MetadataPastPaper.sampleTitle

The percentage uncertainty in resistivity is calculated by summing the percentage uncertainties of the independent variables, with the percentage uncertainty of \(d\) being multiplied by 2 because it is squared. % uncertainty of \(R = (0.1 / 4.5) \times 100 \approx 2.22\%\). % uncertainty of \(d = (0.01 / 0.35) \times 100 \approx 2.86\%\). % uncertainty of \(L = (0.02 / 1.20) \times 100 \approx 1.67\%\). Therefore, total percentage uncertainty = \(2.22\% + 2 \times 2.86\% + 1.67\% = 9.61\%\), which rounds to \(9.6\%\).

1 mark for correct choice C. Award 1 mark for correct percentage uncertainty calculation: % uncertainty of \(R\) is 2.22%, % uncertainty of \(d\) is 2.86% (contribution is 5.71%), % uncertainty of \(L\) is 1.67%, leading to a total percentage uncertainty of 9.6%.

The equation of displacement in simple harmonic motion is \(x = A \sin(\omega t)\), where the angular frequency \(\omega = 2\pi f = 2\pi \times 2.5 = 5\pi\text{ rad s}^{-1}\). Substituting \(t = 0.15\text{ s}\), we get \(x = 4.0 \sin(5\pi \times 0.15) = 4.0 \sin(0.75\pi)\). Since \(0.75\pi\text{ rad}\) is equal to \(135^\circ\), \(\sin(0.75\pi) = \frac{1}{\sqrt{2}} \approx 0.707\). Thus, the displacement is \(x = 4.0 \times 0.707 \approx 2.8\text{ cm}\).

1 mark for the correct choice C. Method: Calculate angular frequency \(\omega = 5\pi\text{ rad s}^{-1}\). Apply SHM equation \(x = A \sin(\omega t)\) to find \(x \approx 2.8\text{ cm}\).

First, calculate the redshift \(z = \frac{\Delta \lambda}{\lambda_0} = \frac{672.1 - 656.3}{656.3} \approx 0.02407\). The recessional velocity \(v\) is \(v = z c = 0.02407 \times 3.00 \times 10^8\text{ m s}^{-1} \approx 7.22 \times 10^6\text{ m s}^{-1}\). Using Hubble's law \(v = H_0 d\), the distance is \(d = \frac{v}{H_0} = \frac{7.22 \times 10^6}{2.2 \times 10^{-18}} \approx 3.3 \times 10^{24}\text{ m}\).

1 mark for the correct choice B. Award 1 mark for correct calculations of redshift, velocity, and distance.

From the equation of state \(PV = nRT\), pressure is given by \(P = \frac{nRT}{V}\). When the mass is halved, the number of moles becomes \(0.5n\). The new pressure is \(P' = \frac{(0.5n) R (1.5T)}{0.75V} = \frac{0.75}{0.75} \frac{nRT}{V} = P\). Hence, the pressure remains unchanged.

1 mark for correct choice B. Award 1 mark for using the ideal gas equation and finding that the factors cancel out exactly.

Using \(s = ut + \frac{1}{2} at^2\) with upward as positive: \(s = -25\text{ m}\), \(u = +15\text{ m s}^{-1}\), and \(a = -9.81\text{ m s}^{-2}\). This gives \(-25 = 15t - 4.905t^2\), which rearranges to \(4.905t^2 - 15t - 25 = 0\). Solving the quadratic formula for the positive root yields \(t = \frac{15 + \sqrt{15^2 - 4 \times 4.905 \times (-25)}}{2 \times 4.905} \approx 4.3\text{ s}\).

1 mark for correct choice C. Award 1 mark for setting up the quadratic equation and correctly solving for the positive time.

After \(24.0\text{ hours}\), the number of elapsed half-lives is \(n = \frac{24.0}{8.0} = 3\). The fraction of nuclei remaining undecayed is \((1/2)^3 = 1/8\), so \(N_{\text{undecayed}} = \frac{1}{8} N_0\). The number of decayed nuclei is \(N_{\text{decayed}} = N_0 - \frac{1}{8} N_0 = \frac{7}{8} N_0\). The ratio is \(\frac{N_{\text{decayed}}}{N_{\text{undecayed}}} = \frac{7/8}{1/8} = 7.0\).

1 mark for correct choice C. Award 1 mark for calculating the correct fraction of undecayed and decayed nuclei and finding their ratio.

The fringe spacing formula is \(w = \frac{\lambda D}{a}\). Since \(a\) is constant, \(w \propto \lambda D\). The ratio of the new fringe width to the original is \(\frac{w'}{x} = \frac{\lambda' D'}{\lambda D} = \frac{480\text{ nm} \times 2D}{640\text{ nm} \times D} = \frac{960}{640} = 1.50\). Therefore, the new fringe width is \(1.50 x\).

1 mark for correct choice D. Award 1 mark for using the proportional relationship between fringe width, wavelength, and distance.

The total resistance in the circuit is \(R + r\). Thus, the current \(I\) is given by \(I = \frac{E}{R+r}\). The power \(P\) dissipated in the external resistor is \(P = I^2 R = \left(\frac{E}{R+r}\right)^2 R = \frac{E^2 R}{(R+r)^2}\).

1 mark for correct choice B. Award 1 mark for correctly applying Ohm's law and the electrical power formula.

The equation can be rearranged to find the units of \(k\): \(k = \frac{P x}{A \Delta \theta}\). The SI unit of power \(P\) is the watt (\text{W}), which is equivalent to joules per second (\text{J s}^{-1}). In terms of SI base units, \text{J} = \text{kg m}^2 \text{s}^{-2}, which means \text{W} = \text{kg m}^2 \text{s}^{-3}. The unit of thickness \(x\) is \text{m}, the unit of area \(A\) is \text{m}^2, and the unit of temperature difference \(\Delta \theta\) is \text{K}. Substituting these base units into the expression for \(k\) gives: \(\frac{(\text{kg m}^2 \text{s}^{-3}) \cdot \text{m}}{\text{m}^2 \cdot \text{K}} = \text{kg m s}^{-3} \text{K}^{-1}\).

Award 1 mark for the correct option B. (Method: 1 mark for identifying the correct SI base unit representation of power and rearranging the equation to solve for the units of k).

Let upwards be the positive direction. The initial velocity of the stone is \(u_i = +u\), and the final velocity when it hits the sea is \(u_f = -3u\). Since acceleration due to gravity is constant, the average velocity \(v_{\text{avg}}\) during the entire time of flight \(T\) is given by: \(v_{\text{avg}} = \frac{u_i + u_f}{2} = \frac{u + (-3u)}{2} = -u\). The vertical displacement \(s\) of the stone over the interval \(T\) is: \(s = v_{\text{avg}} T = -uT\). Since the displacement is \(s = -H\) (measured downwards from the launch point), we have \(-H = -uT\), which simplifies to \(H = uT\).

Award 1 mark for the correct option A. (Method: 1 mark for calculating the constant-acceleration average velocity and using it to express displacement).

First, calculate the redshift \(z\): \(z = \frac{\Delta \lambda}{\lambda_0} = \frac{680 - 656}{656} = \frac{24}{656} \approx 0.0366\). The recession speed \(v\) of the galaxy is: \(v = zc = 0.0366 \times 3.00 \times 10^8\text{ m s}^{-1} \approx 1.10 \times 10^7\text{ m s}^{-1}\). According to Hubble's Law, \(v = H_0 d\), where \(d\) is the distance: \(d = \frac{v}{H_0} = \frac{1.10 \times 10^7}{2.4 \times 10^{-18}}\text{ m} \approx 4.6 \times 10^{24}\text{ m}\).

Award 1 mark for the correct option B. (Method: 1 mark for calculating redshift and using Hubble's Law to find the correct order of magnitude for distance).

From the kinetic theory of gases, the pressure is given by \(p = \frac{1}{3} \frac{N m \langle c^2 \rangle}{V}\), which can be written as \(pV = \frac{1}{3} N m \langle c^2 \rangle\). Since the number of molecules \(N\) and the mass \(m\) of each molecule are constant, \(\langle c^2 \rangle \propto pV\). Initially, the product of pressure and volume is \(pV\). In the final state, the product is \(\left(\frac{1}{2}p\right) \times (3V) = 1.5 pV\). Therefore, the new mean-square speed is \(1.5 \langle c^2 \rangle\).

Award 1 mark for the correct option C. (Method: 1 mark for using the kinetic theory of gases equation to determine the proportional relationship between mean-square speed and the product of pressure and volume).

The displacement of a particle undergoing simple harmonic motion is given by \(x = A \sin(\omega t)\), where the angular frequency is \(\omega = 2\pi f\). The maximum speed of the particle is \(v_{\text{max}} = \omega A = 2\pi f A\). The maximum kinetic energy \(E_{\text{k, max}}\) is: \(E_{\text{k, max}} = \frac{1}{2} m v_{\text{max}}^2 = \frac{1}{2} m (2\pi f A)^2 = \frac{1}{2} m (4\pi^2 f^2 A^2) = 2 \pi^2 m f^2 A^2\).

Award 1 mark for the correct option B. (Method: 1 mark for identifying the correct expression for maximum velocity in simple harmonic motion and substituting it into the kinetic energy equation).

The initial charge stored is \(Q_0 = C V\). The initial stored energy is \(E = \frac{Q_0^2}{2C}\). When connected in parallel with an uncharged capacitor of capacitance \(3C\), the total capacitance becomes \(C_{\text{total}} = C + 3C = 4C\). By conservation of charge, the total charge remains \(Q_{\text{total}} = Q_0\). The new total energy stored is \(E_{\text{final}} = \frac{Q_{\text{total}}^2}{2 C_{\text{total}}} = \frac{Q_0^2}{2(4C)} = \frac{1}{4} \left(\frac{Q_0^2}{2C}\right) = \frac{1}{4}E\).

Award 1 mark for the correct option B. (Method: 1 mark for using the conservation of charge and the formula for energy in terms of charge and capacitance).

The two first-order diffracted beams lie symmetrically on opposite sides of the central zero-order maximum. Therefore, the angle of diffraction \(\alpha\) relative to the normal for each first-order beam is half of the total angle between them: \(\alpha = \frac{\theta}{2}\). Using the diffraction grating equation \(d \sin\alpha = n\lambda\) for first-order (\(n = 1\)), we get: \(d \sin\left(\frac{\theta}{2}\right) = \lambda \implies d = \frac{\lambda}{\sin(\theta/2)}\).

Award 1 mark for the correct option B. (Method: 1 mark for correctly identifying that the diffraction angle is half of the total angle between the two first-order maxima, and substituting this into the grating equation).

The gravitational potential energy of the satellite is \(E_p = -\frac{GMm}{r}\). For a circular orbit, the centripetal force is provided by the gravitational force: \(\frac{m v^2}{r} = \frac{GMm}{r^2}\), which gives \(m v^2 = \frac{GMm}{r}\). The kinetic energy of the satellite is \(E_k = \frac{1}{2} m v^2 = \frac{GMm}{2r}\). The total mechanical energy \(E\) of the satellite is the sum of its kinetic and potential energies: \(E = E_k + E_p = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}\).

Award 1 mark for the correct option A. (Method: 1 mark for linking centripetal force to gravitational force to find the kinetic energy expression, and summing it with the negative potential energy).

We determine the SI base units of each quantity in the equation:

1. \(n\) is number density (number per unit volume): \([n] = \text{m}^{-3}\)
2. \(e\) is elementary charge: since \(I = Q/t\), we have \([e] = \text{A s}\)
3. \(\tau\) is time: \([\tau] = \text{s}\)
4. \(m\) is mass: \([m] = \text{kg}\)

Substituting these into the formula for \(\sigma\):

$$\sigma = \frac{\text{m}^{-3} \cdot (\text{A s})^2 \cdot \text{s}}{\text{kg}} = \frac{\text{m}^{-3} \cdot \text{A}^2 \text{s}^3}{\text{kg}} = \text{A}^2 \text{s}^3 \text{kg}^{-1} \text{m}^{-3}$$

This matches option A.

1 mark for substituting the correct SI base units for each quantity in the formula and correctly simplifying the powers.

Using the equation of motion \(s = ut + \frac{1}{2}at^2\) with \(u = 0\) and \(a = g\):

For the first distance \(d\) in time \(t_1\):
$$d = \frac{1}{2}g t_1^2 \implies t_1 = \sqrt{\frac{2d}{g}}$$

For the total distance \(2d\) in total time \(t_1 + t_2\):
$$2d = \frac{1}{2}g (t_1 + t_2)^2 \implies t_1 + t_2 = \sqrt{\frac{4d}{g}} = \sqrt{2} \sqrt{\frac{2d}{g}} = \sqrt{2} t_1$$

Subtracting \(t_1\) from both sides:
$$t_2 = \sqrt{2}t_1 - t_1 = (\sqrt{2} - 1)t_1$$

Therefore, the ratio is:
$$\frac{t_2}{t_1} = \sqrt{2} - 1$$

This matches option A.

1 mark for setting up the equations of motion for distance d and 2d, and correctly solving for the ratio t_2 / t_1.

Let the original direction of motion of the spacecraft be positive.
Initial momentum of the system is:
$$p_i = M v$$

Let \(v_A\) and \(v_B\) be the velocities of fragment A and fragment B in the original frame of reference.
Since fragment A is ejected backwards relative to fragment B with speed \(2.0v\):
$$v_B - v_A = 2.0v \implies v_A = v_B - 2.0v$$

By conservation of linear momentum:
$$M v = 0.20M v_A + 0.80M v_B$$

Dividing both sides by \(M\):
$$v = 0.20 (v_B - 2.0v) + 0.80 v_B$$
$$v = 0.20 v_B - 0.40v + 0.80 v_B$$
$$v = 1.0 v_B - 0.40v$$
$$1.40v = v_B$$

Thus, the speed of fragment B in the original frame is \(1.4v\).

1 mark for formulating the relative velocity equation and momentum conservation equation, then solving for v_B.

Let the left end of the rod be \(L\) and the right end be \(R\). The weight of the uniform rod (\(60\text{ N}\)) acts at its midpoint, \(0.60\text{ m}\) from either end. Taking moments about the left end \(L\):

$$\text{Sum of counter-clockwise moments} = \text{Sum of clockwise moments}$$
$$T_R \times 1.2 = W \times 0.30 + W_{\text{rod}} \times 0.60$$

Since \(T_R = 40\text{ N}\) and \(W_{\text{rod}} = 60\text{ N}\):
$$40 \times 1.2 = 0.30 W + 60 \times 0.60$$
$$48 = 0.30 W + 36$$
$$12 = 0.30 W$$
$$W = \frac{12}{0.30} = 40\text{ N}$$

This matches option B.

1 mark for setting up a correct moment equation about a pivot and solving for the unknown weight W.

First, calculate the grating spacing \(d\):
$$d = \frac{1}{500 \times 10^3\text{ m}^{-1}} = 2.0 \times 10^{-6}\text{ m} = 2000\text{ nm}$$

The diffraction grating equation is:
$$d \sin\theta = n \lambda$$

For a maximum to be observable on a screen, the angle \(\theta\) must satisfy \(\sin\theta \le 1\). Therefore:
$$n \le \frac{d}{\lambda} = \frac{2000\text{ nm}}{600\text{ nm}} \approx 3.33$$

Since \(n\) must be an integer, the highest order observable is \(n = 3\).

The total number of maxima includes the central maximum (\(n = 0\)) and the orders on both sides (\(n = \pm 1, \pm 2, \pm 3\)):
$$\text{Total number of maxima} = 2n + 1 = 2(3) + 1 = 7$$

This matches option D.

1 mark for calculating the grating spacing, finding the maximum integer order, and doubling it then adding one to find the total number of maxima.

The resistance of a wire is given by \(R = \frac{\rho L}{A}\).

Let the original volume of the wire be \(V = A L\). Since the volume remains constant when the wire is stretched, the new cross-sectional area \(A'\) is:
$$A' \times 1.25L = A L \implies A' = \frac{A}{1.25} = 0.80 A$$

The new resistance \(R'\) is:
$$R' = \frac{\rho (1.25L)}{0.80A} = \frac{1.25}{0.80} \left(\frac{\rho L}{A}\right) = 1.5625 R \approx 1.56 R$$

This matches option C.

1 mark for finding the correct new cross-sectional area from volume conservation and using it to calculate the correct new resistance.

According to the ideal gas law \(pV = nRT\), at a constant volume \(V\) and constant amount of gas \(n\), the pressure is directly proportional to its absolute temperature \(T\) (in kelvins):
$$p \propto T$$

First, convert the initial temperature to kelvins:
$$T_1 = 27 + 273 = 300\text{ K}$$

Since the pressure is tripled, the absolute temperature is also tripled:
$$T_2 = 3 T_1 = 3 \times 300\text{ K} = 900\text{ K}$$

Convert this absolute temperature back to degrees Celsius:
$$\theta_2 = 900 - 273 = 627^\circ\text{C}$$

This matches option C.

1 mark for converting Celsius to kelvin, using the linear proportionality of pressure and absolute temperature, and correctly converting back to Celsius.

The displacement \(x\) of an object in simple harmonic motion starting from maximum displacement \(x = A\) at \(t = 0\) is modeled by:
$$x = A \cos(\omega t)$$
where \(\omega = \frac{2\pi}{T}\).

We need to find the minimum time \(t\) when the displacement is half the amplitude (\(x = A/2\)):
$$\frac{A}{2} = A \cos(\omega t) \implies \cos(\omega t) = \frac{1}{2}$$

The smallest positive angle that satisfies this is:
$$\omega t = \frac{\pi}{3}$$

Substituting \(\omega = \frac{2\pi}{T}\):
$$\frac{2\pi}{T} t = \frac{\pi}{3} \implies t = \frac{T}{6}$$

This matches option C.

1 mark for using the correct cosine equation of simple harmonic motion, substituting the displacement condition, and solving for time t in terms of T.

The electrical resistivity \(\rho\) of a cylindrical wire of resistance \(R\), diameter \(d\), and length \(L\) is given by the formula:

\(\rho = \frac{R A}{L} = \frac{\pi R d^2}{4 L}\)

The fractional uncertainty in \(\rho\) is given by adding the fractional uncertainties of the independent quantities (with the power of 2 for \(d\) acting as a multiplier):

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

First, calculate the individual percentage uncertainties:

\(\frac{\Delta R}{R} \times 100\% = \frac{0.5}{25.0} \times 100\% = 2.0\%\)

\(\frac{\Delta d}{d} \times 100\% = \frac{0.01}{0.40} \times 100\% = 2.5\%\)

\(\frac{\Delta L}{L} \times 100\% = \frac{0.03}{1.50} \times 100\% = 2.0\%\)

Now, sum these percentage uncertainties, doubling the contribution from the diameter:

\(\text{Percentage uncertainty in } \rho = 2.0\% + 2(2.5\%) + 2.0\% = 9.0\%\)

1 mark for the correct calculation of the percentage uncertainty of resistivity:
- Identifies the formula for resistivity and the contribution of uncertainties: \(\Delta \rho / \rho = \Delta R / R + 2 \Delta d / d + \Delta L / L\) (Method mark)
- Computes the correct percentage uncertainty of 9.0% (Accuracy mark)

Let the downward direction be positive, and let the edge of the cliff be the origin.

For the upward throw, the displacement when hitting the ground is \(h\), the initial velocity is \(-u\), and the time of flight is \(T\):

\(h = -u T + \frac{1}{2} g T^2 \implies u T = \frac{1}{2} g T^2 - h \implies u = \frac{1}{2} g T - \frac{h}{T}\) (Equation 1)

For the downward throw, the displacement is \(h\), the initial velocity is \(u\), and the time of flight is \(t\):

\(h = u t + \frac{1}{2} g t^2\) (Equation 2)

Substitute Equation 1 into Equation 2 to eliminate \(u\):

\(h = \left( \frac{1}{2} g T - \frac{h}{T} \right) t + \frac{1}{2} g t^2\)

\(h = \frac{1}{2} g T t - h \frac{t}{T} + \frac{1}{2} g t^2\)

Rearranging to group terms in \(h\):

\(h \left( 1 + \frac{t}{T} \right) = \frac{1}{2} g t ( T + t )\)

\(h \left( \frac{T + t}{T} \right) = \frac{1}{2} g t ( T + t )\)

Since \(T + t \neq 0\), we can divide both sides by \(T + t\):

\(\frac{h}{T} = \frac{1}{2} g t \implies h = \frac{1}{2} g T t\)

1 mark for setting up the equations of motion for both cases and correctly solving for \(h\) by eliminating the initial speed \(u\).

The gravitational potential energy \(U\) of a mass \(m\) in a gravitational field of a planet of mass \(M\) at distance \(r\) from its center is given by:

\(U = -\frac{G M m}{r}\)

Initially, at radius \(r_1 = 3R\):

\(U_1 = -\frac{G M m}{3R}\)

Finally, at radius \(r_2 = 4R\):

\(U_2 = -\frac{G M m}{4R}\)

The increase in gravitational potential energy \(\Delta U\) is:

\(\Delta U = U_2 - U_1 = -\frac{G M m}{4R} - \left( -\frac{G M m}{3R} \right) = G M m \left( \frac{1}{3R} - \frac{1}{4R} \right) = \frac{G M m}{12R}\)

Note: If the question asked for the increase in total energy (which includes both kinetic and potential energy), the answer would be \(\frac{G M m}{24R}\), which corresponds to option B.

1 mark for using the correct expression for gravitational potential energy to find the difference between final and initial energy states.

Let \(E_k\) be the kinetic energy, \(E_p\) be the potential energy, and \(E\) be the total mechanical energy of the simple harmonic oscillator.

We are given:

\(E_k = 3 E_p\)

The total energy is the sum of the kinetic and potential energies:

\(E = E_k + E_p = 3 E_p + E_p = 4 E_p\)

The total energy \(E\) is given by:

\(E = \frac{1}{2} m \omega^2 x_0^2\)

and the potential energy at displacement \(x\) is given by:

\(E_p = \frac{1}{2} m \omega^2 x^2\)

Substituting these expressions into the total energy equation:

\(\frac{1}{2} m \omega^2 x_0^2 = 4 \left( \frac{1}{2} m \omega^2 x^2 \right)\)

\(x_0^2 = 4 x^2\)

Taking the square root of both sides:

\(x = \pm \frac{x_0}{2}\)

1 mark for using the energy relationship of SHM to express displacement in terms of amplitude.

Using the ideal gas equation in terms of the number of molecules:

\(P V = N k T \implies P = \frac{N k T}{V}\)

For the second container, let the pressure be \(P_2\). The parameters are volume \(2V\), molecules \(3N\), and temperature \(2T\):

\(P_2 (2V) = (3N) k (2T)\)

\(2 P_2 V = 6 N k T\)

\(P_2 = 3 \left( \frac{N k T}{V} \right)\)

Since \(P = \frac{N k T}{V}\), we get:

\(P_2 = 3 P\)

1 mark for applying the equation of state for both gases to establish the relationship between \(P_2\) and \(P\).

For a diffraction grating, the maxima occur at angles \(\theta\) given by the grating equation:

\(d \sin \theta = m \lambda\)

For a maximum of \(\lambda_1\) at order \(m_1\) to coincide with a maximum of \(\lambda_2\) at order \(m_2\), they must share the same diffraction angle \(\theta\). Therefore:

\(m_1 \lambda_1 = m_2 \lambda_2\)

Rearranging the equation to find the ratio of the orders:

\(\frac{m_1}{m_2} = \frac{\lambda_2}{\lambda_1} = \frac{450\text{ nm}}{600\text{ nm}} = \frac{3}{4}\)

Since \(m_1\) and \(m_2\) must be integers, the smallest integer values for this ratio are \(m_1 = 3\) and \(m_2 = 4\).

Thus, the lowest order of the \(\lambda_1\) maximum that coincides with a maximum of \(\lambda_2\) is the third order (\(m_1 = 3\)).

1 mark for using the grating equation to show that \(m_1 \lambda_1 = m_2 \lambda_2\) and finding the lowest integer value of \(m_1\).

Let \(\rho_d\) be the density of the material, which is the same for both wires. The mass \(m\) of a wire is given by:

\(m = \text{density} \times \text{volume} = \rho_d A L\)

Since \(m_X = m_Y\) and the material is the same:

\(A_X L_X = A_Y L_Y\)

We are given that \(L_Y = 2 L_X\), so:

\(A_X L_X = A_Y (2 L_X) \implies A_Y = \frac{1}{2} A_X\)

The resistance \(R\) of a wire of resistivity \(\rho\) is:

\(R = \rho \frac{L}{A}\)

Therefore, the ratio of the resistances is:

\(\frac{R_Y}{R_X} = \frac{\rho \frac{L_Y}{A_Y}}{\rho \frac{L_X}{A_X}} = \left( \frac{L_Y}{L_X} \right) \times \left( \frac{A_X}{A_Y} \right)\)

Substitute the known ratios:

\(\frac{R_Y}{R_X} = (2) \times (2) = 4\)

1 mark for relating the mass to cross-sectional area and length, then correctly combining this with the formula for resistance to find the ratio.

First, calculate the redshift \(z\):

\(z = \frac{\Delta \lambda}{\lambda_0} = \frac{672.7\text{ nm} - 656.3\text{ nm}}{656.3\text{ nm}} = \frac{16.4}{656.3} \approx 0.02499\)

The recession speed \(v\) of the galaxy is:

\(v = z c = 0.02499 \times 3.00 \times 10^8\text{ m s}^{-1} \approx 7.497 \times 10^6\text{ m s}^{-1}\)

According to Hubble's law:

\(v = H_0 d\)

where \(d\) is the distance to the galaxy. Thus:

\(d = \frac{v}{H_0} = \frac{7.497 \times 10^6\text{ m s}^{-1}}{2.3 \times 10^{-18}\text{ s}^{-1}} \approx 3.26 \times 10^{24}\text{ m} \approx 3.3 \times 10^{24}\text{ m}\)

1 mark for calculating the redshift \(z\), using it to find the velocity \(v\), and applying Hubble's law to find the correct distance.

The percentage uncertainty in \( h \) is:
\( \frac{\Delta h}{h} \times 100\% = \frac{0.02}{1.50} \times 100\% \approx 1.33\% \)

The percentage uncertainty in \( t \) is:
\( \frac{\Delta t}{t} \times 100\% = \frac{0.01}{0.55} \times 100\% \approx 1.82\% \)

Since \( g = \frac{2h}{t^2} \), the percentage uncertainty in \( g \) is:
\( \frac{\Delta g}{g} \times 100\% = \frac{\Delta h}{h} \times 100\% + 2 \times \left( \frac{\Delta t}{t} \times 100\% \right) \)
\( \frac{\Delta g}{g} \times 100\% = 1.33\% + 2 \times 1.82\% = 4.97\% \approx 5.0\% \)

1 mark for the correct calculation of percentage uncertainty to 2 significant figures (5.0%).

We use the equations of motion. Let the vertically upward direction be positive.
Initial velocity \( u = +15 \text{ m s}^{-1} \)
Acceleration \( a = -9.81 \text{ m s}^{-2} \)
Time \( t = 4.2 \text{ s} \)

Using \( s = ut + \frac{1}{2}at^2 \):
\( s = (15 \times 4.2) + \frac{1}{2}(-9.81)(4.2)^2 \)
\( s = 63.0 - 4.905 \times 17.64 \)
\( s = 63.0 - 86.5 = -23.5 \text{ m} \)

The displacement \( s \) is \( -23.5 \text{ m} \), which means the final position is \( 23.5 \text{ m} \) below the starting point. Thus, the height of the cliff \( H \approx 24 \text{ m} \).

1 mark for correctly applying the kinematic equation and calculating the displacement to 2 significant figures.

The mass of water hitting the wall per second (mass flow rate) is given by:
\( \frac{dm}{dt} = \rho A v \)
Where:
\( \rho = 1000 \text{ kg m}^{-3} \)
\( A = 2.0 \times 10^{-4} \text{ m}^2 \)
\( v = 12 \text{ m s}^{-1} \)

\( \frac{dm}{dt} = 1000 \times (2.0 \times 10^{-4}) \times 12 = 2.4 \text{ kg s}^{-1} \)

The force \( F \) is equal to the rate of change of momentum of the water:
\( F = \frac{dp}{dt} = \frac{dm}{dt} \times \Delta v \)
Since the water comes to rest, \( \Delta v = v = 12 \text{ m s}^{-1} \).

\( F = 2.4 \times 12 = 28.8 \text{ N} \approx 29 \text{ N} \).

1 mark for calculating the mass flow rate and multiplying by speed to find the force of 29 N.

Let the distance of the centre of gravity from the left-hand end be \( x \).
Taking moments about the left-hand end:
Clockwise moments = Counter-clockwise moments
\( W \times x = T_Y \times L \)
Where:
\( W = 150 \text{ N} \)
\( T_Y = 90 \text{ N} \)
\( L = 4.0 \text{ m} \)

\( 150 \times x = 90 \times 4.0 \)
\( 150x = 360 \)
\( x = \frac{360}{150} = 2.4 \text{ m} \)

1 mark for using the principle of moments to find the correct distance of 2.4 m.

The diffraction grating equation is:
\( d \sin\theta = n\lambda \)

Given:
\( \lambda = 600 \times 10^{-9} \text{ m} \)
\( n = 3 \)
\( \theta = 35^\circ \)

\( d = \frac{n\lambda}{\sin\theta} = \frac{3 \times 600 \times 10^{-9}}{\sin(35^\circ)} \approx 3.14 \times 10^{-6} \text{ m} \)

The number of lines per millimetre \( N \) is:
\( N = \frac{1 \times 10^{-3} \text{ m}}{d} = \frac{10^{-3}}{3.14 \times 10^{-6}} \approx 319 \text{ lines mm}^{-1} \approx 320 \text{ lines mm}^{-1} \).

1 mark for calculating the grating spacing \( d \) and converting it to the number of lines per millimetre (320).

The resistance of a wire of resistivity \( \rho \), length \( l \), and cross-sectional area \( A \) is:
\( R = \frac{\rho l}{A} = \frac{\rho l}{\pi (D/2)^2} = \frac{4\rho l}{\pi D^2} \)

For wire X:
\( R_X = \frac{4\rho L}{\pi d^2} = R \)

For wire Y:
\( R_Y = \frac{4\rho (2L)}{\pi (3d)^2} = \frac{8\rho L}{9\pi d^2} = \frac{2}{9} \left( \frac{4\rho L}{\pi d^2} \right) = \frac{2}{9}R \)

1 mark for expressing resistance in terms of length and diameter and determining the ratio of resistances.

Using the ideal gas equation \( pV = nRT \). Since volume \( V \) and the amount of gas \( n \) are constant:
\( \frac{p_1}{T_1} = \frac{p_2}{T_2} \)

Convert temperatures to Kelvin:
\( T_1 = 27 + 273 = 300 \text{ K} \)
\( T_2 = 177 + 273 = 450 \text{ K} \)

Now calculate \( p_2 \):
\( p_2 = p_1 \times \frac{T_2}{T_1} = 1.2 \times 10^5 \times \frac{450}{300} = 1.8 \times 10^5 \text{ Pa} \)

1 mark for converting temperatures to Kelvin and correctly applying the pressure law to find 1.8 x 10^5 Pa.

First, find the redshift \( z \) of the galaxy:
\( z = \frac{\Delta \lambda}{\lambda} = \frac{663.2 - 656.3}{656.3} = \frac{6.9}{656.3} \approx 0.01051 \)

The recessional velocity \( v \) is:
\( v = zc = 0.01051 \times 3.00 \times 10^8 \approx 3.15 \times 10^6 \text{ m s}^{-1} \)

According to Hubble's law:
\( v = H_0 d \)

Therefore, the distance \( d \) is:
\( d = \frac{v}{H_0} = \frac{3.15 \times 10^6}{2.2 \times 10^{-18}} \approx 1.43 \times 10^{24} \text{ m} \approx 1.4 \times 10^{24} \text{ m} \)

1 mark for calculating the redshift, then using it to calculate the velocity, and finally dividing by Hubble's constant to find 1.4 x 10^24 m.

(a)
1. Start by rearranging the formula for dynamic viscosity \( \eta \):
\( \eta = \frac{F_D}{6 \pi r v} \)

2. Determine the SI base units of each quantity in the equation:
- Force \( F_D \) has units of newtons (\( \text{N} \)), which in SI base units is \( \text{kg}\ \text{m}\ \text{s}^{-2} \) (since \( F = ma \)).
- \( 6\pi \) is a dimensionless constant and has no units.
- Radius \( r \) has the SI base unit of meter (\( \text{m} \)).
- Velocity \( v \) has the SI base unit of meter per second (\( \text{m}\ \text{s}^{-1} \)).

3. Substitute the base units into the expression for \( \eta \):
\( \text{Units of } \eta = \frac{\text{kg}\ \text{m}\ \text{s}^{-2}}{\text{m} \times \text{m}\ \text{s}^{-1}} \)

4. Simplify the expression:
\( \text{Units of } \eta = \frac{\text{kg}\ \text{m}\ \text{s}^{-2}}{\text{m}^2\ \text{s}^{-1}} = \text{kg}\ \text{m}^{1 - 2}\ \text{s}^{-2 - (-1)} = \text{kg}\ \text{m}^{-1}\ \text{s}^{-1} \).
This is the required result.

(b)
1. Estimate a reasonable mass \( m \) for a domestic cat. A standard value is around \( 4.0\ \text{kg} \) (accept any value from \( 3.0\ \text{kg} \) to \( 6.0\ \text{kg} \)).
2. Estimate a reasonable maximum running speed \( v \) for a domestic cat. A standard value is around \( 10\ \text{m}\ \text{s}^{-1} \) (accept any value from \( 8.0\ \text{m}\ \text{s}^{-1} \) to \( 12\ \text{m}\ \text{s}^{-1} \)).
3. Calculate the estimated kinetic energy using the formula:
\( E_k = \frac{1}{2} m v^2 \)
Using our values:
\( E_k = \frac{1}{2} \times 4.0\ \text{kg} \times (10\ \text{m}\ \text{s}^{-1})^2 = 200\ \text{J} \).
If another reasonable set of estimates is used, such as \( m = 5.0\ \text{kg} \) and \( v = 8.0\ \text{m}\ \text{s}^{-1} \):
\( E_k = \frac{1}{2} \times 5.0 \times 64 = 160\ \text{J} \).

(a)
- C1: Identifies base unit of force as \( \text{kg}\ \text{m}\ \text{s}^{-2} \) (either by stating \( F = ma \) or showing derivation).
- C1: Identifies base units of \( r \) as \( \text{m} \) and \( v \) as \( \text{m}\ \text{s}^{-1} \).
- C1: Shows correct algebraic setup of units: \( \frac{\text{kg}\ \text{m}\ \text{s}^{-2}}{\text{m} \times \text{m}\ \text{s}^{-1}} \).
- A1: Simplifies units correctly to arrive at the final given base units: \( \text{kg}\ \text{m}^{-1}\ \text{s}^{-1} \).

(b)
- B1: States a reasonable mass estimate for a domestic cat (accept \( 3.0 \text{ kg} \le m \le 6.0 \text{ kg} \)).
- B1: States a reasonable speed estimate for a running cat (accept \( 5.0 \text{ m s}^{-1} \le v \le 15 \text{ m s}^{-1} \)).
- C1: Recall and substitute into \( E_k = \frac{1}{2} m v^2 \).
- A1.5: Calculates a mathematically consistent kinetic energy value based on stated estimates, expressed with appropriate unit (\( \text{J} \) or \( \text{kg}\ \text{m}^2\ \text{s}^{-2} \)).

(a)
When the ball is moving upwards, the gravitational force (weight) acts vertically downwards.
The air resistance (drag force) always opposes motion, so it also acts vertically downwards on its way up.
Since both forces act in the same downward direction, the net force acting on the ball is \( F_{\text{net}} = W + F_R \), which is greater than the weight \( W \) alone.
By Newton's second law, \( a = \frac{F_{\text{net}}}{m} = g + \frac{F_R}{m} \), which is greater than \( g \).

(b)(i)
Using the equations of motion for uniform acceleration:
\( v^2 = u^2 + 2as \)
At maximum height, \( v = 0 \).
Taking upwards as positive:
\( u = 24.0\ \text{m}\ \text{s}^{-1} \)
\( a = -g = -9.81\ \text{m}\ \text{s}^{-2} \)
\( 0 = (24.0)^2 + 2(-9.81)s \)
\( 19.62 s = 576 \)
\( s = \frac{576}{19.62} \approx 29.36\ \text{m} \approx 29.4\ \text{m} \).

(b)(ii)
We can use the work-energy principle.
Initial kinetic energy \( E_k = \frac{1}{2} m u^2 \):
\( E_k = 0.5 \times 0.0580\ \text{kg} \times (24.0\ \text{m}\ \text{s}^{-1})^2 = 16.704\ \text{J} \).

Final gravitational potential energy at maximum height \( E_p = mgh \):
\( E_p = 0.0580\ \text{kg} \times 9.81\ \text{m}\ \text{s}^{-2} \times 22.1\ \text{m} = 12.575\ \text{J} \).

By energy conservation, the work done against the average resistive force \( F_R \) is the loss in mechanical energy:
\( W_{\text{resistive}} = E_k - E_p \)
\( W_{\text{resistive}} = 16.704\ \text{J} - 12.575\ \text{J} = 4.129\ \text{J} \).

Since \( W_{\text{resistive}} = F_R \times d \) (where \( d = 22.1\ \text{m} \) is the distance traveled):
\( F_R = \frac{4.129\ \text{J}}{22.1\ \text{m}} \approx 0.187\ \text{N} \).

(a)
- B1: States that both weight and air resistance act in the downward direction (during ascent).
- B1: States that the total downward net force is greater than weight alone.
- B0.5: Concludes that since \( a = F_{\text{net}}/m \), the acceleration (deceleration) must be greater than \( g \).

(b)(i)
- C1: Uses \( v^2 = u^2 + 2as \) with appropriate values (\( v=0 \), \( a = 9.81 \text{ m s}^{-2} \)).
- A1: Obtains \( 29.4\ \text{m} \) (or \( 29\ \text{m} \) if \( g = 9.8 \) is used).

(b)(ii)
- C1: Calculates initial kinetic energy \( E_k = 16.7\ \text{J} \) (accept substitution with mass in kg).
- C1: Calculates final gravitational potential energy \( E_p = 12.6\ \text{J} \).
- C1: Relates work done against resistance to loss of mechanical energy: \( W = E_k - E_p \).
- A1: Obtains \( 0.187\ \text{N} \) (accept range \( 0.185\text{--}0.190\ \text{N} \) due to rounding).

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

(b)(i)
Let the direction to the right be positive.
Initial velocity of glider A: \( u_A = +2.40\ \text{m}\ \text{s}^{-1} \)
Initial velocity of glider B: \( u_B = -1.20\ \text{m}\ \text{s}^{-1} \) (since it moves to the left)

According to conservation of momentum:
\( m_A u_A + m_B u_B = (m_A + m_B) v \)

Substitute the values:
\( (0.350 \times 2.40) + (0.150 \times (-1.20)) = (0.350 + 0.150) v \)
\( 0.840 - 0.180 = 0.500 v \)
\( 0.660 = 0.500 v \)
\( v = \frac{0.660}{0.500} = 1.32\ \text{m}\ \text{s}^{-1} \)

Since the velocity is positive, it is directed to the right.

(b)(ii)
Calculate the initial kinetic energy (\( E_{k,i} \)):
\( E_{k,i} = \frac{1}{2} m_A u_A^2 + \frac{1}{2} m_B u_B^2 \)
\( E_{k,i} = \frac{1}{2}(0.350)(2.40)^2 + \frac{1}{2}(0.150)(-1.20)^2 \)
\( E_{k,i} = 1.008 + 0.108 = 1.116\ \text{J} \)

Calculate the final kinetic energy (\( E_{k,f} \)):
\( E_{k,f} = \frac{1}{2}(m_A + m_B) v^2 \)
\( E_{k,f} = \frac{1}{2}(0.500)(1.32)^2 \)
\( E_{k,f} = 0.4356\ \text{J} \)

Calculate the loss in kinetic energy:
\( \text{Loss in } KE = E_{k,i} - E_{k,f} = 1.116 - 0.4356 = 0.6804\ \text{J} \approx 0.680\ \text{J} \)

Since the final kinetic energy is less than the initial kinetic energy, the collision is inelastic.

(a)
- B1: States that the total momentum of a system is constant (or conserved).
- B1: Clarifies this holds for an isolated system or in the absence of external resultant forces.

(b)(i)
- C1: Uses conservation of momentum equation: \( m_A u_A + m_B u_B = (m_A + m_B)v \).
- C1: Substitutes correct signs for velocities (e.g., \( -1.20 \text{ m s}^{-1} \)).
- A1: Obtains \( 1.32\ \text{m}\ \text{s}^{-1} \) (to the right).

(b)(ii)
- C1: Calculates initial KE correctly as \( 1.116\ \text{J} \) (or shows correct substitution).
- C1: Calculates final KE correctly as \( 0.436\ \text{J} \).
- A1.5: Calculates difference to find loss of KE as \( 0.680\ \text{J} \) (accept \( 0.68\ \text{J} \)) and explicitly states that since kinetic energy is lost / not conserved, the collision is inelastic.

(a)
The torque of a couple is defined as the product of one of the forces and the perpendicular distance between the lines of action of the two forces.

(b)(i)
The diagram must show:
1. Weight \( W = 180\ \text{N} \) acting vertically downwards from the exact midpoint (center of gravity) of the beam, which is at a distance of \( 1.2\ \text{m} \) from P.
2. Tension \( T \) acting from end Q, pointing upwards and to the left (towards the wall) at an angle of \( 35^\circ \) to the horizontal beam.
3. The reaction force \( R \) at the hinge P. Since the horizontal component of tension points to the left, \( R \) must point to the right. Since weight points down and tension has an upward component, \( R \) must also have a vertical component (likely upwards to share the load). Hence, \( R \) is directed upwards and to the right.

(b)(ii)
For the beam to be in rotational equilibrium, the sum of the moments about any pivot must be zero.
Taking moments about the hinge P:
- The clockwise moment is caused by the weight of the beam:
\( \text{Clockwise Moment} = W \times \frac{L}{2} = 180\ \text{N} \times 1.2\ \text{m} = 216\ \text{N}\ \text{m} \).

- The anticlockwise moment is caused by the vertical component of the tension \( T \):
\( T_{\text{vertical}} = T \sin(35^\circ) \).
This force acts at a distance of \( 2.4\ \text{m} \) from P.
\( \text{Anticlockwise Moment} = T \sin(35^\circ) \times 2.4\ \text{m} \).

Setting the moments equal:
\( T \sin(35^\circ) \times 2.4 = 216 \)
\( 1.3766 T = 216 \)
\( T = \frac{216}{2.4 \sin(35^\circ)} \)
\( T = \frac{90}{\sin(35^\circ)} \approx \frac{90}{0.57358} \approx 156.9\ \text{N} \approx 157\ \text{N} \).

(a)
- B1: Product of one of the forces (of the couple).
- B1: And the perpendicular distance between their lines of action.

(b)(i)
- B1: Weight represented by a vertical arrow pointing downwards from the center of the beam.
- B1: Tension represented by an arrow at end Q pointing upwards and towards the wall, making an angle of roughly 35 degrees with the beam.
- B0.5: Hinge force represented by an arrow at P pointing upwards and to the right.

(b)(ii)
- C1: States principle of moments or equates clockwise and anticlockwise moments.
- C1: Identifies perpendicular distance of weight is \( 1.2\ \text{m} \) (or writes \( 180 \times 1.2 \)).
- C1: Correct expression for moment of tension: \( T \sin(35^\circ) \times 2.4 \).
- A1: Correctly calculates \( T = 157\ \text{N} \) (accept range \( 156\text{--}158\ \text{N} \)).

(a)
Elastic deformation: the material returns to its original shape/dimensions when the deforming force (or stress) is removed.
Plastic deformation: the material does not return to its original shape/dimensions when the deforming force is removed, resulting in permanent deformation.

(b)(i)
The diameter of the wire is \( d = 0.68\ \text{mm} = 0.68 \times 10^{-3}\ \text{m} \).
The radius \( r \) is \( \frac{0.68 \times 10^{-3}}{2} = 0.34 \times 10^{-3}\ \text{m} \).

The cross-sectional area \( A \) of the wire is:
\( A = \pi r^2 = \pi \times (0.34 \times 10^{-3}\ \text{m})^2 \)
\( A = \pi \times 1.156 \times 10^{-7} \approx 3.631 \times 10^{-7}\ \text{m}^2 \).

(b)(ii)
From the definition of Young modulus \( E \):
\( E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{x / L} = \frac{F L}{A x} \)

where:
- \( F \) is the tension force in the wire. Here, \( F = mg = 4.50\ \text{kg} \times 9.81\ \text{m}\ \text{s}^{-2} = 44.145\ \text{N} \).
- \( L \) is the original length of the wire, \( 1.80\ \text{m} \).
- \( A \) is the cross-sectional area, \( 3.631 \times 10^{-7}\ \text{m}^2 \).
- \( E \) is the Young modulus, \( 2.0 \times 10^{11}\ \text{Pa} \).
- \( x \) is the extension.

Rearranging the formula for the extension \( x \):
\( x = \frac{F L}{A E} \)

Substitute the values:
\( x = \frac{44.145\ \text{N} \times 1.80\ \text{m}}{(3.631 \times 10^{-7}\ \text{m}^2) \times (2.0 \times 10^{11}\ \text{Pa})} \)
\( x = \frac{79.461}{7.262 \times 10^4} \approx 1.094 \times 10^{-3}\ \text{m} \approx 1.09\ \text{mm} \).

(a)
- B1: Defines elastic deformation as returning to original shape when load/force is removed.
- B1: Defines plastic deformation as permanent / not returning to original shape.

(b)(i)
- C1: Uses area formula \( A = \pi r^2 \) or \( A = \frac{\pi d^2}{4} \) with conversion from mm to m.
- A1: Obtains \( 3.63 \times 10^{-7}\ \text{m}^2 \) (accept \( 3.6 \times 10^{-7}\ \text{m}^2 \)).

(b)(ii)
- C1: Calculates the load force as \( F = mg = 44.1\ \text{N} \).
- C1: Recalls and rearranges Young modulus formula: \( x = \frac{FL}{AE} \).
- C1: Substitutes values correctly into formula.
- A1.5: Obtains \( 1.09 \times 10^{-3}\ \text{m} \) (accept \( 1.1 \times 10^{-3}\ \text{m} \) or \( 1.1\ \text{mm} \)).

(a)
Two conditions for a stable interference pattern are:
1. The sources must be coherent, meaning they have a constant phase difference (and therefore the same frequency/wavelength).
2. The waves should have approximately the same amplitude (to ensure complete or near-complete destructive interference, yielding high-contrast fringes).

(b)(i)
We use the double-slit interference formula:
\( x = \frac{\lambda D}{a} \)
where:
- \( \lambda = 632.8\ \text{nm} = 6.328 \times 10^{-7}\ \text{m} \)
- \( D = 1.50\ \text{m} \)
- \( a = 0.250\ \text{mm} = 2.50 \times 10^{-4}\ \text{m} \)

Substitute the values:
\( x = \frac{(6.328 \times 10^{-7}\ \text{m}) \times 1.50\ \text{m}}{2.50 \times 10^{-4}\ \text{m}} \)
\( x = 3.7968 \times 10^{-3}\ \text{m} \approx 3.80\ \text{mm} \).

(b)(ii)
The distance between the central bright fringe (\( n = 0 \)) and the fifth bright fringe (\( n = 5 \)) is given as \( 1.62\ \text{cm} = 1.62 \times 10^{-2}\ \text{m} \).
This distance represents 5 fringe widths:
\( 5 x_{\text{new}} = 1.62 \times 10^{-2}\ \text{m} \)
\( x_{\text{new}} = \frac{1.62 \times 10^{-2}}{5} = 3.24 \times 10^{-3}\ \text{m} \).

Using the formula again to find \( \lambda_{\text{new}} \):
\( x_{\text{new}} = \frac{\lambda_{\text{new}} D}{a} \implies \lambda_{\text{new}} = \frac{x_{\text{new}} a}{D} \)
Substitute the values:
\( \lambda_{\text{new}} = \frac{(3.24 \times 10^{-3}\ \text{m}) \times (2.50 \times 10^{-4}\ \text{m})}{1.50\ \text{m}} \)
\( \lambda_{\text{new}} = 5.40 \times 10^{-7}\ \text{m} = 540\ \text{nm} \).

(a)
- B1: Coherent sources / constant phase difference.
- B1: Same polarisation / same type of wave / comparable amplitudes (for high contrast).

(b)(i)
- C1: Recalls and uses \( x = \frac{\lambda D}{a} \).
- C1: Substitutes with correct unit conversions (e.g., \( \lambda = 6.328 \times 10^{-7} \), \( a = 2.50 \times 10^{-4} \)).
- A1: Obtains \( 3.80 \times 10^{-3}\ \text{m} \) (or \( 3.80\ \text{mm} \)).

(b)(ii)
- C1: Relates the 5th bright fringe distance to 5 times the fringe width (\( 5x = 1.62\ \text{cm} \)).
- C1: Calculates \( x_{\text{new}} = 3.24 \times 10^{-3}\ \text{m} \).
- A1.5: Obtains \( 5.40 \times 10^{-7}\ \text{m} \) (or \( 540\ \text{nm} \)) with correct units.

(a)
As the light intensity increases, the resistance of the light-dependent resistor (LDR) decreases.
This happens because light energy frees charge carriers (electrons) within the semiconductor material, making it more conductive. The decrease in resistance is non-linear (it decreases rapidly at first and then more slowly at higher light levels).

(b)(i)
Let \( R_1 = 4.7\ \text{k}\Omega \) be the resistance of the fixed resistor and \( R_{\text{LDR}} = 15\ \text{k}\Omega \) be the resistance of the LDR in the dark.
The total resistance in the series circuit is:
\( R_{\text{total}} = R_1 + R_{\text{LDR}} = 4.7\ \text{k}\Omega + 15\ \text{k}\Omega = 19.7\ \text{k}\Omega \).

Using the potential divider formula, the potential difference across the LDR is:
\( V_{\text{LDR}} = V_{\text{in}} \times \left( \frac{R_{\text{LDR}}}{R_{\text{total}}} \right) \)
\( V_{\text{LDR}} = 9.0\ \text{V} \times \left( \frac{15\ \text{k}\Omega}{19.7\ \text{k}\Omega} \right) \)
\( V_{\text{LDR}} = 9.0 \times 0.7614 = 6.85\ \text{V} \approx 6.9\ \text{V} \).

(b)(ii)
If the potential difference across the fixed resistor \( R_1 \) is \( V_1 = 6.0\ \text{V} \), the potential difference across the LDR is:
\( V_{\text{LDR}} = 9.0\ \text{V} - 6.0\ \text{V} = 3.0\ \text{V} \).

Since the resistors are in series, the ratio of their potential differences equals the ratio of their resistances:
\( \frac{R_{\text{LDR}}}{R_1} = \frac{V_{\text{LDR}}}{V_1} \)
\( R_{\text{LDR}} = R_1 \times \left( \frac{V_{\text{LDR}}}{V_1} \right) \)
\( R_{\text{LDR}} = 4.7\ \text{k}\Omega \times \left( \frac{3.0\ \text{V}}{6.0\ \text{V}} \right) \)
\( R_{\text{LDR}} = 4.7\ \text{k}\Omega \times 0.5 = 2.35\ \text{k}\Omega \approx 2.4\ \text{k}\Omega \) (or \( 2350\ \Omega \)).

(a)
- B1: States that resistance decreases as light intensity increases.
- B1: Mentions that the change is non-linear.
- B0.5: Explains in terms of charge carriers being released by light energy.

(b)(i)
- C1: Calculates total resistance: \( R_{\text{total}} = 19.7\ \text{k}\Omega \).
- C1: Uses correct potential divider formula.
- A1: Obtains \( 6.9\ \text{V} \) (accept range \( 6.8\text{--}6.9\ \text{V} \)).

(b)(ii)
- C1: Calculates potential difference across the LDR as \( 3.0\ \text{V} \).
- C1: Equates ratio of resistances to ratio of voltages: \( R_{\text{LDR}} = 4.7\ \text{k}\Omega \times \frac{3.0}{6.0} \).
- A1: Obtains \( 2.4\ \text{k}\Omega \) (or \( 2.35\ \text{k}\Omega \) or \( 2350\ \Omega \)).

**Sample Experimental Readings:**
- Constant mass \(M = 100\text{ g}\)
- Number of oscillations timed: \(N = 20\)

**Table of Results:**

| \(L/\text{m}\) | \(t_1/\text{s}\) | \(t_2/\text{s}\) | \(t_{\text{mean}}/\text{s}\) | \(T/\text{s}\) | \(\ln(L/\text{m})\) | \(\ln(T/\text{s})\) |
|---|---|---|---|---|---|---|
| 0.500 | 7.32 | 7.28 | 7.30 | 0.365 | -0.693 | -1.008 |
| 0.580 | 9.15 | 9.09 | 9.12 | 0.456 | -0.545 | -0.785 |
| 0.660 | 11.12 | 11.20 | 11.16 | 0.558 | -0.416 | -0.583 |
| 0.740 | 13.25 | 13.31 | 13.28 | 0.664 | -0.301 | -0.409 |
| 0.800 | 14.90 | 14.98 | 14.94 | 0.747 | -0.223 | -0.292 |
| 0.850 | 16.35 | 16.41 | 16.38 | 0.819 | -0.163 | -0.199 |

**Graph Plotting:**
- The plotted graph of \(\ln(T/\text{s})\) on the y-axis against \(\ln(L/\text{m})\) on the x-axis shows a linear trend.
- Gradient \(b\) is calculated using a large triangle from \(x_1 = -0.60\) to \(x_2 = -0.20\):
\(b = \frac{-0.199 - (-1.008)}{-0.163 - (-0.693)} = \frac{0.809}{0.530} \approx 1.53\)
- y-intercept \(\ln a\) is calculated from \(y = mx + c\):
\(-0.199 = 1.53 \times (-0.163) + \ln a \implies \ln a = 0.050 \implies a = e^{0.050} = 1.05\)
- Units for constants: \(b\) is dimensionless. Since \(T = a L^b\), the unit of \(a\) is \(\text{s m}^{-b}\) (or \(\text{s m}^{-1.53}\)).

**Uncertainty Calculation (for L = 0.750 m):**
- If absolute uncertainty in stopwatch timing \(\Delta t = 0.2\text{ s}\) (accounting for human reaction time):
\(\% \text{ uncertainty in } T = \frac{0.2}{13.48} \times 100\% \approx 1.5\%\).

**Data Collection (6 Marks):**
- **1 Mark:** Successful collection of at least 6 sets of readings of \(L\) and \(t\) showing correct trend (as \(L\) increases, \(t\) increases).
- **1 Mark:** Range of \(L\) must span at least \(0.30\text{ m}\).
- **1 Mark:** Correct column headings with units in format "\(\text{quantity/unit}\)" (e.g. \(L/\text{m}\), \(t/\text{s}\), \(T/\text{s}\), \(\ln(L/\text{m})\), \(\ln(T/\text{s})\)).
- **1 Mark:** Consistency: All raw \(L\) values recorded to nearest millimetre (\(0.001\text{ m}\)).
- **1 Mark:** All values of \(\ln(T/\text{s})\) calculated to 3 or 4 decimal places.
- **1 Mark:** Repeats: Measurement of \(t\) repeated for each value of \(L\) and a mean calculated.

**Graph (5 Marks):**
- **1 Mark:** Axes: Linear scale, sensible choice of scales (no awkward factors of 3, 7, etc.). Axes labeled with quantities (ln terms are dimensionless).
- **1 Mark:** Plotting: Points plotted accurately to within half a small grid square.
- **1 Mark:** Best-fit line: Balanced line drawn with points evenly distributed about it.
- **1 Mark:** Line thickness: Thin, sharp line (less than half a small square).
- **1 Mark:** Quality: No significant scatter of points around the best-fit line.

**Analysis of Graph (2 Marks):**
- **1 Mark:** Gradient: Large triangle used (hypotenuse length must be at least 50% of the drawn line length).
- **1 Mark:** Intercept: Correct calculation using \(y = mx + c\) with a point on the line, or direct reading from \(x=0\) if appropriate.

**Constants and Units (3 Marks):**
- **1 Mark:** Constant \(b\) equated to gradient with no unit.
- **1 Mark:** Constant \(a\) calculated correctly as \(e^{\text{intercept}}\).
- **1 Mark:** Correct unit for \(a\) corresponding to the value of \(b\) (e.g., \(\text{s m}^{-1.5}\)).

**Uncertainty and Precision (4 Marks):**
- **1 Mark:** Practical steps described to maximize precision (e.g., counting down "3-2-1-0" to start stopwatch, using a fiducial marker at equilibrium).
- **1 Mark:** Absolute uncertainty in \(t\) estimated sensibly (usually between \(0.1\text{ s}\) and \(0.3\text{ s}\)) with justification.
- **2 Marks:** Correct percentage uncertainty calculation in \(T\).

**Sample Experimental Readings and Calculation:**

- **Trial 1:**
\(h_1 = 0.200\text{ m}\)
Suspended mass to initiate movement \(m_1 = 86\text{ g}\)
\(k_1 = \frac{86}{0.200} = 430\text{ g m}^{-1}\)

- **Trial 2:**
\(h_2 = 0.400\text{ m}\)
Suspended mass to initiate movement \(m_2 = 156\text{ g}\)
\(k_2 = \frac{156}{0.400} = 390\text{ g m}^{-1}\)

- **Percentage Uncertainty in \(h_1\):**
Estimating absolute uncertainty in height measurements as \(\Delta h = 3\text{ mm} = 0.003\text{ m}\) due to alignment difficulties at the bench and ruler placement:
\(\% \text{ uncertainty} = \frac{0.003}{0.200} \times 100\% = 1.5\%\).

- **Comparison of \(k\) values:**
\(\% \text{ difference} = \frac{|430 - 390|}{(430 + 390)/2} \times 100\% = \frac{40}{410} \times 100\% = 9.8\%\).
Since the percentage difference (9.8%) is within the typical criteria threshold of 10%, the experimental results support the suggested relationship within the limits of experimental accuracy.

**Measurement & Calculations (8 Marks):**
- **1 Mark:** Correctly recorded value of \(m_1\) to the nearest gram (or equivalent in kg).
- **1 Mark:** Statement of a sensible absolute uncertainty in height \(h\) (e.g., 2 mm to 5 mm) with physical reason.
- **1 Mark:** Correct calculation of percentage uncertainty in \(h_1\).
- **1 Mark:** Correctly recorded value of \(m_2\) such that \(m_2 > m_1\).
- **1 Mark:** Correct calculation of \(k_1\) and \(k_2\) with matching units (e.g., \(\text{g m}^{-1}\) or \(\text{kg m}^{-1}\)).
- **1 Mark:** Correct calculation of percentage difference between \(k_1\) and \(k_2\).
- **2 Marks:** Conclusion based on comparison: 1 mark for calculating the percentage difference correctly, and 1 mark for clearly stating whether the relationship is supported by comparing the percentage difference to a defined criterion (e.g. 10% or the estimated percentage uncertainty).

**Evaluation - Limitations and Improvements (8 Marks Total):**
- **Up to 4 Marks (1 per limitation):**
1. Friction in the pulley resists motion, meaning measured mass is systematically higher.
2. Difficult to judge the precise instant that motion begins (sudden movement vs. static friction stick-slip behavior).
3. Large mass increments (e.g., 10g steps) limit the precision of the mass determination.
4. Only two values of \(h\) are used, which is insufficient to robustly establish a mathematical law.
5. Board may bend or flex under the weight of the block, changing the angle of incline.
- **Up to 4 Marks (1 per corresponding improvement):**
1. Use a high-quality low-friction ball-bearing pulley (or measure pulley friction and adjust values).
2. Tap the board lightly before adding mass to overcome static friction, or use an electronic force sensor.
3. Use smaller fractional masses (e.g., 1g) or add water/sand to a container and weigh it to obtain continuous mass values.
4. Take multiple readings of \(h\) and plot a graph of \(m\) against \(h\).
5. Use a stiffer metal track or support the board from below using a rigid block to prevent bending.

(a) Under simple harmonic motion: 1. Acceleration is directly proportional to displacement from equilibrium. 2. Acceleration is directed towards the equilibrium position (opposite to displacement).

(b)(i) Use \( \omega = \sqrt{\frac{k}{m}} \):
\( \omega = \sqrt{\frac{180}{0.45}} = \sqrt{400} = 20\text{ rad s}^{-1} \).

(ii) Maximum speed \( v_0 = \omega x_0 \):
\( v_0 = 20 \times 0.060 = 1.2\text{ m s}^{-1} \).

(iii) Kinetic energy is given by \( E_k = \frac{1}{2} m \omega^2 (x_0^2 - x^2) \):
\( E_k = \frac{1}{2} \times 0.45 \times 20^2 \times (0.060^2 - 0.030^2) \)
\( E_k = 90 \times (0.0036 - 0.0009) = 90 \times 0.0027 = 0.243\text{ J} \approx 0.24\text{ J} \).

(a)
- Acceleration is proportional to displacement [1]
- Acceleration is directed towards a fixed point / opposite in sign to displacement [1]

(b)(i)
- Formula \( \omega = \sqrt{k/m} \) seen or used [1]
- Correct calculation leading to \( 20\text{ rad s}^{-1} \) [1]

(b)(ii)
- Formula \( v_0 = \omega x_0 \) seen or used [1]
- Answer \( 1.2\text{ m s}^{-1} \) [1]

(b)(iii)
- Use of \( E_k = \frac{1}{2} m \omega^2 (x_0^2 - x^2) \) or subtraction of PE from total energy [1.1]
- Substitution of correct values: \( 0.5 \times 0.45 \times 400 \times (0.060^2 - 0.030^2) \) [1]
- Final answer: \( 0.24\text{ J} \) (or \( 0.243\text{ J} \)) [1]

(a) A standard candle is an astronomical object of known luminosity (such as a Cepheid variable star). By measuring its apparent brightness (flux \( F \)) at Earth, its distance \( d \) can be calculated using the inverse square law \( F = \frac{L}{4\pi d^2} \).

(b)(i) Redshift \( z = \frac{\Delta \lambda}{\lambda_0} \):
\( z = \frac{670.8 - 656.3}{656.3} = \frac{14.5}{656.3} = 0.022093 \approx 0.0221 \).

(ii) The recessional velocity \( v \) is given by \( v = zc \):
\( v = 0.022093 \times 3.00 \times 10^8 = 6.628 \times 10^6\text{ m s}^{-1} \approx 6.63 \times 10^6\text{ m s}^{-1} \).

(iii) Hubble's Law states \( v = H_0 d \). Therefore:
\( d = \frac{v}{H_0} = \frac{6.628 \times 10^6}{2.2 \times 10^{-18}} = 3.01 \times 10^{24}\text{ m} \approx 3.0 \times 10^{24}\text{ m} \).

(a)
- Defines standard candle as an object of known luminosity [1]
- Explains that measuring its radiant flux (or apparent brightness) allows distance to be calculated using \( F = L / (4\pi d^2) \) [1]

(b)(i)
- Formula \( z = \Delta \lambda / \lambda_0 \) seen or implied [1]
- Correct calculation to give \( 0.0221 \) (or \( 0.022 \)) [1]

(b)(ii)
- Uses \( v = zc \) with \( c = 3.00 \times 10^8\text{ m s}^{-1} \) [1.1]
- Correct calculation to give \( 6.63 \times 10^6\text{ m s}^{-1} \) [1]

(b)(iii)
- Uses \( v = H_0 d \) [1]
- Correct substitution of values [1]
- Correct calculation to give \( 3.0 \times 10^{24}\text{ m} \) (accept \( 3.0 \times 10^{24} \) to \( 3.01 \times 10^{24} \)) [1]

(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) \( \phi = -\frac{GM}{R} = -\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{6.4 \times 10^6} = -6.253 \times 10^7\text{ J kg}^{-1} \approx -6.3 \times 10^7\text{ J kg}^{-1} \).

(ii) By conservation of energy, the total energy of the probe remains constant:
\( E_{\text{total}} = E_k + E_p = E_{k, i} + m\phi_i \)
\( E_{\text{total}} = 5.2 \times 10^{10} + 1200 \times (-6.253 \times 10^7) \)
\( E_{\text{total}} = 5.2 \times 10^{10} - 7.504 \times 10^{10} = -2.304 \times 10^{10}\text{ J} \).

At the maximum distance \( r \), the velocity is zero, so \( E_k = 0 \):
\( E_{\text{total}} = -\frac{GMm}{r} \)
\( -2.304 \times 10^{10} = -\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 1200}{r} \)
\( -2.304 \times 10^{10} = -\frac{4.8024 \times 10^{17}}{r} \)
\( r = \frac{4.8024 \times 10^{17}}{2.304 \times 10^{10}} \approx 2.084 \times 10^7\text{ m} \approx 2.1 \times 10^7\text{ m} \).

(a)
- Work done per unit mass [1]
- In bringing a mass from infinity to the point [1]

(b)(i)
- Formula \( \phi = -GM/R \) stated [1]
- Correct substitution and calculation to at least 3 s.f. (i.e. \( -6.25 \times 10^7\text{ J kg}^{-1} \)) [1]

(b)(ii)
- Uses conservation of energy: \( E_k + E_p = \text{constant} \) [1.1]
- Calculates initial potential energy as \( -7.5 \times 10^{10}\text{ J} \) [1]
- Finds total mechanical energy as \( -2.30 \times 10^{10}\text{ J} \) [1]
- Equates total mechanical energy at peak to \( -GMm/r \) [1]
- Solves for \( r \) to get \( 2.1 \times 10^7\text{ m} \) (accept range \( 2.08 \times 10^7 \) to \( 2.10 \times 10^7\text{ m} \)) [1]

(a) Basic assumptions:
1. A gas consists of a very large number of molecules in rapid, continuous, random motion.
2. The total volume of the molecules is negligible compared to the volume of the container.
3. All collisions between molecules and with the walls are perfectly elastic.
4. Intermolecular forces are negligible except during collisions.
5. The duration of collisions is negligible compared to the time between collisions.

(b)(i) Use \( pV = nRT \):
First, convert temperature to kelvins: \( T = 27 + 273 = 300\text{ K} \).
\( n = \frac{pV}{RT} = \frac{1.5 \times 10^5 \times 0.045}{8.31 \times 300} = \frac{6750}{2493} = 2.707\text{ mol} \approx 2.7\text{ mol} \).

(ii) Since the cylinder is rigid, the volume \( V \) is constant. Therefore:
\( \frac{p_1}{T_1} = \frac{p_2}{T_2} \implies T_2 = T_1 \times \frac{p_2}{p_1} \)
\( T_2 = 300 \times \frac{2.4 \times 10^5}{1.5 \times 10^5} = 300 \times 1.6 = 480\text{ K} \).
Convert back to Celsius: \( T_2 = 480 - 273 = 207^\circ\text{C} \).

(iii) The average translational kinetic energy of a single molecule is:
\( E_k = \frac{3}{2} k_B T \)
\( E_k = 1.5 \times 1.38 \times 10^{-23} \times 480 = 9.936 \times 10^{-21}\text{ J} \approx 9.9 \times 10^{-21}\text{ J} \).

(a)
- Any three correct assumptions from: large number of molecules, negligible volume of molecules, elastic collisions, no intermolecular forces, continuous random motion [3, 1 mark for each]

(b)(i)
- Convert temperature to \( 300\text{ K} \) [0.5]
- Use of \( pV = nRT \) [0.6]
- Final answer: \( 2.7\text{ mol} \) (or \( 2.71\text{ mol} \)) [1]

(b)(ii)
- States or uses constant volume condition: \( p_1/T_1 = p_2/T_2 \) or calculates new temperature in Kelvin as \( 480\text{ K} \) [1]
- Subtracts \( 273 \) to give final answer \( 207^\circ\text{C} \) [1]

(b)(iii)
- Formula \( E_k = \frac{3}{2} k_B T \) seen or used with new temperature in Kelvin [1]
- Correct calculation leading to \( 9.9 \times 10^{-21}\text{ J} \) (accept \( 9.94 \times 10^{-21}\text{ J} \)) [1]

(a) The electric field exerts a force \( F_E = qE \) on the ions, and the magnetic field exerts a force \( F_B = qvB \). If the fields are oriented such that these forces act in opposite directions, they can balance. For the forces to balance, \( qE = qvB \), which simplifies to \( v = \frac{E}{B} \). Only ions with this specific speed pass through undeflected; those with other speeds are deflected.

(b)(i) The magnetic force provides the centripetal force for circular motion:
\( F_B = F_c \implies qvB = \frac{mv^2}{r} \).
Rearranging for \( r \) gives:
\( r = \frac{mv}{qB} \).

(ii) Substitute the given values into the equation:
\( r = \frac{6.6 \times 10^{-26} \times 2.5 \times 10^5}{1.6 \times 10^{-19} \times 0.18} \)
\( r = \frac{1.65 \times 10^{-20}}{2.88 \times 10^{-20}} = 0.573\text{ m} \approx 0.57\text{ m} \).

(iii) An isotope with a larger mass number has a larger mass \( m \). Since \( r \propto m \) (as speed \( v \), charge \( q \), and magnetic flux density \( B \) remain the same), the radius of the path will increase.

(a)
- Electric force is given by \( qE \) and magnetic force by \( qvB \) [1]
- These forces act in opposite directions [1]
- For no deflection, forces balance, so \( qE = qvB \) meaning only ions with speed \( v = E/B \) pass undeflected [1]

(b)(i)
- Equates magnetic force to centripetal force: \( qvB = mv^2 / r \) [1]
- Correct rearrangement to show \( r = mv / (qB) \) [1]

(b)(ii)
- Correct substitution of values into the formula [1.1]
- Calculates path radius as \( 0.57\text{ m} \) (or \( 0.573\text{ m} \)) [1]

(b)(iii)
- Path radius increases [1]
- Because mass \( m \) is larger and \( r \propto m \) (for constant \( v \), \( q \), and \( B \)) [1]

(a) The decay constant is the probability of decay of a nucleus per unit time.

(b)(i) Activity is related to the number of active nuclei by \( A = \lambda N \). Therefore:
\( \lambda = \frac{A_0}{N_0} = \frac{8.5 \times 10^{12}}{3.4 \times 10^{18}} = 2.5 \times 10^{-6}\text{ s}^{-1} \).

(ii) The half-life is given by:
\( t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{2.5 \times 10^{-6}} = 277259\text{ s} \).
Convert to hours:
\( t_{1/2} = \frac{277259}{3600} = 77.02\text{ hours} \approx 77\text{ hours} \).

(iii) The activity after \( 5.0\text{ hours} \) is given by \( A = A_0 e^{-\lambda t} \).
First, convert time to seconds:
\( t = 5.0 \times 3600 = 18000\text{ s} \).
\( A = (8.5 \times 10^{12}) e^{-(2.5 \times 10^{-6} \times 18000)} \)
\( A = (8.5 \times 10^{12}) e^{-0.045} = (8.5 \times 10^{12}) \times 0.955997 = 8.126 \times 10^{12}\text{ Bq} \approx 8.1 \times 10^{12}\text{ Bq} \).

(a)
- Probability of decay of a nucleus [1]
- Per unit time [1]

(b)(i)
- Formula \( A = \lambda N \) used [1.1]
- Value \( 2.5 \times 10^{-6} \) calculated [1]
- Unit: \( \text{s}^{-1} \) (or per second) [1]

(b)(ii)
- Use of \( t_{1/2} = \ln 2 / \lambda \) [1]
- Converts seconds to hours to give \( 77\text{ hours} \) (accept range \( 77 \) to \( 77.1\text{ hours} \)) [1]

(b)(iii)
- Converts \( 5.0\text{ hours} \) to \( 18000\text{ s} \) and uses \( A = A_0 e^{-\lambda t} \) (or equivalent ratio method) [1]
- Final answer: \( 8.1 \times 10^{12}\text{ Bq} \) (accept \( 8.13 \times 10^{12}\text{ Bq} \)) [1]

(a) Acoustic impedance \( Z = \rho c \), where \( \rho \) is the density of the medium and \( c \) is the speed of the wave in the medium.

(b) The reflection coefficient \( \alpha \) (fraction of reflected intensity) is given by:
\( \alpha = \frac{(Z_2 - Z_1)^2}{(Z_2 + Z_1)^2} = \frac{(6.4 \times 10^6 - 1.7 \times 10^6)^2}{(6.4 \times 10^6 + 1.7 \times 10^6)^2} \)
\( \alpha = \frac{(4.7 \times 10^6)^2}{(8.1 \times 10^6)^2} = \frac{22.09}{65.61} = 0.3367 \approx 0.34 \).

(c)(i) The relation is \( I = I_0 e^{-\mu x} \).

(ii) We require \( I / I_0 = 0.45 \).
\( 0.45 = e^{-23 x} \)
Take the natural logarithm of both sides:
\( \ln(0.45) = -23 x \)
\( -0.7985 = -23 x \)
\( x = \frac{0.7985}{23} = 0.0347\text{ m} \approx 0.035\text{ m} \) (or \( 3.5\text{ cm} \)).

(a)
- Acoustic impedance is product of density and speed of wave [1]
- Formula \( Z = \rho c \) with symbols defined correctly [1]

(b)
- Uses \( \alpha = (Z_2 - Z_1)^2 / (Z_2 + Z_1)^2 \) [1.1]
- Correct calculation leading to \( 0.34 \) (or \( 0.337 \)) [1]

(c)(i)
- State \( I = I_0 e^{-\mu x} \) [1]

(c)(ii)
- Sets \( I / I_0 = 0.45 \) [1]
- Takes ln of both sides correctly: \( -23 x = \ln(0.45) \) [1]
- Calculation of \( x \) to give \( 0.035\text{ m} \) (or \( 3.5\text{ cm} \)) [2]

(a) The time constant is the time taken for the charge, potential difference, or current in a discharging capacitor-resistor circuit to fall to \( 1/e \) (approximately \( 37\% \)) of its initial value. Alternatively, it is given by \( \tau = RC \).

(b)(i) The time constant \( \tau = RC \):
\( \tau = 15 \times 10^3 \times 220 \times 10^{-6} = 3.3\text{ s} \).

(ii) The initial charge on the capacitor is:
\( Q_0 = C V_0 = 220 \times 10^{-6} \times 12.0 = 2.64 \times 10^{-3}\text{ C} \).

The charge remaining after \( t = 5.0\text{ s} \) is:
\( Q = Q_0 e^{-t/RC} = 2.64 \times 10^{-3} e^{-5.0 / 3.3} \)
\( Q = 2.64 \times 10^{-3} e^{-1.515} = 2.64 \times 10^{-3} \times 0.2198 = 5.80 \times 10^{-4}\text{ C} \approx 5.8 \times 10^{-4}\text{ C} \).

(iii) The energy remaining in the capacitor is:
\( E = \frac{Q^2}{2C} = \frac{(5.80 \times 10^{-4})^2}{2 \times 220 \times 10^{-6}} = \frac{3.364 \times 10^{-7}}{4.40 \times 10^{-4}} = 7.65 \times 10^{-4}\text{ J} \approx 7.6 \times 10^{-4}\text{ J} \).
Alternatively, using \( E = E_0 e^{-2t/\tau} \) where \( E_0 = \frac{1}{2} C V_0^2 = 0.01584\text{ J} \):
\( E = 0.01584 e^{-10.0 / 3.3} = 7.65 \times 10^{-4}\text{ J} \).

(a)
- Time for charge / potential difference / current to decrease [1]
- To \( 1/e \) (or \( 37\% \)) of its initial value [1]

(b)(i)
- Formula \( \tau = RC \) seen [1]
- Correct calculation leading to \( 3.3\text{ s} \) [1]

(b)(ii)
- Uses \( Q_0 = CV_0 \) to find \( 2.64 \times 10^{-3}\text{ C} \) [1]
- Uses \( Q = Q_0 e^{-t/RC} \) [1.1]
- Correct calculation to give \( 5.8 \times 10^{-4}\text{ C} \) (accept range \( 5.8 \times 10^{-4} \) to \( 5.82 \times 10^{-4}\text{ C} \)) [1]

(b)(iii)
- Use of \( E = \frac{1}{2} Q^2/C \) or \( E = \frac{1}{2} C V^2 \) with new values [1]
- Correct calculation to give \( 7.6 \times 10^{-4}\text{ J} \) (accept \( 7.6 \times 10^{-4} \) to \( 7.7 \times 10^{-4}\text{ J} \)) [1]

(a) The gravitational force of attraction between two point masses is directly proportional to the product of their masses and inversely proportional to the square of their separation.

(b) For a circular orbit, the gravitational force provides the centripetal force:
\(F_g = F_c\)
\(G M m / r^2 = m v^2 / r\)
Since \(v = 2 \pi r / T\), substituting this gives:
\(G M / r = 4 \pi^2 r^2 / T^2\)
Rearranging for \(T^2\) gives:
\(T^2 = 4 \pi^2 r^3 / (GM)\)

(c) (i) Using \(T = \sqrt{4 \pi^2 r^3 / (GM)}\):
\(T = \sqrt{\frac{4 \pi^2 \times (4.20 \times 10^6)^3}{(6.67 \times 10^{-11}) \times (6.42 \times 10^{23})}}\)
\(T = \sqrt{\frac{2.925 \times 10^{21}}{4.282 \times 10^{13}}} = \sqrt{6.831 \times 10^7} = 8260\text{ s}\).

(ii) Kinetic energy \(E_k = \frac{1}{2} m v^2 = \frac{GMm}{2r}\):
\(E_k = \frac{(6.67 \times 10^{-11}) \times (6.42 \times 10^{23}) \times 450}{2 \times (4.20 \times 10^6)}\)
\(E_k = 2.29 \times 10^9\text{ J}\).

(a) Force proportional to product of masses and inversely proportional to square of separation: 1 mark. Point masses specified: 1 mark.
(b) Gravitational force equated to centripetal force: 1 mark. Substitution of v = 2*pi*r / T and correct algebraic steps showing final equation: 1 mark.
(c) (i) Substitution of values into formula: 1 mark. Correct calculation of GM and r^3: 1 mark. Final answer of 8260 s (or 8300 s): 1 mark.
(ii) Formula for kinetic energy (GMm/2r or 0.5mv^2): 1 mark. Correct value of 2.29 x 10^9 J: 1.1 marks.

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

(b) (i) \(E = h f = \frac{h c}{\lambda}\)
\(E = \frac{(6.63 \times 10^{-34}) \times (3.00 \times 10^8)}{320 \times 10^{-9}} = 6.216 \times 10^{-19}\text{ J}\)
In eV: \(E = \frac{6.216 \times 10^{-19}}{1.60 \times 10^{-19}} = 3.88\text{ eV}\).
(ii) Since the photon energy (3.88 eV) is greater than the work function energy (2.36 eV), photoelectric emission will occur.

(c) Maximum kinetic energy of electrons: \(E_{k,\text{max}} = E - \Phi = 3.88\text{ eV} - 2.36\text{ eV} = 1.52\text{ eV}\).
In Joules: \(E_k = 1.52 \times 1.60 \times 10^{-19} = 2.43 \times 10^{-19}\text{ J}\).
De Broglie wavelength: \(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m_e E_k}}\)
\(\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times (9.11 \times 10^{-31}) \times (2.43 \times 10^{-19})}}\)
\(\lambda = \frac{6.63 \times 10^{-34}}{6.656 \times 10^{-25}} = 9.96 \times 10^{-10}\text{ m}\).

(a) Minimum energy: 1 mark. Required to remove an electron from metal surface: 1 mark.
(b) (i) Use of E = hc/\u03bb: 1 mark. Conversion to eV: 1 mark. Correct answer of 3.88 eV: 1 mark.
(ii) Statement that emission occurs: 1 mark. Reason stating photon energy (3.88 eV) is greater than work function (2.36 eV): 1 mark.
(c) Calculating max KE in Joules: 1 mark. Correct de Broglie wavelength calculation to give 9.96 x 10^-10 m (accept range 9.9 x 10^-10 m to 1.0 x 10^-9 m): 1.1 marks.

(a) Resonance is the condition where a system is driven at its natural frequency, resulting in maximum amplitude of oscillation.

(b) Damping is the loss of mechanical energy from an oscillating system due to resistive forces (such as friction or air resistance).

(c) (i) For light damping, resonant frequency is equal to the natural frequency: \(f_0 = 4.5\text{ Hz}\).
Using \(f_0 = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\):
\(k = 4 \pi^2 f_0^2 m = 4 \times \pi^2 \times 4.5^2 \times 0.350\)
\(k = 4 \times 9.870 \times 20.25 \times 0.350 = 280\text{ N m}^{-1}\).
(ii) Increased damping causes:
1. The maximum amplitude (peak value) of the oscillations to decrease.
2. The peak of the resonance curve to become broader and shift slightly towards a lower frequency.

(a) System driven at/near its natural frequency: 1 mark. Resulting in maximum amplitude: 1 mark.
(b) Loss of mechanical energy / amplitude decrease: 1 mark. Due to resistive/frictional forces: 1 mark.
(c) (i) Recall of formula f = (1/2pi) * sqrt(k/m) or omega = sqrt(k/m): 1 mark. Correct substitution of values: 1 mark. Correct value of 280 N m^-1 (or 279.8 N m^-1): 1 mark.
(ii) Peak amplitude decreases: 1 mark. Resonant peak becomes broader or shifts to a lower frequency: 1.1 marks.

### Method of Data Collection
- **Setup:** Clamp a steel rule (cantilever) firmly to a heavy bench using a G-clamp and wooden blocks. Attach a small neodymium magnet securely to the free end of the cantilever.
- **Coil Placement:** Position a copper coil on a separate retort stand so that the magnet passes exactly through the center of the coil as it oscillates. Ensure there is no physical contact between the magnet and the coil during oscillations.
- **Electrical Circuit:** Connect the coil in series with a decade resistance box (to vary \(R\)) using connecting wires. Use a digital multimeter (configured as an ohmmeter) to measure the internal resistance of the coil \(R_c\) when disconnected from the circuit.
- **Data Recording:** Position a motion sensor directly above or below the free end of the cantilever, connected to a datalogger/computer to record its vertical position \(x\) as a function of time \(t\).

### Procedure
1. Measure and record the coil's internal resistance \(R_c\).
2. Disconnect the coil circuit (equivalent to \(R \to \infty\)) and displace the cantilever vertically by a fixed distance. Release it and record the position-time data to determine the natural damping constant \(b_0\).
3. Connect the coil to the decade resistance box set to a known value of \(R\).
4. Displace the cantilever by the same initial distance, release it, and record the amplitude decay over time.
5. Repeat the procedure for at least six different values of \(R\) spanning a wide range.

### Analysis of Data
- **Determining \(b\):** For each run, extract the peak amplitudes \(A_1, A_2, A_3, \dots\) at times \(t_1, t_2, t_3, \dots\). Since the envelope of the oscillations decays exponentially according to \(A = A_0 e^{-bt}\), plot a graph of \(\ln A\) against \(t\). The gradient of this straight line is equal to \(-b\).
- **Testing the Relationship:** Rearrange the proposed equation:
\[b - b_0 = \frac{k}{R + R_c} \implies \frac{1}{b - b_0} = \frac{1}{k} R + \frac{R_c}{k}\]
Plot a graph of \(\frac{1}{b - b_0}\) against \(R\).
- **Determining Constants:** A straight line with a positive y-intercept confirms the relationship.
- The gradient of the graph is \(m = \frac{1}{k} \implies k = \frac{1}{m}\).
- The y-intercept is \(c = \frac{R_c}{k} \implies R_c = c \cdot k = \frac{c}{m}\).

### Safety Precautions
- Wear safety goggles to protect eyes from high-speed flying objects in case the cantilever snaps or the magnet detaches.
- Use a G-clamp to firmly secure the retort stand to the bench, ensuring it cannot topple over during large amplitude oscillations.

**Defining the Problem (3 marks):**
- [1] \(R\) is the independent variable and \(b\) is the dependent variable.
- [1] Keep the length of the cantilever (or mass of the magnet, or initial displacement) constant.
- [1] Measure the natural damping constant \(b_0\) with the coil circuit open (isolated).

**Methods of Data Collection (5 marks):**
- [1] Complete diagram showing the cantilever clamped, magnet attached to the free end, and the coil positioned around the path of the magnet.
- [1] Diagram showing the coil connected in a complete closed loop with a variable resistor/resistance box.
- [1] Use of a motion sensor/datalogger or high-speed video camera with a vertical millimeter scale to record position-time data.
- [1] Use of an ohmmeter/multimeter to measure the resistance of the coil \(R_c\) directly.
- [1] Method to determine the resistance of the variable resistor \(R\) (e.g., using a calibrated decade resistance box).

**Method of Analysis (3 marks):**
- [1] Plot a graph of \(\ln A\) against time \(t\) to determine \(b\) from the gradient (\(b = -\text{gradient}\)).
- [1] Plot a graph of \(\frac{1}{b - b_0}\) against \(R\).
- [1] Identify that \(k = 1/\text{gradient}\) and \(R_c = \text{y-intercept}/\text{gradient}\).

**Safety and Additional Details (4 marks):**
- [1] Safety precaution: Wear eye protection/goggles to protect against snap failure of the cantilever, or clamp the stand securely.
- [1] Keep the initial displacement constant for each trial to ensure consistent starting conditions.
- [1] Perform preliminary runs to establish an appropriate range of \(R\) so that the motion is underdamped and clearly decays.
- [1] Repeat the decay process for each value of \(R\) to obtain a mean value of \(b\).

### Part (a) Data Table
\[\begin{array}{|c|c|c|}
\hline
D / \text{cm} & 1/D / \text{cm}^{-1} & v / \text{cm s}^{-1} \\
\hline
1.5 & 0.667 & 9.3 \pm 0.3 \\
2.0 & 0.500 & 10.1 \pm 0.3 \\
3.0 & 0.333 & 10.9 \pm 0.3 \\
4.0 & 0.250 & 11.3 \pm 0.3 \\
5.0 & 0.200 & 11.5 \pm 0.3 \\
8.0 & 0.125 & 11.9 \pm 0.3 \\
\hline
\end{array}\]

### Part (b) Graph Plotting
- **Axes:** Scale the vertical axis to start from \(8.0 \text{ cm s}^{-1}\) to \(13.0 \text{ cm s}^{-1}\) and the horizontal axis from \(0\) to \(0.70 \text{ cm}^{-1}\).
- **Plotting:** All points are plotted accurately, and vertical error bars of length \(\pm 0.3\) units are added.
- **Best-fit line:** A straight line passing exactly through the coordinates is drawn.
- **Worst acceptable line:** A shallower or steeper line passing through all error bars (e.g., from the top of the error bar at \(1/D = 0.667\) to the bottom of the error bar at \(1/D = 0.125\)).

### Part (c) Gradient and Intercept Calculations
- **Best-fit Gradient (\(m\)):**
Using points \((0.125, 11.90)\) and \((0.667, 9.30)\):
\[m = \frac{9.30 - 11.90}{0.667 - 0.125} = \frac{-2.60}{0.542} = -4.80 \text{ cm}^2 \text{ s}^{-1}\]
- **Worst Gradient (\(m_{\text{worst}}\)):**
Using points \((0.125, 11.60)\) and \((0.667, 9.60)\):
\[m_{\text{worst}} = \frac{9.60 - 11.60}{0.667 - 0.125} = -3.69 \text{ cm}^2 \text{ s}^{-1}\]
- **Uncertainty in Gradient (\(\Delta m\)):**
\[\Delta m = |m_{\text{best}} - m_{\text{worst}}| = |-4.80 - (-3.69)| = 1.11 \text{ cm}^2 \text{ s}^{-1}\]
- **Best-fit y-intercept (\(c\)):**
\[11.90 = -4.80(0.125) + c \implies c = 12.50 \text{ cm s}^{-1}\]
- **Worst y-intercept (\(c_{\text{worst}}\)):**
\[11.60 = -3.69(0.125) + c_{\text{worst}} \implies c_{\text{worst}} = 12.06 \text{ cm s}^{-1}\]
- **Uncertainty in y-intercept (\(\Delta c\)):**
\[\Delta c = |c_{\text{best}} - c_{\text{worst}}| = |12.50 - 12.06| = 0.44 \approx 0.40 \text{ cm s}^{-1}\]

### Part (d) Determining \(v_0\) and \(k\)
- **Value of \(v_0\):**
\[v_0 = c = 12.5 \pm 0.4 \text{ cm s}^{-1}\]
- **Value of \(k\):**
From the model equation \(v = v_0 - \frac{v_0 k d}{D}\), we have:
\[\text{gradient } m = -v_0 k d \implies k = \frac{-m}{v_0 d}\]
Given \(d = 3.18 \text{ mm} = 0.318 \text{ cm}\):
\[k = \frac{4.80}{12.50 \times 0.318} = 1.207 \approx 1.21\]
- **Uncertainty in \(k\):**
Using fractional uncertainties:
\[\frac{\Delta k}{k} = \frac{\Delta m}{|m|} + \frac{\Delta v_0}{v_0} + \frac{\Delta d}{d}\]
\[\frac{\Delta k}{k} = \frac{1.11}{4.80} + \frac{0.44}{12.50} + \frac{0.005}{0.318} = 0.231 + 0.035 + 0.016 = 0.282\]
\[\Delta k = 1.21 \times 0.282 = 0.34\]
Therefore, \(k = 1.2 \pm 0.3\) (dimensionless).

**Part (a) [1 mark]:**
- [1] All \(1/D\) values calculated correctly to 3 significant figures: \(0.667\), \(0.500\), \(0.333\), \(0.250\), \(0.200\), \(0.125\).

**Part (b) [3 marks]:**
- [1] Points plotted correctly with error bars of size \(\pm 0.3\) vertical units.
- [1] Best-fit line drawn straight, with balanced distribution of points.
- [1] Worst acceptable line drawn as a straight line passing through all error bars, clearly labelled.

**Part (c) [4 marks]:**
- [1] Correct method for gradient using a large triangle spanning at least half of the line length.
- [1] Gradient of best-fit line \(m = -4.80 \pm 0.20 \text{ cm}^2 \text{ s}^{-1}\) with uncertainty calculated from difference with worst line (approx. \(\pm 1.1\)).
- [1] Correct extraction of best-fit y-intercept value: \(12.50 \pm 0.30 \text{ cm s}^{-1}\).
- [1] Uncertainty in y-intercept correctly determined from the difference between best-fit and worst acceptable lines (approx. \(\pm 0.4\)).

**Part (d) [7 marks]:**
- [1] Correctly identifies \(v_0 = y\text{-intercept}\).
- [1] Value of \(v_0\) given as \(12.5 \pm 0.4 \text{ cm s}^{-1}\) (or \(0.125 \pm 0.004 \text{ m s}^{-1}\)) with appropriate units.
- [1] Correct algebraic relation for \(k\): \(k = \frac{-m}{v_0 d}\).
- [1] Correct conversion of unit of \(d\) to cm (\(0.318 \text{ cm}\)) or variables to SI units.
- [1] Calculated value of \(k = 1.2\) or \(1.21\).
- [1] Correct summing of fractional uncertainties: \(\frac{\Delta k}{k} = \frac{\Delta m}{|m|} + \frac{\Delta v_0}{v_0} + \frac{\Delta d}{d}\).
- [1] Absolute uncertainty of \(k\) calculated as \(\pm 0.3\) or \(\pm 0.34\), with \(k\) correctly stated as dimensionless.