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First, calculate the resistance \(R\) of the resistor using Ohm's law:
\(R = \frac{V}{I} = \frac{6.0\text{ V}}{0.20\text{ A}} = 30\ \Omega\).

Second, calculate the new current when the voltage is increased to \(9.0\text{ V}\):
\(I_{\text{new}} = \frac{V_{\text{new}}}{R} = \frac{9.0\text{ V}}{30\ \Omega} = 0.30\text{ A}\).

Third, calculate the power dissipated at this new voltage:
\(P = V_{\text{new}} \times I_{\text{new}} = 9.0\text{ V} \times 0.30\text{ A} = 2.7\text{ W}\).

1.5 marks total:
- 0.5 marks for calculating the correct resistance of 30 ohms.
- 0.5 marks for calculating the correct new current of 0.30 A.
- 0.5 marks for calculating the correct power dissipation of 2.7 W and selecting option A.

First, find the electrical energy consumed in joules:
\text{Electrical energy} = \frac{\text{Thermal energy required}}{\text{COP}} = \frac{90 \times 10^6\text{ J}}{3.0} = 30 \times 10^6\text{ J} = 30\text{ MJ}.

Second, convert this energy from joules to kilowatt-hours (kWh):
Since \(1\text{ kWh} = 3.6 \times 10^6\text{ J}\):
\text{Energy in kWh} = \frac{30 \times 10^6\text{ J}}{3.6 \times 10^6\text{ J/kWh}} \approx 8.33\text{ kWh} \approx 8.3\text{ kWh}.

1.5 marks total:
- 0.5 marks for calculating the electrical energy required in joules (30 MJ).
- 0.5 marks for using the conversion factor (1 kWh = 3.6 MJ) correctly.
- 0.5 marks for obtaining 8.3 kWh and selecting option B.

Using the equations of motion to find acceleration (deceleration):
\(v^2 - u^2 = 2as\)
Where \(u = 4.0\text{ m/s}\), \(v = 0\text{ m/s}\), and \(s = 2.0\text{ m}\).
\(0^2 - (4.0)^2 = 2 \times a \times 2.0\)
\(-16 = 4a \implies a = -4.0\text{ m/s}^2\).

Now, calculate the braking force using Newton's Second Law:
\(F = m \times a = 0.50\text{ kg} \times (-4.0\text{ m/s}^2) = -2.0\text{ N}\).

The magnitude of the braking force is \(2.0\text{ N}\).

1.5 marks total:
- 0.5 marks for calculating the deceleration of 4 m/s² using equations of motion.
- 0.5 marks for applying F = ma to find the force.
- 0.5 marks for correctly stating the magnitude as 2.0 N and selecting option B.

First, calculate the initial gravitational potential energy (GPE) of the child at the top of the slide:
\(E_p = mgh = 30\text{ kg} \times 10\text{ N/kg} \times 3.0\text{ m} = 900\text{ J}\).

Second, calculate the final kinetic energy (KE) of the child at the bottom of the slide:
\(E_k = \frac{1}{2}mv^2 = 0.5 \times 30\text{ kg} \times (5.0\text{ m/s})^2 = 15 \times 25 = 375\text{ J}\).

Third, apply the conservation of energy principle to find the work done against friction (energy transferred to thermal store):
\(W_{\text{friction}} = E_p - E_k = 900\text{ J} - 375\text{ J} = 525\text{ J}\).

1.5 marks total:
- 0.5 marks for calculating GPE = 900 J.
- 0.5 marks for calculating KE = 375 J.
- 0.5 marks for finding the difference (525 J) as work done against friction and selecting option B.

When a wave transitions from one medium to another, its frequency remains unchanged. Thus, the frequency in the shallow water is still \(5.0\text{ Hz}\).

Using the wave equation \(v = f \lambda\) for the shallow water:
\(\lambda_{\text{shallow}} = \frac{v_{\text{shallow}}}{f} = \frac{4.0\text{ m/s}}{5.0\text{ Hz}} = 0.80\text{ m}\).

1.5 marks total:
- 0.5 marks for stating or showing that the frequency of the wave remains constant at 5.0 Hz.
- 0.5 marks for rearranging the wave equation to find wavelength.
- 0.5 marks for calculating 0.80 m and selecting option A.

Using the formula for elastic potential energy, \(E = \frac{1}{2} k x^2\):

Initial elastic potential energy stored:
\(E_{\text{initial}} = \frac{1}{2} \times 150\text{ N/m} \times (0.10\text{ m})^2 = 75 \times 0.01 = 0.75\text{ J}\).

Final elastic potential energy stored:
\(E_{\text{final}} = \frac{1}{2} \times 150\text{ N/m} \times (0.20\text{ m})^2 = 75 \times 0.04 = 3.00\text{ J}\).

Increase in stored elastic potential energy:
\(\Delta E = E_{\text{final}} - E_{\text{initial}} = 3.00\text{ J} - 0.75\text{ J} = 2.25\text{ J}\).

1.5 marks total:
- 0.5 marks for calculating the initial energy of 0.75 J.
- 0.5 marks for calculating the final energy of 3.00 J.
- 0.5 marks for calculating the difference of 2.25 J and selecting option C.

The redshift \(z\) is calculated as:
\(z = \frac{\Delta \lambda}{\lambda_0}\)
Where \(\Delta \lambda = 412\text{ nm} - 400\text{ nm} = 12\text{ nm}\), and \(\lambda_0 = 400\text{ nm}\).

\(z = \frac{12\text{ nm}}{400\text{ nm}} = 0.03\).

Because the observed wavelength has increased (shifted towards the red end of the spectrum), the light source is moving away from the observer on Earth.

1.5 marks total:
- 0.5 marks for calculating the correct change in wavelength (12 nm).
- 0.5 marks for dividing by initial wavelength to obtain z = 0.03.
- 0.5 marks for stating that redshift implies moving away from Earth and selecting option B.

First, find the fraction of activity remaining:
\(\frac{\text{Activity after 15 hours}}{\text{Initial activity}} = \frac{100\text{ Bq}}{800\text{ Bq}} = \frac{1}{8}\).

Since \(\frac{1}{8} = \left(\frac{1}{2}\right)^3\), this means that 3 half-lives have elapsed in total.

Now, calculate the length of one half-life:
\(3 \times t_{1/2} = 15\text{ hours} \implies t_{1/2} = 5\text{ hours}\).

1.5 marks total:
- 0.5 marks for establishing that the activity has halved 3 times (reduced to 1/8).
- 0.5 marks for dividing the total time (15 hours) by 3.
- 0.5 marks for calculating the half-life of 5 hours and selecting option B.

First, calculate the acceleration using Newton's Second Law: \(a = \frac{F}{m} = \frac{3600\text{ N}}{1200\text{ kg}} = 3\text{ m/s}^2\). Next, use the equation of motion: \(v^2 = u^2 + 2as\). Since the spacecraft starts from rest, \(u = 0\), which gives: \(v^2 = 0 + 2 \times 3\text{ m/s}^2 \times 150\text{ m} = 900\). Taking the square root gives: \(v = 30\text{ m/s}\).

1 mark for calculating the acceleration of \(3\text{ m/s}^2\) using \(a = \frac{F}{m}\). 0.5 marks for correctly calculating the final velocity as \(30\text{ m/s}\) using \(v^2 = u^2 + 2as\).

A thermistor is a semiconductor device. As the temperature of the thermistor increases, its resistance decreases. Because the resistance decreases, more current can flow through the circuit for a given potential difference. This happens because the extra thermal energy liberates more charge carriers (electrons) within the semiconductor material.

1 mark for identifying that resistance decreases and current increases as temperature rises. 0.5 marks for the correct physical explanation that more charge carriers are released.

First, calculate the total input power by dividing the kinetic energy input by the time in seconds: \(\text{Input Power} = \frac{12000\text{ J}}{60\text{ s}} = 200\text{ W}\). Next, use the efficiency formula to find the useful power output: \(\text{Useful Power Output} = \text{Input Power} \times \text{Efficiency} = 200\text{ W} \times 0.35 = 70\text{ W}\).

1 mark for converting the energy per minute to input power in watts (\(200\text{ W}\)). 0.5 marks for calculating the correct useful power output (\(70\text{ W}\)).

When a wave crosses a boundary and undergoes refraction, its frequency remains constant. Therefore, we can find the frequency in deep water and know it is the same in shallow water. Using the wave equation: \(v = f \lambda\), we get \(f = \frac{v}{\lambda} = \frac{12\text{ cm/s}}{2.4\text{ cm}} = 5.0\text{ Hz}\). The frequency in shallow water is also 5.0 Hz.

1 mark for calculating the deep-water frequency of \(5.0\text{ Hz}\) using the wave speed equation. 0.5 marks for stating that the frequency does not change during refraction, giving a final frequency of \(5.0\text{ Hz}\).

First, convert the extension to meters: \(x = 1.5\text{ mm} = 0.0015\text{ m}\). Calculate the spring constant: \(k = \frac{F}{x} = \frac{40\text{ N}}{0.0015\text{ m}} \approx 2.67 \times 10^4\text{ N/m}\). Next, calculate the stored elastic potential energy: \(E_e = \frac{1}{2} F x = 0.5 \times 40\text{ N} \times 0.0015\text{ m} = 0.03\text{ J}\).

1 mark for calculating the spring constant as \(2.7 \times 10^4\text{ N/m}\) (allowing rounding). 0.5 marks for calculating the correct elastic potential energy as \(0.03\text{ J}\).

First decay (alpha emission): The uranium nucleus loses an alpha particle (2 protons and 2 neutrons). The nucleon number decreases by 4 (\(238 - 4 = 234\)) and the proton number decreases by 2 (\(92 - 2 = 90\)), forming \(^{234}_{90}\text{Th}\). Second decay (beta-minus emission): A neutron in the Thorium nucleus decays into a proton and an electron. The nucleon number remains unchanged (234), while the proton number increases by 1 (\(90 + 1 = 91\)), forming \(^{234}_{91}\text{Pa}\).

1 mark for determining the correct mass number (234) and atomic number (90) of the intermediate Thorium nucleus. 0.5 marks for correctly applying the beta decay rules to obtain a mass number of 234 and an atomic number of 91.

The initial gravitational potential energy is: \(E_p = mgh = 4.0\text{ kg} \times 10\text{ N/kg} \times 3.0\text{ m} = 120\text{ J}\). The final kinetic energy is: \(E_k = \frac{1}{2}mv^2 = 0.5 \times 4.0\text{ kg} \times (6.0\text{ m/s})^2 = 72\text{ J}\). The energy lost is the work done against friction and air resistance: \(\text{Work Done} = E_p - E_k = 120\text{ J} - 72\text{ J} = 48\text{ J}\).

1 mark for calculating the initial gravitational potential energy (\(120\text{ J}\)) and the final kinetic energy (\(72\text{ J}\)). 0.5 marks for calculating the difference (\(48\text{ J}\)) as the work done against resistive forces.

First, use the equation linking potential difference, current, and resistance: \(V = I \times R\). Rearrange the equation to solve for resistance: \(R = V / I\). Substitute the given values: \(R = 6.0\text{ V} / 0.25\text{ A} = 24\ \Omega\). Next, explain how resistance changes with temperature: as the temperature of the filament increases, the metal ions vibrate more, which increases collisions with flowing electrons, thus increasing the resistance.

1 mark for the formula \(R = V / I\). 1 mark for substitution \(6.0 / 0.25\). 1 mark for correct value and unit \(24\ \Omega\). 0.5 marks for stating that the resistance increases as temperature increases.

First, calculate the acceleration of the car using the formula: \(a = (v - u) / t\). Substituting the values: \(a = (4.5\text{ m/s} - 0\text{ m/s}) / 1.5\text{ s} = 3.0\text{ m/s}^2\). Next, calculate the resultant force using Newton's second law: \(F = m \times a\). Substituting the values: \(F = 0.80\text{ kg} \times 3.0\text{ m/s}^2 = 2.4\text{ N}\). The direction of the resultant force must be in the same direction as the acceleration, which is forward (in the direction of motion).

1 mark for calculating the acceleration as \(3.0\text{ m/s}^2\). 1 mark for recalling and using \(F = m \times a\). 1 mark for the correct force of \(2.4\text{ N}\). 0.5 marks for stating that the force acts forward or in the direction of motion.

First, calculate the extension (\(x\)) of the wire: \(x = 2.06\text{ m} - 2.0\text{ m} = 0.06\text{ m}\). Next, apply Hooke's Law: \(F = k \times x\), where \(F\) is force and \(k\) is the spring constant. Rearranging the equation gives \(k = F / x\). Substitute the values: \(k = 45\text{ N} / 0.06\text{ m} = 750\). The unit of spring constant is newtons per meter (\(N/m\)).

1 mark for calculating the extension as \(0.06\text{ m}\). 1 mark for the rearranged formula \(k = F / x\). 1 mark for calculating the value of \(750\). 0.5 marks for the correct unit \(N/m\) (or \(N\ m^{-1}\)).

First, convert the power rating of the kettle from kilowatts to watts: \(P = 2.2\text{ kW} = 2200\text{ W}\). Calculate the total electrical energy input: \(E_{\text{input}} = P \times t = 2200\text{ W} \times 150\text{ s} = 330,000\text{ J}\). Now, use the efficiency formula to find the useful energy output: \(\text{Useful energy} = \text{Total energy input} \times \text{efficiency} = 330,000\text{ J} \times 0.85 = 280,500\text{ J}\) (or \(280.5\text{ kJ}\)).

0.5 marks for converting power to \(2200\text{ W}\). 1 mark for calculating total energy input as \(330,000\text{ J}\). 1 mark for multiplying the total energy by \(0.85\). 1 mark for the correct final answer of \(280,500\text{ J}\) or \(280.5\text{ kJ}\) with unit.

First, convert frequency to hertz: \(f = 95\text{ MHz} = 95 \times 10^6\text{ Hz}\). Use the wave equation: \(v = f \times \lambda\). Rearrange to find wavelength: \(\lambda = v / f\). Substitute the values: \(\lambda = (3.0 \times 10^8\text{ m/s}) / (95 \times 10^6\text{ Hz}) \approx 3.16\text{ m}\) (or \(3.2\text{ m}\) to two significant figures). If the wave enters a medium where its speed decreases and frequency remains constant, since \(\lambda = v / f\), the wavelength must also decrease proportionally.

0.5 marks for converting frequency to \(95 \times 10^6\text{ Hz}\). 1 mark for rearranging the wave equation to \(\lambda = v / f\). 1 mark for calculating wavelength as \(3.2\text{ m}\) (accept \(3.16\text{ m}\)). 1 mark for stating that the wavelength decreases because speed decreases while frequency is constant.

First, find the number of half-lives that have passed: \(24\text{ days} / 8.0\text{ days} = 3\) half-lives. After 1 half-life, the activity is \(1600 / 2 = 800\text{ Bq}\). After 2 half-lives, the activity is \(800 / 2 = 400\text{ Bq}\). After 3 half-lives, the activity is \(400 / 2 = 200\text{ Bq}\). Alpha radiation is the most ionizing type of nuclear radiation because of its large mass and positive charge (\(+2\)).

1 mark for calculating that 3 half-lives have passed. 1 mark for showing a clear halving sequence or method. 1 mark for the correct activity of \(200\text{ Bq}\). 0.5 marks for identifying alpha radiation as the most ionizing.

First, calculate the gravitational potential energy (\(E_p\)) of the diver at the top: \(E_p = m \times g \times h = 60\text{ kg} \times 10\text{ N/kg} \times 10\text{ m} = 6000\text{ J}\). Since air resistance is negligible, conservation of energy applies, meaning all gravitational potential energy is converted into kinetic energy: \(E_k = E_p = 6000\text{ J}\). As the diver falls, energy is shifted from the gravitational store to the kinetic store via the mechanical pathway.

1 mark for recalling the formula \(E_p = m \times g \times h\). 1 mark for calculating gravitational potential energy as \(6000\text{ J}\). 1 mark for applying conservation of energy to state that kinetic energy is \(6000\text{ J}\). 0.5 marks for identifying the mechanical pathway.

Redshift is the increase in wavelength of light from distant galaxies, shifting it towards the red end of the spectrum. This Doppler-like effect indicates that the galaxy is moving away from Earth. The cosmic evidence discovered in 1965 is Cosmic Microwave Background Radiation (CMBR), which is the remaining thermal radiation from the Big Bang, filling the universe almost uniformly in all directions.

1 mark for explaining that redshift represents an increase in wavelength. 1 mark for explaining that this means the galaxy is moving away from Earth. 1.5 marks for identifying Cosmic Microwave Background Radiation (CMBR) (accept CMBR or Cosmic Microwave Background).

First, calculate the total resistance of the wire using Ohm's Law: \(R = \frac{V}{I}\). Substituting the values: \(R = \frac{3.0\text{ V}}{1.2\text{ A}} = 2.5\ \Omega\). Next, calculate the resistance per unit length by dividing the total resistance by the length of the wire: \(\text{Resistance per unit length} = \frac{2.5\ \Omega}{0.40\text{ m}} = 6.25\ \Omega/\text{m}\).

- Recall and use \(R = V/I\) to find total resistance: \(R = \frac{3.0}{1.2} = 2.5\ \Omega\) (1.5 marks total: 1 mark for correct substitution/working, 0.5 mark for correct value). - Divide resistance by the length of the wire: \(\frac{2.5}{0.40} = 6.25\) (1.5 marks total: 1 mark for correct method, 0.5 mark for correct value). - State the correct unit as \(\Omega/\text{m}\) (0.5 marks).

First, calculate the useful gravitational potential energy (GPE) transferred: \(\text{GPE} = 4.5\text{ J} \times 0.80 = 3.6\text{ J}\). Then, state the equation for gravitational potential energy: \(\text{GPE} = mgh\). Rearranging this formula for height gives \(h = \frac{\text{GPE}}{mg}\). Substituting the known values into the equation: \(h = \frac{3.6}{0.15 \times 10} = \frac{3.6}{1.5} = 2.4\text{ m}\).

- Calculate the gravitational potential energy transferred: \(4.5 \times 0.80 = 3.6\text{ J}\) (1 mark). - State the formula \(\text{GPE} = mgh\) or rearrange for \(h\) (1 mark). - Substitute correct values and calculate the numerical value of height: \(h = \frac{3.6}{1.5} = 2.4\) (1 mark). - Give the correct unit (\(\text{m}\)) (0.5 marks).

First, calculate the thermal energy required to heat the water using \(Q = m c \Delta T\):
\(Q = 10\text{ kg} \times 4200\text{ J/(kg}^\circ\text{C)} \times (45 - 15)^\circ\text{C} = 10 \times 4200 \times 30 = 1,260,000\text{ J}\).

Second, calculate the total power received by the solar panel from the Sun:
\(P_{\text{input}} = \text{intensity} \times \text{area} = 700\text{ W/m}^2 \times 2.0\text{ m}^2 = 1400\text{ W}\).

Third, calculate the useful power output using the efficiency of \(50\%\):
\(P_{\text{useful}} = 1400\text{ W} \times 0.50 = 700\text{ W}\).

Fourth, calculate the time taken in seconds:
\(t = \frac{Q}{P_{\text{useful}}} = \frac{1,260,000\text{ J}}{700\text{ W}} = 1800\text{ seconds}\).

Finally, convert the time into minutes:
\(\text{Time in minutes} = \frac{1800}{60} = 30\text{ minutes}\).

1 mark: Correct calculation of the energy required to heat the water (\(1,260,000\text{ J}\)).
1 mark: Correct calculation of the useful output power of the solar panel (\(700\text{ W}\)).
1 mark: Correct application of \(t = E/P\) to find the time in seconds (\(1800\text{ s}\)).
1 mark: Correct conversion of time from seconds to minutes (\(30\text{ minutes}\)).

First, find the resistance of the second wire. Resistance \(R\) is proportional to length \(L\) and inversely proportional to cross-sectional area \(A\):
\(R = \rho \frac{L}{A}\).
Since the second wire is twice as long and has half the area:
\(R_2 = R_1 \times \frac{2}{0.5} = 8.0\ \Omega \times 4 = 32\ \Omega\).

Second, calculate the power dissipated by the second wire:
\(P = \frac{V^2}{R_2} = \frac{12^2}{32} = \frac{144}{32} = 4.5\text{ W}\).

Third, convert the time from minutes to seconds:
\(t = 5.0\text{ minutes} \times 60\text{ s/min} = 300\text{ s}\).

Fourth, calculate the energy transferred:
\(E = P \times t = 4.5\text{ W} \times 300\text{ s} = 1350\text{ J}\).

1 mark: Correct calculation of the resistance of the second wire (\(32\ \Omega\)).
1 mark: Correct calculation of the electrical power of the second wire (\(4.5\text{ W}\)).
1 mark: Correct conversion of time to seconds (\(300\text{ s}\)).
1 mark: Correct calculation of total energy transferred (\(1350\text{ J}\)).

Method 1: Using work-energy equivalence:
First, calculate the initial kinetic energy of the car:
\(E_k = \frac{1}{2} m v^2 = 0.5 \times 1200\text{ kg} \times (15\text{ m/s})^2 = 600 \times 225 = 135,000\text{ J}\).

Second, equate kinetic energy to the work done by the braking force to stop the car:
\(W = F \times d\)
\(135,000\text{ J} = 3600\text{ N} \times d\).

Third, solve for the stopping distance \(d\):
\(d = \frac{135,000}{3600} = 37.5\text{ m}\).

Method 2: Using Newton's Second Law and equations of motion:
First, calculate deceleration:
\(a = \frac{F}{m} = \frac{-3600\text{ N}}{1200\text{ kg}} = -3.0\text{ m/s}^2\).

Second, use \(v^2 = u^2 + 2as\):
\(0 = 15^2 + 2(-3)s\)
\(6s = 225\)
\(s = 37.5\text{ m}\).

1 mark: Correct calculation of the car's initial kinetic energy (\(135,000\text{ J}\)) OR correct deceleration (\(3.0\text{ m/s}^2\)).
1 mark: Stating a correct formula relating energy, force and distance (\(W = F \times d\)) OR equations of motion (\(v^2 = u^2 + 2as\)).
1 mark: Correct substitution of values into chosen formula (e.g. \(135,000 = 3600 \times d\) or \(0 = 225 - 6s\)).
1 mark: Correct calculation of the stopping distance (\(37.5\text{ m}\)).

First, calculate the gravitational potential energy (GPE) of the car at the top of the hill:
\(E_p = m g h = 200\text{ kg} \times 10\text{ N/kg} \times 10\text{ m} = 20,000\text{ J}\).

Second, calculate the kinetic energy (KE) of the car at the bottom by subtracting the energy transferred to the surroundings:
\(E_k = E_p - E_{\text{lost}} = 20,000\text{ J} - 5600\text{ J} = 14,400\text{ J}\).

Third, equate this to the kinetic energy formula to solve for velocity:
\(E_k = \frac{1}{2} m v^2\)
\(14,400\text{ J} = 0.5 \times 200\text{ kg} \times v^2\)
\(14,400 = 100 v^2\)
\(v^2 = 144\).

Fourth, calculate the square root of 144 to find the velocity:
\(v = \sqrt{144} = 12\text{ m/s}\).

1 mark: Correct calculation of the initial GPE (\(20,000\text{ J}\)).
1 mark: Correct calculation of the final KE (\(14,400\text{ J}\)).
1 mark: Correct algebraic substitution into \(E_k = \frac{1}{2} m v^2\) showing \(v^2 = 144\).
1 mark: Correct speed of the car (\(12\text{ m/s}\)).

First, calculate the speed of the wave using distance and time:
\(v = \frac{\text{distance}}{\text{time}} = \frac{1.2\text{ m}}{3.0\text{ s}} = 0.4\text{ m/s}\).

Second, use the wave equation to find the wavelength in metres:
\(v = f \lambda \implies \lambda = \frac{v}{f}\)
\
\lambda = \frac{0.4\text{ m/s}}{8.0\text{ Hz}} = 0.05\text{ m}\).

Third, convert the wavelength from metres to centimetres:
\(\lambda = 0.05\text{ m} \times 100\text{ cm/m} = 5.0\text{ cm}\).

1 mark: Correct calculation of wave speed (\(0.4\text{ m/s}\)).
1 mark: Correct rearrangement of the wave equation (\(\lambda = v / f\)).
1 mark: Correct calculation of the wavelength in metres (\(0.05\text{ m}\)).
1 mark: Correct conversion of wavelength to centimetres (\(5.0\text{ cm}\)).

First, calculate the initial extension \(x_1\) under the load of \(6.0\text{ N}\):
\(x_1 = 18.0\text{ cm} - 15.0\text{ cm} = 3.0\text{ cm} = 0.03\text{ m}\).

Second, calculate the spring constant \(k\) using Hooke's Law:
\(F = k x_1 \implies k = \frac{F}{x_1} = \frac{6.0\text{ N}}{0.03\text{ m}} = 200\text{ N/m}\).

Third, calculate the new extension \(x_2\) when the spring is stretched to a total length of \(22.0\text{ cm}\):
\(x_2 = 22.0\text{ cm} - 15.0\text{ cm} = 7.0\text{ cm} = 0.07\text{ m}\).

Fourth, calculate the energy stored using the elastic potential energy formula:
\(E_e = \frac{1}{2} k x_2^2 = 0.5 \times 200\text{ N/m} \times (0.07\text{ m})^2 = 100 \times 0.0049 = 0.49\text{ J}\).

1 mark: Correct calculation of the spring constant (\(200\text{ N/m}\)) showing conversion of extension to metres.
1 mark: Correct determination of the new extension as \(0.07\text{ m}\) (or \(7.0\text{ cm}\)).
1 mark: Correct use of the formula \(E_e = \frac{1}{2} k x^2\) with substitution.
1 mark: Correct final stored energy (\(0.49\text{ J}\)).

First, calculate the total resistance of the parallel combination using Ohm's Law:
\(R_{\text{total}} = \frac{V}{I} = \frac{12\text{ V}}{2.0\text{ A}} = 6.0\ \Omega\).

Second, write down the formula for the equivalent resistance of two parallel resistors:
\(\frac{1}{R_{\text{total}}} = \frac{1}{R_{\text{A}}} + \frac{1}{R_{\text{B}}}\).

Third, substitute the known values into the equation:
\(\frac{1}{6} = \frac{1}{10} + \frac{1}{R}\).

Fourth, rearrange and solve for \(R\):
\(\frac{1}{R} = \frac{1}{6} - \frac{1}{10} = \frac{5}{30} - \frac{3}{30} = \frac{2}{30} = \frac{1}{15}\).
Therefore, \(R = 15\ \Omega\).

1 mark: Correct calculation of total equivalent resistance (\(6.0\ \Omega\)).
1 mark: Stating the correct formula for parallel resistors (\(\frac{1}{R_{\text{total}}} = \frac{1}{R_{\text{A}}} + \frac{1}{R_{\text{B}}}\)).
1 mark: Correct substitution into the parallel resistor formula (\(\frac{1}{6} = \frac{1}{10} + \frac{1}{R}\)).
1 mark: Correct evaluation of resistance \(R = 15\ \Omega\).

First, find the fraction of the initial activity remaining:
\(\text{Fraction} = \frac{25\text{ Bq}}{800\text{ Bq}} = \frac{1}{32}\).

Second, determine how many half-lives correspond to this fraction:
Since \(\left(\frac{1}{2}\right)^5 = \frac{1}{32}\), the sample has undergone 5 half-lives.

Third, use the total time elapsed and the number of half-lives to find the duration of one half-life:
\(\text{Half-life} = \frac{\text{Total time}}{\text{Number of half-lives}} = \frac{15\text{ days}}{5}\).

Fourth, calculate the final value:
\(\text{Half-life} = 3.0\text{ days}\).

1 mark: Correct calculation of the fraction of activity remaining (\(1/32\) or \(0.03125\)).
1 mark: Deducing that the sample has undergone 5 half-lives.
1 mark: Setting up the calculation for half-life duration (\(15 / 5\)).
1 mark: Correct final answer with unit (\(3.0\text{ days}\)).

When light intensity increases, the LDR absorbs more light energy. This absorbed energy is transferred to the semiconductor material, releasing more free charge carriers (electrons). This increase in the concentration of conduction electrons decreases the resistance of the LDR. Since the potential difference supplied by the battery remains constant, the decrease in resistance causes the current in the circuit to increase, according to the relationship \(I = \frac{V}{R}\).

1. States that current increases when light intensity increases (1 mark)
2. Explains that the LDR absorbs light energy (1 mark)
3. Explains that this energy releases/liberates extra charge carriers (electrons) in the semiconductor (1 mark)
4. Identifies that this increase in charge carriers reduces the resistance of the LDR (1 mark)
5. Connects the reduced resistance to increased current using Ohm's law / constant potential difference (1 mark)

Immediately after jumping, the only force acting on the skydiver is weight, causing them to accelerate downwards. As their speed increases, air resistance (drag) increases in the opposite direction. Eventually, air resistance increases until it is equal and opposite to the weight. At this point, the resultant force is zero, and according to Newton's First Law, the skydiver travels at a constant speed called terminal velocity. When the parachute is opened, the large surface area dramatically increases air resistance, making it much larger than weight. The upward resultant force causes the skydiver to decelerate (Newton's Second Law). As their speed decreases, the air resistance decreases until it once again equals the weight. A new, lower terminal velocity is then reached where the forces are balanced again.

1. Explains that increasing speed causes air resistance to increase until it equals weight, resulting in zero resultant force and first terminal velocity (1 mark)
2. Identifies that opening the parachute increases surface area, significantly increasing air resistance (1 mark)
3. States that the upward air resistance is now greater than weight, causing a net upward force (1 mark)
4. Explains that this resultant force causes deceleration / decrease in speed (1 mark)
5. Explains that as speed decreases, air resistance decreases until it equals weight again, establishing a lower terminal velocity (1 mark)

Cavity wall insulation filled with a foam or fiber material contains trapped pockets of air. Since air is a very poor conductor, and the solid fibers/foam are also poor conductors, this significantly reduces energy transfer by conduction. Because the air is trapped in tiny pockets, it cannot circulate and set up convection currents, which effectively stops energy transfer by convection across the cavity. The reflective foil backing on the insulation has a shiny surface that reflects infrared radiation (heat) back towards the inner wall of the house rather than absorbing it. Additionally, shiny, light surfaces are poor emitters of radiation, which further reduces radiative energy transfer across any remaining air gaps to the outer wall.

1. Explains that conduction is reduced because the insulation material and/or trapped air are poor conductors (1 mark)
2. Explains that convection is reduced because the trapped air pockets prevent the air from moving/circulating to form convection currents (1 mark)
3. Explains that radiation is reduced because the shiny foil backing reflects infrared radiation back into the house (1 mark)
4. Explains that the shiny foil is a poor emitter of radiation, reducing transfer across gaps (1 mark)
5. Demonstrates a clear and structured connection between each transfer mechanism (conduction, convection, radiation) and the corresponding feature of the insulation (1 mark)

Glass is an optically denser medium than air, which causes the speed of the light wave to decrease as it enters the glass block. Because the light wave approaches the boundary at an angle, different parts of each wavefront reach the glass at different times. The side of the wavefront that enters the glass first slows down immediately, while the rest of the wavefront still in the air continues at the higher speed. This difference in speed across the wavefront causes the wave to pivot and change direction, bending towards the normal. Since the frequency of the light wave remains constant when crossing boundaries, the decrease in speed leads to a corresponding decrease in wavelength inside the glass, in accordance with the wave equation \(v = f \lambda\).

1. Identifies that light travels slower in glass than in air (1 mark)
2. Explains that because the wave enters at an angle, one side of each wavefront enters the glass block before the other (1 mark)
3. Explains that the side entering first slows down first, while the other side is still moving faster (1 mark)
4. Explains that this difference in speed across the wavefront causes the wave to pivot/bend towards the normal (1 mark)
5. States that frequency is unchanged, so the wavelength decreases inside the glass (1 mark)

During elastic deformation, the applied stretching force pulls the metal atoms slightly further apart from their equilibrium positions, stretching the chemical bonds between them. Crucially, the bonds do not break. When the force is removed, the attractive forces between the atoms pull them back to their original positions, so the wire returns to its original length. During plastic deformation, the force applied is larger than the elastic limit of the material. This large force causes planes of atoms to slide past one another. In this process, atomic bonds are broken and new bonds are formed in different positions. When the force is removed, the atoms do not return to their original positions, resulting in a permanent change in the shape and length of the wire.

1. Explains that in elastic deformation, atoms are pulled apart and bonds stretch but do not break (1 mark)
2. States that when the force is removed, atoms return to their original equilibrium positions with no permanent change (1 mark)
3. Explains that in plastic deformation, planes of atoms slide past one another (1 mark)
4. Explains that in plastic deformation, atomic bonds are broken and reform in new positions (1 mark)
5. States that when the force is removed, atoms remain in their new positions, resulting in permanent deformation (1 mark)

Alpha radiation is highly ionizing because of its large mass and double positive charge, but it has very low penetrating power, being absorbed by a few centimeters of air or the outer layer of dead skin cells. When an alpha source is outside the body, the alpha particles cannot penetrate the skin to reach living tissue, posing minimal risk. However, if inhaled or swallowed, the alpha source is in direct contact with delicate internal living cells and organs; because alpha is highly ionizing, it causes severe damage, cell mutation, or death. In contrast, gamma radiation is highly penetrating and can easily pass through human skin and tissue. Therefore, a gamma source outside the body can easily penetrate to damage internal organs. If swallowed, gamma radiation is still hazardous because it is highly penetrating, although many gamma photons will pass straight out of the body due to their low ionizing power, those that are absorbed will still damage internal organs.

1. States that alpha radiation is highly ionizing but has low penetrating power / is stopped by skin (1 mark)
2. Explains that outside the body, alpha is not hazardous because it cannot penetrate the dead outer layer of skin (1 mark)
3. Explains that inside the body, alpha is highly hazardous because it is in direct contact with living cells and causes intense local ionization/DNA damage (1 mark)
4. States that gamma radiation is highly penetrating and can pass through skin and tissue (1 mark)
5. Explains that gamma is hazardous outside the body because it can penetrate to reach internal organs, and remains hazardous inside the body for the same reason (1 mark)

To conduct this investigation, wrap equal thicknesses of bubble wrap, newspaper, and cotton wool around three identical beakers. Use a fourth beaker without any insulation as a control. Pour equal volumes of hot water at the same starting temperature into each beaker. Cover each beaker with a lid to minimize heat loss from the top. Insert a thermometer through a small hole in each lid.

Start a stopwatch and record the initial temperature of the water in each beaker. Record the temperature of the water at regular intervals (e.g., every minute) for 10 minutes.

To ensure accuracy:
1. Ensure the thermometer bulb is completely submerged in the water and does not touch the bottom or sides of the beaker.
2. Read the thermometer at eye level to avoid parallax error.
3. Stir the water gently before taking a reading to ensure uniform temperature.

The beaker with the smallest temperature drop over the 10-minute period is the best thermal insulator.

Key control variables:
- The volume of hot water used in each beaker (measured using a measuring cylinder).
- The initial temperature of the water in each beaker.
- The material, size, and shape of the beakers.
- The room temperature (conduct all trials in the same room away from drafts).

This is an extended response question where the quality of written communication is assessed.

**Level 3 (6-7 marks)**:
A detailed, coherent, and logical experimental description. The candidate includes a complete list of apparatus, a clear explanation of how to measure the temperature drop over a set time, detailed accuracy steps (such as using a lid or avoiding parallax error), and identifies at least three control variables with a practical method to keep them constant.

**Level 2 (3-5 marks)**:
A reasonable description of the method. The candidate mentions most of the necessary apparatus, explains how to measure temperature changes, and identifies at least two control variables, though details on accuracy or specific control methods may be lacking.

**Level 1 (1-2 marks)**:
A simple description of wrapping beakers and measuring temperature. There is limited mention of control variables, and no clear explanation of how to ensure accuracy or a fair test.

**0 marks**:
No response or completely irrelevant response.

**Key points to look for:**
- **Apparatus**: Beakers, thermometer, stopwatch, measuring cylinder, three insulating materials, lid.
- **Measurements**: Initial and final temperature (or temperature at fixed intervals) over a set time (e.g., 10 minutes).
- **Accuracy**: Use of a lid; thermometer bulb fully submerged and not touching sides; reading at eye level.
- **Control Variables**: Same volume of water, same initial water temperature, same type/size of beaker, same thickness of insulation.

To investigate how resistance varies with length:

1. **Circuit Setup**: Set up a series circuit containing a low-voltage DC power supply, an ammeter, a switch, and a test wire taped to a metre ruler. Connect a voltmeter in parallel across the portion of the test wire being measured using crocodile clips.

2. **Measurements**: Attach the first crocodile clip at the 0 cm mark on the ruler and the second crocodile clip at 10 cm. Close the switch, record the current \(I\) from the ammeter and the potential difference \(V\) from the voltmeter. Open the switch immediately after taking the readings. Move the second crocodile clip to 20 cm, 30 cm, up to 100 cm, repeating the process to get at least five different lengths.

3. **Analysis**: For each length \(L\), calculate the resistance using Ohm's law: \(R = \frac{V}{I}\). Plot a graph of resistance on the y-axis against length on the x-axis. A straight line passing through the origin demonstrates that the resistance of the wire is directly proportional to its length.

4. **Safety Precaution**: The wire can become hot when current passes through it, which is a burn hazard and also increases the resistance of the wire (affecting the results). To prevent this, use a low current/voltage and open the switch between readings to allow the wire to cool.

This is an extended response question where the quality of written communication is assessed.

**Level 3 (6-7 marks)**:
A comprehensive circuit description with a clear, step-by-step practical procedure. Explains how to measure length, current, and potential difference. Clearly states how to calculate resistance using \(R = \frac{V}{I}\), describes plotting a graph to find the relationship, and provides a correct safety precaution with scientific justification.

**Level 2 (3-5 marks)**:
A good description of the circuit and method. The candidate describes taking current and voltage measurements at different lengths, calculates resistance, but may provide a less detailed analysis (e.g., omits graphing) or lacks a fully explained safety precaution.

**Level 1 (1-2 marks)**:
A simple circuit description or method. Mentions connecting components and changing the wire length, but lacks details on how resistance is calculated or analyzed, with little or no safety consideration.

**0 marks**:
No response or completely irrelevant response.

**Key points to look for:**
- **Circuit**: Ammeter in series, voltmeter in parallel with the test wire, power supply, switch.
- **Method**: Use crocodile clips to adjust the length of the wire on a metre ruler.
- **Measurements**: Measure length \(L\), potential difference \(V\), and current \(I\).
- **Analysis**: Calculate \(R = \frac{V}{I}\) and plot a graph of resistance vs. length to show direct proportionality.
- **Safety**: Switch off current between readings to prevent overheating (burn risk / resistance change).

To investigate the force-extension relationship of a spring:

1. **Apparatus & Setup**: Attach a spring to a clamp hanging from a stand. Securely clamp a metre ruler vertically next to the spring, ensuring the zero mark is at the top. Attach a horizontal pointer to the bottom of the spring so it aligns with the ruler scale.

2. **Measurements & Accuracy**:
- Measure the original position of the pointer on the ruler with no masses attached (the unstretched length).
- Add a 100 g mass hanger (which exerts a force of 1.0 N) to the spring. Record the new position of the pointer.
- Calculate the extension: \(\text{extension} = \text{new position} - \text{original position}\).
- Add more 100 g masses one at a time up to a maximum (e.g., 600 g), recording the pointer position at each step.
- To ensure accuracy and avoid parallax error, ensure the eye is at the same horizontal level as the pointer when reading the scale. Use a set square to ensure the ruler is perfectly vertical.

3. **Determining the Spring Constant**:
- Convert the masses to forces in Newtons (using \(W = mg\), where \(g \approx 10\text{ N/kg}\), so 100 g = 1.0 N).
- Plot a graph with Force on the y-axis and Extension on the x-axis.
- Draw a line of best fit. Within the limit of proportionality, this will be a straight line through the origin.
- Determine the spring constant \(k\) by finding the gradient of the linear region of the graph: \(k = \frac{\Delta F}{\Delta x}\).

This is an extended response question where the quality of written communication is assessed.

**Level 3 (6-7 marks)**:
Clear and logical description of the experimental setup and step-by-step procedure. Detailed explanation of how to calculate extension and avoid parallax/systematic errors. Precise explanation of how to use a force-extension graph and calculate the spring constant from its gradient.

**Level 2 (3-5 marks)**:
A good description of the method. Explains adding weights and measuring extension. Describes plotting a graph but may lack precision on avoiding errors or details of how the gradient relates to the spring constant.

**Level 1 (1-2 marks)**:
A basic experimental setup is described. Mentions adding masses and measuring the spring, but lacks details on how extension is calculated or how to find the spring constant.

**0 marks**:
No response or completely irrelevant response.

**Key points to look for:**
- **Setup**: Stand, clamp, spring, vertical ruler, pointer, masses.
- **Method**: Measure original length, add masses, measure extended length, calculate extension \((\text{extension} = \text{length} - \text{original length})\).
- **Accuracy**: Read at eye level (avoid parallax), ensure ruler is vertical, use a pointer.
- **Analysis**: Plot force vs. extension. Find the gradient of the straight-line section to determine the spring constant \(k\) (since \(F = kx\)).

To investigate the refraction of light:

1. **Setup & Tracing**:
- Place a rectangular glass block on a sheet of white paper and trace its outline using a sharp pencil.
- Use a ray box to shine a single, narrow beam of light at an angle into one of the long sides of the block.
- Mark the path of the incident ray and the emergent ray by placing small crosses (at least two for each ray) on the paper.
- Remove the block. Draw straight lines through the crosses to represent the incident and emergent rays. Draw a straight line connecting the point where the incident ray entered the block to the point where the emergent ray left it; this represents the path of the refracted ray inside the block.

2. **Measurements**:
- Draw a normal line (a line perpendicular to the surface of the block, at \(90^\circ\)) at the point of entry.
- Use a protractor to measure the angle of incidence \(i\) (between the normal and the incident ray) and the angle of refraction \(r\) (between the normal and the refracted ray inside the block).
- Repeat this procedure for several different angles of incidence (e.g., \(15^\circ, 30^\circ, 45^\circ, 60^\circ\)).

3. **Analysis & Conclusion**:
- Compare the angles of incidence \(i\) and refraction \(r\).
- Refraction is demonstrated because the light changes direction at the boundary. As light enters the optically denser glass block, it slows down and bends towards the normal, meaning that the angle of refraction \(r\) is always less than the angle of incidence \(i\) (for \(i > 0^\circ\)).

This is an extended response question where the quality of written communication is assessed.

**Level 3 (6-7 marks)**:
A comprehensive explanation of tracing the light rays using a ray box. Clear, logical steps on how to mark the path of the light inside the glass block. Detailed description of how to measure the angles of incidence and refraction accurately using a protractor. Correctly explains how the direction of the ray changes (bends towards the normal) and how the data shows that refraction has occurred.

**Level 2 (3-5 marks)**:
A good description of tracing the ray and measuring angles. Identifies the correct angles but may lack detail on how to mark the path inside the block or how to draw clear conclusions about the direction change.

**Level 1 (1-2 marks)**:
A simple description of shining light through glass. Mentions tracing or measuring angles but lacks clear instructions or systematic procedure.

**0 marks**:
No response or completely irrelevant response.

**Key points to look for:**
- **Setup**: Glass block on paper, trace outline, use ray box.
- **Tracing**: Mark rays with dots/crosses, remove block, draw lines to show the path inside the block.
- **Measurements**: Draw a normal line. Measure angle of incidence \(i\) and angle of refraction \(r\) using a protractor.
- **Analysis**: Contrast \(i\) and \(r\). Explain that light bends towards the normal as it enters the glass because glass is denser than air, so \(r < i\).

### 1. Payback Time Calculations:
* **Method A (Cavity Wall Insulation):**
$$\text{Payback Time} = \frac{\text{Installation Cost}}{\text{Annual Saving}} = \frac{1200}{240} = 5\text{ years}$$
* **Method B (Double Glazing):**
$$\text{Payback Time} = \frac{\text{Installation Cost}}{\text{Annual Saving}} = \frac{4000}{160} = 25\text{ years}$$

### 2. Mechanisms of Energy Transfer Reduction:
* **Cavity Wall Insulation:**
* An empty cavity allows heat to transfer via convection because air can circulate inside the gap, carrying heat from the inner wall to the outer wall.
* Filling the cavity with foam or mineral wool traps the air in tiny pockets, preventing it from circulating and thus eliminating convection currents.
* Air and the insulating material are poor conductors, which also significantly reduces heat loss by conduction.
* **Double Glazing:**
* Consists of two glass panes separated by a sealed gap containing a vacuum, dry air, or an inert gas (like argon).
* Glass is a poor thermal conductor, and the gas/vacuum gap has very low thermal conductivity, which greatly reduces conduction.
* If a vacuum is used, conduction and convection are completely eliminated because no particles are present. If a narrow gas gap is used, it is too narrow for convection currents to easily form, minimizing convection.

### 3. Recommendation:
* The homeowner should prioritize **Method A (Cavity Wall Insulation)** first.
* It has a much shorter payback time (5 years vs 25 years), meaning the initial investment is recovered 5 times faster.
* It also provides a greater absolute financial saving per year (£240 vs £160) for a much lower upfront cost (£1200 vs £4000).

**Level 3 (5–6 marks):**
* Correctly calculates both payback times (5 years and 25 years) with clear working shown.
* Explains in detail how both methods reduce energy transfer, specifically mentioning conduction and the role of trapped air/convection in cavity insulation, and the low-conductivity gap/vacuum in double glazing.
* Offers a clear, reasoned recommendation to prioritize Method A based on both cost-effectiveness (shorter payback time and lower initial cost) and annual savings.

**Level 2 (3–4 marks):**
* Calculates at least one payback time correctly, or both with minor calculation errors.
* Provides a basic explanation of how one or both methods reduce energy loss, mentioning conduction or convection.
* Recommends Method A with some basic financial justification.

**Level 1 (1–2 marks):**
* Attempts a payback calculation with significant errors, or simply states which is cheaper.
* Makes simple statements about insulating materials keeping a house warm without detailed explanation of mechanisms.
* Shows limited structure and clarity of explanation.

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

### 1. Analysis of Results and Radiation Identification
To find the actual radiation from the source, we subtract the background count rate of 20 cpm:
* **Net count rate with no barrier:** $420 - 20 = 400\text{ cpm}$
* **Net count rate with thin paper:** $422 - 20 = 402\text{ cpm}$
* *Deduction:* Since there is virtually no change in the count rate, **no alpha radiation** is emitted. Alpha radiation is highly ionizing and would have been absorbed/stopped by the thin paper, causing a noticeable drop.
* **Net count rate with thin aluminium:** $218 - 20 = 198\text{ cpm}$
* *Deduction:* The count rate drops by approximately half (from ~400 cpm to ~200 cpm). Since aluminium stops beta radiation, this indicates that **beta radiation is present** in the emission.
* **Net count rate with thick lead:** $22 - 20 = 2\text{ cpm}$ (effectively zero net counts above background)
* *Deduction:* The remaining radiation (~200 cpm) which passed through aluminium is completely absorbed by the lead block. Since gamma radiation is highly penetrating and absorbed by thick lead, this indicates that **gamma radiation is present**.
* **Conclusion:** The source emits **beta** and **gamma** radiation, but not alpha.

### 2. Safety Precautions
* **Distance:** Use long tongs or forceps to handle the source, maintaining as much distance as possible to reduce intensity/exposure.
* **Direction:** Point the source away from the body, especially eyes and vital organs.
* **Time:** Minimize the time that the radioactive source is out of its protective storage.
* **Shielding:** Store the source in its lead-lined wooden box immediately after use.

### 3. Accounting for Background Radiation
* Measure the background count rate in the room *before* bringing the radioactive source into the area.
* Subtract this background count rate (20 cpm) from all subsequent experimental measurements to obtain the true (net) activity of the source.

**Level 3 (5–6 marks):**
* Correctly identifies that the source emits **beta** and **gamma** radiation, but **no alpha** radiation.
* Fully justifies these deductions using the data, explicitly referencing the absorption properties of paper (does not stop any, so no alpha), aluminium (stops beta, hence the drop to ~200 cpm net), and lead (stops gamma, hence the drop to background level).
* Describes key safety precautions (e.g., handling with tongs, storage in lead container) and explains how to account for background radiation by subtracting it from the measured counts.

**Level 2 (3–4 marks):**
* Correctly identifies at least one correct radiation type with some justification from the data.
* Mentions safety precautions or explains how background radiation is handled, though the description may lack detail or fail to explicitly show the subtraction of 20 cpm.

**Level 1 (1–2 marks):**
* Identifies a radiation type but with incorrect or missing justification.
* Lists general safety rules with little reference to the specific procedure, or makes a simple mention of background radiation without explaining how it is used in calculations.

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