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(a) Use the formula: \(a = \frac{v - u}{t}\) where \(v = 8.0\text{ m/s}\), \(u = 0\text{ m/s}\), and \(t = 5.0\text{ s}\). \(a = \frac{8.0 - 0}{5.0} = 1.6\text{ m/s}^2\). (b) Total distance is the area under the velocity-time graph. The shape can be split into three sections: Area 1 (first triangle) = \(\frac{1}{2} \times 5.0 \times 8.0 = 20\text{ m}\); Area 2 (rectangle) = \(12.0 \times 8.0 = 96\text{ m}\); Area 3 (second triangle) = \(\frac{1}{2} \times 3.0 \times 8.0 = 12\text{ m}\). Total distance = \(20 + 96 + 12 = 128\text{ m}\). (c) Average speed = \(\frac{\text{total distance}}{\text{total time}} = \frac{128\text{ m}}{20.0\text{ s}} = 6.4\text{ m/s}\).

(a) M1: recall and substitution into \(a = \frac{v-u}{t}\) (1); A1: 1.6 (1); B1: correct unit \(\text{m/s}^2\) (1). (b) M1: recognition that distance is the area under the graph (1); M2: correct calculation of at least one component area (1); M3: summing the component areas correctly (1); A1: 128 m (1). (c) M1: recall of average speed equation (1); M2: substitution of values: 128 / 20 (1); A1: 6.4 m/s (1).

(a) The equation is: \(\text{momentum} = \text{mass} \times \text{velocity}\) (or \(p = mv\)). (b) Change in momentum \(\Delta p = m \times (v - u) = 1.5\text{ kg} \times (0 - 6.0\text{ m/s}) = -9.0\text{ kg m/s}\) (magnitude is \(9.0\text{ kg m/s}\)). The unit is \(\text{kg m/s}\) or \(\text{N s}\). (c) Use the equation: \(\text{force} = \frac{\text{change in momentum}}{\text{time taken}}\). \(F = \frac{9.0\text{ kg m/s}}{0.080\text{ s}} = 112.5\text{ N}\) (accept \(110\text{ N}\)). (d) Bubble wrap is compressible, which increases the time taken for the parcel to come to rest during the collision. Since the change in momentum is constant, a longer collision time reduces the rate of change of momentum. Because force is equal to the rate of change of momentum, this reduces the average force experienced by the fragile items, preventing damage.

(a) B1: correct equation in words or symbols (1). (b) M1: substitution of mass and velocity (1); A1: 9.0 (1); B1: correct unit \(\text{kg m/s}\) or \(\text{N s}\) (1). (c) M1: recall of \(F = \frac{\Delta p}{t}\) (1); M2: substitution of values (1); A1: 112.5 N (1). (d) B1: bubble wrap increases the collision/impact time (1); B1: same change in momentum occurs over a longer time (1); B1: this reduces the force experienced by the fragile items (1).

(a) Half-life is the time taken for the activity of a radioactive sample to halve, or the time taken for the number of radioactive nuclei in a sample to halve. (b) 1. Initial source count rate = \(410 - 24 = 386\text{ counts per minute}\). 2. Final source count rate = \(120 - 24 = 96\text{ counts per minute}\). 3. Determine number of half-lives: after 1 half-life: \(386 / 2 = 193\); after 2 half-lives: \(193 / 2 = 96.5\) (approximately 96). Thus, 2 half-lives have passed in 12.0 hours. 4. Half-life = \(12.0 / 2 = 6.0\text{ hours}\). (c) Two natural sources are: radon gas from rocks/soil, and cosmic rays from space. (d) One safety precaution is to use tongs to handle the source to keep a safe distance.

(a) B1: time taken for activity/count rate to halve (1); B1: or time taken for number of radioactive nuclei to halve (1). (b) M1: subtract background from initial count: \(410 - 24 = 386\) (1); M2: subtract background from final count: \(120 - 24 = 96\) (1); M3: recognise 2 half-lives have passed (1); M4: set up equation: \(2 \times t_{1/2} = 12.0\text{ hours}\) (1); A1: \(6.0\text{ hours}\) (1). (c) B1: radon gas (1); B1: cosmic rays (or food/rocks) (1). (d) B1: suitable precaution (e.g., use tongs, lead container, keep distance) (1).

(a) The equation is \(n = \frac{\sin i}{\sin r}\). (b) \(n = \frac{\sin(38.0^\circ)}{\sin(24.0^\circ)} = \frac{0.61566}{0.40674} \approx 1.51\). (c) The relationship is \(\sin c = \frac{1}{n}\). (d) \(\sin c = \frac{1}{1.51} \approx 0.66225\), so \(c = \sin^{-1}(0.66225) \approx 41.5^\circ\). Since the angle of incidence (\(50.0^\circ\)) is greater than the critical angle (\(41.5^\circ\)), total internal reflection (TIR) occurs. The ray of light does not escape into the air and is entirely reflected back into the glass block at an angle of reflection of \(50.0^\circ\).

(a) B1: \(n = \frac{\sin i}{\sin r}\) (1). (b) M1: substitution of correct angles (1); M2: evaluation of sines (1); A1: 1.51 (must be 3 s.f.) (1). (c) B1: \(\sin c = \frac{1}{n}\) (1). (d) M1: substitution of calculated n value (1); A1: \(c = 41.5^\circ\) (allow \(41.0^\circ - 42.0^\circ\)) (1); B1: state that angle of incidence is greater than critical angle (1); B1: state that total internal reflection (TIR) occurs (1); B1: state that light reflects inside the block (1).

(a) The equation is \(\text{GPE} = m \times g \times h\). (b) \(\text{GPE} = 45.0\text{ kg} \times 10\text{ N/kg} \times 12.0\text{ m} = 5400\text{ J}\). The useful work done is equal to this increase in GPE, which is \(5400\text{ J}\) (or \(5.4\text{ kJ}\)). (c) \(\text{efficiency} = \frac{\text{useful energy output}}{\text{total energy input}}\). Given \(\text{total energy input} = 7.20\text{ kJ} = 7200\text{ J}\), \(\text{efficiency} = \frac{5400}{7200} = 0.75\) (or \(75\%\)). (d) The wasted energy is transferred to the surroundings as thermal energy (by heating) and sound energy. This is caused by friction in the moving parts of the motor and electrical resistance in the windings. This energy becomes dissipated (spread out) and is less useful.

(a) B1: \(\text{GPE} = mgh\) (1). (b) M1: substitution of values: \(45 \times 10 \times 12\) (1); A1: 5400 (1); B1: correct unit \(\text{J}\) or Joules (1). (c) B1: recall of efficiency equation (1); M1: conversion of 7.20 kJ and substitution: \(\frac{5400}{7200}\) (1); A1: 0.75 or 75% (1). (d) B1: thermal energy/sound produced (1); B1: dissipated/spread out to surroundings (1); B1: due to friction/electrical resistance (1).

(a) The equation is \(V = I \times R\). (b) (i) In series, total resistance \(R_{\text{total}} = R_1 + R_2 = 12\ \Omega + 18\ \Omega = 30\ \Omega\). (ii) \(I = \frac{V}{R_{\text{total}}} = \frac{9.0\text{ V}}{30\ \Omega} = 0.30\text{ A}\). The unit is Amperes (\(\text{A}\)). (c) The voltage across the thermistor is \(V_{\text{thermistor}} = I \times R_{\text{thermistor}} = 0.30\text{ A} \times 18\ \Omega = 5.4\text{ V}\). (d) When the temperature of the thermistor is increased, its resistance decreases. This reduces the total resistance of the circuit, causing the current flowing through the circuit to increase. Since the resistance of the fixed resistor remains constant and \(V = I \times R\), the increased current results in a larger voltage across the fixed resistor, so the voltmeter reading increases.

(a) B1: \(V = IR\) (1). (b) (i) B1: \(30\ \Omega\) (1); (ii) M1: substitution of values: \(\frac{9.0}{30}\) (1); A1: 0.30 (1); B1: unit: \(\text{A}\) or Ampere (1). (c) M1: substitution: \(0.30 \times 18\) (1); A1: \(5.4\text{ V}\) (1). (d) B1: thermistor resistance decreases, so total current increases (1); B1: fixed resistor has a constant resistance (1); B1: voltage across fixed resistor increases (since \(V=IR\)) (1).

(a) In a solid, molecules are closely packed in a regular, ordered pattern and vibrate about fixed positions. Upon melting to a liquid, the arrangement becomes random/irregular, and the molecules, while still close together, can now slide and move past each other. (b) During melting, the thermal energy supplied is used to break or overcome the intermolecular forces of attraction between the molecules, rather than increasing their average kinetic energy. Since temperature is a measure of the average kinetic energy of the molecules, the temperature remains constant during the state change. (c) 1. Energy supplied: \(E = \text{Power} \times \text{time} = 120\text{ W} \times (4.0 \times 60\text{ s}) = 120 \times 240 = 28,800\text{ J}\). 2. Temperature change: \(\Delta \theta = 42^\circ\text{C} - 18^\circ\text{C} = 24^\circ\text{C}\). 3. Use formula: \(Q = m \times c \times \Delta \theta \Rightarrow 28,800 = 0.50 \times c \times 24 \Rightarrow 28,800 = 12 \times c \Rightarrow c = 2400\text{ J/(kg }^\circ\text{C)}\).

(a) B1: solid has a regular pattern/lattice and vibrates (1); B1: liquid has a random/irregular arrangement (1); B1: liquid molecules can slide/move past each other (1). (b) B1: thermal energy breaks bonds/overcomes forces of attraction between molecules (1); B1: average kinetic energy of molecules does not change (1); B1: temperature depends on kinetic energy, so stays constant (1). (c) M1: calculation of energy supplied: \(120 \times 240 = 28,800\text{ J}\) (1); M2: calculation of temperature change: \(24^\circ\text{C}\) (1); A1: substitution and calculation: \(c = 2400\) (1); B1: unit: \(\text{J/(kg }^\circ\text{C)}\) or \(\text{J/kg K}\) (1).

(a) To investigate the field, place a thick wire vertically through a piece of card. Connect the wire to a low-voltage, high-current power supply. Either sprinkle iron filings on the card and tap it gently to reveal concentric circular field lines, or place several plotting compasses around the wire and mark the direction of their North pointers to show the direction of the field. (b) (i) Reversing the current direction will reverse the direction of the force (it will act downwards). (ii) Increasing the magnetic field strength will increase the magnitude of the force. (c) The rule is Fleming's Left-Hand Rule. To apply it, hold the thumb, first finger, and second finger of the left hand mutually perpendicular to each other. Point the First finger in the direction of the magnetic Field (from North to South) and the seCond finger in the direction of the Current (from positive to negative). The thuMb then points in the direction of the Force/Motion (which is upwards in this case).

(a) B1: use iron filings (to show shape) or plotting compasses (to show direction) (1); B1: pass a current through the wire (1); B1: tap card (for filings) or mark compass needle directions (1); B1: identify concentric circles centered on the wire (1). (b) (i) B1: force acts in opposite direction/downwards (1); (ii) B1: force increases (1). (c) B1: Fleming's Left-Hand Rule (1); B1: thumb, first finger, second finger held mutually at right angles (1); B1: First finger = Field, seCond finger = Current (1); B1: thuMb = Motion/Force (1).

(a) Use the equation of motion:
\(v = u + at\)
\(v = 0 + (5.0\text{ m/s}^2 \times 4.0\text{ s}) = 20\text{ m/s}\).

(b) For the second part of the motion (decelerating under gravity):
Initial velocity, \(u = 20\text{ m/s}\)
Final velocity, \(v = 0\text{ m/s}\) (at maximum height)
Acceleration, \(a = -g = -10\text{ m/s}^2\)
\(v = u + at\)
\(0 = 20 + (-10)t_{\text{decel}}\)
\(t_{\text{decel}} = 2.0\text{ s}\)
Total time from launch = \(t_{\text{accel}} + t_{\text{decel}} = 4.0\text{ s} + 2.0\text{ s} = 6.0\text{ s}\).

(c) The velocity-time graph consists of two straight-line segments forming a triangle:
- A straight line with a positive gradient starting at the origin \((0, 0)\) and ending at \((4.0\text{ s}, 20\text{ m/s})\).
- A straight line with a negative gradient starting at \((4.0\text{ s}, 20\text{ m/s})\) and ending at the time-axis at \((6.0\text{ s}, 0\text{ m/s})\).

(d) The maximum height is represented by the total area under the velocity-time graph:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\)
\(\text{Height} = \frac{1}{2} \times 6.0\text{ s} \times 20\text{ m/s} = 60\text{ m}\).
(Alternatively, calculating the distance for each stage: \(s_1 = \frac{1}{2} \times 5.0 \times 4.0^2 = 40\text{ m}\); \(s_2 = 20 \times 2.0 - \frac{1}{2} \times 10 \times 2.0^2 = 20\text{ m}\); \(\text{Total} = 40\text{ m} + 20\text{ m} = 60\text{ m}\)).

(a) [2 marks]
- 1 mark for substituting correct values into \(v = u + at\) or correct working.
- 1 mark for correct final velocity with unit (\(20\text{ m/s}\)).

(b) [3 marks]
- 1 mark for setting up deceleration stage equation, e.g., \(0 = 20 - 10t\).
- 1 mark for finding deceleration time of \(2.0\text{ s}\).
- 1 mark for adding the two time intervals to find total time of \(6.0\text{ s}\).

(c) [3 marks]
- 1 mark for describing a straight-line increase from \(0\) to \(20\text{ m/s}\) at \(4\text{ s}\).
- 1 mark for describing a straight-line decrease from \(20\text{ m/s}\) to \(0\) at \(6\text{ s}\).
- 1 mark for correctly identifying all key coordinates: \((0,0)\), \((4,20)\), and \((6,0)\).

(d) [2 marks]
- 1 mark for using the area under the graph or calculating distances for both phases.
- 1 mark for the correct final answer of \(60\text{ m}\).

(a) The normal must be drawn perpendicular (at \(90^\circ\)) to the flat boundary of the glass block where the light ray enters.

(b) Snell's Law: \(n = \frac{\sin(i)}{\sin(r)}\)

(c) Substitute the given angles into the formula:
\(n = \frac{\sin(35^\circ)}{\sin(22^\circ)}\)
\(n = \frac{0.5736}{0.3746} = 1.531...\)
To 3 significant figures, \(n = 1.53\).

(d) Use the critical angle formula:
\(\sin(c) = \frac{1}{n}\)
\(\sin(c) = \frac{1}{1.531} \approx 0.6531\) (using unrounded value of \(n\))
\(c = \sin^{-1}(0.6531) = 40.78^\circ\)
To the nearest whole degree, the critical angle is \(41^\circ\) (accept \(41^\circ\) or \(40.8^\circ\) if rounded from \(1.53\)).

(e) Since the angle of incidence inside the glass (\(45^\circ\)) is greater than the critical angle (\(41^\circ\)), total internal reflection (TIR) occurs. The ray does not exit into the air but reflects completely back inside the glass block at an angle of reflection of \(45^\circ\).

(a) [1 mark]
- 1 mark for stating that the normal is perpendicular / at \(90^\circ\) to the interface.

(b) [1 mark]
- 1 mark for \(n = \sin(i) / \sin(r)\).

(c) [3 marks]
- 1 mark for correct substitution: \(\sin(35) / \sin(22)\).
- 1 mark for calculating intermediate sines: \(0.574\) and \(0.375\).
- 1 mark for correct final refractive index value to 3 s.f. (\(1.53\)).

(d) [3 marks]
- 1 mark for using the correct formula \(\sin(c) = 1/n\).
- 1 mark for calculation of \(\sin(c) \approx 0.653\).
- 1 mark for correct final critical angle to nearest degree (\(41^\circ\) or \(41\)).

(e) [2 marks]
- 1 mark for stating that total internal reflection (TIR) occurs.
- 1 mark for explaining that this happens because the angle of incidence (\(45^\circ\)) is greater than the critical angle (\(41^\circ\)).

(a) Gas particles are in constant, random motion. They collide with the internal walls of the container. During these collisions, the particles undergo a change in momentum, exerting a force on the walls. The sum of these forces over the surface area of the cylinder walls results in pressure (pressure = force / area).

(b) To convert Celsius to kelvin:
\(T\text{ (K)} = T\text{ (}^\circ\text{C)} + 273\)
\(T = 20 + 273 = 293\text{ K}\).

(c) Since the volume and mass of the gas are constant (rigid sealed cylinder), we use the pressure law:
\(\frac{P_1}{T_1} = \frac{P_2}{T_2}\)
where \(T\) must be in kelvin.
\(P_1 = 1.2 \times 10^5\text{ Pa}\)
\(T_1 = 293\text{ K}\)
\(P_2 = 1.8 \times 10^5\text{ Pa}\)

Rearranging for \(T_2\):
\(T_2 = T_1 \times \frac{P_2}{P_1} = 293 \times \frac{1.8 \times 10^5}{1.2 \times 10^5}\)
\(T_2 = 293 \times 1.5 = 439.5\text{ K}\)

Convert back to Celsius:
\(T_2\text{ (}^\circ\text{C)} = 439.5 - 273 = 166.5^\circ\text{C}\) (or \(167^\circ\text{C}\) to 3 significant figures).

(d) Any two assumptions from:
1. The particles have negligible volume compared to the container volume.
2. There are no attractive or repulsive intermolecular forces between the particles (except during collisions).
3. All collisions between particles and the walls are perfectly elastic.
4. The time duration of a collision is negligible compared to the time between collisions.

(a) [3 marks]
- 1 mark for stating that particles are in constant random motion and collide with the walls.
- 1 mark for mentioning that collisions cause a change in momentum / exert a force on the walls.
- 1 mark for linking force to pressure (pressure = force/area).

(b) [1 mark]
- 1 mark for correct conversion: \(293\text{ K}\).

(c) [4 marks]
- 1 mark for stating or selecting the correct relationship: \(P_1/T_1 = P_2/T_2\) or \(P \propto T\).
- 1 mark for substituting temperature in Kelvin (\(293\text{ K}\)) and correct pressures.
- 1 mark for finding intermediate temperature \(T_2 = 439.5\text{ K}\).
- 1 mark for converting back to Celsius to get \(166.5^\circ\text{C}\) or \(167^\circ\text{C}\) (accept \(166.5\) to \(167\)).

(d) [2 marks]
- 1 mark each for any two valid assumptions of ideal gases (e.g., negligible volume of particles, no intermolecular forces, elastic collisions, random motion).

Part (a): Use the formula a = (v - u) / t. Initial velocity u = 0, final velocity v = \(12\text{ m/s}\), and time t = \(3.0\text{ s}\). Therefore, a = 12 / 3.0 = \(4.0\text{ m/s}^2\). Part (b): Total distance is the area under the velocity-time graph. 1) Acceleration phase: distance = 0.5 * 3.0 * 12 = \(18\text{ m}\). 2) Constant velocity phase: distance = 5.0 * 12 = \(60\text{ m}\). 3) Deceleration phase: distance = 0.5 * 2.0 * 12 = \(12\text{ m}\). Total distance = 18 + 60 + 12 = \(84\text{ m}\).

Part (a) [3 marks]: 1 mark for correct acceleration formula or substitution, 1 mark for correct calculation (4.0), 1 mark for correct unit (\(m/s^2\)). Part (b) [5.75 marks]: 1 mark for showing distance is represented by the area under the graph, 1 mark for calculating first phase distance (18), 1 mark for calculating second phase distance (60), 1 mark for third phase distance (12), 1.75 marks for correct final sum (84) with correct unit (m).

Part (a): Total initial momentum = 0. According to conservation of momentum, the total final momentum must also equal zero. Let right be the positive direction. Momentum of Skater A = mass * velocity = 60 * (-1.5) = \(-90\text{ kg m/s}\). Momentum of Skater B = mass * velocity = 45 * v. Therefore, -90 + 45 * v = 0, which gives v = \(2.0\text{ m/s}\) to the right. Part (b): Since no external forces act on the system of skaters, total momentum is conserved. The skaters start from rest (zero momentum), so their final momenta must be equal in magnitude but opposite in direction to sum to zero.

Part (a) [4.75 marks]: 1 mark for total initial momentum = 0, 1 mark for p = m * v, 1 mark for calculating Skater A's momentum as -90 kg m/s, 1 mark for setting up conservation equation (45 * v = 90), 0.75 marks for final velocity magnitude of 2.0 m/s and direction. Part (b) [4 marks]: 1 mark for stating no external forces act, 1 mark for defining conservation of momentum, 1 mark for noting initial momentum is zero, 1 mark for explaining final momenta are equal and opposite.

Part (a): Subtract background radiation to find the corrected count rates. Corrected initial rate = 510 - 30 = \(480\text{ cpm}\). Corrected rate after 24 mins = 90 - 30 = \(60\text{ cpm}\). Part (b): Determine the number of half-lives that have elapsed to decrease from 480 to 60. 480 -> 240 (1 half-life) -> 120 (2 half-lives) -> 60 (3 half-lives). 3 half-lives have occurred in \(24\text{ minutes}\). Therefore, half-life = 24 / 3 = \(8.0\text{ minutes}\).

Part (a) [3 marks]: 1.5 marks for corrected initial count (480 cpm), 1.5 marks for corrected final count (60 cpm). Part (b) [5.75 marks]: 2 marks for showing step-by-step halving of 480 to reach 60, 1 mark for identifying that 3 half-lives have passed, 1.75 marks for calculation (24 / 3) and correct final answer (8.0 minutes) with unit.

(a) Place the semi-circular glass block on a sheet of paper and trace its boundary. Direct a narrow ray of light from a ray box through the curved face so it hits the center of the flat face. (b) Adjust the angle of the incident ray at the flat face until the refracted ray travels exactly along the flat boundary. Measure the angle of incidence (the angle between the incident ray and the normal) using a protractor; this is the critical angle c. (c) Use the equation \(n = 1 / \sin(c)\) to find the refractive index.

Part (a) [3 marks]: 1 mark for tracing/positioning block, 1 mark for directing ray through curved boundary, 1 mark for aiming at the center of the flat edge. Part (b) [3 marks]: 1 mark for adjusting ray until refraction is along the boundary (90 degrees), 1 mark for measuring angle of incidence, 1 mark for using a protractor. Part (c) [2.75 marks]: 1.75 marks for correct formula n = 1 / sin(c), 1 mark for identifying c as critical angle.

Part (a): Energy (E) = Power (P) * time (t). E = 50 * 120 = \(6000\text{ J}\). Part (b): Use the formula Q = m * L, where Q is heat energy, m is mass melted, and L is specific latent heat of fusion. Here, Q = \(6000\text{ J}\) and m = \(0.018\text{ kg}\). L = 6000 / 0.018 = \(3.3\times 10^5\text{ J/kg}\) (or \(333,333\text{ J/kg}\)). Assumption: All thermal energy from the heater was transferred directly to the solid to melt it, with no heat lost to the surroundings.

Part (a) [3 marks]: 1 mark for formula E = P * t, 1 mark for correct substitution, 1 mark for 6000 J. Part (b) [4 marks]: 1 mark for formula Q = m * L, 1 mark for correct rearrangement L = Q / m, 1 mark for correct calculation (approx 330,000 J/kg), 1 mark for unit J/kg. Part (c) [1.75 marks]: 1.75 marks for stating a valid assumption (e.g. no heat loss to surroundings).

Part (a): Convert temperatures to Kelvin: T1 = 20 + 273 = \(293\text{ K}\). T2 = 80 + 273 = \(353\text{ K}\). Use the pressure law: P1 / T1 = P2 / T2. 100 / 293 = P2 / 353. P2 = (100 * 353) / 293 = \(120.5\text{ kPa}\) (which rounds to \(120\text{ kPa}\)). Part (b): As temperature increases, the average kinetic energy of the gas molecules increases, so they move faster. They collide with the container walls more frequently and with greater force, resulting in a higher pressure.

Part (a) [4.75 marks]: 1 mark for converting both temperatures to Kelvin, 1 mark for the correct formula P1 / T1 = P2 / T2, 1 mark for correct substitution, 1.75 marks for correct final answer (120 or 120.5 kPa). Part (b) [4 marks]: 1 mark for linking temperature to higher kinetic energy/speed, 1 mark for stating more frequent collisions, 1 mark for stating harder/more forceful collisions with walls, 1 mark for linking force to pressure.

Part (a): When a current flows through the coil, a magnetic field is created around the wires. This field interacts with the magnetic field of the permanent magnets. This interaction produces forces on the sides of the coil perpendicular to the field (Fleming's Left-Hand Rule). Because current flows in opposite directions on each side, the forces on opposite sides are in opposite directions (one up, one down), creating a turning moment that rotates the coil. Part (b): The split-ring commutator reverses the direction of the current in the coil every half turn. This ensures that the forces on each side of the coil reverse direction when passing the vertical position, maintaining continuous rotation in the same direction.

Part (a) [4.75 marks]: 1 mark for current creating a magnetic field, 1 mark for interaction of fields creating a force, 1 mark for forces acting in opposite directions on each side, 1.75 marks for explaining that this creates a moment/turning effect. Part (b) [4 marks]: 1 mark for stating commutator reverses current, 1 mark for stating this happens every half turn (180 degrees), 2 marks for explaining this keeps the forces in the same rotational direction to ensure continuous rotation.

After the main sequence phase, a massive star runs out of hydrogen in its core, causing the core to contract while the outer layers expand and cool to form a red supergiant. Helium and heavier elements undergo fusion in the core. When fusion stops, the core collapses rapidly, triggering a massive explosion called a supernova, which ejects outer layers. The remaining dense core is compressed. Depending on the star's initial mass, it collapses into either a neutron star or a black hole. In contrast, the Sun (a low-mass star) will not explode as a supernova; it will expand to a red giant and then end as a white dwarf.

Stages [5.75 marks]: 1 mark for identifying expansion into a red supergiant, 1 mark for mentioning fusion of heavier elements, 1 mark for describing the supernova explosion, 1.75 marks for stating the final core remnant is a neutron star or black hole, 1 mark for explaining core collapse when fusion ceases. Comparison [3 marks]: 1.5 marks for stating the Sun's final state is a white dwarf, 1.5 marks for noting the Sun does not undergo a supernova/does not form a black hole due to its lower mass.