2022 Edexcel GCSE Physics (1PH0) 模擬試題連答案詳解

Gamma rays are high-energy, high-frequency electromagnetic waves emitted from the nucleus of an unstable atom during radioactive decay.

1 mark for the correct option (C).

Using the equation: \(\text{distance} = \text{speed} \times \text{time}\), \(\text{distance} = 6\text{ m/s} \times 20\text{ s} = 120\text{ m}\).

1 mark for the correct option (C).

A massive star expands into a red supergiant, explodes as a supernova, and the remaining core collapses into either a neutron star or a black hole, depending on its mass.

1 mark for the correct option (C).

Infrared radiation is emitted by warm objects like people. Thermal imaging cameras detect this radiation to create an image, even in total darkness.

1 mark for the correct option (A).

Efficiency is calculated using: \(\text{efficiency} = \frac{\text{useful energy transferred}}{\text{total energy supplied}}\). Substituting the values: \(\text{efficiency} = \frac{140\text{ J}}{200\text{ J}} = 0.70\).

1 mark for the correct option (C).

Use the equation: \( \text{average speed} = \frac{\text{distance}}{\text{time}} \). Substituting the given values: \( \text{average speed} = \frac{150}{30} = 5 \). The unit for speed is metres per second (m/s).

- 1 mark for the correct equation: speed = distance / time (or \( v = d/t \)).
- 1 mark for the correct numerical calculation: 5.
- 0.5 marks for the correct unit: m/s (or metres per second).

Use the equation: \( \text{efficiency} = \frac{\text{useful energy transfer}}{\text{total energy input}} \). Substituting the given values: \( \text{efficiency} = \frac{80}{200} = 0.4 \) (or 40%).

- 1 mark for the correct formula: efficiency = useful energy / total energy input.
- 1 mark for the correct calculation: 0.4.
- 0.5 marks for expressing the final answer correctly as a decimal (0.4) or as a percentage (40%).

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

- 1 mark for selecting/recalling the wave equation: wave speed = frequency \( \times \) wavelength.
- 1 mark for correct substitution and numerical calculation: 4.
- 0.5 marks for the correct unit: m/s (or metres per second).

In the electromagnetic spectrum, ultraviolet radiation has a shorter wavelength and a higher frequency than visible light, whereas infrared radiation has a longer wavelength and a lower frequency than visible light. Therefore, ultraviolet has the higher frequency and infrared has the longer wavelength.

- 1 mark for stating that ultraviolet has a higher frequency than infrared (or infrared has a lower frequency than ultraviolet).
- 1 mark for stating that infrared has a longer wavelength than ultraviolet (or ultraviolet has a shorter wavelength than infrared).
- 0.5 marks for writing a clear, coherent comparison.

Determine the number of half-lives that have elapsed: \( 800 \rightarrow 400 \) (1 half-life), \( 400 \rightarrow 200 \) (2 half-lives), \( 200 \rightarrow 100 \) (3 half-lives). The total time is 24 hours. Therefore, 3 half-lives = 24 hours. Dividing the total time by the number of half-lives gives: \( 24 \div 3 = 8 \text{ hours} \).

- 1 mark for showing that the sample activity halves 3 times to get from 800 Bq to 100 Bq.
- 1 mark for dividing the total time of 24 hours by 3.
- 0.5 marks for the correct final answer of 8 hours, including the correct unit.

The geocentric model describes the Earth as being at the centre of the Solar System, with the Sun and other planets orbiting it. The heliocentric model describes the Sun as being at the centre, with the Earth and other planets orbiting the Sun. Scientists such as Nicolaus Copernicus or Galileo Galilei provided crucial historical support for the heliocentric model.

- 1 mark for identifying that the geocentric model is Earth-centred and the heliocentric model is Sun-centred.
- 1 mark for correctly naming an associated historical scientist (e.g., Copernicus, Galileo, Kepler).
- 0.5 marks for clear scientific explanation of the differences.

Use Newton's second law: \( F = m \times a \). Rearranging for acceleration gives: \( a = \frac{F}{m} \). Substituting the given values: \( a = \frac{1.5}{0.5} = 3 \text{ m/s}^2 \).

- 1 mark for selecting/rearranging the equation: \( a = F / m \) (or \( F = m \times a \)).
- 1 mark for correct calculation: 3.
- 0.5 marks for the correct unit of acceleration: \( \text{m/s}^2 \) (or metres per second squared).

An alpha particle is identical to a helium nucleus and consists of 2 protons and 2 neutrons. Due to its larger mass and charge (+2), an alpha particle has a much higher ionizing power than a beta-minus particle, which is a fast-moving electron with a charge of -1 and very small mass.

- 1 mark for stating that an alpha particle consists of 2 protons and 2 neutrons (or is a helium-4 nucleus).
- 1 mark for stating that alpha particles have higher ionizing power than beta-minus particles.
- 0.5 marks for logical connection of structure to ionization properties.

1. Shape of orbit: Planets orbit the Sun in almost circular paths, whereas comets have highly elliptical (elongated or oval-shaped) orbits.
2. Orbital speed: A planet's speed remains relatively constant throughout its orbit, whereas a comet's speed varies greatly, speeding up as it gets closer to the gravitational pull of the Sun and slowing down as it moves away.

Award 1 mark for describing the difference in orbit shapes (circular vs elliptical).
Award 1 mark for describing the difference in speed variation (constant speed vs varying speed/faster near Sun).
Award 0.5 marks for correct scientific vocabulary (e.g. elliptical, circular).

Using the equation for distance:
\(\text{distance} = \text{speed} \times \text{time}\)

Substitute the given values into the equation:
\(\text{distance} = 6.0\text{ m/s} \times 45\text{ s}\)
\(\text{distance} = 270\text{ m}\)

Award 1 mark for recalling or substituting into the correct formula: \(d = v \times t\).
Award 1 mark for the correct numerical calculation of 270.
Award 0.5 marks for including the correct unit (m or metres).

An alpha particle is identical to a helium nucleus. It is composed of 2 protons and 2 neutrons tightly bound together. Because it has 2 protons and no orbiting electrons, it carries a relative electric charge of \(+2\).

Award 1 mark for stating that it consists of 2 protons and 2 neutrons (or is a helium nucleus).
Award 1 mark for identifying the charge as +2 (or positive).
Award 0.5 marks for identifying its overall mass number as 4.

Use the efficiency formula:
\(\text{Efficiency} = \frac{\text{Useful energy transferred}}{\text{Total energy supplied}} \times 100\%\)

Substitute the values:
\(\text{Efficiency} = \frac{75}{200} \times 100\%\)
\(\text{Efficiency} = 0.375 \times 100\% = 37.5\%\)

Award 1 mark for substituting values into the correct formula: \(75 / 200\).
Award 1 mark for evaluating the calculation to 0.375 or 37.5.
Award 0.5 marks for expressing the final answer as a percentage with the unit %.

Frequency is defined as the number of complete wave cycles or oscillations that pass a given point per second. A frequency of \(4.0\text{ Hz}\) specifically means that exactly 4 complete waves pass a fixed point in one second.

Award 1 mark for defining frequency as the number of waves/oscillations per unit of time.
Award 1 mark for specifying the quantity '4 waves' and 'per second' (or 'every second').
Award 0.5 marks for specifying 'complete' waves or 'cycles' of oscillation.

Ultraviolet radiation carries high energy which can damage cellular DNA. This damage can cause sunburn, premature skin aging, or lead to mutations resulting in skin cancer (melanoma). It can also cause eye damage. To reduce this risk, you can apply sunscreen with a high SPF rating, wear UV-blocking sunglasses, or cover up with clothing and hats when outdoors.

Award 1 mark for a correct health hazard (e.g., skin cancer, cataracts, sunburn, cellular DNA damage).
Award 1 mark for a correct preventive measure (e.g., sunscreen, wearing protective sunglasses, hats, avoiding midday sun).
Award 0.5 marks for explaining how the preventive measure reduces exposure (e.g., blocks/absorbs the UV rays).

1. The standard SI unit for distance is the metre (\(\text{m}\)).
2. Since \(1\text{ km} = 1000\text{ m}\), multiply the distance by 1000:
\(8.5 \times 1000 = 8500\text{ m}\)
3. Convert 8500 into standard form (scientific notation):
\(8500 = 8.5 \times 10^{3}\text{ m}\)

Award 1 mark for converting kilometers to meters correctly (8500).
Award 1 mark for writing the number in correct standard form (\(8.5 \times 10^3\)).
Award 0.5 marks for including the correct unit 'm' or 'metres'.

Use the equation for acceleration:
\(a = \frac{v - u}{t}\)
Where:
\(v = 15\text{ m/s}\) (final velocity),
\(u = 0\text{ m/s}\) (initial velocity),
\(t = 6.0\text{ s}\) (time taken).

Substitute the values:
\(a = \frac{15 - 0}{6.0} = 2.5\text{ m/s}^2\)

Award 1 mark for substituting values into the acceleration formula: \(\frac{15}{6.0}\).
Award 1 mark for the correct calculation: 2.5.
Award 0.5 marks for the correct unit of acceleration (\(\text{m/s}^2\) or \(\text{m/s/s}\)).

To find the number of remaining nuclei, we first determine the number of half-lives that have passed in 45 minutes:

\(\text{Number of half-lives} = \frac{45\text{ minutes}}{15\text{ minutes}} = 3\)

Next, we halve the original number of unstable nuclei 3 times:
- After 1 half-life: \(8000 \div 2 = 4000\)
- After 2 half-lives: \(4000 \div 2 = 2000\)
- After 3 half-lives: \(2000 \div 2 = 1000\)

Therefore, 1000 unstable nuclei remain.

- **1 mark**: For calculating the number of half-lives (3).
- **1 mark**: For showing a process of halving the starting value three times (e.g., 8000 -> 4000 -> 2000 -> 1000).
- **0.5 marks**: For the correct final answer (1000).

We use the wave speed equation:

\(v = f \times \lambda\)

Where:
- \(f = 5.0\text{ Hz}\)
- \(\lambda = 0.12\text{ m}\)

\(v = 5.0 \times 0.12 = 0.6\text{ m/s}\)

- **1 mark**: For selecting/recalling the correct formula: wave speed = frequency x wavelength (\(v = f \lambda\)).
- **1 mark**: For substituting the numbers correctly: \(5.0 \times 0.12\).
- **0.5 marks**: For the correct final answer of 0.6 with the correct unit (m/s).

When a main sequence star of comparable mass to the Sun runs out of hydrogen fuel in its core, the inward force of gravity becomes greater than the outward pressure from fusion, causing the core to collapse. This increases the core temperature, triggering fusion in the outer shell. The outer layers of the star expand and cool, causing the star to transition into a red giant.

- **1 mark**: For mentioning that the core collapses / fusion of hydrogen stops.
- **1 mark**: For describing that the outer layers of the star expand and cool down.
- **0.5 marks**: For naming the next stage as a 'red giant'.

Glass is more optically dense than air. As a result, when light enters the glass, its speed decreases. Because the ray hits the boundary at an angle, the change in speed causes the wave boundary to rotate, bending the light ray towards the normal line.

- **1 mark**: For stating that the speed of the light ray decreases.
- **1 mark**: For stating that the ray bends towards the normal.
- **0.5 marks**: For linking the bending of the light to the decrease in speed at the interface.

Using Newton's second law:

\(F = m \times a\)

Rearranging for acceleration:

\(a = \frac{F}{m} = \frac{36\text{ N}}{12\text{ kg}} = 3\text{ m/s}^2\)

- **1 mark**: For rearranging Newton's second law to make acceleration the subject: \(a = F / m\).
- **1 mark**: For substituting the correct values: \(36 / 12 = 3\).
- **0.5 marks**: For the correct unit of acceleration (\(\text{m/s}^2\) or \(\text{m s}^{-2}\)).

Efficiency is calculated using the formula:

\(\text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}}\)

Substituting the given values:

\(\text{Efficiency} = \frac{150\text{ J}}{250\text{ J}} = 0.6\) (or \(60\%\))

- **1 mark**: For recalling the correct efficiency formula: useful energy / total energy.
- **1 mark**: For substituting the values correctly: \(150 / 250\).
- **0.5 marks**: For the correct final efficiency value of 0.6 (or 60%).

An alpha particle is highly ionizing because it has a large mass and a charge of +2. Because of this high rate of ionization, it loses energy quickly and has a low penetration depth (stopped by a single sheet of paper). A beta-minus particle is less ionizing (moderate ionizing power) due to its smaller size and -1 charge, allowing it to penetrate further through materials (requiring a few millimeters of aluminum to stop it).

- **1 mark**: For stating that alpha particles have higher ionizing power than beta-minus particles (or vice versa).
- **1 mark**: For stating that beta-minus particles have greater penetration ability than alpha particles (or vice versa).
- **0.5 marks**: For mentioning appropriate shielding materials as context (e.g., alpha stopped by paper/skin, beta stopped by aluminium).

In transverse waves, the oscillations of the particles or field are perpendicular (at right angles) to the direction of energy transfer (or wave propagation). In longitudinal waves, the oscillations are parallel (in the same direction) to the direction of energy transfer.

- **1 mark**: For stating that transverse waves oscillate perpendicular to the direction of energy transfer.
- **1 mark**: For stating that longitudinal waves oscillate parallel to the direction of energy transfer.
- **0.5 marks**: For explicitly referencing 'energy transfer' or 'wave direction' in the comparisons.

Use the formula: weight = mass \(\times\) gravitational field strength. Weight = \(75\text{ kg} \times 1.6\text{ N/kg} = 120\text{ N}\).

1 mark for selecting the correct equation and substituting the values: \(75 \times 1.6\). 1 mark for the correct calculation: 120. 0.5 marks for the correct unit: N (or Newtons).

Use the formula: efficiency = useful energy transfer / total energy transfer. Efficiency = \(48 / 120 = 0.4\). To express this as a percentage: \(0.4 \times 100 = 40\%\).

1 mark for substituting the correct values into the efficiency equation: \(48 / 120\). 1 mark for calculating the correct numerical value: 0.4 (or 40). 0.5 marks for clearly stating the correct unit or format (% or decimal).

Use the wave speed equation: wave speed = frequency \(\times\) wavelength. Wave speed = \(4\text{ Hz} \times 0.5\text{ m} = 2\text{ m/s}\).

1 mark for recall of formula or substituting correct values: \(4 \times 0.5\). 1 mark for correct calculation: 2. 0.5 marks for correct unit: m/s (or metres per second).

Infrared is commonly used in remote controls, thermal imaging cameras, heaters, or cooking. Overexposure to infrared radiation can cause heating of skin tissues, which can lead to skin burns.

1 mark for a valid use (such as thermal imaging, remote controls, cooking, or heating). 1 mark for identifying the main danger of skin burns or heating of cells. 0.5 marks for clear explanation linking the danger to thermal effects.

Number of half-lives = \(30\text{ minutes} / 10\text{ minutes} = 3\) half-lives. After 1 half-life: \(800 / 2 = 400\). After 2 half-lives: \(400 / 2 = 200\). After 3 half-lives: \(200 / 2 = 100\) counts per minute.

1 mark for identifying that 3 half-lives have passed (\(30 / 10 = 3\)). 1 mark for halving the initial activity three times (800 to 400 to 200 to 100). 0.5 marks for writing the correct unit: counts per minute (or cpm).

In the geocentric model, the Earth is at the center of the solar system, whereas in the heliocentric model, the Sun is at the center. The gravitational force maintains the orbits of planets.

1 mark for defining the difference (Earth at center for geocentric versus Sun at center for heliocentric). 1 mark for identifying gravity or gravitational force. 0.5 marks for clear communication and organization of the two parts.

Use Newton's second law: Force = mass \(\times\) acceleration. Rearranging for acceleration gives: acceleration = Force / mass = \(3\text{ N} / 0.5\text{ kg} = 6\text{ m/s}^2\).

1 mark for rearranging the formula or substitution (\(3 / 0.5\)). 1 mark for the correct value (6). 0.5 marks for the correct unit (\(\text{m/s}^2\) or metres per second squared).

Unwanted energy transfers can be reduced by using lubrication (such as oil or grease) to reduce friction between moving parts, and by streamlining or polishing surfaces. The main type of wasted energy produced by friction is thermal (heat) energy.

1 mark for identifying lubrication or another valid friction-reduction method. 1 mark for identifying thermal or heat energy. 0.5 marks for identifying a second valid reduction method (such as using rollers, bearings, or streamlining).

To demonstrate the penetrating power:
1. **Background Radiation:** Set up the GM tube and counter. Measure the background count rate for a few minutes without any sources present. Subtract this value from subsequent readings.
2. **Testing Alpha (\(\alpha\)):** Place the alpha source near the GM tube. Note the high count rate. Place a sheet of paper between the source and the detector. The count rate drops back to background levels, showing alpha is absorbed by paper.
3. **Testing Beta (\(\beta\)):** Place the beta source near the GM tube. A sheet of paper will not significantly decrease the count rate, but placing a thin sheet of aluminium (2-5 mm) will reduce the count rate to background, showing beta is absorbed by aluminium.
4. **Testing Gamma (\(\gamma\)):** Place the gamma source near the GM tube. Paper and aluminium have little effect. Placing several centimetres of lead (or thick concrete) will significantly decrease the count rate, showing gamma is highly penetrating and only reduced by dense materials.

**Safety Precautions:**
- Handle radioactive sources only with long tongs to increase distance.
- Keep sources stored in lead-lined boxes when not in use.
- Point sources away from students and the teacher's own body at all times.
- Do not eat, drink, or smoke in the laboratory to avoid internal contamination.
- Wear safety goggles and keep the exposure time as short as possible.

**Level 1 (1–2 marks):**
- Identifies some equipment used (e.g., GM tube, paper, lead) or mentions a basic safety rule (e.g., use tongs).
- The answer lacks structure and shows limited understanding of how to systematically test the sources.

**Level 2 (3–4 marks):**
- Describes how to test at least two of the radiation types using correct absorbers (e.g., paper for alpha, aluminium for beta).
- Mentions at least one safety precaution and the need to measure background radiation.
- The explanation is mostly structured and uses some scientific terms correctly.

**Level 3 (5–6 marks):**
- Provides a detailed and clear experimental procedure for testing all three radiation types (alpha, beta, and gamma) with their correct absorbers (paper, aluminium, and lead respectively).
- Explains how to account for background radiation (measuring and subtracting it).
- Includes multiple relevant safety precautions (e.g., using tongs, keeping distance, lead storage, direction of sources).
- The explanation is well-structured, logical, and uses appropriate scientific terminology throughout.

1. **Wave Types and Properties:**
- **P-waves (Primary waves):** Longitudinal waves where particles vibrate parallel to the direction of wave travel. They are faster than S-waves and can travel through both solids and liquids.
- **S-waves (Secondary waves):** Transverse waves where particles vibrate perpendicular to the direction of wave travel. They are slower and can only travel through solid materials.

2. **Evidence for the Mantle and Outer Core:**
- As seismic waves travel through the Earth, their paths bend (refract) gradually because the density of the mantle increases with depth, causing the wave speed to change continuously.
- **S-wave shadow zone:** There is a large region on the side of the Earth opposite the earthquake's focus where no S-waves are detected. Because S-waves cannot travel through liquids, this shadow zone proves that the Earth has a liquid outer core.
- **P-wave shadow zone:** P-waves are detected on the opposite side, but there is a band (between approx. 103 and 142 degrees) where no P-waves are detected. This is because P-waves refract sharply at the boundary between the solid mantle and the liquid outer core due to a sudden decrease in speed, creating a shadow zone.

3. **Evidence for the Inner Core:**
- Weak P-waves detected within the P-wave shadow zone, along with detailed analysis of wave travel times, show that P-waves speed up again as they pass through the very centre of the Earth, providing evidence that the inner core is solid.

**Level 1 (1–2 marks):**
- Identifies at least one difference between P-waves and S-waves (e.g., longitudinal vs transverse, or different speeds).
- States that they travel through the Earth or are used to study its structure, but with limited detail or explanation of the shadow zones.

**Level 2 (3–4 marks):**
- Explains that S-waves cannot travel through liquids while P-waves can.
- Links this property to the existence of a liquid core or the S-wave shadow zone.
- Explains that wave paths bend/refract inside the Earth due to changing density/speed.

**Level 3 (5–6 marks):**
- Clearly distinguishes between P-waves (longitudinal, travel through solids and liquids) and S-waves (transverse, only travel through solids).
- Explains in detail how the S-wave shadow zone proves the outer core is liquid.
- Explains how refraction of P-waves at the mantle-core boundary creates a P-wave shadow zone, showing a distinct boundary between the solid mantle and liquid core.
- The response is highly detailed, logically structured, and uses accurate scientific terms (longitudinal, transverse, refraction, shadow zone).

**Experimental Setup and Measurements:**
1. Set up the small electric motor to lift a known mass vertically using a string wound around its axle.
2. Measure the **mass** (\(m\)) of the weight using a mass balance.
3. Connect an **ammeter** in series with the motor to measure the current (\(I\)) and a **voltmeter** in parallel with the motor to measure the potential difference (\(V\)).
4. Measure the vertical **height** (\(h\)) the weight is lifted using a metre ruler.
5. Use a **stopwatch** to measure the time (\(t\)) taken for the weight to be lifted through this height.

**Calculations:**
- **Useful energy output (gravitational potential energy gained by the mass):**
\(E_{\text{out}} = m \times g \times h\)
(where \(g = 10\text{ N/kg}\) or \(9.8\text{ N/kg}\))
- **Total electrical energy input:**
\(E_{\text{in}} = V \times I \times t\)
- **Efficiency:**
\(\text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}} = \frac{m \times g \times h}{V \times I \times t}\)
To express this as a percentage, multiply the result by 100.

**Improving Accuracy:**
- Measure the height from a clear starting point to a clear finishing point, keeping the ruler vertical and reading at eye level to avoid parallax error.
- Repeat the experiment multiple times with the same mass and height, and calculate the mean current, voltage, and time to reduce random errors.
- Minimize friction by ensuring the string is wound smoothly and does not slip on the motor spindle.

**Level 1 (1–2 marks):**
- Identifies some of the basic quantities to measure (e.g., mass, height, or time) or names some basic equipment (e.g., ruler, stopwatch).
- Gives a simple formula for efficiency or work done, but lacks details on how to get all the terms.

**Level 2 (3–4 marks):**
- Describes a workable experiment that includes measuring the mechanical aspect (mass, height) and the electrical aspect (current, voltage, time) using correct instruments.
- Outlines how to calculate either the useful energy output or the total energy input correctly.
- Suggests a basic way to improve accuracy.

**Level 3 (5–6 marks):**
- Provides a complete, logical experimental procedure detailing all required measurements (mass, height, voltage, current, time) and the correct equipment for each.
- Shows full calculations for both useful energy output (\(mgh\)) and electrical energy input (\(VIt\) or \(Pt\)), and the correct efficiency formula.
- Offers a clear, practical method to improve accuracy (such as repeating and averaging, avoiding parallax error, or ensuring no string slippage).
- Uses precise scientific vocabulary throughout.

**1. Position on the Electromagnetic Spectrum:**
- **Microwaves:** Located between radio waves and infrared waves. They have long wavelengths (typically millimetres to centimetres) and low frequencies.
- **Ultraviolet (UV):** Located between visible light and X-rays. They have much shorter wavelengths and much higher frequencies than microwaves.
- Therefore, UV waves carry significantly more energy per photon than microwaves.

**2. Effects on Human Body Tissue:**
- **Microwaves:** Penetrate deeper into body tissues. They interact primarily with water molecules, causing them to vibrate and heat up (heating from the inside).
- **Ultraviolet:** Do not penetrate deeply; they are mostly absorbed by the outer layers of the skin and eyes. Because UV has high energy, it can cause chemical changes in cells and ionize atoms/molecules.

**3. Hazards of Overexposure:**
- **Microwaves:** Overexposure can cause internal heating of body tissues, potentially leading to deep tissue burns and organ damage without the surface skin showing immediate signs.
- **Ultraviolet:** Overexposure causes sunburn (cell damage), premature aging of the skin (wrinkling), and damage to the eyes (such as cataracts). Critically, UV is ionizing or near-ionizing, meaning it can damage DNA in skin cells, leading to mutations that cause skin cancer.

**Level 1 (1–2 marks):**
- Identifies at least one difference in spectrum position (e.g., UV has higher frequency, or microwaves have longer wavelength).
- Mentions a basic effect or hazard of either wave (e.g., UV causes sunburn or microwaves heat food/water).

**Level 2 (3–4 marks):**
- Compares the positions of both waves on the EM spectrum correctly in terms of wavelength or frequency.
- Describes the basic effect/hazard of both waves (e.g., microwaves cause internal heating; UV causes skin damage/cancer).
- The response is structured but may lack depth in explaining *why* the hazards occur (e.g., connection to energy/penetration/ionization).

**Level 3 (5–6 marks):**
- Provides a detailed comparison of their positions on the spectrum (both wavelength and frequency relationships clearly stated).
- Explains the differing physical effects on tissue (microwaves penetrate and heat water internally; UV is absorbed on the skin surface and causes ionization/chemical changes).
- Explicitly connects the hazards to these physical effects (microwaves cause internal burns; UV causes DNA damage/mutation leading to skin cancer, as well as cataracts and skin aging).
- The explanation is highly structured, clear, and uses accurate physical terminology throughout.

An LDR (light-dependent resistor) has a resistance that decreases as the intensity of light falling on it increases. In bright light, its resistance is low, while in the dark, its resistance is very high.

1 mark for selecting option A.

Charging by friction involves the transfer of negatively charged electrons. Protons are bound in the nucleus and cannot move. Since the rod becomes negatively charged, it must have gained electrons from the cloth.

1 mark for selecting option A.

Absolute zero is the lowest possible temperature, defined as \(0\text{ K}\). On the Celsius scale, this temperature is equal to \(-273\ ^{\circ}\text{C}\).

1 mark for selecting option A.

Using the equation: force = spring constant \(\times\) extension. Calculating: \(25\text{ N/m} \times 0.20\text{ m} = 5.0\text{ N}\).

1 mark for selecting option A.

Using the equation for work done: work done = force \(\times\) distance moved in direction of force. Calculating: \(15\text{ N} \times 8.0\text{ m} = 120\text{ J}\).

1 mark for selecting option C.

By convention, magnetic field lines outside a permanent bar magnet point from the north-seeking pole (north pole) to the south-seeking pole (south pole).

1 mark for selecting option B.

In a parallel circuit, the potential difference across each parallel branch is the same as the potential difference of the source. Therefore, each resistor has the full potential difference of \(6.0\text{ V}\) across it.

1 mark for selecting option B.

Rubbing the rod with the cloth causes friction. This friction transfers negatively charged electrons from the dry cloth to the plastic rod. Because the rod gains electrons, it becomes negatively charged.

1 mark for stating that electrons are transferred.
1 mark for identifying that the transfer is from the cloth to the rod.
(Do not accept transfer of positive charge or protons).

First, calculate the total resistance of the series circuit:
\(R_{\text{total}} = R_1 + R_2 = 15\ \Omega + 25\ \Omega = 40\ \Omega\).
Next, use Ohm's law to find the current:
\(I = \frac{V}{R} = \frac{12\text{ V}}{40\ \Omega} = 0.3\text{ A}\).

1 mark for calculating the total resistance of the series circuit: \(40\ \Omega\).
1 mark for substituting values into Ohm's law: \(I = \frac{12}{40}\).
1 mark for the correct final answer of \(0.3\text{ A}\) (or with unit \(\text{A}\)).

The magnetic field of a solenoid can be made stronger by either increasing the electrical current flowing through the wire, increasing the number of turns (coils) of the wire, or placing a magnetically soft iron core inside the solenoid.

1 mark for any one of: increase the current, increase the voltage, increase the number of turns on the coil, add an iron core.
1 mark for a second correct method from the list.

First, calculate the density in \(\text{g/cm}^3\):
\(\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{240\text{ g}}{30\text{ cm}^3} = 8\text{ g/cm}^3\).
Then convert \(\text{g/cm}^3\) to \(\text{kg/m}^3\) by multiplying by \(1000\):
\(8 \times 1000 = 8000\text{ kg/m}^3\).
Alternatively, convert the mass and volume to SI units first:
\(m = 0.24\text{ kg}\) and \(V = 0.00003\text{ m}^3\), so \(\text{Density} = \frac{0.24}{0.00003} = 8000\text{ kg/m}^3\).

1 mark for using the density formula correctly: \(\text{density} = \frac{\text{mass}}{\text{volume}}\).
1 mark for calculating the density in intermediate units (e.g., \(8\text{ g/cm}^3\)) or showing correct unit conversion factors.
1 mark for the correct final answer: \(8000\text{ kg/m}^3\).

First, calculate the extension of the spring:
\(x = 20\text{ cm} - 12\text{ cm} = 8\text{ cm}\).
Convert the extension to meters:
\(x = 0.08\text{ m}\).
Next, use Hooke's law (\(F = k \cdot x\)) rearranged to solve for the spring constant (\(k\)):
\(k = \frac{F}{x} = \frac{6.0\text{ N}}{0.08\text{ m}} = 75\text{ N/m}\).

1 mark for calculating the extension of the spring: \(8\text{ cm}\) (or \(0.08\text{ m}\)).
1 mark for rearranging the formula to make \(k\) the subject and substituting values: \(k = \frac{6.0}{0.08}\).
1 mark for the correct final answer of \(75\text{ N/m}\). (Accept \(0.75\text{ N/cm}\) only if the unit is explicitly stated by the candidate).

Use the work done equation:
\(\text{Work done} = \text{force} \times \text{distance}\)
\(W = 150\text{ N} \times 12\text{ m} = 1800\text{ J}\) (or \(1.8\text{ kJ}\)).

1 mark for substituting the values into the work done equation: \(150 \times 12\).
1 mark for the correct evaluation: \(1800\text{ J}\) (or \(1.8\text{ kJ}\)).

Use the equation for thermal energy:
\(\Delta Q = m \cdot c \cdot \Delta \theta\)
Rearrange the equation to solve for the change in temperature (\(\Delta \theta\)):
\(\Delta \theta = \frac{\Delta Q}{m \cdot c}\)
Substitute the values:
\(\Delta \theta = \frac{3850}{0.50 \times 385} = \frac{3850}{192.5} = 20\text{ }^\circ\text{C}\).

1 mark for correct rearrangement of the formula or substitution of values into the formula: \(3850 = 0.50 \times 385 \times \Delta \theta\).
1 mark for calculating the product of mass and specific heat capacity: \(192.5\).
1 mark for the correct calculation of temperature rise: \(20\text{ }^\circ\text{C}\).

Use the electrical power formula:
\(P = I \times V\)
Substitute the given values:
\(P = 10\text{ A} \times 230\text{ V} = 2300\text{ W}\) (or \(2.3\text{ kW}\)).

1 mark for substituting values into the electrical power equation: \(10 \times 230\).
1 mark for the correct power value with units: \(2300\text{ W}\) (or \(2.3\text{ kW}\)).

Using the equation:
\( I = \frac{V}{R} \)

Substituting the values:
\( I = \frac{6.0}{15} = 0.4\ \text{A} \)

1.0 mark for correct substitution or rearrangement: \( \frac{6.0}{15} \)
1.3 marks for correct evaluation: \( 0.4\ \text{A} \) (accept 0.4)

When the rod is rubbed with the woollen cloth, friction causes electrons to move. Since electrons carry a negative charge, their transfer from the cloth to the rod results in the rod gaining a net negative charge.

1.0 mark for stating that electrons are transferred from the cloth to the rod.
1.3 marks for explaining that electrons have a negative charge, so the addition of electrons causes the rod to become negatively charged.

Using the density equation:
\( \text{density} = \frac{\text{mass}}{\text{volume}} \)

Substituting the values:
\( \text{density} = \frac{240}{300} = 0.8\ \text{g/cm}^3 \)

1.0 mark for the correct substitution: \( \frac{240}{300} \)
1.3 marks for the correct evaluation: \( 0.8 \) (accept \( 0.8\ \text{g/cm}^3 \))

Using Hooke's Law equation:
\( F = k \times x \)

Rearranging for the spring constant \( k \):
\( k = \frac{F}{x} \)

Substituting the values:
\( k = \frac{2.5}{0.05} = 50\ \text{N/m} \)

1.0 mark for the correct rearrangement or substitution: \( \frac{2.5}{0.05} \)
1.3 marks for the correct evaluation: \( 50 \) (accept \( 50\ \text{N/m} \))

Using the work done equation:
\( W = F \times d \)

Rearranging for force \( F \):
\( F = \frac{W}{d} \)

Substituting the values:
\( F = \frac{4500}{15} = 300\ \text{N} \)

1.0 mark for the correct rearrangement or substitution: \( \frac{4500}{15} \)
1.3 marks for the correct evaluation: \( 300 \) (accept \( 300\ \text{N} \))

Place the bar magnet on a sheet of paper. Place a plotting compass near one of its poles. Mark dots on the paper at both ends of the compass needle. Move the compass so that one end of the needle points to the second dot, then mark the other end. Repeat this process until you reach the other pole, and connect the dots with a smooth line.

1.0 mark for placing the compass near a pole and marking the direction of the needle/plotting dots.
1.3 marks for moving the compass from dot to dot and connecting the plotted points to draw a complete magnetic field line.

When the temperature of a gas is increased, thermal energy is transferred to the kinetic energy store of the gas molecules. Since kinetic energy is directly proportional to the square of the speed, the average speed of the gas molecules increases.

1.0 mark for stating that the average speed increases.
1.3 marks for explaining that increasing temperature increases the average kinetic energy of the gas molecules.

Using the electrical power equation:
\( P = I \times V \)

Substituting the values:
\( P = 8.0 \times 230 = 1840\ \text{W} \)

1.0 mark for the correct substitution: \( 8.0 \times 230 \)
1.3 marks for the correct evaluation: \( 1840 \) (accept \( 1840\ \text{W} \) or \( 1.84\ \text{kW} \))

First, convert the time from minutes to seconds: \(2.0\text{ minutes} = 2.0 \times 60 = 120\text{ s}\). Next, use the charge formula: \(Q = I \times t\). Substitute the known values: \(Q = 1.5\text{ A} \times 120\text{ s} = 180\text{ C}\).

1 mark for converting minutes to seconds (120 s). 1 mark for substitution of values: \(1.5 \times 120\). 1 mark for correct final answer of \(180\text{ C}\).

Use the formula: \(\text{density} = \frac{\text{mass}}{\text{volume}}\). Substitute the values: \(\text{density} = \frac{360\text{ g}}{45\text{ cm}^3} = 8\text{ g/cm}^3\).

1 mark for correct substitution into density formula: \(\frac{360}{45}\). 1 mark for the calculated value of 8. 1 mark for the correct unit: \(\text{g/cm}^3\).

Use Hooke's Law formula: \(F = k \times x\). Substitute the values: \(F = 25\text{ N/m} \times 0.12\text{ m} = 3\text{ N}\).

1 mark for substitution: \(25 \times 0.12\). 1 mark for correct final answer: \(3\text{ N}\) (or \(3.0\text{ N}\)).

Place the plotting compass near one of the poles of the bar magnet and mark the direction of the compass needle with a dot. Move the compass so that the opposite end of its needle aligns with the dot, and make a new mark. Repeat this process until you reach the other pole, then connect the dots to map the magnetic field line.

1 mark for placing compass near a pole and marking the direction of the needle (e.g., drawing a dot). 1 mark for moving the compass to the next position / repeating to trace a complete line from North to South.

Friction between the cloth and the plastic rod transfers electrons from the cloth to the rod. Since electrons carry a negative charge, this excess of electrons results in the rod gaining a net negative charge.

1 mark for stating that electrons are transferred from the cloth to the rod (or rod gains electrons). 1 mark for explaining that electrons have a negative charge, leading to a negative overall charge on the rod. Do not accept 'protons move'.

Use the work done formula: \(\text{work done} = \text{force} \times \text{distance}\). Substitute the values: \(\text{work done} = 60\text{ N} \times 4.5\text{ m} = 270\text{ J}\).

1 mark for correct substitution: \(60 \times 4.5\). 1 mark for correct answer: \(270\text{ J}\) (accept \(270\text{ N m}\)).

Calculate the total resistance first: \(R = 15 + 30 = 45\,\Omega\). Then use Ohm's law to find the current: \(I = \frac{V}{R} = \frac{9.0\text{ V}}{45\,\Omega} = 0.2\text{ A}\).

1 mark for calculating the total resistance as \(45\,\Omega\). 1 mark for correct substitution into the current equation: \(\frac{9.0}{45}\). 1 mark for correct answer: \(0.2\text{ A}\) (or \(0.20\text{ A}\)).

Use the formula: \(E = m \times L\). Substitute the given values: \(E = 0.35\text{ kg} \times 3.34 \times 10^5\text{ J/kg} = 116,900\text{ J}\) (or \(1.17 \times 10^5\text{ J}\)).

1 mark for correct substitution: \(0.35 \times 3.34 \times 10^5\). 1 mark for correct final answer: \(116,900\text{ J}\) (accept \(117,000\text{ J}\) or \(1.17 \times 10^5\text{ J}\)).

Use the equation \(V = I \times R\). Rearranging for resistance gives \(R = \frac{V}{I}\). Substituting the values: \(R = \frac{6.0}{0.4} = 15\ \Omega\).

1 mark for correct substitution: \(R = \frac{6.0}{0.4}\). 1 mark for correct value with unit: \(15\ \Omega\) (accept ohms).

Heating the gas increases its thermal energy store, which increases the average kinetic energy of the particles. This causes them to move faster. As a result, they collide with the walls of the cylinder more frequently and with greater force, increasing the pressure.

1 mark for stating that particles gain kinetic energy or move faster. 1 mark for stating that particles collide with the walls of the cylinder more frequently. 1 mark for stating that particles collide with greater force or change of momentum.

First, calculate the extension \(x\) of the spring: \(x = 18\text{ cm} - 12\text{ cm} = 6\text{ cm} = 0.06\text{ m}\). Then, use Hooke's Law \(F = k \times x\) and rearrange to find the spring constant: \(k = \frac{F}{x} = \frac{4.5\text{ N}}{0.06\text{ m}} = 75\text{ N/m}\).

1 mark for calculating the extension in metres: \(0.06\text{ m}\). 1 mark for correct calculation of spring constant: \(75\text{ N/m}\).

When the plastic rod is rubbed with the dry cloth, friction causes electrons to move. Protons cannot move as they are bound in the nucleus. Electrons are transferred from the cloth to the plastic rod. Because electrons are negatively charged, the plastic rod ends up with an excess of negative charge.

1 mark for stating that electrons are transferred from the cloth to the rod. 1 mark for identifying that gaining electrons gives the rod a net negative charge.

The force on a current-carrying wire in a magnetic field is determined by \(F = B \times I \times l\). To increase this force, you can: 1) Increase the current in the wire; 2) Use stronger magnets to increase the magnetic field strength; 3) Increase the length of the wire inside the magnetic field.

1 mark for: increase the current (or voltage/potential difference). 1 mark for: use stronger magnets (or increase magnetic field strength). 1 mark for: increase the length of the wire in the field (or increase the number of turns/coils).

The work done against gravity is equal to the gain in gravitational potential energy: \(E_p = m \times g \times h\). Substituting the values: \(E_p = 15\text{ kg} \times 10\text{ N/kg} \times 4.0\text{ m} = 600\text{ J}\).

1 mark for correct substitution: \(15 \times 10 \times 4.0\). 1 mark for correct evaluation with units: \(600\text{ J}\) (accept 600).

1. Pressure is caused by gas particles in rapid, random motion colliding with the inside walls of the container.
2. Each collision exerts a force on the wall of the container. The overall pressure is the sum of these forces divided by the total area of the walls.
3. When the gas is heated, thermal energy is transferred to the kinetic energy of the particles, causing them to move faster on average.
4. Since the volume is constant, the faster-moving particles collide with the walls more frequently.
5. They also hit the walls with greater force during each collision.
6. As a result, the total force exerted on the walls increases, which increases the pressure.

Level 1 (1-2 marks): Identifies that particles hit the walls or that higher temperature causes higher pressure. Minimal scientific terminology used.
Level 2 (3-4 marks): Explains either how pressure is produced (collisions exerting force) or how temperature increases pressure (faster particles colliding more frequently/harder).
Level 3 (5-6 marks): Gives a full and detailed explanation linking particle collisions to pressure, and explains both increased frequency and force of collisions due to higher temperature at constant volume.

1. Set up a circuit containing a thermistor, an ammeter in series, and a power supply. Connect a voltmeter in parallel across the thermistor.
2. Place the thermistor in a beaker of water (water bath) along with a thermometer.
3. Measure the initial temperature of the water, the current \( I \) from the ammeter, and the potential difference \( V \) from the voltmeter.
4. Gently heat the beaker of water using a Bunsen burner or electric heater. Measure the temperature, current, and potential difference at regular intervals (e.g., every \( 10\,^{o}\text{C} \)) up to \( 80\,^{o}\text{C} \).
5. For each temperature, calculate the resistance \( R \) of the thermistor using the equation \( R = V / I \).
6. Plot a graph of resistance against temperature to show the relationship (resistance decreases as temperature increases).

Level 1 (1-2 marks): Simple description of setup or mentions measuring temperature and resistance. Major gaps in practical detail.
Level 2 (3-4 marks): Describes a clear method including a circuit setup to measure current and potential difference, and a way to change the temperature of the thermistor.
Level 3 (5-6 marks): Detailed and logical method including correct electrical connections (ammeter series, voltmeter parallel), controlled temperature range (using a water bath up to \( 80\,^{o}\text{C} \)), specific data to record, and clear instruction on calculating resistance using \( R = V / I \).

1. The nozzle of the spray gun gives the paint droplets a specific electrostatic charge (e.g., positive).
2. Because all the paint droplets carry the same charge, they repel each other. This causes the spray to spread out into a very fine, even mist.
3. The metal bicycle frame is given an opposite charge (e.g., negative) or is earthed.
4. The charged paint droplets are electrostatically attracted to the oppositely charged metal frame.
5. This attraction is strong enough to pull paint droplets towards the back and sides of the frame (the 'shadow' zones), ensuring complete and even coverage.
6. Very little paint misses the target or floats away, which significantly reduces the amount of paint wasted.

Level 1 (1-2 marks): States that static electricity or charges attract the paint to the frame. Simple, isolated statements.
Level 2 (3-4 marks): Explains either the repulsion between droplets (leading to even spray) or the attraction between droplets and the frame (leading to coverage).
Level 3 (5-6 marks): Comprehensive explanation of both effects: like charges repelling to produce a fine mist, and opposite charges attracting to ensure full coverage (even on hidden surfaces) and minimize paint waste.

1. Set up a clamp stand with a boss and clamp. Hang the helical spring from the clamp, and set up a vertical meter ruler close to the spring.
2. Record the initial position of the bottom of the spring on the ruler (or use a pointer to read the initial length of the unstretched spring).
3. Add a known mass (e.g., \( 100\,\text{g} \), which exerts a force of approximately \( 1\,\text{N} \)) to the mass hanger at the bottom of the spring.
4. Read the new position of the bottom of the spring and calculate the extension (extension = stretched length minus original length).
5. Repeat this process by adding further masses one at a time, recording the new extension for each load.
6. Plot a graph of force (weight) against extension. If the line is straight and passes through the origin, the extension is directly proportional to the force. To check for elastic behavior, ensure that when all weights are removed, the spring returns to its original length.

Level 1 (1-2 marks): Identifies basic apparatus or describes adding weights and measuring length. Lacks systematic detail on calculating extension.
Level 2 (3-4 marks): Describes a clear experimental procedure, including measuring original and final lengths to find extension, repeating for different weights, and plotting a graph.
Level 3 (5-6 marks): Detailed, logical method including vertical alignment of ruler/pointer, calculation of extension, plotting a force-extension graph, and clearly explaining how to verify elastic behavior (linear graph and returning to original length).