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The redshift of light indicates that galaxies are moving away from us, which means the universe is expanding. This is a key piece of evidence supporting the Big Bang theory.

1 mark for correct option B.

The total pressure is the sum of atmospheric pressure and the pressure due to the water column: \(P_{\text{total}} = P_{\text{atm}} + \rho g h\). Substituting the given values: \(3.0 \times 10^5\text{ Pa} = 1.0 \times 10^5\text{ Pa} + (1000\text{ kg/m}^3 \times 10\text{ N/kg} \times h)\). Subtracting atmospheric pressure: \(2.0 \times 10^5\text{ Pa} = 10000 \times h\). Solving for \(h\): \(h = \frac{2.0 \times 10^5}{10000} = 20\text{ m}\).

1 mark for the correct calculation showing depth of 20 m (B).

For an NTC thermistor, resistance decreases as temperature increases. In a series potential divider circuit, a decrease in the thermistor's resistance means it takes a smaller share of the source potential difference, which increases the share of the potential difference across the fixed resistor.

1 mark for identifying both the decrease in thermistor resistance and the increase in potential difference across the fixed resistor (D).

The frequency of a wave remains constant when it changes medium. Therefore, the frequency in shallow water is still \(6.0\text{ Hz}\). Using the wave equation \(v = f \lambda\) rearranged for wavelength: \(\lambda = \frac{v}{f} = \frac{1.8\text{ m/s}}{6.0\text{ Hz}} = 0.30\text{ m}\).

1 mark for the correct calculation of wavelength as 0.30 m (B).

First, calculate the useful gravitational potential energy gained: \(E_p = mgh = 50\text{ kg} \times 10\text{ N/kg} \times 4.0\text{ m} = 2000\text{ J}\). Since efficiency is \(80\%\), \(\text{Total energy input} = \frac{\text{Useful energy output}}{\text{Efficiency}} = \frac{2000\text{ J}}{0.80} = 2500\text{ J}\).

1 mark for the correct calculation of total electrical energy as 2500 J (C).

A larger redshift (\(z\)) indicates a higher recessional velocity. According to Hubble's Law, the speed at which a galaxy is receding is directly proportional to its distance from Earth. Since Galaxy Z has the largest redshift (\(z = 0.12\)), it has the greatest recessional velocity and is the furthest away.

1 mark for correctly identifying Galaxy Z (D).

At the bottom of the ramp, all of the car's energy is kinetic: \(E_k = \frac{1}{2}mv^2 = 0.5 \times 0.25 \times 4.0^2 = 2.0\text{ J}\). At maximum height, all kinetic energy has been transferred to gravitational potential energy: \(E_p = mgh\). Therefore, \(2.0 = 0.25 \times 10 \times h \Rightarrow 2.0 = 2.5h \Rightarrow h = 0.8\text{ m}\).

1 mark: Calculate initial kinetic energy of \(2.0\text{ J}\).
1 mark: State or show that kinetic energy at the bottom equals gravitational potential energy at the top.
1.5 marks: Rearrange and calculate height \(h = 0.8\text{ m}\).

Since temperature is kept constant, we use Boyle's Law: \(p_1 V_1 = p_2 V_2\). Substituting the known variables: \(1.2 \times 10^5 \times 0.040 = p_2 \times 0.015\). This gives \(4800 = p_2 \times 0.015\). Solving for \(p_2\): \(p_2 = \frac{4800}{0.015} = 320,000\text{ Pa} = 3.2 \times 10^5\text{ Pa}\).

1 mark: Recall or state the equation \(p_1 V_1 = p_2 V_2\).
1 mark: Correctly substitute values into the equation.
1.5 marks: Correctly calculate the final pressure as \(320,000\text{ Pa}\) or \(3.2 \times 10^5\text{ Pa}\).

Resistance \(R\) is given by \(R = \frac{\rho L}{A}\). Doubling the length \(L\) doubles the resistance to \(0.80\ \Omega\). Halving the cross-sectional area \(A\) doubles the resistance again. Therefore, the new resistance is \(0.80 \times 2 = 1.6\ \Omega\).

1 mark: State that resistance is proportional to length and inversely proportional to area.
1 mark: Determine that doubling length increases resistance to \(0.80\ \Omega\).
1.5 marks: Determine that halving cross-sectional area doubles this value, yielding \(1.6\ \Omega\).

The frequency of the wave remains constant across media. First, find the frequency: \(f = \frac{v}{\lambda} = \frac{6.0}{15} = 0.40\text{ Hz}\). Then, use this frequency to find the wavelength in shallow water: \(10\text{ m} = \frac{4.0}{0.40}\).

1 mark: State or imply that the frequency of the wave is constant.
1 mark: Correctly calculate the frequency as \(0.40\text{ Hz}\).
1.5 marks: Correctly calculate the new wavelength as \(10\text{ m}\).

As the coil spins, the side of the coil that moved up must move down once it passes the vertical position to maintain rotation. The split-ring commutator switches the electrical contacts every half-turn (180 degrees). This reverses the direction of the current flowing through each side of the coil, which in turn reverses the direction of the magnetic force acting on each side, maintaining continuous rotation.

1 mark: Explain that the commutator reverses the direction of the current.
1 mark: State that this reversal occurs every half-turn.
1.5 marks: Explain that this reverses the magnetic forces on each side of the coil, keeping the torque acting in the same rotational direction.

First, calculate the weight of the rocket acting downwards: \(W = mg = 1200 \times 10 = 12000\text{ N}\). Next, calculate the resultant force: \(F_{\text{net}} = 18000\text{ N} - 12000\text{ N} = 6000\text{ N}\). Finally, use Newton's second law to find the acceleration: \(a = \frac{F_{\text{net}}}{m} = \frac{6000}{1200} = 5\text{ m/s}^2\).

1 mark: Correctly calculate weight of the rocket as \(12000\text{ N}\).
1 mark: Calculate the resultant force of \(6000\text{ N}\).
1.5 marks: Use \(F = ma\) to correctly calculate the acceleration of \(5\text{ m/s}^2\).

Alpha particles have low penetrating power and are stopped by a few centimeters of air or by the outer layer of dead skin cells, meaning they cannot reach living tissue from outside. Inside the body, however, there is no protective skin barrier. Because alpha particles are highly ionizing, they deposit concentrated energy directly into living cells, damaging DNA and causing mutation or cell death.

1 mark: Explain that outside the body alpha particles are stopped by dead skin cells due to low penetrating power.
1 mark: State that there is no protective barrier inside the body.
1.5 marks: Explain that alpha is highly ionizing and causes severe damage or mutations to cells and DNA once inside.

Redshift shows that light wavelengths have been stretched, meaning the galaxy is moving away from us. When we observe that almost all distant galaxies show redshift, it indicates the entire Universe is expanding. If the Universe is expanding today, it must have been smaller and denser in the past, suggesting it began from a single hot, dense point (the Big Bang).

1 mark: Explain that redshift means light wavelengths are stretched as the galaxy moves away.
1 mark: Connect redshift observations to the expansion of the Universe.
1.5 marks: Explain that an expanding Universe implies it must have started from a single dense point (the Big Bang).

1. First, find the total resistance of the series circuit using Ohm's Law:
\(R_{\text{total}} = \frac{V}{I} = \frac{6.0\text{ V}}{0.015\text{ A}} = 400\ \Omega\).

2. In a series circuit, total resistance is the sum of the individual resistances:
\(R_{\text{total}} = R_{\text{thermistor}} + R_{\text{fixed}}\).

3. Substitute the known values:
\(400\ \Omega = R_{\text{thermistor}} + 120\ \Omega\).

4. Rearrange to find the thermistor's resistance:
\(R_{\text{thermistor}} = 400\ \Omega - 120\ \Omega = 280\ \Omega\).

- Mark 1 (Method): Use of \(R = \frac{V}{I}\) with correct substitution to find total resistance \(400\ \Omega\).
- Mark 2 (Method): Use of series resistance formula \(R_{\text{total}} = R_1 + R_2\) to subtract \(120\ \Omega\) from total resistance.
- Mark 3 (Accuracy): Correct final answer of \(280\ \Omega\).

1. Use the constant temperature gas relationship (Boyle's Law): \(p_1 V_1 = p_2 V_2\).

2. Rearrange to solve for the new pressure \(p_2\):
\(p_2 = \frac{p_1 V_1}{V_2}\).

3. Substitute the given values:
\(p_2 = \frac{1.5 \times 10^5\text{ Pa} \times 0.040\text{ m}^3}{0.012\text{ m}^3} = 5.0 \times 10^5\text{ Pa}\).

4. Explanation: Reducing the volume reduces the space between particles. The gas particles collide with the walls of the container more frequently (per unit area). This increases the average net force exerted on the walls per unit area, hence increasing the pressure.

- Mark 1 (Method): Recall and use of \(p_1 V_1 = p_2 V_2\).
- Mark 2 (Accuracy): Correct calculation of the final pressure as \(5.0 \times 10^5\text{ Pa}\) (or \(500,000\text{ Pa}\)).
- Mark 3 (Explanation): State that the particles collide with the container walls more frequently / more often.
- Mark 4 (Explanation): Link more frequent collisions to a larger total force per unit area on the walls.

1. Calculate the initial gravitational potential energy (\(E_p\)) at the top of the hill:
\(E_p = m \cdot g \cdot h = 450\text{ kg} \times 10\text{ N/kg} \times 24\text{ m} = 108,000\text{ J}\).

2. Calculate the kinetic energy (\(E_k\)) at the bottom of the hill:
\(E_k = \frac{1}{2} m v^2 = 0.5 \times 450\text{ kg} \times (18\text{ m/s})^2 = 72,900\text{ J}\).

3. Apply conservation of energy. The difference between the initial gravitational potential energy and the final kinetic energy is the energy transferred to thermal energy:
\(E_{\text{thermal}} = E_p - E_k = 108,000\text{ J} - 72,900\text{ J} = 35,100\text{ J}\) (or \(35.1\text{ kJ}\)).

- Mark 1 (Method): Correct calculation of gravitational potential energy: \(108,000\text{ J}\).
- Mark 2 (Method): Correct calculation of kinetic energy: \(72,900\text{ J}\).
- Mark 3 (Method): Subtracting kinetic energy from gravitational potential energy to find the thermal energy transfer.
- Mark 4 (Accuracy): Correct final value of \(35,100\text{ J}\) (or \(35.1\text{ kJ}\)).

1. Take the initial direction of motion as positive:
Initial velocity \(u = +25\text{ m/s}\)
Final velocity \(v = -15\text{ m/s}\) (since it moves in the opposite direction).

2. Calculate the change in momentum (\(\Delta p\)) of the ball:
\(\Delta p = m(v - u) = 0.060\text{ kg} \times (-15\text{ m/s} - 25\text{ m/s}) = 0.060 \times (-40) = -2.4\text{ kg m/s}\).

3. Calculate the average force using Newton's Second Law:
\(F = \frac{\Delta p}{t} = \frac{-2.4\text{ kg m/s}}{0.050\text{ s}} = -48\text{ N}\).

4. The magnitude of the force is therefore \(48\text{ N}\).

- Mark 1 (Method): Accounts for the change in direction to calculate the change in momentum magnitude as \(2.4\text{ kg m/s}\) (accept \(0.060 \times 40\)).
- Mark 2 (Method): Recalls and applies the equation \(F = \frac{\Delta p}{t}\) using the calculated change in momentum and \(0.050\text{ s}\).
- Mark 3 (Accuracy): Correct final magnitude of \(48\text{ N}\).

1. Use the wave equation \(v = f \lambda\) to find the frequency in air:
\(f = \frac{v}{\lambda} = \frac{3.0 \times 10^8\text{ m/s}}{6.5 \times 10^{-7}\text{ m}} \approx 4.62 \times 10^{14}\text{ Hz}\).

2. The frequency of a wave does not change when it moves from one medium to another. Therefore, the frequency in the glass is also \(4.62 \times 10^{14}\text{ Hz}\).

3. Use the wave equation with the speed in glass to find the new wavelength:
\(\lambda_{\text{glass}} = \frac{v_{\text{glass}}}{f} = \frac{2.0 \times 10^8\text{ m/s}}{4.62 \times 10^{14}\text{ Hz}} \approx 4.33 \times 10^{-7}\text{ m}\).

(Alternative ratio method: \(\lambda_{\text{glass}} = \lambda_{\text{air}} \times \frac{v_{\text{glass}}}{v_{\text{air}}} = 6.5 \times 10^{-7} \times \frac{2.0 \times 10^8}{3.0 \times 10^8} = 4.33 \times 10^{-7}\text{ m}\))

- Mark 1 (Method): Rearranging and substituting into \(f = \frac{v}{\lambda}\) for light in air.
- Mark 2 (Accuracy): Correct frequency calculation of \(4.6 \times 10^{14}\text{ Hz}\) (or \(4.62 \times 10^{14}\text{ Hz}\)).
- Mark 3 (Method): Showing that the frequency remains the same in the glass, or using the correct ratio of wave speeds.
- Mark 4 (Accuracy): Correct wavelength in glass of \(4.3 \times 10^{-7}\text{ m}\) (accept \(4.33 \times 10^{-7}\text{ m}\)).

1. As the coil rotates, it cuts through the magnetic field lines between the permanent magnets. This relative motion induces an alternating potential difference (voltage) across the ends of the coil.

2. To increase the peak voltage produced by the generator, the engineer can:
- Increase the speed of rotation of the coil (so field lines are cut faster).
- Increase the number of turns of wire on the coil (increasing total flux linkage).
- Use stronger magnets to produce a stronger magnetic field.
- Increase the cross-sectional area of the coil.

- Mark 1 (Explanation of induction): State that the rotating coil cuts magnetic field lines (or experiences a changing magnetic field/flux).
- Mark 2 (Modification 1): State one valid change to increase voltage (e.g., spin faster, more turns, stronger magnets, larger coil area).
- Mark 3 (Modification 2): State a second distinct valid change from the list.

1. Alpha radiation has very low penetration and is highly ionising. If injected, it would be absorbed entirely by the surrounding tissue, causing significant local cell damage, and would not be able to pass through the body to be detected outside.

2. A half-life of 8 days is far too long for a diagnostic tracer. The radioactive source would remain active inside the patient's body for several weeks, delivering an unnecessarily high and harmful cumulative radiation dose.

3. A half-life of 6 hours is long enough to complete the medical examination but short enough that the radioactivity decays quickly to a safe level, minimizing the long-term risk to the patient.

- Mark 1 (Alpha limitation): State that alpha radiation cannot penetrate body tissue to be detected externally OR is too highly ionising/damaging inside the body.
- Mark 2 (Long half-life limitation): State that an 8-day half-life is too long, resulting in the patient being exposed to harmful radiation for an extended period.
- Mark 3 (Short half-life advantage): State that a 6-hour half-life allows enough time for the medical scan but decays rapidly afterward, reducing radiation dose.

1. Light from distant galaxies contains spectral lines that are shifted toward the red end of the spectrum, which corresponds to longer wavelengths. This is known as redshift.

2. Redshift occurs because of the Doppler effect as galaxies move away from us.

3. Observations show that the further away a galaxy is, the greater its redshift, meaning more distant galaxies are receding faster.

4. This proportional relationship indicates that the entire fabric of space is expanding. If the Universe is currently expanding, it must have been much smaller, hotter, and denser in the past, pointing back to a single starting point at the Big Bang.

- Mark 1 (Definition): Explain that redshift is the shifting of spectral lines to longer wavelengths (or the red end of the spectrum).
- Mark 2 (Motion): State that redshift indicates that distant galaxies are moving away from us.
- Mark 3 (Hubble's relationship): State that more distant galaxies have a greater redshift / are moving away faster.
- Mark 4 (Big Bang connection): Explain that this expansion implies the Universe originated from a single hot, dense point in the past.

First, calculate the initial kinetic energy: \(E_k = 0.5 \times m \times v^2 = 0.5 \times 0.25 \times 8.0^2 = 8.0\text{ J}\). At the maximum height, this kinetic energy is fully converted into gravitational potential energy: \(\Delta E_p = m \times g \times \Delta h = 8.0\text{ J}\). Rearranging for change in height: \(\Delta h = \frac{8.0}{0.25 \times 10} = 3.2\text{ m}\). Finally, add the initial height to find the maximum height above the ground: \(1.5\text{ m} + 3.2\text{ m} = 4.7\text{ m}\).

1 mark: Calculate initial kinetic energy (8.0 J) or correctly recall equations \(E_k = 0.5mv^2\) and \(E_p = mgh\). 1 mark: Calculate change in height (3.2 m). 1 mark: Add initial height to obtain final height (4.7 m). 0.5 mark: State correct unit (m).

First, calculate the total resistance of the wire using Ohm's Law: \(R = \frac{V}{I} = \frac{3.0\text{ V}}{1.5\text{ A}} = 2.0\ \Omega\). Next, calculate the resistance per unit length by dividing the resistance by the length of the wire: \(\text{Resistance per unit length} = \frac{2.0\ \Omega}{0.80\text{ m}} = 2.5\ \Omega/\text{m}\).

1 mark: Calculate the total resistance of the wire (2.0 ohms). 1 mark: Divide the resistance by the length (2.0 / 0.80). 1 mark: State correct numerical value (2.5). 0.5 mark: Correct unit (ohms per meter or \(\Omega/\text{m}\)).

At constant temperature, Boyle's Law applies: \(p_1 V_1 = p_2 V_2\). Substituting the known values into the equation: \((1.0 \times 10^5\text{ Pa}) \times (3.0 \times 10^{-5}\text{ m}^3) = p_2 \times (1.2 \times 10^{-5}\text{ m}^3)\). Rearranging to solve for the final pressure: \(p_2 = \frac{3.0}{1.2 \times 10^{-5}} = 2.5 \times 10^5\text{ Pa}\).

1 mark: Recall the formula \(p_1 V_1 = p_2 V_2\). 1 mark: Substitute the given values correctly into the equation. 1 mark: Calculate the correct numerical answer (\(2.5 \times 10^5\)). 0.5 mark: Correct unit (Pa or \(\text{N/m}^2\)).

First, convert the frequency from MHz to Hz: \(120\text{ MHz} = 120 \times 10^6\text{ Hz} = 1.2 \times 10^8\text{ Hz}\). Next, use the wave equation \(v = f \lambda\) and rearrange it to solve for wavelength: \(\lambda = \frac{v}{f} = \frac{3.0 \times 10^8\text{ m/s}}{1.2 \times 10^8\text{ Hz}} = 2.5\text{ m}\).

1 mark: Convert frequency to hertz (\(1.2 \times 10^8\text{ Hz}\)). 1 mark: Correctly recall and rearrange the wave equation to \(\lambda = v / f\). 1 mark: Calculate the correct wavelength (2.5). 0.5 mark: State the correct unit (m).

Alpha particles are highly ionizing but have a very low penetrating power, meaning they can only travel a few centimeters in air and are stopped by thin barriers such as the outer layer of dead skin cells. Therefore, when outside the body, alpha radiation cannot penetrate to damage living organs. However, if inhaled or swallowed, there is no protective skin barrier, and the highly ionizing particles can directly collide with and damage DNA inside delicate living cells, causing mutations and cancer.

1 mark: Mention that alpha particles have very low penetrating power and are stopped by skin or travel only a short distance in air. 1 mark: State that alpha particles are highly ionizing. 1 mark: Explain that inside the body, they can directly contact living cells and damage DNA. 0.5 mark: Identify the resulting risk such as cancer or mutation.

First, convert the charging time from hours to seconds: \(2.5\text{ hours} = 2.5 \times 3600\text{ s} = 9000\text{ s}\). Next, use the charge equation \(Q = I \times t\) to calculate the charge: \(Q = 1.8\text{ A} \times 9000\text{ s} = 16200\text{ C}\).

1 mark: Convert hours to seconds correctly (9000 s). 1 mark: Recall the formula \(Q = I \times t\). 1 mark: Calculate correct charge value (16200). 0.5 mark: Give the correct unit (C or Coulombs).

### Method and Measurements
1. Measure the mass \(m\) of the toy car using an electronic balance.
2. Set up a ramp with two light gates connected to a data logger. Place one light gate near the top of the ramp and the second near the bottom.
3. Attach a card of known length \(d\) to the top of the car.
4. Measure the vertical height \(h_1\) of the first light gate above the bench, and the vertical height \(h_2\) of the second light gate above the bench. The vertical drop is \(h = h_1 - h_2\).
5. Release the car from rest above the first light gate.
6. The data logger records the time taken for the card to pass through each light gate (\(t_1\) and \(t_2\)).

### Calculations
- Calculate the initial velocity \(u = \frac{d}{t_1}\) and the final velocity \(v = \frac{d}{t_2}\).
- Calculate the loss in gravitational potential energy: \(\Delta E_p = m g h\), where \(g = 10\text{ N/kg}\) (or \(9.8\text{ N/kg}\)).
- Calculate the gain in kinetic energy: \(\Delta E_k = \frac{1}{2} m v^2 - \frac{1}{2} m u^2\).

### Analysis and Evaluation
- Compare \(\Delta E_p\) and \(\Delta E_k\). Ideally, if energy is conserved, \(\Delta E_p = \Delta E_k\).
- In practice, \(\Delta E_k < \Delta E_p\) because some energy is transferred to the surroundings as thermal energy due to work done against friction and air resistance.
- Errors can be minimized by: using a low-friction ramp/bearings, ensuring the card passes cleanly through the light gates, and repeating the experiment to find a mean.

**Level 3 (5–6 marks)**
Provides a detailed, logical description of a workable procedure. Specifies how to measure mass, vertical height change, and velocities (ideally using light gates and a card of known length). Correctly states the equations for calculating both \(\Delta E_p\) and \(\Delta E_k\). Correctly identifies that \(\Delta E_k\) will be less than \(\Delta E_p\) due to work done against friction, and suggests at least one sensible way to reduce friction or experimental error.

**Level 2 (3–4 marks)**
Describes a largely correct procedure. Identifies most of the necessary measurements (e.g., mass, heights, speeds/times). States at least one correct energy equation. Acknowledges that energy is lost as heat/friction, but may lack detail on how to calculate both energies precisely or how to minimize errors.

**Level 1 (1–2 marks)**
Offers a basic outline of the experiment. Recognizes that mass and height or speed need to be measured. May mention gravitational potential energy transferring to kinetic energy, but calculations or apparatus descriptions are incomplete or contain significant errors.

**Level 0 (0 marks)**
No relevant content.

### Experimental Setup and Procedure
1. Connect the 12 V filament lamp in series with an ammeter, a variable resistor (or connect to a variable DC power supply), and a switch.
2. Connect a voltmeter in parallel across the filament lamp.
3. Close the switch and adjust the variable resistor (or power supply) so that the voltmeter reads a low value (e.g., 1.0 V). Record the reading on the voltmeter (\(V\)) and the ammeter (\(I\)).
4. Vary the resistance of the variable resistor to change the potential difference in steps of 1.0 V or 2.0 V, up to 12.0 V, recording the current at each step.
5. Switch off the circuit between readings to prevent the lamp from overheating before the next measurement.
6. Calculate the resistance at each point using \(R = \frac{V}{I}\).

### Expected Relationship and Particle Explanation
- As the potential difference increases, the current increases, but at a decreasing rate. The graph of current against potential difference is a curve that gets less steep at higher voltages.
- This indicates that the resistance of the filament increases as the temperature of the filament increases.
- At a particle level, as potential difference increases, more electrons flow through the metal filament per second (larger current).
- These moving electrons collide with the positive metal ions in the lattice structure of the filament, transferring kinetic energy to them. This causes the ions to vibrate with greater amplitude (the temperature increases).
- The increased vibration of the metal ions makes it more difficult for the free electrons to pass through, increasing the rate of collisions and thus increasing the electrical resistance.

**Level 3 (5–6 marks)**
Provides a detailed and correct description of the circuit setup (voltmeter in parallel with the lamp, ammeter in series, and a method of varying potential difference such as a variable resistor or variable power supply). Explains how to obtain a wide range of readings systematically up to 12 V. Correctly describes that current increases at a decreasing rate (non-linear relationship) and provides a clear, detailed particle-level explanation (electrons colliding with vibrating metal ions, causing greater vibration/heat and resisting electron flow).

**Level 2 (3–4 marks)**
Describes a workable circuit with correct placement of the meters. Outlines a method to vary the voltage and take readings. States that resistance increases with temperature/voltage, but the explanation in terms of particles is incomplete (e.g., mentions collisions but not the increased vibration of the ions, or vice versa).

**Level 1 (1–2 marks)**
Identifies the basic components needed (ammeter, voltmeter, lamp). States that readings of current and voltage should be taken. May mention that resistance changes or that the lamp gets hot, but fails to explain why in terms of particles.

**Level 0 (0 marks)**
No relevant content.

1. Calculate initial Gravitational Potential Energy (GPE): \(GPE = m \times g \times h = 250 \times 9.8 \times 18 = 44100\text{ J}\). 2. Calculate final Kinetic Energy (KE): \(KE = \frac{1}{2} m v^2 = 0.5 \times 250 \times 15^2 = 28125\text{ J}\). 3. Apply conservation of energy: \\text{Energy lost to surroundings} = \\text{Initial GPE} - \\text{Final KE} = 44100 - 28125 = 15975\\text{ J}\.

1 mark: Correct calculation of initial gravitational potential energy (44100 J). 1 mark: Correct calculation of final kinetic energy (28125 J). 1 mark: Identification that the energy difference equals energy transferred to surroundings. 1 mark: Correct final answer of 15975 J (allow 16000 J or 16 kJ).

1. Calculate pressure exerted by water: \(P_{\text{water}} = \rho g h = 1000 \times 9.8 \times 1.2 = 11760\text{ Pa}\). 2. Calculate total pressure: \(P_{\text{total}} = P_{\text{water}} + P_{\text{atmosphere}} = 11760 + 101000 = 112760\text{ Pa}\). 3. Calculate total force: \(F = P \times A = 112760 \times 0.05 = 5638\text{ N}\).

1 mark: Correct calculation of pressure due to water column (11760 Pa). 1 mark: Addition of atmospheric pressure to obtain total pressure (112760 Pa). 1 mark: Rearranging pressure formula to F = P x A. 1 mark: Correct calculation of total force (5638 N or 5.64 kN).

1. Place the thermistor in a beaker of water (water bath) along with a thermometer. 2. Set up an electrical circuit containing a power supply, an ammeter connected in series with the thermistor, and a voltmeter connected in parallel across the thermistor. 3. Heat the water slowly using an electric heater. 4. At regular temperature intervals (e.g. every \(10^\circ\text{C}\)), record the temperature, potential difference (voltmeter reading), and current (ammeter reading). 5. Calculate the resistance at each temperature using the equation \(R = \frac{V}{I}\).

1 mark: Mention of using a water bath and thermometer to control and measure temperature. 1 mark: Correct circuit setup showing ammeter in series and voltmeter in parallel with the thermistor. 1 mark: Explaining the collection of current and voltage readings at regular temperature intervals. 1 mark: Use of the equation R = V / I to compute the resistance.

1. Calculate refractive index of glass: \(n = \frac{\text{speed in air}}{\text{speed in glass}} = \frac{3.00 \times 10^8}{1.97 \times 10^8} \approx 1.523\). 2. Use Snell's Law: \(n_{\text{air}} \sin(\theta_{\text{air}}) = n_{\text{glass}} \sin(\theta_{\text{glass}})\). 3. \(1.00 \times \sin(42^\circ) = 1.523 \times \sin(\theta_{\text{glass}})\). 4. \(\sin(\theta_{\text{glass}}) = \frac{\sin(42^\circ)}{1.523} = \frac{0.6691}{1.523} \approx 0.4393\). 5. \(\theta_{\text{glass}} = \arcsin(0.4393) \approx 26.1^\circ\).

1 mark: Calculation of the glass refractive index (1.52). 1 mark: Recalling or writing down Snell's law equation. 1 mark: Correct substitution of values into the formula. 1 mark: Correct calculation of the final angle of refraction (26.1 degrees).

1. Find total mass of the system: \(m_{\text{total}} = 1800 + 800 = 2600\text{ kg}\). 2. Find net force on the whole system: \(F_{\text{net}} = 5200 - 1200 - 600 = 3400\text{ N}\). 3. Calculate acceleration: \(a = \frac{F_{\text{net}}}{m_{\text{total}}} = \frac{3400}{2600} \approx 1.308\text{ m/s}^2\). 4. Analyze forces on the trailer to find tension \(T\): \(F_{\text{net, trailer}} = T - 600\). By Newton's second law: \(T - 600 = 800 \times 1.308 = 1046.4\text{ N}\). Hence, \(T = 1046.4 + 600 = 1646.4\text{ N}\).

1 mark: Calculation of the overall system acceleration (approx 1.31 m/s^2). 1 mark: Formulation of the force equation for either the trailer or the van. 1 mark: Correct substitution of values. 1 mark: Correct final calculation of tension (1646 N; allow answers in range 1640 N - 1650 N depending on rounding).

1. Setup: Place a beta source, aluminium sheets of varying thickness, and a Geiger-Muller (GM) tube connected to a counter in a straight line with a constant distance between source and detector. 2. Procedure: First, measure the background radiation count rate over a set time (e.g., 5 minutes) without the source present. Then place the source and record the count rate with no absorber. Insert aluminium sheets of known, increasing thicknesses between the source and GM tube, and record the count rate for each. 3. Data processing: Subtract the background count rate from each reading to obtain the corrected count rate. 4. Safety: Use long-handled tongs to move the source, keep the source directed away from the body, and store the source in a lead-lined container when not in use.

1 mark: Method to measure and subtract background radiation to get corrected readings. 1 mark: Describing a clear physical setup with constant distance between the beta source and the GM detector. 1 mark: Systematically varying the thickness of aluminium sheets and recording count rates. 1 mark: Identifying key safety measures (tongs, direction of source, lead storage).

1. Identify that frequency remains constant at 512 Hz during transition. 2. Calculate wavelength in air: \(\lambda_{\text{air}} = \frac{v_{\text{air}}}{f} = \frac{340}{512} \approx 0.664\text{ m}\). 3. Calculate wavelength in water: \(\lambda_{\text{water}} = \frac{v_{\text{water}}}{f} = \frac{1480}{512} \approx 2.891\text{ m}\). 4. Calculate change in wavelength: \(\Delta \lambda = \lambda_{\text{water}} - \lambda_{\text{air}} = 2.891 - 0.664 = 2.227\text{ m}\).

1 mark: Stating or using the principle that the frequency of the wave does not change. 1 mark: Correctly calculating the wavelength in air (0.66 m) or water (2.89 m). 1 mark: Calculating both wavelengths correctly. 1 mark: Finding the correct difference (2.23 m; accept 2.2 m).

1. Recall the magnetic force formula for a single wire: \(F = B I L\). 2. Since there are \(N\) turns, the force acting on one side of the coil is multiplied by \(N\): \(F = N B I L\). 3. Substitute values: \(F = 50 \times 0.15 \times 2.4 \times 0.08\). 4. Calculate final force: \(F = 7.5 \times 2.4 \times 0.08 = 1.44\text{ N}\).

1 mark: Correctly recalling or applying the formula F = B I L. 1 mark: Realizing that the force must be multiplied by the number of turns N = 50. 1 mark: Correct substitution of all quantities. 1 mark: Correct final calculation of force (1.44 N).

Using Boyle's Law: \(P_1 V_1 = P_2 V_2\). Substitute the given values: \(1.01 \times 10^5\text{ Pa} \times 24.0\text{ cm}^3 = P_2 \times 15.0\text{ cm}^3\). Solve for \(P_2\): \(P_2 = \frac{1.01 \times 10^5 \times 24.0}{15.0} = 1.616 \times 10^5\text{ Pa}\). Rounded to 3 significant figures, this is \(1.62 \times 10^5\text{ Pa}\). Pushing the piston slowly is necessary because work done on the gas transfers energy to its thermal store, which would raise its temperature. Pushing it slowly allows time for this thermal energy to be transferred to the surroundings, keeping the temperature constant.

1 mark for selecting and rearranging the formula \(P_1 V_1 = P_2 V_2\). 1 mark for correct calculation of \(1.62 \times 10^5\text{ Pa}\). 1 mark for stating that compressing the gas transfers energy to its thermal store (raising temperature). 1.33 marks for explaining that slow movement allows thermal energy to dissipate to the surroundings to maintain constant temperature.

First, calculate the initial gravitational potential energy (GPE): \(E_p = mgh = 350 \times 10 \times 22 = 77,000\text{ J}\). Next, calculate the final kinetic energy (KE): \(E_k = \frac{1}{2}mv^2 = 0.5 \times 350 \times 18^2 = 56,700\text{ J}\). Finally, calculate efficiency: \(\text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}} = \frac{56,700}{77,000} \approx 0.7364\). Expressed as a percentage to 2 significant figures, this is \(74\%\).

1 mark for calculating GPE as \(77,000\text{ J}\). 1 mark for calculating KE as \(56,700\text{ J}\). 1 mark for utilizing the efficiency formula. 1.33 marks for the final correct answer of \(74\%\) (or \(0.74\)) to 2 significant figures.

Calculate total electrical power: \(P = I \times V = 1.5\text{ A} \times 6.0\text{ V} = 9.0\text{ W}\). Since \(20\%\) is transferred as light, the remaining \(80\%\) is transferred as thermal energy. The rate of thermal energy transfer (thermal power) is: \(9.0\text{ W} \times 0.80 = 7.2\text{ W}\) (or \(7.2\text{ J/s}\)).

1 mark for recalling and using \(P = I \times V\). 1 mark for calculating total power as \(9.0\text{ W}\). 1 mark for identifying that \(80\%\) of the power is wasted as thermal energy. 1.33 marks for the final rate of \(7.2\text{ W}\) (or \(7.2\text{ J/s}\)).

First, find the refractive index \(n\) using Snell's Law: \(n = \frac{\sin(i)}{\sin(r)} = \frac{\sin(42^\circ)}{\sin(26^\circ)} = \frac{0.6691}{0.4384} \approx 1.526\). Next, calculate the speed of light in glass: \(v = \frac{c}{n} = \frac{3.0 \times 10^8\text{ m/s}}{1.526} \approx 1.966 \times 10^8\text{ m/s}\). To 3 significant figures, this is \(1.97 \times 10^8\text{ m/s}\).

1 mark for calculating the refractive index as \(1.53\) (or \(1.526\)). 1 mark for using \(v = \frac{c}{n}\). 2.33 marks for the final calculated speed of \(1.97 \times 10^8\text{ m/s}\) (accept \(1.96 \times 10^8\text{ m/s}\) to \(1.97 \times 10^8\text{ m/s}\) depending on rounding steps).

Use Newton's second law to find the deceleration: \(F = ma \implies a = \frac{F}{m} = \frac{-4800\text{ N}}{1200\text{ kg}} = -4.0\text{ m/s}^2\). (Deceleration magnitude is \(4.0\text{ m/s}^2\)). To find the braking distance \(s\), use the equation of motion: \(v^2 - u^2 = 2as\). Substituting the values: \(0^2 - 20^2 = 2 \times (-4.0) \times s \implies -400 = -8.0 \times s \implies s = 50\text{ m}\).

1 mark for calculating deceleration using \(F = ma\) to get \(4.0\text{ m/s}^2\). 1 mark for choosing a correct equation of motion (e.g., \(v^2 - u^2 = 2as\)). 1 mark for correct substitution of values. 1.33 marks for calculating the final braking distance of \(50\text{ m}\).

The distance from the first to the sixth wavefront corresponds to 5 full wavelengths. Therefore, wavelength \(\lambda = \frac{8.5\text{ cm}}{5} = 1.7\text{ cm} = 0.017\text{ m}\). Using the wave equation: \(v = f \lambda = 15\text{ Hz} \times 0.017\text{ m} = 0.255\text{ m/s}\). Rounded to 2 significant figures, this is \(0.26\text{ m/s}\).

1 mark for stating that the distance between 1st and 6th wavefronts is 5 wavelengths. 1 mark for calculating wavelength as \(1.7\text{ cm}\) (or \(0.017\text{ m}\)). 1 mark for using the wave speed equation \(v = f \lambda\). 1.33 marks for the final wave speed of \(0.26\text{ m/s}\) (or \(0.255\text{ m/s}\)).

Transformers work via electromagnetic induction: an alternating current in the primary coil creates a continuously changing magnetic field in the iron core, which induces an alternating potential difference across the secondary coil. For an ideal transformer, input electrical power equals output electrical power: \(V_p I_p = V_s I_s\). Substituting the values: \(25,000\text{ V} \times 4,000\text{ A} = 400,000\text{ V} \times I_s\). Therefore, \(I_s = \frac{1.0 \times 10^8\text{ W}}{400,000\text{ V}} = 250\text{ A}\).

1 mark for mentioning electromagnetic induction / changing magnetic field inducing potential difference. 1 mark for recognizing that primary power equals secondary power for an ideal transformer. 1 mark for correct substitution into \(V_p I_p = V_s I_s\). 1.33 marks for the correct transmission current of \(250\text{ A}\).

Each half-value thickness reduces the intensity by half: \(100\% \rightarrow 50\% \rightarrow 25\% \rightarrow 12.5\%\). This process requires exactly 3 half-value thicknesses. Since each thickness is \(2.0\text{ mm}\), the total thickness required is \(3 \times 2.0\text{ mm} = 6.0\text{ mm}\). Other safety practices include standing behind a protective screen, maximizing the distance from the X-ray tube (using the inverse-square law), or using a personal dosimeter badge to monitor exposure.

1 mark for identifying that reduction to \(12.5\%\) represents 3 half-value thicknesses (or \((0.5)^3\)). 1 mark for calculating the thickness of lead as \(6.0\text{ mm}\). 1 mark for stating a valid non-shielding safety practice (e.g., maximizing distance from the source or minimizing exposure time). 1.33 marks for a clear explanation of how the safety practice lowers radiation dose.

First, calculate the final internal pressure (p2) using Boyle's Law: p1 * V1 = p2 * V2. Substituting the values gives 101 kPa * 20.0 cm^3 = p2 * 12.5 cm^3, which solves to p2 = 161.6 kPa (or 161,600 Pa). Second, calculate the pressure difference across the plunger: delta p = p2 - p_atm = 161.6 kPa - 101 kPa = 60.6 kPa (or 60,600 Pa). Third, calculate the external force required to balance this pressure difference using F = delta p * A: F = 60,600 Pa * 4.0 * 10^-4 m^2 = 24.24 N. Rounded to 2 or 3 significant figures, this is 24 N or 24.2 N.

1 mark for calculating the final internal pressure of 161.6 kPa (or 161,600 Pa) using Boyle's Law. 1 mark for finding the pressure difference of 60.6 kPa (or 60,600 Pa). 1 mark for substituting the pressure difference and area into the force formula (F = p * A). 1.33 marks for the final calculated force of 24 N or 24.2 N (accept 24.24 N).

First, calculate the gravitational potential energy lost by the car: GPE = m * g * h = 450 kg * 10 m/s^2 * 28 m = 126,000 J. Second, calculate the kinetic energy gained by the car at the bottom: KE = 0.5 * m * v^2 = 0.5 * 450 kg * (21 m/s)^2 = 99,225 J. Third, calculate the work done against resistive forces (energy transferred to the surroundings as heat and sound): Work = GPE - KE = 126,000 J - 99,225 J = 26,775 J. Fourth, use the work equation (Work = Force * distance) to find the average resistive force: Force = Work / distance = 26,775 J / 55 m = 486.8 N, which rounds to 487 N (or 490 N to 2 significant figures).

1 mark for calculating the initial gravitational potential energy as 126,000 J. 1 mark for calculating the final kinetic energy as 99,225 J. 1 mark for calculating the work done against resistance (energy loss) as 26,775 J. 1.33 marks for the final average resistive force of 487 N or 486.8 N or 490 N (accept 441 N if g = 9.8 m/s^2 is used instead).