2024 Edexcel IGCSE Physics 模擬試題連答案詳解

Using the formula: \(\text{momentum} = \text{mass} \times \text{velocity}\). Substituting the given values: \(\text{momentum} = 0.25\text{ kg} \times 1.2\text{ m/s} = 0.30\text{ kg m/s}\).

1 mark for calculating the correct value of 0.30 (or 0.3). 1 mark for the correct unit of kg m/s (accept kg.m/s or Ns).

Using Snell's law: \(n = \frac{\sin(i)}{\sin(r)}\). Substituting the angles: \(n = \frac{\sin(35^\circ)}{\sin(22^\circ)} = \frac{0.5736}{0.3746} \approx 1.531\). Rounded to 2 significant figures, the refractive index is 1.5.

1 mark for correct substitution of values into the refractive index formula. 1 mark for final calculated answer of 1.5 (accept 1.53).

First convert time into seconds: \(t = 3.0 \times 60 = 180\text{ s}\). Then use the formula: \(Q = I \times t\). Substitute the values: \(Q = 0.45\text{ A} \times 180\text{ s} = 81\text{ C}\).

1 mark for converting time to seconds (180 s). 1 mark for the correct final answer of 81 C (accept 81).

Determine the number of half-lives that have elapsed: \(800 \rightarrow 400 \rightarrow 200 \rightarrow 100\), which is 3 half-lives. Divide the total time by the number of half-lives: \(\frac{24\text{ hours}}{3} = 8\text{ hours}\).

1 mark for showing that 3 half-lives have passed. 1 mark for the correct final answer of 8 hours (or 8).

Using the formula for pressure in a liquid: \(p = \rho \times g \times h\). Substituting the given values: \(p = 1030\text{ kg/m}^3 \times 10\text{ N/kg} \times 15\text{ m} = 154,500\text{ Pa}\) (or \(1.55 \times 10^5\text{ Pa}\)).

1 mark for correct substitution into the formula. 1 mark for the correct calculation to 154,500 Pa (accept 150,000 Pa if g = 9.8 N/kg is used instead, or 1.55 x 10^5 Pa with correct unit).

During the main sequence phase, a star is in stable equilibrium. The inward gravitational force pulling the matter together is balanced by the outward radiation pressure (and gas pressure) from nuclear fusion in the core.

1 mark for identifying the inward gravitational force (gravity). 1 mark for identifying the outward radiation pressure (or thermal expansion / gas pressure).

As temperature increases, the average kinetic energy of the gas particles increases, so they move faster. They collide with the walls of the container more frequently and with greater force, resulting in a higher total force per unit area (pressure).

1 mark for stating that gas particles gain kinetic energy or move faster. 1 mark for stating they collide more frequently or with greater force with the walls of the container.

Using the transformer ratio equation: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\). Rearranging to solve for \(V_s\): \(V_s = V_p \times \frac{N_s}{N_p} = 12 \times \frac{3000}{150} = 240\text{ V}\).

1 mark for correct substitution or rearrangement of the formula. 1 mark for the correct output voltage of 240 V.

1. As temperature increases, the average kinetic energy of the gas particles increases, so they move faster on average.
2. This means the particles collide with the walls of the container more frequently and with a greater force (a larger rate of change of momentum per collision), resulting in an increased average force per unit area on the walls and hence greater pressure.

MP1: State that particles have more kinetic energy OR move faster (at higher temperature) [1]
MP2: State that they collide with the container walls more frequently OR with more force (larger change in momentum per collision) [1]

1. Determine the fraction of initial activity remaining:
\(\frac{150}{1200} = \frac{1}{8}\)

2. Determine the number of half-lives that have passed:
Since \(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\), exactly 3 half-lives have elapsed.

3. Calculate the duration of one half-life:
\(\text{Half-life} = \frac{15\text{ hours}}{3} = 5\text{ hours}\)

MP1: Determination that 3 half-lives have elapsed (e.g. showing halving steps: \(1200 \rightarrow 600 \rightarrow 300 \rightarrow 150\) or stating activity is reduced to \(1/8\)) [1]
MP2: Calculation of half-life as \(5\text{ hours}\) [1] (accept '5 h' or '300 minutes', but reject '5' on its own without correct unit)

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

Rearrange the equation to solve for \(\sin r\):
\(\sin r = \frac{\sin i}{n} = \frac{\sin(42^\circ)}{1.45}\)

Substitute the value of \(\sin(42^\circ) \approx 0.6691\):
\(\sin r = \frac{0.6691}{1.45} \approx 0.4615\)

Calculate \(r\):
\(r = \arcsin(0.4615) \approx 27.48^\circ\)

To 2 significant figures, the angle of refraction is \(27^\circ\).

MP1: Correct substitution into Snell's Law or rearrangement to find \(\sin r\) (e.g. \(\sin r = \frac{\sin(42^\circ)}{1.45}\)) [1]
MP2: Correct value of \(27^\circ\) (accept range \(27^\circ\) to \(27.5^\circ\)) [1]

1. Recall the relation between speed, distance, and time:
\(\text{speed} = \frac{\text{distance}}{\text{time}}\)

2. Convert the distance from kilometers to meters:
\(3.9\text{ km} = 3900\text{ m}\)

3. Calculate the time in seconds:
\(\text{time} = \frac{3900\text{ m}}{6.5\text{ m/s}} = 600\text{ s}\)

4. Convert the time from seconds to minutes:
\(\text{time in minutes} = \frac{600}{60} = 10\text{ minutes}\)

MP1: Correct conversion of distance to meters (\(3900\text{ m}\)) OR correct calculation of time in seconds (\(600\text{ s}\)) [1]
MP2: Correct calculation of final time in minutes (\(10\text{ minutes}\)) [1]

During a collision, the cyclist's head undergoes a specific change in momentum as its velocity is reduced to zero. The formula relating force, momentum, and time is \(F = \frac{\Delta p}{t}\). By using a deformable polystyrene foam layer, the distance and therefore the time \(t\) over which this collision occurs is increased. As the collision time \(t\) increases, the rate of change of momentum \(\frac{\Delta p}{t}\) decreases. Consequently, the average impact force \(F\) acting on the cyclist's skull and brain is greatly reduced, decreasing the risk of severe injury.

1. MP1: Identifying that the foam layer deforms/collapses, which increases the time duration of the impact (1 mark). 2. MP2: Explaining that this reduces the rate of change of momentum (1 mark). 3. MP3: Referencing the formula \(F = \frac{\Delta p}{t}\) or stating that force is equal to the rate of change of momentum (1 mark). 4. MP4: Concluding that the average force acting on the head is reduced (0.5 marks).

Inside the optical fibre, light rays travel through the central core. When they hit the boundary with the outer cladding, they undergo total internal reflection (TIR). This keeps the light trapped inside the core, allowing it to travel long distances. Two conditions are necessary for TIR to occur: first, the light must travel in a medium of higher refractive index (the core) towards a medium of lower refractive index (the cladding). Second, the angle of incidence of the light ray at the boundary must be greater than the critical angle for that boundary.

1. MP1: State that light undergoes total internal reflection (TIR) at the core-cladding boundary (1 mark). 2. MP2: Explain that the core has a higher refractive index than the cladding (1 mark). 3. MP3: Explain that the angle of incidence at the boundary must be greater than the critical angle (1 mark). 4. MP4: State that light must be travelling in the more optically dense medium (0.5 marks).

When an alternating current (a.c.) flows through the primary coil of the transformer, it generates an alternating (continuously changing) magnetic field. The soft iron core easily magnetises and demagnetises, linking this changing magnetic field directly to the secondary coil. As the magnetic field lines continuously cut through the turns of the secondary coil, a changing magnetic flux is experienced by the secondary coil. This induces an alternating potential difference (voltage) across the terminals of the secondary coil via electromagnetic induction.

1. MP1: State that alternating current in the primary coil produces a changing/alternating magnetic field (1 mark). 2. MP2: State that the soft iron core transfers/links this magnetic field to the secondary coil (1 mark). 3. MP3: State that the changing magnetic field cuts through the secondary coil (1 mark). 4. MP4: State that this induces an alternating voltage via electromagnetic induction (0.5 marks).

When the temperature of the gas is increased, the average kinetic energy of the gas molecules increases, which means they move at higher average speeds. Since the volume of the canister is constant, these faster-moving molecules collide with the inner walls of the canister more frequently. Furthermore, because they are moving faster, each collision involves a larger change in momentum, resulting in a larger force during each impact. The total average force exerted on the walls increases, and since pressure is defined as force divided by area (\(p = \frac{F}{A}\)), the pressure increases.

1. MP1: State that increasing temperature increases the average kinetic energy or speed of the gas molecules (1 mark). 2. MP2: State that molecules collide with the canister walls more frequently (1 mark). 3. MP3: State that molecules collide with the walls with greater force or undergo a greater change in momentum per collision (1 mark). 4. MP4: Reference \(p = \frac{F}{A}\) to conclude that the increased total force per unit area results in increased pressure (0.5 marks).

Once a massive star finishes fusing hydrogen in its core, it leaves the main sequence and expands, cooling at its outer surface to become a red supergiant. Inside, heavier elements are fused until iron forms in the core. Without outer radiation pressure to support it, the core suddenly collapses under gravity, triggering a massive shockwave and explosion known as a supernova, which ejects the outer layers of the star. The remnant core is compressed into an incredibly dense neutron star. If the remaining mass is exceptionally large, gravity is strong enough to collapse it further into a black hole.

1. MP1: Explain that the star expands and cools to become a red supergiant (1 mark). 2. MP2: Explain that the core collapses rapidly, resulting in a supernova explosion that ejects outer layers (1 mark). 3. MP3: State that the remaining core collapses to form a neutron star (1 mark). 4. MP4: Mention that if the mass is extremely large, it collapses into a black hole (0.5 marks).

To minimize radiation exposure, the teacher should follow several safety practices: 1. Use long-handled tongs when holding the sources. Since intensity decreases with distance (following the inverse square law), this greatly reduces the dose received. 2. Store the radioactive sources in a sealed, lead-lined wooden box immediately after use, which absorbs the radiation and prevents exposure when the sources are idle. 3. Point the source apertures away from all people (both the teacher and the students) and avoid looking directly into the source container, ensuring that highly ionizing radiation does not enter sensitive organs like the eyes.

1. MP1: Explain the use of long-handled tongs to increase distance from the source (1 mark). 2. MP2: Explain the use of a lead-lined container for safe storage when not in use to absorb/shield radiation (1 mark). 3. MP3: Explain the precaution of pointing the source away from people to avoid direct exposure paths (1 mark). 4. MP4: Mention keeping the exposure time as short as possible / wearing a film badge to monitor dose (0.5 marks).

Heat loss occurs through three mechanisms: conduction, convection, and radiation. The vacuum between the two glass walls contains no matter, completely preventing conduction and convection because both require a medium of particles to transfer thermal energy. The plastic lid, being a poor thermal conductor (insulator), reduces conduction through the top, and by closing the container, it prevents warm air currents from rising and escaping (convection). The silvered inner surfaces of the glass walls are poor emitters and good reflectors of infrared radiation, reflecting radiant energy back into the hot liquid.

1. MP1: Explain that the vacuum prevents conduction and convection because these processes require particles/a medium (1 mark). 2. MP2: Explain that the plastic lid reduces conduction (as plastic is an insulator) and prevents convection currents from escaping the top (1 mark). 3. MP3: Explain that shiny/silvered walls reflect infrared radiation back into the liquid (or are poor emitters/absorbers) (1 mark). 4. MP4: Mention that glass is a poor conductor, further reducing conduction (0.5 marks).

When an electrical fault occurs and the live wire makes contact with the metal casing of the kettle, the casing becomes live. However, the casing is connected to the earth wire, which provides a path of extremely low resistance to the ground. This low resistance causes a massive surge in current to flow from the live wire, through the casing and the earth wire, to the ground. Because this current exceeds the fuse's current rating, the fuse wire in the plug heats up rapidly and melts ('blows'). This breaks the circuit, isolating the appliance from the high-voltage live supply, which ensures that anyone touching the kettle will not receive a dangerous electric shock.

1. MP1: State that the current flows from the live wire through the metal casing to the earth wire (1 mark). 2. MP2: State that this low resistance path causes a very large current surge (1 mark). 3. MP3: Explain that the large current exceeds the fuse rating, causing the fuse wire to melt and break the circuit (1 mark). 4. MP4: Conclude that this disconnects the live supply and prevents the user from receiving an electric shock (0.5 marks).

When a fault occurs and the live wire touches the metal casing, a low-resistance path to the ground is provided by the earth wire. This causes a huge surge in current to flow from the live wire through the casing and down the earth wire. This high current exceeds the fuse's current rating, causing the wire inside the fuse to heat up and melt. This breaks the circuit, disconnecting the live supply and preventing the metal casing from remaining at a high voltage, thereby protecting the user from an electric shock.

1 mark: State that the earth wire provides a low resistance path to earth (or that current flows to earth). 1 mark: State that this causes a large current / surge of current to flow. 1 mark: Explain that this high current melts the fuse, which breaks the circuit / cuts off the live supply.

Heating the canister transfers thermal energy to the gas, increasing the average kinetic energy of the gas particles. Since their kinetic energy is higher, the particles move faster. This means they collide with the inner walls of the rigid canister more frequently (more collisions per second). Additionally, they collide with a greater force because of a larger change in momentum per collision. Since pressure is defined as force per unit area, \( p = F/A \), the increased overall force exerted on the constant surface area of the walls results in an increased pressure.

1 mark: Identify that gas particles gain kinetic energy / move faster. 1 mark: State that particles collide with the walls of the container more frequently / more often. 1 mark: State that particles collide with more force / greater change in momentum (resulting in a larger total force on the walls). Reject: particles expand / particles vibrate more.

Use the equation: \(v^2 = u^2 + 2as\). Given: initial velocity, \(u = 0\text{ m/s}\); final velocity, \(v = 15\text{ m/s}\); distance, \(s = 45\text{ m}\). Rearranging for acceleration \(a\): \(a = \frac{v^2 - u^2}{2s}\). Substituting the values: \(a = \frac{15^2 - 0^2}{2 \times 45} = \frac{225}{90} = 2.5\text{ m/s}^2\).

1 mark for selecting the correct equation: \(v^2 = u^2 + 2as\). 1 mark for correct rearrangement or substitution. 1.8 marks for the correct final answer with the unit \(\text{m/s}^2\).

Use conservation of momentum: total momentum before collision = total momentum after collision. \(m_1 v_1 + m_2 v_2 = (m_1 + m_2) v\). Substituting the given values: \((1.2 \times 3.5) + (0.8 \times 0) = (1.2 + 0.8) \times v\). \(4.2 + 0 = 2.0 \times v\). Therefore, \(v = \frac{4.2}{2.0} = 2.1\text{ m/s}\).

1 mark for using conservation of momentum. 1 mark for calculating momentum before collision as \(4.2\text{ kg m/s}\). 1.8 marks for correct final velocity \(2.1\text{ m/s}\).

To exert the maximum pressure, the block must rest on its smallest surface area. Smallest area, \(A = 0.10\text{ m} \times 0.20\text{ m} = 0.020\text{ m}^2\). Weight of the block, \(F = m \times g = 6.0\text{ kg} \times 10\text{ N/kg} = 60\text{ N}\). Pressure, \(P = \frac{F}{A} = \frac{60\text{ N}}{0.020\text{ m}^2} = 3000\text{ Pa}\) (or \(3000\text{ N/m}^2\)).

1 mark for identifying the smallest surface area of \(0.020\text{ m}^2\). 1 mark for calculating the weight as \(60\text{ N}\). 1.8 marks for the correct pressure of \(3000\text{ Pa}\) (or \(3000\text{ N/m}^2\)).

Convert time to seconds: \(t = 5.0\text{ minutes} = 5.0 \times 60 = 300\text{ s}\). Use the formula for energy transferred: \(E = I \times V \times t\). Substituting the values: \(E = 1.5\text{ A} \times 12\text{ V} \times 300\text{ s} = 5400\text{ J}\).

1 mark for converting minutes to seconds (\(300\text{ s}\)). 1 mark for using the correct formula \(E = I \times V \times t\). 1.8 marks for the correct final answer of \(5400\text{ J}\) (or \(5.4\text{ kJ}\)).

Use Snell's Law for refractive index: \(n = \frac{\sin(i)}{\sin(r)}\). Here, \(i = 42^\circ\) and \(r = 26^\circ\). \(n = \frac{\sin(42^\circ)}{\sin(26^\circ)} = \frac{0.66913}{0.43837} \approx 1.5264\). Rounded to 2 decimal places, \(n = 1.53\).

1 mark for using Snell's Law \(n = \frac{\sin(i)}{\sin(r)}\). 1 mark for substituting the values correctly. 1.8 marks for the correct value of \(1.53\) (no units required).

Convert the temperature to Kelvin: \(T_1 = 27 + 273 = 300\text{ K}\). Use the pressure law formula (volume constant): \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\). Rearranging for \(T_2\): \(T_2 = T_1 \times \frac{P_2}{P_1}\). Substituting the values: \(T_2 = 300 \times \frac{1.8 \times 10^5}{1.2 \times 10^5} = 300 \times 1.5 = 450\text{ K}\). Convert back to Celsius: \(T_2 = 450 - 273 = 177^\circ\text{C}\).

1 mark for converting temperature to Kelvin (\(300\text{ K}\)). 1 mark for rearranging and calculating \(T_2 = 450\text{ K}\). 1.8 marks for converting back to Celsius to get \(177^\circ\text{C}\).

Use the latent heat formula: \(Q = m \times L\). The mass that was vaporised is \(m = 0.15\text{ kg}\) (not the total starting mass of \(0.80\text{ kg}\)). Substituting the values: \(Q = 0.15\text{ kg} \times 2.26 \times 10^6\text{ J/kg} = 339,000\text{ J}\) (or \(3.39 \times 10^5\text{ J}\)).

1 mark for identifying the correct mass of water vaporised (\(0.15\text{ kg}\)). 1 mark for using the formula \(Q = m \times L\). 1.8 marks for the correct energy calculation of \(339,000\text{ J}\) (or \(339\text{ kJ}\) or \(3.39 \times 10^5\text{ J}\)).

First calculate the work done (which is the gain in gravitational potential energy): \(W = m \times g \times h\). \(W = 450 \times 10 \times 12 = 54,000\text{ J}\). Next, calculate the power output: \(P = \frac{W}{t}\). \(P = \frac{54,000}{15} = 3600\text{ W}\) (or \(3.6\text{ kW}\)).

1 mark for calculating work done (\(54,000\text{ J}\)). 1 mark for using the formula \(P = \frac{W}{t}\). 1.8 marks for the correct power calculation of \(3600\text{ W}\) (or \(3.6\text{ kW}\)).

First, calculate the volume of the wooden block:
\(V = 0.20\text{ m} \times 0.15\text{ m} \times 0.40\text{ m} = 0.012\text{ m}^3\)

Next, calculate the mass of the block using \(\text{mass} = \text{density} \times \text{volume}\):
\(m = 650\text{ kg/m}^3 \times 0.012\text{ m}^3 = 7.8\text{ kg}\)

Calculate the weight (force exerted downwards) using \(W = m \times g\):
\(F = 7.8\text{ kg} \times 10\text{ N/kg} = 78\text{ N}\)

To find the maximum pressure, we must divide this force by the minimum surface area. The minimum area is the product of the two smallest dimensions:
\(A_{\text{min}} = 0.20\text{ m} \times 0.15\text{ m} = 0.03\text{ m}^2\)

Finally, calculate the maximum pressure using \(P = \frac{F}{A}\):
\(P_{\text{max}} = \frac{78\text{ N}}{0.03\text{ m}^2} = 2600\text{ Pa}\)

M1: Calculation of volume \(0.012\text{ m}^3\) and mass \(7.8\text{ kg}\) (allow 1 mark for correct method if calculation error occurs).
M2: Calculation of force/weight of block \(78\text{ N}\) (accept \(76.44\text{ N}\) if \(g = 9.8\text{ N/kg}\) or \(76.5\text{ N}\) if \(g = 9.81\text{ N/kg}\) is used).
M3: Identifying and calculating the minimum surface area \(0.03\text{ m}^2\) and substituting into \(P = \frac{F}{A}\).
A1: Correct final answer of \(2600\text{ Pa}\) (or \(2600\text{ N/m}^2\)) with matching unit. (Accept \(2548\text{ Pa}\) to \(2551\text{ Pa}\) if \(g = 9.8\) or \(9.81\text{ N/kg}\) is used).

Distance travelled can be found by calculating the total area under the velocity-time graph. The area is made up of a triangle and a rectangle. 1) Area of the triangle (from 0 to 4 s): \(\text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = 0.5 \times 4.0\text{ s} \times 6.0\text{ m/s} = 12.0\text{ m}\). 2) Area of the rectangle (from 4 to 10 s): \(\text{Area}_2 = \text{width} \times \text{height} = (10.0\text{ s} - 4.0\text{ s}) \times 6.0\text{ m/s} = 6.0\text{ s} \times 6.0\text{ m/s} = 36.0\text{ m}\). 3) Total distance = \(\text{Area}_1 + \text{Area}_2 = 12.0\text{ m} + 36.0\text{ m} = 48.0\text{ m}\). Alternatively, using the trapezium area formula: \(\text{Distance} = \frac{(6.0 + 10.0)}{2} \times 6.0 = 48\text{ m}\).

1 mark: For identifying that the total distance is the area under the velocity-time graph. 1 mark: For the correct calculation of individual areas (12 m and 36 m) or correct substitution into the trapezium formula. 1 mark: For the correct final value of 48 with the correct unit (m).

We use Snell's Law to calculate the refractive index: \(n = \frac{\sin(i)}{\sin(r)}\), where \(i = 40.0^\circ\) and \(r = 25.0^\circ\). Substituting these values: \(n = \frac{\sin(40.0^\circ)}{\sin(25.0^\circ)}\). Since \(\sin(40.0^\circ) \approx 0.6428\) and \(\sin(25.0^\circ) \approx 0.4226\), we get: \(n = \frac{0.6428}{0.4226} \approx 1.52\).

1 mark: State the correct formula \(n = \frac{\sin(i)}{\sin(r)}\). 1 mark: Correct substitution of values into the formula. 1 mark: Correct calculation of refractive index to 2 decimal places (1.52).

At 10 minutes, the substance is undergoing a change of state (melting), so it is a mixture of solid ice and liquid water. Between 5 and 15 minutes, the temperature remains constant at \(0^\circ\text{C}\) because the thermal energy supplied is being used to overcome (break) the intermolecular forces (bonds) holding the water molecules together in the solid lattice. This means the average kinetic energy of the molecules does not increase during this phase change, so the temperature remains constant.

1 mark: State that the substance is a mixture of solid and liquid (or is melting). 1 mark: State that the energy supplied is used to break intermolecular bonds/forces between molecules. 1 mark: State that the kinetic energy of the particles remains constant during this time (hence no temperature rise).

As the voltage across the filament lamp increases, the current increases, which causes the temperature of the metal filament to rise. The resistance of the filament lamp increases as a result. At higher temperatures, the positive metal ions in the filament lattice vibrate more rapidly and with a larger amplitude. This increases the frequency of collisions between the flowing conduction electrons and the vibrating metal ions, which resists the flow of charge and thus increases the resistance.

1 mark: State that the resistance increases as the voltage increases. 1 mark: Explain that the increased current causes the temperature of the filament to rise, which increases the vibrations of the metal ions. 1 mark: Explain that this increases the collision frequency between conduction electrons and the vibrating ions.

First, convert the extension from centimeters to meters: \(x = 12.0\text{ cm} = 0.120\text{ m}\). Using Hooke's Law in the linear region: \(F = k \times x\), where \(F = 8.0\text{ N}\). Rearrange to solve for the spring constant: \(k = \frac{F}{x} = \frac{8.0\text{ N}}{0.120\text{ m}} \approx 66.7\text{ N/m}\) (or \(67\text{ N/m}\)). The point at \(8.0\text{ N}\) where the graph ceases to be a straight line represents the limit of proportionality (or elastic limit).

1 mark: Convert the extension to meters (\(0.12\text{ m}\)) or show correct use of \(k = \frac{F}{x}\). 1 mark: Correct calculation of spring constant with correct units (\(66.7\text{ N/m}\) or \(67\text{ N/m}\)). 1 mark: Identify the point as the limit of proportionality or elastic limit.

Nuclear fusion involves forcing two positively charged nuclei close enough together so that they can fuse. Since both nuclei have a positive charge, they experience a strong electrostatic repulsion. High temperature is required to give the nuclei high kinetic energy, enabling them to move fast enough to overcome this repulsive force. High pressure is required to keep the nuclei close together, increasing the collision rate and therefore the probability of fusion occurring.

1 mark: Reference to giving nuclei high kinetic energy to overcome electrostatic repulsion. 0.8 marks: Reference to high pressure keeping nuclei close together to increase the frequency of collisions.

We use the transformer relationship: \(V_p / V_s = N_p / N_s\). Substituting the given values: \(230 / V_s = 120 / 3000\). Rearranging for \(V_s\): \(V_s = 230 \times (3000 / 120) = 5750\) V.

1 mark: Correct recall and substitution into the transformer equation: \(230 / V_s = 120 / 3000\). 0.8 marks: Correct evaluation of output voltage to give 5750 V with unit.

After the main sequence, a massive star expands and cools to become a red supergiant. It then undergoes a massive explosion called a supernova. The remaining core collapses to form either a very dense neutron star or, if the mass is extremely large, a black hole.

1 mark: Mentioning the expansion into a red supergiant and the supernova explosion. 0.8 marks: Identifying the final remnant as either a neutron star or a black hole depending on the mass.

Using the pressure law: \(P_1 / T_1 = P_2 / T_2\), temperatures must be converted to Kelvin: \(T_1 = 20 + 273 = 293\) K, and \(T_2 = 80 + 273 = 353\) K. Substituting the values: \(101 / 293 = P_2 / 353\). Rearranging for \(P_2\): \(P_2 = 101 \times (353 / 293) \approx 121.7\) kPa, which rounds to 122 kPa.

1 mark: Converting temperatures to Kelvin (293 K and 353 K) and substituting into the formula. 0.8 marks: Calculating the final pressure of 122 kPa with correct unit.

Using the conservation of momentum: Total momentum before = Total momentum after. \(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v\). Substituting the values: \(0.45 \times 2.0 + 0 = (0.45 + 0.75) \times v\). This simplifies to \(0.90 = 1.20 v\). Solving for \(v\): \(v = 0.90 / 1.20 = 0.75\) m/s.

1 mark: Calculating initial momentum (0.90 kg m/s) and equating to final total momentum expression (1.20 * v). 0.8 marks: Calculating the correct final velocity of 0.75 m/s with unit.

The critical angle \(c\) is calculated using \(\sin c = n_2 / n_1\), where \(n_1\) is the refractive index of the denser medium (core) and \(n_2\) is the refractive index of the less dense medium (cladding). \(\sin c = 1.41 / 1.48 \approx 0.9527\). Taking the inverse sine: \(c = \arcsin(0.9527) \approx 72.3^{\circ}\).

1 mark: Recalling and substituting into critical angle formula \(\sin c = 1.41 / 1.48\). 0.8 marks: Correct calculation of critical angle as 72.3 degrees.

Applying conservation of energy: Heat lost by copper = Heat gained by water. \(m_c \times c_c \times \Delta T_c = m_w \times c_w \times \Delta T_w\). Here, \(\Delta T_c = 100 - 25 = 75\) \(^{\circ}\)C, and \(\Delta T_w = 25 - 20 = 5\) \(^{\circ}\)C. Substituting the values: \(0.15 \times c_c \times 75 = 0.20 \times 4200 \times 5\). This gives \(11.25 \times c_c = 4200\). Solving for \(c_c\): \(c_c = 4200 / 11.25 \approx 373.3\) J/kg\(^{\circ}\)C.

1 mark: Calculating the heat gained by water (4200 J) and setting up the conservation equation. 0.8 marks: Correctly calculating the specific heat capacity of copper as 373 J/kg°C with correct unit.

Using the Doppler shift formula: \(\Delta \lambda / \lambda_0 = v / c\). First, calculate the change in wavelength: \(\Delta \lambda = 662 - 656 = 6\) nm. Substituting the values: \(6 / 656 = v / (3.0 \times 10^8)\). Rearranging to solve for \(v\): \(v = (6 / 656) \times 3.0 \times 10^8 \approx 2.74 \times 10^6\) m/s.

1 mark: Calculating the change in wavelength (6 nm) and substituting correctly into the Doppler equation. 0.8 marks: Evaluating the correct recessional speed of \(2.7 \times 10^6\) m/s with unit.

Nuclear fusion involves joining two light nuclei (e.g., hydrogen isotopes) to form a heavier nucleus. Since nuclei are positively charged, they experience a very strong electrostatic force of repulsion when brought close together. High temperatures are required because they provide the nuclei with high kinetic energies, allowing them to overcome this electrostatic repulsion. High pressures are necessary to increase the collision rate, ensuring that the nuclei come close enough for the strong nuclear force to bind them.

Award 1 mark for identifying that nuclei are positively charged and experience electrostatic repulsion. Award 1 mark for explaining that high temperatures provide enough kinetic energy to overcome this repulsion. Award 1 mark for explaining that high pressure increases the collision rate / brings nuclei close enough for the strong nuclear force to act.

As the magnet falls through the copper tube, its moving magnetic field cuts through the conducting walls of the tube, inducing an electromotive force (e.m.f.) and a current in the copper. According to Lenz's law, this induced current creates its own magnetic field that opposes the motion of the falling magnet, resulting in an upward magnetic force. This upward force reduces the net downward force on the magnet, slowing it down. In contrast, plastic is an electrical insulator, so no current is induced and no opposing magnetic force is created.

Award 1 mark for stating that the falling magnet induces an e.m.f. / current in the conducting copper tube. Award 1 mark for explaining that this current creates a magnetic field that opposes the magnet's motion (Lenz's law). Award 1 mark for stating that this opposing force reduces the net downward force, whereas no current is induced in the insulating plastic tube.

Increasing the temperature of the cylinder increases the average kinetic energy and speed of the gas molecules. Because the molecules are moving faster, they collide with the walls of the rigid cylinder more frequently. Additionally, they collide with a greater force because each collision involves a larger change in momentum. Since pressure is force per unit area, these more frequent and energetic collisions result in an increased overall pressure.

Award 1 mark for stating that higher temperature increases the average kinetic energy / speed of the gas particles. Award 1 mark for explaining that particles collide with the walls of the container more frequently. Award 1 mark for explaining that collisions occur with greater force / greater change in momentum, resulting in higher pressure.

When a person lands on a safety cushion, the cushion compresses, which increases the time duration \(\Delta t\) over which the person comes to a complete stop. The total change in momentum \(\Delta p\) required to stop the person is constant. Since the average force experienced is given by the rate of change of momentum \(F = \frac{\Delta p}{\Delta t}\), increasing the duration of the impact significantly reduces the average force exerted on the person, minimizing injury.

Award 1 mark for stating that the cushion increases the time taken for the person to stop. Award 1 mark for stating that the change in momentum is constant. Award 1 mark for using the formula \(F = \frac{\Delta p}{\Delta t}\) to explain that a larger time results in a smaller average force.

Alpha particles have a relatively large mass and a positive charge of \(+2e\). This high charge and mass mean they interact strongly with orbital electrons of atoms they pass, causing frequent ionizations. Because they ionize so many atoms, they lose their kinetic energy rapidly and can only travel a few centimeters in air. Gamma rays are electromagnetic waves with no mass and no charge, making them highly penetrating and weakly ionizing because they have a very low probability of interacting with atoms, allowing them to travel great distances in air.

Award 1 mark for explaining that alpha particles are highly ionizing due to their large charge and mass, causing rapid energy loss and a short range. Award 1 mark for explaining that gamma rays are uncharged and massless, giving them a very low ionization probability. Award 1 mark for relating the low rate of interaction of gamma rays to their long range in air.

The critical angle is the angle of incidence in an optically denser medium for which the angle of refraction is exactly \(90^\circ\) (refracting along the boundary). For total internal reflection to occur, two conditions must be met: first, the light must travel from a medium with a higher refractive index (more optically dense) to one with a lower refractive index (less optically dense). Second, the angle of incidence must be greater than the critical angle.

Award 1 mark for defining critical angle as the angle of incidence where the angle of refraction is \(90^\circ\). Award 1 mark for stating that the light must travel from a more dense to a less dense medium. Award 1 mark for stating that the angle of incidence must exceed the critical angle.

If a fault causes the live wire to touch the metal casing, the low-resistance earth wire provides a safe path for the current to flow directly to the ground instead of through a user. This extremely low resistance path causes a very large current surge to flow through the live wire. This high current exceeds the rating of the fuse, causing the fuse wire to heat up and melt (blow), which breaks the circuit and isolates the appliance from the mains.

Award 1 mark for stating that the low-resistance earth wire provides a safe path to ground. Award 1 mark for explaining that this low-resistance path causes a large current surge. Award 1 mark for explaining that the large current exceeds the fuse rating, melting the fuse and breaking the circuit.

When hydrogen fuel in the core of a main sequence star is depleted, nuclear fusion in the core ceases, reducing outward thermal and radiation pressure. Gravity causes the core to contract and heat up. This heating triggers the fusion of hydrogen in a shell surrounding the core. The massive energy release from this shell fusion increases the outward pressure, causing the outer layers of the star to expand dramatically and cool, resulting in a red giant.

Award 1 mark for stating that core fusion stops, causing the core to contract and heat up under gravity. Award 1 mark for stating that fusion begins in a shell surrounding the core. Award 1 mark for stating that this shell fusion causes the outer layers of the star to expand and cool.

1. Convert contact time to seconds:
\(t = 8.0\text{ ms} = 0.0080\text{ s}\)

2. Calculate the change in momentum (taking the initial direction as positive):
Initial momentum: \(p_i = m u = 0.058\text{ kg} \times 32\text{ m/s} = +1.856\text{ kg m/s}\)
Final momentum: \(p_f = m v = 0.058\text{ kg} \times (-24\text{ m/s}) = -1.392\text{ kg m/s}\)
Change in momentum: \(\Delta p = p_f - p_i = -1.392 - 1.856 = -3.248\text{ kg m/s}\)
Magnitude of momentum change = \(3.248\text{ kg m/s}\)

3. Calculate the average force:
\(F = \frac{\Delta p}{t} = \frac{3.248\text{ kg m/s}}{0.0080\text{ s}} = 406\text{ N}\)

Rounding to 2 significant figures gives \(410\text{ N}\).

M1: Substitution of values into the momentum formula or calculation of either initial or final momentum (e.g., 1.9 kg m/s or 1.4 kg m/s)
M1: Correctly calculating the total change in momentum by adding the magnitudes of initial and final momentum (3.25 kg m/s or 3.2 kg m/s)
M1: Correct calculation of force by dividing change in momentum by time in seconds (0.0080 s)
A0.6: Correct final value of 406 N or 410 N with appropriate unit (N or kg m/s²)

1. Recall the transformer relationship for turns and current for a 100% efficient transformer:
\(\frac{I_p}{I_s} = \frac{N_s}{N_p}\)

2. Rearrange the formula to solve for primary current (\(I_p\)):
\(I_p = I_s \times \frac{N_s}{N_p}\)

3. Substitute the values:
\(I_p = 1.5\text{ A} \times \frac{80}{1200}\)
\(I_p = 1.5 \times 0.0667\)
\(I_p = 0.10\text{ A}\)

M1: Recognition and recall of transformer efficiency relation (V_p * I_p = V_s * I_s) or transformer ratio equation (I_p / I_s = N_s / N_p)
M1: Rearrangement of equation to make I_p the subject
M1: Substitution of values into the correct equation
A0.6: Correct final current value of 0.10 A (or 0.1 A) with the unit A

1. Convert temperatures from Celsius to Kelvin:
\(T_1 = 20 + 273 = 293\text{ K}\)
\(T_2 = 85 + 273 = 358\text{ K}\)

2. State the pressure-temperature relationship for a constant volume gas (Pressure Law):
\(\frac{p_1}{T_1} = \frac{p_2}{T_2}\)

3. Rearrange the formula to solve for the final pressure (\(p_2\)):
\(p_2 = p_1 \times \frac{T_2}{T_1}\)

4. Substitute the values:
\(p_2 = 101\text{ kPa} \times \frac{358}{293} \approx 123.41\text{ kPa}\)

To 3 significant figures, the final pressure is \(123\text{ kPa}\).

M1: Conversion of both temperatures to Kelvin (293 K and 358 K)
M1: Correct use of the pressure-temperature relationship (p_1 / T_1 = p_2 / T_2)
M1: Correct rearrangement and substitution to find p_2
A0.6: Final answer of 123 kPa (or 123000 Pa) with unit

1. Calculate the change in wavelength (\(\Delta \lambda\)):
\(\Delta \lambda = 672.1\text{ nm} - 656.3\text{ nm} = 15.8\text{ nm}\)

2. Recall the Doppler redshift formula:
\(\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}\)
where \(\lambda_0\) is the reference wavelength (\(656.3\text{ nm}\)).

3. Rearrange the formula to solve for recessional velocity (\(v\)):
\(v = c \times \frac{\Delta \lambda}{\lambda_0}\)

4. Substitute the values:
\(v = 3.0 \times 10^8\text{ m/s} \times \frac{15.8\text{ nm}}{656.3\text{ nm}}\)
\(v \approx 3.0 \times 10^8 \times 0.024073\)
\(v \approx 7.22 \times 10^6\text{ m/s}\)

To 2 significant figures, the velocity is \(7.2 \times 10^6\text{ m/s}\).

M1: Calculation of change in wavelength, delta lambda = 15.8 nm
M1: Correct use of Doppler equation (change in wavelength / reference wavelength = v / c)
M1: Rearrangement of the formula and substitution of known variables
A0.6: Correct recessional velocity of 7.2 * 10^6 m/s (or 7.22 * 10^6 m/s) with unit m/s

1. Convert power and time to SI units:
Power, \(P = 2.2\text{ kW} = 2200\text{ W}\)
Time, \(t = 3.0\text{ minutes} = 3.0 \times 60 = 180\text{ s}\)

2. Calculate the total thermal energy transferred (\(Q\)):
\(Q = P \times t = 2200\text{ W} \times 180\text{ s} = 396000\text{ J}\)

3. Use the latent heat formula to find the mass (\(m\)):
\(Q = m \times L_v\)
\(m = \frac{Q}{L_v} = \frac{396000\text{ J}}{2.26 \times 10^6\text{ J/kg}} \approx 0.1752\text{ kg}\)

To 2 significant figures, the mass of water vaporized is \(0.18\text{ kg}\) (or \(180\text{ g}\)).

M1: Conversion of time to seconds (180 s) and/or power to watts (2200 W)
M1: Calculation of energy transferred using E = P * t (yielding 396 kJ)
M1: Rearrangement of latent heat formula Q = mL to find mass (m = Q / L)
A0.6: Correct mass of 0.18 kg (or 180 g) with unit

1. Convert charge capacity from \(\text{mAh}\) to coulombs (\(\text{C}\)):
\(1\text{ mAh} = 10^{-3}\text{ A} \times 3600\text{ s} = 3.6\text{ C}\)
\(10000\text{ mAh} = 10\text{ Ah}\)
\(Q = 10\text{ A} \times 3600\text{ s} = 36000\text{ C}\)

2. Recall the relationship between energy, charge, and voltage:
\(E = Q \times V\)

3. Calculate the energy in joules:
\(E = 36000\text{ C} \times 5.0\text{ V} = 180000\text{ J}\)

4. Convert joules to megajoules (\(\text{MJ}\)):
\(E = \frac{180000}{1000000} = 0.18\text{ MJ}\)

M1: Conversion of charge from mAh or Ah to Coulombs (36000 C)
M1: Use of E = Q * V (or equivalent formula combining power and time)
M1: Conversion of the final energy from Joules into Megajoules (dividing by 1,000,000)
A0.6: Correct final answer of 0.18 MJ with unit MJ

1. Use the critical angle formula to calculate the refractive index (\(n\)) of the glass:
\(\sin c = \frac{1}{n}\)
\(n = \frac{1}{\sin(41.8^{\circ})} = \frac{1}{0.6665} \approx 1.50\)

2. Write down Snell's law for refraction at the boundary:
\(n_{\text{glass}} \sin i = n_{\text{air}} \sin r\)
Since \(n_{\text{air}} \approx 1\):
\(n \sin i = \sin r\)

3. Substitute the values of \(n\) and the angle of incidence \(i = 30.0^{\circ}\):
\(1.50 \times \sin(30.0^{\circ}) = \sin r\)
\(1.50 \times 0.500 = \sin r\)
\(\sin r = 0.750\)

4. Calculate the angle of refraction \(r\):
\(r = \sin^{-1}(0.750) \approx 48.59^{\circ}\)

To 3 significant figures, the angle of refraction is \(48.6^{\circ}\).

M1: Recall and use of sin c = 1 / n to find the refractive index of glass as 1.5
M1: Correct use of Snell's law (n_1 * sin i = n_2 * sin r)
M1: Rearrangement to solve for sin r (yielding sin r = 0.75)
A0.6: Correct angle of refraction of 48.6 degrees with units

To find the total distance travelled from a velocity-time graph, we calculate the total area under the graph. 1. First part of the journey (0 to 4 seconds) is a triangle: \(\text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4\text{ s} \times 6\text{ m/s} = 12\text{ m}\). 2. Second part of the journey (4 to 10 seconds) is a rectangle: \(\text{Area}_2 = \text{base} \times \text{height} = (10 - 4)\text{ s} \times 6\text{ m/s} = 6\text{ s} \times 6\text{ m/s} = 36\text{ m}\). 3. Total distance: \(\text{Total Distance} = \text{Area}_1 + \text{Area}_2 = 12\text{ m} + 36\text{ m} = 48\text{ m}\).

1. [1 mark] Recalling that distance is equal to the area under a velocity-time graph (or writing down correct area equations). 2. [1 mark] Correctly calculating either individual area: triangle area = \(12\text{ m}\) or rectangle area = \(36\text{ m}\). 3. [1 mark] Correct total sum of \(48\). 4. [0.5 marks] Stating the correct unit (\(\text{m}\) or meters).

1. The melting point is the constant temperature during the first change of state (solid to liquid), which is \(80^\circ\text{C}\). 2. Between \(t = 3\text{ min}\) and \(t = 8\text{ min}\), the substance is melting. The thermal energy supplied is absorbed to break or weaken the intermolecular bonds (forces of attraction) holding the solid particles together, rather than increasing their average kinetic energy (so temperature remains constant). 3. The arrangement of the particles changes from a regular lattice structure (as a solid) to a random, closely packed arrangement (as a liquid).

1. [1 mark] Identifying the melting point as \(80^\circ\text{C}\) (must include unit). 2. [1 mark] Explaining that the thermal energy supplied is used to break/overcome intermolecular bonds/forces of attraction. 3. [1 mark] Describing the transition of particle arrangement from regular/ordered to random/unordered. 4. [0.5 marks] Stating that the temperature or average kinetic energy of the particles remains constant during this phase change.

1. The vertical distance from crest to trough is twice the amplitude. \(\text{Amplitude} = \frac{\text{vertical distance}}{2} = \frac{0.16\text{ m}}{2} = 0.08\text{ m}\). 2. The distance from the 1st crest to the 3rd crest represents exactly 2 full wavelengths (\(2\lambda\)). \(2\lambda = 0.60\text{ m}\), so \(\lambda = \frac{0.60\text{ m}}{2} = 0.30\text{ m}\).

1. [1 mark] Showing a valid calculation or explanation for amplitude (e.g., dividing vertical distance by 2). 2. [1 mark] Correct value of amplitude: \(0.08\text{ m}\) (accept \(8\text{ cm}\)). 3. [1 mark] Recalling/showing that the distance between 1st and 3rd crest is equivalent to 2 wavelengths (e.g., \(2\lambda = 0.60\text{ m}\)). 4. [0.5 marks] Correct value of wavelength: \(0.30\text{ m}\) (accept \(30\text{ cm}\)).

1. In the linear region, Hooke's Law is obeyed: \(F = kx\), where \(k\) is the stiffness (spring constant). 2. First, convert the extension from millimetres to metres: \(x = 3.0\text{ mm} = 3.0 \times 10^{-3}\text{ m} = 0.003\text{ m}\). 3. Calculate \(k\): \(k = \frac{F}{x} = \frac{15\text{ N}}{0.003\text{ m}} = 5000\text{ N/m}\). 4. The point on the graph where the linear relationship ends and the graph begins to curve is called the limit of proportionality (or elastic limit).

1. [1 mark] Converting the extension unit correctly: \(3.0\text{ mm} = 0.003\text{ m}\). 2. [1 mark] Recalling and using \(F = kx\) (or \(k = F/x\)). 3. [1 mark] Correct calculation of stiffness: \(5000\text{ N/m}\) (or \(5\text{ kN/m}\)). 4. [0.5 marks] Identifying the point as the limit of proportionality (accept 'elastic limit').