2023 Edexcel IGCSE Physics 模擬試題連答案詳解

To find the average force, we use the relationship between force, change in momentum, and time:

\(F = \frac{\Delta p}{t}\)

First, define the direction. Let the initial direction of travel be positive.
Initial velocity, \(u = +30\text{ m/s}\)
Final velocity, \(v = -40\text{ m/s}\)

Calculate the change in momentum (\(\Delta p\)):
\(\Delta p = m(v - u)\)
\(\Delta p = 0.060\text{ kg} \times (-40\text{ m/s} - 30\text{ m/s})\)
\(\Delta p = 0.060 \times (-70) = -4.2\text{ kg m/s}\)

The magnitude of the change in momentum is \(4.2\text{ kg m/s}\).

Now, calculate the average force:
\(F = \frac{4.2\text{ kg m/s}}{0.012\text{ s}} = 350\text{ N}\)

[0.5 marks] For identifying and writing down the correct equation relating force and change in momentum: \(F = \frac{\Delta p}{t}\).
[0.5 marks] For calculating the correct change in momentum, taking into account the opposite directions (\(4.2\text{ kg m/s}\)).
[0.5 marks] For dividing by the contact time to get the correct force value of \(350\text{ N}\).

The relationship between refractive index (\(n\)) and critical angle (\(c\)) is given by:

\(\sin(c) = \frac{1}{n}\)

Substitute the refractive index into the equation:
\(\sin(c) = \frac{1}{1.52} \approx 0.6579\)

Now, find the angle \(c\):
\(c = \sin^{-1}(0.6579) \approx 41.1^{\circ}\)

[0.5 marks] For stating the equation linking critical angle and refractive index: \(\sin(c) = \frac{1}{n}\).
[0.5 marks] For substituting values correctly to find \(\sin(c) = 0.658\).
[0.5 marks] For calculating the correct angle: \(41.1^{\circ}\).

First, calculate the useful work output done in lifting the load:
\(\text{Work done} = \text{Force} \times \text{distance}\)
\(\text{Work done} = 400\text{ N} \times 6.0\text{ m} = 2400\text{ J}\)

Next, calculate the useful power output:
\(\text{Power output} = \frac{\text{Work done}}{\text{time}}\)
\(\text{Power output} = \frac{2400\text{ J}}{5.0\text{ s}} = 480\text{ W}\)

Now use the efficiency formula to find the total power input:
\(\text{Efficiency} = \frac{\text{Useful power output}}{\text{Total power input}}\)
\(0.75 = \frac{480\text{ W}}{\text{Power input}}\)
\(\text{Power input} = \frac{480}{0.75} = 640\text{ W}\)

[0.5 marks] For calculating the useful work output or useful power output (\(2400\text{ J}\) or \(480\text{ W}\)).
[0.5 marks] For using the efficiency formula: \(\text{Power input} = \frac{\text{Useful power}}{\text{Efficiency}}\).
[0.5 marks] For obtaining the correct input power of \(640\text{ W}\).

Determine the number of half-lives that have elapsed by halving the initial activity until reaching the final activity:

1st half-life: \(800\text{ Bq} \to 400\text{ Bq}\)
2nd half-life: \(400\text{ Bq} \to 200\text{ Bq}\)
3rd half-life: \(200\text{ Bq} \to 100\text{ Bq}\)

Since \(3\) half-lives elapsed over a total time of \(15\text{ hours}\), we calculate the duration of one half-life (\(T_{1/2}\)):
\(3 \times T_{1/2} = 15\text{ hours}\)
\(T_{1/2} = \frac{15}{3} = 5.0\text{ hours}\)

[0.5 marks] For demonstrating understanding that the activity halves each half-life.
[0.5 marks] For determining that \(3\) half-lives have elapsed (from \(800\) to \(100\)).
[0.5 marks] For dividing total time by the number of half-lives to find \(5.0\text{ hours}\).

First, calculate the equivalent total resistance (\(R_p\)) of the parallel network:
\(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}\)
\(\frac{1}{R_p} = \frac{1}{6.0\ \Omega} + \frac{1}{12.0\ \Omega}\)
\(\frac{1}{R_p} = \frac{2}{12.0\ \Omega} + \frac{1}{12.0\ \Omega} = \frac{3}{12.0\ \Omega}\)
\(R_p = \frac{12.0}{3} = 4.0\ \Omega\)

Next, use Ohm's law to find the total current (\(I\)) drawn from the supply:
\(I = \frac{V}{R_p}\)
\(I = \frac{9.0\text{ V}}{4.0\ \Omega} = 2.25\text{ A}\)

[0.5 marks] For using the parallel resistor formula correctly to find the equivalent resistance (\(4.0\ \Omega\)).
[0.5 marks] For applying Ohm's law (\(I = V/R\)) to the total system.
[0.5 marks] For calculating the correct total current of \(2.25\text{ A}\).

The orbital speed formula is:
\(v = \frac{2 \pi r}{T}\)

First, find the total orbital radius \(r\) measured from the center of the Earth:
\(r = R_{\text{Earth}} + \text{altitude}\)
\(r = 6400\text{ km} + 800\text{ km} = 7200\text{ km}\)

Now, convert the orbital period \(T\) from minutes to seconds:
\(T = 100\text{ minutes} = 100 \times 60 = 6000\text{ s}\)

Calculate the orbital speed \(v\):
\(v = \frac{2 \pi \times 7200\text{ km}}{6000\text{ s}}\)
\(v = \frac{14400 \pi}{6000} \approx 7.54\text{ km/s}\)

This rounds to approximately \(7.5\text{ km/s}\).

[0.5 marks] For adding the Earth's radius and height to find the true orbital radius (\(7200\text{ km}\)).
[0.5 marks] For converting time into seconds (\(6000\text{ s}\)) and utilizing the correct orbital speed formula.
[0.5 marks] For arriving at the correct numerical value of \(7.5\text{ km/s}\).

The total pressure at depth is given by:
\(p_{\text{total}} = p_{\text{atmospheric}} + \rho g h\)

We need to find the pressure due only to the water column (\(p_{\text{water}}\)):
\(p_{\text{water}} = p_{\text{total}} - p_{\text{atmospheric}}\)
\(p_{\text{water}} = 250\text{ kPa} - 101\text{ kPa} = 149\text{ kPa} = 149,000\text{ Pa}\)

Now use the pressure in liquids formula: \(p = \rho g h\)
\(149,000\text{ Pa} = 1025\text{ kg/m}^3 \times 10\text{ m/s}^2 \times h\)
\(149,000 = 10250 \times h\)
\(h = \frac{149,000}{10250} \approx 14.5\text{ m}\)

[0.5 marks] For calculating the pressure difference (excluding atmospheric pressure) to find \(p_{\text{water}} = 149\text{ kPa}\).
[0.5 marks] For recalling and rearranging \(p = \rho g h\) to make \(h\) the subject.
[0.5 marks] For evaluating the final depth correctly as \(14.5\text{ m}\).

When charging by friction, only electrons (which are negatively charged and located on the outer shells of atoms) can move. Since the polythene rod becomes negatively charged, it must have gained electrons from the cloth. The wool cloth, having lost negatively charged electrons, is left with a net positive charge.

[0.5 marks] For recognizing that only electrons (not protons) are transferred during charging by friction.
[0.5 marks] For identifying that electrons move from the cloth to the polythene rod to make it negative.
[0.5 marks] For concluding that the wool cloth becomes positively charged due to a loss of electrons.

At terminal velocity, the air resistance acting upwards on the parachutist balances her weight acting downwards. Because these forces are equal in magnitude and opposite in direction, the resultant force is zero. Consequently, her acceleration is zero, and she falls at a constant speed.

[1 mark] For identifying that the resultant force acting on the parachutist is zero at terminal velocity.
[0.5 marks] For stating that the weight is equal in magnitude to the air resistance.

First, find the equivalent resistance of the two parallel resistors using the parallel resistance formula:
\(\frac{1}{R_p} = \frac{1}{6.0} + \frac{1}{6.0} = \frac{2}{6.0}\)
\(R_p = 3.0\ \Omega\)

Next, add this combination resistance to the third resistor which is in series:
\(R_{\text{total}} = R_p + R = 3.0\ \Omega + 6.0\ \Omega = 9.0\ \Omega\)

[1 mark] For calculating the correct equivalent resistance of the parallel branch (\(3.0\ \Omega\)).
[0.5 marks] For adding the series resistor to find the correct total resistance (\(9.0\ \Omega\)).

First, calculate how many half-lives have elapsed in 60 hours:
\(\text{Number of half-lives} = \frac{60\text{ hours}}{15\text{ hours}} = 4\)

Next, halve the activity 4 times:
- After 1 half-life: \(320 \div 2 = 160\text{ Bq}\)
- After 2 half-lives: \(160 \div 2 = 80\text{ Bq}\)
- After 3 half-lives: \(80 \div 2 = 40\text{ Bq}\)
- After 4 half-lives: \(40 \div 2 = 20\text{ Bq}\)

[1 mark] For correctly determining that 4 half-lives have elapsed.
[0.5 marks] For correctly halving the activity 4 times to find the final activity of \(20\text{ Bq}\).

First, convert the speed to meters per second:
\(v = 7.0\text{ km/s} = 7000\text{ m/s}\)

Use the orbital speed formula:
\(v = \frac{2\pi r}{T}\)

Rearrange the formula to solve for time period \(T\):
\(T = \frac{2\pi r}{v}\)

Substitute the given values:
\(T = \frac{2 \times \pi \times (8.0 \times 10^6\text{ m})}{7000\text{ m/s}} \approx 7180.7\text{ s}\)

Rounding to 2 significant figures gives \(7.2 \times 10^3\text{ s}\).

[1 mark] For converting speed to \(7000\text{ m/s}\) and correctly substituting values into the rearranged orbital speed equation.
[0.5 marks] For calculating the final time period to 2 significant figures.

First, define the initial direction of travel as positive. This means:
- Initial velocity, \(u = +5.0\text{ m/s}\)
- Final velocity, \(v = -2.5\text{ m/s}\) (since it rebounds in the opposite direction)

Calculate the change in momentum of the toy car:
\(\Delta p = m(v - u)\)
\(\Delta p = 0.60\text{ kg} \times (-2.5\text{ m/s} - 5.0\text{ m/s})\)
\(\Delta p = 0.60 \times (-7.5) = -4.5\text{ kg m/s}\)

Now, use the relationship between force, change in momentum, and time:
\(Force = \frac{\text{change in momentum}}{\text{time}}\)
\(F = \frac{4.5\text{ kg m/s}}{0.15\text{ s}} = 30\text{ N}\)

- 1 mark for calculating the change in momentum (4.5 kg m/s) with a correct formula.
- 1 mark for dividing change in momentum by the time interval (0.15 s).
- 1.2 marks for the correct final force of 30 N with correct units.

First, calculate the electrical power of the heater using:
\(P = \frac{V^2}{R}\)
\(P = \frac{24^2}{12} = \frac{576}{12} = 48\text{ W}\)

Alternatively, find the current first:
\(I = \frac{V}{R} = \frac{24}{12} = 2.0\text{ A}\)
Then find power:
\(P = I \times V = 2.0 \times 24 = 48\text{ W}\)

Convert time to seconds:
\(t = 5.0\text{ minutes} = 5.0 \times 60\text{ s} = 300\text{ s}\)

Calculate the total energy transferred:
\(E = P \times t\)
\(E = 48\text{ W} \times 300\text{ s} = 14400\text{ J}\) (or \(14.4\text{ kJ}\))

- 1 mark for finding power (48 W) or current (2.0 A).
- 1 mark for converting time to seconds (300 s).
- 1.2 marks for the correct final energy value with units (14400 J or 14.4 kJ).

First, determine the total orbital radius, \(r\), from the center of the Earth:
\(r = \text{Earth radius} + \text{altitude} = 6400\text{ km} + 650\text{ km} = 7050\text{ km}\)

Convert the time period, \(T\), into seconds:
\(T = 98\text{ minutes} \times 60\text{ s/minute} = 5880\text{ s}\)

Use the orbital speed formula:
\(v = \frac{2 \pi r}{T}\)
\(v = \frac{2 \times \pi \times 7050}{5880}\)
\(v \approx 7.5333\text{ km/s}\)

Rounding to 3 significant figures gives \(7.53\text{ km/s}\).

- 1 mark for calculating correct orbital radius (7050 km).
- 1 mark for converting time to seconds (5880 s) and substituting into the orbital speed formula.
- 1.2 marks for correct final answer of 7.53 km/s (accept answers in the range 7.5 to 7.54).

Calculate the pressure exerted by the seawater column using:
\(p_{\text{water}} = \rho \times g \times h\)
\(p_{\text{water}} = 1025\text{ kg/m}^3 \times 10\text{ N/kg} \times 25\text{ m} = 256250\text{ Pa}\)

Calculate the total pressure by adding the atmospheric pressure acting on the surface of the water:
\(p_{\text{total}} = p_{\text{water}} + p_{\text{atmospheric}}\)
\(p_{\text{total}} = 256250\text{ Pa} + 101000\text{ Pa} = 357250\text{ Pa}\) (or \(3.57 \times 10^5\text{ Pa}\))

- 1 mark for correct substitution into the hydrostatic pressure formula to find water pressure (256250 Pa).
- 1 mark for adding the atmospheric pressure to the calculated water pressure.
- 1.2 marks for the correct final total pressure of 357250 Pa (or equivalent 3.57 * 10^5 Pa).

First, calculate the useful work output (gravitational potential energy gained by the crate):
\(\text{Work done} = m \times g \times h = 45\text{ kg} \times 10\text{ N/kg} \times 12\text{ m} = 5400\text{ J}\)

Next, calculate the total electrical energy input over the 8.0 seconds:
\(\text{Energy input} = \text{Power} \times t = 900\text{ W} \times 8.0\text{ s} = 7200\text{ J}\)

Now calculate the efficiency using the energy values:
\(\text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}} = \frac{5400\text{ J}}{7200\text{ J}} = 0.75\) (or \(75\%\))

Alternatively, calculate the useful power output:
\(\text{Power output} = \frac{\text{Work done}}{\text{time}} = \frac{5400\text{ J}}{8.0\text{ s}} = 675\text{ W}\)
\(\text{Efficiency} = \frac{\text{Power output}}{\text{Power input}} = \frac{675\text{ W}}{900\text{ W}} = 0.75\)

- 1 mark for calculating useful work done (5400 J) or useful power output (675 W).
- 1 mark for calculating total electrical energy input (7200 J) or setting up the correct efficiency ratio.
- 1.2 marks for the correct final efficiency of 0.75 or 75%.

The process has two parts:
1. Melting the ice at \(0\ ^\circ\text{C}\):
\(Q_1 = m \times L_f = 0.20\text{ kg} \times 3.3 \times 10^5\text{ J/kg} = 66000\text{ J}\)

2. Warming the water from \(0\ ^\circ\text{C}\) to \(15\ ^\circ\text{C}\):
\(Q_2 = m \times c \times \Delta\theta\)
\(Q_2 = 0.20\text{ kg} \times 4200\text{ J/(kg }^\circ\text{C)} \times (15 - 0)\ ^\circ\text{C}\)
\(Q_2 = 0.20 \times 4200 \times 15 = 12600\text{ J}\)

3. Calculate the total energy required:
\(Q_{\text{total}} = Q_1 + Q_2\)
\(Q_{\text{total}} = 66000\text{ J} + 12600\text{ J} = 78600\text{ J}\) (or \(78.6\text{ kJ}\))

- 1 mark for calculating the energy required to melt the ice (66000 J).
- 1 mark for calculating the energy required to heat the water (12600 J).
- 1.2 marks for the correct total energy with unit (78600 J or 78.6 kJ).

First, calculate the frequency of the waves:
\(f = \frac{\text{number of waves}}{\text{time}}\)
\(f = \frac{18}{45\text{ s}} = 0.40\text{ Hz}\)

The wavelength \(\lambda\) is the distance between adjacent wave crests:
\(\lambda = 2.5\text{ m}\)

Calculate the wave speed using the wave equation:
\(v = f \times \lambda\)
\(v = 0.40\text{ Hz} \times 2.5\text{ m} = 1.0\text{ m/s}\)

- 1 mark for calculating the correct frequency (0.40 Hz) or wave period (2.5 s).
- 1 mark for using the correct wave equation relationship (speed = frequency * wavelength).
- 1.2 marks for correct speed with unit (1.0 m/s).

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

Rearrange to make \(\sin r\) the subject:
\(\sin r = \frac{\sin i}{n}\)

Substitute the given values:
\(\sin r = \frac{\sin(48^\circ)}{1.52}\)
\(\sin r \approx \frac{0.7431}{1.52} \approx 0.4889\)

Calculate \(r\):
\(r = \sin^{-1}(0.4889) \approx 29.3^\circ\)

To the nearest degree, \(r = 29^\circ\).

- 1 mark for using/stating the correct formula for refractive index (Snell's law).
- 1 mark for rearranging and calculating the correct sine of the angle of refraction (approx. 0.49).
- 1.2 marks for correct final angle (29 degrees; accept 29 to 29.3 degrees).

First, calculate the work done using the formula:

\(\text{work done} = \text{force} \times \text{distance}\)

\(\text{Work done} = 35\text{ N} \times 18\text{ m} = 630\text{ J}\)

Next, calculate the average power using the formula:

\(\text{power} = \frac{\text{work done}}{\text{time taken}}\)

\(\text{Power} = \frac{630\text{ J}}{15\text{ s}} = 42\text{ W}\)

- **M1**: Correct calculation of work done: \(35 \times 18 = 630\text{ (J)}\) (1.0 mark)
- **M2**: Correct recall and substitution into power equation: \(\text{power} = \frac{\text{work done}}{\text{time}}\) leading to \(\frac{630}{15}\) (1.0 mark)
- **A1**: Correct final value of \(42\) with correct unit \(\text{W}\) or \(\text{J/s}\) (1.2 marks)
- *Note: Allow full marks for correct answer with unit shown without working. Deduct 0.2 marks if the correct value of 42 is given but unit is missing or incorrect.*

To avoid parallax error, the student must position their eye level horizontally in line with the scale of the measuring cylinder and read from the bottom of the water meniscus.

First, find the volume of the rock:
\(V = 68.5\text{ cm}^{3} - 45.0\text{ cm}^{3} = 23.5\text{ cm}^{3}\)

Next, use the density formula:
\(\text{Density} = \frac{\text{mass}}{\text{volume}}\)
\(\text{Density} = \frac{63.2\text{ g}}{23.5\text{ cm}^{3}} \approx 2.69\text{ g/cm}^{3}\)

Part (a):
- Position eye level horizontally in line with the scale to avoid parallax error (1 mark)
- Read the measurement at the bottom of the meniscus (1 mark)

Part (b):
- Correct calculation of the rock's volume: 23.5 cm\(^{3}\) (1 mark)
- Correct calculation of density: 2.69 (0.5 marks)
- Correct unit: g/cm\(^{3}\) (or g cm\(^{-3}\)) (1 mark)

To accurately mark the light rays, the student should place at least two small pencil dots exactly in the center of the incident beam, and another two dots in the center of the emergent beam. After removing the block, they can use a ruler to connect these dots to draw the complete rays.

To find the refractive index, \(n\):
\(n = \frac{\sin(i)}{\sin(r)}\)
\(n = \frac{\sin(40^{\circ})}{\sin(25^{\circ})} \approx \frac{0.6428}{0.4226} \approx 1.52\)

Part (a):
- Place at least two spaced-out pencil dots along each ray (1 mark)
- Make sure dots are aligned in the center of the light beam, and use a ruler to join them (1 mark)

Part (b):
- State the correct formula: \(n = \frac{\sin(i)}{\sin(r)}\) (0.5 marks)
- Substitute values correctly: \(\frac{\sin(40^{\circ})}{\sin(25^{\circ})}\) (1 mark)
- Calculate final answer: 1.52 (accept answers from 1.5 to 1.52) (1 mark)

To ensure the ruler is vertical, the student can use a plumb line suspended parallel and near the ruler, or position a set square on the flat laboratory bench surface held against the ruler.

First, calculate the extension of the spring in meters:
\(x = 22.2\text{ cm} - 15.0\text{ cm} = 7.2\text{ cm}\)
\(x = 0.072\text{ m}\)

Using Hooke's Law:
\(F = k x\)
\(k = \frac{F}{x} = \frac{4.5\text{ N}}{0.072\text{ m}} = 62.5\text{ N/m}\)

Part (a):
- Use a plumb line suspended near the ruler OR a set square resting on the horizontal bench (1 mark)

Part (b):
- Calculate extension as 7.2 cm (1 mark)
- Convert extension to 0.072 m (0.5 marks)
- State or use correct formula: \(k = \frac{F}{x}\) (1 mark)
- Calculate final value: 62.5 (1 mark)

For safety, the student must keep all electrical connections, wires, and power supplies dry and away from the beaker of water to avoid potential short circuits or electrical shocks. Alternatively, the student should use a heatproof mat to protect the lab bench from hot water spills.

To calculate the resistance of the thermistor at 20°C:
\(R = \frac{V}{I} = \frac{4.5\text{ V}}{0.015\text{ A}} = 300\ \Omega\)

For an NTC thermistor, the resistance decreases as the temperature increases. Therefore, at 80°C, the resistance will be lower.

Part (a):
- Suitable safety precaution (e.g. keep water away from electrical connections, wear goggles, use a heatproof mat, do not touch hot beaker) (1 mark)

Part (b):
- Use of \(R = V/I\) formula (0.5 marks)
- Correct value of resistance: 300 (1 mark)
- Correct unit: \(\Omega\) or ohms (1 mark)
- Correctly stating that the resistance decreases at higher temperatures (1 mark)

To minimize heat loss, the outer surface of the aluminium block should be wrapped with an insulating material, such as bubble wrap, cotton wool, or foam. To improve thermal contact, a small volume of oil should be poured into the thermometer and heater holes to fill any insulating air gaps.

First, calculate the change in temperature:
\(\Delta\theta = 22.9\text{°C} - 18.5\text{°C} = 4.4\text{°C}\)

Using the specific heat capacity formula:
\(Q = m c \Delta\theta\)
\(c = \frac{Q}{m \Delta\theta}\)
\(c = \frac{4800\text{ J}}{1.2\text{ kg} \times 4.4\text{°C}} \approx 909\text{ J/(kg °C)}\)

Part (a):
- Wrap the block in insulating material to reduce heat loss to the environment (1 mark)
- Put a drop of oil in the holes to improve thermal contact between the heater/thermometer and the block (1 mark)

Part (b):
- Calculate the temperature change as 4.4°C (0.5 marks)
- Correct formula rearrangement and substitution: \(4800 / (1.2 \times 4.4)\) (1 mark)
- Final value: 909 (accept 910) (0.5 marks)
- Correct unit: \(\text{J/(kg °C)}\) or \(\text{J/kg K}\) (0.5 marks)

When a car undergoes a collision, it must change its momentum from its initial value to zero, represented by \(\Delta p = m \Delta v\). The force experienced by the passengers is given by the rate of change of momentum, represented by \(F = \frac{\Delta p}{t}\). By incorporating a crumple zone, the front structure of the car undergoes plastic deformation, which increases the time interval \(t\) over which this change in momentum occurs. Because \(t\) is larger, the resulting force \(F\) on the passengers is significantly smaller, reducing the severity of injuries.

1 mark: State that the crumple zone increases the time \(t\) taken for the car to come to a stop. 1 mark: State that the change in momentum \(\Delta p\) of the passengers is constant for the collision. 1 mark: Reference the relationship \(F = \frac{\Delta p}{t}\) or state that force is the rate of change of momentum. 1 mark: Conclude that a larger time results in a smaller average force acting on the passengers, decreasing injury risk.

For a thickness gauge to function, the radiation must pass through the material and be partially absorbed. Alpha particles are highly ionizing and would be completely absorbed by even thin sheets of paper, making it impossible to detect any variation in thickness. Beta particles have medium penetrating power and can pass through paper with some absorption, meaning changes in paper thickness will result in measurable changes in the count rate. A source with a short half-life of 2 hours would decay very quickly, causing the count rate to drop rapidly and requiring continuous recalibration. A half-life of 30 years ensures a stable and nearly constant activity over the lifetime of the equipment.

1 mark: Explain why alpha is unsuitable because it is completely absorbed by the paper, meaning no count rate would be detected. 1 mark: Explain why beta is suitable because it is partially absorbed/penetrates paper, allowing thickness variations to change the detected count rate. 1 mark: Explain why a short half-life of 2 hours is unsuitable because the activity drops too quickly, requiring frequent recalibration. 1 mark: Explain why a long half-life of 30 years is suitable because the activity remains relatively constant over a long period, ensuring stable measurements.

When the coil is rotated, its sides cut across the magnetic field lines. This changes the magnetic flux linkage through the coil, which induces an electromotive force (voltage) across the ends of the coil according to Faraday's law. Because the coil is connected to a complete external circuit via slip rings and brushes, this induced voltage causes an induced current to flow. As the coil rotates through \(180^{\circ}\) (half a turn), the direction in which the wire moves relative to the magnetic field reverses. This reverses the direction of the induced voltage and current, resulting in a continuous alternating current (AC).

1 mark: State that the rotating coil cuts magnetic field lines (or experiences a changing magnetic flux). 1 mark: State that this induces an electromotive force or voltage. 1 mark: Explain that because there is a complete circuit, an induced current flows. 1 mark: Explain that the direction of movement of the coil relative to the field reverses every half turn, which reverses the direction of the induced current (producing alternating current).

Once a main sequence star of solar mass runs out of hydrogen in its core, nuclear fusion of hydrogen stops. The core contracts, causing the outer layers of the star to expand and cool, turning the star into a red giant. Helium fusion then begins in the core. Eventually, helium fusion stops and the star becomes unstable. It ejects its outer layers of gas and dust into space as a planetary nebula. The hot, dense core remains behind, cooling over time to become a white dwarf.

1 mark: State that when hydrogen runs out, the star expands and cools to become a red giant. 1 mark: Mention that helium fusion occurs in the core of the red giant. 1 mark: State that the outer layers are eventually ejected into space as a planetary nebula. 1 mark: State that the remaining hot, dense core is left behind as a white dwarf.

The total initial momentum is \(0\text{ kg m/s}\) because both the cannon and the shell are at rest. According to the principle of conservation of momentum, the total final momentum must also be \(0\text{ kg m/s}\). Therefore: \(0 = (m_{\text{cannon}} \times v_{\text{cannon}}) + (m_{\text{shell}} \times v_{\text{shell}})\). Substituting the given values: \(0 = (1.5 \times v_{\text{cannon}}) + (0.050 \times 12)\). This simplifies to: \(1.5 \times v_{\text{cannon}} = -0.60\), which yields \(v_{\text{cannon}} = -0.40\text{ m/s}\). The negative sign indicates that the direction of the recoil is opposite to the direction of the shell (to the left). Thus, the recoil velocity of the cannon is \(0.40\text{ m/s}\) to the left.

1 mark for calculating the final momentum of the shell as \(0.60\text{ kg m/s}\) or setting up the conservation of momentum equation. 1 mark for calculating the final velocity of the cannon as \(0.40\text{ m/s}\) and identifying the direction as to the left.

To find the recessional velocity of the galaxy, use the Doppler shift equation: \(\frac{\lambda - \lambda_0}{\lambda_0} = \frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}\). First, calculate the change in wavelength: \(\Delta \lambda = 672.7\text{ nm} - 656.3\text{ nm} = 16.4\text{ nm}\). Next, substitute the values into the equation: \(\frac{16.4}{656.3} = \frac{v}{3.0 \times 10^8}\). Rearranging to solve for the recessional velocity \(v\): \(v = \frac{16.4}{656.3} \times 3.0 \times 10^8 \approx 0.024987 \times 3.0 \times 10^8 = 7.5 \times 10^6\text{ m/s}\).

1 mark for calculating the change in wavelength \(\Delta \lambda = 16.4\text{ nm}\). 1 mark for substituting values into the Doppler equation and calculating the velocity correctly as \(7.5 \times 10^6\text{ m/s}\).

For a transformer that is \(100\%\) efficient, the input electrical power equals the output electrical power: \(V_p \times I_p = V_s \times I_s\). According to the transformer equation, we also know that \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\). Combining these relationships gives: \(\frac{I_p}{I_s} = \frac{N_s}{N_p}\). Substituting the known quantities into this formula: \(\frac{I_p}{0.15} = \frac{3000}{150}\). This simplifies to: \(\frac{I_p}{0.15} = 20\). Therefore, the primary current is: \(I_p = 20 \times 0.15 = 3.0\text{ A}\).

1 mark for stating or applying the inverse relationship between turns ratio and current ratio, or calculating the secondary voltage as \(240\text{ V}\). 1 mark for the final primary current calculation of \(3.0\text{ A}\).

An alpha particle is identical to a helium nucleus, represented as \(^4_2\text{He}\). During an alpha decay, the parent nucleus loses 2 protons and 2 neutrons. Therefore, the mass number (nucleon number) decreases by 4, and the atomic number (proton number) decreases by 2. New mass number: \(226 - 4 = 222\). New atomic number: \(88 - 2 = 86\). The resulting Radon nucleus is represented as \(^{222}_{86}\text{Rn}\).

1 mark for recalling the mass and atomic numbers of an alpha particle. 1 mark for correctly determining the mass number as 222 and the atomic number as 86.

First, calculate the total thermal energy supplied by the heater: \(E = P \times t = 45\text{ W} \times 240\text{ s} = 10800\text{ J}\). Next, use the specific latent heat formula for a state change: \(Q = m \times L\). Rearrange the equation to solve for the mass: \(m = \frac{Q}{L}\). Substituting the values: \(m = \frac{10800}{3.3 \times 10^5} \approx 0.0327\text{ kg}\). When rounded to two significant figures, this equals \(0.033\text{ kg}\).

1 mark for calculating the thermal energy transferred as \(10800\text{ J}\). 1 mark for substituting values into \(Q = mL\) and finding the mass as \(0.033\text{ kg}\).

First, convert all units to SI units:
- Mass, \(m = 45\text{ g} = 0.045\text{ kg}\)
- Time, \(\Delta t = 1.2\text{ ms} = 0.0012\text{ s}\)

Next, calculate the change in momentum (\(\Delta p\)) of the golf ball:
\(\Delta p = m(v - u) = 0.045\text{ kg} \times (64\text{ m/s} - 0\text{ m/s}) = 2.88\text{ kg m/s}\)

Finally, use the formula relating force and change in momentum:
\(F = \frac{\Delta p}{\Delta t}\)
\(F = \frac{2.88\text{ kg m/s}}{0.0012\text{ s}} = 2400\text{ N}\)

- 1 mark for converting units: \(45\text{ g} = 0.045\text{ kg}\) and \(1.2\text{ ms} = 0.0012\text{ s}\).
- 1 mark for recalling and substituting into \(F = \frac{mv - mu}{t}\) (or finding change in momentum as \(2.88\text{ kg m/s}\)).
- 1 mark for correct calculation of force (2400).
- 0.5 mark for the correct unit (N).

First, convert the time from minutes to seconds:
\(t = 2.5\text{ minutes} = 2.5 \times 60 = 150\text{ s}\)

Next, recall the formula for electrical energy transferred:
\(E = I \times V \times t\)

Substitute the given values into the formula:
\(E = 4.5\text{ A} \times 230\text{ V} \times 150\text{ s} = 155250\text{ J}\)

Rounding to 2 significant figures gives \(160000\text{ J}\) (or \(160\text{ kJ}\)).

- 1 mark for converting time to seconds (150 s).
- 1 mark for recalling and substituting into \(E = I V t\).
- 1 mark for calculating energy correctly (accept 155250 J, 160000 J, 155 kJ, or 160 kJ).
- 0.5 mark for correct unit (J or kJ).

First, calculate the energy (\(Q\)) required to boil away 0.35 kg of water using the specific latent heat formula:
\(Q = m \times L\)
\(Q = 0.35\text{ kg} \times 2.26 \times 10^6\text{ J/kg} = 791000\text{ J}\)

Next, use the relationship between energy, power, and time:
\(E = P \times t \implies t = \frac{E}{P}\)

Substitute the calculated energy and the given power into the formula:
\(t = \frac{791000\text{ J}}{2200\text{ W}} \approx 359.5\text{ s}\)

Rounding to 2 significant figures gives \(360\text{ s}\).

- 1 mark for recalling and substituting into \(Q = mL\).
- 1 mark for calculating energy correctly (791000 J).
- 1 mark for rearranging \(P = E/t\) and calculating time (accept 359.5 s, 360 s).
- 0.5 mark for correct unit (s).

First, calculate the pressure difference due to depth (hydrostatic pressure) using the formula:
\(p = h \times \rho \times g\)
\(p = 120\text{ m} \times 1030\text{ kg/m}^3 \times 10\text{ N/kg} = 1236000\text{ Pa}\)

Next, convert this pressure to kilopascals (kPa):
\(1236000\text{ Pa} = 1236\text{ kPa}\)

Finally, add the atmospheric pressure at the surface to find the total pressure:
\(p_{\text{total}} = p_{\text{hydrostatic}} + p_{\text{atmospheric}}\)
\(p_{\text{total}} = 1236\text{ kPa} + 101\text{ kPa} = 1337\text{ kPa}\)

Rounding to 3 significant figures gives \(1340\text{ kPa}\) (or \(1.34 \times 10^6\text{ Pa}\)).

- 1 mark for recalling and substituting into \(p = h \rho g\) to get \(1.236 \times 10^6\text{ Pa}\).
- 1 mark for converting pressure values to consistent units (either Pa or kPa).
- 1 mark for adding atmospheric pressure to find the total pressure (1337 kPa).
- 0.5 mark for the correct unit (kPa or Pa, matching the calculated value).

First, calculate the total orbital radius (\(r\)) from the center of the Earth:
\(r = R_{\text{Earth}} + \text{altitude} = 6400\text{ km} + 35800\text{ km} = 42200\text{ km}\)

Next, convert the time period (\(T\)) from hours to seconds:
\(T = 24\text{ hours} = 24 \times 3600\text{ s} = 86400\text{ s}\)

Now, use the orbital speed formula:
\(v = \frac{2 \pi r}{T}\)
\(v = \frac{2 \times \pi \times 42200\text{ km}}{86400\text{ s}} \approx 3.0688\text{ km/s}\)

Rounding to 3 significant figures gives \(3.07\text{ km/s}\).

- 1 mark for calculating the correct orbital radius (42200 km).
- 1 mark for converting the orbital period to seconds (86400 s).
- 1 mark for recalling and substituting into \(v = \frac{2\pi r}{T}\).
- 0.5 mark for the correct unit (km/s).

First, calculate the useful work done in lifting the crate (which is equal to the gravitational potential energy gained):
\(\text{Work done} = m \times g \times h = 85\text{ kg} \times 10\text{ N/kg} \times 12\text{ m} = 10200\text{ J}\)

Next, calculate the useful power output of the winch:
\(P_{\text{out}} = \frac{\text{Work done}}{t} = \frac{10200\text{ J}}{15\text{ s}} = 680\text{ W}\)

Using the efficiency formula:
\(\text{efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}}\)
\(0.72 = \frac{680\text{ W}}{P_{\text{in}}}\)
\(P_{\text{in}} = \frac{680}{0.72} \approx 944.4\text{ W}\)

Rounding to 2 significant figures gives \(940\text{ W}\).

- 1 mark for calculating the useful work done / GPE gained (10200 J).
- 1 mark for calculating the useful power output (680 W).
- 1 mark for using the efficiency equation to calculate total power input (944 W or 940 W).
- 0.5 mark for the correct unit (W or kW).

Part (a):
As volume increases by 10 \(\text{cm}^3\), mass increases by 8 \(\text{g}\). Extrapolating back to 0 \(\text{cm}^3\) volume: mass = 43 \(\text{g}\) - 8 \(\text{g}\) = 35 \(\text{g}\). Therefore, the mass of the empty cylinder is 35 \(\text{g}\).

Part (b):
On a graph of total mass (y-axis) against volume (x-axis), the gradient represents the density of the liquid because density = mass / volume, and the gradient equals the change in mass divided by the change in volume.
Using two points from the table, (10, 43) and (50, 75):
Gradient = \(\frac{75 - 43}{50 - 10} = \frac{32}{40} = 0.8\text{ g/cm}^3\).

Part (a) [1.5 marks total]:
- Correct method of extrapolation or finding mass change per 10 \(\text{cm}^3\) (e.g., 8 \(\text{g}\)) [1 mark]
- Correct mass of empty cylinder: 35 \(\text{g}\) [0.5 marks]

Part (b) [4 marks total]:
- Stating that density is the gradient of the graph [1 mark]
- Correct substitution of values into gradient formula: \(\frac{75 - 43}{50 - 10}\) [1 mark]
- Calculation of gradient as 0.8 [1 mark]
- Correct unit: \(\text{g/cm}^3\) (or \(\text{g cm}^{-3}\)) [1 mark]

Part (a):
- The independent variable is the applied force (or weight) because it is the variable changed by the experimenter.
- The dependent variable is the length of the spring (or extension) because it is the variable measured for each change.

Part (b):
- First, calculate the extension (length - unstretched length) for each force:
- 1.0 N: 14.3 - 12.5 = 1.8 cm
- 2.0 N: 16.1 - 12.5 = 3.6 cm
- 3.0 N: 17.9 - 12.5 = 5.4 cm
- 4.0 N: 19.7 - 12.5 = 7.2 cm
- 5.0 N: 21.8 - 12.5 = 9.3 cm
- From 0 to 4.0 N, the extension increases by exactly 1.8 cm for every 1.0 N of force added, showing a linear relationship (direct proportionality).
- At 5.0 N, the extension increases to 9.3 cm (an increase of 2.1 cm from the previous step instead of 1.8 cm).
- Since the extension is no longer directly proportional to the force applied at 5.0 N, the spring has exceeded its limit of proportionality.

Part (a) [1.5 marks total]:
- Correctly identifying the independent variable as applied force / load [0.75 marks]
- Correctly identifying the dependent variable as length / extension [0.75 marks]

Part (b) [4 marks total]:
- Calculating extensions correctly for at least three loads [1 mark]
- Stating that extension increases by 1.8 cm per Newton up to 4.0 N [1 mark]
- Pointing out that at 5.0 N, the extension increase is 2.1 cm (or total extension is 9.3 cm, which deviates from the linear trend) [1 mark]
- Concluding that the limit of proportionality has been exceeded because force and extension are no longer directly proportional [1 mark]

Part (a):
1. Plotting a graph allows anomalous results (outliers) to be easily identified and ignored when drawing the line of best fit.
2. Drawing a line of best fit averages out random errors, providing a more accurate overall value for the refractive index than a single calculation.

Part (b):
- According to Snell's Law, \(n = \frac{\sin(i)}{\sin(r)}\).
- On a graph of \(\sin(i)\) on the y-axis against \(\sin(r)\) on the x-axis, the gradient represents the refractive index \(n\).
- Gradient = \(n = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0.60 - 0.30}{0.40 - 0.20}\)
- \(n = \frac{0.30}{0.20} = 1.5\) (or 1.50).

Part (a) [2 marks total]:
- Identifying that a graph allows anomalous points to be easily spotted/identified [1 mark]
- Stating that a line of best fit averages out random errors / improves accuracy [1 mark]

Part (b) [3.5 marks total]:
- Relating the refractive index to the gradient of the graph (e.g., stating \(n = \text{gradient}\) or showing Snell's Law arrangement matching \(y = mx\)) [1 mark]
- Correct substitution of the coordinates into the gradient formula [1 mark]
- Calculation of the gradient value as 1.5 (or 1.50) [1 mark]
- Stating the final answer with no units [0.5 marks]

When fuel flows through the delivery hose, friction between the fuel and the pipe walls causes a transfer of electrons. This leads to a build-up of electrostatic charge on both the fuel and the aircraft. If the charge is allowed to accumulate, it can create a large potential difference between the nozzle and the fuel tank. This potential difference can cause a spark, which has enough energy to ignite volatile fuel vapors, resulting in a dangerous fire or explosion. To prevent this hazard, a copper bonding (earthing) wire is connected between the aircraft and the fueling tanker (and the ground) before refueling begins. This provides a low-resistance path, allowing the charge to flow safely to the ground and preventing any potential difference from building up.

MP1: Friction between the flowing fuel and the delivery hose transfers electrons, causing a build-up of static charge. [1]
MP2: A high accumulation of charge creates a large potential difference, which can result in an electrostatic spark. [1]
MP3: The spark can ignite volatile fuel vapors, causing a fire or explosion. [1]
MP4: An earthing (bonding) wire connected between the aircraft and tanker/ground allows the excess charge to flow safely to the earth, preventing charge build-up and potential differences. [1]

Nuclear fusion involves joining two light, positively charged nuclei (such as hydrogen isotopes) to form a heavier nucleus. Since both nuclei carry a positive charge, they experience a very strong electrostatic force of repulsion when they are brought close together. To overcome this mutual electrostatic repulsion, the nuclei must have extremely high kinetic energies, which require extremely high temperatures. High pressures are also required to squeeze the nuclei close together, increasing the frequency of collisions. In contrast, nuclear fission is initiated when a heavy, stable nucleus absorbs a neutron. Because neutrons carry no electric charge, they experience no electrostatic repulsion as they approach the positive nucleus, allowing fission to occur at relatively low temperatures.

MP1: Identifies that nuclear fusion involves bringing two positively charged nuclei together. [1]
MP2: Explains that these positive nuclei experience strong electrostatic repulsion. [1]
MP3: States that high temperatures provide the high kinetic energies needed to overcome this electrostatic repulsion, while high pressure increases the collision rate. [1]
MP4: Explains that fission involves neutrons, which are neutral (uncharged) and thus do not experience electrostatic repulsion, allowing the reaction to occur at low temperatures. [1]

The red-shift of light from distant galaxies provides evidence that the universe is expanding. When we analyze light from distant galaxies, the wavelengths are shifted toward the red (longer wavelength) end of the spectrum, indicating they are moving away from us. Furthermore, the further away a galaxy is, the greater its red-shift, showing that the rate of expansion is increasing. This supports the Big Bang theory as it suggests that tracing the expansion backward in time leads to a single, infinitely small starting point. Cosmic microwave background (CMB) radiation is the microwave radiation detected uniformly from all directions in space. According to the Big Bang theory, the early universe was extremely hot and dense, filled with high-energy gamma radiation. As space expanded, this radiation stretched to much longer wavelengths, cooling down to the microwave spectrum we detect today. This provides strong evidence that the universe originated from a hot, dense state.

MP1: Explains that red-shift shows the light from distant galaxies is shifted to longer wavelengths because they are moving away from us. [1]
MP2: Explains that further galaxies have greater red-shifts (moving faster), which indicates the universe is expanding from a single point of origin. [1]
MP3: Describes CMB radiation as low-energy microwave radiation received uniformly from all directions in space. [1]
MP4: Explains that CMB radiation is the cooled and stretched remnant of high-energy (gamma) radiation from the early, hot, and dense phase of the universe, confirming a hot Big Bang start. [1]