2024 Edexcel IGCSE Physics Practice Paper with Answers

The electromagnetic spectrum is ordered by increasing frequency as: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. Therefore, gamma rays have the highest frequency among the given options.

[1 mark] D - Gamma rays
- Award 1 mark for identifying that gamma rays have the highest frequency.

High-mass stars begin as a nebula, evolve into a main sequence star, then expand into a red supergiant. They eventually explode in a supernova, leaving behind either a neutron star or a black hole.

[1 mark] B - Nebula -> main sequence star -> red supergiant -> supernova -> neutron star
- Award 1 mark for selecting the correct pathway for a massive star.

Since the sphere is made of metal, free electrons are able to move. When a negatively charged rod is brought near the sphere, it repels the free electrons to the far side of the sphere, leaving the near side with a net positive charge. This process is called electrostatic induction.

[1 mark] A - Positive charges on the side near the rod, negative charges on the side far from the rod.
- Award 1 mark for identifying the correct distribution due to electrostatic induction.

For three identical resistors connected in parallel, the total equivalent resistance \(R_{\text{total}}\) is given by:
\(\frac{1}{R_{\text{total}}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R}\)
Thus, \(R_{\text{total}} = \frac{R}{3}\).
Using Ohm's law, the total current \(I\) drawn from the supply is:
\(I = \frac{V}{R_{\text{total}}} = \frac{V}{\frac{R}{3}} = \frac{3V}{R}\).

[1 mark] D - \(\frac{3V}{R}\)
- Award 1 mark for finding the correct expression for total current.

Newton's first law of motion states that an object will remain at rest or continue to move at a constant velocity unless acted upon by a net (resultant) external force. Since the car is travelling at a constant velocity, its acceleration is zero, which means the resultant force acting on it must be 0 N.

[1 mark] A - 0 N
- Award 1 mark for identifying that a constant velocity implies a resultant force of zero.

First, convert the time taken into seconds:
\(1 \text{ minute and } 30 \text{ seconds} = 60 \text{ s} + 30 \text{ s} = 90 \text{ s}\).
Now, calculate the average speed using the formula:
\(\text{average speed} = \frac{\text{distance}}{\text{time}} = \frac{450 \text{ m}}{90 \text{ s}} = 5.0 \text{ m/s}\).

[1 mark] C - 5.0 m/s
- Award 1 mark for calculating the correct average speed.

We use Snell's Law:
\(n = \frac{\sin(i)}{\sin(r)}\)
Rearranging to solve for \(\sin(r)\):
\(\sin(r) = \frac{\sin(i)}{n} = \frac{\sin(35^\circ)}{1.5} \approx \frac{0.5736}{1.5} \approx 0.3824\)
\(r = \sin^{-1}(0.3824) \approx 22.48^\circ\).
To 2 significant figures, this is \(22^\circ\).

[1 mark] A - \(22^\circ\)
- Award 1 mark for the correct calculation and rounding of the angle of refraction.

According to Gay-Lussac's Law (or the pressure law) for an ideal gas, the pressure \(P\) of a gas is directly proportional to its absolute temperature \(T\) when the volume is kept constant (rigid container):
\(P \propto T\).
Therefore, if the absolute temperature is doubled, the pressure of the gas is also doubled.

[1 mark] C - It doubles.
- Award 1 mark for identifying the correct proportional relationship between pressure and temperature.

The relationship between total charge \(Q\) and the number of excess electrons \(n\) is given by:

\(Q = n \times e\)

Rearranging to solve for \(n\):

\(n = \frac{Q}{e}\)

Substitute the given values into the equation:

\(n = \frac{4.8 \times 10^{-16}\text{ C}}{1.6 \times 10^{-19}\text{ C}} = 3000\)

Therefore, the dust particle has acquired 3000 excess electrons.

- **1 mark**: Recalling and using the relationship \(Q = n \times e\) or equivalent.
- **1 mark**: Correct substitution of the values: \(\frac{4.8 \times 10^{-16}}{1.6 \times 10^{-19}}\).
- **0.5 marks**: Correct final value of 3000 (or \(3.0 \times 10^3\)).

The wave equation is:

\(v = f \times \lambda\)

Rearranging for wavelength (\(\lambda\)):

\(\lambda = \frac{v}{f}\)

Substituting the given values:

\(\lambda = \frac{3.0 \times 10^{8}\text{ m/s}}{1.5 \times 10^{8}\text{ Hz}} = 2.0\text{ m}\)

- **1 mark**: Recalling the wave equation \(v = f \lambda\).
- **1 mark**: Correct substitution and calculation resulting in 2.0.
- **0.5 marks**: Correct unit (meters, m).

The relationship between energy transferred (\(E\)), charge (\(Q\)), and voltage (\(V\)) is:

\(E = Q \times V\)

Substitute the values into the equation:

\(E = 15\text{ C} \times 9.0\text{ V} = 135\text{ J}\)

Note that the time value of \(40\text{ s}\) is extra information and is not required for this calculation.

- **1 mark**: Recalling the relationship \(E = Q \times V\).
- **1 mark**: Substituting the correct values to find 135.
- **0.5 marks**: Stating the correct unit of energy (Joules, J).

A high-mass star leaves the stable main sequence stage when it runs out of hydrogen fuel in its core. It expands and cools to become a red supergiant. When nuclear fusion ceases entirely, the core collapses rapidly, resulting in a cataclysmic explosion called a supernova. Depending on the remaining mass of the core, the remnant left behind is either a highly dense neutron star or, if the mass is sufficiently high, a black hole.

- **1 mark**: Identifies expansion into a 'red supergiant'.
- **1 mark**: Identifies the 'supernova' explosion stage.
- **0.5 marks**: Identifies the final state as a 'neutron star' or 'black hole'.

Use Snell's law:

\(n = \frac{\sin(i)}{\sin(r)}\)

Substitute the given values:

\(n = \frac{\sin(42^{\circ})}{\sin(25^{\circ})}\)

Using a calculator:

\(\sin(42^{\circ}) \approx 0.6691\)
\(\sin(25^{\circ}) \approx 0.4226\)

\(n = \frac{0.6691}{0.4226} \approx 1.583\)

Rounding to two decimal places gives \(1.58\).

- **1 mark**: Recalling and stating Snell's law \(n = \frac{\sin(i)}{\sin(r)}\).
- **1 mark**: Correct substitution of the angles.
- **0.5 marks**: Correct evaluation of the refractive index to two decimal places (1.58).

According to the principle of conservation of momentum:

Total initial momentum = Total final momentum

\(m_{\text{car}} u_{\text{car}} + m_{\text{truck}} u_{\text{truck}} = (m_{\text{car}} + m_{\text{truck}}) v\)

Substitute the given values:

\((0.50 \times 4.0) + (m_{\text{truck}} \times 0) = (0.50 + m_{\text{truck}}) \times 1.6\)

\(2.0 = 0.80 + 1.6 m_{\text{truck}}\)

\(1.2 = 1.6 m_{\text{truck}}\)

\(m_{\text{truck}} = \frac{1.2}{1.6} = 0.75\text{ kg}\)

- **1 mark**: Recalling and applying the conservation of momentum equation: \(m_1 u_1 = (m_1 + m_2) v\).
- **1 mark**: Correct rearrangement of the equation to solve for the unknown mass.
- **0.5 marks**: Correct calculation of the mass (0.75) with unit (kg).

Using the equation of motion:

\(s = u t + \frac{1}{2} a t^2\)

Since the drone starts from rest, the initial velocity \(u = 0\):

\(s = \frac{1}{2} a t^2\)

Rearranging the equation to solve for acceleration (\(a\)):

\(a = \frac{2s}{t^2}\)

Substitute the given values:

\(a = \frac{2 \times 45}{(6.0)^2}\)

\(a = \frac{90}{36} = 2.5\text{ m/s}^{2}\)

- **1 mark**: Selecting and writing the equation of motion \(s = u t + \frac{1}{2} a t^2\) or \(s = \frac{1}{2} a t^2\).
- **1 mark**: Substituting the correct values: \(45 = 0.5 \times a \times 36\).
- **0.5 marks**: Correct final numerical value (2.5) and unit (\(\text{m/s}^2\)).

First, convert temperatures from Celsius to Kelvin:

\(T_1 = 20 + 273 = 293\text{ K}\)

\(T_2 = 80 + 273 = 353\text{ K}\)

For a fixed mass of gas at constant volume:

\(\frac{P_1}{T_1} = \frac{P_2}{T_2}\)

Rearranging the equation to solve for the final pressure (\(P_2\)):

\(P_2 = P_1 \times \frac{T_2}{T_1}\)

Substitute the values:

\(P_2 = 100\text{ kPa} \times \frac{353}{293} \approx 120.48\text{ kPa}\)

Rounding to the nearest whole number gives \(120\text{ kPa}\).

- **1 mark**: Converting both Celsius temperatures into Kelvin (293 K and 353 K).
- **1 mark**: Using the formula \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\) with correct substitutions.
- **0.5 marks**: Correct final value (120) with unit (kPa).

1. Calculate the extension of the spring: extension = stretched length - unstretched length = 19.0 cm - 15.0 cm = 4.0 cm. 2. Convert extension to meters: 4.0 cm = 0.04 m. 3. Use the formula for Hooke's Law: F = k * x, which rearranges to k = F / x. 4. Substitute the values: k = 12.0 N / 0.04 m = 300 N/m.

1 mark for calculating correct extension of 0.04 m (or 4.0 cm); 1 mark for correct substitution into rearranged formula k = F / x; 0.5 marks for correct final value of 300 with correct unit (N/m).

1. Convert the time from minutes to seconds: t = 2.0 * 60 = 120 s. 2. Use the charge formula: Q = I * t, which rearranges to I = Q / t. 3. Substitute the values: I = 1.8 C / 120 s = 0.015 A.

1 mark for converting time to 120 seconds; 1 mark for correct rearrangement and substitution of formula I = Q / t; 0.5 marks for correct calculation of 0.015 A.

1. Convert the wavelength from centimeters to meters: wavelength = 1.2 cm = 0.012 m. 2. Use the wave speed equation: v = f * wavelength, which rearranges to f = v / wavelength. 3. Substitute the values: f = (3.0 * 10^8 m/s) / 0.012 m = 2.5 * 10^10 Hz.

1 mark for converting wavelength to 0.012 meters; 1 mark for correct rearrangement and substitution of formula f = v / wavelength; 0.5 marks for correct calculation of 2.5 * 10^10 Hz.

1. Use the energy and charge relationship: E = Q * V, where E is energy in joules, Q is charge in coulombs, and V is potential difference in volts. 2. Rearrange the formula to solve for V: V = E / Q. 3. Substitute the known values: V = 7200 J / 1500 C = 4.8 V.

1 mark for identifying the correct formula E = Q * V or V = E / Q; 1 mark for correct substitution of values; 0.5 marks for correct final answer of 4.8 V.

1. Identify the relationship between mass and gravity: A star with higher mass experiences stronger gravitational forces, compressing its core. 2. Explain the thermal effect: This core compression results in a significantly higher core temperature and pressure. 3. Explain the fusion rate: The higher temperature and pressure vastly increases the rate of nuclear fusion of hydrogen into helium. 4. Conclude on lifespan: Despite having more hydrogen fuel, the massive star consumes its fuel at a disproportionately faster rate, ending its main sequence life much sooner than a lower-mass star.

1 mark for mentioning stronger gravitational forces/pressures in the core; 1 mark for linking this to higher core temperature and a much faster rate of nuclear fusion; 0.5 marks for explaining that the fuel is consumed much faster, leading to a shorter main sequence lifespan.

1. State Snell's Law: n = sin(i) / sin(r). 2. Rearrange the equation to solve for sin(r): sin(r) = sin(i) / n. 3. Substitute the given values: sin(r) = sin(45 degrees) / 1.52 = 0.7071 / 1.52 = 0.4652. 4. Calculate r by taking the inverse sine: r = arcsin(0.4652) = 27.7 degrees, which rounds to 28 degrees.

1 mark for using Snell's Law n = sin(i) / sin(r); 1 mark for rearranging and substituting to find sin(r) = 0.465; 0.5 marks for final angle of 28 degrees.

1. Identify the appropriate equation of motion: v^2 = u^2 + 2*a*s. 2. Identify the known variables: initial velocity u = 0 m/s (starts from rest), acceleration a = 2.5 m/s^2, distance s = 80 m. 3. Substitute the values into the equation: v^2 = 0^2 + 2 * 2.5 * 80 = 400. 4. Solve for v by taking the square root: v = sqrt(400) = 20 m/s.

1 mark for choosing the formula v^2 = u^2 + 2as; 1 mark for correct substitution (2 * 2.5 * 80 = 400); 0.5 marks for correct square root calculation to obtain 20 m/s with correct units.

1. Convert the temperatures from Celsius to Kelvin: T1 = 27 + 273 = 300 K, T2 = 127 + 273 = 400 K. 2. State the Pressure Law for a constant volume: P1 / T1 = P2 / T2. 3. Rearrange the equation to solve for P2: P2 = P1 * (T2 / T1). 4. Substitute the values into the equation: P2 = 120 kPa * (400 K / 300 K) = 160 kPa.

1 mark for converting both temperatures to Kelvin (300 K and 400 K); 1 mark for correct substitution into the constant-volume gas law P1/T1 = P2/T2; 0.5 marks for correct final pressure of 160 kPa.

1. Identify that the total momentum before launch is zero because both the car and ball are at rest: \(p_{\text{initial}} = 0\). 2. Use the principle of conservation of momentum: \(p_{\text{initial}} = p_{\text{final}} = m_{\text{car}} v_{\text{car}} + m_{\text{ball}} v_{\text{ball}} = 0\). 3. Substitute the given values: \(0.80 \text{ kg} \times v_{\text{car}} + 0.050 \text{ kg} \times 12 \text{ m/s} = 0\). 4. Solve for the recoil velocity: \(0.80 v_{\text{car}} + 0.60 = 0 \implies v_{\text{car}} = -0.75 \text{ m/s}\). The magnitude of the velocity is 0.75 m/s.

1 mark: Correct substitution into momentum equation \(0.050 \times 12\) to find ball's momentum (0.60 kg m/s). 1 mark: Application of momentum conservation showing \(0.80 \times v = -0.60\). 0.5 mark: Correct calculation of magnitude (0.75) with correct unit (m/s).

1. Express the relationship between total charge and the number of elementary charges: \(Q = n \times e\), where \(Q\) is the total charge, \(n\) is the number of electrons, and \(e\) is the charge of one electron. 2. Rearrange the formula to solve for \(n\): \(n = \frac{Q}{e}\). 3. Substitute the values: \(n = \frac{-4.8 \times 10^{-9} \text{ C}}{-1.6 \times 10^{-19} \text{ C}}\). 4. Calculate the final answer: \(n = 3.0 \times 10^{10}\).

1 mark: Recalling or writing the formula \(n = Q / e\). 1 mark: Correct substitution of values \((-4.8 \times 10^{-9}) / (-1.6 \times 10^{-19})\). 0.5 mark: Correct numerical value (\(3.0 \times 10^{10}\) or \(3 \times 10^{10}\)).

1. State the wave equation relating speed, frequency, and wavelength: \(v = f \lambda\). 2. Convert the wavelength from centimeters to meters: \(15 \text{ cm} = 0.15 \text{ m}\). 3. Rearrange the wave equation to solve for frequency: \(f = \frac{v}{\lambda}\). 4. Substitute the values into the equation: \(f = \frac{3.0 \times 10^8 \text{ m/s}}{0.15 \text{ m}} = 2.0 \times 10^9 \text{ Hz}\). 5. Convert hertz to megahertz (MHz) by dividing by \(10^6\): \(2.0 \times 10^9 \text{ Hz} = 2000 \text{ MHz}\).

1 mark: Converting wavelength to meters (0.15 m) and recalling wave speed equation rearrangement \(f = v / \lambda\). 1 mark: Substitution and calculation of frequency in Hz (\(2.0 \times 10^9 \text{ Hz}\)). 0.5 mark: Correct final answer converted to MHz with correct unit (2000 MHz).

1. Identify the formula for electrical energy transferred: \(E = V \times I \times t\). 2. Convert the time from minutes to seconds: \(t = 15 \text{ minutes} \times 60 \text{ s/minute} = 900 \text{ s}\). 3. Substitute the values of voltage, current, and time into the energy formula: \(E = 12 \text{ V} \times 4.5 \text{ A} \times 900 \text{ s}\). 4. Perform the calculation: \(E = 48,600 \text{ J}\). 5. Convert the energy from Joules to kilojoules (kJ) by dividing by 1000: \(E = 48.6 \text{ kJ}\).

1 mark: Converting 15 minutes to 900 seconds and identifying energy formula \(E = V I t\) (or using \(P = V I\) followed by \(E = P t\)). 1 mark: Substituting the correct values to find energy in Joules (48,600 J). 0.5 mark: Converting the calculated energy to kJ with correct units (48.6 kJ).

1. During the main sequence stage, a star is stable because the inward gravitational force (pulling matter towards the center) is balanced by the outward thermal and radiation pressure resulting from nuclear fusion in the core. 2. When the hydrogen fuel in the core runs out, nuclear fusion ceases, causing the temperature and outward pressure to drop. 3. Because the outward pressure is reduced, the inward force of gravity dominates, causing the core of the star to compress and contract.

1 mark: Stating the two balanced forces in a main sequence star (inward gravitational force and outward radiation/thermal pressure). 1 mark: Stating that these two opposing forces are balanced/equal. 0.5 mark: Explaining that when fusion stops, outward pressure decreases, allowing gravity to cause the core to contract/collapse.

1. State the formula relating critical angle \(c\) and refractive index \(n\) when moving from a medium into air: \(\sin(c) = \frac{1}{n}\). 2. Rearrange the formula to solve for \(c\): \(c = \sin^{-1}\left(\frac{1}{n}\right)\). 3. Substitute the given refractive index: \(c = \sin^{-1}\left(\frac{1}{1.55}\right)\). 4. Calculate the ratio: \(\frac{1}{1.55} \approx 0.64516\). 5. Compute the inverse sine: \(c \approx 40.178^{\circ}\). Rounded to 3 significant figures, \(c = 40.2^{\circ}\).

1 mark: Stating the correct formula \(\sin(c) = 1 / n\). 1 mark: Rearranging and substituting values: \(c = \sin^{-1}(1 / 1.55)\). 0.5 mark: Calculating correct final angle value to 3 s.f. (40.2 degrees or \(40.2^{\circ}\); accept 40 degrees).

1. Identify the given values: initial velocity \(u = 0 \text{ m/s}\), acceleration \(a = 1.2 \text{ m/s}^2\), and distance \(s = 60 \text{ m}\). 2. Recall the relevant SUVAT equation linking \(v\), \(u\), \(a\), and \(s\): \(v^2 = u^2 + 2as\). 3. Substitute the values: \(v^2 = 0^2 + 2 \times 1.2 \times 60\). 4. Calculate the intermediate value: \(v^2 = 144\). 5. Solve for \(v\) by taking the square root: \(v = \sqrt{144} = 12 \text{ m/s}\).

1 mark: Selecting and writing down the formula \(v^2 = u^2 + 2as\). 1 mark: Substituting the correct values into the formula: \(v^2 = 2 \times 1.2 \times 60 = 144\). 0.5 mark: Calculating the correct final square root value with correct units (12 m/s).

1. Convert the initial and final temperatures from Celsius to Kelvin: \(T_1 = 27 + 273 = 300 \text{ K}\) and \(T_2 = 127 + 273 = 400 \text{ K}\). 2. State the pressure law for a constant volume of gas: \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\). 3. Rearrange the formula to solve for final pressure \(P_2\): \(P_2 = P_1 \times \frac{T_2}{T_1}\). 4. Substitute the known values: \(P_2 = 120 \text{ kPa} \times \frac{400 \text{ K}}{300 \text{ K}}\). 5. Calculate the result: \(P_2 = 120 \times 1.333 = 160 \text{ kPa}\).

1 mark: Converting both temperatures to Kelvin (300 K and 400 K). 1 mark: Using pressure law relation \(P_1/T_1 = P_2/T_2\) and substituting values correctly. 0.5 mark: Correct calculation of final pressure with unit (160 kPa).

1. Calculate the extension \(x\) of the spring: \(x = 15.0\text{ cm} - 12.0\text{ cm} = 3.0\text{ cm}\). 2. Convert the extension into metres: \(x = 0.030\text{ m}\). 3. Use Hooke's Law formula: \(F = kx\). 4. Rearrange to find \(k\): \(k = \frac{F}{x}\). 5. Substitute the values: \(k = \frac{6.0\text{ N}}{0.030\text{ m}} = 200\text{ N/m}\).

[1 mark] for calculating the extension of the spring in metres (\(0.030\text{ m}\)). [1 mark] for recalling \(F = kx\) and rearranging it correctly to \(k = \frac{F}{x}\). [0.5 marks] for the correct numerical value with the correct unit (\(200\text{ N/m}\)).

When the plastic rod is rubbed with the dry cloth, friction causes negatively charged electrons to move from the cloth to the rod. Since the rod gains these electrons, it acquires a net negative charge. Consequently, the cloth loses electrons and is left with an equal and opposite positive charge.

[1 mark] for stating that electrons are transferred from the cloth to the rod. [1 mark] for specifying that the transferred particles are electrons (no marks for 'positive charges moving'). [0.5 marks] for identifying that the cloth is left with a positive charge.

1. Use the wave equation: \(v = f \lambda\). 2. Rearrange the equation to solve for wavelength: \(\lambda = \frac{v}{f}\). 3. Substitute the given values: \(\lambda = \frac{3.0 \times 10^8\text{ m/s}}{4.5 \times 10^{14}\text{ Hz}} = 6.67 \times 10^{-7}\text{ m}\). 4. Since the wavelength is between \(4 \times 10^{-7}\text{ m}\) and \(7 \times 10^{-7}\text{ m}\), it falls within the visible light region of the electromagnetic spectrum.

[1 mark] for recalling and rearranging the wave equation \(v = f \lambda\). [1 mark] for the correct calculation of wavelength as \(6.7 \times 10^{-7}\text{ m}\) (or \(6.67 \times 10^{-7}\text{ m}\)). [0.5 marks] for correctly identifying the electromagnetic region as visible light.

1. Convert the time from minutes into seconds: \(t = 5.0 \times 60 = 300\text{ s}\). 2. Use the electrical energy formula: \(E = I \times V \times t\). 3. Substitute the values into the formula: \(E = 0.25\text{ A} \times 12\text{ V} \times 300\text{ s}\). 4. Perform the calculation: \(E = 3.0\text{ W} \times 300\text{ s} = 900\text{ J}\).

[1 mark] for converting the time into seconds (\(300\text{ s}\)). [1 mark] for recalling the electrical energy formula \(E = I V t\) (or calculating power \(P = IV = 3\text{ W}\) and then using \(E = Pt\)). [0.5 marks] for the correct final energy value with the unit (\(900\text{ J}\) or \(0.9\text{ kJ}\)).

A high-mass star in the main sequence eventually runs out of hydrogen fuel in its core, causing it to expand and cool to form a red supergiant. When nuclear fusion in the core stops completely, the core collapses rapidly, resulting in a giant explosion called a supernova. The remnants of the core collapse further to form either an extremely dense neutron star or, if the mass is large enough, a black hole.

[1 mark] for describing the expansion and cooling of the star to become a red supergiant when hydrogen runs out. [1 mark] for describing the massive explosion known as a supernova. [0.5 marks] for identifying the final remnant state as either a neutron star or a black hole.

1. Use Snell's law equation: \(n = \frac{\sin(i)}{\sin(r)}\). 2. Substitute the angles into the equation: \(n = \frac{\sin(42.0^\circ)}{\sin(25.4^\circ)}\). 3. Calculate the sine values: \(\sin(42.0^\circ) \approx 0.6691\) and \(\sin(25.4^\circ) \approx 0.4289\). 4. Divide the values: \(n = \frac{0.6691}{0.4289} \approx 1.56\). Refractive index has no units.

[1 mark] for stating the equation \(n = \frac{\sin(i)}{\sin(r)}\). [1 mark] for correct substitution of angle values into the sine functions. [0.5 marks] for the correct refractive index value of \(1.56\) (accept \(1.6\)).

1. Identify the given values: initial velocity \(u = 0\text{ m/s}\), final velocity \(v = 8.5\text{ m/s}\), and distance \(s = 34\text{ m}\). 2. Use the equation of motion: \(v^2 = u^2 + 2as\). 3. Rearrange the equation to solve for acceleration \(a\): \(a = \frac{v^2 - u^2}{2s}\). 4. Substitute the values: \(a = \frac{8.5^2 - 0^2}{2 \times 34} = \frac{72.25}{68} \approx 1.0625\text{ m/s}^2\). Rounded to three significant figures, this is \(1.06\text{ m/s}^2\).

[1 mark] for recalling the equation \(v^2 = u^2 + 2as\). [1 mark] for correct substitution and rearrangement of the equation. [0.5 marks] for the correct final answer with units (\(1.1\text{ m/s}^2\) or \(1.06\text{ m/s}^2\)).

1. Place a bar magnet on a sheet of paper and place a plotting compass near one of its poles. 2. Mark dots on the paper at both ends of the compass needle. 3. Move the compass so that the tail end of the needle is over the dot previously made at the tip end, and mark a new dot at the tip end. 4. Repeat this process until reaching the other pole of the magnet, then connect the dots with a smooth curve. 5. Add arrows to the lines pointing from the north pole to the south pole to show the field direction.

[1 mark] for describing the method of marking dots at the ends of the needle and moving the compass progressively. [1 mark] for mentioning connecting the dots to draw continuous field lines. [0.5 marks] for stating that arrows must point from North to South to indicate field direction.

To find the final velocity, we use the principle of conservation of momentum:

\(\text{Total momentum before collision} = \text{Total momentum after collision}\)

\(p_{\text{initial}} = m_1 u_1 + m_2 u_2\)
\(p_{\text{initial}} = (0.50 \text{ kg} \times 1.2 \text{ m/s}) + (0.30 \text{ kg} \times 0 \text{ m/s}) = 0.60 \text{ kg m/s}\)

Let \(v\) be the common velocity after the collision:
\(p_{\text{final}} = (m_1 + m_2) v\)
\(0.60 \text{ kg m/s} = (0.50 \text{ kg} + 0.30 \text{ kg}) \times v\)
\(0.60 = 0.80 \times v\)
\(v = \frac{0.60}{0.80} = 0.75 \text{ m/s}\)

• 1 mark: Recall and substitution into momentum formula: \(p = mv\) or statement of conservation of momentum (e.g., \(0.50 \times 1.2 = (0.50 + 0.30) \times v\))
• 1 mark: Calculation of total initial momentum (\(0.60 \text{ kg m/s}\)) or rearrangement to find \(v = \frac{0.60}{0.80}\)
• 0.5 mark: Correct final velocity value with correct unit (\(0.75 \text{ m/s}\))

First, convert the time from minutes to seconds:
\(t = 3.0 \text{ minutes} = 3.0 \times 60 = 180 \text{ s}\)

Next, use the electrical energy equation:
\(E = I \times V \times t\)
\(E = 2.5 \text{ A} \times 12 \text{ V} \times 180 \text{ s}\)
\(E = 30 \text{ W} \times 180 \text{ s} = 5400 \text{ J}\) (or \(5.4 \text{ kJ}\))

• 1 mark: Recall and use of the equation relating energy, voltage, current, and time: \(E = I \times V \times t\) (or separate calculation of power \(P = IV\) and energy \(E = P \times t\))
• 0.5 mark: Conversion of time into seconds (\(3.0 \times 60 = 180 \text{ s}\))
• 1 mark: Correct calculation of final energy with a valid unit (\(5400 \text{ J}\) or \(5.4 \text{ kJ}\)) [0.5 mark for numerical value, 0.5 mark for unit]

1. Identify the given values and sign convention:
Let the direction towards the wall be positive (+).
Mass of the ball, \(m = 0.12\text{ kg}\)
Initial velocity, \(u = +15\text{ m/s}\)
Final velocity, \(v = -12\text{ m/s}\) (negative because it is in the opposite direction)
Time of contact, \(t = 0.040\text{ s}\)

2. Calculate the change in momentum (\(\Delta p\)):
\(\Delta p = m(v - u)\)
\(\Delta p = 0.12\text{ kg} \times (-12\text{ m/s} - 15\text{ m/s}) = 0.12 \times (-27) = -3.24\text{ kg m/s}\).
The magnitude is \(3.24\text{ kg m/s}\) and the direction is away from the wall.

3. Calculate the average force (\(F\)):
\(F = \frac{\Delta p}{\Delta t}\)
\(F = \frac{-3.24\text{ kg m/s}}{0.040\text{ s}} = -81\text{ N}\).
The magnitude of the force is \(81\text{ N}\) and its direction is away from the wall.

Part (a) [4.0 Marks]:
- Award 1 mark for stating the momentum formula: \(p = mv\) or \(\Delta p = m(v-u)\).
- Award 1 mark for correct substitution of values including opposite signs for velocity: \(0.12 \times (-12 - 15)\) or equivalent.
- Award 1 mark for the correct numerical value: \(3.24\text{ kg m/s}\) (or \(\text{N s}\)).
- Award 1 mark for the correct direction: 'away from the wall' or indicating negative sign if initial velocity was positive.

Part (b) [2.5 Marks]:
- Award 1 mark for stating the force-momentum relationship: \(F = \frac{\Delta p}{t}\) or \(F = \frac{m(v-u)}{t}\).
- Award 1 mark for correct calculation of magnitude: \(81\text{ N}\).
- Award 0.5 marks for stating the direction is away from the wall (or negative sign consistent with part a).

(a) To charge the polythene rod, the student should rub it vigorously with the dry woollen cloth. Due to friction, work is done to overcome electrostatic attraction, and electrons (which are negatively charged particles) are transferred. Electrons move from the woollen cloth onto the surface of the polythene rod. Because the polythene rod now has a surplus of electrons, it becomes negatively charged, while the cloth is left with a positive charge of equal magnitude.

(b) When the negatively charged rod is brought near the stream of water, the stream bends towards the rod. This occurs because water molecules are polar, meaning they have a permanent charge separation (asymmetric distribution of charge). The negative charge on the polythene rod repels the negative charges (electrons) in the water molecules and attracts the positive charges. As a result, the side of the water stream closest to the rod becomes positively charged. Since opposite charges attract, and these positive charges are closer to the rod than the repelled negative charges, a net attractive electrostatic force pulls the water stream towards the rod.

Part (a) [3.5 Marks]:
- Award 1 mark for stating that the rod is rubbed with the cloth (friction).
- Award 1 mark for identifying that electrons are the charged particles that transfer.
- Award 1.5 marks for explaining the direction of transfer: electrons transfer from the woollen cloth to the polythene rod (creating a net negative charge on the rod).

Part (b) [3.0 Marks]:
- Award 1 mark for stating that the stream of water bends or is attracted towards the rod.
- Award 1 mark for explaining that charge redistribution / polarisation occurs in the water stream (or water molecules are polar).
- Award 1 mark for explaining that the opposite (positive) charges are closer to the rod, resulting in a net attractive electrostatic force.

(a) Both X-rays and microwaves belong to the electromagnetic spectrum, meaning they share the following properties: they are transverse waves, they can travel through a vacuum, they travel at the same speed in a vacuum (\(3 \times 10^8\text{ m/s}\)), and they transfer energy and information.

(b) Comparing their wave parameters: X-rays have a much shorter wavelength than microwaves. Since the speed of all electromagnetic waves in a given medium is constant, and \(v = f\lambda\), X-rays have a much higher frequency than microwaves. Since photon energy is directly proportional to frequency (\(E = hf\)), X-rays also have a much higher photon energy than microwaves.

(c) Because of their high frequency and photon energy, X-rays are a form of ionizing radiation. This means they possess sufficient energy to remove tightly bound electrons from atoms, creating ions. This ionization can break chemical bonds within biological molecules, damaging DNA and leading to cell mutations or cancer. In contrast, microwaves have much lower frequency and energy; they are non-ionizing and can only cause heating of body tissue by vibrating polar molecules like water.

Part (a) [2.0 Marks]:
- Award 1 mark each for any two correct common properties (e.g., both are transverse waves, both travel at the speed of light in a vacuum / \(3 \times 10^8\text{ m/s}\), both can travel through a vacuum, both transfer energy).

Part (b) [3.0 Marks]:
- Award 1 mark for stating that X-rays have a shorter wavelength than microwaves (or vice versa).
- Award 1 mark for stating that X-rays have a higher frequency than microwaves (or vice versa).
- Award 1 mark for stating that X-rays have a higher photon energy than microwaves (or vice versa).

Part (c) [1.5 Marks]:
- Award 1 mark for stating that X-rays are ionizing radiation while microwaves are non-ionizing.
- Award 0.5 marks for explaining that ionization can damage DNA / cause mutations / cause cancer, whereas microwaves only cause heating.

To find the common velocity after the collision, we use the principle of conservation of momentum: total initial momentum equals total final momentum. Initial momentum = \(m_1 v_1 + m_2 v_2 = (0.50\text{ kg} \times 4.0\text{ m/s}) + (0.30\text{ kg} \times 0\text{ m/s}) = 2.0\text{ kg m/s}\). Total mass after the collision = \(0.50\text{ kg} + 0.30\text{ kg} = 0.80\text{ kg}\). Final velocity = \(\frac{2.0\text{ kg m/s}}{0.80\text{ kg}} = 2.5\text{ m/s}\).

Award 1 mark for the correct answer (C).

For a star with a mass much larger than the Sun, the life cycle begins with a nebula which collapses to form a main sequence star. After the main sequence phase, the star expands and cools to become a red supergiant. The red supergiant then explodes in a supernova, leaving behind either a highly dense neutron star or a black hole. This matches sequence B.

Award 1 mark for the correct answer (B).

We use the formula linking charge, current, and time: \(Q = I \times t\). Rearranging for current: \(I = \frac{Q}{t}\). Substituting the values: \(I = \frac{1.2 \times 10^{-6}\text{ C}}{0.050\text{ s}} = 2.4 \times 10^{-5}\text{ A}\).

1 mark for stating or using the correct equation \(I = Q/t\). 1 mark for the correct calculation value of \(2.4 \times 10^{-5}\) (or \(0.000024\)). 0.5 marks for stating the unit as Amperes (A).

When the hydrogen fuel in the core of a main sequence star runs out, nuclear fusion in the core decreases, so the core collapses due to gravity. This contraction heats up the core and the region immediately surrounding it, triggering the fusion of hydrogen in a shell around the core. The immense energy released increases the outward radiation pressure, causing the outer layers of the star to expand and cool.

1 mark: Core runs out of hydrogen / core fusion decreases, leading to core collapse under gravity. 1 mark: Core contracts and heats up, starting hydrogen shell fusion around the core. 0.5 marks: The resulting increased radiation pressure forces the outer layers to expand.

Using the principle of conservation of momentum: total initial momentum = total final momentum. \(m_A \times u_A + m_B \times u_B = (m_A + m_B) \times v\). Substituting the known values: \(0.25 \times 1.2 + 0.15 \times 0 = (0.25 + 0.15) \times v\). \(0.30 = 0.40 \times v\). Rearranging for v: \(v = \frac{0.30}{0.40} = 0.75\text{ m/s}\).

1 mark for using conservation of momentum to set up the equation: \(m_A u_A = (m_A + m_B) v\) or showing total initial momentum is \(0.30\text{ kg m/s}\). 1 mark for calculating total mass as \(0.40\text{ kg}\) and rearranging. 0.5 marks for correct final velocity \(0.75\text{ m/s}\).

First, convert both temperatures from Celsius to Kelvin: \(T_1 = 20 + 273 = 293\text{ K}\) and \(T_2 = 80 + 273 = 353\text{ K}\). Using the pressure law for constant volume: \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\). Substituting the values: \(\frac{101}{293} = \frac{P_2}{353}\). Solving for \(P_2\): \(P_2 = 101 \times \frac{353}{293} \approx 121.68\text{ kPa}\). Rounding to 3 significant figures gives \(122\text{ kPa}\).

1 mark for converting both temperatures to Kelvin (\(293\text{ K}\) and \(353\text{ K}\)). 1 mark for correct formula or substitution of Kelvin values. 0.5 marks for correct final answer of \(122\text{ kPa}\) (accept \(121.7\text{ kPa}\) or \(121.68\text{ kPa}\)).

The relationship between refractive index and critical angle is given by: \(\sin(c) = \frac{1}{n}\). Substituting \(n = 1.55\) gives: \(\sin(c) = \frac{1}{1.55} \approx 0.64516\). Taking the inverse sine: \(c = \sin^{-1}(0.64516) \approx 40.2^\circ\).

1 mark for stating the correct equation: \(\sin(c) = 1/n\). 1 mark for calculating \(\sin(c) \approx 0.645\) or substituting values correctly. 0.5 marks for correct final angle: \(40.2^\circ\) (accept \(40^\circ\) or \(40.18^\circ\)).

Place the bar magnet on a piece of paper. Place the plotting compass near one of the poles of the magnet and make a pencil dot on the paper at the position of the pointer head of the compass needle. Move the compass so that the tail of the needle aligns with the dot just made, and then mark the new position of the pointer head with a new dot. Repeat this process step-by-step until the other pole of the magnet is reached. Connect the dots with a smooth line to map the field line, adding an arrow from North to South.

1 mark: Place compass near one pole and mark the needle's pointer direction with a dot. 1 mark: Move compass so the tail of the needle aligns with the previous dot and mark the new pointer direction. 0.5 marks: Repeat this process to trace a line from North to South and connect the dots.

The relationship between energy transferred, charge, and potential difference is: \(E = Q \times V\). Substituting the given values: \(E = 15\text{ C} \times 2.2\text{ V} = 33\text{ J}\).

1 mark for the correct formula \(E = Q \times V\) or substitution. 1 mark for the correct calculated value of \(33\). 0.5 marks for stating the unit as Joules or \(J\).

First, convert the frequency from MHz to Hz: \(f = 95.8\text{ MHz} = 95.8 \times 10^6\text{ Hz}\). Use the wave speed equation: \(v = f \times \lambda\). Rearranging for wavelength: \(\lambda = \frac{v}{f}\). Substituting the values: \(\lambda = \frac{3.0 \times 10^8\text{ m/s}}{95.8 \times 10^6\text{ Hz}} \approx 3.13\text{ m}\).

1 mark for converting frequency to Hz: \(95.8 \times 10^6\text{ Hz}\). 1 mark for rearranging to \(\lambda = v / f\) and substituting values. 0.5 marks for the correct wavelength: \(3.1\text{ m}\) or \(3.13\text{ m}\) (accept \(3.132\text{ m}\)).

1. Let the initial velocity to the right be positive: \(v_i = +3.0\text{ m/s}\). 2. The final velocity to the left is negative: \(v_f = -2.0\text{ m/s}\). 3. The initial momentum is \(p_i = m \times v_i = 0.25\text{ kg} \times 3.0\text{ m/s} = 0.75\text{ kg m/s}\). 4. The final momentum is \(p_f = m \times v_f = 0.25\text{ kg} \times (-2.0\text{ m/s}) = -0.50\text{ kg m/s}\). 5. The change in momentum is \(\Delta p = p_f - p_i = -0.50\text{ kg m/s} - 0.75\text{ kg m/s} = -1.25\text{ kg m/s}\). This is a magnitude of \(1.25\text{ kg m/s}\) to the left.

[1 mark] Recall and use of momentum formula \(p = mv\) or change in momentum \(\Delta p = m(v-u)\), with opposite signs assigned to opposite directions. [1 mark] Correct calculation of the magnitude (1.25). [0.5 marks] Correct unit (kg m/s or N s) and direction (to the left or negative).

During refueling, the flow of fuel through the hose causes friction, which leads to a build-up of electrostatic charge on the aircraft. Because the aircraft is insulated from the ground by its rubber tires, this charge cannot escape. If a large amount of charge accumulates, a high potential difference is created, which could cause a spark. A spark in the presence of highly flammable fuel vapor can cause an explosion. Connecting a copper bonding wire (a good electrical conductor) provides a safe, low-resistance path for the charge to discharge, preventing any electrostatic build-up and ensuring no sparks can occur.

[1 mark] Explanation that fuel flow/friction causes a build-up of electrostatic charge. [1 mark] Explanation that the copper wire is a conductor that allows this charge to safely discharge/neutralize (earthing), preventing charge build-up. [0.5 marks] Stating that this prevents a spark which could ignite the flammable fuel vapor.

We use the wave equation: \(v = f \lambda\), where \(v\) is the speed of light, \(f\) is the frequency, and \(\lambda\) is the wavelength. Rearranging for wavelength: \(\lambda = \frac{v}{f}\). Substitute the values: \(\lambda = \frac{3.00 \times 10^8\text{ m/s}}{2.45 \times 10^9\text{ Hz}} \approx 0.1224\text{ m}\). To 3 significant figures, the wavelength is \(0.122\text{ m}\) (or \(12.2\text{ cm}\)).

[1 mark] Recall and rearrangement of the wave equation: \(v = f \lambda\) into \(\lambda = \frac{v}{f}\). [1 mark] Correct substitution and calculation to at least 2 significant figures (0.12 or 0.122). [0.5 marks] Correct unit (m or cm with consistent value conversion).

First, convert time from minutes to seconds: \(t = 45 \times 60\text{ s} = 2700\text{ s}\). Next, recall the formula for electrical energy: \(E = I \times V \times t\). Substitute the given values: \(E = 8.5\text{ A} \times 230\text{ V} \times 2700\text{ s} = 5,278,500\text{ J}\). Convert this energy to megajoules (MJ) by dividing by \(1,000,000\): \(E = 5.2785\text{ MJ}\). Rounding to 2 significant figures gives \(5.3\text{ MJ}\).

[1 mark] Correct conversion of time from minutes to seconds (\(2700\text{ s}\)). [1 mark] Recall and substitution into the energy formula \(E = I \times V \times t\) (or using \(P = IV\) to find \(P = 1955\text{ W}\) first, then \(E = Pt\)). [0.5 marks] Correct final calculation in megajoules (5.3 MJ or 5.28 MJ) with correct unit.

After leaving the main sequence: 1. A star with a mass similar to the Sun expands and cools to become a red giant. It eventually sheds its outer layers of gas as a planetary nebula, leaving behind a hot, dense core called a white dwarf. 2. A star with a mass much greater than the Sun expands to become a red supergiant. When it runs out of fuel, it undergoes a rapid gravitational collapse and explodes in a supernova. The remaining core collapses to form either a neutron star or, if the mass is high enough, a black hole.

[1 mark] Correctly describing the path of a low-mass star (red giant to planetary nebula and white dwarf). [1 mark] Correctly describing the path of a high-mass star (red supergiant to supernova and neutron star/black hole). [0.5 marks] Stating a clear, correct comparison or distinction between the endpoints of the two stellar paths.

The relationship between refractive index (\(n\)) and critical angle (\(c\)) is given by: \(n = \frac{1}{\sin(c)}\). Substitute \(c = 41.8^\circ\): \(n = \frac{1}{\sin(41.8^\circ)}\). Calculate the sine of the angle: \(\sin(41.8^\circ) \approx 0.6665\). Thus, \(n = \frac{1}{0.6665} \approx 1.50\). The refractive index is a ratio and has no units.

[1 mark] Recall of the formula relating refractive index and critical angle: \(n = \frac{1}{\sin(c)}\) or \(\sin(c) = \frac{1}{n}\). [1 mark] Correct substitution of \(41.8^\circ\) and calculation of the denominator (\(\sin(41.8^\circ) \approx 0.67\)). [0.5 marks] Correct final value of 1.50 (accept answers in range 1.49 to 1.51) and no units specified.

Average speed is defined as the total distance traveled divided by the total time taken: \(\text{average speed} = \frac{\text{total distance}}{\text{total time}}\). 1. Calculate the total distance: \(\text{total distance} = 250\text{ m} + 150\text{ m} = 400\text{ m}\). 2. Calculate the total time: \(\text{total time} = 30\text{ s} + 22\text{ s} = 52\text{ s}\). 3. Calculate average speed: \(\text{average speed} = \frac{400\text{ m}}{52\text{ s}} \approx 7.69\text{ m/s}\). To 2 significant figures, this is \(7.7\text{ m/s}\).

[1 mark] Recall of the formula for average speed: \(\text{average speed} = \frac{\text{total distance}}{\text{total time}}\). [1 mark] Correct calculation of total distance (400 m) and total time (52 s), and their division to yield approximately 7.7. [0.5 marks] Correct unit (m/s or m s^-1).

(a) The pesticide droplets all acquire the same positive charge. Since like charges repel, the droplets exert a repulsive force on each other, causing them to spread out and form a fine, even mist.

(b) The charged droplets induce an opposite (negative) charge on the surfaces of the plant leaves as they approach. The attractive force between the positively charged droplets and the induced negative charge on the leaves pulls the droplets towards the leaves, including the undersides which would otherwise be shielded.

(c) Use the equation: \(Q = I \times t\)
Convert time to seconds: \(t = 2.5 \times 60 = 150\text{ s}\)
Substitute the values: \(Q = 3.2 \times 10^{-5}\text{ A} \times 150\text{ s} = 4.8 \times 10^{-3}\text{ C}\) (or \(0.0048\text{ C}\)).

Part (a) [2 marks]:
- Droplets have the same charge (1)
- Repel each other (causing them to spread out) (1)

Part (b) [2 marks]:
- Charge on droplets induces opposite charge on leaves (1)
- Opposite charges attract (causing droplets to bend around and coat undersides) (1)

Part (c) [3.5 marks]:
- Recall of \(Q = I \times t\) (1)
- Conversion of time: \(2.5\text{ min} = 150\text{ s}\) (1)
- Calculation: \(4.8 \times 10^{-3}\) (1)
- Unit: C / Coulomb (0.5)

(a) A star on the main sequence is in hydrostatic equilibrium. There is an inward gravitational force (due to the star's mass) pulling the gas inward, which is balanced by an outward force due to thermal expansion and radiation pressure from nuclear fusion reactions in the core. Because these forces are equal and opposite, the star remains at a stable size.

(b) When the Sun runs out of hydrogen in its core, nuclear fusion ceases, and the core collapses under gravity, heating up. This causes the outer layers to expand and cool, transforming the Sun into a red giant. Eventually, fusion in the core stops completely, and the outer layers of gas are ejected into space as a planetary nebula. The remaining hot, dense core cools down over time, leaving behind a white dwarf.

Part (a) [3.5 marks]:
- Gravitational force pulls inwards (1)
- Thermal/radiation pressure acts outwards (1)
- Outward pressure is caused by nuclear fusion in the core (1)
- The forces are balanced/equal (0.5)

Part (b) [4 marks]:
- Core runs out of hydrogen / collapses AND outer layers expand (to form a red giant) (1)
- Red giant cools/outer layers are ejected (1)
- Planetary nebula formed (1)
- Leaving behind a white dwarf as the dense core (1)

(a) The principle of conservation of momentum states that the total momentum before a collision is equal to the total momentum after the collision, provided no external forces act on the system.

(b) Calculate initial momentum:
\(p_{\text{initial}} = (m_1 \times v_1) + (m_2 \times v_2)\)
\(p_{\text{initial}} = (0.25\text{ kg} \times 4.0\text{ m/s}) + (0.15\text{ kg} \times 0\text{ m/s}) = 1.0\text{ kg m/s}\)
After collision, they stick together:
\(p_{\text{final}} = (m_1 + m_2) \times v = (0.25 + 0.15) \times v = 0.40 \times v\)
Using conservation of momentum:
\(1.0 = 0.40 \times v\)
\(v = \frac{1.0}{0.40} = 2.5\text{ m/s}\)

(c) The average force is given by the rate of change of momentum:
\(F = \frac{\Delta p}{t}\)
For the stationary car, its change in momentum is:
\(\Delta p = m_2 \times v - m_2 \times 0 = 0.15\text{ kg} \times 2.5\text{ m/s} = 0.375\text{ kg m/s}\)
Calculate the force:
\(F = \frac{0.375\text{ kg m/s}}{0.080\text{ s}} = 4.6875\text{ N} \approx 4.7\text{ N}\)

Part (a) [1.5 marks]:
- Total momentum before (collision) = total momentum after (collision) (1)
- Provided no external forces act / in a closed system (0.5)

Part (b) [3 marks]:
- Calculation of initial momentum as \(1.0\text{ kg m/s}\) (1)
- Equation equating initial and final momentum: \(1.0 = 0.40 \times v\) (1)
- Calculation of velocity \(2.5\text{ m/s}\) (1)

Part (c) [3 marks]:
- Formula: \(F = \frac{\Delta p}{t}\) or \(F = \frac{mv - mu}{t}\) (1)
- Calculation of momentum change for the second car (\(0.375\text{ kg m/s}\)) (1)
- Calculation of force (\(4.6875\text{ N}\) or \(4.7\text{ N}\)) (1)

(a) Wavelength: UV waves have a much shorter wavelength than radio waves.
Frequency: UV waves have a much higher frequency than radio waves.
Safety risks: UV waves are ionizing and carry more energy, which can cause damage to skin cells, sunburn, skin cancer, and cataracts (damage to eyes). Radio waves have very low energy, are non-ionizing, and have no known harmful effects at standard exposure levels.

(b) Use the wave equation:
\(v = f \lambda\)
Rearrange for frequency:
\(f = \frac{v}{\lambda}\)
Substitute the values:
\(f = \frac{3.0 \times 10^8\text{ m/s}}{3.2 \times 10^{-7}\text{ m}} = 9.375 \times 10^{14}\text{ Hz}\) (or \(9.4 \times 10^{14}\text{ Hz}\))

Part (a) [3.5 marks]:
- UV has a shorter wavelength than radio waves (1)
- UV has a higher frequency than radio waves (1)
- UV is ionizing / can damage cells / cause skin cancer / cause cataracts (1)
- Radio waves are non-ionizing / safe / do not damage cells (0.5)

Part (b) [4 marks]:
- Recall of \(v = f \lambda\) (1)
- Rearrangement \(f = \frac{v}{\lambda}\) (1)
- Correct substitution (1)
- Correct calculation: \(9.4 \times 10^{14}\text{ Hz}\) or \(9.375 \times 10^{14}\text{ Hz}\) (1) (accept correct unit for the mark)