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Velocity is a vector quantity, which has both magnitude (speed) and direction. Although the speed of the satellite is constant, its direction of motion is continuously changing as it travels along the circular orbit, meaning its velocity is constantly changing. Kinetic energy, orbital period, and mass are all scalar quantities and remain constant.

1 mark for identifying that velocity changes due to the change in direction of motion.

Using the equation for motion from rest, \(s = \frac{1}{2}at^2\), we can test the options.
For \(t = 1\text{ s}\), \(2 = \frac{1}{2}a(1)^2 \implies a = 4\text{ m/s}^2\).
For \(t = 2\text{ s}\), \(8 = \frac{1}{2}a(2)^2 \implies a = 4\text{ m/s}^2\).
For \(t = 3\text{ s}\), \(18 = \frac{1}{2}a(3)^2 \implies a = 4\text{ m/s}^2\).
For \(t = 4\text{ s}\), \(32 = \frac{1}{2}a(4)^2 \implies a = 4\text{ m/s}^2\).
Therefore, the car has a constant acceleration of \(4\text{ m/s}^2\).

1 mark for the correct calculation showing acceleration is constant at \(4\text{ m/s}^2\).

To find the maximum pressure, we use \(P = \frac{F}{A}\). The force \(F\) is the weight of the block: \(F = m \times g = 6.0\text{ kg} \times 10\text{ N/kg} = 60\text{ N}\). The maximum pressure occurs when the contact area \(A\) is at its minimum. The smallest face area is \(0.10\text{ m} \times 0.20\text{ m} = 0.02\text{ m}^2\). Thus, \(P_{\text{max}} = \frac{60\text{ N}}{0.02\text{ m}^2} = 3000\text{ Pa}\).

1 mark for calculating weight as 60 N, finding the minimum area of 0.02 m^2, and correctly computing the pressure as 3000 Pa.

Total internal reflection occurs when the angle of incidence is greater than the critical angle, and the light is travelling from a more optically dense medium (water) to a less optically dense medium (air). Since the angle of incidence (\(50^\circ\)) is greater than the critical angle (\(48^\circ\)), the light ray undergoes total internal reflection, reflecting back into the water at an angle of \(50^\circ\) to the normal.

1 mark for identifying that total internal reflection occurs because the angle of incidence is greater than the critical angle.

Determine the number of half-lives that have elapsed: \(1600 \rightarrow 800 \rightarrow 400 \rightarrow 200\). This is exactly 3 half-lives. Since the total time is \(15\text{ hours}\), the duration of one half-life is \(\frac{15\text{ hours}}{3} = 5\text{ hours}\).

1 mark for identifying that 3 half-lives have elapsed and dividing 15 hours by 3 to get 5 hours.

First, calculate the total resistance of the circuit using Ohm's law: \(R_{\text{total}} = \frac{V}{I} = \frac{12\text{ V}}{0.3\text{ A}} = 40\ \Omega\). Since the two resistors are identical and connected in series, the total resistance is the sum of the individual resistances: \(R_{\text{total}} = R + R = 2R\). Therefore, \(2R = 40\ \Omega \implies R = 20\ \Omega\).

1 mark for calculating the total resistance of 40 ohms and halving it to find the resistance of each identical resistor as 20 ohms.

By conservation of momentum, the total momentum before the collision equals the total momentum after the collision.
\(p_{\text{initial}} = (m_1 \times u_1) + (m_2 \times u_2) = (2.0 \times 6.0) + (4.0 \times 0) = 12.0\text{ kg m/s}\).
After the collision, the combined mass is \(m_1 + m_2 = 2.0 + 4.0 = 6.0\text{ kg}\).
\(p_{\text{final}} = (m_1 + m_2) \times v = 6.0 \times v\).
Setting them equal: \(6.0 \times v = 12.0 \implies v = 2.0\text{ m/s}\).

1 mark for applying conservation of momentum to find the combined velocity of 2.0 m/s.

Use the equation:
\(v^2 = u^2 + 2as\)

Identify the known values:
- Initial velocity, \(u = 0\text{ m/s}\) (since it starts from rest)
- Acceleration, \(a = 1.5\text{ m/s}^2\)
- Distance, \(s = 12\text{ m}\)

Substitute the values into the equation:
\(v^2 = 0^2 + 2 \times 1.5 \times 12\)
\(v^2 = 36\)
\(v = \sqrt{36} = 6.0\text{ m/s}\)

- Recall of formula \(v^2 = u^2 + 2as\) (1)
- Substitution of values into formula (1)
- Correct calculation of final velocity with units (1)

Accept: \(6\text{ m/s}\)
Reject: \(6\text{ m/s}^2\)

Use the equation:
\(E = Q \times V\)

Identify the known values:
- Voltage, \(V = 9.0\text{ V}\)
- Charge, \(Q = 15\text{ C}\)

Substitute the values into the equation:
\(E = 15 \times 9.0 = 135\text{ J}\)

- Recall of formula \(E = Q \times V\) (1)
- Correct substitution of values (1)
- Correct final answer with units (1)

Accept: \(140\text{ J}\) (to 2 s.f.)
Reject: lower case \(j\) for units

First, calculate the volume of the block:
\(V = 0.15 \times 0.080 \times 0.050 = 0.0006\text{ m}^3\) (or \(6.0 \times 10^{-4}\text{ m}^3\))

Next, use the density equation:
\(\text{density} = \frac{\text{mass}}{\text{volume}}\)

Substitute the values:
\(\text{density} = \frac{0.48}{0.0006} = 800\text{ kg/m}^3\)

- Correct calculation of volume with unit (1)
- Correct substitution into density equation (1)
- Correct final density with units (1)

Accept: \(0.80\text{ g/cm}^3\) if volume and mass are correctly converted.
Reject: \(800\text{ kg/m}^2\)

Find the number of half-lives that have elapsed by halving the initial activity until reaching the final activity:
- Start: \(3200\text{ Bq}\)
- 1 half-life: \(1600\text{ Bq}\)
- 2 half-lives: \(800\text{ Bq}\)
- 3 half-lives: \(400\text{ Bq}\)
- 4 half-lives: \(200\text{ Bq}\)
- 5 half-lives: \(100\text{ Bq}\)

So, 5 half-lives have elapsed in \(45\text{ hours}\).

Calculate the length of one half-life:
\(\text{half-life} = \frac{45\text{ hours}}{5} = 9\text{ hours}\)

- Correct determination that 5 half-lives have passed (1)
- Division of total time by the number of half-lives (1)
- Correct final answer with unit (1)

Accept: \(540\text{ minutes}\)
Reject: \(5\text{ hours}\)

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

Identify the angles:
- Angle of incidence, \(i = 42^\circ\)
- Angle of refraction, \(r = 26^\circ\)

Substitute the values into the equation:
\(n = \frac{\sin(42^\circ)}{\sin(26^\circ)}\)
\(n = \frac{0.66913}{0.43837}\)
\(n \approx 1.526\)

Rounded to 2 decimal places, the refractive index is \(1.53\).

- Recall of Snell's Law equation (1)
- Correct substitution of values (1)
- Correct calculation to 2 decimal places (1)

Accept: \(1.53\)
Reject: any units (refractive index is a ratio)

Use the principle of conservation of momentum:
\(\text{total initial momentum} = \text{total final momentum}\)

Calculate total initial momentum:
\(p_{\text{initial}} = (m_1 \times u_1) + (m_2 \times u_2)\)
\(p_{\text{initial}} = (1.2 \times 3.0) + (0.80 \times 0) = 3.6\text{ kg m/s}\)

Calculate total final mass:
\(m_{\text{total}} = 1.2 + 0.80 = 2.0\text{ kg}\)

Set initial momentum equal to final momentum:
\(3.6 = 2.0 \times v\)
\(v = \frac{3.6}{2.0} = 1.8\text{ m/s}\)

- Calculation of initial momentum or statement of conservation law (1)
- Correct addition of masses to find total mass (1)
- Correct calculation of final velocity with unit (1)

Accept: \(1.8\text{ m/s}\)
Reject: \(1.8\text{ m/s}^2\)

First, convert both temperatures from Celsius to Kelvin:
- \(T_1 = 27 + 273 = 300\text{ K}\)
- \(T_2 = 127 + 273 = 400\text{ K}\)

Use the pressure law equation for a fixed volume of gas:
\(\frac{p_1}{T_1} = \frac{p_2}{T_2}\)

Substitute the values:
\(\frac{150}{300} = \frac{p_2}{400}\)

Rearrange to solve for \(p_2\):
\(p_2 = \frac{150 \times 400}{300} = 200\text{ kPa}\)

- Convert both temperatures to Kelvin (1)
- Recall and correct substitution into \(p_1/T_1 = p_2/T_2\) (1)
- Correct calculation of final pressure with unit (1)

Accept: \(200,000\text{ Pa}\)
Reject: \(705\text{ kPa}\) (obtained by using temperatures in Celsius)

First, calculate the work done (or energy transferred) in lifting the weight:
\(W = F \times d\)
\(W = 45 \times 2.0 = 90\text{ J}\)

Next, use the power equation:
\(P = \frac{W}{t}\)

Substitute the values:
\(P = \frac{90}{3.0} = 30\text{ W}\)

- Correct calculation of work done as \(90\text{ J}\) (1)
- Correct substitution into power equation (1)
- Correct calculation of power with unit (1)

Accept: \(30\text{ J/s}\)
Reject: \(30\text{ J}\)

First, calculate the acceleration of the athlete:
\( a = \frac{v - u}{t} \)
\( a = \frac{9.0 - 0}{1.5} = 6.0\text{ m/s}^2 \)

Next, use Newton's second law to calculate the average force:
\( F = m \times a \)
\( F = 75 \times 6.0 = 450\text{ N} \)

1. Recall and use of \( a = \frac{v - u}{t} \) or \( F = \frac{m(v - u)}{t} \) (1 mark)
2. Substitution of values into formula: \( F = 75 \times \frac{9.0}{1.5} \) (1 mark)
3. Correct evaluation with unit: \( 450\text{ N} \) (1 mark)

Use the pressure difference formula in a liquid:
\( p = h \times \rho \times g \)

Rearrange to solve for depth \( h \):
\( h = \frac{p}{\rho \times g} \)

Substitute the values:
\( h = \frac{1.25 \times 10^5}{1000 \times 10} \)
\( h = \frac{125\,000}{10\,000} = 12.5\text{ m} \)

1. Recall of \( p = h \times \rho \times g \) (1 mark)
2. Rearrangement and substitution: \( h = \frac{125\,000}{1000 \times 10} \) (1 mark)
3. Evaluation with correct unit: \( 12.5\text{ m} \) (1 mark)

Use Snell's Law to calculate the refractive index \( n \):
\( n = \frac{\sin i}{\sin r} \)

Substitute the angles:
\( n = \frac{\sin 48^\circ}{\sin 29^\circ} \)
\( n = \frac{0.743}{0.485} \approx 1.53 \)

1. Recall of Snell's law: \( n = \frac{\sin i}{\sin r} \) (1 mark)
2. Correct substitution of angles: \( n = \frac{\sin 48^\circ}{\sin 29^\circ} \) (1 mark)
3. Correct evaluation to 2 decimal places: \( 1.53 \) (1 mark)

First, convert the time from minutes to seconds:
\( t = 4.0\text{ minutes} = 4.0 \times 60\text{ s} = 240\text{ s} \)

Next, use the electrical energy equation:
\( E = I \times V \times t \)

Substitute the values:
\( E = 1.5\text{ A} \times 6.0\text{ V} \times 240\text{ s} = 2160\text{ J} \) (or \( 2.16\text{ kJ} \))

1. Conversion of time to seconds: \( 240\text{ s} \) (1 mark)
2. Recall/use of energy formula: \( E = I \times V \times t \) (1 mark)
3. Correct evaluation with unit: \( 2160\text{ J} \) (or \( 2.16\text{ kJ} \)) (1 mark)

1. The light enters the inner core and strikes the boundary with the outer cladding.
2. Total internal reflection occurs, preventing light from escaping through the sides of the fibre.
3. This happens because the refractive index of the core is greater than the refractive index of the cladding.
4. Additionally, the angle of incidence at the boundary must be greater than the critical angle for the two materials.

M1: State that light undergoes total internal reflection (TIR) at the boundary. [1]
M2: Explain that the core has a higher refractive index than the cladding (or is optically denser). [1]
M3: State that the angle of incidence must be greater than the critical angle. [1]
M4: Explain that this keeps the light trapped inside the core, allowing it to travel long distances. [1]

1. The weight of the person is the downward force, which remains constant.
2. Snowshoes have a much larger surface area in contact with the snow compared to ordinary boots.
3. Pressure is defined by the formula: \(P = \frac{F}{A}\).
4. Because the area is larger, the pressure exerted on the snow is much lower, so the snow can support the weight without collapsing.

M1: Identify that the downward force is the weight of the person (which is constant in both cases). [1]
M2: State that snowshoes have a much larger surface area in contact with the snow than ordinary boots. [1]
M3: State or use the relationship Pressure = Force / Area. [1]
M4: Explain that a larger area results in a lower pressure exerted on the snow, preventing the person from sinking. [1]

1. Alpha particles consist of 2 protons and 2 neutrons, giving them a large mass and a charge of +2, whereas beta particles are single, light, fast-moving electrons with a charge of -1.
2. Because of their larger charge and mass, alpha particles are much more likely to interact with and remove electrons from atoms they pass, leading to high ionizing power.
3. Each ionization event transfers energy from the alpha particle to the atom.
4. Due to the high frequency of these energy-transferring interactions, the alpha particle loses its kinetic energy very quickly, resulting in a short range of only a few centimetres in air.

M1: Describe alpha particles as having a larger mass and/or larger charge (+2) compared to beta particles (-1). [1]
M2: Explain that this larger charge/mass makes alpha particles highly likely to collide with/ionize air atoms (high ionizing power). [1]
M3: State that energy is transferred from the radiation to the air molecules during each ionization. [1]
M4: Explain that because they ionize so frequently, they lose all their kinetic energy rapidly, resulting in a short range in air. [1]

1. A current-carrying conductor placed in an external magnetic field experiences a force (the motor effect).
2. The current flows in opposite directions along the two parallel sides of the rectangular coil that are perpendicular to the magnetic field.
3. According to Fleming's Left-Hand Rule, this causes equal and opposite forces on these two sides (one side experiences an upward force, the other a downward force).
4. These opposing forces create a turning moment (torque) about the central axis, causing the coil to rotate.

M1: State that the current in the coil creates a magnetic field that interacts with the permanent magnet's field (or experiences a force). [1]
M2: Explain that the current flows in opposite directions on opposite sides of the coil. [1]
M3: State that this produces forces in opposite directions on these two sides (e.g., one force up, one force down). [1]
M4: Explain that these opposing forces create a turning effect/moment that rotates the coil. [1]

1. An NTC thermistor has a resistance that decreases as its temperature increases.
2. Because the thermistor is connected in series with a fixed resistor, the total resistance of the circuit is the sum of their individual resistances.
3. Therefore, as the temperature increases, the total resistance of the circuit decreases.
4. Since the voltage of the power supply is constant, according to Ohm's law (\(I = \frac{V}{R}\)), the decrease in total resistance causes the current to increase, resulting in a higher reading on the ammeter.

M1: State that the resistance of the NTC thermistor decreases as temperature increases. [1]
M2: Explain that because the components are in series, the total resistance of the circuit decreases. [1]
M3: Recall and use the relationship \(I = V / R\) where voltage \(V\) is constant. [1]
M4: Conclude that the current increases, hence the ammeter reading increases. [1]

1. Molecules in a liquid have a distribution of different kinetic energies and move at different speeds.
2. The fastest-moving molecules (those with the highest kinetic energy) near the surface have enough energy to break the intermolecular forces holding them in the liquid and escape as a gas.
3. As these high-energy molecules leave, the average kinetic energy of the remaining molecules in the liquid decreases.
4. Since temperature is proportional to the average kinetic energy of the molecules, the temperature of the remaining liquid falls.

M1: State that molecules in the liquid have a range of kinetic energies/speeds. [1]
M2: Explain that only the highest kinetic energy (fastest) molecules near the surface can escape the liquid. [1]
M3: State that this escape reduces the average kinetic energy of the remaining molecules. [1]
M4: Link average kinetic energy to temperature to explain why the liquid cools down. [1]

1. At the moment of jumping, the only force acting is weight, creating a large downward resultant force and a high initial acceleration.
2. As the skydiver's velocity increases, the upward air resistance (drag) acting on them increases.
3. The resultant force (weight minus air resistance) decreases, which reduces the rate of acceleration, although the speed continues to rise.
4. Eventually, the air resistance increases until it is equal in magnitude and opposite in direction to the weight. The forces are balanced, the resultant force becomes zero, and the skydiver falls at a constant terminal velocity.

M1: State that initially weight is the dominant force (or the only force), causing a downward acceleration. [1]
M2: Explain that as speed increases, the upward air resistance (drag) also increases. [1]
M3: Explain that this reduces the resultant force, decreasing the acceleration (but speed still increases). [1]
M4: State that when air resistance equals weight, the resultant force is zero, leading to a constant terminal velocity. [1]

1. In the main sequence, outward radiation pressure from nuclear fusion balances the inward pull of gravity.
2. When the hydrogen in the core is depleted, nuclear fusion in the core slows down, causing the outward radiation pressure to decrease.
3. The core contracts under gravity and heats up, which triggers the fusion of hydrogen in a shell surrounding the core (and eventually helium fusion in the core).
4. The increased energy release from these new fusion reactions creates a massive outward radiation pressure, causing the outer layers of the star to expand significantly and cool down, turning the star into a red giant.

M1: State that the hydrogen fuel in the core runs out, causing core fusion to slow down/stop. [1]
M2: Explain that the core contracts under gravity and heats up because inward gravitational force is now greater than outward radiation pressure. [1]
M3: State that this heating triggers hydrogen fusion in a shell surrounding the core (or helium fusion in the core). [1]
M4: Explain that the increased radiation pressure causes the outer layers of the star to expand and cool, forming a red giant. [1]

To find the refractive index \(n\) for each trial, use Snell's Law:
\(n = \frac{\sin(i)}{\sin(r)}\)

For Trial 1:
\(n_1 = \frac{\sin(30^\circ)}{\sin(19^\circ)} = \frac{0.5000}{0.3256} \approx 1.536\)

For Trial 2:
\(n_2 = \frac{\sin(45^\circ)}{\sin(28^\circ)} = \frac{0.7071}{0.4695} \approx 1.506\)

For Trial 3:
\(n_3 = \frac{\sin(60^\circ)}{\sin(35^\circ)} = \frac{0.8660}{0.5736} \approx 1.510\)

Now, calculate the mean value:
\(\text{Mean } n = \frac{1.536 + 1.506 + 1.510}{3} = \frac{4.552}{3} \approx 1.517\)

Rounding to 3 significant figures gives \(1.52\).

- Recall of Snell's Law equation \(n = \sin(i) / \sin(r)\) (1 mark)
- Correct calculations of individual refractive indices (at least two correct) (1 mark)
- All three individual refractive indices correct (1.54, 1.51, 1.51) (1 mark)
- Calculation of mean using student's values (1 mark)
- Correct final answer rounded to 3 significant figures (1.52) (1 mark)

1. Calculate the volume change (\(\Delta V\)) for each measurement to find the volume of the bolt:
- Measurement 1: \(53.5 - 45.0 = 8.5\text{ cm}^3\)
- Measurement 2: \(53.7 - 45.0 = 8.7\text{ cm}^3\)
- Measurement 3: \(53.3 - 45.0 = 8.3\text{ cm}^3\)

2. Calculate the mean volume of the bolt:
\(\text{Mean volume} = \frac{8.5 + 8.7 + 8.3}{3} = 8.5\text{ cm}^3\)

3. Use the density formula:
\(\text{Density} = \frac{\text{mass}}{\text{volume}}\)

4. Calculate the density:
\(\text{Density} = \frac{78.4\text{ g}}{8.5\text{ cm}^3} \approx 9.2235\text{ g/cm}^3\)

5. Round to 3 significant figures:
\(9.22\text{ g/cm}^3\)

- Recall of \(\text{density} = \text{mass} / \text{volume}\) (1 mark)
- Determination of individual volume changes (8.5, 8.7, 8.3) or mean final level (53.5) (1 mark)
- Determination of mean volume (\(8.5\text{ cm}^3\)) (1 mark)
- Substitution of mass and mean volume into the density formula (1 mark)
- Correct final density value with units to 3 s.f. (\(9.22\text{ g/cm}^3\)) (1 mark)

(i) The natural length of the spring is \(15.0\text{ cm}\) (at \(0.0\text{ N}\)).
For each addition of \(1.0\text{ N}\) of force, the length increases by \(2.0\text{ cm}\) (from \(15.0\text{ cm} \rightarrow 17.0\text{ cm} \rightarrow 19.0\text{ cm}\)).
At \(3.0\text{ N}\), the length is \(22.0\text{ cm}\) (an increase of \(3.0\text{ cm}\) from the previous step), which does not fit the linear pattern. At \(4.0\text{ N}\), the length is \(23.0\text{ cm}\) (which is \(15.0 + 4 \times 2.0 = 23.0\text{ cm}\)). Therefore, the anomalous length measurement is \(22.0\text{ cm}\) (at \(3.0\text{ N}\)).

(ii) Excluding the anomaly:
Extension (\(x\)) for \(1.0\text{ N}\) force = \(17.0 - 15.0 = 2.0\text{ cm} = 0.02\text{ m}\).
Using Hooke's Law, \(F = kx\):
\(k = \frac{F}{x} = \frac{1.0\text{ N}}{0.02\text{ m}} = 50\text{ N/m}\).

Alternatively, using the maximum valid point (\(5.0\text{ N}\)):
Extension = \(25.0 - 15.0 = 10.0\text{ cm} = 0.10\text{ m}\).
\(k = \frac{5.0\text{ N}}{0.10\text{ m}} = 50\text{ N/m}\).

- Identifies anomalous length as \(22.0\text{ cm}\) (at \(3.0\text{ N}\)) (1 mark)
- Calculates extension correctly for any valid point (e.g. \(10.0\text{ cm}\) at \(5.0\text{ N}\) or \(2.0\text{ cm}\) at \(1.0\text{ N}\)) (1 mark)
- Converts extension from cm to m (e.g. \(10.0\text{ cm} = 0.10\text{ m}\)) (1 mark)
- Recall and rearrangement of Hooke's Law: \(k = F/x\) (1 mark)
- Correct calculation of spring constant with units (\(50\text{ N/m}\)) (1 mark)

Nuclear fusion requires joining two positively charged nuclei. Since like charges repel, there is a strong electrostatic repulsive force between them. High temperatures give the nuclei sufficient kinetic energy to move fast enough to overcome this repulsion, and high pressure increases the collision rate, allowing them to get close enough for the strong nuclear force to bind them together.

1 mark for selecting correct option B.

Using Boyle's Law for a constant temperature process: \(p_1 V_1 = p_2 V_2\). Rearranging to solve for the final pressure gives \(p_2 = \frac{p_1 V_1}{V_2}\). Substituting the values: \(p_2 = \frac{1.5 \times 10^5\text{ Pa} \times 0.20\text{ m}^3}{0.050\text{ m}^3} = 6.0 \times 10^5\text{ Pa}\).

1 mark for selecting correct option C.

The expansion of the universe is supported by the red-shift of distant galaxies, which shows they are moving away from us, and the cosmic microwave background (CMB) radiation, which is the cosmic expansion-cooled remnant of the Big Bang.

1 mark for selecting correct option A.

First, calculate the energy supplied by the heater: \(E = P \times t\) which gives \(E = 150 \text{ W} \times 550 \text{ s} = 82500 \text{ J}\). Next, use the formula for thermal energy change during a change of state: \(Q = m \times L\). Rearrange the formula to solve for the specific latent heat of fusion: \(L = \frac{Q}{m}\). Substituting the values gives \(L = \frac{82500 \text{ J}}{0.25 \text{ kg}} = 330000 \text{ J/kg}\).

1 mark: For calculating the total energy supplied by the heater using \(E = P \times t\) to get \(82500 \text{ J}\). 1 mark: For rearranging the change of state equation \(Q = m \times L\) to make \(L\) the subject. 1 mark: For the correct final value of \(330000 \text{ J/kg}\) (or \(3.3 \times 10^5 \text{ J/kg}\)) with the correct unit.

The relationship between the refractive index (\(n\)) and the critical angle (\(c\)) is given by: \(\sin(c) = \frac{1}{n}\). Substitute the given refractive index: \(\sin(c) = \frac{1}{1.62} \approx 0.61728\). Calculate the critical angle: \(c = \sin^{-1}(0.61728) \approx 38.118^\circ\). Rounded to 3 significant figures, the critical angle is \(38.1^\circ\).

1 mark: For recalling and stating the formula \(\sin(c) = 1/n\). 1 mark: For substituting the value of refractive index to calculate \(\sin(c) \approx 0.617\). 1 mark: For calculating the correct critical angle of \(38.1^\circ\) (accept \(38^\circ\) to \(38.1^\circ\)).

First, convert the power output from megawatts to watts: \(24 \text{ MW} = 2.4 \times 10^7 \text{ J/s}\). Next, find the number of reactions per second by dividing the total power by the energy released per reaction: \(\text{Number of reactions per second} = \frac{\text{Total energy per second}}{\text{Energy per reaction}}\). This gives \(\frac{2.4 \times 10^7 \text{ J/s}}{3.2 \times 10^{-12} \text{ J}} = 7.5 \times 10^{18}\).

1 mark: For converting power from MW to W (\(2.4 \times 10^7 \text{ W}\) or \(\text{J/s}\)). 1 mark: For setting up the division of total power (or energy per second) by the energy of a single reaction. 1 mark: For the correct calculation leading to \(7.5 \times 10^{18}\).

Since the temperature remains constant, use Boyle's Law: \(P_1 \times V_1 = P_2 \times V_2\). Rearrange the formula to solve for the final volume (\(V_2\)): \(V_2 = \frac{P_1 \times V_1}{P_2}\). Substitute the given values: \(V_2 = \frac{100 \text{ kPa} \times 3.5 \text{ m}^3}{28 \text{ kPa}} = 12.5 \text{ m}^3\).

1 mark: For stating/identifying the formula \(P_1 V_1 = P_2 V_2\). 1 mark: For correctly substituting the known values into the equation. 1 mark: For calculating the correct volume of \(12.5 \text{ m}^3\) with the appropriate unit.

First, calculate the change in wavelength: \(\Delta \lambda = 608 \text{ nm} - 600 \text{ nm} = 8 \text{ nm}\). Now, use the redshift formula: \(\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}\). Rearranging to solve for speed gives \(v = c \times \frac{\Delta \lambda}{\lambda_0}\). Substitute the values: \(v = 3.0 \times 10^8 \text{ m/s} \times \frac{8 \text{ nm}}{600 \text{ nm}} = 4.0 \times 10^6 \text{ m/s}\).

1 mark: For calculating the change in wavelength (\(\Delta \lambda = 8 \text{ nm}\)). 1 mark: For recalling and rearranging the redshift formula \(\frac{\Delta \lambda}{\lambda} = \frac{v}{c}\). 1 mark: For evaluating the correct recessional velocity as \(4.0 \times 10^6 \text{ m/s}\).

The relationship between the primary and secondary voltages and turn ratios is given by: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\). Rearrange the formula to find the secondary voltage (\(V_s\)): \(V_s = V_p \times \frac{N_s}{N_p}\). Substituting the given values: \(V_s = 12 \text{ V} \times \frac{3750}{150} = 300 \text{ V}\).

1 mark: For selecting the correct transformer formula \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\). 1 mark: For correct substitution of values. 1 mark: For correct calculation of \(300 \text{ V}\).

First, calculate the pressure exerted by the seawater column: \(p_{\text{fluid}} = \rho \times g \times h = 1020 \text{ kg/m}^3 \times 10 \text{ N/kg} \times 40 \text{ m} = 408000 \text{ Pa} = 408 \text{ kPa}\). To find the total pressure, add the atmospheric pressure: \(p_{\text{total}} = p_{\text{atmospheric}} + p_{\text{fluid}} = 101 \text{ kPa} + 408 \text{ kPa} = 509 \text{ kPa}\).

1 mark: For calculating fluid pressure using \(p = \rho g h\) to get \(408000 \text{ Pa}\) (or \(408 \text{ kPa}\)). 1 mark: For adding the atmospheric pressure (\(101 \text{ kPa}\)) to their calculated fluid pressure. 1 mark: For the correct final value of \(509 \text{ kPa}\) (or \(509000 \text{ Pa}\)).

We use the equation of motion: \(v^2 = u^2 + 2as\). Rearranging to solve for acceleration (\(a\)) gives: \(a = \frac{v^2 - u^2}{2s}\). Substituting the given values: \(a = \frac{28^2 - 12^2}{2 \times 160} = \frac{784 - 144}{320} = \frac{640}{320} = 2.0 \text{ m/s}^2\).

1 mark: For choosing and stating the equation \(v^2 = u^2 + 2as\). 1 mark: For correct substitution or algebraic rearrangement. 1 mark: For a correct acceleration value of \(2.0 \text{ m/s}^2\) (accept 2).

When the temperature of a gas is increased, the average kinetic energy of the gas molecules increases, which means they move at higher average speeds. As a result, the molecules collide with the internal walls of the container more frequently. Furthermore, because they are moving faster, each collision involves a greater change in momentum, meaning they exert a larger force on the walls. Since pressure is force per unit area \(P = F/A\) and the container's volume (and thus surface area) is constant, this increase in force leads to an increase in pressure.

1 mark (MP1): Mention that higher temperature increases the average kinetic energy or speed of the gas molecules.
1 mark (MP2): Mention that molecules collide with the container walls more frequently (more collisions per unit time).
1 mark (MP3): Mention that molecules collide with the walls with greater force / greater change in momentum.
1 mark (MP4): Link the increased force to increased pressure because pressure is force per unit area \(P = F/A\).

An alternating current (a.c.) is supplied to the primary coil, which creates a continuously changing magnetic field around it. This changing magnetic field is guided through the magnetically soft iron core to the secondary coil. As the magnetic field cuts through the secondary coil, it induces an alternating voltage across its terminals through electromagnetic induction. Since a step-up transformer has more turns on its secondary coil than on its primary coil (\(N_s > N_p\)), the induced voltage in the secondary coil is larger than the input voltage.

1 mark (MP1): State that an alternating current in the primary coil creates a changing magnetic field.
1 mark (MP2): State that the iron core guides/transfers this changing magnetic field to the secondary coil.
1 mark (MP3): State that this changing magnetic field cuts the secondary coil and induces a voltage (electromagnetic induction).
1 mark (MP4): State that because there are more turns on the secondary coil (\(N_s > N_p\)), the induced voltage is higher.

During a collision, a car and its passengers experience a fixed change in momentum (\(\Delta p\)) to come to a complete stop from their initial speed. The crumple zone is designed to fold and collapse systematically, which increases the time (\(\Delta t\)) taken for the car to come to a halt. Since force is equal to the rate of change of momentum, given by the equation \(F = \frac{\Delta p}{\Delta t}\), increasing the duration of the impact greatly reduces the average force experienced by the passengers, thereby reducing the risk and severity of injuries.

1 mark (MP1): State that there is a fixed/constant change in momentum to bring the passenger to a stop.
1 mark (MP2): State that the crumple zone increases the collision time / time taken to stop.
1 mark (MP3): State or use the formula relating force to rate of change of momentum: \(F = \frac{\Delta p}{\Delta t}\).
1 mark (MP4): Explain that a larger time interval results in a significantly reduced force acting on the passengers.

Light signals enter the optical fiber and hit the boundary between the inner core and outer cladding at a very shallow angle. Consequently, the angle of incidence at this boundary is greater than the critical angle (\(i > c\)). Because the core of the fiber has a higher refractive index than the outer cladding, total internal reflection occurs. This means no light refracts out of the fiber; instead, 100% of the light is reflected internally. This reflection happens repeatedly down the length of the curved fiber, ensuring the light remains trapped and reaches the destination.

1 mark (MP1): State that the angle of incidence is greater than the critical angle (\(i > c\)).
1 mark (MP2): State that the refractive index of the core is greater than that of the cladding.
1 mark (MP3): Mention that total internal reflection (TIR) occurs at the boundary.
1 mark (MP4): Explain that this reflection occurs repeatedly along the fiber, trapping the light within.

Alpha particles have a strong positive charge (\(+2e\)) and are relatively massive, meaning they easily interact with and strip electrons from nearby atoms, making them highly ionizing. However, because they ionize atoms so frequently, they transfer their kinetic energy very rapidly to the air molecules, which limits their range to only a few centimeters in air. In contrast, gamma rays are high-energy electromagnetic waves with no electric charge and no mass. This makes them much less likely to interact with and ionize atoms, rendering them weakly ionizing. Because they lose energy very slowly through these rare interactions, they can travel hundreds of meters in air.

1 mark (MP1): State that alpha particles are highly ionizing because they have a large charge/mass and easily interact with atoms.
1 mark (MP2): Explain that this frequent ionization causes alpha particles to lose energy quickly, resulting in a short range.
1 mark (MP3): State that gamma rays are weakly ionizing because they have no charge/mass and are much less likely to interact with atoms.
1 mark (MP4): Explain that because they ionize atoms infrequently, gamma rays lose energy slowly, giving them a long range.

According to the Big Bang theory, the Universe began from an extremely hot, dense singular state, which would have emitted high-energy, short-wavelength radiation (such as gamma and X-rays). As the Universe expanded over billions of years, space itself stretched. This expansion stretched the wavelength of the original cosmic radiation, shifting it from high-energy radiation into the microwave region of the electromagnetic spectrum. Today, this radiation is detected uniformly coming from all directions in the sky (isotropic). The existence of this CMBR, and its specific temperature of about \(2.7\text{ K}\), matches the exact predictions of the Big Bang model and cannot be easily explained by alternative static universe theories.

1 mark (MP1): State that the Big Bang theory predicts the early Universe was extremely hot and dense, releasing high-energy, short-wavelength radiation.
1 mark (MP2): Explain that the expansion of the Universe caused space to stretch, which stretched the wavelength of this radiation.
1 mark (MP3): Identify that this radiation has been redshifted into the microwave region (hence CMBR).
1 mark (MP4): State that the uniform presence of CMBR in all directions of space confirms a common, hot cosmic origin.

1. **Mass Measurement**: Use an electronic balance to find the mass of a group of steel ball bearings (e.g., 20 bearings). Let this mass be \(m\). Measuring multiple bearings reduces the percentage uncertainty in the mass of a single bearing.
2. **Initial Volume**: Fill a graduated measuring cylinder with a known volume of water and record this initial volume, \(V_1\). Read the volume at eye level from the bottom of the meniscus to avoid parallax error.
3. **Final Volume**: Gently lower the ball bearings into the measuring cylinder (using a thread or by tilting the cylinder) to avoid splashing and prevent trapping air bubbles. Record the new volume of the water and bearings, \(V_2\).
4. **Volume of Bearings**: Calculate the volume of the ball bearings, \(V\), using the formula:
\(V = V_2 - V_1\)
5. **Density Calculation**: Calculate the density, \(\rho\), of the steel using the formula:
\(\rho = \frac{m}{V}\)

- **MP1 (Mass)**: Measure the mass of the ball bearings using an electronic balance.
- **MP2 (Initial Volume & Accuracy)**: Measure the initial volume of water in a measuring cylinder, ensuring the reading is taken at eye level / at the bottom of the meniscus.
- **MP3 (Final Volume)**: Completely submerge the ball bearings in the water, avoiding splashing / trapping air bubbles, and record the new volume.
- **MP4 (Volume Calculation)**: Calculate the volume of the bearings by subtracting the initial volume from the final volume (\(V = V_2 - V_1\)).
- **MP5 (Density Calculation)**: State the formula \(\text{density} = \frac{\text{mass}}{\text{volume}}\) to determine the density.

1. **Apparatus**: Set up a sealed syringe containing a fixed mass of air, connected to a digital pressure sensor or pressure gauge.
2. **Procedure**: Record the initial volume of air from the scale on the syringe and the corresponding pressure from the pressure gauge.
3. **Varying Volume**: Slowly push or pull the plunger to a new position to change the volume of the air. Record the new volume and the corresponding pressure.
4. **Constant Temperature**: Move the plunger very slowly (or wait a short time after moving the plunger before taking the reading) to allow any heat generated or lost during compression/expansion to equilibrate with the surroundings, keeping the temperature constant.
5. **Analysis**: Repeat the procedure to obtain at least 5 different pairs of pressure and volume readings. Plot a graph of pressure (\(p\)) on the y-axis against \(1/\text{Volume}\) (\(1/V\)) on the x-axis. A straight line passing through the origin confirms that pressure is inversely proportional to volume (\(p \propto 1/V\)). Alternatively, calculate \(p \times V\) for each reading and show that the product is constant.

- **MP1**: Mention using a sealed syringe connected to a pressure gauge / pressure sensor.
- **MP2**: Record both the volume (from the syringe scale) and the corresponding pressure (from the gauge).
- **MP3**: State that the plunger must be moved slowly (or wait between readings) to keep the temperature of the gas constant.
- **MP4**: State that the measurement should be repeated for a range of different volumes (at least 5 different volumes).
- **MP5**: Explain how to analyze the results: either by plotting a graph of \(p\) against \(1/V\) and checking for a straight line through the origin, or by showing that \(p \times V\) is constant for all data points.