2025 Edexcel IAL Physics (YPH11) 模擬試題連答案詳解

Using the diffraction grating equation: \(d \sin\theta = n\lambda\), which can be written in terms of the number of lines per mm \(N\) as \(\sin\theta = n\lambda N\).

For the first setup: \(\sin\theta = 3 \lambda N\)
For the second setup: \(\sin\theta = 2 (0.8\lambda) N' = 1.6\lambda N'\)

Equating the two expressions for \(\sin\theta\):
\(3 \lambda N = 1.6\lambda N'\)

Solving for \(N'\):
\(N' = \frac{3}{1.6} N = 1.875 N \approx 1.88 N\)

1 mark for the correct answer. No partial marks.

The drift velocity equation is given by \(I = nAvq\), where \(A = \frac{\pi d^2}{4}\).

Rearranging for drift velocity \(v\):
\(v = \frac{4I}{n \pi d^2 q}\)

Since both wires are made of the same metal, the number density of conduction electrons \(n\) and the elementary charge \(q\) are constant. Therefore:
\(v \propto \frac{I}{d^2}\)

Using the ratios:
\(\frac{v_X}{v_Y} = \left(\frac{I_X}{I_Y}\right) \times \left(\frac{d_Y}{d_X}\right)^2 = 3 \times \left(\frac{1}{2}\right)^2 = 0.75\)

1 mark for the correct answer. No partial marks.

Using Einstein's photoelectric equation:
\(E_k = hf - \Phi\)

Rearranging for the original photon energy:
\(hf = E_k + \Phi\)

When the frequency is doubled, the new maximum kinetic energy \(E'_k\) is:
\(E'_k = h(2f) - \Phi = 2(hf) - \Phi\)

Substituting \(hf = E_k + \Phi\) into this equation:
\(E'_k = 2(E_k + \Phi) - \Phi = 2E_k + 2\Phi - \Phi = 2E_k + \Phi\)

1 mark for the correct answer. No partial marks.

From the terminal potential difference equations:
1) \(\mathcal{E} = V_1 + I_1 r = V_1 \left(1 + \frac{r}{R_1}\right)\)
2) \(\mathcal{E} = V_2 + I_2 r = V_2 \left(1 + \frac{r}{2R_1}\right)\)

Equating the expressions for \(\mathcal{E}\):
\(V_1 \left(1 + \frac{r}{R_1}\right) = V_2 \left(1 + \frac{r}{2R_1}\right)\)

Expanding the terms:
\(V_1 + \frac{V_1 r}{R_1} = V_2 + \frac{V_2 r}{2R_1}\)

Rearranging to group the terms with \(r\):
\(\frac{V_1 r}{R_1} - \frac{V_2 r}{2R_1} = V_2 - V_1\)

\(\frac{r}{2R_1} (2V_1 - V_2) = V_2 - V_1\)

Solving for \(r\):
\(r = \frac{2R_1(V_2 - V_1)}{2V_1 - V_2}\)

1 mark for the correct answer. No partial marks.

For any standing wave on a string, all particles between two adjacent nodes move in phase (phase difference is zero) and reach their maximum displacements at the same time. Adjacent loops are out of phase by \(\pi\) radians. The amplitude varies along the string, and the wave does not transfer energy.

1 mark for the correct answer. No partial marks.

Let the resistance of each resistor be \(R\).
The total current from the source is \(I\). This current passes completely through \(R_3\).
The current splits equally between the two identical parallel resistors \(R_1\) and \(R_2\). Therefore, the current through \(R_1\) is \(I_1 = \frac{I}{2}\).

The power dissipated in \(R_1\) is:
\(P_1 = I_1^2 R = \left(\frac{I}{2}\right)^2 R = \frac{1}{4}I^2 R\)

The power dissipated in \(R_3\) is:
\(P_3 = I^2 R\)

The ratio of power is:
\(\frac{P_1}{P_3} = \frac{\frac{1}{4}I^2 R}{I^2 R} = 0.25\)

1 mark for the correct answer. No partial marks.

Total internal reflection (TIR) can only occur when light travels from a medium of higher refractive index to a medium of lower refractive index. Since \(1.52 > 1.33\), the incident ray must travel in the glass block.

Using Snell's Law for the critical angle \(\theta_c\):
\(\sin\theta_c = \frac{n_{\text{water}}}{n_{\text{glass}}} = \frac{1.33}{1.52} \approx 0.875\)
\(\theta_c = \sin^{-1}(0.875) \approx 61.0^\circ\)

1 mark for the correct answer. No partial marks.

For a negative temperature coefficient (NTC) thermistor, a decrease in temperature causes its resistance to increase.

In a potential divider circuit, as the resistance of the thermistor increases relative to the fixed resistor, it takes a larger proportion of the total potential difference of the supply. Therefore, the potential difference across the thermistor increases.

1 mark for the correct answer. No partial marks.

According to the photoelectric equation, \(E_k = hf - \phi\), where \(\phi\) is the work function of the metal. If the frequency is doubled to \(2f\), the new maximum kinetic energy of the photoelectrons is given by \(E_k' = h(2f) - \phi = 2hf - \phi\). Rearranging the first equation gives \(hf = E_k + \phi\). Substituting this into the second equation yields \(E_k' = 2(E_k + \phi) - \phi = 2E_k + \phi\). Since the work function \(\phi\) must be a positive quantity for photoemission to occur, it follows that the new maximum kinetic energy \(E_k'\) must be greater than \(2E_k\).

1 mark: Correctly identifies that \(E_k' > 2E_k\) (Option B).

Because the wires are connected in series, the electric current \(I\) flowing through both wires is identical. The transport equation is \(I = nAvq\), where \(n\) is the charge carrier density, \(A\) is the cross-sectional area, \(v\) is the drift velocity, and \(q\) is the charge of an electron. Since both wires are made of copper, \(n\) is identical, meaning drift velocity is inversely proportional to cross-sectional area: \(v \propto \frac{1}{A}\). The cross-sectional area is proportional to the square of the diameter \(d\) of the wire, so \(v \propto \frac{1}{d^2}\). Given that \(d_X = 2d_Y\), we can calculate the ratio of drift velocities: \(\frac{v_X}{v_Y} = \frac{d_Y^2}{d_X^2} = \frac{d_Y^2}{(2d_Y)^2} = \frac{1}{4} = 0.25\).

1 mark: Correctly determines the ratio as 0.25 (Option D).

(a) The ammeter and variable resistor must be connected in series with the cell. The voltmeter must be connected in parallel across the terminals of the cell.

(b) Using the equation \(V = \varepsilon - Ir\):
For the first set of data: \(1.30 = \varepsilon - 0.40r\) (Equation 1)
For the second set of data: \(1.10 = \varepsilon - 0.80r\) (Equation 2)
Subtracting Equation 2 from Equation 1:
\(0.20 = 0.40r\)
\(r = 0.50\ \Omega\)
Substitute \(r\) back into Equation 1:
\(1.30 = \varepsilon - 0.40(0.50)\)
\(\varepsilon = 1.30 + 0.20 = 1.50\text{ V}\)

(c) Precaution: Switch off the circuit between readings.
Explanation: This prevents the temperature of the cell and wires from increasing. A change in temperature would alter the internal resistance of the cell and the resistance of the circuit components, affecting the reliability of the measurements.

(a)
- Voltmeter connected in parallel across the cell (or across both cell and variable resistor) (1)
- Ammeter and variable resistor connected in series with the cell (1)

(b)
- Use of \(V = \varepsilon - Ir\) to set up two simultaneous equations (1)
- Subtracting equations to find \(r = 0.50\ \Omega\) (1)
- Substituting \(r\) to find \(\varepsilon = 1.50\text{ V}\) (1)
- Correct units for both answers (1)

(c)
- Disconnect/switch off circuit between readings (1)
- To prevent heating of the cell / circuit (1)
- Avoids changes in internal resistance during the experiment (0.75)

(a) A progressive wave is generated by the vibration generator and travels along the wire. This wave reflects at the fixed end. The reflected wave travels in the opposite direction and superposes (interferes) with the continuous incident wave. Where they interfere constructively (in phase), antinodes are formed; where they interfere destructively (out of phase by \(180^\circ\)), nodes are formed.

(b) For the first harmonic, the wavelength is \(\lambda = 2L = 2 \times 0.65\text{ m} = 1.30\text{ m}\).
Using the wave equation:
\(v = f\lambda\)
\(v = 45\text{ Hz} \times 1.30\text{ m} = 58.5\text{ m s}^{-1}\)

(c) The speed of a transverse wave on a string is given by \(v = \sqrt{\frac{T}{\mu}}\). Since the tension \(T\) is quadrupled, the speed \(v\) is doubled (\(v_{new} = 2v\)). Since the length of the string remains constant, the wavelength for the first harmonic remains the same (\(\lambda = 1.30\text{ m}\)). Therefore, the new frequency is:
\(f_{new} = \frac{v_{new}}{\lambda} = 2 \times 45\text{ Hz} = 90\text{ Hz}\).

(a)
- Progressive wave reflects at the fixed boundary/end (1)
- Incident and reflected waves travel in opposite directions and superpose/interfere (1)
- Node is formed where destructive interference occurs, and antinode where constructive interference occurs (1)

(b)
- Recall that for first harmonic \(\lambda = 2L = 1.30\text{ m}\) (1)
- Use of \(v = f\lambda\) (1)
- Correct value with unit: \(58.5\text{ m s}^{-1}\) or \(59\text{ m s}^{-1}\) (1)

(c)
- Identifies that wave speed is proportional to \(\sqrt{T}\) (1)
- Identifies that doubling the speed doubles the frequency (since wavelength is constant) (1)
- Correct final frequency: \(90\text{ Hz}\) (0.75)

(a) Using \(R = \frac{\rho L}{A}\):
\(R = \frac{1.7 \times 10^{-8}\ \Omega\text{ m} \times 2.5\text{ m}}{4.5 \times 10^{-7}\text{ m}^2} = 0.0944\ \Omega\)

(b) Using the drift velocity equation \(I = nAvq\), where \(q = e = 1.60 \times 10^{-19}\text{ C}\):
\(v = \frac{I}{nAe}\)
\(v = \frac{3.2}{8.5 \times 10^{28} \times 4.5 \times 10^{-7} \times 1.60 \times 10^{-19}}\)
\(v = \frac{3.2}{6120} = 5.23 \times 10^{-4}\text{ m s}^{-1}\)

(c) When the temperature increases, the positive lattice ions within the metallic lattice vibrate with greater amplitude. This increased vibration increases the frequency of collisions between the moving conduction electrons and the lattice ions. Consequently, the drift velocity is reduced for a given electric field, and the resistance increases.

(a)
- Use of \(R = \frac{\rho L}{A}\) (1)
- Correct answer: \(0.094\ \Omega\) or \(0.0944\ \Omega\) (1)

(b)
- Use of \(I = nAvq\) (1)
- Correct substitution of electron charge \(q = 1.60 \times 10^{-19}\text{ C}\) (1)
- Correct answer: \(5.2 \times 10^{-4}\text{ m s}^{-1}\) (or \(5.23 \times 10^{-4}\text{ m s}^{-1}\)) (1)

(c)
- Lattice ions vibrate with greater amplitude/kinetic energy (1)
- Rate/frequency of collisions between conduction electrons and lattice ions increases (1)
- This hinders the flow of electrons (1)
- Resulting in an increase in resistance (0.75)

(a) \(d = \frac{1}{N} = \frac{1}{500 \times 10^3\text{ m}^{-1}} = 2.00 \times 10^{-6}\text{ m}\).

(b) Using the diffraction grating equation \(d \sin\theta = n\lambda\):
For the second-order maximum (\(n = 2\)):
\(2.00 \times 10^{-6} \sin\theta = 2 \times 632.8 \times 10^{-9}\)
\(\sin\theta = \frac{1.2656 \times 10^{-6}}{2.00 \times 10^{-6}} = 0.6328\)
\ heta = \arcsin(0.6328) = 39.3^\circ\)

(c) The maximum theoretical angle of diffraction is \(\theta = 90^\circ\), which gives \(\sin\theta = 1\).
Using \(n = \frac{d \sin\theta}{\lambda}\):
\(n = \frac{2.00 \times 10^{-6} \times 1}{632.8 \times 10^{-9}} = 3.16\)
Since the order \(n\) must be an integer, the highest order maximum that can be observed is \(n = 3\).

(d) Since \(d \sin\theta = n\lambda\), a shorter wavelength \(\lambda\) means that \(\sin\theta\) (and thus the angle \(\theta\)) will be smaller for the first-order maximum (\(n = 1\)). The first-order maximum will therefore move closer to the central zero-order maximum.

(a)
- Shows \(d = \frac{1}{500 \times 10^3} = 2.00 \times 10^{-6}\text{ m}\) (1)

(b)
- Use of \(d \sin\theta = n\lambda\) with \(n=2\) (1)
- Correct calculation of \(\sin\theta = 0.6328\) (1)
- Correct angle: \(39.3^\circ\) (1)

(c)
- Uses \(\sin\theta \le 1\) (or \(\theta \le 90^\circ\)) (1)
- Calculates non-integer order: \(n = 3.16\) (1)
- Truncates to integer value: \(n = 3\) (1)

(d)
- Angle \(\theta\) decreases (maximum moves closer to center) (1)
- Because \(\sin\theta\) is directly proportional to \(\lambda\) (for constant \(d\) and \(n\)) (0.75)

(a) Using the potential divider formula:
\(V_{\text{out}} = V_{\text{in}} \times \left(\frac{R_{\text{LDR}}}{R + R_{\text{LDR}}}\right)\)
\(V_{\text{out}} = 9.0\text{ V} \times \left(\frac{300}{1200 + 300}\right) = 9.0 \times \left(\frac{300}{1500}\right) = 1.8\text{ V}\)

(b) In darkness, \(R_{\text{LDR}} = 8000\ \Omega\):
\(V_{\text{out}} = 9.0\text{ V} \times \left(\frac{8000}{1200 + 8000}\right) = 9.0 \times \left(\frac{8000}{9200}\right) = 7.83\text{ V}\)

(c) The potential divider outputs a higher voltage as light levels drop (in darkness). To adapt the system to turn on a heater when the temperature drops, the LDR should be replaced by a negative temperature coefficient (NTC) thermistor. As temperature drops, the resistance of the NTC thermistor increases, which increases the voltage output across it. This high voltage output can then trigger a relay/switch to activate the heater.

(a)
- Use of potential divider equation (1)
- Correct substitution: \(9.0 \times \frac{300}{1500}\) (1)
- Correct answer: \(1.8\text{ V}\) (1)

(b)
- Converts \(8.0\text{ k}\Omega\) to \(8000\ \Omega\) (1)
- Correct substitution: \(9.0 \times \frac{8000}{9200}\) (1)
- Correct answer: \(7.8\text{ V}\) or \(7.83\text{ V}\) (1)

(c)
- Replace LDR with a thermistor (1)
- Resistance of NTC thermistor increases as temperature decreases (1)
- Output voltage across thermistor increases when cold (which triggers heater) (0.75)

(a) \(\Phi = 4.3\text{ eV} \times 1.60 \times 10^{-19}\text{ J eV}^{-1} = 6.88 \times 10^{-19}\text{ J}\), which is approximately \(6.9 \times 10^{-19}\text{ J}\).

(b) Using Einstein's photoelectric equation: \(h f = \Phi + E_{k,\text{max}}\)
\(E_{k,\text{max}} = hf - \Phi\)
\(E_{k,\text{max}} = (6.63 \times 10^{-34}\text{ J s} \times 1.25 \times 10^{15}\text{ Hz}) - 6.88 \times 10^{-19}\text{ J}\)
\(E_{k,\text{max}} = 8.2875 \times 10^{-19}\text{ J} - 6.88 \times 10^{-19}\text{ J} = 1.41 \times 10^{-19}\text{ J}\)

(c) Using \(E_{k,\text{max}} = \frac{1}{2}mv^2\):
\(v = \sqrt{\frac{2 E_{k,\text{max}}}{m}}\)
\(v = \sqrt{\frac{2 \times 1.41 \times 10^{-19}\text{ J}}{9.11 \times 10^{-31}\text{ kg}}}\)
\(v = \sqrt{3.095 \times 10^{11}} = 5.56 \times 10^5\text{ m s}^{-1}\)

(d) Light behaves as packets of energy called photons. One photon interacts with exactly one electron on the metal surface. If the energy of the photon (\(E = hf\)) is less than the work function (\(\Phi\)), the electron cannot escape the surface. Increasing the intensity increases the number of photons incident per second, but does not increase the energy of individual photons, so still no electrons are released.

(a)
- Shows multiplication by \(1.60 \times 10^{-19}\) to yield \(6.88 \times 10^{-19}\text{ J}\) (1)

(b)
- Recall of \(E = hf\) or calculation of photon energy: \(8.29 \times 10^{-19}\text{ J}\) (1)
- Einstein photoelectric equation: \(E_{k,\text{max}} = hf - \Phi\) (1)
- Correct calculation: \(1.41 \times 10^{-19}\text{ J}\) (accept \(1.4 \times 10^{-19}\text{ J}\)) (1)

(c)
- Equating \(E_{k,\text{max}}\) to \(\frac{1}{2}mv^2\) (1)
- Correct rearrangement for \(v\) (1)
- Correct answer: \(5.56 \times 10^5\text{ m s}^{-1}\) (or \(5.6 \times 10^5\text{ m s}^{-1}\)) (1)

(d)
- Photons have one-to-one interaction with electrons (1)
- If single photon energy is less than work function, no emission occurs; intensity only increases rate of photons, not energy per photon (0.75)

(a) Plane-polarized light has oscillations of the electric field vector restricted to a single plane, which contains the direction of wave propagation.

(b) (i) Malus's law: \(I_2 = I_1 \cos^2\theta\)

(ii) Given \(I_2 = 0.25 I_1\):
\(0.25 = \cos^2\theta\)
\(\cos\theta = \sqrt{0.25} = 0.50\)
\(\theta = \arccos(0.50) = 60^\circ\)

(c) Light reflected off horizontal surfaces such as water is partially horizontally polarised. Polarising sunglasses have a vertical transmission axis. Since the horizontally polarised glare is perpendicular to the sunglasses' transmission axis, the glare is absorbed/blocked, reducing the intensity of light entering the eyes.

(a)
- Oscillations are in a single plane (1)
- Which is perpendicular to the direction of propagation (1)

(b) (i)
- Correct formula: \(I_2 = I_1 \cos^2\theta\) (1)

(b) (ii)
- Sets up \(\cos^2\theta = 0.25\) (1)
- Determines \(\cos\theta = 0.50\) (1)
- Obtains \(\theta = 60^\circ\) (1)

(c)
- Reflected glare is horizontally polarised (1)
- Sunglasses have a vertical polarization axis (1)
- Hence, the horizontally polarised reflected light is blocked/absorbed (0.75)

(a) Using the formula for the critical angle:
\(\sin\theta_c = \frac{n_2}{n_1} = \frac{1.45}{1.52}\)
\(\sin\theta_c = 0.9539\)
\(\theta_c = \arcsin(0.9539) = 72.5^\circ\)

(b) (i) Since the angle of incidence \(68^\circ\) is less than the critical angle \(72.5^\circ\), the ray will undergo refraction. It will bend away from the normal as it enters the cladding and leak out of the core.
(ii) Since the angle of incidence \(75^\circ\) is greater than the critical angle \(72.5^\circ\), the ray will undergo total internal reflection (TIR) and remain entirely within the core.

(c) Advantages:
1. Cladding protects the core surface from scratches or dirt, which could otherwise allow light to leak out / scatter.
2. It prevents 'crosstalk' / transfer of light between adjacent optical fibres in a bundle.
3. It ensures a constant difference in refractive indices to maintain TIR.

(a)
- Use of \(\sin\theta_c = \frac{n_2}{n_1}\) (1)
- Correct substitution: \(\frac{1.45}{1.52}\) (1)
- Correct critical angle: \(72.5^\circ\) (1)

(b)
- For \(68^\circ\): Refracts into cladding because angle of incidence is less than the critical angle (1.5)
- For \(75^\circ\): Undergoes total internal reflection because angle of incidence is greater than the critical angle (1.5)

(c)
- Prevents light leaking out/damage to core boundary (1)
- Prevents crosstalk between adjacent fibres (1)
- Mention of maintaining TIR conditions (0.75)

Before the collision, the total kinetic energy is \(E_{\text{initial}} = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2\). By conservation of momentum, the components of the initial momentum along two perpendicular axes are \(mv\) and \(mv\). The magnitude of the total initial momentum is \(p = \sqrt{(mv)^2 + (mv)^2} = \sqrt{2}mv\). After the collision, the combined particle of mass \(2m\) moves with speed \(V_f\) such that \(2mV_f = \sqrt{2}mv\), which gives \(V_f = \frac{v}{\sqrt{2}}\). The final kinetic energy is \(E_{\text{final}} = \frac{1}{2}(2m)V_f^2 = m\left(\frac{v}{\sqrt{2}}\right)^2 = \frac{1}{2}mv^2\). The kinetic energy lost is \(E_{\text{initial}} - E_{\text{final}} = mv^2 - \frac{1}{2}mv^2 = \frac{1}{2}mv^2\).

1 mark for the correct answer B. Reject other options. Method involves finding the total initial kinetic energy, applying conservation of momentum to find the final speed of the combined mass, calculating the final kinetic energy, and subtracting to find the energy lost.

Let \(\theta\) be the angle that the string makes with the vertical. The vertical forces on the object are in equilibrium, so \(T \cos\theta = mg \implies T = \frac{mg}{\cos\theta}\). From the geometry of the cone, the height is \(h = \sqrt{L^2 - r^2}\), so \(\cos\theta = \frac{h}{L} = \frac{\sqrt{L^2 - r^2}}{L}\). Substituting this into the tension equation gives \(T = \frac{mgL}{\sqrt{L^2 - r^2}}\).

1 mark for the correct answer C. Reject other options. Method involves resolving forces vertically and using the geometric relationship for \(\cos\theta\) in terms of \(L\) and \(r\).

The electric field strength between the plates is \(E = \frac{V}{d}\). The force on the electron is \(F = eE = \frac{eV}{d}\). The acceleration of the electron is \(a = \frac{F}{m} = \frac{eV}{md}\). Using the equation of motion \(s = ut + \frac{1}{2}at^2\) with \(u = 0\) and \(s = d\), we get \(d = \frac{1}{2} a t^2 \implies t = \sqrt{\frac{2d}{a}} = \sqrt{\frac{2d^2 m}{eV}} = d \sqrt{\frac{2m}{eV}}\).

1 mark for the correct answer B. Reject other options. Method involves expressing the acceleration of the electron in terms of \(V\), \(d\), \(e\), and \(m\), and then applying equations of uniform acceleration to solve for time.

The energy \(E\) stored in a capacitor at any time is given by \(E = \frac{1}{2}CV^2\), where the potential difference \(V\) decays over time according to \(V = V_0 e^{-t/RC}\). Substituting this into the energy formula gives \(E = \frac{1}{2}C(V_0 e^{-t/RC})^2 = \frac{1}{2}CV_0^2 e^{-2t/RC} = E_0 e^{-2t/RC}\). At \(t = 2RC\), the fraction of energy remaining is \(\frac{E}{E_0} = e^{-2(2RC)/RC} = e^{-4}\).

1 mark for the correct answer B. Reject other options. Method involves relating stored energy to the square of potential difference and applying the exponential decay formula for voltage at \(t = 2RC\).

Initially, the magnetic flux linkage of the coil is \(\Phi_1 = NBA\). After rotating the coil by \(180^\circ\), its orientation is reversed relative to the field, so the final magnetic flux linkage is \(\Phi_2 = -NBA\). The change in magnetic flux linkage is \(\Delta\Phi = \Phi_1 - \Phi_2 = NBA - (-NBA) = 2NBA\). According to Faraday's law of electromagnetic induction, the average induced e.m.f. is \(\varepsilon = \frac{\Delta\Phi}{\Delta t} = \frac{2NBA}{\Delta t}\).

1 mark for the correct answer C. Reject other options. Method involves calculating the initial and final flux linkages, taking their difference to find the change in flux linkage, and dividing by \(\Delta t\) as per Faraday's law.

The total mass-energy of each particle is the sum of its rest energy and its kinetic energy, which is \(E_{\text{total}} = m_e c^2 + E_k\). Since there are two identical particles colliding head-on, the total initial energy is \(2(m_e c^2 + E_k)\). Due to conservation of momentum, the two produced photons must have equal energy. Therefore, each photon carries away half of the total energy: \(E_{\text{photon}} = h f = m_e c^2 + E_k\). Solving for frequency gives \(f = \frac{m_e c^2 + E_k}{h}\).

1 mark for the correct answer D. Reject other options. Method involves expressing the total energy of a single colliding particle (rest energy + kinetic energy) and equating it to the energy of one of the two identical photons produced.

An up quark \(\text{u}\) has charge \(+\frac{2}{3}e\) and baryon number \(+\frac{1}{3}\). A down quark \(\text{d}\) has charge \(-\frac{1}{3}e\) and baryon number \(+\frac{1}{3}\), which means an anti-down quark \(\bar{\text{d}}\) has charge \(+\frac{1}{3}e\) and baryon number \(-\frac{1}{3}\). The total charge of the \(\text{u}\bar{\text{d}}\) meson is \(Q = +\frac{2}{3}e + \frac{1}{3}e = +1e\). The total baryon number is \(B = +\frac{1}{3} - \frac{1}{3} = 0\).

1 mark for the correct answer A. Reject other options. Method involves summing the individual quark and anti-quark charges and baryon numbers.

The centripetal force is provided by the magnetic force: \(\frac{mv^2}{R} = qvB \implies R = \frac{mv}{qB} = \frac{p}{qB}\), where \(p\) is the momentum of the particle. The kinetic energy \(E_k\) is related to momentum by \(E_k = \frac{p^2}{2m} \implies p = \sqrt{2mE_k}\). Thus, the radius is \(R = \frac{\sqrt{2mE_k}}{qB}\). Since \(R \propto \sqrt{E_k}\), doubling the kinetic energy increases the radius by a factor of \(\sqrt{2}\).

1 mark for the correct answer B. Reject other options. Method involves establishing the relationship between radius and momentum, relating momentum to kinetic energy, and determining the proportionality factor.

The radius r of a circular path for a charged particle in a magnetic field is given by: r = mv / (qB). Since kinetic energy E_k = 0.5 * m * v^2, we can express momentum p = mv as p = sqrt(2 * m * E_k). Substituting this into the radius formula gives: r = sqrt(2 * m * E_k) / (q * B). For the proton (mass m_p = m, charge q_p = e): r_p = sqrt(2 * m * E_k) / (e * B). For the alpha particle (mass m_alpha = 4m, charge q_alpha = 2e): r_alpha = sqrt(2 * 4m * E_k) / (2e * B) = 2 * sqrt(2 * m * E_k) / (2e * B) = sqrt(2 * m * E_k) / (e * B). Therefore, r_p = r_alpha.

1 mark for selecting option b. All other options are incorrect.

Let v_1 be the speed of the particle moving at 30 degrees and v_2 be the speed of the particle moving at 60 degrees. By conservation of momentum perpendicular to the initial direction: 0 = m * v_1 * sin(30) - m * v_2 * sin(60) which simplifies to v_1 * sin(30) = v_2 * sin(60), leading to v_1 * 0.5 = v_2 * (sqrt(3)/2), so v_1 = sqrt(3) * v_2. By conservation of momentum parallel to the initial direction: m * v = m * v_1 * cos(30) + m * v_2 * cos(60) which simplifies to v = v_1 * cos(30) + v_2 * cos(60). Substituting v_1 = sqrt(3) * v_2 into this equation: v = (sqrt(3) * v_2) * (sqrt(3)/2) + v_2 * 0.5 = 1.5 * v_2 + 0.5 * v_2 = 2.0 * v_2. Thus, v_2 = 0.50 * v.

1 mark for selecting option b. All other options are incorrect.

(a) When there is no sideways frictional force, the components of the normal contact force \(R\) provide the necessary centripetal force and balance the weight of the car:
Horizontal component: \(R \sin\theta = \frac{mv^2}{r}\)
Vertical component: \(R \cos\theta = mg\)

Dividing the two equations gives:
\(\tan\theta = \frac{v^2}{rg}\)

Substitute the given values:
\(\tan\theta = \frac{24^2}{85 \times 9.81} = \frac{576}{833.85} \approx 0.6908\)
\(\theta = \arctan(0.6908) \approx 34.6^\circ \approx 35^\circ\)

(b) At speeds greater than \(24\text{ m s}^{-1}\), the required centripetal force \(\frac{mv^2}{r}\) increases and exceeds the horizontal component of the normal contact force \(R \sin\theta\).
To prevent the car from sliding up the track, a sideways frictional force must act downwards along the slope.
This frictional force has a horizontal component pointing towards the center of the circular path, which increases the net centripetal force, and a vertical component pointing downwards, which increases the normal contact force and allows the car to maintain its circular path without slipping.

(a)
- Equating horizontal component of normal contact force to centripetal force: \(R \sin\theta = mv^2/r\) (1)
- Equating vertical component to weight: \(R \cos\theta = mg\) (1)
- Deducing \(\tan\theta = v^2/rg\) (1)
- Correct calculation of angle \(\theta = 35^\circ\) (accept \(34.6^\circ\)) (1)

(b)
- States that higher speed requires a larger centripetal force (1)
- Identifies that the frictional force must act down the slope (1)
- Explains that the horizontal component of this frictional force acts towards the center of the curve, providing the additional centripetal force (1)
- Mentions that the limit to safe speed depends on the maximum available static friction (1)

(a) The potential difference during capacitor discharge is given by:
\(V = V_0 e^{-\frac{t}{RC}}\)

Rearranging for \(R\):
\(\frac{V}{V_0} = e^{-\frac{t}{RC}}\)
\(\ln\left(\frac{V}{V_0}\right) = -\frac{t}{RC}\)
\(R = -\frac{t}{C \ln(V/V_0)}\)

Substitute the values:
\(R = -\frac{8.5}{470 \times 10^{-6} \times \ln(3.0/12.0)}\)
\(R = -\frac{8.5}{470 \times 10^{-6} \times \ln(0.25)}\)
\(R = -\frac{8.5}{470 \times 10^{-6} \times (-1.3863)} \approx 13045\ \Omega\)
\(R \approx 1.3 \times 10^4\ \Omega\) (or \(13\text{ k}\Omega\))

(b) The energy delivered to the resistor is equal to the decrease in the energy stored by the capacitor:
\(\Delta E = E_{\text{initial}} - E_{\text{final}} = \frac{1}{2} C V_0^2 - \frac{1}{2} C V^2\)
\(\Delta E = \frac{1}{2} C (V_0^2 - V^2)\)

Substitute the values:
\(\Delta E = 0.5 \times 470 \times 10^{-6} \times (12.0^2 - 3.0^2)\)
\(\Delta E = 2.35 \times 10^{-4} \times (144 - 9)\)
\(\Delta E = 2.35 \times 10^{-4} \times 135 = 0.0317\text{ J}\)
\(\Delta E \approx 3.2 \times 10^{-2}\text{ J}\)

(a)
- Recall of \(V = V_0 e^{-t/RC}\) (1)
- Correct use of logarithms to make \(R\) or \(RC\) the subject (1)
- Substitution of \(C = 470 \times 10^{-6}\text{ F}\) and values (1)
- Correct calculation of \(R = 1.3 \times 10^4\ \Omega\) (accept \(1.3 \times 10^4\ \Omega\) to \(1.31 \times 10^4\ \Omega\)) (1)

(b)
- Recall of \(E = \frac{1}{2}CV^2\) (1)
- Calculation of initial energy \(0.0338\text{ J}\) or final energy \(0.0021\text{ J}\) (1)
- Subtracting final energy from initial energy (1)
- Correct calculation of energy change \(3.2 \times 10^{-2}\text{ J}\) (accept \(0.032\text{ J}\) or \(0.0317\text{ J}\)) (1)

(a) To find the quark composition of \(\Xi^0\), we apply conservation laws:
1. Charge (\(Q\)) conservation:
\(Q(\text{K}^-) + Q(\text{p}) = Q(\Xi^0) + Q(\text{K}^0)\)
\((-1) + (+1) = Q(\Xi^0) + 0 \implies Q(\Xi^0) = 0\)

2. Baryon number (\(B\)) conservation:
\(B(\text{K}^-) + B(\text{p}) = B(\Xi^0) + B(\text{K}^0)\)
\(0 + (+1) = B(\Xi^0) + 0 \implies B(\Xi^0) = +1\) (So \(\Xi^0\) is a baryon composed of 3 quarks).

3. Strangeness (\(S\)) conservation:
\(S(\text{K}^-) + S(\text{p}) = S(\Xi^0) + S(\text{K}^0)\)
For \(\text{K}^-\), \(S = -1\). For \(\text{p}\), \(S = 0\). For \(\text{K}^0\), \(S = +1\).
\((-1) + 0 = S(\Xi^0) + (+1) \implies S(\Xi^0) = -2\).

Since \(\Xi^0\) has \(S = -2\), it must contain two strange (\(s\)) quarks (each has \(S = -1\), charge \(-1/3\)).
Let the third quark be \(q\).
Total charge \(Q = 2(-1/3) + Q_q = 0 \implies Q_q = +2/3\).
The quark with charge \(+2/3\) and strangeness \(0\) is the up (\(u\)) quark.
Therefore, the quark composition of \(\Xi^0\) is \(uss\).

(b) The strong interaction is responsible for this reaction because strangeness is conserved in this process (initial strangeness \(S = -1\); final strangeness \(S = -2 + 1 = -1\)). Weak interactions are the only ones that can violate strangeness conservation (by \(\pm 1\)). Since strangeness is conserved, the reaction proceeds via the strong force which has a much higher probability and occurs over a much shorter characteristic timescale (typically \(10^{-23}\text{ s}\) to \(10^{-24}\text{ s}\)) compared to weak interactions (typically \(10^{-10}\text{ s}\) to \(10^{-8}\text{ s}\)).

(a)
- Conserves charge to show \(Q(\Xi^0) = 0\) AND baryon number to show \(B(\Xi^0) = +1\) (1)
- Conserves strangeness to show \(S(\Xi^0) = -2\) (1)
- Identifies that \(S = -2\) requires two strange (\(s\)) quarks (1)
- Deduces that the third quark must be an up (\(u\)) quark to yield a neutral baryon, giving composition \(uss\) (1)

(b)
- States that strangeness is conserved in the reaction (1)
- Explains that the strong interaction must be responsible because it conserves strangeness, whereas weak interactions can change strangeness (1)
- States that the strong interaction occurs on a much shorter timescale than the weak interaction (1)
- Quantifies the typical timescale for strong interactions as approximately \(10^{-23}\text{ s}\) (accept range \(10^{-24}\) to \(10^{-20}\text{ s}\)) (1)

(a) The kinetic energy gained by the ion of charge \(q\) when accelerated through a potential difference \(V\) is equal to the work done on it:
\(E_k = qV = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2qV}{m}}\)

In the magnetic field, the magnetic force acts as the centripetal force:
\(F_B = F_c \implies qvB = \frac{mv^2}{r} \implies r = \frac{mv}{qB}\)

Substituting \(v\) into the equation for \(r\):
\(r = \frac{m}{qB}\sqrt{\frac{2qV}{m}}\)
\(r = \sqrt{\frac{m^2}{q^2B^2} \cdot \frac{2qV}{m}} = \sqrt{\frac{2mV}{qB^2}}\)

(b) For \(^{24}\text{Mg}^{2+}\):
\(m = 3.98 \times 10^{-26}\text{ kg}\)
\(q = 2e = 2 \times 1.60 \times 10^{-19}\text{ C} = 3.20 \times 10^{-19}\text{ C}\)
\(V = 2.50\text{ kV} = 2500\text{ V}\)
\(B = 0.380\text{ T}\)

Substitute these values into the derived equation:
\(r = \sqrt{\frac{2 \times (3.98 \times 10^{-26}) \times 2500}{3.20 \times 10^{-19} \times 0.380^2}}\)
\(r = \sqrt{\frac{1.99 \times 10^{-22}}{3.20 \times 10^{-19} \times 0.1444}}\)
\(r = \sqrt{\frac{1.99 \times 10^{-22}}{4.6208 \times 10^{-20}}}\)
\(r = \sqrt{4.3066 \times 10^{-3}} \approx 0.06562\text{ m}\)

\(r \approx 0.0656\text{ m}\) (or \(6.56\text{ cm}\))

(a)
- Equating kinetic energy to electrical work done: \(qV = \frac{1}{2}mv^2\) (1)
- Equating magnetic force to centripetal force: \(qvB = mv^2/r\) (1)
- Clear algebraic steps to substitute and show the given expression (1)

(b)
- Calculation of the charge of the ion: \(q = 2 \times 1.60 \times 10^{-19}\text{ C} = 3.20 \times 10^{-19}\text{ C}\) (1)
- Substitution of all values correctly into the formula (1)
- Calculation of intermediate values (e.g., numerator \(1.99 \times 10^{-22}\) or denominator \(4.62 \times 10^{-20}\)) (1)
- Correct calculation of numerical value: \(0.0656\text{ m}\) (accept \(0.066\text{ m}\)) (1)
- Correct units (\ ext{m}\ or \\text{cm}\) and appropriate 3 significant figures (1)

(a) The thrust force \(F\) produced by the ejected gases is equal to the rate of change of momentum of the gas:
\(F = v_{\text{ex}} \frac{\Delta m}{\Delta t}\)
Where \(v_{\text{ex}} = 3200\text{ m s}^{-1}\) and \(\frac{\Delta m}{\Delta t} = 2.40\text{ kg s}^{-1}\).

\(F = 3200 \times 2.40 = 7680\text{ N}\)

Using Newton's second law, the initial acceleration \(a\) of the spacecraft of mass \(M = 3500\text{ kg}\) is:
\(a = \frac{F}{M} = \frac{7680}{3500} \approx 2.194\text{ m s}^{-2}\)

\(a \approx 2.19\text{ m s}^{-2}\)

(b) According to the principle of conservation of linear momentum, the total momentum of the closed system (spacecraft + fuel) must remain constant. Ejecting exhaust gases backwards creates backward momentum. To conserve total momentum, the spacecraft must gain an equal and opposite amount of forward momentum, resulting in a forward thrust force.
As the thrusters continue to fire, the mass of the spacecraft decreases because fuel is being consumed and ejected. Since the rate of mass ejection and exhaust velocity remain constant, the thrust force \(F\) remains constant. However, since \(a = F/M\) and the mass \(M\) is decreasing, the acceleration of the spacecraft will increase over time.

(a)
- Use of \(F = v (\Delta m/\Delta t)\) to calculate the thrust force (1)
- Correct calculation of force \(F = 7680\text{ N}\) (1)
- Use of \(a = F/M\) (1)
- Correct calculation of acceleration \(a = 2.19\text{ m s}^{-2}\) (accept \(2.2\text{ m s}^{-2}\)) (1)

(b)
- States that total momentum of the system is conserved (1)
- Explains that backward momentum of the gas must be balanced by an equal and opposite forward momentum of the spacecraft (1)
- Identifies that the thrust force remains constant because \(dm/dt\) and \(v\) are constant (1)
- Concludes that acceleration increases because the total mass of the spacecraft decreases (1)

(a) Area of the coil, \(A = 0.050 \times 0.080 = 4.0 \times 10^{-3}\text{ m}^2\).

Initial flux linkage when the plane is perpendicular to \(B\) (angle with normal is \(0^\circ\)):
\(N\Phi_1 = N B A = 150 \times 0.12 \times (4.0 \times 10^{-3}) = 0.072\text{ Wb-turns}\)

Final flux linkage when the coil is rotated by \(90^\circ\) (plane is parallel to \(B\), angle with normal is \(90^\circ\)):
\(N\Phi_2 = 0\)

Change in magnetic flux linkage:
\(\Delta(N\Phi) = 0.072\text{ Wb-turns}\)

Using Faraday's Law, the average induced e.m.f. \(E\) is:
\(E = \left| \frac{\Delta(N\Phi)}{\Delta t} \right| = \frac{0.072}{0.060} = 1.2\text{ V}\)

(b) According to Faraday's Law, the induced e.m.f. is proportional to the rate of change of magnetic flux linkage (\(E \propto d(N\Phi)/dt\)). Increasing the frequency of rotation reduces the time \(\Delta t\) taken for the same change in flux linkage, thereby increasing the rate of change of flux linkage and increasing the maximum induced e.m.f.
According to Lenz's law, the direction of the induced e.m.f. (and current) is such that it opposes the change in magnetic flux that produces it. Since the flux is changing faster, the induced current will still flow in a direction to create a magnetic field that opposes this faster change, reversing its direction every half cycle as the flux increases and decreases.

(a)
- Calculation of coil area \(A = 4.0 \times 10^{-3}\text{ m}^2\) (1)
- Calculation of initial flux linkage \(N\Phi = 0.072\text{ Wb-turns}\) (1)
- Recall of Faraday's law \(E = \Delta(N\Phi)/\Delta t\) (1)
- Correct calculation of e.m.f. \(E = 1.2\text{ V}\) (1)

(b)
- States that higher frequency of rotation means flux change occurs in a shorter time (1)
- Cites Faraday's law to explain why a higher rate of change of flux results in a larger peak/maximum e.m.f. (1)
- Cites Lenz's law to state that the induced current/e.m.f. always opposes the change in flux (1)
- Explains that the direction of the induced current changes direction/alternates twice per cycle (1)

(a) In a linac, protons are accelerated only in the gaps between the drift tubes by the electric field. Inside the drift tubes, there is no electric field, so the protons travel at a constant velocity (they 'drift').
To ensure that the protons are always accelerated when they reach each gap, the alternating potential difference must reverse its polarity exactly when the protons are inside a drift tube. This means the time spent inside each drift tube must be constant and equal to half the period of the alternating voltage (\(T/2\)).
Since the protons accelerate in each gap, their velocity increases as they move along the linac. Because the time spent in each tube must remain constant (\(t = T/2\)), the length of the tubes \(L\) must increase along the path according to \(L = v \times t\) where \(v\) is the speed in that tube.

(b) The frequency of the alternating potential difference is \(f = 25.0\text{ MHz} = 2.50 \times 10^7\text{ Hz}\).
The period \(T\) is:
\(T = \frac{1}{f} = \frac{1}{2.50 \times 10^7\text{ Hz}} = 4.00 \times 10^{-8}\text{ s}\)

The time spent in one drift tube is:
\(t = \frac{T}{2} = 2.00 \times 10^{-8}\text{ s}\)

Given the velocity of the protons \(v = 3.50 \times 10^7\text{ m s}^{-1}\), the length of the drift tube is:
\(L = v \times t = (3.50 \times 10^7\text{ m s}^{-1}) \times (2.00 \times 10^{-8}\text{ s}) = 0.700\text{ m}\)

(a)
- Protons accelerate only in the gaps / travel at constant speed inside tubes (1)
- The AC supply has a constant frequency/period (1)
- Protons must spend a constant time interval (equal to half-period, \(T/2\)) inside each tube (1)
- Since \(L = v t\), as velocity \(v\) increases, the tube length \(L\) must increase to keep time \(t\) constant (1)

(b)
- Recall of \(T = 1/f\) and calculation of period \(T = 4.00 \times 10^{-8}\text{ s}\) (1)
- Time inside the tube identified as \(t = T/2 = 2.00 \times 10^{-8}\text{ s}\) (1)
- Use of \(L = v t\) (1)
- Correct calculation of length \(L = 0.700\text{ m}\) (1)

(a) Let the midpoint be at distance \(r = 0.075\text{ m}\) from both charges.
The electric field strength \(E_1\) due to \(Q_1\) acts away from the positive charge (towards \(Q_2\)):
\(E_1 = \frac{k |Q_1|}{r^2} = \frac{8.99 \times 10^9 \times 4.5 \times 10^{-6}}{0.075^2} = 7.192 \times 10^6\text{ V m}^{-1}\)

The electric field strength \(E_2\) due to \(Q_2\) acts towards the negative charge (towards \(Q_2\)):
\(E_2 = \frac{k |Q_2|}{r^2} = \frac{8.99 \times 10^9 \times 9.0 \times 10^{-6}}{0.075^2} = 1.4384 \times 10^7\text{ V m}^{-1}\)

Since both fields point in the same direction (from \(Q_1\) to \(Q_2\)), we add them:
\(E = E_1 + E_2 = 7.192 \times 10^6 + 1.4384 \times 10^7 \approx 2.16 \times 10^7\text{ V m}^{-1}\)
Direction: Towards the negative charge (\(Q_2\)).

(b) Let the point where potential \(V = 0\) be at a distance \(x\) from \(Q_1\) between the charges. The distance from \(Q_2\) is \(0.15 - x\).
\(V = \frac{k Q_1}{x} + \frac{k Q_2}{0.15 - x} = 0\)
\(\frac{k (4.5 \times 10^{-6})}{x} = -\frac{k (-9.0 \times 10^{-6})}{0.15 - x}\)
\(\frac{4.5}{x} = \frac{9.0}{0.15 - x}\)
\(4.5(0.15 - x) = 9.0x\)
\(0.15 - x = 2x\)
\(3x = 0.15 \implies x = 0.050\text{ m}\)

(Alternatively, looking outside to the left of \(Q_1\) at distance \(y\):
\(\frac{4.5}{y} = \frac{9.0}{y + 0.15} \implies y + 0.15 = 2y \implies y = 0.15\text{ m}\) to the left of \(Q_1\). Either position is correct.)

(a)
- Recall of electric field formula \(E = kQ/r^2\) (1)
- Correct calculation of both individual field strengths at the midpoint (1)
- States that both fields point in the same direction and adds them (1)
- Correct calculation of total field magnitude \(2.2 \times 10^7\text{ V m}^{-1}\) (or \(2.16 \times 10^7\text{ V m}^{-1}\)) AND specifies direction towards the negative charge (1)

(b)
- Recall of potential formula \(V = kQ/r\) (1)
- Sets up equation for total potential \(V_1 + V_2 = 0\) (1)
- Algebraic solution steps (1)
- Obtains either \(0.050\text{ m}\) from \(Q_1\) (between them) OR \(0.15\text{ m}\) from \(Q_1\) (on the side away from \(Q_2\)) with appropriate description of position (1)

(a) For a collision to be perfectly elastic:
1. Total momentum of the system is conserved.
2. Total kinetic energy of the system is conserved.

(b) Let the initial direction of the proton be along the positive x-axis.
Initial momentum:
\(p_{i,x} = m_p u_p = (1.67 \times 10^{-27}\text{ kg}) \times (4.50 \times 10^6\text{ m s}^{-1}) = 7.515 \times 10^{-21}\text{ kg m s}^{-1}\)
\(p_{i,y} = 0\)

After the collision:
Proton's final momentum components:
\(p_{p,x} = m_p v_p \cos(60.0^\circ) = (1.67 \times 10^{-27}\text{ kg}) \times (2.00 \times 10^6\text{ m s}^{-1}) \times 0.500 = 1.670 \times 10^{-21}\text{ kg m s}^{-1}\)
\(p_{p,y} = m_p v_p \sin(60.0^\circ) = (1.67 \times 10^{-27}\text{ kg}) \times (2.00 \times 10^6\text{ m s}^{-1}) \times 0.8660 = 2.892 \times 10^{-21}\text{ kg m s}^{-1}\)

By conservation of momentum in the x-direction:
\(p_{i,x} = p_{p,x} + p_{\alpha,x}\)
\(7.515 \times 10^{-21} = 1.670 \times 10^{-21} + p_{\alpha,x}\)
\(p_{\alpha,x} = 5.845 \times 10^{-21}\text{ kg m s}^{-1}\)

By conservation of momentum in the y-direction:
\(0 = p_{p,y} + p_{\alpha,y}\)
\(p_{\alpha,y} = -2.892 \times 10^{-21}\text{ kg m s}^{-1}\)

Calculate components of velocity of helium-4 nucleus:
\(v_{\alpha,x} = \frac{p_{\alpha,x}}{m_\alpha} = \frac{5.845 \times 10^{-21}\text{ kg m s}^{-1}}{6.64 \times 10^{-27}\text{ kg}} = 8.803 \times 10^5\text{ m s}^{-1}\)
\(v_{\alpha,y} = \frac{p_{\alpha,y}}{m_\alpha} = \frac{-2.892 \times 10^{-21}\text{ kg m s}^{-1}}{6.64 \times 10^{-27}\text{ kg}} = -4.355 \times 10^5\text{ m s}^{-1}\)

Magnitude of velocity:
\(v_{\alpha} = \sqrt{v_{\alpha,x}^2 + v_{\alpha,y}^2} = \sqrt{(8.803 \times 10^5)^2 + (-4.355 \times 10^5)^2} = 9.82 \times 10^5\text{ m s}^{-1}\)

Direction of velocity:
\(\theta = \tan^{-1}\left(\frac{|v_{\alpha,y}|}{v_{\alpha,x}}\right) = \tan^{-1}\left(\frac{4.355 \times 10^5}{8.803 \times 10^5}\right) = 26.3^\circ\text{ or }26.4^\circ\) below the proton's initial direction of travel.

(a) (Total 2 marks):
- Conservation of momentum stated (1 mark)
- Conservation of kinetic energy stated (1 mark)

(b) (Total 6 marks):
- Use of \(p = mv\) to calculate initial momentum of proton AND final momentum components of proton (1 mark)
- Applies conservation of momentum in the horizontal direction to find \(p_{\alpha,x} = 5.85 \times 10^{-21}\text{ kg m s}^{-1}\) (or velocity component \(8.80 \times 10^5\text{ m s}^{-1}\)) (1 mark)
- Applies conservation of momentum in the vertical direction to find \(p_{\alpha,y} = -2.89 \times 10^{-21}\text{ kg m s}^{-1}\) (or velocity component \(-4.36 \times 10^5\text{ m s}^{-1}\)) (1 mark)
- Uses Pythagoras theorem to combine components to find magnitude of momentum or velocity (1 mark)
- Correct calculation of final velocity magnitude \(9.82 \times 10^5\text{ m s}^{-1}\) (accept range \(9.8 \times 10^5\) to \(9.9 \times 10^5\)) (1 mark)
- Correct direction stated as \(26.3^\circ\) or \(26.4^\circ\) below initial line of motion (1 mark)

(a)
- Draw parallel, equally spaced straight vertical lines going from the top plate to the bottom plate (1 mark).
- Add arrows pointing downwards (from positive to negative) (1 mark).
- Draw curved field lines bowing outwards at the left and right ends of the plates (fringing effect), and label the top plate (+) and bottom plate (-) (1 mark).

(b)
1. Calculate the electric field strength \(E\) between the plates:
\(E = \frac{V}{d} = \frac{150\text{ V}}{0.020\text{ m}} = 7500\text{ V m}^{-1}\)

2. Calculate the electric force \(F\) on the electron:
\(F = eE = (1.60 \times 10^{-19}\text{ C}) \times 7500\text{ V m}^{-1} = 1.20 \times 10^{-15}\text{ N}\)

3. Calculate the vertical acceleration \(a_y\):
\(a_y = \frac{F}{m_e} = \frac{1.20 \times 10^{-15}\text{ N}}{9.11 \times 10^{-31}\text{ kg}} = 1.317 \times 10^{15}\text{ m s}^{-2}\)

4. Since the electron is injected midway, the vertical distance to the top plate is:
\(s_y = \frac{d}{2} = 0.010\text{ m}\)

5. Find the time \(t\) taken to travel this vertical distance:
\(s_y = \frac{1}{2} a_y t^2 \implies t = \sqrt{\frac{2 s_y}{a_y}} = \sqrt{\frac{2 \times 0.010}{1.317 \times 10^{15}}} = 3.897 \times 10^{-9}\text{ s}\)

6. The horizontal distance traveled in this time must be equal to the length of the plates \(L = 0.080\text{ m}\) to just exit:
\(v = \frac{L}{t} = \frac{0.080\text{ m}}{3.897 \times 10^{-9}\text{ s}} = 2.05 \times 10^7\text{ m s}^{-1} \approx 2.1 \times 10^7\text{ m s}^{-1}\)
This is approximately \(2 \times 10^7\text{ m s}^{-1}\).

(a) (Total 3 marks):
- Uniform electric field represented by straight, parallel, equally-spaced lines (1 mark)
- Arrows on lines pointing downwards, with plates clearly labeled positive (top) and negative/ground (bottom) (1 mark)
- Edge effects shown with curved lines bulging outwards at both ends of the plates (1 mark)

(b) (Total 5 marks):
- Use of \(E = V/d\) to find field strength \(E = 7500\text{ V m}^{-1}\) (1 mark)
- Use of \(F = eE\) and \(F = ma\) to determine vertical acceleration \(a = 1.32 \times 10^{15}\text{ m s}^{-2}\) (1 mark)
- Use of equations of motion with vertical displacement \(y = 0.010\text{ m}\) (accept standard practice of using half-separation) to find time \(t\) (1 mark)
- Use of \(v = L/t\) with \(L = 0.080\text{ m}\) (1 mark)
- Correct final calculated value of \(2.05 \times 10^7\text{ m s}^{-1}\) or \(2.06 \times 10^7\text{ m s}^{-1}\) showing it rounds to \(2 \times 10^7\text{ m s}^{-1}\) (1 mark)

Using the ideal gas equation \(pV = nRT\), the number of moles is given by \(n = \frac{pV}{RT}\). For container A: \(n_A = \frac{pV}{RT}\). For container B: \(n_B = \frac{(2p)(2V)}{R(3T)} = \frac{4pV}{3RT}\). Taking the ratio: \(\frac{n_A}{n_B} = \frac{pV/RT}{4pV/3RT} = \frac{3}{4} = 0.75\).

1 mark for the correct option (B).

The maximum velocity in simple harmonic motion is given by \(v_{\text{max}} = \omega A\), where \(\omega = \sqrt{\frac{k}{m}}\). Let the original maximum velocity be \(v = \sqrt{\frac{k}{m}} A\). The new angular frequency is \(\omega' = \sqrt{\frac{2k}{0.5m}} = \sqrt{\frac{4k}{m}} = 2\omega\). Since the amplitude \(A\) remains constant, the new maximum velocity is \(2v\), which is double the original.

1 mark for the correct option (C).

In a fission reaction, the daughter nuclei have a higher binding energy per nucleon than the original heavy parent nucleus. This means they are more tightly bound and have a lower total mass than the parent. The mass defect is converted into energy according to \(\Delta E = \Delta m c^2\).

1 mark for the correct option (B).

From Wien's displacement law, \(\lambda_{\text{max}} T = \text{constant}\), so the temperature of Star Y is \(T_Y = 0.5 T_X\). According to the Stefan-Boltzmann law, luminosity is given by \(L = 4\pi R^2 \sigma T^4\). Therefore, \(\frac{L_X}{L_Y} = \frac{R_X^2 T_X^4}{R_Y^2 T_Y^4} = \frac{R^2 T_X^4}{(3R)^2 (0.5 T_X)^4} = \frac{1}{9 \times (1/16)} = \frac{16}{9} \approx 1.78\).

1 mark for the correct option (C).

During a phase change, the energy supplied is given by \(\Delta E = m L_f\), where \(L_f\) is the specific latent heat of fusion. The energy supplied by the heater is \(\Delta E = P \Delta t = 100\text{ W} \times 120\text{ s} = 12000\text{ J}\). Thus, \(L_f = \frac{\Delta E}{m} = \frac{12000\text{ J}}{0.50\text{ kg}} = 2.4 \times 10^4\text{ J kg}^{-1}\).

1 mark for the correct option (A).

Critical damping is the specific level of damping that returns an oscillating system to its equilibrium position in the shortest possible time without overshoot or subsequent oscillation. Underdamping would allow oscillations, and overdamping would take a longer time to return to equilibrium.

1 mark for the correct option (B).

The activity is given by \(A = \lambda N\), where the decay constant is \(\lambda = \frac{\ln 2}{T_{1/2}}\). The initial activity of X is \(A_X = \frac{\ln 2}{T_{1/2}} N_0\). The initial activity of Y is \(A_Y = \frac{\ln 2}{3T_{1/2}} (2N_0) = \frac{2}{3} \frac{\ln 2}{T_{1/2}} N_0\). The ratio is \(\frac{A_X}{A_Y} = \frac{1}{2/3} = 1.5\).

1 mark for the correct option (B).

Using the Doppler redshift formula, \(z = \frac{\Delta \lambda}{\lambda} = \frac{v}{c}\). The change in wavelength is \(\Delta \lambda = 672.5\text{ nm} - 656.3\text{ nm} = 16.2\text{ nm}\). Thus, the recessional velocity is \(v = c \frac{\Delta \lambda}{\lambda} = 3.00 \times 10^8\text{ m s}^{-1} \times \frac{16.2\text{ nm}}{656.3\text{ nm}} \approx 7.40 \times 10^6\text{ m s}^{-1}\).

1 mark for the correct option (A).

According to the Stefan-Boltzmann law, the luminosity \(L\) of a star behaving as a black body radiator is given by \(L = 4\pi R^2 \sigma T^4\), where \(R\) is the radius and \(T\) is the absolute surface temperature. For Star X, the luminosity is \(L_X = 4\pi R^2 \sigma T^4\). For Star Y, the luminosity is \(L_Y = 4\pi (4R)^2 \sigma (0.5T)^4 = 4\pi (16R^2) \sigma (0.0625 T^4) = 4\pi R^2 \sigma T^4\). Therefore, the ratio of their luminosities is \(\frac{L_X}{L_Y} = \frac{4\pi R^2 \sigma T^4}{4\pi R^2 \sigma T^4} = 1.0\). This corresponds to option C.

C is the correct answer. 1 mark for selecting C.

For a fixed mass of an ideal gas at constant volume, pressure is directly proportional to absolute temperature (Kelvin), which is represented by Amontons' Law: \(\frac{p_1}{T_1} = \frac{p_2}{T_2}\). First, convert the temperatures from Celsius to Kelvin: \(T_1 = 27 + 273 = 300\text{ K}\) and \(T_2 = 177 + 273 = 450\text{ K}\). Next, calculate the new pressure: \(p_2 = p \times \frac{T_2}{T_1} = p \times \frac{450}{300} = 1.5p\). This corresponds to option B.

B is the correct answer. 1 mark for selecting B.

(a) First, convert temperature to Kelvin: \( T = 27.0 + 273.15 = 300.15\text{ K} \). Using the ideal gas equation: \( pV = N k T \), which gives \( N = \frac{pV}{kT} = \frac{(1.20 \times 10^5\text{ Pa}) \times (3.50 \times 10^{-3}\text{ m}^3)}{(1.38 \times 10^{-23}\text{ J K}^{-1}) \times 300.15\text{ K}} \). Thus, \( N = 1.014 \times 10^{23} \approx 1.0 \times 10^{23} \). (b) The root-mean-square speed is given by \( v_{\text{rms}} = \sqrt{\frac{3 R T}{M}} \). Substituting the values: \( v_{\text{rms}} = \sqrt{\frac{3 \times 8.31\text{ J mol}^{-1}\text{ K}^{-1} \times 300.15\text{ K}}{4.00 \times 10^{-3}\text{ kg mol}^{-1}}} = \sqrt{1.871 \times 10^6} = 1368\text{ m s}^{-1} \approx 1.37 \times 10^3\text{ m s}^{-1} \).

Part (a) (3 marks): - Use of \( T = t + 273 \) (1) - Use of \( pV = NkT \) (1) - Correct calculation showing \( N \approx 1.0 \times 10^{23} \) (1) Part (b) (4 marks): - Use of \( v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \) (1) - Correct substitution of values (1) - Correct value under square root (1) - Correct final answer with units (accept 1360 to 1370) (1)

(a) Energy supplied by the heater: \( E = P \times t = 45\text{ W} \times (8.0 \times 60\text{ s}) = 21600\text{ J} \). Using \( Q = m L_f \), we find \( L_f = \frac{Q}{m} = \frac{21600\text{ J}}{0.25\text{ kg}} = 8.64 \times 10^4\text{ J kg}^{-1} \). Assumption: No thermal energy is lost to the surroundings. (b) Energy supplied during heating of liquid: \( E = P \times t = 45\text{ W} \times (3.0 \times 60\text{ s}) = 8100\text{ J} \). Using \( Q = m c \Delta T \), we get \( 8100 = 0.25 \times c \times 15 \), which leads to \( c = \frac{8100}{3.75} = 2160\text{ J kg}^{-1}\text{ K}^{-1} \).

Part (a) (4 marks): - Use of \( E = P t \) with time converted to seconds (1) - Use of \( Q = m L_f \) (1) - Correct calculation of \( L_f = 8.64 \times 10^4\text{ J kg}^{-1} \) (1) - Valid assumption stated (1) Part (b) (3 marks): - Correct calculation of energy supplied as \( 8100\text{ J} \) (1) - Use of \( Q = m c \Delta T \) (1) - Correct calculation of \( c = 2160\text{ J kg}^{-1}\text{ K}^{-1} \) (1)

(a) Convert the half-life from years to seconds: \( t_{1/2} = 5730 \times 365.25 \times 24 \times 3600\text{ s} = 1.808 \times 10^{11}\text{ s} \) (or \( 1.807 \times 10^{11}\text{ s} \) using 365 days). Using the decay constant formula: \( \lambda = \frac{\ln(2)}{t_{1/2}} = \frac{0.69315}{1.808 \times 10^{11}\text{ s}} = 3.83 \times 10^{-12}\text{ s}^{-1} \). (b) The activity \( A \) is given by: \( A = \lambda N \). Substituting the values: \( A = (3.83 \times 10^{-12}\text{ s}^{-1}) \times (2.4 \times 10^{10}) = 0.0919\text{ Bq} \) (or \( 0.092\text{ Bq} \)).

Part (a) (3 marks): - Converts half-life to seconds (1) - Use of \( \lambda = \frac{\ln 2}{t_{1/2}} \) (1) - Correct calculation of \( \lambda = 3.83 \times 10^{-12}\text{ s}^{-1} \) (1) Part (b) (4 marks): - Use of \( A = \lambda N \) (1) - Substitution of values (1) - Correct calculation of activity (1) - Correct unit \(\text{Bq}\) or \(\text{s}^{-1}\) (1)

(a) Total mass of reactants = \( 2.014102 + 3.016049 = 5.030151\text{ u} \). Total mass of products = \( 4.002603 + 1.008665 = 5.011268\text{ u} \). Mass defect \( \Delta m = 5.030151 - 5.011268 = 0.018883\text{ u} \). In kg: \( \Delta m = 0.018883 \times 1.66 \times 10^{-27}\text{ kg} = 3.135 \times 10^{-29}\text{ kg} \approx 3.13 \times 10^{-29}\text{ kg} \). (b) Using Einstein's equation: \( E = \Delta m c^2 = (3.135 \times 10^{-29}\text{ kg}) \times (3.00 \times 10^8\text{ m s}^{-1})^2 = 2.822 \times 10^{-12}\text{ J} \). Converting to MeV: \( E = \frac{2.822 \times 10^{-12}\text{ J}}{1.60 \times 10^{-13}\text{ J MeV}^{-1}} = 17.6\text{ MeV} \). (Alternatively, using \( 1\text{ u} = 931.5\text{ MeV} \), \( E = 0.018883 \times 931.5 = 17.6\text{ MeV} \)).

Part (a) (4 marks): - Calculates reactant and product mass sum (1) - Obtains mass defect as \( 0.018883\text{ u} \) (1) - Uses \( 1.66 \times 10^{-27}\text{ kg} \) conversion (1) - Obtains mass defect in kg as \( 3.13 \times 10^{-29}\text{ kg} \) (accept \( 3.13 \times 10^{-29} \) to \( 3.14 \times 10^{-29} \)) (1) Part (b) (3 marks): - Use of \( E = \Delta m c^2 \) or \( 1\text{ u} = 931.5\text{ MeV} \) (1) - Conversion to MeV (1) - Correct value of \( 17.6\text{ MeV} \) (accept \( 17.5 \) to \( 17.7 \)) (1)

(a) Comparing the given equation to the general form \( x = A \cos(\omega t) \): Amplitude \( A = 0.15\text{ m} \). Angular frequency \( \omega = 4.5\text{ rad s}^{-1} \). Using \( \omega = 2 \pi f \), we get \( f = \frac{\omega}{2 \pi} = \frac{4.5}{2 \pi} = 0.716\text{ Hz} \approx 0.72\text{ Hz} \). (b) Maximum velocity \( v_{\text{max}} \) is given by \( v_{\text{max}} = \omega A = 4.5 \times 0.15 = 0.675\text{ m s}^{-1} \approx 0.68\text{ m s}^{-1} \). Acceleration magnitude is given by \( a = \omega^2 x \). At \( x = 0.080\text{ m} \), \( a = (4.5)^2 \times 0.080 = 20.25 \times 0.080 = 1.62\text{ m s}^{-2} \).

Part (a) (3 marks): - Identifies amplitude as \( 0.15\text{ m} \) (1) - Identifies \( \omega = 4.5\text{ rad s}^{-1} \) (1) - Uses \( f = \frac{\omega}{2\pi} \) to show \( f \approx 0.72\text{ Hz} \) (1) Part (b) (4 marks): - Use of \( v_{\text{max}} = \omega A \) (1) - Correct value for maximum velocity \( 0.68\text{ m s}^{-1} \) (1) - Use of \( a = \omega^2 x \) (1) - Correct value for acceleration \( 1.62\text{ m s}^{-2} \) (1)

(a) Free oscillations occur when an oscillating system is displaced and released, allowing it to oscillate at its natural frequency without any periodic external forces. Forced oscillations occur when a periodic external driving force is applied, compelling the system to oscillate at the frequency of the driver. (b) As damping increases: 1. The maximum amplitude of the oscillations (the peak of the resonance curve) decreases. 2. The peak of the curve becomes broader and less sharp. 3. The frequency at which the maximum amplitude occurs (the resonant frequency) shifts slightly to a lower frequency.

Part (a) (3 marks): - Free oscillations: system oscillates at natural frequency / no external driver (1) - Forced oscillations: system oscillates due to external periodic driver (1) - Forced oscillations occur at the frequency of the driver (1) Part (b) (4 marks): - Peak/maximum amplitude of curve decreases (1) - Curve peak becomes broader / wider (1) - Resonant frequency shifts to a lower frequency (1) - Logical structure and correct physics terminology used (1)

(a) From Wien's displacement law: \( \lambda_{\text{max}} T = 2.898 \times 10^{-3}\text{ m K} \). Therefore, \( T = \frac{2.898 \times 10^{-3}\text{ m K}}{2.4 \times 10^{-7}\text{ m}} = 12075\text{ K} \approx 12000\text{ K} \). (b) Using the Stefan-Boltzmann law: \( L = 4 \pi r^2 \sigma T^4 \), which gives \( r^2 = \frac{L}{4 \pi \sigma T^4} \). Substituting the calculated temperature \( T = 12075\text{ K} \): \( r^2 = \frac{4.6 \times 10^{31}}{4 \pi \times (5.67 \times 10^{-8}\text{ W m}^{-2}\text{ K}^{-4}) \times (12075\text{ K})^4} = \frac{4.6 \times 10^{31}}{1.515 \times 10^{10}} = 3.036 \times 10^{21}\text{ m}^2 \). Finding the square root yields \( r = 5.51 \times 10^{10}\text{ m} \) (or \( 5.58 \times 10^{10}\text{ m} \) if using \( T = 12000\text{ K} \)).

Part (a) (3 marks): - Selects and recalls Wien's displacement law equation (1) - Correct substitution of values (1) - Calculation showing \( T \approx 12000\text{ K} \) (1) Part (b) (4 marks): - Use of \( L = 4\pi r^2\sigma T^4 \) (1) - Rearranging equation for \( r \) or \( r^2 \) (1) - Correct substitution of values (either 12075 K or 12000 K) (1) - Correct final answer for radius with unit (accept range \( 5.5 \times 10^{10}\text{ m} \) to \( 5.6 \times 10^{10}\text{ m} \)) (1)

(a) Using the Doppler redshift relation: \( z = \frac{\Delta \lambda}{\lambda} = \frac{v}{c} \). The change in wavelength is \( \Delta \lambda = 672.1\text{ nm} - 656.3\text{ nm} = 15.8\text{ nm} \). The redshift parameter is \( z = \frac{15.8\text{ nm}}{656.3\text{ nm}} = 0.02407 \). The velocity is \( v = z c = 0.02407 \times (3.00 \times 10^8\text{ m s}^{-1}) = 7.22 \times 10^6\text{ m s}^{-1} \). (b) Using Hubble's law: \( v = H_0 d \), we find \( d = \frac{v}{H_0} = \frac{7.22 \times 10^6\text{ m s}^{-1}}{2.2 \times 10^{-18}\text{ s}^{-1}} = 3.28 \times 10^{24}\text{ m} \).

Part (a) (4 marks): - Calculates \( \Delta \lambda = 15.8\text{ nm} \) (1) - Recalls and uses \( \frac{\Delta \lambda}{\lambda} = \frac{v}{c} \) (1) - Substitutes values correctly (1) - Obtains \( v = 7.22 \times 10^6\text{ m s}^{-1} \) (accept \( 7.2 \times 10^6\text{ m s}^{-1} \)) (1) Part (b) (3 marks): - Use of \( v = H_0 d \) (1) - Correct substitution of values (1) - Calculates distance \( d \) as \( 3.28 \times 10^{24}\text{ m} \) (accept range \( 3.2 \times 10^{24}\text{ m} \) to \( 3.3 \times 10^{24}\text{ m} \)) (1)

Part (a):
First, convert the temperature from Celsius to Kelvin:
\(T = 120 + 273.15 = 393.15\text{ K}\) (or \(393\text{ K}\))

The root-mean-square speed is given by:
\(c_{\text{rms}} = \sqrt{\frac{3RT}{M}}\)

Substitute the values:
\(c_{\text{rms}} = \sqrt{\frac{3 \times 8.31 \times 393}{4.00 \times 10^{-3}}} = \sqrt{2.45 \times 10^6} \approx 1565\text{ m s}^{-1}\)

Part (b):
As the temperature increases, the mean kinetic energy of the helium atoms increases, meaning they move with higher mean speed. This results in a higher rate of collisions with the walls of the container. Additionally, because the atoms are moving faster, each collision involves a greater change in momentum. Since force is the rate of change of momentum (\(F = \frac{\Delta p}{\Delta t}\)), the average force exerted on the walls increases. Because pressure is force per unit area, the pressure increases at constant volume.

Part (a) [3 marks]:
- MP1: Conversion of temperature to Kelvin: \(120\,^\circ\text{C} = 393\text{ K}\) (allow \(393.15\text{ K}\))
- MP2: Use of \(c_{\text{rms}} = \sqrt{\frac{3RT}{M}}\) or \(\frac{1}{2}m\langle c^2 \rangle = \frac{3}{2}kT\)
- MP3: Correct calculation of r.m.s. speed to 3 significant figures: \(1.56 \times 10^3\text{ m s}^{-1}\) to \(1.57 \times 10^3\text{ m s}^{-1}\)

Part (b) [4 marks]:
- MP4: Higher temperature means greater mean kinetic energy / speed of the gas molecules.
- MP5: Leads to a greater frequency of collisions with the walls.
- MP6: Greater change in momentum during each collision.
- MP7: Links increased rate of change of momentum to increased force and thus increased pressure (\(p = F/A\)).

Part (a):
The frequency of a mass-spring system is given by:
\(f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\)
Rearranging for stiffness \(k\):
\(k = 4\pi^2 m f^2\)
Substitute the values:
\(k = 4 \times \pi^2 \times 0.350 \times (1.80)^2 = 44.8\text{ N m}^{-1}\)
This is approximately \(45\text{ N m}^{-1}\).

Part (b):
The maximum kinetic energy is equal to the total energy of the oscillator:
\(E_{\text{k,max}} = \frac{1}{2} k A^2\)
Substitute the values using the calculated \(k = 44.8\text{ N m}^{-1}\):
\(E_{\text{k,max}} = 0.5 \times 44.8 \times (0.045)^2 = 0.0454\text{ J}\)
(If using the rounded value of \(45\text{ N m}^{-1}\), \(E_{\text{k,max}} = 0.5 \times 45 \times (0.045)^2 = 0.0456\text{ J}\))

Part (c):
Under light damping, the total energy of the oscillating system decreases exponentially over time. This is because the system does work against resistive forces such as air resistance, which dissipates mechanical energy (kinetic and potential energy) into thermal energy in the surroundings.

Part (a) [2 marks]:
- MP1: Use of \(f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\) or \(\omega = 2\pi f\) and \(\omega = \sqrt{\frac{k}{m}}\)
- MP2: Correct calculation to show \(k = 44.8\text{ N m}^{-1}\) (must show at least 3 s.f. to justify the 'show that')

Part (b) [2 marks]:
- MP3: Use of \(E_k = \frac{1}{2} k A^2\) or \(v_{\text{max}} = \omega A\) and \(E_k = \frac{1}{2} m v^2\)
- MP4: Correct calculation of maximum kinetic energy as \(0.045\text{ J}\) to \(0.046\text{ J}\)

Part (c) [3 marks]:
- MP5: Total energy decreases (exponentially) over time.
- MP6: Work is done by the system against resistive forces (e.g. air resistance / friction).
- MP7: Mechanical energy is transferred / dissipated into thermal energy in the surroundings.

Part (a):
Radiation flux is given by:
\(F = \frac{L}{4\pi d^2}\)
Rearranging for distance \(d\):
\(d = \sqrt{\frac{L}{4\pi F}}\)
Substitute the values:
\(d = \sqrt{\frac{1.50 \times 10^{30}}{4\pi \times 3.15 \times 10^{-18}}} = 1.95 \times 10^{23}\text{ m}\)
This is approximately \(2 \times 10^{23}\text{ m}\).

Part (b):
Redshift is defined as:
\(z = \frac{\Delta\lambda}{\lambda} = \frac{\lambda' - \lambda}{\lambda}\)
Calculate redshift:
\(z = \frac{660.8 - 656.3}{656.3} = \frac{4.5}{656.3} = 6.856 \times 10^{-3}\)
Using the doppler shift equation for recessional velocity (where \(v \ll c\)):
\(v = zc = 6.856 \times 10^{-3} \times 3.00 \times 10^8\text{ m s}^{-1} = 2.06 \times 10^6\text{ m s}^{-1}\)

Part (c):
According to Hubble's Law:
\(v = H_0 d \implies H_0 = \frac{v}{d}\)
Using the calculated values:
\(H_0 = \frac{2.06 \times 10^6\text{ m s}^{-1}}{1.95 \times 10^{23}\text{ m}} = 1.06 \times 10^{-17}\text{ s}^{-1}\)
(If using the rounded distance of \(2.0 \times 10^{23}\text{ m}\), \(H_0 = 1.03 \times 10^{-17}\text{ s}^{-1}\))

Part (a) [2 marks]:
- MP1: Use of \(F = \frac{L}{4\pi d^2}\)
- MP2: Correct calculation showing \(d = 1.95 \times 10^{23}\text{ m}\) (must show at least 3 s.f.)

Part (b) [3 marks]:
- MP3: Use of \(z = \frac{\Delta\lambda}{\lambda}\)
- MP4: Use of \(v = zc\)
- MP5: Correct calculation of recessional velocity \(v = 2.06 \times 10^6\text{ m s}^{-1}\) (accept \(2.1 \times 10^6\text{ m s}^{-1}\))

Part (c) [2 marks]:
- MP6: Use of \(v = H_0 d\)
- MP7: Correct calculation of \(H_0\) in the range \(1.0 \times 10^{-17}\text{ s}^{-1}\) to \(1.1 \times 10^{-17}\text{ s}^{-1}\) (allow ecf from (a) and (b))

### (1) Labelled Diagram
Draw a diagram showing:
- A U-shaped magnet sitting on a digital balance.
- A stiff horizontal copper wire suspended between the poles of the magnet without touching them.
- A series circuit containing the wire, a variable DC power supply, an ammeter, and a variable resistor (or variable supply alone).

### (2) Measurements
- Measure the length \(L\) of the wire that lies within the magnetic field using a vernier caliper or a millimetre ruler.
- Record the initial mass reading on the balance when there is no current (or tare the balance to zero).
- Measure the current \(I\) using the digital ammeter.
- Record the new balance reading for various currents \(I\). Calculate the change in mass \(m\).
- Repeat the measurements of mass change for each current as the current is varied, and calculate mean values.

### (3) Analysis
- The magnetic force is given by \(F = BIL\).
- This force acts upwards or downwards on the wire, and by Newton's third law, an equal and opposite force acts on the magnet, changing the reading on the balance by \(F = mg\), where \(g\) is the acceleration of free fall.
- Therefore, \(mg = BIL\), which rearranges to:
\(m = \left(\frac{BL}{g}\right)I\)
- Plot a graph of mass change \(m\) on the y-axis against current \(I\) on the x-axis.
- The graph should be a straight line passing through the origin.
- The gradient of this graph is \(m/I = \frac{BL}{g}\).
- Calculate \(B\) using: \(B = \frac{\text{gradient} \times g}{L}\).

### (4) Minimising Uncertainties
- **Zero Error on Balance:** Tare the digital balance to zero before turning on the power supply so that only the magnetic force is measured.
- **Wire Alignment:** Use a set square to ensure that the wire is aligned exactly perpendicular (at \(90^\circ\)) to the magnetic field lines. This minimises cosine/sine alignment error.
- **Heating Effects:** Keep the current relatively low and switch off the circuit between readings to prevent the wire from heating up, which would alter its resistance and could cause convection currents affecting the balance.

- **Diagram (3.0 marks):**
- [1 mark] Circuit diagram showing DC power supply, ammeter, and stiff wire in series.
- [1 mark] Setup showing the magnet on a digital balance with the wire suspended between the poles.
- [1 mark] Proper labelling of all components (balance, wire, magnet, ammeter, power supply).

- **Measurements (3.0 marks):**
- [1 mark] Identify that the length \(L\) of the wire within the magnetic field is measured using a vernier caliper or ruler.
- [1 mark] Identify that current \(I\) is measured using an ammeter and the mass change \(m\) is recorded from the digital balance.
- [1 mark] Mention varying the current to obtain at least 5 different values, repeating each measurement to find a mean.

- **Analysis (4.0 marks):**
- [1 mark] State the governing physics formula: \(F = BIL\) and relate force to mass change: \(F = mg\).
- [1 mark] State that a graph of \(m\) against \(I\) (or \(F\) against \(I\)) should be plotted.
- [1 mark] Identify that the gradient of this graph is \(\frac{BL}{g}\) (or \(BL\) if force is plotted).
- [1 mark] Correctly rearrange to express \(B\) in terms of the gradient: \(B = \frac{\text{gradient} \times g}{L}\).

- **Uncertainties and Safety (2.5 marks):**
- [1 mark] Awarded for taring the balance to zero before starting to prevent zero errors.
- [1 mark] Awarded for switching off the current between readings to prevent heating of the wire.
- [0.5 mark] Awarded for using a set square to ensure the wire is perpendicular to the magnetic field.

### (a) Background Count Rate and Uncertainty
- Background counts \(N_b = 48\)
- Time \(t_b = 4.0 \times 60 = 240\text{ s}\)
- Background count rate \(R_b = \frac{48}{240} = 0.20\text{ s}^{-1}\)
- The uncertainty in count number is \(\Delta N_b = \sqrt{48} \approx 6.93\)
- Uncertainty in background count rate \(\Delta R_b = \frac{\sqrt{48}}{240} = 0.029\text{ s}^{-1}\)
- Therefore, background rate = \(0.20 \pm 0.03\text{ s}^{-1}\)

### (b) Corrected Count Rate at \(t = 0\)
- Measured counts at \(t=0\) in 10 s is \(N = 154\)
- Raw count rate \(R_{\text{raw}} = \frac{154}{10} = 15.4\text{ s}^{-1}\)
- Corrected count rate \(R = R_{\text{raw}} - R_b = 15.4 - 0.20 = 15.2\text{ s}^{-1}\)
- Uncertainty in raw counts \(\Delta N = \sqrt{154} \approx 12.41\)
- Uncertainty in raw rate \(\Delta R_{\text{raw}} = \frac{12.41}{10} = 1.24\text{ s}^{-1}\)
- Uncertainty in corrected rate \(\Delta R = \sqrt{(\Delta R_{\text{raw}})^2 + (\Delta R_b)^2} = \sqrt{1.24^2 + 0.029^2} \approx 1.24\text{ s}^{-1}\) (Accept addition: \(1.24 + 0.03 = 1.27\text{ s}^{-1}\))
- Percentage uncertainty in \(R\) = \(\frac{1.24}{15.2} \times 100\% \approx 8.2\%\) (or \(8.4\%\) if using absolute addition)

### (c) Derivation of Straight Line Equation
- Take natural logarithms of both sides of \(R = R_0 e^{-\lambda t}\):
\(\ln(R) = \ln(R_0 e^{-\lambda t})\)
\(\ln(R) = \ln(R_0) - \lambda t\)
- Rearranging into \(y = mx + c\) format:
\(\ln(R) = (-\lambda)t + \ln(R_0)\)
- This represents a straight line with a gradient equal to \(-\lambda\). Thus, \(\lambda = -\text{gradient}\).

### (d) Half-Life and Uncertainty
- Best fit gradient \(m_{\text{best}} = -0.0094\text{ s}^{-1} \implies \lambda_{\text{best}} = 0.0094\text{ s}^{-1}\)
- Half-life \(t_{1/2} = \frac{\ln(2)}{\lambda_{\text{best}}} = \frac{0.69315}{0.0094\text{ s}^{-1}} = 73.74\text{ s}\)
- Absolute uncertainty in \(\lambda\):
\(\Delta \lambda = |(-0.0102) - (-0.0094)| = 0.0008\text{ s}^{-1}\)
- Percentage uncertainty in \(\lambda\):
\(\%\Delta \lambda = \frac{0.0008}{0.0094} \times 100\% \approx 8.51\%\)
- Since \(t_{1/2} \propto \frac{1}{\lambda}\), the percentage uncertainty in \(t_{1/2}\) is also \(8.51\%\).
- Absolute uncertainty in \(t_{1/2}\):
\(\Delta t_{1/2} = 73.74 \times 0.0851 \approx 6.28\text{ s}\)
- State answer with appropriate significant figures: \(t_{1/2} = 74\text{ s} \pm 6\text{ s}\) (or \(73.7 \pm 6.3\text{ s}\)).

- **(a) Background (2.5 marks):**
- [1 mark] Correct calculation of rate: \(0.20\text{ s}^{-1}\).
- [1 mark] Uses \(\sqrt{N}\) to estimate count uncertainty (\(\sqrt{48}\)).
- [0.5 mark] Correctly divides by time to get \(0.03\text{ s}^{-1}\).

- **(b) Corrected rate (3.0 marks):**
- [1 mark] Correct calculation of raw rate (\(15.4\text{ s}^{-1}\)) and corrected rate (\(15.2\text{ s}^{-1}\)).
- [1 mark] Correct calculation of absolute uncertainty in raw count rate (\(1.24\text{ s}^{-1}\)).
- [1 mark] Correct percentage uncertainty calculation to 2 s.f. (\(8.2\%\) to \(8.4\%\)).

- **(c) Derivation (2.0 marks):**
- [1 mark] Shows log conversion correctly resulting in \(\ln(R) = -\lambda t + \ln(R_0)\).
- [1 mark] Explicitly relates \(y = mx + c\) showing gradient \(= -\lambda\).

- **(d) Half-life & uncertainty (5.0 marks):**
- [1 mark] Correct calculation of half-life (\(74\text{ s}\) or \(73.7\text{ s}\)).
- [1 mark] Correct calculation of absolute uncertainty in gradient (\(0.0008\text{ s}^{-1}\)).
- [1 mark] Uses percentage uncertainty of gradient correctly for the percentage uncertainty of half-life (\(8.5\%\)).
- [1 mark] Correct absolute uncertainty in half-life (\(6\text{ s}\) or \(6.3\text{ s}\)).
- [1 mark] Expresses final answer and uncertainty to a consistent and appropriate number of decimal places/significant figures.

### (a) Experimental Procedure & Graphical Analysis
- **Setup and Method:**
- Clamp the spring securely to a rigid stand.
- Hang a mass holder and initial mass \(M\) from the spring.
- Displace the mass vertically and release so it oscillates. Measure the time \(t\) for 20 oscillations using a stopwatch. Repeat this measurement twice more and find the average time. Divide by 20 to get the period \(T\).
- Change the mass \(M\) and repeat to get at least 5 different values of mass \(M\).
- **Graphical Analysis:**
- Square the equation: \(T^2 = \frac{4\pi^2}{k}\left(M + \frac{m_s}{3}\right) = \left(\frac{4\pi^2}{k}\right)M + \frac{4\pi^2 m_s}{3k}\).
- Plot a graph of \(T^2\) (y-axis) against \(M\) (x-axis).
- The gradient \(m = \frac{4\pi^2}{k} \implies k = \frac{4\pi^2}{\text{gradient}}\).
- The y-intercept \(c = \frac{4\pi^2 m_s}{3k} \implies m_s = 3 \times k \times \frac{c}{4\pi^2} = 3 \times \frac{c}{\text{gradient}}\).

### (b) Mean Period and Uncertainty
- Mean time for 20 oscillations: \(t_{\text{mean}} = \frac{14.22 + 14.15 + 14.31}{3} = 14.227\text{ s}\)
- Range of times = \(14.31 - 14.15 = 0.16\text{ s}\)
- Absolute uncertainty in time: \(\Delta t = \frac{\text{range}}{2} = \frac{0.16}{2} = 0.08\text{ s}\)
- Mean period \(T = \frac{14.227\text{ s}}{20} = 0.711\text{ s}\)
- Percentage uncertainty in \(T\) (same as percentage uncertainty in \(t\)):
\(\%\Delta T = \frac{0.08}{14.227} \times 100\% \approx 0.56\%\) (Accept \(0.6\%\))

### (c) Percentage Uncertainty in \(M\)
- \(\%\Delta M = \frac{1\text{ g}}{500\text{ g}} \times 100\% = 0.20\%\)

### (d) Use of Fiducial Marker
- Place the fiducial marker (such as a pin or card with a mark) at the center of the oscillation (equilibrium position).
- This is because the mass moves at its maximum velocity at the equilibrium position, which minimises the human reaction time error as it crosses the marker sharply.

### (e) Choice of Graph Variables
- Plotting \(T^2\) against \(M\) yields a straight-line equation of the form \(y = mx + c\) with a non-zero intercept.
- Plotting \(T\) against \(\sqrt{M}\) does not yield a straight line because of the additional \(\frac{m_s}{3}\) term inside the square root (unless \(m_s = 0\)), making linear gradient analysis impossible.

- **(a) Experimental Plan (4.5 marks):**
- [1 mark] Measure the time for a fixed number of oscillations (e.g., 20) and repeat to find the mean period \(T\).
- [1 mark] Repeat for at least 5 different masses \(M\).
- [1 mark] Identify that \(T^2\) should be plotted against \(M\).
- [1 mark] Identify that \(k = \frac{4\pi^2}{\text{gradient}}\).
- [0.5 mark] Identify that \(m_s = 3 \times \frac{\text{intercept}}{\text{gradient}}\).

- **(b) Period calculation & Uncertainty (3.0 marks):**
- [1 mark] Correct calculation of mean time (\(14.227\text{ s}\)) and period (\(0.711\text{ s}\)).
- [1 mark] Correctly estimates absolute uncertainty in total time as half the range (\(0.08\text{ s}\)).
- [1 mark] Correct percentage uncertainty in \(T\) (\(0.56\%\) or \(0.6\%\)).

- **(c) Mass Uncertainty (1.0 mark):**
- [1 mark] Correct percentage uncertainty calculation (\(0.20\%\)).

- **(d) Fiducial Marker (2.0 marks):**
- [1 mark] State that it is placed at the equilibrium position.
- [1 mark] Explain that the mass has maximum speed at this position, reducing reaction time uncertainty.

- **(e) Graph Preference (2.0 marks):**
- [1 mark] States that \(T^2\) vs \(M\) gives a straight line because it matches the linear equation \(y = mx + c\).
- [1 mark] Explains that \(T\) vs \(\sqrt{M}\) is non-linear because of the \(m_s/3\) term under the root, meaning gradient and intercept cannot be easily interpreted.

### (a) Obtaining Data
- Set up a circuit with a DC power supply, a switch, the capacitor in parallel with a high-resistance digital voltmeter, and the resistor \(R\) in parallel.
- Close the switch to fully charge the capacitor to \(V_0\).
- Open the switch to begin discharge through the resistor, simultaneously starting a stopwatch.
- Record the potential difference \(V\) across the capacitor at fixed intervals (e.g., every 10 seconds up to at least 60 seconds).

### (b) Equation and Gradient
- Start with \(V = V_0 e^{-t/RC}\).
- Take natural logs on both sides:
\(\ln(V) = \ln(V_0 e^{-t/RC})\)
\(\ln(V) = \ln(V_0) - \frac{t}{RC}\)
- Rearrange to the form \(y = mx + c\):
\(\ln(V) = \left(-\frac{1}{RC}\right)t + \ln(V_0)\)
- The gradient of a graph of \(\ln(V)\) vs \(t\) is indeed \(-\frac{1}{RC}\).

### (c) Logarithm Calculations and Uncertainty
- At \(t = 0\text{ s}\), \(\ln(9.0) = 2.197\) (or \(2.20\))
- At \(t = 40\text{ s}\), \(\ln(4.1) = 1.411\) (or \(1.41\))
- To estimate the uncertainty in \(\ln(V)\) at \(V = 4.1\text{ V}\) with \(\Delta V = 0.1\text{ V}\):
- Maximum value of \(\ln(V)\) is \(\ln(4.2) = 1.435\)
- Minimum value of \(\ln(V)\) is \(\ln(4.0) = 1.386\)
- The absolute uncertainty is approximately \(\frac{1.435 - 1.386}{2} \approx 0.025\) (Accept the alternative approximation: \(\Delta(\ln V) \approx \frac{\Delta V}{V} = \frac{0.1}{4.1} = 0.024\))
- Absolute uncertainty is \(\pm 0.02\) (or \(\pm 0.024\)).

### (d) Calculation of \(C\)
- Gradient \(m = -0.0195\text{ s}^{-1} \implies -0.0195 = -\frac{1}{RC}\)
- \(C = \frac{1}{R \times 0.0195} = \frac{1}{120 \times 10^3\ \Omega \times 0.0195\text{ s}^{-1}} = 4.2735 \times 10^{-4}\text{ F} \approx 4.3 \times 10^{-4}\text{ F}\) (or \(427\ \mu\text{F}\))

### (e) Uncertainties in \(C\)
- \(C = \frac{1}{R \cdot |m|}\)
- Percentage uncertainty in \(C\) is the sum of percentage uncertainties in \(R\) and the gradient \(m\):
\(\%\Delta C = \%\Delta R + \%\Delta m = 1\% + 4\% = 5\%\)
- Absolute uncertainty in \(C\):
\(\Delta C = 5\% \times 4.2735 \times 10^{-4}\text{ F} = 0.2137 \times 10^{-4}\text{ F} \approx 0.2 \times 10^{-4}\text{ F}\) (or \(21\ \mu\text{F}\))
- Final formatted value: \(C = (4.3 \pm 0.2) \times 10^{-4}\text{ F}\)

- **(a) Procedure (3.0 marks):**
- [1 mark] Charge capacitor and state that discharge begins by opening/closing a switch.
- [1 mark] State that potential difference \(V\) is recorded using a voltmeter simultaneously with a stopwatch.
- [1 mark] Specify taking readings at regular intervals (e.g., every 10 seconds).

- **(b) Straight Line Derivation (1.5 marks):**
- [1 mark] Logarithmic expansion to show \(\ln(V) = \ln(V_0) - \frac{t}{RC}\).
- [0.5 mark] Compare directly to the equation of a straight line, showing gradient \(= -\frac{1}{RC}\).

- **(c) Log Calculations & Uncertainty (3.0 marks):**
- [1 mark] Correctly computes \(\ln(9.0) = 2.20\) and \(\ln(4.1) = 1.41\).
- [1 mark] Demonstrates a method to calculate uncertainty in log value (either through worst-case differences or \(\Delta V / V\)).
- [1 mark] Obtains uncertainty value in the range \(0.02\) to \(0.03\).

- **(d) Capacitance Calculation (2.0 marks):**
- [1 mark] Use of \(C = \frac{1}{R \times \text{gradient}}\) with correct substitution of powers of ten (\(120\text{ k}\Omega = 120,000\ \Omega\)).
- [1 mark] Correct calculation of \(C = 4.3 \times 10^{-4}\text{ F}\) (or \(427\ \mu\text{F}\)).

- **(e) Uncertainty in Capacitance (3.0 marks):**
- [1 mark] Correct calculation of percentage uncertainty by summing percentages (\(5\%\)).
- [1 mark] Correct calculation of absolute uncertainty (\(0.2 \times 10^{-4}\text{ F}\) or \(21\ \mu\text{F}\)).
- [1 mark] Final answer written with consistent and appropriate significant figures (e.g. \(4.3 \pm 0.2 \times 10^{-4}\text{ F}\)).