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A star with a mass much larger than the Sun will evolve from the main sequence to a red supergiant, then explode as a supernova, leaving behind either a neutron star or a black hole. Option B is correct because it shows this progression.

1 mark for the correct answer (B).

Using the formula: \(\text{distance} = \text{speed} \times \text{time}\). Substituting the given values: \(\text{distance} = 5.0\text{ m/s} \times 12\text{ s} = 60\text{ m}\).

1 mark for the correct answer (C).

All electromagnetic waves travel at the same speed in a vacuum, which is approximately \(3 \times 10^8\text{ m/s}\). Their frequency, wavelength, and amplitude vary depending on the type of wave.

1 mark for the correct answer (C).

An alpha particle consists of two protons and two neutrons, which is identical to a helium-4 nucleus. Beta particles are high-speed electrons or positrons, and gamma rays are electromagnetic waves.

1 mark for the correct answer (A).

Efficiency is calculated using the formula: \(\text{efficiency} = \frac{\text{useful energy transferred}}{\text{total energy transferred}}\). Substituting the given values: \(\text{efficiency} = \frac{150\text{ J}}{500\text{ J}} = 0.30\).

1 mark for the correct answer (A).

In the electromagnetic spectrum, gamma rays have the highest frequency and the shortest wavelength.

1 mark for the correct answer (B).

After 1 half-life, the activity is halved: \( 160 / 2 = 80\text{ Bq} \). After 2 half-lives, the activity is halved again: \( 80 / 2 = 40\text{ Bq} \). After 3 half-lives, the activity is halved once more: \( 40 / 2 = 20\text{ Bq} \).

1 mark for dividing the activity by 2 three times (or showing the sequence 160 -> 80 -> 40 -> 20). 1 mark for the correct final activity of 20 (Bq).

Use the equation: \( \text{distance} = \text{speed} \times \text{time} \). Substitute the values into the equation: \( \text{distance} = 5.5\text{ m/s} \times 40\text{ s} = 220\text{ m} \).

1 mark for the correct substitution: \( 5.5 \times 40 \). 1 mark for the correct calculation of 220 (m).

Use the wave speed equation: \( v = f \lambda \). Substitute the given values: \( v = 5.0\text{ Hz} \times 0.8\text{ m} = 4.0\text{ m/s} \).

1 mark for the correct substitution: \( 5.0 \times 0.8 \). 1 mark for the correct final answer of 4.0 (m/s) (or 4).

Use the equation: \( \Delta GPE = m \times g \times \Delta h \). Substitute the values: \( \Delta GPE = 15\text{ kg} \times 10\text{ N/kg} \times 4.0\text{ m} = 600\text{ J} \).

1 mark for the correct substitution: \( 15 \times 10 \times 4.0 \). 1 mark for the correct calculation of 600 (J).

Radio waves and gamma rays are both electromagnetic waves, but radio waves have a much longer wavelength and a much lower frequency compared to gamma rays.

1 mark for stating radio waves have a longer wavelength (or gamma rays have a shorter wavelength). 1 mark for stating radio waves have a lower frequency / energy (or gamma rays have a higher frequency / energy).

A main sequence star is in a stable state because the inward force of gravity is perfectly balanced by the outward force of radiation pressure (or gas pressure) caused by nuclear fusion in the star's core.

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

Use Newton's second law: \( F = m \times a \). Rearranging the formula for mass gives: \( m = \frac{F}{a} \). Substitute the values: \( m = \frac{18\text{ N}}{3.0\text{ m/s}^2} = 6.0\text{ kg} \).

1 mark for rearranging the formula or correct substitution: \( 18 / 3.0 \). 1 mark for the correct mass value of 6.0 (kg).

An alpha particle is identical to a helium-4 nucleus. It contains exactly 2 protons and 2 neutrons.

1 mark for identifying that it has 2 protons. 1 mark for identifying that it has 2 neutrons.

The gravitational force acts towards the Sun, providing a centripetal force. This force constantly changes the direction of the planet's velocity, keeping it in a circular path.

1 mark for identifying that the gravitational force acts towards the centre or Sun. 1 mark for explaining that this force changes the direction of velocity but not the speed.

800 Bq to 400 Bq is 1 half-life. 400 Bq to 200 Bq is 2 half-lives. 200 Bq to 100 Bq is 3 half-lives. Therefore, 3 half-lives have passed in 12 days. Half-life = 12 / 3 = 4 days.

1 mark for identifying that 3 half-lives have passed. 1 mark for the correct calculation: 4 days.

Use the formula: acceleration = force / mass. Acceleration = 1.5 / 0.5 = 3 m/s^2.

1 mark for correct substitution into formula: 1.5 / 0.5. 1 mark for the correct answer: 3 m/s^2.

Efficiency = useful energy output / total energy input. Efficiency = 140 / 200 = 0.7 (or 70%).

1 mark for correct substitution into the efficiency equation: 140 / 200. 1 mark for correct answer: 0.7 or 70%.

Use the wave speed equation: wave speed = frequency x wavelength. Wave speed = 5 x 0.4 = 2 m/s.

1 mark for correct substitution into the formula: 5 x 0.4. 1 mark for correct answer: 2 m/s.

All electromagnetic waves travel at the same speed in a vacuum (300,000,000 m/s) and they are all transverse waves.

1 mark for stating they travel at the same speed or speed of light. 1 mark for stating they are all transverse waves.

In beta-minus decay, a neutron changes into a proton and an electron. Therefore, the number of protons increases by 1, and the number of neutrons decreases by 1.

1 mark for stating the number of protons increases by 1. 1 mark for stating the number of neutrons decreases by 1.

Thinking distance = speed x reaction time. Thinking distance = 15 x 0.6 = 9 m.

1 mark for correct substitution: 15 x 0.6. 1 mark for correct calculation of thinking distance: 9 m.

Use the formula:
\(\text{acceleration} = \frac{\text{change in velocity}}{\text{time taken}}\)

\(\text{change in velocity} = 6\text{ m/s} - 0\text{ m/s} = 6\text{ m/s}\)

\(\text{acceleration} = \frac{6\text{ m/s}}{4\text{ s}} = 1.5\text{ m/s}^2\)

- 1 mark for correct substitution: \(\frac{6}{4}\)
- 1 mark for correct evaluation with units: \(1.5\text{ (m/s}^2\text{)}\)

Use the equation for gravitational potential energy (GPE):
\(\Delta GPE = m \times g \times \Delta h\)

\(\Delta GPE = 0.5\text{ kg} \times 10\text{ N/kg} \times 2.0\text{ m} = 10\text{ J}\)

- 1 mark for correct substitution: \(0.5 \times 10 \times 2.0\)
- 1 mark for the correct evaluation: \(10\text{ (J)}\)

Use the wave equation:
\(\text{wave speed} = \text{frequency} \times \text{wavelength}\)

\(\text{wave speed} = 5\text{ Hz} \times 0.4\text{ m} = 2\text{ m/s}\)

- 1 mark for correct substitution: \(5 \times 0.4\)
- 1 mark for correct evaluation: \(2\text{ (m/s)}\)

Ultraviolet radiation has a higher frequency than visible light and can damage living cells. Exposure can cause sunburn, premature aging of the skin, and can damage cell DNA, leading to skin cancer or cataracts in the eyes.

1 mark for each correct harmful effect mentioned (up to 2):
- Sunburn / premature aging of skin
- Skin cancer
- Cataracts / eye damage
(Do not accept general 'cancer' unless specified as skin cancer)

An alpha particle is identical to a helium nucleus. It is made up of two protons and two neutrons bound together.

- 1 mark for stating it consists of protons and neutrons.
- 1 mark for specifying there are two of each (or stating it is a helium-4 nucleus).

The two primary pieces of evidence for the Big Bang are:
1. The red-shift of light from distant galaxies, showing that the Universe is expanding.
2. Cosmic Microwave Background Radiation (CMBR), which is the leftover thermal radiation from the early Universe.

- 1 mark for red-shift of distant galaxies (accept cosmological red-shift).
- 1 mark for Cosmic Microwave Background Radiation (accept CMBR).

A scalar quantity has magnitude (size) only, such as distance or speed. A vector quantity has both magnitude and a specific direction, such as displacement, velocity, acceleration, or force.

- 1 mark for stating that a vector has direction as well as magnitude, while a scalar has only magnitude.
- 1 mark for a correct example of a vector (e.g., velocity, force, weight, acceleration, displacement, momentum).

The activity of a sample halves with every half-life that passes.

- Initial activity = \(800\text{ Bq}\)
- After 1 half-life: \(800 \div 2 = 400\text{ Bq}\)
- After 2 half-lives: \(400 \div 2 = 200\text{ Bq}\)

- 1 mark for calculating the activity after one half-life (\(400\text{ Bq}\)).
- 1 mark for the correct final answer of \(200\text{ (Bq)}\).

Use the acceleration formula: acceleration = (final velocity - initial velocity) / time taken. Given that initial velocity u = 0 m/s, final velocity v = 6.0 m/s, and time t = 4.0 s, we substitute these values into the formula: acceleration = (6.0 - 0) / 4.0 = 1.5 m/s^2.

1 mark for recalling or using the equation: acceleration = change in velocity / time taken. 1 mark for the correct calculation: 6.0 / 4.0 = 1.5. 1 mark for the correct unit: m/s^2 (or m/s/s).

Use the efficiency formula: efficiency = useful energy output / total energy input. Substitute the given values into the equation: efficiency = 120 / 300 = 0.4. This can also be expressed as a percentage: 0.4 * 100% = 40%.

1 mark for recalling or using the equation: efficiency = useful energy output / total energy input. 1 mark for substituting the correct values: 120 / 300. 1 mark for the correct final answer: 0.4 (or 40%).

Use the wave speed equation: wave speed = frequency * wavelength. Substitute the given values: wave speed = 2.5 Hz * 1.6 m = 4.0 m/s.

1 mark for recalling or using the equation: wave speed = frequency * wavelength. 1 mark for the correct calculation: 2.5 * 1.6 = 4.0. 1 mark for the correct unit: m/s.

Ultraviolet radiation has a higher frequency and a shorter wavelength than radio waves. Additionally, ultraviolet radiation is ionizing whereas radio waves are non-ionizing. Exposure to ultraviolet radiation can cause sunburn, skin cancer, or premature aging of the skin.

1 mark for stating one difference (e.g., UV has higher frequency or shorter wavelength than radio waves). 1 mark for stating another difference (e.g., UV is ionizing whereas radio waves are non-ionizing). 1 mark for describing a valid health risk of UV (e.g., causes sunburn, skin cancer, premature skin aging, or eye cataracts).

1) In beta-minus decay, the mass number remains unchanged, so A = 14. 2) The atomic number increases by 1 because a neutron changes into a proton, so Z = 6 + 1 = 7. 3) Inside the nucleus, a neutron decays/changes into a proton (and an electron is emitted as a beta particle).

1 mark for A = 14. 1 mark for Z = 7. 1 mark for describing that a neutron changes/turns into a proton (and an electron).

1) Comet orbits are highly elliptical (elongated) while planet orbits are near-circular. 2) Comets have orbital speeds that vary greatly (moving fastest near the Sun and slowest far away) while planets have relatively constant orbital speeds. 3) Comets generally have much longer orbital periods than planets.

1 mark for each of any three valid points: Comet orbits are highly elliptical / elongated whereas planet orbits are near-circular. Comets have much longer orbital periods than planets. Comets experience a large change in speed during their orbit (very fast near the Sun, very slow far away), whereas planets have near-constant orbital speeds. The Sun is located near one focus of the comet's highly stretched orbit, whereas the Sun is near the center of a planet's orbit.

First, determine the number of half-lives that have passed. The activity halves from 800 to 400 (1 half-life), then to 200 (2 half-lives), and then to 100 (3 half-lives). Thus, 3 half-lives equal 24 hours. The duration of one half-life is 24 hours / 3 = 8 hours.

1 mark for identifying that 3 half-lives have elapsed (800 to 400 to 200 to 100). 1 mark for dividing the total time by 3 (24 / 3). 1 mark for the correct half-life: 8 hours.

First, calculate the total mass: mass = 60 kg + 15 kg = 75 kg. Next, use Newton's second law: force = mass * acceleration, which rearranges to acceleration = force / mass. Substitute the values: acceleration = 150 N / 75 kg = 2.0 m/s^2.

1 mark for finding the total mass: 60 + 15 = 75 kg. 1 mark for using acceleration = force / mass (e.g. 150 / 75). 1 mark for the correct value: 2 (m/s^2).

First, state the equation that relates force, mass, and acceleration: \( \text{force} = \text{mass} \times \text{acceleration} \). Next, substitute the given values into the equation: \( 1.5 = 0.5 \times a \). Rearrange the equation to solve for acceleration: \( a = \frac{1.5}{0.5} \). Calculate the final value: \( a = 3 \text{ m/s}^2 \).

1 mark for stating the equation: force = mass x acceleration. 1 mark for correct substitution: 1.5 = 0.5 x a (or 1.5 / 0.5). 1 mark for the correct value of 3.

First, calculate how many half-lives have passed in 45 minutes: \( \frac{45}{15} = 3 \text{ half-lives} \). Next, halve the initial activity three times: after 1 half-life, the activity is \( \frac{800}{2} = 400 \text{ Bq} \); after 2 half-lives, the activity is \( \frac{400}{2} = 200 \text{ Bq} \); after 3 half-lives, the activity is \( \frac{200}{2} = 100 \text{ Bq} \).

1 mark for identifying that 3 half-lives have passed. 1 mark for a correct halving process shown (e.g., 800 -> 400 -> 200). 1 mark for the correct final answer of 100 (Bq).

First, state the equation for efficiency: \( \text{efficiency} = \frac{\text{useful energy transfer}}{\text{total energy transfer}} \). Next, substitute the given values into the equation: \( \text{efficiency} = \frac{120}{200} \). Calculate the final efficiency: \( 0.6 \) (or \( 60\% \)).

1 mark for stating the equation: efficiency = useful energy transfer / total energy transfer. 1 mark for correct substitution: 120 / 200. 1 mark for the correct final calculation of 0.6 (or 60%).

An effective answer must include: Safety precautions: Use tongs to increase distance from the source. Point the source away from the teacher and students. Keep source exposure time to a minimum and store sources in a lead-lined container immediately after use. Experimental procedure: 1. Measure the background radiation count rate without any source present. 2. Set up the radiation source at a fixed distance from the GM tube connected to a counter. 3. Measure the count rate with no absorber. 4. Place a sheet of paper between the source and the detector. If the count rate falls back to the background level, the source is an alpha emitter. 5. If the radiation passes through paper, replace the paper with a thin sheet of aluminium (a few mm thick). If the count rate now falls to background, the source is a beta emitter. 6. If the radiation passes through both paper and aluminium, place a thick sheet of lead in front of the source. If a significant reduction in count rate is only observed with lead, the source is a gamma emitter.

This is a 6-mark level of response question. Level 1 (1-2 marks): Demonstrates isolated knowledge of safety (e.g. use tongs) or radiation properties (e.g. alpha stopped by paper), but lacks a structured method. Level 2 (3-4 marks): Describes a partially complete method, including some safety precautions and explains how to identify at least two types of radiation using absorbers. Level 3 (5-6 marks): Explains a fully detailed and safe method. Includes measuring background radiation, safe handling of sources, and systematically using paper, aluminium, and lead to distinguish between all three types of radiation.

Comparison of life cycles: 1. Formation (common to both): Stars form from a cloud of dust and gas called a nebula. Gravity pulls this together to form a protostar. As temperature and pressure rise, hydrogen nuclei fuse to form helium (nuclear fusion), and the star enters the stable main sequence stage. 2. Star like the Sun (low-mass path): When hydrogen runs out, the star expands and cools to become a red giant. The outer layers are eventually ejected to form a planetary nebula, leaving behind a hot, dense core called a white dwarf. This eventually cools to become a black dwarf. 3. High-mass star (high-mass path): A much larger star expands to become a red supergiant. When it runs out of fuel, it collapses rapidly and explodes in a massive explosion called a supernova. The remaining core collapses to form either a neutron star or, if the mass is extremely large, a black hole from which not even light can escape.

This is a 6-mark level of response question. Level 1 (1-2 marks): Identifies some stages of a star's life cycle (e.g., nebula, main sequence, red giant) with little or no comparison between the two pathways. Level 2 (3-4 marks): Describes the complete life cycle of either a Sun-like star or a high-mass star, or provides a partial comparison of both pathways with minor omissions. Level 3 (5-6 marks): Provides a clear, detailed, and accurate comparison of both life cycle pathways from formation through to the distinct final stages (white dwarf vs neutron star/black hole), using correct scientific terminology throughout.

When a polythene rod is rubbed with a dry cloth, friction causes negatively charged electrons to move. The electrons move from the cloth onto the rod. Since the rod gains electrons, it becomes negatively charged. Protons are bound tightly inside the nuclei of the atoms and do not move during charging by friction.

[1] A - Electrons have transferred from the cloth to the rod. Award 1 mark for the correct option.

To find the density of an irregular solid, the mass is measured using a balance, and the volume is found using a displacement (eureka) can and a measuring cylinder. When the stone is lowered into the filled displacement can, it displaces a volume of water equal to its own volume, which is collected and measured in the measuring cylinder.

[1] B - A displacement (eureka) can and a measuring cylinder. Award 1 mark for identifying the correct equipment to find the volume of an irregular solid.

Use the equation relating charge (Q), current (I), and time (t): \(Q = I \times t\). Substituting the given values: \(Q = 0.5\text{ A} \times 40\text{ s} = 20\text{ C}\).

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

The extension of a spring is the increase in its length when a force is applied: \(\text{Extension} = \text{total length} - \text{original length}\). This gives: \(\text{Extension} = 11.0\text{ cm} - 5.0\text{ cm} = 6.0\text{ cm}\).

[1] C - 6.0 cm. Award 1 mark for calculating the extension of the spring correctly.

The main magnetic elements are iron, nickel, and cobalt. Materials like copper, aluminium, zinc, gold, plastic, and paper are non-magnetic.

[1] B - Iron, nickel, cobalt. Award 1 mark for the correct list of magnetic materials.

Work done is calculated using the equation: \(\text{Work done} = \text{force} \times \text{distance moved in the direction of the force}\). Substituting the values: \(\text{Work done} = 20\text{ N} \times 4.0\text{ m} = 80\text{ J}\).

[1] D - 80.0 J. Award 1 mark for the correct calculation of work done.

To find the density, use the equation: \(\text{density} = \frac{\text{mass}}{\text{volume}}\). Substituting the given values: \(\text{density} = \frac{240\text{ g}}{300\text{ cm}^3} = 0.8\text{ g/cm}^3\).

1 mark for correct substitution: \(240 / 300\). 1 mark for correct value and unit: \(0.8\text{ g/cm}^3\) (accept \(0.8\) with correct unit).

To find the charge, use the equation: \(Q = I \times t\). Substituting the given values: \(Q = 0.5\text{ A} \times 40\text{ s} = 20\text{ C}\).

1 mark for correct substitution: \(0.5 \times 40\). 1 mark for correct calculation of charge: \(20\) (accept \(20\text{ C}\) / coulombs).

Use the equation: \(F = k \times x\), where \(F\) is force, \(k\) is the spring constant, and \(x\) is the extension. Rearrange to solve for extension: \(x = \frac{F}{k} = \frac{5.0}{25} = 0.2\text{ m}\).

1 mark for correct rearrangement or substitution: \(5.0 = 25 \times x\) or \(5.0 / 25\). 1 mark for correct evaluation: \(0.2\) (accept \(0.2\text{ m}\) or \(20\text{ cm}\)).

Use the work done formula: \(E = F \times d\). Substituting the given values: \(E = 80\text{ N} \times 15\text{ m} = 1200\text{ J}\).

1 mark for correct substitution: \(80 \times 15\). 1 mark for correct answer: \(1200\) (accept \(1200\text{ J}\) or \(1.2\text{ kJ}\)).

Use the change of state thermal energy formula: \(Q = m \times L\). Substituting the given values: \(Q = 0.50\text{ kg} \times 334,000\text{ J/kg} = 167,000\text{ J}\).

1 mark for correct substitution: \(0.50 \times 334,000\). 1 mark for correct calculation: \(167,000\) (accept \(167\text{ kJ}\)).

The magnetic field strength of a solenoid can be increased by: 1) Increasing the electrical current flowing through the wire, 2) Increasing the number of turns/loops on the coil, or 3) Adding an iron core inside the coil.

1 mark for each correct way listed, up to a maximum of 2 marks: Increase the current (or voltage/potential difference), Increase the number of turns/coils, Add a soft iron core.

Frictional force during rubbing transfers electrons from the dry cloth to the plastic rod. Since electrons are negatively charged, the excess of electrons on the rod gives it a overall negative charge.

1 mark for mentioning that electrons are transferred / rubbed off. 1 mark for stating the direction of transfer: from the cloth to the rod (accept: rod gains electrons).

Use the power formula: \(P = I \times V\). Substituting the given values: \(P = 8.0\text{ A} \times 230\text{ V} = 1840\text{ W}\).

1 mark for correct substitution: \(8.0 \times 230\). 1 mark for correct evaluation: \(1840\) (accept \(1840\text{ W}\) or \(1.84\text{ kW}\)).

First, calculate the mass of the liquid: 114 g - 50 g = 64 g. Then, use the density formula: density = mass / volume. Density = 64 g / 80 cm\(^{3}\) = 0.8 g/cm\(^{3}\).

1 mark for calculating the mass of the liquid as 64 g (or correctly substituting into the density formula). 1 mark for the correct final density of 0.8 (g/cm\(^{3}\)).

Use Hooke's Law: force = spring constant * extension (\(F = k \times x\)). Rearrange the equation to solve for extension: \(x = \frac{F}{k}\). Extension = 12 N / 25 N/m = 0.48 m.

1 mark for rearranging the formula or correct substitution: 12 / 25. 1 mark for the correct final answer of 0.48.

First, convert the time from minutes to seconds: 5 minutes = 5 * 60 = 300 seconds. Then use the charge equation: charge = current * time (\(Q = I \times t\)). Charge = 0.4 A * 300 s = 120 C.

1 mark for converting 5 minutes to 300 seconds. 1 mark for the correct final charge of 120 (C).

First, convert the distance from centimetres to metres: 40 cm = 0.4 m. Use the formula: moment of a force = force * perpendicular distance from pivot. Moment = 3 N * 0.4 m = 1.2 Nm.

1 mark for converting 40 cm to 0.4 m (or correctly substituting 3 * 0.4). 1 mark for the correct final moment of 1.2 (Nm).

During friction, electrons are transferred from the dry cloth to the plastic rod. Because the cloth loses negatively charged electrons, it is left with an overall positive charge.

1 mark for stating that the cloth loses electrons. 1 mark for stating that the cloth becomes positively charged.

Use the formula: work done = force * distance. Work done = 45 N * 6.0 m = 270 J.

1 mark for correct substitution: 45 * 6. 1 mark for the correct final work done of 270 (J).

First, convert power to watts: 1.5 kW = 1500 W. Use the formula: power = current * voltage (\(P = I \times V\)), which rearranges to current = power / voltage. Current = 1500 W / 230 V = 6.52 A, which rounds to 6.5 A to 2 significant figures.

1 mark for converting power to 1500 W or correct rearrangement: 1500 / 230. 1 mark for the correct current of 6.5 (A) rounded to 2 significant figures.

Use the specific latent heat formula: change in thermal energy = mass * specific latent heat (\(Q = m \times L\)). Energy = 0.25 kg * 334000 J/kg = 83500 J.

1 mark for correct substitution: 0.25 * 334000. 1 mark for the correct energy of 83500 (J).

To find the density, substitute the values into the formula:
$$\text{density} = \frac{0.54\text{ kg}}{0.000060\text{ m}^3} = 9000\text{ kg/m}^3$$

- **1 mark**: Correct substitution into the equation: \(\frac{0.54}{0.000060}\)
- **1 mark**: Correct evaluation: \(9000\text{ (kg/m}^3)\)

Using the equation:
$$\text{force} = 25\text{ N/m} \times 0.12\text{ m} = 3.0\text{ N}$$

- **1 mark**: Correct substitution into the equation: \(25 \times 0.12\)
- **1 mark**: Correct evaluation: \(3\) or \(3.0\text{ (N)}\)

To calculate the charge, multiply the current by the time:
$$\text{charge} = 0.8\text{ A} \times 120\text{ s} = 96\text{ C}$$

- **1 mark**: Correct substitution: \(0.8 \times 120\)
- **1 mark**: Correct evaluation: \(96\text{ (C)}\)

Using the formula:
$$\text{work done} = 60\text{ N} \times 4.5\text{ m} = 270\text{ J}$$

- **1 mark**: Correct substitution: \(60 \times 4.5\)
- **1 mark**: Correct evaluation: \(270\text{ (J)}\)

As temperature increases, the gas particles gain kinetic energy and move faster. This causes them to collide with the container walls more frequently and with greater force, which increases the total pressure.

- **1 mark**: State that the particles move faster or gain kinetic energy.
- **1 mark**: State that particles collide with the walls of the container more frequently OR with greater force.

1. Place the plotting compass near one pole of the bar magnet and draw a dot on the paper where the needle points.
2. Move the compass so that the opposite end of the needle is over the dot, make a new mark at the pointing end, and repeat this process to plot a line from one pole to the other.

- **1 mark**: Place the compass near the magnet and mark the direction of the needle (using dots/pencil).
- **1 mark**: Move the compass to the new mark / repeat the process to trace the field lines from pole to pole.

Friction between the rod and the cloth causes electrons (which are negatively charged subatomic particles) to be transferred. Since electrons move from the cloth to the rod, the rod gains a surplus of negative charge.

- **1 mark**: Identification that electrons are transferred.
- **1 mark**: Correct direction of transfer (from the cloth to the rod). Reject any mention of protons moving.

A scalar quantity has magnitude (size) only. A vector quantity has both magnitude and direction. An example of a vector quantity is force, velocity, displacement, or acceleration.

- **1 mark**: Correctly defining the difference (vectors have magnitude and direction, scalars have only magnitude/size).
- **1 mark**: A valid example of a vector quantity (e.g., force, weight, velocity, displacement, acceleration, momentum).

Using the formula: \(R = \frac{V}{I}\). Substituting the given values: \(R = \frac{6.0}{0.15} = 40\ \Omega\).

1. State or recall the equation \(V = I \times R\) or \(R = \frac{V}{I}\) (1 mark). 2. Correct substitution of values: \(\frac{6.0}{0.15}\) (1 mark). 3. Correct final answer of \(40\ \Omega\) (1 mark).

Using the formula: \(\text{density} = \frac{\text{mass}}{\text{volume}}\). Substituting the given values: \(\text{density} = \frac{405}{150} = 2.7\text{ g/cm}^3\).

1. Recall the formula \(\text{density} = \frac{\text{mass}}{\text{volume}}\). (1 mark) 2. Correct substitution and calculation: \(\frac{405}{150} = 2.7\) (1 mark). 3. Correct unit: \(\text{g/cm}^3\) (1 mark).

First, find the extension of the spring: \(x = 20\text{ cm} - 12\text{ cm} = 8\text{ cm} = 0.08\text{ m}\). Then, use Hooke's Law: \(F = k \times x\), so \(k = \frac{F}{x}\). Substituting the values: \(k = \frac{6.0}{0.08} = 75\text{ N/m}\).

1. Calculate the extension and convert to metres: \(0.08\text{ m}\) (1 mark). 2. Rearrange the formula \(F = k \times x\) to \(k = \frac{F}{x}\) and substitute: \(\frac{6.0}{0.08}\) (1 mark). 3. Correct evaluation: \(75\text{ N/m}\) (1 mark).

When the plastic rod is rubbed with the cloth, friction causes electrons to be transferred. Since electrons are negatively charged particles, and they move from the cloth to the rod, the rod gains a net negative charge.

1. Friction/rubbing occurs (1 mark). 2. Electrons are transferred (1 mark). 3. Electrons move from the cloth to the rod / the rod gains electrons (1 mark).

Place the bar magnet on a piece of paper. Place the plotting compass near one of the poles. Draw a dot on the paper where the needle points. Move the compass so that the opposite end of the needle points to the dot just drawn, and make a new dot at the other end. Repeat this process until you reach the other pole, then join the dots with a smooth line.

1. Place the compass near a pole of the magnet and mark the needle's direction with a dot (1 mark). 2. Move the compass so that the tail of the needle aligns with the previous dot, and mark the new position (1 mark). 3. Repeat the steps to plot multiple points and connect them with a continuous line (1 mark).

Using the formula: \(\text{Work done} = \text{force} \times \text{distance}\). Substituting the values: \(\text{Work done} = 4.5\text{ N} \times 8.0\text{ m} = 36\text{ J}\).

1. Recall of formula \(\text{work done} = \text{force} \times \text{distance}\) (1 mark). 2. Correct calculation: \(4.5 \times 8.0 = 36\) (1 mark). 3. Correct unit: \(\text{Joules}\) or \(\text{J}\) (1 mark).

When heated, the gas particles gain kinetic energy and move faster. This causes them to collide with the walls of the container more frequently and with greater force. This increase in force per unit area results in an increase in gas pressure.

1. Gas particles gain kinetic energy / move faster (1 mark). 2. Particles collide with the container walls more frequently (1 mark). 3. Collisions are harder / exert more force (1 mark).

Using the formula: \(\text{Power} = \text{current} \times \text{potential difference}\). Rearranging for current: \(\text{current} = \frac{\text{Power}}{\text{potential difference}}\). Substituting the values: \(I = \frac{48}{12} = 4.0\text{ A}\).

1. Recall or rearrangement of \(P = I \times V\) to \(I = \frac{P}{V}\) (1 mark). 2. Correct substitution: \(\frac{48}{12}\) (1 mark). 3. Correct evaluation: \(4.0\text{ A}\) (accept \(4\)) (1 mark).

To determine the volume of an irregular solid:
1. Fill the displacement can with water until the water is level with the bottom of the spout and allow any excess water to drain.
2. Place an empty measuring cylinder under the spout.
3. Lower the rock gently into the displacement can, ensuring it is fully submerged.
4. Read the volume of water collected in the measuring cylinder. This volume of water is equal to the volume of the rock.

1 mark: Fill the displacement can with water until it is level with the bottom of the spout.
1 mark: Submerge the rock gently in the water (using a thread/string).
1 mark: Measure the volume of the displaced water collected in the measuring cylinder.

First, find the extension of the spring in metres:
Extension \(x = 15.0\text{ cm} - 12.0\text{ cm} = 3.0\text{ cm}\)
Convert extension to metres:
\(x = 3.0 / 100 = 0.03\text{ m}\)
Next, use the formula for force on a spring:
\(F = k \times x\)
Rearrange the formula to solve for the spring constant \(k\):
\(k = \frac{F}{x}\)
Substitute the values:
\(k = \frac{6.0\text{ N}}{0.03\text{ m}} = 200\text{ N/m}\)

1 mark: Calculate the extension of the spring in metres, \(0.03\text{ m}\) (allow 1 mark for calculating \(3.0\text{ cm}\) if conversion is not done).
1 mark: Recall and rearrange the formula: \(k = \frac{F}{x}\).
1 mark: Correct calculation of the spring constant: \(200\text{ (N/m)}\). (If extension was kept in cm, giving 2 N/cm, award 2 marks total).

First, convert the time from minutes to seconds:
\(t = 1.5\text{ minutes} = 1.5 \times 60 = 90\text{ s}\)
Next, use the formula relating charge, current, and time:
\(Q = I \times t\)
Rearrange the formula to solve for current \(I\):
\(I = \frac{Q}{t}\)
Substitute the values:
\(I = \frac{45\text{ C}}{90\text{ s}} = 0.5\text{ A}\)

1 mark: Convert time to seconds: \(90\text{ s}\).
1 mark: Recall and rearrange the formula: \(I = \frac{Q}{t}\).
1 mark: Correct calculation of current: \(0.5\text{ (A)}\) (allow 2 marks for \(30\text{ A}\) which arises from not converting time to seconds).

Level 3 (5-6 marks): A detailed description of the experimental setup, measurements, and calculations is given. The student explains how to set up a clamp stand, spring, and ruler, and details the measurement of both the unstretched (original) length and the stretched length for multiple known masses. The student clearly shows how to calculate extension: \(\text{extension} = \text{new length} - \text{original length}\). At least two experimental techniques to ensure accuracy are included (e.g. reading at eye level to avoid parallax, using a fiducial marker or pointer, ensuring the ruler is vertical). Level 2 (3-4 marks): A line of reasoning is present. The student describes how to set up the apparatus and measure the length of the spring for different loads. Some explanation of calculating extension is given, but accuracy details may be missing. Level 1 (1-2 marks): The explanation is simple. The student mentions adding masses and measuring the spring but lacks structure and key details of the procedure or calculation.

Level 1 (1-2 marks): Identifies basic equipment such as a spring, ruler, and masses. States that masses are added to the spring. Level 2 (3-4 marks): Describes setting up the spring vertically and measuring its length. Mentions adding masses systematically and measuring the new length. Suggests subtracting the original length. Level 3 (5-6 marks): Provides a complete method including: measuring the original length, adding known masses to vary force, measuring the new length, calculating extension using the formula \(\text{extension} = \text{stretched length} - \text{original length}\), and describes specific techniques for accuracy (e.g. using a pointer, reading at eye level to avoid parallax error, or repeating and averaging).

Level 3 (5-6 marks): A detailed and logically structured description is provided. The student correctly identifies all circuit components (power supply, ammeter, voltmeter, filament lamp, and variable resistor) and their correct connections (ammeter in series, voltmeter in parallel across the lamp). The student explains how to take readings of current and potential difference, and how to use the variable resistor to vary these values. The student also mentions a method to expand the range or improve reliability, such as reversing the power supply to get negative values or repeating and averaging. Level 2 (3-4 marks): A correct circuit setup is described, including the ammeter in series and voltmeter in parallel with the lamp. The student states that current and potential difference are measured, but the method for varying the voltage may be vague or the description lacks detail on obtaining a full range of results. Level 1 (1-2 marks): The description is basic. The student mentions measuring current and voltage but may describe incorrect connections or omit the variable resistor entirely.

Level 1 (1-2 marks): Identifies the need to use an ammeter and voltmeter. Simple description of connecting a circuit. Level 2 (3-4 marks): Correctly details connecting the ammeter in series with the lamp and the voltmeter in parallel across the lamp. Explains that current and potential difference are measured. Level 3 (5-6 marks): Provides a comprehensive method including: complete circuit with variable resistor, detailed explanation of adjusting the variable resistor to obtain a range of current and voltage readings, and additional detail such as reversing the power supply to obtain negative readings or repeating the experiment to find averages.