2023 Edexcel GCSE Physics (1PH0) Practice Paper with Answers

We use the speed equation: average speed = distance / time. Substituting the given values: speed = 150 m / 6.0 s = 25 m/s.

1 mark for the correct calculation of average speed, leading to option b.

Alpha radiation consists of alpha particles. An alpha particle is made up of two protons and two neutrons, which is identical to the nucleus of a helium atom.

1 mark for identifying alpha particles as the correct answer.

Efficiency is calculated using the formula: efficiency = useful energy output / total energy input. Substituting the values: efficiency = 120 J / 400 J = 0.30 (or 30%).

1 mark for the correct calculation of efficiency, leading to option a.

In the electromagnetic spectrum, waves ordered from longest to shortest wavelength are: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. Therefore, radio waves have the longest wavelength among the options.

1 mark for selecting radio waves as having the longest wavelength.

We use the wave equation: wave speed = frequency * wavelength. Substituting the values: wave speed = 5.0 Hz * 0.80 m = 4.0 m/s.

1 mark for the correct calculation of wave speed, leading to option b.

A massive star starts as a nebula, becomes a main sequence star, expands into a red supergiant, explodes in a supernova, and leaves behind either a neutron star or a black hole.

1 mark for identifying the correct lifecycle stages of a high-mass star, leading to option b.

We use the acceleration equation: acceleration = change in velocity / time taken. Since starting from rest, the change in velocity is 6.0 m/s - 0 m/s = 6.0 m/s. Acceleration = 6.0 m/s / 4.0 s = 1.5 m/s^2.

1 mark for the correct calculation of acceleration, leading to option a.

The number of half-lives that have elapsed in 30 minutes is 30 / 10 = 3 half-lives. After 1 half-life, the activity is 400 Bq. After 2 half-lives, the activity is 200 Bq. After 3 half-lives, the activity is 100 Bq.

1 mark for the correct calculation of the remaining activity, leading to option d.

Alpha particles are helium nuclei consisting of two protons and two neutrons.
Beta-minus particles are high-speed electrons emitted from the nucleus during radioactive decay.
Gamma rays are high-frequency electromagnetic waves emitted from the nucleus.

Award 1 mark for each correct line drawn:
- Alpha (\(\alpha\)) particle matched to 'A helium nucleus consisting of two protons and two neutrons' (1)
- Beta-minus (\(\beta^-\)) particle matched to 'A high-speed electron emitted from the nucleus' (1)
- Gamma (\(\gamma\)) ray matched to 'High-frequency electromagnetic radiation' (1)

First, determine the number of half-lives that have passed. \(\text{Number of half-lives} = \frac{30 \text{ minutes}}{15 \text{ minutes}} = 2\). After 1 half-life, the activity is \(\frac{800}{2} = 400\) Bq. After 2 half-lives, the activity is \(\frac{400}{2} = 200\) Bq.

1 mark for identifying that 30 minutes is 2 half-lives OR showing one halving step (e.g., finding 400 Bq). 1 mark for the correct final answer of 200 (Bq).

Using Newton's second law: \(F = m \times a\). Rearranging for acceleration gives \(a = \frac{F}{m}\). Substituting the given values: \(a = \frac{4.5 \text{ N}}{1.5 \text{ kg}} = 3 \text{ m/s}^2\).

1 mark for substituting the values into the correct formula: \(\frac{4.5}{1.5}\). 1 mark for the correct calculation: 3 (m/s²).

The geocentric model describes the Earth as being at the centre of the solar system with the Sun and other planets orbiting it. In contrast, the heliocentric model correctly identifies the Sun as being at the centre, with the Earth and other planets orbiting around it.

1 mark for stating that the Earth is at the centre of the geocentric model. 1 mark for stating that the Sun is at the centre of the heliocentric model.

The formula for efficiency is: \(\text{Efficiency} = \frac{\text{Useful energy transfer}}{\text{Total energy input}}\). Substituting the values: \(\text{Efficiency} = \frac{150 \text{ J}}{250 \text{ J}} = 0.6\) (or 60%).

1 mark for correct substitution: \(\frac{150}{250}\). 1 mark for correct final value of 0.6 (or 60%).

Using the wave equation: \(v = f \times \lambda\), where \(v\) is wave speed, \(f\) is frequency, and \(\lambda\) is wavelength. Substituting the values: \(v = 5 \text{ Hz} \times 0.8 \text{ m} = 4 \text{ m/s}\).

1 mark for correct substitution: \(5 \times 0.8\). 1 mark for correct calculation: 4 (m/s).

Ultraviolet (UV) radiation is commonly used commercially for fluorescent light bulbs, security ink, sterilising water, or in sunbeds. Exposure to UV radiation carries hazards, primarily causing sunburn, premature skin aging, cataracts (eye damage), or increasing the risk of skin cancer.

1 mark for a valid use (e.g., security marking, counterfeit detection, sterilization, tanning). 1 mark for a valid hazard (e.g., skin cancer, sunburn, eye damage/cataracts).

During beta-minus decay, a neutron in an unstable nucleus decays into a proton and an electron. The proton remains in the nucleus, and the electron is ejected at high speed as a beta-minus particle.

1 mark for stating that the neutron changes into a proton. 1 mark for stating that an electron (or beta particle) is emitted.

The formula for speed is: \(\text{speed} = \frac{\text{distance}}{\text{time}}\). Substituting the given numbers: \(\text{speed} = \frac{120 \text{ m}}{15 \text{ s}} = 8 \text{ m/s}\).

1 mark for correct substitution: \(\frac{120}{15}\). 1 mark for correct calculation: 8 (m/s).

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

Here, the change in velocity is:
\(6\text{ m/s} - 0\text{ m/s} = 6\text{ m/s}\)

The time taken is \(4\text{ s}\).

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

1 mark for the correct calculated value: 1.5 (or 3/2).
1 mark for the correct unit: m/s^2 (or m/s/s).

First, find the number of half-lives that have passed by repeatedly halving the initial activity:
\(800\text{ Bq} \rightarrow 400\text{ Bq}\) (1 half-life)
\(\rightarrow 200\text{ Bq}\) (2 half-lives)
\(\rightarrow 100\text{ Bq}\) (3 half-lives)

So, 3 half-lives have elapsed in \(15\text{ days}\).

The length of one half-life is:
\(\frac{15\text{ days}}{3} = 5\text{ days}\)

1 mark for showing that 3 half-lives have elapsed (e.g. by writing out the decay chain or dividing 800 by 2 repeatedly).
1 mark for the correct calculation of half-life: 5 (days).

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

Substitute the given values:
\(\text{wave speed} = 4\text{ Hz} \times 0.5\text{ m} = 2\text{ m/s}\)

1 mark for correct substitution: 4 x 0.5.
1 mark for correct calculation: 2 (m/s).

Use the efficiency equation:
\(\text{efficiency} = \frac{\text{useful energy output}}{\text{total energy input}}\)

Substitute the given values:
\(\text{efficiency} = \frac{30}{120} = 0.25\) (or \(25\%\))

1 mark for correct substitution into the efficiency formula: 30 / 120.
1 mark for correct calculation of efficiency: 0.25 (accept 25%).

Red-shift is the increase in wavelength of light from galaxies moving away from us. This shows that distant galaxies are moving away (receding) from Earth. The observation that more distant galaxies have a larger red-shift shows they are moving away faster, which supports the idea that the universe is expanding.

1 mark for stating that the galaxies are moving away from Earth / us.
1 mark for stating that more distant galaxies are moving faster / have a greater red-shift.

Ultraviolet radiation has a higher frequency and shorter wavelength than visible light. Because of this, UV radiation carries more energy per photon. This higher energy allows it to penetrate skin cells and cause damage, such as mutations to DNA, which can lead to skin cancers, whereas visible light does not carry enough energy to cause this type of damage.

1 mark for stating that ultraviolet radiation has a higher frequency / shorter wavelength / more energy (than visible light).
1 mark for stating that this energy can cause damage to cells / mutation of DNA / skin cancer.

1. Find the number of half-lives that have passed: \(\text{Number of half-lives} = \frac{24\text{ days}}{8\text{ days}} = 3\)

2. Halve the activity 3 times:
- After 1 half-life: \(800 \div 2 = 400\text{ Bq}\)
- After 2 half-lives: \(400 \div 2 = 200\text{ Bq}\)
- After 3 half-lives: \(200 \div 2 = 100\text{ Bq}\)

- 1 mark for calculating the number of half-lives (3) or showing a division process.
- 1 mark for carrying out the halving process correctly.
- 0.5 marks for the correct final value of 100.

Use the wave speed equation: \(v = f \times \lambda\)

Substitute the given values: \(v = 250\text{ Hz} \times 1.36\text{ m} = 340\text{ m/s}\)

- 1 mark for recalling/using the correct wave equation: \(\text{wave speed} = \text{frequency} \times \text{wavelength}\).
- 1 mark for correct substitution of values: \(250 \times 1.36\).
- 0.5 marks for the correct final speed of 340.

Use Newton's second law: \(F = m \times a\)

Substitute the given values: \(F = 1200\text{ kg} \times 2.5\text{ m/s}^2 = 3000\text{ N}\)

- 1 mark for recalling/using the correct equation: \(\text{force} = \text{mass} \times \text{acceleration}\).
- 1 mark for correct substitution of values: \(1200 \times 2.5\).
- 0.5 marks for the correct final force of 3000.

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

Substitute the given values: \(\Delta GPE = 15\text{ kg} \times 10\text{ N/kg} \times 4.0\text{ m} = 600\text{ J}\)

- 1 mark for recalling/using the correct equation: \(\text{GPE} = \text{mass} \times g \times \text{height}\).
- 1 mark for correct substitution: \(15 \times 10 \times 4.0\).
- 0.5 marks for the correct final answer of 600.

Use the speed equation: \(\text{average speed} = \frac{\text{distance}}{\text{time}}\)

Substitute the given values: \(\text{average speed} = \frac{400\text{ m}}{80\text{ s}} = 5\text{ m/s}\)

- 1 mark for recalling/using the correct equation: \(\text{speed} = \text{distance} / \text{time}\).
- 1 mark for correct substitution: \(400 / 80\).
- 0.5 marks for the correct final speed of 5.

Use the magnification equation: \(\text{magnification} = \frac{\text{image height}}{\text{actual height}}\)

Substitute the given values: \(\text{magnification} = \frac{6.0\text{ cm}}{1.5\text{ cm}} = 4\)

- 1 mark for recalling/using the magnification formula: \(\text{image height} / \text{actual height}\).
- 1 mark for correct substitution: \(6.0 / 1.5\).
- 0.5 marks for the correct magnification of 4.

Use the acceleration equation: \(a = \frac{v - u}{t}\)

Since the cyclist starts from rest, the initial velocity \(u = 0\text{ m/s}\) and final velocity \(v = 12\text{ m/s}\).

\(a = \frac{12 - 0}{4.0} = 3\text{ m/s}^2\)

- 1 mark for recalling/using the acceleration equation: \(a = \frac{\text{change in velocity}}{\text{time}}\).
- 1 mark for correct substitution: \(\frac{12}{4.0}\).
- 0.5 marks for the correct acceleration of 3.

Use the wave speed equation rearranged for time: \(\text{time} = \frac{\text{distance}}{\text{speed}}\)

Substitute the given values: \(\text{time} = \frac{150,000,000,000\text{ m}}{300,000,000\text{ m/s}} = 500\text{ s}\)

- 1 mark for rearranging the distance speed equation to make time the subject: \(\text{time} = \text{distance} / \text{speed}\).
- 1 mark for correct substitution of values: \(\frac{150,000,000,000}{300,000,000}\).
- 0.5 marks for the correct time of 500.

First, identify the initial velocity (\(u = 0\text{ m/s}\)), final velocity (\(v = 12\text{ m/s}\)), and time taken (\(t = 4.0\text{ s}\)). Use the acceleration formula: \(a = \frac{v - u}{t}\). Substituting the values gives: \(a = \frac{12 - 0}{4.0} = 3.0\text{ m/s}^2\).

1 mark for substituting values into correct equation: \(\frac{12 - 0}{4.0}\). 1 mark for correct evaluation: \(3\) (or \(3.0\)). 0.5 marks for correct unit: \(\text{m/s}^2\).

Use the equation for gravitational potential energy: \(\Delta GPE = m \times g \times \Delta h\). Substituting the given values: \(\Delta GPE = 0.6\text{ kg} \times 10\text{ N/kg} \times 1.5\text{ m} = 9\text{ J}\).

1 mark for substituting values into the correct equation: \(0.6 \times 10 \times 1.5\). 1 mark for correct calculation: \(9\). 0.5 marks for correct unit: \(\text{J}\).

Use the wave speed equation: \(v = f \times \lambda\). Substituting the values: \(v = 4.0\text{ Hz} \times 0.15\text{ m} = 0.6\text{ m/s}\).

1 mark for substituting values into correct equation: \(4.0 \times 0.15\). 1 mark for correct calculation: \(0.6\). 0.5 marks for correct unit: \(\text{m/s}\).

First, find the number of half-lives that have passed: \(\frac{18\text{ hours}}{6\text{ hours}} = 3\text{ half-lives}\). 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 calculating the number of half-lives (\(\frac{18}{6} = 3\)) or showing a process of halving three times. 1 mark for final activity value of \(100\). 0.5 marks for correct unit: \(\text{Bq}\).

Alpha (\(\alpha\)) radiation consists of helium nuclei containing two protons and two neutrons. It has a high ionising power and a low penetrating power, being stopped by a single sheet of paper or a few centimetres of air. Beta (\(\beta\)) radiation consists of high-speed electrons. It has moderate ionising power and moderate penetrating power, being stopped by a few millimetres of aluminium or about a metre of air. Gamma (\(\gamma\)) radiation is a high-energy electromagnetic wave. It has low ionising power and high penetrating power, requiring several centimetres of lead or thick concrete to be significantly reduced. To identify the unknown source: First, measure the background radiation count rate over a set period of time using a Geiger-Muller (GM) tube without the source present. Then, place the unknown source a fixed distance away from the GM tube and measure the new count rate. Next, place a sheet of paper between the source and the GM tube. If the count rate drops significantly back to the background level, the source is alpha. If paper has little effect, replace the paper with a thin sheet of aluminium. If the count rate now drops significantly to the background level, the source is beta. If neither paper nor aluminium significantly reduces the count rate, but a thick sheet of lead does, then the source is gamma.

This is a 6-mark level-of-response question. Indicative content: Nature of radiation: Alpha (helium nucleus), Beta (high-speed electron), Gamma (EM wave). Ionising power: Alpha is highly ionising, Beta is moderately ionising, Gamma is weakly ionising. Penetrating power/Absorbers: Alpha stopped by paper, Beta by aluminium, Gamma by thick lead. Experimental process: Measure background radiation first. Place source near GM tube and measure count rate. Test systematically with paper, aluminium, and lead. Deduce radiation type based on which absorber reduces the count rate back to background level. Level 1 (1-2 marks): Simple description of at least one type of radiation or a basic test. The explanation lacks detail and structure. Level 2 (3-4 marks): Clear comparison of at least two types of radiation in terms of their properties or a partially correct method to identify the source using absorbers. Level 3 (5-6 marks): Comprehensive comparison of all three types of radiation (nature, ionising power, and penetrating power) and a detailed, logical experimental method to identify the source using paper, aluminium, and lead, including the need to account for background radiation.

Conduction is the transfer of thermal energy through a material by the vibrations of particles, which is most effective in solids. Convection is the transfer of thermal energy in fluids (liquids and gases) where warmer, less dense fluid rises and cooler, denser fluid sinks to take its place, creating convection currents. Radiation is the transfer of energy by infrared electromagnetic waves and does not require a medium (it can travel through a vacuum). Double glazing consists of two panes of glass separated by a vacuum or a layer of dry air. A vacuum prevents conduction and convection because there are no particles to transfer the energy. If dry air is used, it reduces conduction because air is an extremely poor thermal conductor, and the narrow gap prevents air movement, reducing convection. Loft insulation consists of fiberglass wool which traps pockets of air. This trapped air is an excellent insulator (reducing conduction) and because the air cannot move, convection currents cannot form. Cavity wall insulation fills the gap between the inner and outer walls of a house with foam. The foam traps air in tiny pockets, which reduces conduction (since air is a poor conductor) and prevents convection by preventing the air from circulating.

This is a 6-mark level-of-response question. Indicative content: Definitions: Conduction involves particle vibrations in solids. Convection involves density changes in moving fluids. Radiation involves infrared electromagnetic waves. Double glazing: Vacuum/air between panes reduces conduction and convection. Loft insulation: Fiberglass traps air, reducing conduction and convection. Cavity wall insulation: Foam traps air in walls, reducing conduction and convection. Level 1 (1-2 marks): Basic description of at least one heat transfer process or one insulation method. The explanation lacks detail. Level 2 (3-4 marks): Clear explanation of at least two insulation methods linked to how they reduce conduction or convection, showing some understanding of the underlying process. Level 3 (5-6 marks): Detailed explanation of all three insulation methods, clearly explaining how they trap air or use a vacuum to reduce conduction and convection, supported by a clear description of the underlying heat transfer processes.

Using the formula:
\[R = \frac{V}{I}\]
where:
- \(V = 6.0\text{ V}\)
- \(I = 0.50\text{ A}\)

\[R = \frac{6.0}{0.50} = 12\text{ }\Omega\]

Therefore, the resistance is 12 \(\Omega\).

1 mark for selecting option B.

The three common magnetic elements are iron, nickel, and cobalt. Copper, aluminium, and zinc are non-magnetic metals.

1 mark for selecting option C.

Use the formula for density:
\[\text{density} = \frac{\text{mass}}{\text{volume}}\]

Substitute the given values:
\[\text{density} = \frac{540\text{ g}}{200\text{ cm}^3} = 2.7\text{ g/cm}^3\]

1 mark for selecting option B.

First, calculate the extension (\(x\)) of the spring:
\[x = 20\text{ cm} - 12\text{ cm} = 8\text{ cm}\]
Convert this extension to metres:
\[x = 0.08\text{ m}\]

Now, use Hooke's Law formula:
\[F = k \times x\]
Rearrange the formula to find the spring constant (\(k\)):
\[k = \frac{F}{x} = \frac{4.0\text{ N}}{0.08\text{ m}} = 50\text{ N/m}\]

1 mark for selecting option C.

When charging by friction, only electrons can move. Because the polythene rod becomes negatively charged, it must have gained electrons. Therefore, negative electrons have moved from the cloth to the rod.

1 mark for selecting option A.

Use the transformer equation:
\[\frac{V_p}{V_s} = \frac{N_p}{N_s}\]

Substitute the given values:
\[\frac{12}{V_s} = \frac{100}{400}\]
\[\frac{12}{V_s} = \frac{1}{4}\]
\[V_s = 12 \times 4 = 48\text{ V}\]

1 mark for selecting option C.

Use the formula for work done:
\[E = F \times d\]

where:
- \(F = 15\text{ N}\)
- \(d = 4.0\text{ m}\)

\[E = 15 \times 4.0 = 60\text{ J}\]

1 mark for selecting option D.

A vector quantity has both magnitude and direction. Velocity has magnitude (speed) and direction, making it a vector. Mass, distance, and energy only have magnitude, so they are scalar quantities.

1 mark for selecting option C.

The correct matches are:
- **Thermistor (1)** matches description **B** (Its resistance decreases when the temperature increases).
- **Light Dependent Resistor (2)** matches description **C** (Its resistance decreases when the light intensity increases).
- **Diode (3)** matches description **A** (Allows current to flow in one direction only).

Award 1 mark for each correct match:
- Thermistor matched with description B (1 mark)
- Light Dependent Resistor (LDR) matched with description C (1 mark)
- Diode matched with description A (1 mark)

The correct matches are:
- **Solid (1)** matches description **Y** (Particles are closely packed in a regular arrangement and vibrate about fixed positions).
- **Liquid (2)** matches description **Z** (Particles are close together but can move past each other in random directions).
- **Gas (3)** matches description **X** (Particles are far apart and move rapidly and randomly in all directions).

Award 1 mark for each correct match:
- Solid matched with description Y (1 mark)
- Liquid matched with description Z (1 mark)
- Gas matched with description X (1 mark)

First, state the relationship between charge, current, and time: \(\text{charge flow} = \text{current} \times \text{time}\) (\(Q = I \times t\)). Then substitute the given values into the equation: \(Q = 0.4\text{ A} \times 30\text{ s}\). Calculating this gives \(Q = 12\text{ C}\).

1 mark for correct substitution: \(0.4 \times 30\). 1 mark for correct final value of \(12\text{ C}\) (accept \(12\)).

First, state the relationship between work done, force, and distance: \(\text{work done} = \text{force} \times \text{distance}\) (\(E = F \times d\)). Next, substitute the values: \(E = 45\text{ N} \times 6.0\text{ m}\). This gives a work done of \(270\text{ J}\).

1 mark for correct substitution: \(45 \times 6.0\). 1 mark for correct calculation: \(270\text{ J}\) (accept \(270\)).

Use the density formula: \(\text{density} = \frac{\text{mass}}{\text{volume}}\). Substitute the given numbers: \(\text{density} = \frac{1.35\text{ kg}}{0.0005\text{ m}^3}\). Doing this calculation yields \(2700\text{ kg/m}^3\).

1 mark for correct substitution: \(\frac{1.35}{0.0005}\). 1 mark for correct density of \(2700\text{ kg/m}^3\) (accept \(2700\)).

The strength of an electromagnet's magnetic field depends on the current and the coil structure. The field can be made stronger by: 1) increasing the electric current passing through the wire, and 2) adding more turns of wire to the coil (or inserting an iron core).

1 mark for each correct method listed up to a maximum of 2 marks: - Increase the current (1) - Increase the number of turns on the coil (1) - Add an iron core (1)

When the plastic rod is rubbed with the dry cloth, friction causes negative electrons to be transferred from the cloth onto the rod. Because the cloth loses these negative electrons, it is left with an overall positive charge.

1 mark for stating that electrons are transferred from the cloth to the rod. 1 mark for stating that the cloth becomes positively charged.

Use the formula for spring force: \(\text{force} = \text{spring constant} \times \text{extension}\) (\(F = k \times x\)). Substitute the values: \(F = 25\text{ N/m} \times 0.12\text{ m}\). This calculates to \(3.0\text{ N}\).

1 mark for correct substitution: \(25 \times 0.12\). 1 mark for correct force of \(3.0\text{ N}\) (accept \(3\)).

Use the transformer equation: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\). Rearranging the formula to solve for the secondary voltage \(V_s\) gives: \(V_s = V_p \times \frac{N_s}{N_p}\). Substituting the numbers: \(V_s = 12\text{ V} \times \frac{360}{120} = 12 \times 3 = 36\text{ V}\).

1 mark for correct rearrangement or substitution: \(12 \times \frac{360}{120}\) or \(\frac{12}{V_s} = \frac{120}{360}\). 1 mark for correct final answer: \(36\text{ V}\) (accept \(36\)).

Use the formula for the moment of a force: \(\text{moment} = \text{force} \times \text{perpendicular distance}\). Substitute the values: \(\text{moment} = 80\text{ N} \times 0.25\text{ m} = 20\text{ N m}\).

1 mark for correct substitution: \(80 \times 0.25\). 1 mark for correct calculation of \(20\text{ N m}\) (accept \(20\)).

To calculate the resistance, use the formula:
\(R = \frac{V}{I}\)
Substitute the given values into the formula:
\(R = \frac{4.5}{0.30}\)
\(R = 15\ \Omega\)

1 mark for correct substitution: \(4.5 = 0.30 \times R\) or rearrangement: \(R = \frac{4.5}{0.30}\)
1 mark for correct calculation: 15 (\(\Omega\))

First, calculate the extension of the spring in metres:
\(x = 18\text{ cm} - 12\text{ cm} = 6\text{ cm} = 0.06\text{ m}\)
Next, use Hooke's Law to find the spring constant (k):
\(F = k \times x\)
\(k = \frac{F}{x} = \frac{3.0}{0.06} = 50\text{ N/m}\)

1 mark for finding the extension in metres: 0.06 (m) (or 6 cm and showing an attempt to convert to metres)
1 mark for correct calculation of spring constant: 50 (N/m)

Use the formula for density:
\(\text{density} = \frac{\text{mass}}{\text{volume}}\)
Substitute the values into the equation:
\(\rho = \frac{2.8}{0.0040} = 700\text{ kg/m}^3\)

1 mark for correct substitution: \(\frac{2.8}{0.0040}\)
1 mark for correct calculation: 700 (kg/m³)

Use the formula for the moment of a force:
\(\text{moment} = \text{force} \times \text{distance}\)
Substitute the values:
\(\text{moment} = 15 \times 0.40 = 6.0\text{ N m}\)

1 mark for correct substitution: \(15 \times 0.40\)
1 mark for correct evaluation: 6 or 6.0 (N m)

Use the formula for work done:
\(\text{work done} = \text{force} \times \text{distance}\)
Substitute the values:
\(W = 2.5 \times 8.0 = 20\text{ J}\)

1 mark for correct substitution: \(2.5 \times 8.0\)
1 mark for correct calculation: 20 (J)

Use the formula relating charge, current, and time:
\(Q = I \times t\) or \(I = \frac{Q}{t}\)
Substitute the values to find the current:
\(I = \frac{0.045}{0.15} = 0.3\text{ A}\)

1 mark for correct substitution or rearrangement: \(\frac{0.045}{0.15}\)
1 mark for correct calculation: 0.3 (A)

First, convert the time from minutes to seconds: \(2\text{ minutes} = 2 \times 60\text{ seconds} = 120\text{ seconds}\). Next, use the charge equation: \(\text{charge } (Q) = \text{current } (I) \times \text{time } (t)\). Substitute the values: \(Q = 0.15\text{ A} \times 120\text{ s} = 18\text{ C}\).

1 mark for converting time to seconds (\(120\text{ s}\)). 1 mark for substitution into the equation (\(0.15 \times 120\)). 0.5 marks for the correct final answer with units (\(18\text{ C}\)).

Use Hooke's law equation: \(\text{force } (F) = \text{spring constant } (k) \times \text{extension } (x)\). Rearrange the equation to solve for extension: \(x = \frac{F}{k}\). Substitute the values: \(x = \frac{1.5\text{ N}}{25\text{ N/m}} = 0.06\text{ m}\).

1 mark for rearranging the equation (\(x = \frac{F}{k}\)). 1 mark for substitution (\(\frac{1.5}{25}\)). 0.5 marks for the correct final answer with units (\(0.06\text{ m}\)).

Use the density equation: \(\text{density} = \frac{\text{mass}}{\text{volume}}\). Substitute the values: \(\text{density} = \frac{160\text{ g}}{200\text{ cm}^3} = 0.8\text{ g/cm}^3\).

1 mark for selecting the density equation. 1 mark for substituting the values (\(\frac{160}{200}\)). 0.5 marks for the correct final answer with units (\(0.8\text{ g/cm}^3\)).

Use the work done equation: \(\text{work done } (E) = \text{force } (F) \times \text{distance } (d)\). Substitute the values: \(E = 45\text{ N} \times 4\text{ m} = 180\text{ J}\).

1 mark for selecting the correct equation. 1 mark for substituting the values (\(45 \times 4\)). 0.5 marks for the correct final answer with units (\(180\text{ J}\)).

Use the moment equation: \(\text{moment of a force} = \text{force} \times \text{distance}\). Substitute the values: \(\text{moment} = 12\text{ N} \times 0.25\text{ m} = 3\text{ N m}\).

1 mark for selecting the correct equation. 1 mark for substituting the values (\(12 \times 0.25\)). 0.5 marks for the correct final answer with units (\(3\text{ N m}\)).

Use the transformer equation: \(\frac{\text{potential difference across primary } (V_p)}{\text{potential difference across secondary } (V_s)} = \frac{\text{number of turns on primary } (N_p)}{\text{number of turns on secondary } (N_s)}\). This can be rearranged to: \(V_s = V_p \times \frac{N_s}{N_p}\). Substitute the values: \(V_s = 240\text{ V} \times \frac{50}{200} = 60\text{ V}\).

1 mark for rearranging the transformer equation. 1 mark for substituting the values (\(240 \times \frac{50}{200}\)). 0.5 marks for the correct final answer with units (\(60\text{ V}\)).

Use the electrical power equation: \(\text{power } (P) = \text{current}^2 (I^2) \times \text{resistance } (R)\). Substitute the values: \(P = (4\text{ A})^2 \times 30\ \Omega = 16 \times 30 = 480\text{ W}\).

1 mark for selecting the correct equation. 1 mark for substituting the values (\(4^2 \times 30\)). 0.5 marks for the correct final answer with units (\(480\text{ W}\)).

Use the change in thermal energy equation: \(\text{change in thermal energy } (\Delta Q) = \text{mass } (m) \times \text{specific heat capacity } (c) \times \text{change in temperature } (\Delta\theta)\). Substitute the values: \(\Delta Q = 0.5\text{ kg} \times 900\text{ J/kg}^\circ\text{C} \times 10\ ^\circ\text{C} = 4500\text{ J}\).

1 mark for selecting the correct equation. 1 mark for substituting the values (\(0.5 \times 900 \times 10\)). 0.5 marks for the correct final answer with units (\(4500\text{ J}\)).

Using the equation: charge = current * time. Charge = 0.4 * 30 = 12 C.

1 mark for recalling or using the equation charge = current * time. 1 mark for the correct substitution: 0.4 * 30. 0.5 marks for the correct final value of 12.

Using the equation: density = mass / volume. Density = 162 / 20 = 8.1 g/cm³.

1 mark for recalling or using the equation density = mass / volume. 1 mark for the correct substitution: 162 / 20. 0.5 marks for the correct final value of 8.1.

Using the equation: work done = force * distance. Work done = 400 * 8.5 = 3400 J.

1 mark for recalling or using the equation work done = force * distance. 1 mark for the correct substitution: 400 * 8.5. 0.5 marks for the correct final value of 3400.

Using the equation: force = spring constant * extension. Rearranging for extension: extension = force / spring constant. Extension = 3 / 25 = 0.12 m.

1 mark for recalling or using the equation force = spring constant * extension. 1 mark for correct rearrangement and substitution: 3 / 25. 0.5 marks for the correct final value of 0.12.

As the gas is heated, thermal energy is transferred to the kinetic energy store of the gas particles, raising the temperature (which is a measure of the average kinetic energy of the particles). The gas particles move faster and with more energy. Because the canister has a fixed volume, the faster particles collide with the inside walls of the container more frequently and with greater force. This increased rate of collision and increased force per collision results in a larger total force exerted on a given area of the walls. Since pressure is defined as force divided by area (\(P = \frac{F}{A}\)), the pressure of the gas increases.

Indicative content: - Temperature is a measure of the average kinetic energy of the particles. - Heating the gas increases the kinetic energy of the particles, causing them to move faster. - The particles collide with the walls of the canister. - Because they are moving faster, they collide with the walls more frequently (more collisions per second). - They also collide with a greater force/impact. - Pressure is force per unit area. - The greater total force on the same surface area results in increased pressure. Level 1 (1-2 marks): A simple explanation is given. For example, stating that particles move faster when heated or that they hit the walls more. Level 2 (3-4 marks): A more detailed explanation linking temperature to kinetic energy/speed, and mentioning collisions with the walls. Level 3 (5-6 marks): A detailed and fully coherent explanation linking temperature to increased kinetic energy/speed of particles, which causes more frequent collisions and harder collisions with the walls, explicitly linking this to an increase in force and therefore pressure.

In a series circuit: The current is the same through all components. The potential difference of 12 V is shared equally between the two identical lamps (each gets 6 V), so they will be dimmer. If one lamp breaks, the circuit is broken, and the other lamp goes out. In a parallel circuit: The potential difference across each lamp is the full 12 V, so they will shine at normal brightness. The total current from the battery splits between the branches. If one lamp breaks, the other lamp remains on because its branch is still a complete circuit. A parallel circuit is much better for room lighting because lamps can be switched on and off independently, they shine brighter, and if one lamp fails, the other stays on.

Indicative content: - Series: Current is the same everywhere. - Series: Potential difference is shared (each lamp gets 6 V). Lamps are dimmer. - Series: If one lamp breaks, the current stops flowing and both go out. - Parallel: Potential difference across each lamp is 12 V. Lamps are brighter. - Parallel: Current splits between branches. - Parallel: If one lamp breaks, the other remains lit. - Choice: Parallel is better for lighting because lamps can be controlled independently and do not both turn off if one fails. Level 1 (1-2 marks): Simple statements comparing series and parallel, or identifying what happens when one lamp breaks. Level 2 (3-4 marks): Explains current or potential difference differences, and correctly identifies the effect of a broken lamp in both circuits. Level 3 (5-6 marks): Comprehensive comparison covering current, potential difference, reliability when one fails, and a clear, justified recommendation for room lighting.