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First, calculate the equivalent resistance of the two 12 \(\Omega\) resistors in parallel: \(R_p = \frac{12 \times 12}{12 + 12} = 6.0\ \Omega\).

Next, calculate the total resistance of the circuit by adding the series resistor: \(R_{\text{total}} = 4.0\ \Omega + 6.0\ \Omega = 10\ \Omega\).

Finally, use Ohm's law to find the current: \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{10\ \Omega} = 1.2\text{ A}\).

1 mark for the correct answer (B).

Use the transformer relationship formula: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\).

Rearranging for the secondary voltage: \(V_s = V_p \times \frac{N_s}{N_p} = 12\text{ V} \times \frac{800}{200} = 48\text{ V}\).

Since the number of turns on the secondary coil is greater than on the primary coil (and the secondary voltage is greater than the primary voltage), it is a step-up transformer.

1 mark for the correct answer (D).

First, find the resultant force acting on the toy car: \(F_{\text{resultant}} = 3.5\text{ N} - 1.5\text{ N} = 2.0\text{ N}\).

Then, use Newton's Second Law (\(F = m a\)) to calculate the acceleration: \(a = \frac{F}{m} = \frac{2.0\text{ N}}{0.50\text{ kg}} = 4.0\text{ m/s}^2\).

1 mark for the correct answer (C).

First, calculate the energy supplied by the heater: \(E = P \times t = 50\text{ W} \times 200\text{ s} = 10,000\text{ J}\).

Next, use the specific latent heat formula: \(E = m \times L_f\).

Rearranging to solve for the specific latent heat of fusion (\(L_f\)): \(L_f = \frac{E}{m} = \frac{10,000\text{ J}}{0.10\text{ kg}} = 100,000\text{ J/kg}\).

1 mark for the correct answer (C).

Use the fluid pressure formula: \(p = \rho \times g \times h\).

Substituting the values: \(p = 1000\text{ kg/m}^3 \times 10\text{ N/kg} \times 20\text{ m} = 200,000\text{ Pa}\).

Convert Pascal to kilopascal: \(200,000\text{ Pa} = 200\text{ kPa}\).

1 mark for the correct answer (C).

When a wave passes from one medium to another, the frequency is determined solely by the source, so it remains unchanged. Since the wave speed decreases according to the wave equation \(v = f \lambda\), the wavelength must decrease to maintain this relationship.

1 mark for the correct answer (B).

First, calculate how many half-lives are in 1 hour (60 minutes): \(\frac{60\text{ minutes}}{15\text{ minutes}} = 4\text{ half-lives}\).

Now, halve the activity 4 times:
- 1st half-life: \(800 \div 2 = 400\text{ Bq}\)
- 2nd half-life: \(400 \div 2 = 200\text{ Bq}\)
- 3rd half-life: \(200 \div 2 = 100\text{ Bq}\)
- 4th half-life: \(100 \div 2 = 50\text{ Bq}\).

1 mark for the correct answer (C).

First, calculate the gravitational force (weight) acting on the crate: \(F = m \times g = 120\text{ kg} \times 10\text{ N/kg} = 1200\text{ N}\).

Then, use the work done formula: \(W = F \times d\), where \(d\) is the distance (height) the crate is lifted: \(W = 1200\text{ N} \times 15\text{ m} = 18,000\text{ J}\).

Converting to kilojoules: \(18,000\text{ J} = 18\text{ kJ}\).

1 mark for the correct answer (C).

To calculate the charge, use the formula: \(Q = I \times t\). Here, the current \(I = 0.4\text{ A}\) and the time \(t = 5\text{ minutes} = 5 \times 60 = 300\text{ seconds}\). Therefore, \(Q = 0.4\text{ A} \times 300\text{ s} = 120\text{ C}\).

1 mark for the correct answer (C).

Newton's third law states that whenever one object exerts a force on another, the second object exerts an equal and opposite force on the first. Therefore, if the skater exerts a force of \(150\text{ N}\) on the wall, the wall exerts a force of \(150\text{ N}\) on the skater in the opposite direction.

1 mark for the correct answer (B).

Beta radiation consists of high-speed electrons (or positrons). It has moderate ionizing power and moderate penetrating ability, typically being stopped by a few millimetres of aluminium. Alpha is highly ionizing and stopped by paper, while gamma is weakly ionizing and highly penetrating.

1 mark for the correct answer (D).

The work done \(W\) in lifting an object is equal to the change in gravitational potential energy: \(W = m \times g \times h\). Substituting the given values: \(W = 250\text{ kg} \times 10\text{ N/kg} \times 12\text{ m} = 30,000\text{ J}\).

1 mark for the correct answer (B).

The thermal energy \(E\) required to change the state of a substance of mass \(m\) is calculated using: \(E = m \times L\), where \(L\) is the specific latent heat of fusion. Here, \(E = 0.50\text{ kg} \times 3.34 \times 10^5\text{ J/kg} = 1.67 \times 10^5\text{ J}\).

1 mark for the correct answer (A).

A star with a mass much larger than the Sun follows the pathway: Nebula/Protostar \(\rightarrow\) Main sequence \(\rightarrow\) Red supergiant \(\rightarrow\) Supernova \(\rightarrow\) Neutron star or Black hole. Option A describes the pathway for a star of similar mass to the Sun.

1 mark for the correct answer (B).

Radio waves have the longest wavelengths and lowest frequencies. Gamma rays have the shortest wavelengths and highest frequencies. Therefore, moving from radio to gamma rays means wavelength decreases and frequency increases. All electromagnetic waves travel at the same constant speed in a vacuum.

1 mark for the correct answer (B).

First, determine the current through each of the two branches using Ohm's law: \(I_1 = \frac{V}{R_1} = \frac{9.0}{15} = 0.6\text{ A}\) and \(I_2 = \frac{V}{R_2} = \frac{9.0}{30} = 0.3\text{ A}\). Since the resistors are connected in parallel, the total current from the battery is the sum of the branch currents: \(I = I_1 + I_2 = 0.6 + 0.3 = 0.9\text{ A}\). Alternatively, calculate the total resistance of the parallel circuit: \(\frac{1}{R} = \frac{1}{15} + \frac{1}{30} = \frac{3}{30}\), which gives \(R = 10\ \Omega\). Then, use Ohm's law to find the total current: \(I = \frac{V}{R} = \frac{9.0}{10} = 0.9\text{ A}\).

1 mark for calculating the currents in each branch as \(0.6\text{ A}\) and \(0.3\text{ A}\) OR finding the total equivalent resistance of \(10\ \Omega\). 1 mark for adding the branch currents (\(0.6 + 0.3\)) OR for substituting the equivalent resistance into Ohm's law (\(\frac{9.0}{10}\)). 1 mark for the correct final answer of \(0.9\text{ A}\) (accept 0.9).

Use the transformer equation: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\). Rearranging the equation to solve for the number of turns in the secondary coil (\(N_s\)) gives: \(N_s = N_p \times \frac{V_s}{V_p}\). Substitute the given values into the equation: \(N_s = 1150 \times \frac{12}{230}\). Simplifying this expression: \(N_s = 5 \times 12 = 60\).

1 mark for recalling the transformer equation or its rearrangement: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\). 1 mark for correct substitution: \(1150 \times \frac{12}{230}\). 1 mark for the correct final answer: 60.

First, calculate the resultant force acting on the crate. The resultant force \(F\) is the forward force minus the opposing frictional force: \(F = 150\text{ N} - 60\text{ N} = 90\text{ N}\). Next, use Newton's second law, \(F = ma\), and rearrange it to solve for acceleration \(a\): \(a = \frac{F}{m}\). Substitute the values: \(a = \frac{90}{45} = 2.0\text{ m/s}^2\).

1 mark for calculating the resultant force: \(150 - 60 = 90\text{ N}\). 1 mark for rearranging the Newton's second law equation: \(a = \frac{F}{m}\). 1 mark for the correct acceleration of \(2.0\text{ m/s}^2\) (accept \(2\text{ m/s}^2\) or 2.0).

First, find the extension of the spring in metres. The extension \(x\) is the final length minus the initial length: \(x = 20.0\text{ cm} - 12.0\text{ cm} = 8.0\text{ cm}\). Convert this extension to metres: \(8.0\text{ cm} = 0.08\text{ m}\). Now apply Hooke's Law, \(F = ke\), where \(F\) is the applied force and \(k\) is the spring constant. Rearrange for \(k\): \(k = \frac{F}{e} = \frac{6.0\text{ N}}{0.08\text{ m}} = 75\text{ N/m}\).

1 mark for calculating the extension in metres: \(0.08\text{ m}\) (or \(8\text{ cm}\) with subsequent division by 100). 1 mark for recalling Hooke's Law and rearranging: \(k = \frac{F}{e}\). 1 mark for correct calculation: \(75\text{ N/m}\) (accept 75).

First, calculate the total thermal energy \(E\) supplied by the heater using the power equation: \(E = \text{Power} \times \text{time} = 120\text{ W} \times 1100\text{ s} = 132000\text{ J}\). Next, use the specific latent heat equation: \(E = mL\), where \(m\) is the mass of ice and \(L\) is the specific latent heat of fusion. Rearranging for \(L\): \(L = \frac{E}{m} = \frac{132000\text{ J}}{0.40\text{ kg}} = 330000\text{ J/kg}\) (or \(3.3 \times 10^5\text{ J/kg}\)).

1 mark for calculating the energy supplied: \(120 \times 1100 = 132,000\text{ J}\). 1 mark for rearranging \(E = mL\) to make \(L\) the subject: \(L = \frac{E}{m}\) and substituting their values. 1 mark for the correct answer: \(330,000\text{ J/kg}\) (accept \(3.3 \times 10^5\text{ J/kg}\) or \(3.3 \times 10^5\)).

Use the equation for hydrostatic pressure in a column of liquid: \(p = h \rho g\), where \(h\) is the depth, \(\rho\) is the density of the fluid, and \(g\) is the gravitational field strength. Substitute the given values: \(p = 120\text{ m} \times 1025\text{ kg/m}^3 \times 9.8\text{ N/kg}\). Calculate the final pressure: \(p = 120 \times 1025 \times 9.8 = 1205400\text{ Pa}\) (or \(1.21 \times 10^6\text{ Pa}\)).

1 mark for recalling the equation: \(p = h \rho g\). 1 mark for correct substitution of the values: \(120 \times 1025 \times 9.8\). 1 mark for correct final calculation: \(1,205,400\text{ Pa}\) (accept \(1.2 \times 10^6\text{ Pa}\) or \(1.21 \times 10^6\text{ Pa}\)).

Determine how many half-lives have passed as the activity falls from 800 Bq to 50 Bq. Each half-life reduces the activity by half: \(800\text{ Bq} \rightarrow 400\text{ Bq}\) (1 half-life), \(400\text{ Bq} \rightarrow 200\text{ Bq}\) (2 half-lives), \(200\text{ Bq} \rightarrow 100\text{ Bq}\) (3 half-lives), \(100\text{ Bq} \rightarrow 50\text{ Bq}\) (4 half-lives). Therefore, 4 half-lives have elapsed over 24 days. To find the duration of one half-life, divide the total time by the number of half-lives: \(\frac{24\text{ days}}{4} = 6\text{ days}\).

1 mark for establishing that the sample has decayed through 4 half-lives. 1 mark for dividing the total time of 24 days by the number of half-lives: \(\frac{24}{4}\). 1 mark for the correct final answer: 6 days (or 6).

The work done by the cyclist is equal to the kinetic energy gained by the cyclist and bicycle combination. The kinetic energy \(E_k\) is given by the formula: \(E_k = \frac{1}{2} m v^2\). Substitute the given values: \(E_k = 0.5 \times 80\text{ kg} \times (6.0\text{ m/s})^2\). Calculate the value: \(E_k = 40 \times 36 = 1440\text{ J}\). Thus, the work done is \(1440\text{ J}\) (or \(1.44\text{ kJ}\)).

1 mark for identifying that the work done is equal to the kinetic energy gained (\(W = \frac{1}{2} m v^2\)). 1 mark for correct substitution: \(0.5 \times 80 \times 6^2\). 1 mark for the correct final answer: 1440 J (accept 1.44 kJ).

First, calculate the extension of the spring in meters:
Extension, \(e = 21.0\text{ cm} - 15.0\text{ cm} = 6.0\text{ cm}\)
In meters, \(e = \frac{6.0}{100} = 0.06\text{ m}\)

Next, use Hooke's Law: \(F = k \times e\)
Rearrange the equation to solve for the spring constant (\(k\)):
\(k = \frac{F}{e}\)

Substitute the values into the equation:
\(k = \frac{4.5\text{ N}}{0.06\text{ m}} = 75\text{ N/m}\)

1 mark: Calculate the extension of the spring and convert to meters (0.06 m).
1 mark: Recall and rearrange Hooke's Law equation (k = F / e).
1 mark: Correct calculation of spring constant (75 N/m).

First, calculate the total thermal energy supplied by the heater:
\(E = P \times t\)
\(E = 80\text{ W} \times 1000\text{ s} = 80,000\text{ J}\)

Next, use the equation for thermal energy required for a change of state:
\(E = m \times L\)

Rearrange the equation to solve for the specific latent heat of fusion (\(L\)):
\(L = \frac{E}{m}\)

Substitute the values into the equation:
\(L = \frac{80,000\text{ J}}{0.25\text{ kg}} = 320,000\text{ J/kg}\)

1 mark: Correct calculation of the energy supplied (80,000 J).
1 mark: Rearrangement of change of state equation (L = E / m) or substitution of values into formula.
1 mark: Correct final calculation of specific latent heat (320,000 J/kg or 3.2 x 10^5 J/kg).

(a) Calculate energy: Energy = Power x time = \(50\text{ W} \times (15 \times 60)\text{ s} = 45\,000\text{ J}\). (b) Calculate latent heat: \(L_f = \frac{E}{m}\). Mass of ice in kg = \(0.15\text{ kg}\). \(L_f = \frac{45\,000\text{ J}}{0.15\text{ kg}} = 3.0 \times 10^5\text{ J/kg}\). (c) The ice absorbs thermal energy from the warmer surroundings (air/beaker). This means more ice melts than would be melted by the heater alone. Since the calculation only uses the electrical energy input, the calculated specific latent heat of fusion will be lower than the actual value.

Part (a) [2 marks]: 1 mark for correct time conversion (900 s) and substitution: \(50 \times 900\), 1 mark for correct answer: \(45\,000\text{ J}\). Part (b) [2 marks]: 1 mark for converting mass to kg (0.15 kg) and substituting into formula, 1 mark for correct answer: \(300\,000\text{ J/kg}\) (or \(3.0 \times 10^5\text{ J/kg}\)). Part (c) [2 marks]: 1 mark for identifying heat absorbed from surroundings / beaker / air, 1 mark for explaining that this extra energy means the calculated latent heat is lower than the true value because more ice melted for the same electrical energy.

(a) Convert resistance of fixed resistor to ohms: \(2.0\text{ k}\Omega = 2000\,\Omega\). Total resistance: \(R_{\text{total}} = 400 + 2000 = 2400\,\Omega\). Output voltage: \(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{fixed}}}{R_{\text{total}}} = 9.0\text{ V} \times \frac{2000}{2400} = 7.5\text{ V}\). (b) As light intensity decreases, the resistance of the LDR increases. Because the LDR resistance is now larger, a larger proportion of the supply voltage is dropped across the LDR. Therefore, the potential difference across the fixed resistor (which is \(V_{\text{out}}\)) must decrease.

Part (a) [3 marks]: 1 mark for finding the total resistance of the series circuit (\(2400\,\Omega\)), 1 mark for the correct formula or ratio substitution (e.g. \(9 \times 2000 / 2400\)), 1 mark for the correct calculation: \(7.5\text{ V}\). Part (b) [3 marks]: 1 mark for stating that LDR resistance increases as light intensity decreases, 1 mark for explaining that LDR takes a larger fraction of the total potential difference (or current in the circuit decreases, so \(V = IR\) across the fixed resistor decreases), 1 mark for stating that \(V_{\text{out}}\) decreases.

(a) Immediately after jumping, the only force is weight: \(F = W = m \times g = 80\text{ kg} \times 10\text{ N/kg} = 800\text{ N}\). Using \(F = m \times a\), the acceleration is \(a = \frac{800\text{ N}}{80\text{ kg}} = 10\text{ m/s}^2\) (equal to \(g\)). (b) As the skydiver accelerates, the air resistance acting upwards increases. Eventually, the upward air resistance becomes equal in size to the downward weight. At this point, the resultant force on the skydiver is zero, so they stop accelerating and fall at a constant terminal velocity. (c) Opening the parachute increases the surface area significantly, which instantly increases the upward air resistance. This upward force is now much larger than the downward weight. The upward resultant force causes the skydiver to decelerate (or accelerate upwards) and slow down.

Part (a) [2 marks]: 1 mark for using \(F = ma\) or noting that initial acceleration equals \(g\), 1 mark for correct calculation: \(10\text{ m/s}^2\) (or \(10\text{ N/kg}\)). Part (b) [2 marks]: 1 mark for explaining that air resistance increases with speed, 1 mark for explaining that terminal velocity is reached when air resistance equals weight (resultant force is zero). Part (c) [2 marks]: 1 mark for explaining that the parachute increases air resistance so it is greater than weight, 1 mark for stating that this creates an upward resultant force causing deceleration.

(a) Convert wavelength to meters: \(12\text{ cm} = 0.12\text{ m}\). Speed: \(v = f \lambda = 5.0\text{ Hz} \times 0.12\text{ m} = 0.60\text{ m/s}\). (b) When entering shallow water, the frequency remains constant at \(5.0\text{ Hz}\). New wavelength: \(\lambda = \frac{v}{f} = \frac{0.45\text{ m/s}}{5.0\text{ Hz}} = 0.09\text{ m}\) (or \(9\text{ cm}\)). (c) As the waves enter shallow water and slow down, their direction of travel bends towards the normal (refraction). Because the speed decreases but frequency is constant, the wavelength decreases, meaning the wavefronts become closer together.

Part (a) [2 marks]: 1 mark for converting wavelength to \(0.12\text{ m}\) and substituting into formula, 1 mark for correct answer: \(0.60\text{ m/s}\) (accept \(60\text{ cm/s}\) if unit is clearly stated). Part (b) [2 marks]: 1 mark for identifying that frequency remains unchanged (\(5.0\text{ Hz}\)) and substituting, 1 mark for correct answer: \(0.09\text{ m}\) (or \(9\text{ cm}\)). Part (c) [2 marks]: 1 mark for stating that the waves bend towards the normal, 1 mark for stating that the wavefronts become closer together / wavelength decreases.

(a) Mass number balance: \(222 = A + 4 \implies A = 218\). Atomic number balance: \(86 = Z + 2 \implies Z = 84\). (b) Determine the number of half-lives: \(1600 \to 800\) (1 half-life), \(800 \to 400\) (2 half-lives), \(400 \to 200\) (3 half-lives). It takes exactly 3 half-lives. Total time = \(3 \times 3.8\text{ days} = 11.4\text{ days}\). (c) Radioactive decay is a completely random process, meaning it is impossible to predict when any individual nucleus will decay.

Part (a) [2 marks]: 1 mark for \(A = 218\), 1 mark for \(Z = 84\). Part (b) [2 marks]: 1 mark for determining that 3 half-lives are required, 1 mark for correct final answer: \(11.4\text{ days}\). Part (c) [2 marks]: 1 mark for identifying that decay is random / a matter of chance, 1 mark for explaining that we can only predict the probability of a nucleus decaying per unit time, not which individual nucleus will decay next.

(a) A massive star spends most of its life in the stable main sequence stage fusing hydrogen into helium. Once hydrogen runs out, the core contracts, the outer layers expand and cool, forming a red supergiant. The star then undergoes a rapid collapse and violent explosion called a supernova. The remaining core collapses further to become either a dense neutron star or, if the initial mass was extremely high, a black hole. (b) Red-shift is the observed increase in the wavelength of light from distant galaxies, moving it towards the red end of the spectrum. This occurs because the galaxies are moving away from us (recession). Observations show that more distant galaxies have a greater red-shift, meaning they are receding faster. This proportional relationship indicates that space itself is expanding in all directions, supporting the Big Bang theory.

Part (a) [3 marks]: 1 mark for explaining expansion into a red supergiant when hydrogen runs out, 1 mark for describing the supernova explosion, 1 mark for identifying the remnants as a neutron star or black hole. Part (b) [3 marks]: 1 mark for explaining that red-shift is an increase in wavelength of light because galaxies are moving away, 1 mark for stating that more distant galaxies have a greater red-shift (moving away faster), 1 mark for concluding that this shows the Universe / space is expanding.

The two parallel resistors of value \(R\) have an equivalent resistance of \(R_p = \frac{R \times R}{R + R} = 0.5R\). This combination is connected in series with the third resistor of value \(R\), so the total resistance is \(R_{total} = R + 0.5R = 1.5R\). Therefore, the correct option is C.

[1 mark] - Award 1 mark for the correct option C.

First, calculate the equivalent resistance of the two parallel 6.0 \(\Omega\) resistors: \(R_p = \frac{6.0 \times 6.0}{6.0 + 6.0} = 3.0\ \Omega\). Next, find the total resistance of the series circuit: \(R_{total} = 3.0\ \Omega + 3.0\ \Omega = 6.0\ \Omega\). The total current delivered by the 12 V power supply is: \(I_{total} = \frac{V}{R_{total}} = \frac{12\text{ V}}{6.0\ \Omega} = 2.0\text{ A}\). Since the two parallel resistors are identical, the total current of 2.0 A splits equally between the two branches, giving a current of 1.0 A through each 6.0 \(\Omega\) resistor.

1 mark: Correctly identifies the current through one of the branches as 1.0 A (Option A).

According to Newton's Second Law, \(F = m a\), where \(F\) is the resultant force. Calculate the resultant force: \(F_{resultant} = 3200\text{ N} - 800\text{ N} = 2400\text{ N}\). Now, calculate the acceleration: \(a = \frac{F_{resultant}}{m} = \frac{2400\text{ N}}{1200\text{ kg}} = 2.0\text{ m/s}^2\).

1 mark: Correctly calculates the acceleration as 2.0 m/s\(^2\) (Option B).

First, calculate the thermal energy supplied by the heater: \(E = P \times t = 50\text{ W} \times 330\text{ s} = 16,500\text{ J}\). Using the formula for specific latent heat: \(E = m L\), where \(m\) is the mass and \(L\) is the specific latent heat of fusion. Rearrange for mass: \(m = \frac{E}{L} = \frac{16,500\text{ J}}{3.3 \times 10^5\text{ J/kg}} = 0.05\text{ kg}\).

1 mark: Correctly identifies the mass of the melted ice as 0.05 kg (Option A).

First, calculate the pressure due to the water column: \(p_{water} = h \rho g = 25\text{ m} \times 1000\text{ kg/m}^3 \times 10\text{ N/kg} = 2.5 \times 10^5\text{ Pa}\). To find the total pressure, add the atmospheric pressure: \(p_{total} = p_{water} + p_{atmospheric} = 2.5 \times 10^5\text{ Pa} + 1.0 \times 10^5\text{ Pa} = 3.5 \times 10^5\text{ Pa}\).

1 mark: Correctly determines the total pressure as 3.5 \(\times 10^5\) Pa (Option C).

When a wave transitions from one medium to another, its frequency remains constant. Therefore, the frequency in shallow water is still 4.0 Hz. Using the wave equation: \(v = f \lambda\), rearrange for wavelength: \(\lambda = \frac{v}{f} = \frac{0.80\text{ m/s}}{4.0\text{ Hz}} = 0.20\text{ m}\).

1 mark: Correctly calculates the wavelength in shallow water as 0.20 m (Option B).

Alpha particles are stopped by paper. Beta-minus particles are highly penetrating compared to alpha and can pass through paper, but are stopped by a few millimetres of aluminium. Gamma rays are highly penetrating and can easily pass through thin aluminium, requiring thick lead or concrete to stop them. Thus, the correct option is beta-minus particles.

1 mark: Correctly identifies the radiation as beta-minus particles (Option B).

A star with approximately the same mass as the Sun follows the path: Main sequence \(\rightarrow\) Red giant \(\rightarrow\) White dwarf \(\rightarrow\) Black dwarf. High-mass stars form red supergiants, leading to supernovas and neutron stars or black holes. Protostars exist before the main sequence phase.

1 mark: Correctly identifies the lifecycle sequence for a solar-mass star (Option B).

First, calculate the total resistance of the parallel combination using the formula \(\frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2}\). This gives \(\frac{1}{R_t} = \frac{1}{4.0} + \frac{1}{12.0} = \frac{3}{12.0} + \frac{1}{12.0} = \frac{4}{12.0} = \frac{1}{3.0}\), so \(R_t = 3.0\ \Omega\). Then, use Ohm's law to calculate the total current: \(I = \frac{V}{R_t} = \frac{12}{3.0} = 4.0\text{ A}\). Alternatively, calculate individual currents: \(I_1 = \frac{12}{4.0} = 3.0\text{ A}\) and \(I_2 = \frac{12}{12.0} = 1.0\text{ A}\). Total current \(I = I_1 + I_2 = 3.0 + 1.0 = 4.0\text{ A}\).

1 mark for calculating total resistance of 3.0 \(\Omega\) or calculating individual branch currents (3.0 A and 1.0 A). 1 mark for correctly substituting into the current formula \(I = V/R\) or \(I = I_1 + I_2\). 1 mark for the correct final current of 4.0 A (accept 4 A) with unit.

First, calculate the force (weight) exerted by the block: \(F = m \times g = 4.8\text{ kg} \times 10\text{ N/kg} = 48\text{ N}\). For maximum pressure, the block must rest on the face with the minimum surface area. Minimum area \(A = 0.20\text{ m} \times 0.30\text{ m} = 0.06\text{ m}^2\). Calculate maximum pressure using \(P = \frac{F}{A} = \frac{48}{0.06} = 800\text{ Pa}\).

1 mark for calculating weight of 48 N. 1 mark for calculating minimum area of 0.06 m\(^2\) and stating formula \(P = F/A\). 1 mark for the correct calculation of pressure as 800 Pa (or N/m\(^2\)).

First, find the number of half-lives that have elapsed by halving the activity: \(800\text{ Bq} \to 400\text{ Bq}\) (1 half-life) \(\to 200\text{ Bq}\) (2 half-lives) \(\to 100\text{ Bq}\) (3 half-lives). Therefore, 3 half-lives have passed in 45 minutes. The half-life \(T_{1/2}\) is calculated by dividing total time by the number of half-lives: \(T_{1/2} = \frac{45\text{ minutes}}{3} = 15\text{ minutes}\).

1 mark for showing that 3 half-lives have elapsed (or activity decreased to 1/8). 1 mark for setting up the calculation \(45 / 3\). 1 mark for the correct answer of 15 minutes.

First, calculate the distance traveled in one full orbit using the formula for circumference: \(d = 2 \pi r = 2 \times 3.14 \times 8.0 \times 10^6\text{ m} = 5.024 \times 10^7\text{ m}\). Next, calculate the orbital speed using \(v = \frac{d}{t} = \frac{5.024 \times 10^7\text{ m}}{1.0 \times 10^4\text{ s}} = 5024\text{ m/s}\).

1 mark for the orbital distance formula \(2 \pi r\). 1 mark for calculating distance as \(5.024 \times 10^7\text{ m}\) (or \(5.02 \times 10^7\text{ m}\)). 1 mark for correct speed of 5024 m/s (accept 5020 or 5000 m/s depending on rounding used).

First, calculate the upward force required to lift the crate at a constant speed, which is equal to its weight: \(F = m \times g = 250\text{ kg} \times 10\text{ N/kg} = 2500\text{ N}\). Next, calculate the work done using \(W = F \times d = 2500\text{ N} \times 12\text{ m} = 30000\text{ J}\) (or \(30\text{ kJ}\)).

1 mark for calculating weight of 2500 N. 1 mark for substituting values into \(W = F \times d\). 1 mark for the correct work done of 30000 J (accept 30 kJ).

First, convert the time to seconds: \(t = 4.0\text{ minutes} = 4.0 \times 60 = 240\text{ s}\). Calculate the energy supplied by the heater: \(E = P \times t = 150\text{ W} \times 240\text{ s} = 36000\text{ J}\). Next, use the specific latent heat formula \(E = m \times L\) rearranged to solve for mass: \(m = \frac{E}{L} = \frac{36000}{3.3 \times 10^5} \approx 0.109\text{ kg}\) (or \(0.11\text{ kg}\) to two significant figures).

1 mark for calculating energy supplied as 36000 J. 1 mark for rearranging to \(m = E/L\) and substituting. 1 mark for correct final mass of 0.11 kg (accept 0.109 kg or 0.1091 kg).

Use the conservation of momentum: total momentum before collision equals total momentum after collision. Total momentum before = \(m_1 u_1 + m_2 u_2 = 0.5\text{ kg} \times 4.0\text{ m/s} + 0.3\text{ kg} \times 0\text{ m/s} = 2.0\text{ kg m/s}\). Total mass after collision = \(m_1 + m_2 = 0.5 + 0.3 = 0.8\text{ kg}\). Using \(p = m \times v\), we get \(2.0 = 0.8 \times v\). Therefore, \(v = \frac{2.0}{0.8} = 2.5\text{ m/s}\).

1 mark for calculating momentum before collision as 2.0 kg m/s. 1 mark for equating momentum before and after collision: \(2.0 = 0.8 \times v\). 1 mark for calculating correct velocity of 2.5 m/s.

When a wave transitions from one medium to another, its frequency remains unchanged, so the frequency in water is still \(f = 400\text{ Hz}\). Using the wave speed equation \(v = f \times \lambda\), rearrange to solve for wavelength in water: \(\lambda = \frac{v}{f} = \frac{1500\text{ m/s}}{400\text{ Hz}} = 3.75\text{ m}\).

1 mark for stating or implying that the frequency in water remains 400 Hz. 1 mark for rearranging the wave equation to \(\lambda = v/f\) and substituting \(1500 / 400\). 1 mark for correct wavelength of 3.75 m (accept 3.8 m).

First, calculate the energy supplied by the heater using the formula \(E = P \times t\): \(E = 40\text{ W} \times 150\text{ s} = 6000\text{ J}\). Next, convert the mass from grams to kilograms: \(20\text{ g} = 0.02\text{ kg}\). Finally, use the formula for specific latent heat of fusion: \(L = \frac{E}{m}\), which gives \(L = \frac{6000\text{ J}}{0.02\text{ kg}} = 300000\text{ J/kg}\).

1 mark for calculating energy supplied: \(6000\text{ J}\). 1 mark for converting mass to kg: \(0.02\text{ kg}\) (or rearranging the formula to find \(L\)). 1 mark for the correct final answer: \(300000\text{ J/kg}\) (or \(3 \times 10^5\text{ J/kg}\)).

First, calculate the useful work done in lifting the block, which is equal to the gain in gravitational potential energy: \(GPE = m \times g \times h = 0.80\text{ kg} \times 10\text{ N/kg} \times 1.5\text{ m} = 12\text{ J}\). Then, subtract this useful energy from the total work done to find the energy wasted: \(\text{Wasted energy} = \text{Total work done} - \text{Useful work done} = 15\text{ J} - 12\text{ J} = 3\text{ J}\).

1 mark for correctly recalling and substituting into the gravitational potential energy equation: \(0.80 \times 10 \times 1.5\). 1 mark for calculating the useful work as \(12\text{ J}\). 1 mark for calculating the wasted energy as \(3\text{ J}\).

1. **How sound waves are produced:**
- When an alternating current (AC) flows through the coil of the loudspeaker, it creates an alternating/changing magnetic field around the coil.
- This temporary magnetic field interacts with the magnetic field of the permanent magnet.
- This interaction produces a force on the coil (and therefore on the attached cone) due to the motor effect.
- Since the current is alternating (changing direction), the direction of the force also alternates continuously.
- This causes the coil and the attached cone to vibrate back and forth.
- The vibrating cone compresses and rarefies the air particles in front of it, generating longitudinal sound waves.

2. **Modifications to increase loudness (amplitude):**
- **Increase the current:** A larger current increases the strength of the temporary magnetic field, resulting in a larger force and larger amplitude of vibration.
- **Use a stronger permanent magnet:** This increases the magnetic flux density, resulting in a stronger force on the coil.
- **Increase the number of turns on the coil:** More turns increase the overall magnetic force acting on the coil (
\( F = B I l \)), increasing the vibration amplitude.

This is a Level of Response question.

**Level 3 (5–6 marks):**
- A detailed, coherent description of how the alternating current creates an alternating magnetic field which interacts with the permanent magnet.
- Explains how this causes alternating forces (motor effect) to vibrate the cone and create longitudinal waves.
- Gives at least two valid design modifications to increase loudness, with clear scientific explanations.

**Level 2 (3–4 marks):**
- Describes how the current creates a magnetic field and interacts with the magnet to move the cone.
- Mentions that vibrations cause sound waves.
- Suggests at least one valid design modification to increase loudness, but the explanation may be brief.

**Level 1 (1–2 marks):**
- Identifies basic components (coil, magnet, cone).
- Mentions that the speaker vibrates or that current causes movement.
- Identifies a way to increase loudness (e.g., 'more current' or 'stronger magnet') without detailed explanation.

1. **Initial motion:**
- At the moment the skydiver jumps, air resistance is zero.
- The only force acting is weight: \( W = m \times g = 80\text{ kg} \times 10\text{ N/kg} = 800\text{ N} \).
- The resultant force is therefore \( 800\text{ N} \) downwards. According to Newton's Second Law (\( F = ma \)), the skydiver accelerates downwards at \( 10\text{ m/s}^2 \).

2. **As speed increases:**
- As the skydiver falls faster, air resistance (drag) acts upwards and increases with speed.
- This reduces the resultant force (\( F_{\text{resultant}} = \text{Weight} - \text{Air Resistance} \)).
- As the resultant force decreases, the acceleration decreases (Newton's Second Law), meaning the skydiver's velocity still increases but at a slower rate.

3. **Terminal velocity:**
- Eventually, the upward air resistance force increases until it is equal in size to the downward weight (\( 800\text{ N} \)).
- The resultant force becomes zero.
- According to Newton's First Law, when forces are balanced (resultant force is zero), the object stops accelerating and travels at a constant velocity, known as terminal velocity.

4. **Calculation at terminal velocity:**
- At terminal velocity, \( \text{Air Resistance} = \text{Weight} = 800\text{ N} \).

This is a Level of Response question.

**Level 3 (5–6 marks):**
- Full, logical description of the changes in forces from start to terminal velocity.
- Explicitly applies Newton's Second Law (reducing resultant force causes decreasing acceleration) and Newton's First Law (balanced forces mean zero acceleration/constant terminal speed).
- Correctly calculates the weight/air resistance at terminal velocity as \( 800\text{ N} \) (with units).

**Level 2 (3–4 marks):**
- Explains that air resistance increases with speed and eventually balances weight.
- Mentions that acceleration decreases and becomes zero.
- Attempts the calculation of weight with a correct value of \( 800\text{ N} \).

**Level 1 (1–2 marks):**
- Identifies the forces as weight and air resistance.
- Mentions that the skydiver speeds up and then goes at a steady speed (terminal velocity).
- Calculation of weight is incorrect or missing.

(a)
Since the total initial count rate is \( 320\text{ cpm} \) and both isotopes contribute equally:
- Count rate of X = \( 160\text{ cpm} \)
- Count rate of Y = \( 160\text{ cpm} \)

(b)
Calculate decay of isotope X after 12 hours:
- Half-life of X = 2 hours.
- Number of half-lives of X in 12 hours = \( 12 / 2 = 6 \) half-lives.
- Final count rate of X = \( 160 / 2^6 = 160 / 64 = 2.5\text{ cpm} \).

Calculate decay of isotope Y after 12 hours:
- Half-life of Y = 6 hours.
- Number of half-lives of Y in 12 hours = \( 12 / 6 = 2 \) half-lives.
- Final count rate of Y = \( 160 / 2^2 = 160 / 4 = 40\text{ cpm} \).

Total corrected count rate = \( 2.5 + 40 = 42.5\text{ cpm} \).

(c)
- The student can insert a sheet of paper between the source and the GM tube. Since paper completely absorbs/stops alpha radiation but allows beta to pass, a substantial decrease in the count rate confirms the presence of alpha radiation.
- Next, the student can place a thin sheet of aluminum (e.g., 2-5 mm thick) between the source and the detector. Since aluminum absorbs/stops beta radiation, the remaining count rate should drop to the background level, confirming the presence of beta radiation.

(a)
- \( 160\text{ cpm} \) for each isotope [1 mark].

(b)
- Correct calculation of X activity after 6 half-lives: \( 2.5\text{ cpm} \) [1 mark].
- Correct calculation of Y activity after 2 half-lives: \( 40\text{ cpm} \) [1 mark].
- Correct sum of the two activities: \( 42.5\text{ cpm} \) [1 mark].

(c)
- Statement that paper stops alpha, so inserting paper drops the count rate (proves alpha is present) [1 mark].
- Statement that aluminum stops beta, so inserting aluminum stops the remaining radiation / drops it to background levels (proves beta is present) [1 mark].

(a)
Using the equation for pressure in a liquid: \( p = h \rho g \)
\( p_{\text{top}} = 0.20\text{ m} \times 1000\text{ kg/m}^3 \times 10\text{ N/kg} = 2000\text{ Pa} \) (or \( \text{N/m}^2 \))

(b)
\( p_{\text{bottom}} = 0.60\text{ m} \times 1000\text{ kg/m}^3 \times 10\text{ N/kg} = 6000\text{ Pa} \) (or \( \text{N/m}^2 \))

(c)
Using the equation: \( \text{Force} = \text{Pressure} \times \text{Area} \)
- Force on top surface (downwards):
\( F_{\text{top}} = 2000\text{ Pa} \times 0.05\text{ m}^2 = 100\text{ N} \)
- Force on bottom surface (upwards):
\( F_{\text{bottom}} = 6000\text{ Pa} \times 0.05\text{ m}^2 = 300\text{ N} \)

(d)
- Upthrust is the resultant force due to the water:
\( \text{Upthrust} = F_{\text{bottom}} - F_{\text{top}} = 300\text{ N} - 100\text{ N} = 200\text{ N} \)
- Physical cause of upthrust: Pressure in a fluid increases with depth. This means the pressure (and therefore the upward force) acting on the bottom surface of a submerged object is always greater than the pressure (and downward force) acting on the top surface. The difference between these forces creates a net upward force, which is upthrust.

(a)
- Correct calculation of top pressure: \( 2000\text{ Pa} \) (or \( \text{N/m}^2 \)) [1 mark].

(b)
- Correct calculation of bottom pressure: \( 6000\text{ Pa} \) (or \( \text{N/m}^2 \)) [1 mark].

(c)
- Correct calculation of force on top surface: \( 100\text{ N} \) [1 mark].
- Correct calculation of force on bottom surface: \( 300\text{ N} \) [1 mark].

(d)
- Correct calculation of upthrust: \( 200\text{ N} \) [1 mark].
- Explanation that upthrust is caused by the difference in pressure between the top and bottom surfaces due to depth [1 mark].

1. **Main Sequence Star (Massive):**
- Nuclear fusion of hydrogen into helium occurs in the core.
- The star is stable because the outward radiation pressure from fusion balances the inward pull of gravitational force.

2. **Red Supergiant:**
- When the hydrogen runs out, the core collapses under gravity and heats up.
- Fusion of heavier elements (up to iron) begins in the core.
- The outer layers of the star expand greatly and cool, turning the star into a red supergiant.

3. **Supernova:**
- Once fusion can no longer occur (when the core is mostly iron), the star collapses rapidly under gravity.
- This catastrophic collapse causes a massive explosion called a supernova, releasing immense energy and throwing outer layers of dust and heavy elements into space.

4. **Neutron Star or Black Hole (Final Stage):**
- The remaining core collapses. If the remnants are extremely massive, gravity is so strong that it collapses completely into a black hole.
- If the remnants are slightly less massive, it collapses into a highly dense neutron star.

5. **Comparison with a Sun-sized star:**
- A Sun-sized star expands into a Red Giant (not supergiant), and ends with its outer layers drifting away as a planetary nebula, leaving a stable, dense White Dwarf.
- Massive stars have a violent end (supernova) and leave behind a neutron star or black hole, whereas Sun-sized stars do not undergo a supernova.

This is a Level of Response question.

**Level 3 (5–6 marks):**
- Describes the sequence of stages for a high-mass star in correct chronological order: Main Sequence -> Red Supergiant -> Supernova -> Neutron Star/Black Hole.
- Explains the balance of gravitational and radiation forces during the main sequence stage.
- Directly contrasts the final stages with those of a low-mass star (Sun) which ends as a white dwarf without a supernova.

**Level 2 (3–4 marks):**
- Identifies the main stages of a high-mass star, although some details of the processes/forces may be omitted.
- Briefly compares the life cycle to that of a low-mass star.

**Level 1 (1–2 marks):**
- Mentions some correct terms (e.g., supernova, black hole, red supergiant) but the sequence or physical explanations are incomplete or confused.
- Little or no comparison with the Sun.

(a)
First, calculate the equivalent resistance of the parallel branch containing Resistors B and C (\( R_{\text{p}} \)):
\( \frac{1}{R_{\text{p}}} = \frac{1}{R_{\text{B}}} + \frac{1}{R_{\text{C}}} \)
\( \frac{1}{R_{\text{p}}} = \frac{1}{6.0\ \Omega} + \frac{1}{12.0\ \Omega} = \frac{2}{12.0} + \frac{1}{12.0} = \frac{3}{12.0\ \Omega} \)
\( R_{\text{p}} = \frac{12.0}{3} = 4.0\ \Omega \)

Next, calculate the total resistance of the circuit (\( R_{\text{total}} \)) by adding Resistor A, which is in series with the parallel combination:
\( R_{\text{total}} = R_{\text{A}} + R_{\text{p}} = 4.0\ \Omega + 4.0\ \Omega = 8.0\ \Omega \)

(b)
Using Ohm's Law (\( V = I \times R \)) for the total circuit:
\( I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{8.0\ \Omega} = 1.5\text{ A} \)

(c)
- Potential difference across Resistor A (\( V_{\text{A}} \)):
\( V_{\text{A}} = I \times R_{\text{A}} = 1.5\text{ A} \times 4.0\ \Omega = 6.0\text{ V} \)
- Potential difference across the parallel combination of B and C (\( V_{\text{p}} \)):
Since the total potential difference supplied is \( 12\text{ V} \):
\( V_{\text{p}} = V_{\text{total}} - V_{\text{A}} = 12\text{ V} - 6.0\text{ V} = 6.0\text{ V} \)
(Alternatively, \( V_{\text{p}} = I \times R_{\text{p}} = 1.5\text{ A} \times 4.0\ \Omega = 6.0\text{ V} \))

(a)
- Correct formula for parallel resistance and calculation of parallel branch resistance: \( 4.0\ \Omega \) [2 marks].
- Correct addition of series resistor A to find total equivalent resistance: \( 8.0\ \Omega \) [1 mark].

(b)
- Correct calculation of current using total resistance: \( 1.5\text{ A} \) [1 mark].

(c)
- Correct potential difference across Resistor A: \( 6.0\text{ V} \) [1 mark].
- Correct potential difference across parallel combination: \( 6.0\text{ V} \) [1 mark].