]** State linear form: \(\ln V = \ln V_0 - \left(\frac{T}{C}\right) \frac{1}{R}\) (or equivalent).
- **[1 mark]** Correctly identify that \(V_0 = e^{\text{intercept}}\) and \(C = -\frac{T}{\text{gradient}}\).

### Part (b)
- **[1 mark]** At least 5 values of \(1/R\) calculated correctly to 3 significant figures.
- **[1 mark]** All values of \(\ln(V/\text{V})\) calculated correctly to 3 decimal places.
- **[1 mark]** Uncertainties in \(\ln(V/\text{V})\) correctly calculated and tabulated (approx. \(0.02\) to \(0.03\)).

### Part (c)
- **[1 mark]** (i) Plots are correct and scale is sensible. Error bars are plotted correctly for all points.
- **[1 mark]** (ii) Best-fit line and clearly labelled Worst Acceptable Line drawn. WAL must pass through the top of the first error bar and the bottom of the last error bar.
- **[1 mark]** (iii) Gradient of best-fit line calculated with a clear working triangle occupying more than half the line length.
- **[1 mark]** uncertainty in gradient calculated correctly via \(|m_{\text{best}} - m_{\text{worst}}|\).
- **[1 mark]** (iv) y-intercept of best-fit line correctly read or calculated from \(y = mx + c\).
- **[1 mark]** uncertainty in y-intercept calculated correctly via \(|c_{\text{best}} - c_{\text{worst}}|\).

### Part (d)
- **[1 mark]** (i) Value of \(V_0\) calculated correctly with unit (\(\text{V}\)), and absolute uncertainty calculated correctly.
- **[1 mark]** (ii) Value of \(C\) calculated correctly with unit (\(\text{F}\) or \(\mu\text{F}\)) in the range \(2.1 \times 10^{-4}\) to \(2.4 \times 10^{-4}\text{ F}\).
- **[1 mark]** Correct calculation of absolute uncertainty in \(C\) taking into account the uncertainties in both \(T\) and the gradient.