2024 AQA GCSE Physics 8463 Practice Paper with Answers

Use the equation for kinetic energy:
\(E_k = \frac{1}{2} m v^2\)

Substitute the given values:
\(E_k = 0.5 \times 4\text{ kg} \times (5\text{ m/s})^2\)
\(E_k = 2 \times 25\)
\(E_k = 50\text{ J}\)

[1 mark] C - 50 J. Award 1 mark for the correct option.

Use the equation linking charge, current, and time:
\(Q = I \times t\)

Substitute the given values:
\(Q = 0.5\text{ A} \times 40\text{ s} = 20\text{ C}\)

[1 mark] B - 20 C. Award 1 mark for the correct option.

Sublimation is the transition of a substance directly from the solid to the gas state, without passing through the liquid state.

[1 mark] A - Sublimation. Award 1 mark for the correct definition.

An alpha particle is identical to a helium nucleus. It consists of two protons and two neutrons, having a net positive charge of \(+2\).

[1 mark] B - Two protons and two neutrons. Award 1 mark for the correct description of an alpha particle.

Natural gas is a fossil fuel and is finite, meaning it is a non-renewable resource. Wind, geothermal, and solar are renewable resources.

[1 mark] C - Natural gas. Award 1 mark for identifying the non-renewable resource.

A fuse is a safety device designed to melt and break the circuit if the current becomes too high, preventing overheating and potential electrical fires.

[1 mark] B - To break the circuit and stop too much current flowing if there is a fault. Award 1 mark for identifying the correct electrical safety function of a fuse.

The specific latent heat of vaporisation is the amount of energy required to change 1 kg of a substance from liquid to gas with no change in temperature.

[1 mark] B - The energy needed to change 1 kg of a substance from liquid to gas with no change in temperature. Award 1 mark for the correct scientific definition.

In a series circuit, the total potential difference of the power supply is shared between the components. Since the two resistors are identical, the potential difference is shared equally:
\(V = 12\text{ V} / 2 = 6\text{ V}\) across each resistor.

[1 mark] B - 6 V. Award 1 mark for calculating the correct share of potential difference in series.

Use the equation: \(E_p = m g h\). Substitute the given values: \(E_p = 4.0 \times 9.8 \times 3.0 = 117.6\text{ J}\).

1 mark for the correct calculation and answer of 117.6 J.

Use the formula \(V = I R\). Rearranging for resistance gives \(R = \frac{V}{I}\). Substituting the values: \(R = \frac{6.0}{0.5} = 12\text{ }\Omega\).

1 mark for the correct calculation and units to give 12 \(\Omega\).

Nuclear power is highly reliable because it is not dependent on weather conditions and does not burn fuel to produce carbon dioxide during operation.

1 mark for identifying Nuclear as the reliable, carbon-free resource.

The earth wire provides a low-resistance path to the ground. If a fault causes the live wire to touch the metal casing, a large current flows through the earth wire, blowing the fuse and protecting the user from electric shock.

1 mark for identifying that it prevents the metal casing from becoming live.

Use the equation: \(E = m L\). Substituting the values gives: \(E = 0.02 \times 3.34 \times 10^5 = 6,680\text{ J}\).

1 mark for the correct calculation and answer of 6,680 J.

The number of neutrons is found by subtracting the atomic number from the mass number: \(14 - 6 = 8\).

1 mark for the correct number of neutrons: 8.

Gamma radiation is high-energy electromagnetic radiation with very high penetrating power, requiring thick lead or concrete to absorb it.

1 mark for identifying Gamma radiation.

Use the equation: \(P = V I\). Substitute the values: \(P = 12 \times 3.0 = 36\text{ W}\).

1 mark for the correct calculation of 36 W.

The formula for gravitational potential energy is \( E_p = m \times g \times h \). Substituting the given values: \( E_p = 4.0 \times 9.8 \times 2.5 = 98.0\text{ J} \).

1 mark for the correct calculation leading to 98.0 J (C).

In a standard UK three-core cable, the live wire is brown, the neutral wire is blue, and the earth wire is green and yellow striped.

1 mark for identifying Brown (B) as the colour of the live wire.

The relationship between potential difference, current, and resistance is given by \( V = I \times R \). To find resistance, rearrange to \( R = \frac{V}{I} = \frac{12}{1.5} = 8.0\ \Omega \).

1 mark for the correct calculation of 8.0 ohms (B).

Condensation is the change of state from a gas to a liquid.

1 mark for selecting Condensation (B).

An alpha particle consists of two protons and two neutrons, which is identical to a helium nucleus.

1 mark for identifying that an alpha particle consists of two protons and two neutrons (B).

Nuclear fuel (such as uranium) is a finite resource and is classified as non-renewable, whereas biofuel, geothermal, and wind are renewable energy resources.

1 mark for identifying Nuclear fuel (C) as the non-renewable resource.

We use the equation: \(E_k = \frac{1}{2} m v^2\).

Substitute the given values into the equation:
\(E_k = 0.5 \times 0.5\text{ kg} \times (4.0\text{ m/s})^2\)

\(E_k = 0.25 \times 16 = 4.0\text{ J}\).

1 mark for correct substitution: \(0.5 \times 0.5 \times 4.0^2\)
1 mark for correct final answer: \(4.0\text{ J}\) (or \(4\text{ J}\))

Unlike coal-fired power stations, wind turbines do not burn fossil fuels. This means they do not release greenhouse gases such as carbon dioxide into the atmosphere during operation, reducing their contribution to global warming and climate change.

1 mark for stating that wind turbines do not release greenhouse gases / carbon dioxide (or do not pollute the air / do not cause acid rain).
1 mark for linking this to an environmental benefit, e.g., does not contribute to global warming / climate change.

We use the equation: \(V = I \times R\).

Substitute the given values into the equation:
\(V = 0.5\text{ A} \times 8.0\ \Omega = 4.0\text{ V}\).

1 mark for correct substitution: \(0.5 \times 8.0\)
1 mark for correct final answer: \(4.0\text{ V}\) (or \(4\text{ V}\))

In a solid, the particles are closely packed together in a regular, fixed pattern. The particles cannot move from place to place, but instead they vibrate about fixed positions.

1 mark for stating that particles are in a regular pattern or closely packed.
1 mark for stating that they vibrate about fixed positions.

A beta particle is a high-energy, fast-moving electron emitted from the nucleus. Because of its smaller mass and charge compared to an alpha particle, a beta particle is more penetrating than alpha radiation, easily passing through paper but stopped by a few millimetres of aluminium.

1 mark for identifying a beta particle as a fast-moving/high-speed electron.
1 mark for stating that beta radiation has a higher penetrating power than alpha radiation (or is stopped by a thin sheet of aluminium/few millimetres of aluminium whereas alpha is stopped by paper/air).

If the live wire touches the metal case, the case becomes live. Because the earth wire is connected to the metal case, a very large current immediately flows from the live wire, through the case, and down the earth wire to the ground. This large current exceeds the rating of the fuse, causing the fuse wire to melt and break the circuit, disconnecting the live supply.

1 mark for stating that a large current flows through the earth wire / to earth.
1 mark for stating that this large current causes the fuse to melt (blow) and break the circuit (isolating the appliance).

First, calculate the temperature change:
\(\Delta\theta = 30^\circ\text{C} - 20^\circ\text{C} = 10^\circ\text{C}\).

Next, use the specific heat capacity equation:
\(\Delta E = m c \Delta\theta\)

\(\Delta E = 2.0\text{ kg} \times 4200\text{ J/kg}^\circ\text{C} \times 10^\circ\text{C} = 84,000\text{ J}\) (or \(84\text{ kJ}\)).

1 mark for calculating temperature change (\(10^\circ\text{C}\)) and correct substitution: \(2.0 \times 4200 \times 10\)
1 mark for correct final answer: \(84,000\text{ J}\) (or \(84\text{ kJ}\))

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

1 mark for stating that the particles gain kinetic energy / move faster.
1 mark for stating that they collide with the walls more frequently / with more force.

Efficiency is calculated using the formula: Efficiency = Useful energy output / Total energy input. Substituting the values gives: Efficiency = 12000 / 15000 = 0.8 (or 80%).

1 mark for correct substitution: 12000 / 15000. 1 mark for correct evaluation: 0.8 (accept 80% or 4/5).

Using the equation: charge flow = current x time. Charge flow = 0.6 A * 40 s = 24 C.

1 mark for correct calculation: 0.6 * 40 = 24. 1 mark for correct unit: C (or Coulombs).

Using the equation: density = mass / volume. Density = 210 g / 300 cm³ = 0.7 g/cm³.

1 mark for correct calculation: 210 / 300 = 0.7. 1 mark for correct unit: g/cm³.

An alpha particle consists of 2 protons and 2 neutrons, so it has a mass number of 4 and an atomic number of 2. When Radium-226 decays, the mass number decreases by 4 (226 - 4 = 222) and the atomic number decreases by 2 (88 - 2 = 86).

1 mark for mass number = 222. 1 mark for atomic number = 86.

Using the equation: gravitational potential energy = mass x gravitational field strength x height. Gravitational potential energy = 4.0 kg * 9.8 N/kg * 1.5 m = 58.8 J.

1 mark for correct substitution: 4.0 * 9.8 * 1.5. 1 mark for correct final value: 58.8 (J).

Using the equation: power = potential difference x current. Power = 230 V * 5.0 A = 1150 W.

1 mark for correct substitution: 230 * 5.0. 1 mark for correct answer: 1150 (W).

Step 1: Calculate the increase in gravitational potential energy (\(E_p\)) for the first hoist. \(E_{p1} = m \times g \times h\) which gives \(E_{p1} = 50\text{ kg} \times 9.8\text{ N/kg} \times 8.0\text{ m} = 3920\text{ J}\). Step 2: Calculate the increase in gravitational potential energy for the second hoist. \(E_{p2} = 40\text{ kg} \times 9.8\text{ N/kg} \times 11.0\text{ m} = 4312\text{ J}\). Step 3: Compare the values. Since \(4312\text{ J} > 3920\text{ J}\), the second hoist does more work against gravity.

[1 mark] Correct calculation of the gravitational potential energy for the first hoist (\(3920\text{ J}\)). [1 mark] Correct calculation of the gravitational potential energy for the second hoist (\(4312\text{ J}\)). [1 mark] Valid conclusion comparing both values to identify that the second hoist performs more work.

Step 1: Rearrange the equation to find resistance: \(\text{resistance} = \frac{\text{potential difference}}{\text{current}}\). Step 2: Calculate the resistance of resistor P. \(R_P = \frac{6.0\text{ V}}{0.40\text{ A}} = 15\text{ \Omega}\). Step 3: Calculate the resistance of resistor Q to compare. \(R_Q = \frac{9.0\text{ V}}{0.50\text{ A}} = 18\text{ \Omega}\). Since \(18\text{ \Omega} > 15\text{ \Omega}\), resistor Q has the higher resistance.

[1 mark] Rearrangement of the equation or correct substitution to find resistance of P: \(6.0 / 0.40\). [1 mark] Correct calculation of the resistance of P as \(15\text{ \Omega}\). [1 mark] Correct comparison showing resistor Q (\(18\text{ \Omega}\)) has a higher resistance than resistor P.

Step 1: Rearrange the equation to find specific latent heat (\(L\)): \(L = \frac{E}{m}\). Step 2: Calculate the specific latent heat of fusion of sample A. \(L_A = \frac{45,000\text{ J}}{0.50\text{ kg}} = 90,000\text{ J/kg}\). Step 3: Compare with sample B. \(L_B = \frac{55,000\text{ J}}{0.50\text{ kg}} = 110,000\text{ J/kg}\). Since \(110,000\text{ J/kg} > 90,000\text{ J/kg}\), sample B has a higher specific latent heat of fusion.

[1 mark] Rearrangement of the formula or substitution to find \(L_A\): \(45,000 / 0.50\). [1 mark] Correct calculation of specific latent heat for sample A as \(90,000\text{ J/kg}\). [1 mark] Correct identification that sample B has the higher specific latent heat supported by comparing the values.

Step 1: Determine the activity of Source C after \(8\text{ hours}\). \(8\text{ hours}\) is equivalent to 2 half-lives of Source C (\(8 / 4 = 2\)). Initial activity is \(1200\text{ Bq}\). After 1 half-life: \(1200 / 2 = 600\text{ Bq}\). After 2 half-lives: \(600 / 2 = 300\text{ Bq}\). Step 2: Determine the activity of Source D after \(8\text{ hours}\). \(8\text{ hours}\) is equivalent to 1 half-life of Source D (\(8 / 8 = 1\)). Initial activity is \(800\text{ Bq}\). After 1 half-life: \(800 / 2 = 400\text{ Bq}\). Step 3: Compare the activities. Since \(300\text{ Bq} < 400\text{ Bq}\), Source C has the lower activity.

[1 mark] Correct calculation of the activity of Source C after \(8\text{ hours}\) (\(300\text{ Bq}\)). [1 mark] Correct calculation of the activity of Source D after \(8\text{ hours}\) (\(400\text{ Bq}\)). [1 mark] Correct comparison concluding that Source C has the lower activity.

Step 1: Calculate the efficiency of the first motor. \(\text{Efficiency}_1 = \frac{150\text{ J}}{250\text{ J}} = 0.60\) (or \(60\%\)). Step 2: Calculate the efficiency of the second motor. \(\text{Efficiency}_2 = \frac{220\text{ J}}{400\text{ J}} = 0.55\) (or \(55\%\)). Step 3: Compare the efficiencies. Since \(0.60 > 0.55\), the first motor is more efficient.

[1 mark] Correct calculation of the first motor's efficiency (\(0.60\) or \(60\%\)). [1 mark] Correct calculation of the second motor's efficiency (\(0.55\) or \(55\%\)). [1 mark] Correct comparison identifying that the first motor is more efficient.

Step 1: Rearrange the equation to find current: \(\text{current} = \frac{\text{power}}{\text{potential difference}}\). Step 2: Calculate the current for Heater A. \(I_A = \frac{1150\text{ W}}{230\text{ V}} = 5.0\text{ A}\). Step 3: Calculate the current for Heater B. \(I_B = \frac{2300\text{ W}}{230\text{ V}} = 10.0\text{ A}\) (or deduce that double the power at the same voltage requires double the current). Heater B draws a larger current and therefore requires a larger fuse rating.

[1 mark] Rearrangement of the formula or substitution to find current: \(1150 / 230\). [1 mark] Correct calculation of the current for Heater A (\(5.0\text{ A}\)). [1 mark] Correct comparison stating that Heater B requires a larger fuse rating due to its higher current.

Step 1: Rearrange the equation to find specific heat capacity (\(c\)): \(c = \frac{\Delta E}{m \times \Delta \theta}\). Step 2: Calculate specific heat capacity for liquid X. \(c_X = \frac{24,000\text{ J}}{0.80\text{ kg} \times 15\text{ }^\circ\text{C}} = \frac{24,000}{12} = 2000\text{ J/kg }^\circ\text{C}\). Step 3: Calculate specific heat capacity for liquid Y. \(c_Y = \frac{18,000\text{ J}}{0.80\text{ kg} \times 15\text{ }^\circ\text{C}} = \frac{18,000}{12} = 1500\text{ J/kg }^\circ\text{C}\). Liquid X has the higher specific heat capacity (\(2000 > 1500\)).

[1 mark] Correct rearrangement of the equation or substitution for liquid X: \(24,000 / (0.80 \times 15)\). [1 mark] Correct calculation of the specific heat capacity of liquid X as \(2000\text{ J/kg }^\circ\text{C}\). [1 mark] Correct comparison concluding liquid X has a higher specific heat capacity than liquid Y.

Step 1: Convert time into seconds for both generators. Tidal generator: \(1\text{ minute} = 60\text{ seconds}\). Wind turbine: \(1.5\text{ minutes} = 90\text{ seconds}\). Step 2: Calculate the power output of the tidal generator. \(\text{Power}_{\text{tidal}} = \frac{180,000\text{ J}}{60\text{ s}} = 3000\text{ W}\). Step 3: Calculate the power output of the wind turbine. \(\text{Power}_{\text{wind}} = \frac{210,000\text{ J}}{90\text{ s}} \approx 2333\text{ W}\). Comparing the power outputs, the tidal generator (\(3000\text{ W}\)) is higher than the wind turbine (\(2333\text{ W}\)).

[1 mark] Correct conversion of time into seconds (\(60\text{ s}\) or \(90\text{ s}\)). [1 mark] Correct calculation of power for both generators (tidal = \(3000\text{ W}\), wind = \(2333\text{ W}\)). [1 mark] Correct comparison showing the tidal generator has a higher power output.

First, calculate the gravitational potential energy (\(E_p\)) gained by Bag A: \(E_p = m \times g \times h = 3.0\text{ kg} \times 9.8\text{ N/kg} \times 1.2\text{ m} = 35.28\text{ J}\). Next, calculate the gravitational potential energy gained by Bag B: \(E_p = 2.0\text{ kg} \times 9.8\text{ N/kg} \times 1.5\text{ m} = 29.4\text{ J}\). Finally, calculate the difference between the two values: \(35.28\text{ J} - 29.4\text{ J} = 5.88\text{ J}\).

1 mark for calculating the energy of at least one bag correctly (\(35.28\text{ J}\) or \(29.4\text{ J}\)). 1 mark for showing both calculated energy values. 1 mark for finding the correct difference of \(5.88\text{ J}\) (or \(5.9\text{ J}\)).

First, calculate the charge flow (\(Q\)) through Resistor X: \(Q = I \times t = 0.5\text{ A} \times 60\text{ s} = 30\text{ C}\). Next, calculate the charge flow through Resistor Y: \(Q = I \times t = 0.3\text{ A} \times 120\text{ s} = 36\text{ C}\). Finally, find the difference in charge: \(36\text{ C} - 30\text{ C} = 6\text{ C}\).

1 mark for calculating the charge flow of either Resistor X (\(30\text{ C}\)) or Resistor Y (\(36\text{ C}\)) correctly. 1 mark for calculating both charges correctly (\(30\text{ C}\) and \(36\text{ C}\)). 1 mark for finding the correct difference of \(6\text{ C}\).

Method 1: Calculate the mass difference: \(0.25\text{ kg} - 0.15\text{ kg} = 0.10\text{ kg}\). The difference in energy required is: \(E = m \times L = 0.10\text{ kg} \times 3.34 \times 10^5\text{ J/kg} = 33,400\text{ J}\). Method 2: Calculate energy for Block A: \(E_A = 0.15\text{ kg} \times 334,000\text{ J/kg} = 50,100\text{ J}\). Calculate energy for Block B: \(E_B = 0.25\text{ kg} \times 334,000\text{ J/kg} = 83,500\text{ J}\). Difference: \(83,500\text{ J} - 50,100\text{ J} = 33,400\text{ J}\).

1 mark for calculating the mass difference of \(0.10\text{ kg}\) OR calculating the energy needed for one of the blocks (\(50,100\text{ J}\) or \(83,500\text{ J}\)). 1 mark for setting up the multiplication: \(0.10\text{ kg} \times 3.34 \times 10^5\text{ J/kg}\) or showing the subtraction of the two energies. 1 mark for the correct final answer of \(33,400\text{ J}\) (or \(33.4\text{ kJ}\)).

First, calculate the energy transferred by the electric heater: \(E = P \times t = 1500\text{ W} \times 40\text{ s} = 60,000\text{ J}\). Next, calculate the energy transferred by the electric kettle: \(E = P \times t = 2200\text{ W} \times 25\text{ s} = 55,000\text{ J}\). Comparing the values, the electric heater transfers more energy. Calculate the difference: \(60,000\text{ J} - 55,000\text{ J} = 5,000\text{ J}\).

1 mark for calculating the energy of at least one appliance correctly (\(60,000\text{ J}\) or \(55,000\text{ J}\)). 1 mark for identifying that the heater transfers more energy (or showing both calculated energy values). 1 mark for calculating the correct difference of \(5,000\text{ J}\) (or \(5\text{ kJ}\)).

1. Measure the mass of the dry rock using a digital balance. 2. Fill a displacement (Eureka) can with water until it just begins to drip out of the spout into a beaker. Once the dripping stops, place an empty measuring cylinder under the spout. 3. Tie a piece of thin thread around the rock and carefully lower it into the Eureka can until it is completely submerged. 4. Collect all the displaced water in the measuring cylinder. 5. Measure the volume of this displaced water by looking at the cylinder at eye level, reading from the bottom of the meniscus. This volume is equal to the volume of the rock. 6. Use the formula: \(\text{density} = \frac{\text{mass}}{\text{volume}}\) to calculate the density of the rock.

Level 3 (5-6 marks): A detailed, coherent, and logically structured method is described. All key apparatus and measurements are clearly identified, and the calculation of density is correctly explained with the correct equation. Level 2 (3-4 marks): A logical method is proposed. Most of the key steps and apparatus are identified, and there is some explanation of how to calculate density. Level 1 (1-2 marks): Simple, isolated points are made, such as mentioning a balance to measure mass or a measuring cylinder to measure volume, with little to no structure. Indicative content: - Measure mass of rock using a balance. - Fill Eureka can with water up to the level of the spout. - Lower rock into Eureka can using a thread. - Catch the displaced water in a measuring cylinder. - Read the volume from the measuring cylinder at eye level. - State that volume of water equals volume of rock. - Use density = mass / volume to calculate the density.

Displacement is a vector quantity because it has both magnitude and direction. Distance, mass, and speed only have magnitude, so they are scalar quantities.

1 mark for the correct choice of displacement (C).

The limit of proportionality is the point up to which Hooke's Law is obeyed, where force and extension are directly proportional. Beyond this point, they are no longer directly proportional.

1 mark for identifying the limit of proportionality (B).

Using the equation: \(\text{speed} = \frac{\text{distance}}{\text{time}}\), we get \(\text{speed} = \frac{150\text{ m}}{10\text{ s}} = 15\text{ m/s}\).

1 mark for selecting the correct speed of 15 m/s (B).

Sound waves are longitudinal. In longitudinal waves, the particles vibrate parallel to the direction of wave travel or energy transfer.

1 mark for identifying that sound waves are longitudinal and vibrate parallel (A).

Radio waves are located at the end of the electromagnetic spectrum with the longest wavelengths and the lowest frequencies.

1 mark for identifying radio waves as having the longest wavelength and lowest frequency (D).

The magnetic metals are iron, cobalt, and nickel. Steel is an alloy of iron and is also magnetic. Copper, aluminium, brass, and plastic are non-magnetic.

1 mark for selecting the group containing only magnetic materials (B).

The size of the force on a current-carrying wire in a magnetic field is directly proportional to the current. Therefore, increasing the current in the wire increases the size of the force. Reversing the current direction only changes the direction of the force, not its size.

1 mark for identifying that increasing the current increases the force (C).

Gravitational force is an attractive force that acts between any two masses, providing the centripetal force required to keep the planets in orbit around the Sun.

1 mark for selecting gravitational force (C).

First, find the extension \(e\) of the spring by subtracting the unstretched length from the stretched length: \(e = 0.11\text{ m} - 0.05\text{ m} = 0.06\text{ m}\). Next, use the equation linking force, spring constant, and extension: \(F = k \times e\). Rearranging the equation to solve for the spring constant \(k\) gives: \(k = \frac{F}{e}\). Substituting the values into the equation: \(k = \frac{3\text{ N}}{0.06\text{ m}} = 50\text{ N/m}\).

1 mark for the correct calculation of the spring constant (50 N/m).

An induced magnet is a material that becomes a magnet when it is placed in a magnetic field. When the magnetic field is removed, an induced magnet loses most or all of its magnetism quickly.

1 mark for identifying that an induced magnet only becomes magnetic when placed in a magnetic field.

In the electromagnetic spectrum, gamma rays have the shortest wavelength (highest frequency), followed by X-rays, ultraviolet, visible light, infrared, microwaves, and radio waves which have the longest wavelength (lowest frequency).

1 mark for the correct order of the electromagnetic spectrum from shortest to longest wavelength.

Sound waves are longitudinal waves. In longitudinal waves, the vibrations/oscillations of the particles are parallel to the direction of energy transfer.

1 mark for identifying sound as a longitudinal wave with oscillations parallel to the direction of energy transfer.

Use the acceleration equation: \(a = \frac{v - u}{t}\), where \(v = 12\text{ m/s}\), \(u = 0\text{ m/s}\) (since it starts from rest), and \(t = 6\text{ s}\). Substituting these values gives: \(a = \frac{12 - 0}{6} = 2\text{ m/s}^2\).

1 mark for the correct calculation of acceleration.

The gravitational force of attraction between the Earth and the satellite acts as a centripetal force, pulling the satellite towards the Earth and maintaining its circular orbit.

1 mark for identifying the gravitational force of attraction as the orbital force.

Vector quantities have both magnitude and direction. Force, displacement, and velocity are all vector quantities. Distance, speed, time, mass, temperature, and energy are scalar quantities because they only have magnitude.

1 mark for identifying the correct list of vector quantities.

Use the equation for work done: \(W = F \times s\), where \(F = 15\text{ N}\) and \(s = 8\text{ m}\). Calculating this gives: \(W = 15 \times 8 = 120\text{ J}\).

1 mark for the correct calculation of work done.

Television remote controls use infrared radiation (IR) to send control signals to the television receiver.

[1 mark] C - Infrared waves (Correct answer choice)

Iron is a magnetic material. When placed in the magnetic field of a permanent magnet, it becomes an induced magnet. Induced magnetism always causes a force of attraction.

[1 mark] B - The nail becomes an induced magnet and is attracted to the bar magnet (Correct answer choice)

Using the equation: force = spring constant \(\times\) extension. \(F = k \times e\). Substituting the given values: \(F = 50\text{ N/m} \times 0.20\text{ m} = 10\text{ N}\).

[1 mark] A - \(10\text{ N}\) (Correct answer choice)

Sound waves are longitudinal waves because the particles of the medium oscillate parallel to the direction of energy transfer. Light, radio waves, and water ripples are transverse waves.

[1 mark] C - A sound wave travelling through water (Correct answer choice)

Using the equation: speed = distance / time. \(s = d / t = 15\text{ m} / 6.0\text{ s} = 2.5\text{ m/s}\).

[1 mark] B - \(2.5\text{ m/s}\) (Correct answer choice)

1. The friction force always opposes motion, so since the car is moving to the right, the friction force acts horizontally to the left.
2. Since the friction force opposes the motion and is the only horizontal force acting on the car after the push, it decelerates the car, causing its speed to decrease until it stops.

1 mark: for stating the direction of friction is to the left (or opposite to the direction of motion).
1 mark: for stating that it decreases the speed of the car (or slows it down).

1. X-rays can pass through (penetrate) soft tissues like skin and muscle, but they are absorbed by dense materials like bone, which creates a shadow image on the detector.
2. To minimise their exposure to ionising radiation, the radiographer can stand behind a protective lead screen, wear a lead-lined apron, or leave the room while the X-ray machine is active.

1 mark: for stating that X-rays are highly penetrating / pass through soft tissues but are absorbed by bones.
1 mark: for any valid safety precaution, e.g., stand behind a screen / leave the room / wear a lead apron / stand at a distance.

1. First, identify the change in velocity: \(\Delta v = 6\text{ m/s} - 0\text{ m/s} = 6\text{ m/s}\).
2. Use the acceleration formula: \(a = \frac{6\text{ m/s}}{4\text{ s}} = 1.5\text{ m/s}^2\).

1 mark: for correct substitution, i.e., \(\frac{6}{4}\) or \(\frac{6-0}{4}\).
1 mark: for the correct final answer of 1.5 (with or without unit, but accept 1.5 m/s^2).

1. A permanent magnet has fixed north and south poles. When a known magnet's north pole is brought near the other permanent magnet, it will attract one end (south pole) and repel the other end (north pole).
2. An unmagnetised magnetic material (like iron) will be attracted to either pole of the known magnet and will never experience repulsion. Therefore, only the permanent magnet can show repulsion.

1 mark: for stating that the known magnet is brought close to both ends of each bar.
1 mark: for explaining that only the permanent magnet will show repulsion (at one of its ends), whereas the unmagnetised iron bar will only experience attraction.

1. In a transverse wave, the oscillations/vibrations are at right angles (perpendicular) to the direction of travel or energy transfer of the wave.
2. S-waves are transverse waves and can only travel through solids; they cannot travel through liquids, such as the liquid outer core of the Earth.

1 mark: for stating that vibrations are perpendicular (or at right angles / at \(90^\circ\)) to the direction of energy transfer.
1 mark: for stating that S-waves cannot travel through liquids.

1. The gravitational attraction (gravity) between the Sun and the planets provides the centripetal force required to keep them in orbit.
2. Gravity is an inverse-square force; as the distance between the two bodies increases, the strength of the gravitational force decreases.

1 mark: for identifying the force as gravity / gravitational force.
1 mark: for stating that the force decreases (or gets weaker) as distance increases.

1. Rearrange the equation to find the spring constant \(k\): \(k = \frac{F}{e}\).
2. Substitute the values: \(k = \frac{4\text{ N}}{0.2\text{ m}} = 20\text{ N/m}\).

1 mark: for rearranging the formula or correct substitution, i.e., \(\frac{4}{0.2}\).
1 mark: for the correct final answer of 20 (accept 20 N/m).

1. By convention, magnetic field lines outside any magnet always point away from the North pole and towards the South pole.
2. The density or closeness of the magnetic field lines indicates the strength of the field: where the lines are closest together (near the poles), the magnetic field is at its strongest.

1 mark: for stating the direction is from North (pole) to South (pole).
1 mark: for stating that closer/denser lines represent a stronger magnetic field (or wider spacing represents a weaker magnetic field).

To find the magnitude of the resultant force, subtract the smaller force from the larger force because they act in opposite directions: \(8\text{ N} - 6\text{ N} = 2\text{ N}\). Since the larger force acts to the west, the direction of the resultant force is west.

1 mark for the correct magnitude of \(2\text{ N}\). 1 mark for the correct direction of West.

Velocity is a vector quantity calculated as displacement divided by time: \(\text{Velocity} = \frac{\text{Displacement}}{\text{Time}} = \frac{45\text{ m}}{9\text{ s}} = 5\text{ m/s}\). The direction of velocity is the same as the displacement, which is East.

1 mark for the correct calculation of velocity magnitude as \(5\text{ m/s}\). 1 mark for specifying the direction as East.

Opposite magnetic poles attract each other, so the North pole and South pole will experience an attractive force. This means the first magnet will experience a force pulling it towards the second magnet.

1 mark for identifying the force as attractive. 1 mark for stating that the force on the first magnet is directed towards the second magnet.

In a longitudinal wave, the particles of the medium vibrate parallel to (or in the same direction as) the direction of energy transfer. Sound waves are mechanical waves that require a medium to propagate, meaning they cannot travel through a vacuum.

1 mark for stating that vibrations are parallel to (or in the same direction as) the direction of energy transfer. 1 mark for stating that sound waves cannot travel through a vacuum.

All electromagnetic waves travel at the exact same speed in a vacuum, which is the speed of light (\(3 \times 10^8\text{ m/s}\)). In the visible light spectrum, red light has a lower frequency than blue light, which corresponds to a longer wavelength.

1 mark for stating that both beams travel at the same speed. 1 mark for identifying that red light has the longer wavelength.

Work done is calculated using the formula: \(\text{Work done} = \text{Force} \times \text{Distance} = 50\text{ N} \times 8\text{ m} = 400\text{ J}\). Friction acts to oppose motion, so because the box is pushed North, the friction force acts in the South direction.

1 mark for calculating the work done as \(400\text{ J}\) (or \(400\text{ N m}\)). 1 mark for stating the friction force acts South (or opposite to the direction of motion).

An object resting on a surface experiences an upward force known as the normal contact force (or reaction force). Since the table is horizontal and the book rests on it, this force acts vertically upwards, perpendicular to the surface of the table.

1 mark for naming the force as normal contact force or reaction force. 1 mark for stating the direction of the force is upwards (or perpendicular/normal to the surface).

Glass is more optically dense than air, which causes light to slow down when it transitions from air into glass. This reduction in speed causes the path of the light ray to bend towards the normal line.

1 mark for stating that the light ray bends towards the normal. 1 mark for explaining that the speed of light decreases (light slows down) as it enters the glass block.

1. Calculate the extension of the spring: \(e = 0.28\text{ m} - 0.12\text{ m} = 0.16\text{ m}\). 2. Use the equation: \(F = k e\). 3. Substitute the values: \(F = 25\text{ N/m} \times 0.16\text{ m} = 4.0\text{ N}\).

[1 mark] For calculating the extension: \(0.16\text{ m}\).
[1 mark] For using \(F = k e\) (or substituting values correctly).
[1 mark] For the final answer of \(4\) (or \(4.0\) or \(4\text{ N}\)).

1. Calculate the acceleration of the toy car: \(a = \frac{v - u}{t} = \frac{6.0 - 0}{3.0} = 2.0\text{ m/s}^2\). 2. Use Newton's second law: \(F = m a\). 3. Substitute the values: \(F = 0.50\text{ kg} \times 2.0\text{ m/s}^2 = 1.0\text{ N}\).

[1 mark] For calculating the acceleration: \(2.0\text{ m/s}^2\).
[1 mark] For using \(F = m a\) (or substituting values correctly).
[1 mark] For the final answer of \(1\) (or \(1.0\) or \(1\text{ N}\)).

1. Calculate the wave speed: \(v = \frac{\text{distance}}{\text{time}} = \frac{30\text{ m}}{5.0\text{ s}} = 6.0\text{ m/s}\). 2. Use the wave equation: \(v = f \lambda\), rearranged to \(\lambda = \frac{v}{f}\). 3. Substitute the values: \(\lambda = \frac{6.0\text{ m/s}}{15\text{ Hz}} = 0.40\text{ m}\).

[1 mark] For calculating the wave speed: \(6.0\text{ m/s}\).
[1 mark] For rearranging the wave equation or substituting into \(v = f \lambda\).
[1 mark] For the final answer of \(0.4\) (or \(0.40\) or \(0.4\text{ m}\)).

1. Calculate the distance moved using the work done equation: \(W = F s\), rearranged to \(s = \frac{W}{F} = \frac{48\text{ J}}{12\text{ N}} = 4.0\text{ m}\). 2. Use the speed equation: \(\text{speed} = \frac{\text{distance}}{\text{time}}\) or \(v = \frac{s}{t}\). 3. Substitute the values: \(\text{speed} = \frac{4.0\text{ m}}{8.0\text{ s}} = 0.50\text{ m/s}\).

[1 mark] For calculating the distance: \(4.0\text{ m}\).
[1 mark] For using the speed equation.
[1 mark] For the final answer of \(0.5\) (or \(0.50\) or \(0.5\text{ m/s}\)).

1. Calculate the mass of the astronaut on Earth: \(m = \frac{W}{g} = \frac{784\text{ N}}{9.8\text{ N/kg}} = 80\text{ kg}\). 2. Use the weight equation for the Moon: \(W = m g\). 3. Substitute the values: \(W = 80\text{ kg} \times 1.6\text{ N/kg} = 128\text{ N}\).

[1 mark] For calculating the mass of the astronaut: \(80\text{ kg}\).
[1 mark] For using the weight equation with the Moon's gravitational field strength.
[1 mark] For the final answer of \(128\) (or \(128\text{ N}\)).

1. Convert the frequency from kilohertz to hertz: \(1500\text{ kHz} = 1,500,000\text{ Hz}\) (or \(1.5 \times 10^6\text{ Hz}\)). 2. Use the wave equation rearranged for wavelength: \(\lambda = \frac{v}{f}\). 3. Substitute the values: \(\lambda = \frac{300,000,000\text{ m/s}}{1,500,000\text{ Hz}} = 200\text{ m}\).

[1 mark] For converting the frequency to hertz: \(1,500,000\text{ Hz}\).
[1 mark] For rearranging the wave equation or substituting correctly.
[1 mark] For the final answer of \(200\) (or \(200\text{ m}\)).

1. Calculate the thinking distance: \(\text{thinking distance} = \text{speed} \times \text{reaction time} = 15\text{ m/s} \times 0.80\text{ s} = 12\text{ m}\). 2. Use the stopping distance relation: \(\text{stopping distance} = \text{thinking distance} + \text{braking distance}\). 3. Substitute the values: \(\text{stopping distance} = 12\text{ m} + 18\text{ m} = 30\text{ m}\).

[1 mark] For calculating the thinking distance: \(12\text{ m}\).
[1 mark] For adding the calculated thinking distance and the braking distance.
[1 mark] For the final answer of \(30\) (or \(30\text{ m}\)).

1. Calculate the area of the base of the block: \(A = 0.15\text{ m} \times 0.20\text{ m} = 0.03\text{ m}^2\). 2. Use the pressure equation: \(p = \frac{F}{A}\). 3. Substitute the values: \(p = \frac{6.0\text{ N}}{0.03\text{ m}^2} = 200\text{ Pa}\) (or \(200\text{ N/m}^2\)).

[1 mark] For calculating the area of the base: \(0.03\text{ m}^2\).
[1 mark] For using the pressure equation.
[1 mark] For the final answer of \(200\) (or \(200\text{ Pa}\) or \(200\text{ N/m}^2\)).

Step 1: Calculate the speed of the wave using the formula for distance and time: \(\text{wave speed} = \frac{\text{distance}}{\text{time}}\), so \(v = \frac{18}{6.0} = 3.0\text{ m/s}\). Step 2: Use the wave equation to calculate the wavelength: \(\text{wave speed} = \text{frequency} \times \text{wavelength}\). Rearranging to find wavelength gives: \(\text{wavelength} = \frac{\text{wave speed}}{\text{frequency}}\), so \(\lambda = \frac{3.0}{4.0} = 0.75\text{ m}\).

- [1 mark] Calculation of the speed of the wave: \(18 / 6.0 = 3.0\text{ (m/s)}\). - [1 mark] Rearrangement of wave equation: \(\text{wavelength} = \frac{\text{wave speed}}{\text{frequency}}\) or substitution of values into the wave equation: \(3.0 = 4.0 \times \lambda\). - [1 mark] Correct final calculation of wavelength: \(0.75\text{ (m)}\).

Step 1: Calculate the extension of the spring in cm and convert it to meters (m). \(\text{extension} = 15\text{ cm} - 12\text{ cm} = 3\text{ cm}\). Convert cm to m: \(e = \frac{3}{100} = 0.03\text{ m}\). Step 2: Rearrange the spring equation to calculate the spring constant (\(k\)): \(\text{spring constant} = \frac{\text{force applied}}{\text{extension}}\), so \(k = \frac{6.0\text{ N}}{0.03\text{ m}} = 200\text{ N/m}\).

- [1 mark] Calculation of extension: \(15 - 12 = 3\text{ (cm)}\). - [1 mark] Conversion of extension from cm to m: \(0.03\text{ (m)}\). - [1 mark] Correct substitution and calculation to find the spring constant: \(\frac{6.0}{0.03} = 200\text{ (N/m)}\). (Award 2 marks for an answer of 2 N/m, which comes from omitting the cm to m conversion).

1. Identify the values from the question:
- Initial velocity, \(u = 22\text{ m/s}\)
- Final velocity, \(v = 0\text{ m/s}\) (since the car comes to a stop)
- Acceleration, \(a = -3.5\text{ m/s}^2\) (since it is decelerating)

2. Substitute the values into the equation:
\(v^2 - u^2 = 2as\)
\(0^2 - 22^2 = 2 \times (-3.5) \times s\)

3. Simplify the equation:
\(-484 = -7 \times s\)

4. Rearrange to find the distance, \(s\):
\(s = \frac{-484}{-7}\)
\(s = 69.1428...\text{ m}\)

5. Round the answer to 2 significant figures:
\(s = 69\text{ m}\)

- **[1 mark]** Correct identification that final velocity is \(0\text{ m/s}\) and substitution of known values into the equation: \(0^2 - 22^2 = 2 \times (-3.5) \times s\) (or equivalent magnitude calculation: \(22^2 = 2 \times 3.5 \times s\)).
- **[1 mark]** Correct simplification of squared terms and multiplication: \(484 = 7 \times s\) (or \(-484 = -7 \times s\)).
- **[1 mark]** Correct rearrangement to find the distance: \(s = \frac{484}{7}\) or \(69.1\) (or any value that rounds to \(69\) before 2 s.f. rounding).
- **[1 mark]** Correct rounding of their calculated value to 2 significant figures: \(69\) (accept this mark if they correctly round an incorrect calculation to 2 significant figures).

To investigate the relationship between force and extension for a spring:

1. **Setup**: Attach a clamp, boss, and clamp stand to a bench. Suspend the spring from the clamp.
2. **Ruler Alignment**: Set up a metre ruler vertically next to the spring. Ensure the ruler is vertical, for example by using a set square or a plumb line. Adjust the ruler so that the zero mark is level with the top of the spring (or use a pointer at the bottom of the spring).
3. **Initial Measurement**: Measure and record the original length of the unstretched spring using the metre ruler. Ensure you read the ruler at eye level to avoid parallax error.
4. **Adding Force**: Add a 100 g mass hanger (which exerts a force of 1 N) to the bottom of the spring.
5. **Subsequent Measurements**: Allow the spring to come to rest, then measure and record the new length of the spring. Repeat this process by adding further 100 g masses (1 N increments) up to a total of 500 g (5 N), recording the new length after each mass is added.
6. **Calculating Extension**: For each force, calculate the extension of the spring using the formula:
\(\text{Extension} = \text{new length} - \text{original length}\).
7. **Safety Precaution**: Wear safety goggles in case the spring breaks or flies off.

### Level 3 (5–6 marks)
The method is clear, detailed, and logically ordered. It describes how to vary the independent variable (force), measure the dependent variable (length/extension), and includes at least one specific detail to ensure accuracy. The method for calculating extension is clearly explained.

### Level 2 (3–4 marks)
The method described would lead to a usable set of results. It describes how to change the force and measure the length/extension of the spring. Some details on accuracy or calculation may be missing.

### Level 1 (1–2 marks)
Simple, isolated points are made, such as adding weights or measuring the spring with a ruler. The method is incomplete or lacks logical sequence.

### Level 0 (0 marks)
No relevant content.

### Key points to look for:
- Hang the spring from a clamp stand.
- Use a metre ruler to measure length.
- Measure the original length of the spring with no weights attached.
- Add known weights / masses to the spring one at a time.
- Measure the new length of the spring after each weight is added.
- Explain how to calculate extension: \(\text{extension} = \text{new length} - \text{original length}\).
- Accuracy detail: align ruler vertically / read at eye level to avoid parallax / use a pointer (fiducial marker) / ensure spring is stationary.