2024 Edexcel IGCSE Physics Practice Paper with Answers

To find the new pressure of the gas, we use the relationship between pressure and Kelvin temperature for a fixed volume of gas (Gay-Lussac's Law):

1. Convert both temperatures from degrees Celsius to Kelvin:
\( T_1 = 27 + 273 = 300\text{ K} \)
\( T_2 = 327 + 273 = 600\text{ K} \)

2. State the gas law equation:
\( \frac{P_1}{T_1} = \frac{P_2}{T_2} \)

3. Rearrange the formula to solve for the final pressure, \( P_2 \):
\( P_2 = P_1 \times \frac{T_2}{T_1} \)

4. Substitute the known values:
\( P_2 = 100\text{ kPa} \times \frac{600\text{ K}}{300\text{ K}} = 200\text{ kPa} \)

- 1 mark for converting both temperatures correctly into Kelvin (\( 300\text{ K} \) and \( 600\text{ K} \)).
- 1 mark for recalling and substituting into the pressure law equation \( P_1/T_1 = P_2/T_2 \).
- 0.5 marks for calculating the correct final pressure with units (\( 200\text{ kPa} \)).

For massive stars (much larger than the Sun):
- They start their stable lifetime as a main sequence star.
- When hydrogen fuel runs low, they expand and cool to form a red supergiant.
- They eventually undergo a violent explosion called a supernova.
- The highly dense core remaining after the supernova collapses to form either a neutron star or, if massive enough, a black hole.

- 1 mark for identifying red supergiant as the stage immediately following the main sequence for a high-mass star.
- 1 mark for identifying the supernova explosion.
- 0.5 marks for identifying the end products as a neutron star or black hole.

We can determine the final velocity using the principle of conservation of momentum:

1. Calculate total momentum before the collision:
\( p_{\text{initial}} = m_1 u_1 + m_2 u_2 \)
\( p_{\text{initial}} = (1.5\text{ kg} \times 4.0\text{ m/s}) + (2.5\text{ kg} \times 0\text{ m/s}) = 6.0\text{ kg m/s} \)

2. Express total momentum after the collision:
\( p_{\text{final}} = (m_1 + m_2) \times v \)
\( p_{\text{final}} = (1.5\text{ kg} + 2.5\text{ kg}) \times v = 4.0\text{ kg} \times v \)

3. Since momentum is conserved:
\( p_{\text{initial}} = p_{\text{final}} \)
\( 6.0\text{ kg m/s} = 4.0\text{ kg} \times v \)

4. Rearrange to find \( v \):
\( v = \frac{6.0}{4.0} = 1.5\text{ m/s} \)

- 1 mark for calculating the total initial momentum as \( 6.0\text{ kg m/s} \).
- 1 mark for setting up the conservation equation \( p_{\text{initial}} = (m_1 + m_2)v \).
- 0.5 marks for the correct final velocity value with units (\( 1.5\text{ m/s} \)).

We use Snell's Law to calculate the refractive index of the glass block:

1. State the formula for the refractive index:
\( n = \frac{\sin(i)}{\sin(r)} \)

2. Substitute the angle of incidence, \( i = 45^\circ \), and angle of refraction, \( r = 28^\circ \):
\( n = \frac{\sin(45^\circ)}{\sin(28^\circ)} \)

3. Calculate the sine values:
\( \sin(45^\circ) \approx 0.7071 \)
\( \sin(28^\circ) \approx 0.4695 \)

4. Divide the two values:
\( n = \frac{0.7071}{0.4695} \approx 1.51 \)

- 1 mark for stating the correct relationship for Snell's law: \( n = \sin(i) / \sin(r) \).
- 1 mark for substitution of the correct angle values: \( n = \sin(45^\circ) / \sin(28^\circ) \).
- 0.5 marks for obtaining the correct refractive index value (\( 1.51 \)).

**(a)**
(i) The acceleration is equal to the gradient (or slope) of the tangent to the curve at that instant.
(ii) The distance travelled is equal to the area under the velocity-time graph for that time interval.

**(b)**
- At the moment of release (\(t = 0\)), the only force acting is weight downwards (drag is zero), resulting in a high downward acceleration.
- As the speed of the sphere increases, the upward drag force (viscous fluid resistance) increases.
- Since \(\text{Resultant Force} = \text{Weight} - \text{Drag}\), the net downward force decreases, causing the acceleration to decrease.
- Eventually, at \(t = 0.6\text{ s}\), the upward drag force equals the downward weight. The resultant force becomes zero, so the acceleration becomes zero, and the sphere continues at a constant terminal velocity.

**(c)**
- \(\text{Acceleration} = \text{gradient of the tangent} = \frac{\Delta v}{\Delta t}\)
- \(\text{gradient} = \frac{1.45\text{ m/s} - 0.15\text{ m/s}}{0.2\text{ s} - 0\text{ s}} = \frac{1.30}{0.2} = 6.5\text{ m/s}^2\)

**(d)**
- Between \(t = 0.6\text{ s}\) and \(t = 1.5\text{ s}\), the sphere travels at a constant terminal velocity of \(2.2\text{ m/s}\).
- \(\text{Distance} = \text{speed} \times \text{time}\)
- \(\text{time interval} = 1.5\text{ s} - 0.6\text{ s} = 0.9\text{ s}\)
- \(\text{Distance} = 2.2\text{ m/s} \times 0.9\text{ s} = 1.98\text{ m}\)

**(a)(i)**
- Mention of gradient / slope of the graph / tangent (1)

**(a)(ii)**
- Mention of area under the graph (1)

**(b)**
- Identifies weight acts downwards / drag is zero at start (1)
- States that drag increases as speed/velocity increases (1)
- Explains that resultant force decreases, hence acceleration decreases (1)
- Explains that when drag equals weight, resultant force is zero, leading to terminal/constant velocity (1)

**(c)**
- Recall of \(\text{acceleration} = \text{gradient}\) or showing substitution: \(\frac{1.45 - 0.15}{0.2 - 0}\) (1)
- Correct calculation of change in velocity (\(1.3\)) and change in time (\(0.2\)) (1)
- Correct final value with correct unit: \(6.5\text{ m/s}^2\) (0.5)

**(d)**
- State or use: \(\text{distance} = \text{velocity} \times \text{time}\) or area under horizontal line (1)
- Calculation of correct time interval: \(0.9\text{ s}\) (1)
- Correct calculation of distance with correct unit: \(1.98\text{ m}\) (accept \(2.0\text{ m}\) from rounding) (1)

**(a)**
- As the voltage increases, the current increases.
- The relationship is non-linear / the current increases at a decreasing rate (the graph would curve, showing that it does not obey Ohm's Law).

**(b)**
- As voltage increases, current increases, which causes more collisions between electrons and metal ions, increasing the temperature of the filament.
- The metal ions in the filament lattice vibrate with greater amplitude / faster.
- This increased vibration increases the frequency of collisions between the flowing conduction electrons and the lattice ions, obstructing the flow of current and therefore increasing resistance.

**(c)**
- Formula: \(R = \frac{V}{I}\)
(i) At \(4.0\text{ V}\), \(I = 0.85\text{ A}\):
\(R = \frac{4.0\text{ V}}{0.85\text{ A}} \approx 4.71\ \Omega\) (or \(4.7\ \Omega\))
(ii) At \(12.0\text{ V}\), \(I = 1.55\text{ A}\):
\(R = \frac{12.0\text{ V}}{1.55\text{ A}} \approx 7.74\ \Omega\) (or \(7.7\ \Omega\))

**(d)**
(i) Looking at the data table, when the current is \(1.10\text{ A}\), the voltage across the filament lamp is \(6.0\text{ V}\).
(ii)
- The voltage across the \(6.0\ \Omega\) resistor is: \(V_R = I \times R = 1.10\text{ A} \times 6.0\ \Omega = 6.6\text{ V}\).
- The total potential difference of the power supply is the sum of the voltages across the lamp and the resistor (in series):
\(V_{\text{total}} = V_{\text{lamp}} + V_R = 6.0\text{ V} + 6.6\text{ V} = 12.6\text{ V}\).

**(a)**
- Current increases as voltage increases (1)
- Current increases at a decreasing rate / relationship is non-linear / lamp does not obey Ohm's law (1)

**(b)**
- An increase in current/voltage leads to an increase in temperature of the filament (1)
- Metal ions/atoms in the lattice vibrate more/with greater amplitude (1)
- Resulting in more frequent collisions with moving electrons, which resists current flow (1)

**(c)(i)**
- Uses formula \(R = V/I\) with correct values from the table: \(4.0 / 0.85\) (1)
- Correct calculation: \(4.7\ \Omega\) (or \(4.71\ \Omega\)) (1)

**(c)(ii)**
- Correct substitution of values: \(12.0 / 1.55\) (1)
- Correct calculation with the correct unit \(\Omega\) or ohms shown in at least one part of (c): \(7.7\ \Omega\) (or \(7.74\ \Omega\)) (1)

**(d)(i)**
- Identifies \(6.0\text{ V}\) from the table when current is \(1.10\text{ A}\) (1)

**(d)(ii)**
- Calculates potential difference across the resistor: \(1.10\text{ A} \times 6.0\ \Omega = 6.6\text{ V}\) (0.5)
- Adds the two voltages to get total power supply voltage with correct unit: \(12.6\text{ V}\) (1)

Part (a):
1. Calculate the weight of the rocket: \(W = m \times g = 0.25\text{ kg} \times 10\text{ m/s}^2 = 2.5\text{ N}\).
2. Determine the net upward force: \(F_{\text{net}} = T - W = 6.0\text{ N} - 2.5\text{ N} = 3.5\text{ N}\).
3. Calculate acceleration: \(a = \frac{F_{\text{net}}}{m} = \frac{3.5\text{ N}}{0.25\text{ kg}} = 14\text{ m/s}^2\).

Part (b):
1. Use \(v = u + at\) to find the speed: \(v = 0 + (14 \times 1.5) = 21\text{ m/s}\).
2. Use \(s = ut + \frac{1}{2}at^2\) to find the height: \(s = 0 + 0.5 \times 14 \times (1.5)^2 = 7 \times 2.25 = 15.75\text{ m}\).

Part (c):
1. Once the motor stops, the acceleration is \(-10\text{ m/s}^2\) (gravity acting downwards). The rocket has an initial speed of \(21\text{ m/s}\) and a starting height of \(15.75\text{ m}\).
2. Use \(v^2 = u^2 + 2as\) to find the extra height \(s_{\text{extra}}\) reached before stopping (\(v = 0\)):
\(0 = 21^2 + 2 \times (-10) \times s_{\text{extra}}\)
\(20 \times s_{\text{extra}} = 441\)
\(s_{\text{extra}} = 22.05\text{ m}\).
3. Calculate total height: \(h_{\text{max}} = 15.75 + 22.05 = 37.8\text{ m}\).

Part (d):
1. Calculate momentum: \(p = m \times v = 0.25\text{ kg} \times 4.5\text{ m/s} = 1.125\text{ kg m/s}\) downwards.
2. Since the rocket is descending at terminal velocity (constant speed), the acceleration is zero and the forces are balanced. The downward weight of \(2.5\text{ N}\) is equal in magnitude and opposite in direction to the upward drag force of \(2.5\text{ N}\).

Part (a) [3 marks]:
- 1 mark for calculating weight of 2.5 N.
- 1 mark for finding net force of 3.5 N (Thrust - Weight).
- 1 mark for correct division to show acceleration of 14 m/s^2.

Part (b) [3 marks]:
- 1 mark for correct calculation of final speed (21 m/s).
- 1 mark for correct calculation of height (15.75 m).
- 1 mark for correct units (m/s and m) stated.

Part (c) [3 marks]:
- 1 mark for identifying that acceleration is -10 m/s^2 and setting up the v^2 = u^2 + 2as equation.
- 1 mark for calculating the extra height as 22.05 m.
- 1 mark for calculating the correct total maximum height of 37.8 m (accept 38 m).

Part (d) [3 marks]:
- 1 mark for correct calculation of momentum with correct unit (1.125 kg m/s or N s).
- 1 mark for stating that weight and drag/air resistance are acting on the rocket.
- 1 mark for explaining that these forces are equal/balanced because the rocket is at terminal/constant velocity.

Part (a):
(i) \(\text{density} = \frac{\text{mass}}{\text{volume}}\)
(ii) \(V = \frac{m}{\rho} = \frac{1.2\text{ kg}}{600\text{ kg/m}^3} = 0.0020\text{ m}^3\)
(iii) Since the block is floating, the upthrust equals its weight: \(\text{Upthrust} = W = m \times g = 1.2\text{ kg} \times 10\text{ N/kg} = 12\text{ N}\).

Part (b):
(i) \(p = h \times \rho \times g\) (pressure difference = height \(\times\) density \(\times\) g)
(ii) \(p = 0.35\text{ m} \times 1000\text{ kg/m}^3 \times 10\text{ N/kg} = 3500\text{ Pa}\) (or \(3.5\text{ kPa}\)).

Part (c):
(i) For a floating object, the upthrust equals the weight of the fluid displaced (Archimedes' principle).
Weight of water displaced = Upthrust = \(12\text{ N}\).
Mass of water displaced = \(\frac{\text{Weight}}{g} = \frac{12}{10} = 1.2\text{ kg}\).
Volume of water displaced = \(\frac{\text{Mass}}{\text{Density of water}} = \frac{1.2\text{ kg}}{1000\text{ kg/m}^3} = 0.0012\text{ m}^3\).
(ii) The increase in water height \(\Delta h\) is given by the volume of displaced water divided by the base area of the cylinder:
\(\Delta h = \frac{\Delta V}{A} = \frac{0.0012\text{ m}^3}{0.050\text{ m}^2} = 0.024\text{ m}\).
New depth = \(\text{original depth} + \Delta h = 0.35\text{ m} + 0.024\text{ m} = 0.374\text{ m}\).

Part (a) [4 marks]:
- (i) 1 mark for correct equation (density = mass/volume).
- (ii) 1 mark for substitution and 1 mark for correct calculation of volume (0.0020 m^3) with units.
- (iii) 1 mark for stating upthrust is 12 N.

Part (b) [4 marks]:
- (i) 1 mark for correct equation (p = h * rho * g).
- (ii) 1 mark for substitution, 1 mark for correct calculation (3500), and 1 mark for correct unit (Pa or N/m^2).

Part (c) [4 marks]:
- (i) 1 mark for linking weight of displaced water to the upthrust or mass of block, and 1 mark for dividing by density to get 0.0012 m^3.
- (ii) 1 mark for dividing displaced volume by base area to find height increase (0.024 m), and 1 mark for correct final depth of 0.374 m.

Part (a):
(i) In series, resistances add: \(R_{\text{total}} = R_{\text{fixed}} + R_{\text{thermistor}} = 15\ \Omega + 25\ \Omega = 40\ \Omega\).
(ii) Use Ohm's law: \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{40\ \Omega} = 0.30\text{ A}\).
(iii) PD across thermistor: \(V_{\text{thermistor}} = I \times R_{\text{thermistor}} = 0.30\text{ A} \times 25\ \Omega = 7.5\text{ V}\).

Part (b):
(i) Current through the series circuit is the same everywhere. Since the voltage across the fixed resistor is \(9.0\text{ V}\):
\(I = \frac{V_{\text{fixed}}}{R_{\text{fixed}}} = \frac{9.0\text{ V}}{15\ \Omega} = 0.60\text{ A}\).
(ii) The total resistance of the circuit is now: \(R_{\text{total}} = \frac{V_{\text{supply}}}{I} = \frac{12\text{ V}}{0.60\text{ A}} = 20\ \Omega\).
Therefore, \(R_{\text{thermistor}} = R_{\text{total}} - R_{\text{fixed}} = 20\ \Omega - 15\ \Omega = 5.0\ \Omega\).
(Alternatively, voltage across thermistor is \(12\text{ V} - 9.0\text{ V} = 3.0\text{ V}\). \(R_{\text{thermistor}} = \frac{3.0\text{ V}}{0.60\text{ A}} = 5.0\ \Omega\).)

Part (c):
(i) Use \(Q = I \times t\). Time \(t = 3.0\text{ minutes} = 180\text{ s}\). Current \(I = 0.60\text{ A}\).
\(Q = 0.60\text{ A} \times 180\text{ s} = 108\text{ C}\).
(ii) Use \(E = V \times Q\) or \(E = V \times I \times t\):
\(E = 12\text{ V} \times 108\text{ C} = 1296\text{ J}\) (or \(1.3\text{ kJ}\)).

Part (a) [4 marks]:
- (i) 1 mark for total resistance of 40 ohms.
- (ii) 1 mark for substitution into Ohm's law, 1 mark for current of 0.30 A with unit.
- (iii) 1 mark for PD of 7.5 V (or 12 V - (0.3 A * 15 ohms) = 7.5 V).

Part (b) [4 marks]:
- (i) 1 mark for using Ohm's law with fixed resistor values, 1 mark for current of 0.60 A.
- (ii) 1 mark for a correct method to find thermistor resistance (either finding new total resistance or voltage across thermistor), 1 mark for correct thermistor resistance of 5.0 ohms.

Part (c) [4 marks]:
- (i) 1 mark for converting time to seconds (180 s), 1 mark for charge of 108 C with unit.
- (ii) 1 mark for equation (E = VIt or E = VQ), 1 mark for energy of 1296 J (or 1300 J) with unit.

Part (a):
(i) \(\text{GPE} = m \times g \times h\)
(ii) \(\text{GPE} = 45\text{ kg} \times 10\text{ N/kg} \times 8.0\text{ m} = 3600\text{ J}\).
(iii) The minimum work done on the block is equal to the gain in gravitational potential energy, which is \(3600\text{ J}\).

Part (b):
(i) \(\text{Power} = \text{current} \times \text{voltage}\) (\(P = I \times V\))
(ii) \(P = 0.50\text{ A} \times 240\text{ V} = 120\text{ W}\).
(iii) \(\text{Electrical Energy} = P \times t = 120\text{ W} \times 40\text{ s} = 4800\text{ J}\).

Part (c):
(i) \(\text{efficiency} = \frac{\text{useful energy output}}{\text{total energy input}} \times 100\%\) (or without \(100\%\) for decimal fraction).
(ii) \(\text{efficiency} = \frac{3600\text{ J}}{4800\text{ J}} \times 100\% = 75\%\) (or \(0.75\)).
(iii) The remaining \(1200\text{ J}\) of energy is wasted. It is transferred to the thermal energy stores of the motor and surroundings (dissipated/scattered) due to electrical resistance in the windings and friction in the moving parts.

Part (a) [4 marks]:
- (i) 1 mark for correct equation (GPE = mgh).
- (ii) 1 mark for substitution and 1 mark for correct calculation of GPE (3600 J) with units.
- (iii) 1 mark for stating work done = 3600 J.

Part (b) [4 marks]:
- (i) 1 mark for correct equation (P = IV).
- (ii) 1 mark for substitution and 1 mark for correct power of 120 W with units.
- (iii) 1 mark for energy of 4800 J.

Part (c) [4 marks]:
- (i) 1 mark for correct efficiency equation.
- (ii) 1 mark for substitution and 1 mark for 75% or 0.75.
- (iii) 1 mark for explaining that wasted energy is dissipated as thermal energy to the motor/surroundings due to friction/resistance.

### Experimental Method:

1. **Circuit Setup (Ammeter-Voltmeter Method)**:
- Connect the thermistor in series with a power supply (battery or cell), an ammeter, and a switch.
- Connect a voltmeter in parallel directly across the thermistor.
- *(Alternative: Connect an ohmmeter directly across the terminals of the thermistor with no power supply).*

2. **Apparatus & Heating Setup**:
- Place the thermistor in a beaker filled with water.
- Place a thermometer in the beaker to monitor the water temperature.
- Use a Bunsen burner, tripod, and gauze (or an electric hotplate) to heat the water. Use ice to bring the initial temperature down near \(0\text{ }^\circ\text{C}\).

3. **Measurements**:
- Record the initial temperature from the thermometer.
- Close the switch, read the current \(I\) on the ammeter and the voltage \(V\) on the voltmeter, then open the switch immediately.
- Calculate the initial resistance using the formula: \(R = \frac{V}{I}\).
- Heat the water slowly. At regular intervals (e.g., every \(10\text{ }^\circ\text{C}\) up to \(100\text{ }^\circ\text{C}\)), stir the water, read the temperature, and quickly take current and voltage readings to calculate the resistance.

4. **Accuracy and Safety Precautions**:
- **Accuracy**: Stir the water continuously before taking readings to ensure that the temperature measured by the thermometer is the same as the temperature of the thermistor.
- **Accuracy**: Ensure the thermometer is read at eye level to avoid parallax error.
- **Accuracy**: Keep the switch open between readings so that current does not flow continuously, which would cause self-heating of the thermistor and lead to systematic temperature errors.
- **Safety**: Wear safety goggles to protect eyes from splashing hot water; keep electrical wires away from the water to prevent short circuits/shocks.

5. **Data Analysis**:
- Plot a graph with resistance \(R\) on the vertical (y) axis and temperature \(T\) on the horizontal (x) axis.
- Draw a curve of best fit. The graph should show that as the temperature increases, the resistance of the NTC thermistor decreases non-linearly.

Award up to a maximum of 10 marks from the following points:

**Circuit Setup (Max 3 marks):**
- **MP1**: Thermistor connected in series with a power supply / cell / battery and an ammeter (or ohmmeter connected directly across thermistor with no power supply) [1]
- **MP2**: Voltmeter connected in parallel across the thermistor (not required if ohmmeter is used) [1]
- **MP3**: Switch included in the circuit to control current flow [1]

**Experimental Procedure (Max 3 marks):**
- **MP4**: Thermistor placed in a beaker of water (water bath) with a thermometer [1]
- **MP5**: Method described to vary temperature (e.g. heating with Bunsen burner/electric hotplate and/or adding ice to lower starting temperature) [1]
- **MP6**: Describes taking readings of temperature and measuring current & voltage (or reading resistance directly) at regular temperature intervals (e.g. every \(10\text{ }^\circ\text{C}\)) [1]

**Accuracy and Safety (Max 3 marks):**
- **MP7**: Stir water before taking readings to ensure uniform temperature [1]
- **MP8**: Switch off circuit/open switch between readings to prevent self-heating of the thermistor [1]
- **MP9**: Avoid parallax error when reading thermometer (e.g. read at eye level) / keep thermometer bulb from touching the beaker bottom [1]
- **MP10**: Safety precaution mentioned (e.g. use heatproof mat, wear safety goggles, keep electrical wires dry) [1]

**Data Analysis (Max 1 mark):**
- **MP11**: Plot a graph of resistance against temperature and draw a curve of best fit to identify the relationship [1]

Part (a):
(i) The pointer deflects in one direction (e.g., to the right) and then returns to zero when the movement stops.
(ii) The pointer remains at zero.
(iii) The pointer deflects in the opposite direction (e.g., to the left) and then returns to zero.

Part (b):
- Pushing the magnet into the coil causes the magnetic field lines to cut through the turns of the coil.
- This cutting of magnetic flux induces a voltage (electromotive force) across the coil, causing a current to flow and the pointer to deflect.
- When the magnet is stationary, there is no cutting of magnetic flux, so no voltage is induced and the pointer is at zero.
- Pulling the magnet out cuts the flux in the opposite direction, inducing a voltage of opposite polarity, hence the opposite deflection.

Part (c):
Any two from:
- Move the magnet faster (increase speed of insertion/removal).
- Use a stronger magnet.
- Use a coil with more turns of wire.
- Insert a soft iron core inside the coil.

Part (a) [4 marks]:
- Pointer deflects to one side when magnet is pushed in (1)
- Pointer returns to zero when movement stops / magnet is fully inside (1)
- Pointer stays at zero when stationary (1)
- Pointer deflects to the opposite side when pulled out (1)

Part (b) [4 marks]:
- Voltage is induced when there is a change in magnetic field / magnetic field lines cut the wire (1)
- Moving magnet cuts the field lines, inducing voltage (1)
- No movement means no cutting of field lines, so no voltage is induced (1)
- Direction of induced voltage depends on the direction of movement (cutting in opposite direction) (1)

Part (c) [2 marks]:
- Any two valid methods (stronger magnet, faster movement, more turns of wire, soft iron core) for 1 mark each (max 2).

Part (a):
Redshift means an increase in the observed wavelength (or decrease in frequency) of electromagnetic radiation emitted from a source because the source is moving away from the observer.

Part (b):
Using the Doppler equation:
\(\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}\)
where:
\(\Delta \lambda = 672.1\text{ nm} - 656.3\text{ nm} = 15.8\text{ nm}\)
\(\lambda_0 = 656.3\text{ nm}\)
So:
\(\frac{15.8}{656.3} = \frac{v}{3.0 \times 10^8}\)
\(v = \frac{15.8 \times 3.0 \times 10^8}{656.3}\)
\(v \approx 7.22 \times 10^6\text{ m/s}\)

Part (c):
- Observations show that more distant galaxies have a greater redshift.
- This means that more distant galaxies are receding (moving away) at a higher velocity.
- This proportional relationship (Hubble's Law) implies that space itself is expanding.
- Extrapolating back in time, all galaxies/matter must have started from a single, extremely dense and hot point, supporting the Big Bang theory.

Part (a) [2 marks]:
- Observed wavelength increases / frequency decreases (1)
- Because the light source is moving away from the observer (1)

Part (b) [4 marks]:
- Calculate change in wavelength \(\Delta \lambda = 15.8\text{ nm}\) (1)
- Correct substitution into formula: \(v = \frac{\Delta \lambda}{\lambda_0} \times c\) (1)
- Correct numerical evaluation to \(7.22 \times 10^6\) (accept \(7.2 \times 10^6\)) (1)
- Correct unit: \(\text{m/s}\) (or equivalent, e.g., \(\text{m s}^{-1}\) or \(7220\text{ km/s}\)) (1)

Part (c) [4 marks]:
- Redshift increases with distance / further galaxies have greater redshift (1)
- Greater redshift means faster recessional speed (1)
- This shows that the Universe (space) is expanding (1)
- Suggests that the Universe originated from a single hot, dense point (the Big Bang) (1)

Part (a):
- Temperature is a measure of the average kinetic energy of the gas molecules.
- As temperature increases, the speed of the gas molecules increases.
- This leads to more frequent collisions between the molecules and the walls of the canister.
- The change in momentum during each collision is larger, which means each collision exerts a greater force on the walls.
- Since pressure is force per unit area and the volume/area is constant, the total pressure increases.

Part (b):
Convert initial temperature to Kelvin:
\(T_1 = 20^\circ\text{C} = 20 + 273 = 293\text{ K}\)

Using the pressure-temperature law (constant volume):
\(\frac{P_1}{T_1} = \frac{P_2}{T_2}\)
\(\frac{1.5 \times 10^5}{293} = \frac{2.2 \times 10^5}{T_2}\)

Rearranging to solve for \(T_2\):
\(T_2 = \frac{2.2 \times 10^5 \times 293}{1.5 \times 10^5} \approx 429.73\text{ K}\)

Convert back to Celsius:
\(T_2 = 429.73 - 273 \approx 156.73^\circ\text{C}\)
Rounded to a sensible number of significant figures, the maximum temperature is \(157^\circ\text{C}\).

Part (a) [5 marks]:
- Higher temperature means greater average kinetic energy / speed of molecules (1)
- Molecules collide with the walls of the container more frequently (1)
- Collisions are harder / have a larger change in momentum (1)
- This results in a greater average force exerted on the walls (1)
- Pressure increases because pressure is force per unit area (1)

Part (b) [5 marks]:
- Convert initial temperature to Kelvin (\(293\text{ K}\)) (1)
- Use of \(P_1/T_1 = P_2/T_2\) (1)
- Correct substitution of values (1)
- Correct calculation of final temperature in Kelvin (\(430\text{ K}\) or \(429.7\text{ K}\)) (1)
- Correct conversion back to Celsius (\(157^\circ\text{C}\) or \(156.7^\circ\text{C}\)) (1)
*Note: If no conversion to Kelvin is attempted (i.e. using Celsius directly), maximum score is 1 mark.*

1. Identify the relationship for gravitational potential energy (\(GPE\)):
\(GPE = m \times g \times h\)

2. Calculate the work done in lifting the load (increase in \(GPE\)):
\(GPE = 120\text{ kg} \times 10\text{ N/kg} \times 15\text{ m} = 18\,000\text{ J}\)

3. Identify the relationship for power:
\(\text{Power} = \frac{\text{Work Done}}{\text{Time Taken}}\)

4. Calculate the useful power output of the hoist:
\(\text{Power output} = \frac{18\,000\text{ J}}{8.0\text{ s}} = 2250\text{ W}\)

5. Identify the relationship for efficiency:
\(\text{Efficiency} = \frac{\text{Useful power output}}{\text{Total power input}}\)

6. Calculate the total electrical power input:
\(0.75 = \frac{2250\text{ W}}{\text{Power input}}\)

\(\text{Power input} = \frac{2250}{0.75} = 3000\text{ W}\) (or \(3.0\text{ kW}\))

- [1 Mark] State formula for gravitational potential energy: \(GPE = m \times g \times h\)
- [1 Mark] Correct calculation of work done: \(18\,000\text{ J}\)
- [1 Mark] State formula for power: \(\text{Power} = \frac{\text{Work Done}}{\text{Time Taken}}\)
- [1.5 Marks] Correct calculation of useful power output: \(2250\text{ W}\)
- [1 Mark] State formula for efficiency: \(\text{Efficiency} = \frac{\text{Useful power output}}{\text{Total power input}}\)
- [1 Mark] Correct rearrangement and substitution: \(\frac{2250}{0.75}\)
- [1 Mark] Final calculated power input with units: \(3000\text{ W}\) (or \(3.0\text{ kW}\))

(a) 1. Identify the law of conservation of momentum:
Total momentum before collision = Total momentum after collision

2. Calculate total momentum before collision:
\(p_{\text{before}} = (m_1 \times u_1) + (m_2 \times u_2) = (0.35 \times 1.2) + (0.25 \times 0) = 0.42\text{ kg m/s}\)

3. Set up the equation for the combined carriages after collision:
\(p_{\text{after}} = (m_1 + m_2) \times v = (0.35 + 0.25) \times v = 0.60 \times v\)

4. Solve for \(v\):
\(0.60 \times v = 0.42\)

\(v = \frac{0.42}{0.60} = 0.70\text{ m/s}\)

(b) 1. Identify the relationship between force, change in momentum, and time:
\(F = \frac{\Delta p}{t} = \frac{mv - mu}{t}\)

2. Calculate the change in momentum of the second carriage:
\(\Delta p = m_2 v - m_2 u_2 = (0.25 \times 0.70) - 0 = 0.175\text{ kg m/s}\)

3. Calculate the force:
\(F = \frac{0.175}{0.14} = 1.25\text{ N}\)

Part (a) [4 Marks]:
- [1 Mark] Recall of momentum formula \(p = m \times v\) or conservation of momentum principle.
- [1 Mark] Correct calculation of momentum before: \(0.42\text{ kg m/s}\).
- [1 Mark] Setting up conservation equation: \(0.42 = 0.60 \times v\).
- [1 Mark] Correct evaluation of velocity: \(0.70\text{ m/s}\) (accept unit in standard form).

Part (b) [3.5 Marks]:
- [1 Mark] Recall of force formula: \(F = \frac{\Delta p}{t}\) (or \(F = \frac{mv - mu}{t}\)).
- [1 Mark] Correct calculation of change in momentum for second carriage: \(0.175\text{ kg m/s}\).
- [1 Mark] Correct calculation of force: \(1.25\text{ N}\).
- [0.5 Marks] Correct unit (Newton / N).

1. State the relationship between pressure and Kelvin temperature for a gas at constant volume:
\(\frac{p_1}{T_1} = \frac{p_2}{T_2}\)

2. Convert the initial temperature from degrees Celsius to Kelvin:
\(T_1 = 20 + 273 = 293\text{ K}\)

3. Substitute the values into the equation to find \(T_2\):
\(\frac{1.2 \times 10^5}{293} = \frac{2.4 \times 10^5}{T_2}\)

4. Rearrange the formula to solve for \(T_2\):
\(T_2 = \frac{2.4 \times 10^5}{1.2 \times 10^5} \times 293 = 2 \times 293 = 586\text{ K}\)

5. Convert the final temperature back into degrees Celsius:
\(\theta_2 = 586 - 273 = 313^\circ\text{C}\)

6. State assumption:
The volume of the cylinder remains constant (or mass of gas is constant).

- [1 Mark] State formula relating pressure and Kelvin temperature: \(\frac{p_1}{T_1} = \frac{p_2}{T_2}\) or \(p \propto T\) in Kelvin.
- [1.5 Marks] Conversion of initial temperature to Kelvin: \(293\text{ K}\).
- [1.5 Marks] Rearrangement and calculation of final Kelvin temperature: \(586\text{ K}\).
- [1.5 Marks] Conversion of final temperature back to degrees Celsius: \(313^\circ\text{C}\).
- [1 Mark] State assumption (constant volume or constant mass of gas).
- [1 Mark] Correct unit of temperature (\(^\circ\text{C}\)) associated with the final answer.

1. Recall the formula for orbital speed:
\(v = \frac{2 \pi r}{T}\)

2. Calculate the orbital radius \(r\) by adding the Earth's radius and the satellite's altitude:
First, convert altitude to meters:
\(640\text{ km} = 640\,000\text{ m} = 6.4 \times 10^5\text{ m}\)
Then add to Earth's radius:
\(r = 6.4 \times 10^6\text{ m} + 0.64 \times 10^6\text{ m} = 7.04 \times 10^6\text{ m}\)

3. Convert the orbital period \(T\) from minutes to seconds:
\(T = 97 \times 60 = 5820\text{ s}\)

4. Substitute \(r\) and \(T\) into the orbital speed formula:
\(v = \frac{2 \times \pi \times 7.04 \times 10^6\text{ m}}{5820\text{ s}}\)

\(v \approx \frac{4.42336 \times 10^7}{5820} \approx 7600.28\text{ m/s}\)

5. Round to 2 significant figures:
\(v = 7600\text{ m/s}\) (or \(7.6 \times 10^3\text{ m/s}\))

- [1 Mark] Recall formula: \(v = \frac{2 \pi r}{T}\).
- [2 Marks] Correct calculation of orbital radius \(r = 7.04 \times 10^6\text{ m}\) (1 mark for conversion of km to m, 1 mark for adding to Earth's radius).
- [1 Mark] Correct conversion of period to seconds: \(T = 5820\text{ s}\).
- [1.5 Marks] Correct substitution of \(r\) and \(T\) into the formula.
- [1 Mark] Final calculated value of speed: \(7600\text{ m/s}\) (accept \(7.6 \times 10^3\text{ m/s}\)).
- [1 Mark] Final answer given to 2 significant figures with correct units.

a) First, convert the heating time from minutes to seconds: \(t = 10 \times 60 = 600\text{ s}\). Use the formula for electrical energy: \(E = V \times I \times t\). Substitute the values: \(E = 11.5 \times 3.8 \times 600 = 26220\text{ J}\).

b) Calculate the temperature change: \(\Delta\theta = 45.0 - 20.0 = 25.0^\circ\text{C}\). Use the specific heat capacity formula: \(Q = m \times c \times \Delta\theta\). Rearrange for \(c\): \(c = \frac{Q}{m \times \Delta\theta}\). Assuming all electrical energy is transferred to the block (\(Q = E\)): \(c = \frac{26220}{1.0 \times 25.0} = 1048.8\text{ J/kg}^\circ\text{C}\) (rounds to \(1050\text{ J/kg}^\circ\text{C}\)).

c) The experimental value is higher because some thermal energy is transferred to the surrounding air and the bench rather than heating the block. Therefore, more electrical energy is required to produce a given temperature change. Improvements: 1. Wrap the aluminium block in insulating material (e.g., bubble wrap or cotton wool) to reduce heat loss. 2. Put a few drops of oil or water in the thermometer hole to improve thermal contact and speed up heat transfer to the bulb.

d) Heat takes time to conduct through the metal block from the heater to the thermometer (thermal lag). To account for this, the student should continue monitoring the temperature after turning off the heater and use the maximum temperature reached as the final temperature \(\theta_2\).

a) [3 marks total]
- Conversion of time to seconds: \(10\text{ mins} = 600\text{ s}\) (1)
- Use of \(E = VIt\) (1)
- Correct final answer: \(26220\text{ [J]}\) (1)

b) [2 marks total]
- Correct temperature change: \(25.0^\circ\text{C}\) (1)
- Correct substitution and calculation: \(1048.8\text{ [J/kg}^\circ\text{C]}\) or \(1050\text{ [J/kg}^\circ\text{C]}\) (1)

c) [3 marks total]
- Explanation: Heat is lost to the surroundings / heater heats itself, so more input energy is needed for the same temperature rise (1)
- Improvement 1: Insulate the block / wrap in lagging (1)
- Improvement 2: Add oil/water to the thermometer hole to improve thermal contact (1)

d) [2 marks total]
- Explanation of thermal lag / time taken for heat to conduct to the thermometer (1)
- Method adjustment: Record and use the maximum temperature reached after turning off the power (1)