2024 OCR GCSE Gateway Science - Physics A - J249 Practice Paper with Answers

The motion is split into two phases:
1. Acceleration phase: The average speed is \(\frac{0 + 12}{2} = 6\text{ m/s}\). The distance travelled is \(6\text{ m/s} \times 4\text{ s} = 24\text{ m}\).
2. Constant speed phase: The distance travelled is \(12\text{ m/s} \times 6\text{ s} = 72\text{ m}\).

Total distance = \(24\text{ m} + 72\text{ m} = 96\text{ m}\).

[1 mark] C - 96 m is the correct calculation of total area under the velocity-time graph.

When a wave enters a different medium (or different depth of water), its frequency remains constant because it is determined solely by the source of the wave. Since wave speed \(v = f \lambda\), a decrease in speed \(v\) must result in a proportional decrease in wavelength \(\lambda\).

[1 mark] B - Correctly identifies that frequency remains constant and wavelength decreases.

A thermistor is a negative temperature coefficient (NTC) resistor, meaning its resistance decreases as temperature increases. As its resistance decreases, it takes a smaller share of the total potential difference. Because the total potential difference from the source is constant, the potential difference across the fixed resistor must increase.

[1 mark] C - Correctly identifies the inverse relationship between temperature and thermistor resistance, and the resulting change in the potential divider distribution.

For a given transmission power \(P = I V\), increasing the voltage \(V\) reduces the current \(I\). The power wasted as thermal energy in the transmission cables is given by \(P_{\text{loss}} = I^2 R\). Therefore, reducing the current dramatically decreases energy wasted as heat, making the National Grid highly efficient.

[1 mark] B - Correctly identifies that high voltage reduces current to minimize thermal energy loss.

The formula for elastic potential energy is:
\(E_e = \frac{1}{2} k e^2\)

Substitute the given values:
\(E_e = 0.5 \times 150\text{ N/m} \times (0.20\text{ m})^2\)
\(E_e = 75 \times 0.04 = 3.0\text{ J}\).

[1 mark] A - Correct calculation of elastic potential energy using \(E_e = \frac{1}{2} k e^2\).

The formula for pressure in a liquid column is:
\(p = h \rho g\)

Substitute the given values:
\(p = 15\text{ m} \times 1000\text{ kg/m}^3 \times 10\text{ N/kg}\)
\(p = 150,000\text{ Pa} = 150\text{ kPa}\).

[1 mark] C - Correctly calculates pressure using \(p = h \rho g\) and converts pascals (Pa) to kilopascals (kPa).

Beta particles (fast-moving electrons) have medium penetrating power. They can easily pass through air and paper but are stopped by a thin sheet of metal such as aluminium. Alpha particles would be stopped by paper, and gamma rays would pass straight through the aluminium.

[1 mark] B - Correctly identifies beta particles as the radiation stopped by aluminium but not paper.

The force on a current-carrying conductor in a magnetic field is given by \(F = B I L\) when the wire is perpendicular to the field. If the wire is placed parallel to the magnetic field lines, the force acting on it drops to zero. Therefore, this change does not increase the force.

[1 mark] C - Correctly identifies that placing the wire parallel to the field lines reduces the force to zero instead of increasing it.

In a series circuit, the total resistance is the sum of the individual resistances: \( R_{\text{total}} = 6.0\ \Omega + 4.0\ \Omega = 10.0\ \Omega \). The total current from the battery is found using Ohm's Law: \( I = \frac{V}{R} = \frac{12\text{ V}}{10.0\ \Omega} = 1.2\text{ A} \). Since current is the same at all points in a series circuit, the current through the \( 6.0\ \Omega \) resistor is also \( 1.2\text{ A} \).

1 mark for the correct option B.

First, calculate the extension of the spring: \( x = 18.0\text{ cm} - 12.0\text{ cm} = 6.0\text{ cm} = 0.060\text{ m} \). Next, use Hooke's Law \( F = k x \) to solve for the spring constant \( k \): \( k = \frac{F}{x} = \frac{3.0\text{ N}}{0.060\text{ m}} = 50\text{ N/m} \).

1 mark for the correct option C.

First, find the constant acceleration: \( a = \frac{v - u}{t} = \frac{25\text{ m/s} - 0\text{ m/s}}{5.0\text{ s}} = 5.0\text{ m/s}^2 \). The total time of travel is \( t = 5.0\text{ s} + 3.0\text{ s} = 8.0\text{ s} \). Using the distance equation starting from rest: \( s = ut + \frac{1}{2}at^2 \), with \( u = 0 \), we get \( s = \frac{1}{2} \times 5.0\text{ m/s}^2 \times (8.0\text{ s})^2 = 2.5 \times 64 = 160\text{ m} \).

1 mark for the correct option C.

First, convert mass to kilograms: \( 540\text{ g} = 0.54\text{ kg} \). Convert volume to cubic metres: \( 200\text{ cm}^3 = 200 \times 10^{-6}\text{ m}^3 = 0.00020\text{ m}^3 \). Calculate density: \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{0.54\text{ kg}}{0.00020\text{ m}^3} = 2700\text{ kg/m}^3 \).

1 mark for the correct option D.

Use the formula for change of state: \( E = mL \), where \( m = 0.40\text{ kg} \) and \( L = 3.34 \times 10^5\text{ J/kg} \). Calculating this gives: \( E = 0.40 \times 3.34 \times 10^5 = 1.336 \times 10^5\text{ J} = 133.6\text{ kJ} \).

1 mark for the correct option A.

First, find efficiency: \( \text{Efficiency} = \frac{\text{Useful power output}}{\text{Total power input}} = \frac{150\text{ W}}{250\text{ W}} = 0.60 \). Next, find the wasted power: \( \text{Wasted power} = \text{Total power input} - \text{Useful power output} = 250\text{ W} - 150\text{ W} = 100\text{ W} \).

1 mark for the correct option B.

Work done is calculated using the formula: \( W = F \times d \). In this case, the force needed to lift the crate is equal to its weight, \( 1200\text{ N} \), and the distance moved is \( 6.0\text{ m} \). Therefore, \( W = 1200\text{ N} \times 6.0\text{ m} = 7200\text{ J} \) (or \( 7.2\text{ kJ} \)). Note that the time of \( 8.0\text{ s} \) is not used because work done is independent of the time taken.

1 mark for the correct option C.

(a) Acceleration is calculated using the formula:
\(a = \frac{v - u}{t}\)
Given:
\(u = 0\text{ m/s}\)
\(v = 12\text{ m/s}\)
\(t = 8.0\text{ s}\)
\(a = \frac{12 - 0}{8.0} = 1.5\text{ m/s}^2\)

(b) Force is calculated using Newton's second law:
\(F = m \times a\)
Given:
\(m = 75\text{ kg}\)
\(a = 1.5\text{ m/s}^2\)
\(F = 75 \times 1.5 = 112.5\text{ N}\)

(c) Distance travelled under uniform acceleration:
\(s = \frac{u + v}{2} \times t\)
\(s = \frac{0 + 12}{2} \times 8.0 = 6 \times 8.0 = 48\text{ m}\)

Part (a) [3 marks total]:
- Recall and substitution of acceleration formula: \(a = \frac{v-u}{t}\) [1 mark]
- Correct calculation: 1.5 [1 mark]
- Correct unit: \(\text{m/s}^2\) [1 mark]

Part (b) [3 marks total]:
- Recall and substitution of force formula: \(F = m \times a\) [1 mark]
- Correct calculation using their value from (a): \(75 \times 1.5 = 112.5\) [1 mark] (Allow ecf from part a)
- Correct unit: N [1 mark]

Part (c) [3.375 marks total]:
- Recall and substitution into distance formula: \(s = ut + \frac{1}{2}at^2\) or \(s = \frac{u+v}{2}t\) [1.375 marks]
- Correct calculation: 48 [1 mark]
- Correct unit: m [1 mark]

(a) The total time taken for the sound pulse to go to the sea bed and back is 1.6 s. Therefore, the time taken to travel to the sea bed is:
\(t = \frac{1.6}{2} = 0.8\text{ s}\)
Depth is calculated as:
\(\text{distance} = \text{speed} \times \text{time}\)
\(d = 1500\text{ m/s} \times 0.8\text{ s} = 1200\text{ m}\)

(b) The wave equation is:
\(v = f \lambda\)
Rearranging for wavelength (\(\lambda\)):
\(\lambda = \frac{v}{f}\)
\(\lambda = \frac{1500}{40000} = 0.0375\text{ m}\) (or 3.75 cm)

(c) Ultrasound waves have a much higher frequency, and therefore a much shorter wavelength, than audible sound. This means they experience less diffraction (spreading out) and can form narrow, well-defined beams that provide higher resolution imaging of underwater features.

Part (a) [3 marks total]:
- Realisation that time must be halved (0.8 s) [1 mark]
- Correct equation used: \(d = v \times t\) [1 mark]
- Correct calculation: 1200 m [1 mark]

Part (b) [3.375 marks total]:
- Rearrangement of wave equation: \(\lambda = \frac{v}{f}\) [1.375 marks]
- Substitution of values: \(\lambda = \frac{1500}{40000}\) [1 mark]
- Correct calculation with units: 0.0375 m [1 mark]

Part (c) [3 marks total]:
- Explicit mention of high frequency / short wavelength [1 mark]
- Connects short wavelength to less diffraction / narrow beam [1 mark]
- Connects to clearer image / better resolution / accurate measurement [1 mark]

(a) First, calculate the equivalent resistance of the parallel branch (\(R_p\)):
\(\frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6}\)
So, \(R_p = 6.0\ \Omega\).
Now add the series resistor \(R_1\):
\(R_{\text{total}} = R_1 + R_p = 8.0\ \Omega + 6.0\ \Omega = 14.0\ \Omega\).

(b) Using Ohm's Law (\(V = I R\)):
\(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{14.0\ \Omega} \approx 0.857\text{ A}\) (or 0.86 A to 2 s.f.).

(c) The potential difference across \(R_1\) is:
\(V_1 = I \times R_1 = 0.857\text{ A} \times 8.0\ \Omega \approx 6.86\text{ V}\).

Part (a) [3.375 marks total]:
- Correct formula/calculation for the parallel branch: 6.0 \(\Omega\) [1.375 marks]
- Summing the series components: \(8.0 + 6.0\) [1 mark]
- Correct total resistance: 14.0 \(\Omega\) [1 mark]

Part (b) [3 marks total]:
- Recall and substitution into Ohm's Law: \(I = \frac{V}{R}\) [1 mark]
- Correct substitution of values using their answer from (a) [1 mark]
- Correct current: 0.86 A (or 0.857 A) [1 mark]

Part (c) [3 marks total]:
- Recall and substitution into \(V_1 = I \times R_1\) [1 mark]
- Substitution of their current and \(R_1\) resistance [1 mark]
- Correct potential difference calculation: 6.86 V [1 mark] (allow ecf from b)

(a) Total power output = \(15 \times 2.4\text{ MW} = 36\text{ MW}\).
In watts:
\(36\text{ MW} = 36 \times 10^6\text{ W} = 3.6 \times 10^7\text{ W}\).

(b) Average power output = \(30\%\text{ of } 36\text{ MW} = 0.30 \times 36 = 10.8\text{ MW} = 1.08 \times 10^7\text{ W}\).
Time in 1 day = \(24\text{ hours} \times 3600\text{ seconds/hour} = 86,400\text{ s}\).
Average energy produced in 1 day:
\(E = P \times t = 10.8\text{ MW} \times 86,400\text{ s} = 933,120\text{ MJ}\) (or \(9.33 \times 10^5\text{ MJ}\)).

(c) Environmental advantage: No carbon dioxide / greenhouse gases emitted during operation (does not contribute to climate change).
Environmental disadvantage: Large physical footprint / visual impact / danger to birds and bats.

Part (a) [3 marks total]:
- Correct multiplication of turbines: \(15 \times 2.4 = 36\text{ MW}\) [1 mark]
- Unit conversion: MW to W [1 mark]
- Correct standard form: \(3.6 \times 10^7\text{ W}\) [1 mark]

Part (b) [3.375 marks total]:
- Calculate 30% power: 10.8 MW [1 mark]
- Calculate seconds in a day: 86,400 s [1 mark]
- Use \(E = P \times t\) to find energy in MJ: 933,120 MJ (or \(9.33 \times 10^5\text{ MJ}\)) [1.375 marks]

Part (c) [3 marks total]:
- State a clear environmental advantage (no emission of pollutants/carbon footprint reduction) [1.5 marks]
- State a clear environmental disadvantage (visual pollution, noise, local ecosystem disturbance) [1.5 marks]

(a) First, calculate the extension (\(x\)) of the spring:
\(x = \text{stretched length} - \text{unstretched length}\)
\(x = 0.27\text{ m} - 0.12\text{ m} = 0.15\text{ m}\)
Using Hooke's Law:
\(F = k x \implies k = \frac{F}{x}\)
\(k = \frac{6.0\text{ N}}{0.15\text{ m}} = 40\text{ N/m}\)

(b) Elastic potential energy is given by:
\(E_e = \frac{1}{2} k x^2\)
\(E_e = 0.5 \times 40\text{ N/m} \times (0.15\text{ m})^2\)
\(E_e = 20 \times 0.0225 = 0.45\text{ J}\)
Alternatively, using \(E_e = \frac{1}{2} F x\):
\(E_e = 0.5 \times 6.0\text{ N} \times 0.15\text{ m} = 0.45\text{ J}\)

(c) The limit of proportionality is the maximum force that can be applied to the spring such that the extension remains directly proportional to the force applied (i.e., the force-extension graph is a straight line through the origin).

Part (a) [3.375 marks total]:
- Calculate extension: \(0.27 - 0.12 = 0.15\text{ m}\) [1.375 marks]
- Recall Hooke's Law and rearrange: \(k = \frac{F}{x}\) [1 mark]
- Correct calculation with unit: 40 N/m [1 mark]

Part (b) [3 marks total]:
- Recall equation: \(E_e = \frac{1}{2} k x^2\) or \(E_e = \frac{1}{2} F x\) [1 mark]
- Correct substitution of values [1 mark]
- Correct calculation: 0.45 J [1 mark]

Part (c) [3 marks total]:
- Statement that it is the point/limit where force and extension are no longer proportional [2 marks]
- Mentioning that it is the end of the linear region of the graph [1 mark]

(a) Pressure due to a liquid column is:
\(p = h \rho g\)
\(p = 45\text{ m} \times 1025\text{ kg/m}^3 \times 9.8\text{ N/kg} = 452,025\text{ Pa}\)

(b) Total pressure is the sum of water pressure and atmospheric pressure:
\(p_{\text{total}} = p_{\text{water}} + p_{\text{atm}}\)
\(p_{\text{total}} = 452,025\text{ Pa} + 101,000\text{ Pa} = 553,025\text{ Pa}\)

(c) First, calculate the area of the circular watch face:
\(r = 1.8\text{ cm} = 0.018\text{ m}\)
\(A = \pi r^2 = \pi \times (0.018)^2 \approx 1.0179 \times 10^{-3}\text{ m}^2\)
Force is given by:
\(F = p_{\text{total}} \times A\)
\(F = 553,025\text{ Pa} \times 1.0179 \times 10^{-3}\text{ m}^2 \approx 562.9\text{ N}\) (or 563 N to 3 s.f.).

Part (a) [3 marks total]:
- Recall formula: \(p = h \rho g\) [1 mark]
- Substitution: \(45 \times 1025 \times 9.8\) [1 mark]
- Correct calculation: 452,025 Pa (or \(4.52 \times 10^5\text{ Pa}\)) [1 mark]

Part (b) [3 marks total]:
- Realise to add atmospheric pressure [1 mark]
- Substitution: \(452,025 + 101,000\) [1 mark] (Allow ecf from part a)
- Correct calculation: 553,025 Pa [1 mark]

Part (c) [3.375 marks total]:
- Calculate area of circle: \(A = \pi \times (0.018)^2 = 1.018 \times 10^{-3}\text{ m}^2\) [1.375 marks]
- Force formula: \(F = p \times A\) [1 mark]
- Correct calculation: 563 N [1 mark] (Allow ecf from part b)

(a) In a beta decay, a neutron turns into a proton and an electron (beta particle). The mass number remains unchanged (131), and the atomic number increases by 1 (from 53 to 54). The equation is:
\(^{131}_{53}\text{I} \rightarrow ^{131}_{54}\text{Xe} + ^{0}_{-1}\text{e}\)

(b) Work out the number of half-lives that have passed to drop the activity from 800 Bq to 100 Bq:
\(800 \rightarrow 400\) (1 half-life)
\(400 \rightarrow 200\) (2 half-lives)
\(200 \rightarrow 100\) (3 half-lives)
So, 3 half-lives equal 24 days.
\(1\text{ half-life} = \frac{24\text{ days}}{3} = 8\text{ days}\).

(c) A short half-life means that the isotope decays quickly. This ensures that the concentration of active radioactive atoms inside the patient's body drops rapidly after the medical diagnostic procedure is completed, preventing long-term exposure to harmful ionizing radiation.

Part (a) [3.375 marks total]:
- Correct reactant and product mass numbers (131) [1.375 marks]
- Correct atomic number for Xenon (54) [1 mark]
- Correct beta particle notation (\(^{0}_{-1}\text{e}\) or \(\beta^-\)) [1 mark]

Part (b) [3 marks total]:
- Realise that 3 half-lives have passed [1 mark]
- Division of total time by number of half-lives: \(24 / 3\) [1 mark]
- Correct answer: 8 days [1 mark]

Part (c) [3 marks total]:
- States that it decays quickly / loses activity rapidly [1 mark]
- Links this to lower exposure / dose of radiation to the patient [1 mark]
- Concludes this minimizes damage to healthy tissues / cells [1 mark]

(a) Force on a conductor is given by:
\(F = B I L\)
Given:
\(B = 0.15\text{ T}\)
\(I = 2.5\text{ A}\)
\(L = 8.0\text{ cm} = 0.080\text{ m}\)
\(F = 0.15 \times 2.5 \times 0.080 = 0.030\text{ N}\)

(b) The rule is Fleming's Left-Hand Rule.
- First finger represents the Magnetic Field (North to South).
- Second finger represents the Current (positive to negative).
- Thumb represents the Motion/Force.

(c) (i) The force is directly proportional to the current (\(F \propto I\)), so doubling the current doubles the force (to 0.060 N).
(ii) If the wire is parallel to the magnetic field lines, the magnetic force acting on it becomes zero, as the motor effect requires a component of the wire perpendicular to the field.

Part (a) [3.375 marks total]:
- Convert length to meters: 0.080 m [1 mark]
- Substitution into \(F = B I L\): \(0.15 \times 2.5 \times 0.080\) [1.375 marks]
- Correct calculation: 0.030 N (or 0.03 N) [1 mark]

Part (b) [3 marks total]:
- Correctly name Fleming's Left-Hand Rule [1 mark]
- Correctly state what the first finger and second finger represent [1 mark]
- Correctly state what the thumb represents [1 mark]

Part (c) [3 marks total]:
- (i) Correctly states force doubles [1.5 marks]
- (ii) Correctly states force becomes zero [1.5 marks]

1. First, calculate the frequency of the wave in deep water using the wave equation: \(f = \frac{v}{\lambda} = \frac{1.5\text{ m/s}}{3.0\text{ m}} = 0.5\text{ Hz}\).
2. When a wave enters a different medium (from deep to shallow water), its frequency remains constant. Therefore, the frequency in shallow water is also \(0.5\text{ Hz}\).
3. Next, calculate the new wavelength in the shallow water using the same wave equation rearranged: \(\lambda = \frac{v}{f} = \frac{0.5\text{ m/s}}{0.5\text{ Hz}} = 1.0\text{ m}\).

1 mark for the correct answer A.
- Reject other options because frequency must remain constant at \(0.5\text{ Hz}\), and the wavelength must decrease proportionally to the wave speed to \(1.0\text{ m}\).

1. Determine how many half-lives have passed to reduce the activity from \(800\text{ Bq}\) to \(50\text{ Bq}\):
\(800\text{ Bq} \xrightarrow{\text{1 half-life}} 400\text{ Bq} \xrightarrow{\text{2 half-lives}} 200\text{ Bq} \xrightarrow{\text{3 half-lives}} 100\text{ Bq} \xrightarrow{\text{4 half-lives}} 50\text{ Bq}\).
2. This represents exactly 4 half-lives.
3. Divide the total time by the number of half-lives to find the length of one half-life:
\(\text{Half-life} = \frac{12\text{ hours}}{4} = 3.0\text{ hours}\).

1 mark for the correct answer B.
- Award 1 mark for calculating the correct half-life of 3.0 hours based on 4 half-life cycles.

According to the power equation \(P = I \times V\), increasing the potential difference (\(V\)) for a constant transmitted power (\(P\)) significantly decreases the current (\(I\)) flowing through the transmission lines. Since the power lost as thermal energy in the cables is given by \(P_{\text{loss}} = I^2 R\), reducing the current drastically minimizes energy dissipation to the surroundings, making transmission much more efficient.

1 mark for the correct answer A.
- Reject option B because increasing the potential difference does not increase the physical resistance of the wire.
- Reject option C because electrical signal propagation speed remains essentially constant.
- Reject option D because the frequency of the alternating current is kept constant across the grid (at \(50\text{ Hz}\) in the UK).

According to Newton's Third Law, if object A exerts a force on object B, then object B must exert an equal and opposite force on object A of the same type. In this interaction, the tyres (A) push backward on the road surface (B) with a frictional force. The third-law pair is the forward frictional force that the road surface (B) exerts on the tyres (A).

1 mark for the correct answer A.
- Reject options B, C, and D as they do not represent the equal, opposite, and same-type force acting on the interacting bodies.

1. Calculate the change in depth: \(\Delta h = 45\text{ m} - 10\text{ m} = 35\text{ m}\).
2. Use the fluid pressure equation: \(\Delta p = \rho g \Delta h\).
3. Substitute the values: \(\Delta p = 1000\text{ kg/m}^3 \times 10\text{ N/kg} \times 35\text{ m} = 350,000\text{ Pa}\).
4. Convert the answer to kilopascals: \(350,000\text{ Pa} = 350\text{ kPa}\).

1 mark for the correct answer C.
- Reject A: calculated using 10 m depth.
- Reject B: calculated using 35 m depth but miscalculated.
- Reject D: calculated using total depth of 45 m (\(450\text{ kPa}\)).

The maximum induced potential difference (electromotive force) in an AC generator is directly proportional to the strength of the magnetic field, the number of turns in the coil, the cross-sectional area of the coil, and the angular frequency (speed) of rotation. Doubling the speed of rotation while keeping all other parameters constant will exactly double the rate of change of magnetic flux linkage, thereby doubling the maximum induced potential difference.

1 mark for the correct answer C.
- Reject A and B because these modifications would quadruple the maximum potential difference.
- Reject D because halving the turns and doubling the speed of rotation would keep the maximum potential difference the same.

Light from distant galaxies is shifted towards the red end of the spectrum (red-shift), which indicates that they are moving away from us. Observations show that more distant galaxies have a greater red-shift, meaning they are receding faster. This proportional relationship indicates that the space between galaxies is expanding, suggesting that the entire Universe began expanding outward from a single, highly dense point in the past (the Big Bang).

1 mark for the correct answer A.
- Reject B because a larger red-shift for closer galaxies is not observed and would suggest contraction.
- Reject C because red-shift is not uniform for all galaxies.
- Reject D because distant galaxies exhibit red-shift, not blue-shift.

In the electromagnetic spectrum, ordered by decreasing wavelength and increasing frequency: Radio waves, Microwaves, Infrared, Visible light, Ultraviolet, X-rays, Gamma rays. Therefore, Ultraviolet radiation lies at a higher-frequency, shorter-wavelength region of the spectrum compared to Microwaves. Both waves travel at the same speed (the speed of light) in a vacuum.

1 mark for the correct answer A.
- Reject B because UV has a shorter wavelength and higher frequency than microwaves.
- Reject C because all electromagnetic waves travel at the same speed in a vacuum.
- Reject D because higher frequency means UV photons carry higher energy than microwave photons.

When a wave travels from deep water to shallow water, it slows down due to interaction with the seabed. Since the frequency of the wave is determined solely by the source, it remains constant. Using the wave equation v = f * lambda, if the speed decreases and the frequency remains constant, the wavelength must also decrease.

1 mark for the correct option a.

The radiation described is beta-minus radiation because it passes through paper but is stopped by a few millimetres of aluminium. During beta-minus decay, a neutron in the nucleus decays into a proton and an electron (the beta particle). The conversion of a neutron to a proton increases the atomic (proton) number by 1, while the overall mass number remains unchanged.

1 mark for the correct option b.

To find the useful energy: 1. Convert total input energy to Joules: 120,000 kJ = 120,000 * 10^3 J = 1.2 * 10^8 J. 2. Calculate the useful energy output: 0.15 * 1.2 * 10^8 J = 1.8 * 10^7 J. 3. Convert to kWh: 1.8 * 10^7 J / 3.6 * 10^6 J/kWh = 5.0 kWh. This is represented by the formula in option a.

1 mark for the correct option a.

Redshift is directly proportional to the recessional speed of a galaxy. Since Galaxy B has the highest redshift (0.045), it is moving away at the greatest speed. According to Hubble's Law, the recessional velocity of a distant galaxy is directly proportional to its distance from Earth. Thus, the fastest-moving galaxy (B) is also the furthest away.

1 mark for the correct option c.

Alpha particles have very low penetrating power and are stopped by a sheet of paper or a few centimetres of air, so they cannot pass through steel. Gamma radiation is highly penetrating and can pass through several centimetres of steel, allowing a detector on the opposite side to measure the transmitted intensity and identify cracks.

1 mark for the correct option c.

The electromagnetic spectrum in order of increasing frequency (and decreasing wavelength) is: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. Option a lists radio waves, infrared, ultraviolet, and gamma rays, which is in the correct order.

1 mark for the correct option a.

Using Snell's Law, the refractive index n = sin(i) / sin(r) = sin(35) / sin(22) = 0.574 / 0.375 = 1.53. Since the refractive index of glass is higher than that of air (n = 1.00), light travels slower in the glass block, meaning the speed of light decreases.

1 mark for the correct option b.

(a) To measure acceleration:
1. A card of a known length is attached to the top of the trolley.
2. When the trolley passes through a light gate, the card interrupts the light beam.
3. The data logger measures the time duration for which the beam is blocked at each gate, allowing it to calculate the initial velocity \(u\) at Light Gate 1 and final velocity \(v\) at Light Gate 2 using \(\text{velocity} = \frac{\text{length of card}}{\text{time}}\).
4. The data logger also measures the total time interval \(t\) taken for the trolley to travel from the first light gate to the second light gate, and computes the acceleration using \(a = \frac{v - u}{t}\).

(b)(i) Use the equation of motion:
\(v^2 - u^2 = 2as\)
Rearrange to solve for acceleration \(a\):
\(a = \frac{v^2 - u^2}{2s}\)
Substitute the given values:
\(a = \frac{3.5^2 - 1.5^2}{2 \times 1.25}\)
\(a = \frac{12.25 - 2.25}{2.5}\)
\(a = \frac{10}{2.5} = 4.0\text{ m/s}^2\).

(b)(ii) Use Newton's second law:
\(F = ma\)
Substitute mass \(m = 0.80\text{ kg}\) and acceleration \(a = 4.0\text{ m/s}^2\):
\(F = 0.80 \times 4.0 = 3.2\text{ N}\).

Part (a) [4 marks]:
- Card of known length attached to trolley [1]
- Light gates measure the duration the beam is interrupted [1]
- Velocities at each gate are calculated using velocity = length of card / time [1]
- Acceleration calculated from the change in velocity divided by the time taken between the two light gates [1]

Part (b)(i) [3 marks]:
- Correct selection and rearrangement of formula: \(a = \frac{v^2 - u^2}{2s}\) [1]
- Correct substitution: \(a = \frac{3.5^2 - 1.5^2}{2.5}\) [1]
- Correct final answer: \(4.0\text{ m/s}^2\) [1]

Part (b)(ii) [3.7 marks]:
- Correct formula: \(F = ma\) [1]
- Correct substitution: \(F = 0.80 \times 4.0\) [1]
- Correct final calculated answer with unit: \(3.2\text{ N}\) [1.7]

(a) To measure wavelength:
1. Switch on the vibration generator to produce continuous straight waves.
2. Adjust the frequency of the strobe light until the waves appear to be stationary.
3. Place a metre ruler alongside the wave pattern on the screen below the tank.
4. Measure the distance across a large, clear number of wavefronts (e.g., 10 waves) and divide this total distance by the number of waves to find the wavelength of a single wave.

(b)(i) Use the wave equation:
\(v = f \lambda\)
Rearrange to solve for wavelength \(\lambda\):
\(\lambda = \frac{v}{f}\)
Substitute the given values:
\(\lambda = \frac{0.36}{15} = 0.024\text{ m}\)
Convert to centimeters:
\(0.024\text{ m} \times 100 = 2.4\text{ cm}\).

(b)(ii)
1. The frequency remains constant at \(15\text{ Hz}\) because the frequency is determined solely by the source (the vibration generator).
2. Since the wave speed decreases and frequency is constant, the wavelength must decrease: \(\lambda = \frac{v}{f}\).
3. New wavelength: \(\lambda = \frac{0.24}{15} = 0.016\text{ m} = 1.6\text{ cm}\).

Part (a) [4 marks]:
- Use a strobe light to freeze the motion of the waves (adjust frequency to match) [1]
- Place a ruler alongside the wave projection [1]
- Measure across multiple wavefronts (e.g., 10 waves) to minimize error [1]
- Divide total measured distance by the number of wavefronts to find the wavelength of a single wave [1]

Part (b)(i) [3 marks]:
- Correct rearrangement of wave equation: \(\lambda = \frac{v}{f}\) [1]
- Correct calculation: \(0.024\text{ m}\) [1]
- Correct unit conversion to \(2.4\text{ cm}\) [1]

Part (b)(ii) [3.7 marks]:
- State that frequency remains unchanged / constant because it depends on the source [1]
- State that wavelength decreases [1]
- Calculate the new wavelength as \(1.6\text{ cm}\) (or \(0.016\text{ m}\)) [1.7]

(a) When the resistance of the variable resistor increases:
1. The total resistance of the series circuit increases (\(R_{\text{total}} = R_{\text{fixed}} + R_{\text{variable}}\)).
2. Since potential difference remains constant, the total current in the circuit decreases according to \(I = \frac{V}{R_{\text{total}}}\).
3. Because the current decreases, the potential difference across the fixed \(30\ \Omega\) resistor decreases, according to \(V = IR\).
4. Quantitatively: when variable resistance is \(0\ \Omega\), current is \(\frac{12}{30} = 0.4\text{ A}\) and PD is \(12\text{ V}\). When variable resistance is \(60\ \Omega\), total resistance is \(90\ \Omega\), current is \(\frac{12}{90} = 0.133\text{ A}\), and PD is \(0.133 \times 30 = 4.0\text{ V}\). Thus, the PD across the fixed resistor decreases from \(12\text{ V}\) to \(4\text{ V}\).

(b)(i) For two resistors connected in parallel, the total resistance \(R_p\) is given by:
\(\frac{1}{R_p} =
\frac{1}{R_1} + \frac{1}{R_2} =
\frac{1}{40} + \frac{1}{60}\)
Find a common denominator (120):
\(\frac{1}{R_p} = \frac{3}{120} + \frac{2}{120} = \frac{5}{120}\)
Reciprocate both sides:
\(R_p = \frac{120}{5} = 24\ \Omega\).

(b)(ii) Use Ohm's Law:
\(I = \frac{V}{R_p}\)
Substitute the given values:
\(I = \frac{12\text{ V}}{24\ \Omega} = 0.50\text{ A}\).

Part (a) [4 marks]:
- As variable resistance increases, total resistance in the circuit increases [1]
- This causes total current in the circuit to decrease [1]
- Since current decreases, potential difference across the fixed \(30\ \Omega\) resistor decreases [1]
- Mentions correct limit values (decreases from \(12\text{ V}\) to \(4\text{ V}\)) [1]

Part (b)(i) [3.7 marks]:
- Use of parallel resistance formula: \(\frac{1}{R_p} = \frac{1}{40} + \frac{1}{60}\) [1]
- Correct calculation of fraction: \(\frac{1}{R_p} = \frac{5}{120}\) [1]
- Final calculated resistance value with unit: \(24\ \Omega\) [1.7]

Part (b)(ii) [3 marks]:
- Correct equation: \(I = \frac{V}{R}\) [1]
- Correct substitution of values: \(I = \frac{12}{24}\) [1]
- Correct final current with unit: \(0.50\text{ A}\) (or \(0.5\text{ A}\)) [1]

(a)
1. A reliable energy resource is one that can generate electricity continuously on demand regardless of weather conditions (e.g., hydroelectric power, where water release from a reservoir can be precisely controlled to meet electricity demand).
2. A non-reliable (intermittent) resource cannot generate electricity on demand because it depends on unpredictable environmental factors (e.g., wind power, which only produces energy when the wind is blowing at sufficient speeds).

(b)(i) Ensure units are consistent:
Maximum capacity = \(2.5\text{ MW} = 2500\text{ kW}\)
Average power = \(600\text{ kW}\)
Capacity factor = \(\left(\frac{\text{Average power}}{\text{Maximum capacity}}\right) \times 100\)
\(\text{Capacity factor} = \left(\frac{600}{2500}\right) \times 100 = 0.24 \times 100 = 24\%\).

(b)(ii) Calculate total time in seconds for one week:
\(t = 7\text{ days} \times 24\text{ hours/day} \times 3600\text{ seconds/hour} = 604,800\text{ seconds}\)
Calculate energy using \(E = P \times t\) (where average power is \(600\text{ kW} = 600,000\text{ W}\)):
\(E = 600,000\text{ W} \times 604,800\text{ s} = 362,880,000,000\text{ Joules}\)
Convert Joules to Megajoules (MJ) by dividing by \(1,000,000\):
\(E = 362,880\text{ MJ}\) (or \(3.6288 \times 10^5\text{ MJ}\)).

Part (a) [4 marks]:
- Definition of reliable: can produce energy consistently/on demand [1]
- Example: Hydroelectric can store water to release when needed [1]
- Definition of non-reliable: dependent on weather/environmental factors [1]
- Example: Wind turbine output depends on wind speeds which are unpredictable [1]

Part (b)(i) [3.7 marks]:
- Convert units so they are consistent (e.g., \(2.5\text{ MW} = 2500\text{ kW}\)) [1]
- Use correct ratio: \(\frac{600}{2500}\) [1]
- Correct final percentage: \(24\%\) [1.7]

Part (b)(ii) [3 marks]:
- Calculate total seconds in one week: \(604,800\text{ s}\) [1]
- Use \(E = P \times t\) with consistent units [1]
- Correct final calculated value: \(362,880\text{ MJ}\) [1]

(a) Hooke's Law states that the extension of an elastic object is directly proportional to the force applied to it, provided the limit of proportionality is not exceeded (\(F = kx\)).
- Elastic deformation: The object returns to its original length and shape when the deforming force is removed.
- Plastic deformation: The object is permanently deformed and does not return to its original shape when the force is removed.

(b)(i) First calculate extension \(x\):
\(x = \text{stretched length} - \text{unstretched length} = 27.0\text{ cm} - 12.0\text{ cm} = 15.0\text{ cm}\)
Convert extension to metres:
\(x = 15.0 \div 100 = 0.15\text{ m}\)
Use Hooke's Law \(F = kx\) to solve for \(k\):
\(k = \frac{F}{x} = \frac{6.0\text{ N}}{0.15\text{ m}} = 40\text{ N/m}\).

(b)(ii) Use the formula for elastic potential energy:
\(E_e = \frac{1}{2} k x^2\)
Substitute the values \(k = 40\text{ N/m}\) and \(x = 0.15\text{ m}\):
\(E_e = 0.5 \times 40 \times (0.15)^2\)
\(E_e = 20 \times 0.0225 = 0.45\text{ J}\).

Part (a) [4 marks]:
- Statement of Hooke's Law: Extension is proportional to force [1]
- Condition: Provided limit of proportionality is not exceeded [1]
- Elastic deformation: returns to original shape when force is removed [1]
- Plastic deformation: remains permanently deformed when force is removed [1]

Part (b)(i) [3.7 marks]:
- Correct extension calculated: \(15.0\text{ cm}\) or \(0.15\text{ m}\) [1]
- Correct rearrangement: \(k = \frac{F}{x}\) [1]
- Correct final spring constant with unit: \(40\text{ N/m}\) [1.7]

Part (b)(ii) [3 marks]:
- Correct formula: \(E_e = \frac{1}{2} k x^2\) [1]
- Correct substitution of values: \(0.5 \times 40 \times 0.15^2\) [1]
- Correct final energy value with unit: \(0.45\text{ J}\) [1]

(a) Atmospheric pressure is created by air molecules colliding with surfaces.
1. As altitude increases, the density of the air decreases, meaning there are fewer air molecules per unit volume.
2. This results in fewer collisions per second with surfaces, producing less pressure.
3. Furthermore, at higher altitudes, there is a smaller weight of air above pushing down on the column of air below.

(b)(i) Use the equation for fluid pressure:
\(p = h \rho g\)
Substitute depth \(h = 150\text{ m}\), density \(\rho = 1025\text{ kg/m}^3\), and \(g = 10\text{ N/kg}\):
\(p = 150 \times 1025 \times 10\)
\(p = 1,537,500\text{ Pa}\) (or \(1.5375\text{ MPa}\)).

(b)(ii) The total absolute pressure is the sum of the hydrostatic pressure and the atmospheric pressure:
\(p_{\text{total}} = p_{\text{hydrostatic}} + p_{\text{atmospheric}}\)
\(p_{\text{total}} = 1,537,500\text{ Pa} + 101,000\text{ Pa}\)
\(p_{\text{total}} = 1,638,500\text{ Pa}\) (or \(1.6385\text{ MPa}\)).

Part (a) [4 marks]:
- Pressure is due to collisions of air particles with surfaces [1]
- Density of air decreases as altitude increases [1]
- Fewer particles per unit volume leads to fewer collisions [1]
- There is less weight of air acting downwards from above [1]

Part (b)(i) [3.7 marks]:
- Correct formula used: \(p = h \rho g\) [1]
- Correct substitution of values: \(150 \times 1025 \times 10\) [1]
- Correct final calculated pressure with unit: \(1,537,500\text{ Pa}\) [1.7]

Part (b)(ii) [3 marks]:
- Recognize that total pressure is the sum of hydrostatic and atmospheric pressure [1]
- Correct addition: \(1,537,500 + 101,000\) [1]
- Correct total absolute pressure with unit: \(1,638,500\text{ Pa}\) [1]

(a)
- Ionizing power: Alpha has the highest ionizing power, beta has moderate ionizing power, and gamma has the lowest ionizing power.
- Penetrating power: Gamma has the highest penetrating power (requires thick lead/concrete to stop it), beta has moderate penetrating power (stopped by a thin sheet of aluminium), and alpha has the lowest penetrating power (stopped by a sheet of paper or skin).

(b)(i) Determine the number of half-lives that have elapsed:
- Start at \(100\%\) activity.
- After 1 half-life: \(50\%\)
- After 2 half-lives: \(25\%\)
- After 3 half-lives: \(12.5\%\)
So, exactly 3 half-lives have passed.
Calculate the age:
\(\text{Age} = 3 \times 5730\text{ years} = 17,190\text{ years}\).

(b)(ii) In beta-minus decay, a neutron decays into a proton and an electron. The electron is emitted as a beta particle (represented as \(^{0}_{-1}\text{e}\) or \(^{0}_{-1}\beta\)).
The balanced nuclear equation is:
\(^{14}_{6}\text{C} \rightarrow ^{14}_{7}\text{N} + ^{0}_{-1}\text{e}\)

Part (a) [4 marks]:
- Alpha has the highest ionizing power and lowest penetrating power (stopped by paper) [1]
- Beta has medium ionizing power and medium penetrating power (stopped by aluminium) [1]
- Gamma has the lowest ionizing power and highest penetrating power (stopped by lead/concrete) [1]
- Clear comparative terminology used correctly [1]

Part (b)(i) [3.7 marks]:
- Determine that \(12.5\%\) represents 3 half-lives [1]
- Multiplies half-life by 3: \(3 \times 5730\) [1]
- Correct age: \(17,190\text{ years}\) [1.7]

Part (b)(ii) [3 marks]:
- Correct reactant symbol: \(^{14}_{6}\text{C}\) [1]
- Correct beta particle symbol: \(^{0}_{-1}\text{e}\) (or \(^{0}_{-1}\beta\)) [1]
- Fully balanced nuclear equation [1]