2023 Edexcel IGCSE Physics Practice Paper with Answers

First, determine how many half-lives have passed during the activity drop: \( 1600 \rightarrow 800 \rightarrow 400 \rightarrow 200 \rightarrow 100 \). This represents \( 4 \) half-lives. The total time elapsed is \( 24 \text{ hours} \). Therefore, the duration of one half-life is \( \frac{24 \text{ hours}}{4} = 6 \text{ hours} \).

1 mark for correct selection (B).

The relationship between electrical energy transferred \( E \), charge \( Q \), and potential difference \( V \) is given by: \( E = Q \times V \). Substituting the given values: \( E = 15 \text{ C} \times 12 \text{ V} = 180 \text{ J} \). Note that the time \( 5.0 \text{ s} \) is extra information and is not required for this specific calculation.

1 mark for correct selection (C).

First, convert the mass from kilograms to grams: \( m = 0.48 \text{ kg} = 480 \text{ g} \). Next, calculate the volume of the block: \( V = 5.0 \text{ cm} \times 8.0 \text{ cm} \times 15 \text{ cm} = 600 \text{ cm}^3 \). Finally, use the density formula: \( \text{density} = \frac{\text{mass}}{\text{volume}} = \frac{480 \text{ g}}{600 \text{ cm}^3} = 0.80 \text{ g/cm}^3 \).

1 mark for correct selection (B).

First, find the wave speed using the wave equation: \( v = f \times \lambda \), so \( v = 8.0 \text{ Hz} \times 0.15 \text{ m} = 1.2 \text{ m/s} \). Then, use the relationship between distance, speed, and time: \( \text{distance} = v \times t = 1.2 \text{ m/s} \times 5.0 \text{ s} = 6.0 \text{ m} \).

1 mark for correct selection (C).

To find the wavelength, the student needs to freeze the wave motion. By taking a digital photograph of the wave with a ruler placed parallel to the direction of wave travel, the student can use the ruler's scale shown in the image to measure the horizontal distance between two consecutive crests or troughs.

MP1: Take a photograph of the wave next to a ruler to establish a scale (1). MP2: Measure the distance between adjacent crests, troughs, or any two identical consecutive points (1).

When the temperature of a gas is increased, the average kinetic energy of the particles increases, so they move faster. In a container of fixed volume, these faster-moving particles collide with the walls of the container more frequently and with more force, resulting in a higher pressure.

MP1: State that pressure increases because particles gain kinetic energy and move faster (1). MP2: Explain that this leads to more frequent or more forceful collisions with the walls of the container (1).

The current-voltage graph for a filament lamp is a curve that passes through the origin. At low voltages, the curve is straight, but as the voltage and current increase, the filament heats up. This increase in temperature increases the resistance of the filament, causing the gradient of the \(I-V\) graph to decrease (flatten out) at higher voltages.

MP1: Describe the shape as a curve passing through the origin with decreasing gradient / leveling off (1). MP2: Explain that resistance increases as the temperature of the filament increases (1).

A medical tracer must have a half-life long enough for the diagnostic procedure/scan to be successfully completed. However, it should not be much longer than this so that the radioactivity decays quickly to a safe level, minimizing the long-term radiation exposure and risk of damage to the patient's cells and tissues.

MP1: Long enough to allow the medical investigation or scan to be completed (1). MP2: Short enough so that the activity decays quickly to minimize radiation dose/damage to the patient (1).

Hooke's Law states that extension is directly proportional to the applied force. On a graph of force against extension (or extension against force), this relationship is represented by a straight line passing through the origin. If the graph is curved (non-linear), the rubber band does not obey Hooke's Law.

MP1: Identify that direct proportionality / Hooke's law is represented by a straight line passing through the origin (1). MP2: State that if the graph is curved or non-linear, the material does not obey Hooke's law (1).

For an object to remain stationary, the forces acting on it must be balanced. The downward force acting on the block is its weight (gravity), and the upward force is the upthrust from the water. Because these two forces are equal in magnitude and opposite in direction, the resultant force is zero, meaning the block remains at rest.

MP1: State that the upward force (upthrust) is equal to the downward force (weight) (1). MP2: Explain that these balanced forces result in a net or resultant force of zero (1).

According to the conservation of energy, the total energy remains constant. However, during the collision with the floor and due to air resistance, some of the ball's kinetic energy is transferred to the surroundings as thermal energy (heat) and sound energy. Consequently, the ball has less kinetic energy immediately after the bounce, which means it cannot reach its original gravitational potential energy level (height).

MP1: State that energy is dissipated / transferred to the surroundings as thermal energy (heat) or sound energy (1). MP2: Explain that the ball has less kinetic energy after the bounce, so it cannot reach its original gravitational potential energy / height (1).

As the bar magnet moves into the solenoid, its magnetic field lines cut through the turns of the copper wire. This change in magnetic field (or magnetic flux linkage) induces an electromotive force (voltage) across the ends of the coil, which is detected by the voltmeter. The deflection is temporary because the induction only occurs while there is relative motion between the magnet and the coil.

MP1: Explain that the moving magnetic field cuts through the turns of the coil / there is a change in magnetic flux linkage (1). MP2: State that this induces a voltage / electromotive force (e.m.f.) in the coil (1).

The wavelength of a wave is the distance over which the wave's shape repeats. On a displacement-distance graph, the student can determine this by finding the horizontal distance from one peak (crest) to the very next peak, or from one trough to the very next trough.

1 mark: State that the student should measure the distance between two consecutive peaks / crests / troughs. 1 mark: Specify that this is measured along the horizontal axis (distance axis).

Up to the limit of proportionality, force is directly proportional to extension, which is represented by a straight-line section passing through the origin. Beyond this limit, the relationship is no longer linear, so the graph starts to curve. The limit of proportionality is the exact point where this linear relationship ends and the curve begins.

1 mark: State that the initial linear / straight-line section represents proportionality. 1 mark: Explain that the limit of proportionality is the point where the graph starts to curve / bend / deviate from a straight line.

The gradient of a velocity-time graph represents the acceleration. When the acceleration is constant, the gradient of the straight line is calculated using \(\text{gradient} = \Delta v / \Delta t\). If the acceleration is non-constant (the graph is curved), the student must draw a tangent to the curve at the required time and calculate the gradient of that tangent line.

1 mark: State that acceleration is represented by the gradient (slope) of the velocity-time graph. 1 mark: Explain that for a curve (non-constant acceleration), a tangent must be drawn at that point and its gradient calculated.

Direct current (d.c.) flows in a single direction with a constant value, which is displayed as a flat, straight horizontal line at a fixed voltage level. Alternating current (a.c.) regularly reverses its direction and varies in magnitude, which appears on the screen as a repeating sinusoidal wave that goes above and below the central zero-voltage horizontal axis.

1 mark: Describe d.c. as a straight, flat, or horizontal line. 1 mark: Describe a.c. as a repeating wave shape that alternates above and below the zero-voltage line.

To find the half-life from an activity-time graph, first choose a starting activity value on the vertical axis, \(A_0\). Next, divide this value by two to get \(A_0 / 2\). Locate this halved activity value on the vertical axis, project horizontally to the decay curve, and then project vertically down to the horizontal axis to read the corresponding time interval, which represents the half-life.

1 mark: Mention halving any initial activity value (or finding the time taken for activity to fall from \(N\) to \(N/2\)). 1 mark: Explain reading the corresponding time interval on the horizontal (time) axis from the curve.

At a constant temperature, pressure is inversely proportional to volume (Boyle's Law, \(P \propto 1/V\)), which produces a hyperbolic curve that slopes downwards. If the temperature is increased, the gas molecules have more kinetic energy and exert higher pressure at any given volume, shifting the entire curve upwards and to the right, further away from the origin.

1 mark: Describe the shape as a curve sloping downwards / hyperbola / showing an inverse relationship. 1 mark: State that increasing the temperature shifts the curve further away from the origin / upwards and to the right.

In a Sankey diagram, the total energy input is shown as a single wide arrow. This arrow splits into a useful energy output arrow and a wasted energy output arrow. The width of each arrow is directly proportional to the amount of energy it represents. Therefore, the efficiency of the device is represented visually by the ratio of the width of the useful output arrow to the total width of the input arrow.

1 mark: State that the width (or thickness) of the arrows is proportional to the energy flow. 1 mark: Explain that efficiency is shown by how wide the useful energy arrow is compared to the input arrow.

When the generator's coil is rotated at twice the speed, it cuts magnetic field lines at twice the rate, which doubles the maximum induced electromotive force (peak voltage). Additionally, because the coil completes twice as many rotations per second, the frequency of the alternating current doubles, meaning twice as many complete cycles are displayed in the same time interval on the graph.

1 mark: State that the peak voltage / amplitude of the wave doubles. 1 mark: State that the frequency doubles / time period is halved / twice as many cycles are shown in the same time.

Increasing the pitch increases the frequency of the sound wave, meaning more cycles will fit on the screen in the same time period. Loudness is determined by amplitude, so since loudness is constant, the amplitude (height) remains unchanged.

1 mark: stating that more wave cycles are visible / waves are closer together / horizontal compression. 1 mark: stating that the height / amplitude of the waves remains constant.

Because the metal case is connected to the earth wire, a path of very low resistance is provided. A very large current flows from the live wire directly to the earth. This high current exceeds the fuse's current rating, causing the fuse wire to heat up and melt (blow), which isolates the appliance from the live supply.

1 mark: current flows through the earth wire / low resistance path. 1 mark: this high current melts / blows the fuse, breaking the circuit.

The pressure in a liquid is given by \(p = \rho g h\), meaning pressure increases with depth. The water at the bottom hole is under greater pressure, so it experiences a larger force forcing it out, causing it to travel further horizontally than the water from the top hole.

1 mark: stating that the water from the bottom hole travels further horizontally (or spurts out faster). 1 mark: explaining that pressure in a liquid increases with depth / is greater at the bottom.

The gradient (slope) of a velocity-time graph represents the acceleration, calculated as change in velocity divided by time. A constant deceleration means a constant negative acceleration, which is shown as a straight line with a negative gradient.

1 mark: stating that the gradient or slope represents the acceleration. 1 mark: describing a straight line with a negative gradient / sloping downwards.

Alpha particles are highly ionising because they have a \(+2e\) charge and a large mass, meaning they interact frequently with atoms and lose energy quickly. In contrast, beta particles have a \(-1e\) charge and a tiny mass, so they are much less ionising and can penetrate further before losing their energy.

1 mark: beta particles are less ionising / have less charge. 1 mark: beta particles are smaller / have less mass / travel faster.

Stretching the wire beyond its elastic limit means it undergoes plastic deformation. Once the stretching force is removed, the atoms have slid past one another permanently, so the wire remains permanently stretched and does not return to its original length.

1 mark: stating that the wire does not return to its original length / shape when the force is removed. 1 mark: stating that it undergoes plastic / permanent deformation.

Air is a gas, so its particles are far apart, making it an excellent thermal insulator (poor conductor) which reduces conduction. By trapping the air in small, sealed bubbles, the air cannot circulate, which prevents the formation of convection currents.

1 mark: air is a poor conductor / good insulator which reduces conduction. 1 mark: trapping the air prevents it from moving / circulating, reducing convection.

When light intensity increases, the resistance of the LDR decreases. Since the total resistance of the series circuit decreases, the current flowing through the circuit increases. According to \(V = IR\), an increased current through the constant resistance of the fixed resistor results in an increased voltage across it.

1 mark: stating that the voltage across the fixed resistor increases. 1 mark: explaining that LDR resistance decreases (leading to larger current / smaller share of voltage across the LDR).

When water waves transition from deep water to shallow water, they slow down because of increased friction with the bottom of the tank. Since the frequency of the wave source remains constant, and using the wave equation \(v = f \lambda\), a decrease in speed \(v\) results in a proportional decrease in wavelength \(\lambda\).

1 mark for stating that the speed decreases.
1 mark for stating that the wavelength decreases.

Hooke's law states that the extension of an elastic object is directly proportional to the force applied, provided the limit of proportionality is not exceeded. On a graph of force against extension, this is represented by a straight line passing through the origin. For a rubber band, the line is curved (non-linear), indicating that extension is not directly proportional to the applied force.

1 mark for identifying that the graph is non-linear / curved.
1 mark for explaining that this means force and extension are not directly proportional.

When the paint is sprayed, every droplet acquires a positive charge from the nozzle. Because like charges repel each other, the positively charged droplets experience a electrostatic force pushing them away from one another. This mutual repulsion prevents the droplets from clumping together and causes them to spread out into a very fine and even mist.

1 mark for stating that the paint droplets all gain the same (positive) charge.
1 mark for stating that like charges repel each other, causing the droplets to spread out.

Atmospheric pressure is created by the weight of the air molecules in the atmosphere pulling downwards due to gravity. At higher altitudes, there are fewer air molecules above (the air is less dense), which means there is less weight of air pushing down per unit area, resulting in lower pressure.

1 mark for stating that there are fewer air molecules / air density is lower at higher altitudes.
1 mark for identifying that there is less weight of air pushing down on a unit area.

Background radiation is always present in the environment from sources like radon gas and cosmic rays. To find the true activity of the radioactive sample itself, the student must first measure the background count rate without the source present, and then subtract this background count rate from the total measured count rate when the source is present.

1 mark for stating that background radiation must be measured to identify the ambient level of radiation.
1 mark for stating that the background count rate must be subtracted from the total measured count rate.

Double insulation means that the appliance has two independent layers of electrical insulation between the live internal parts and the outside environment. Since the outer casing is made of plastic (which is an insulator), it cannot conduct electricity. Therefore, even if an internal electrical fault occurs and a live wire comes loose and touches the inside of the casing, the casing cannot become live, eliminating the risk of electric shock and the need for an earth wire.

1 mark for stating that the plastic casing is an insulator / does not conduct electricity.
1 mark for explaining that the casing cannot become live / cannot cause an electric shock if an internal fault occurs.

According to Faraday's law of electromagnetic induction, the size of the induced voltage is directly proportional to the rate of cutting of magnetic flux. To increase this rate, the student can: 1. Move the magnet faster. 2. Use a stronger magnet (stronger magnetic field). 3. Increase the number of turns on the solenoid coil, which increases the total amount of wire cutting the magnetic field lines.

1 mark for stating any one valid method (e.g., move magnet faster / use a stronger magnet / increase the number of turns on the coil).
1 mark for stating a second, different valid method.

When a gas is heated, the average kinetic energy of its particles increases, meaning they move at higher average speeds. Because the container is rigid and its volume is constant, these faster-moving particles collide with the inner walls of the container more frequently and with greater force per collision. This results in an increased average force exerted on the walls, and since pressure is force per unit area, the gas pressure increases.

1 mark for stating that particles gain kinetic energy / move faster at higher temperatures.
1 mark for explaining that they collide with the container walls more frequently or with greater force.

Using the equation for uniform acceleration: \(v^2 = u^2 + 2as\). Since the car starts from rest, the initial velocity \(u = 0\text{ m/s}\). Substitute the given values into the equation: \(v^2 = 0^2 + 2 \times 1.5\text{ m/s}^2 \times 12\text{ m}\), which gives \(v^2 = 36\). Taking the square root gives \(v = \sqrt{36} = 6\text{ m/s}\).

1 mark for selecting the correct equation: \(v^2 = u^2 + 2as\). 1 mark for correct substitution: \(v^2 = 0 + 2 \times 1.5 \times 12\). 1 mark for correct final velocity with correct unit: \(6\text{ m/s}\) (or \(6.0\text{ m/s}\)).

First, convert the time from minutes to seconds: \(t = 5.0\text{ minutes} = 5.0 \times 60\text{ s} = 300\text{ s}\). Use the equation for electrical energy transferred: \(E = I \times V \times t\). Substitute the values into the equation: \(E = 0.35\text{ A} \times 12\text{ V} \times 300\text{ s} = 1260\text{ J}\).

1 mark for converting time to seconds: \(5.0 \times 60 = 300\text{ s}\). 1 mark for using the correct relationship: \(E = I \times V \times t\) (or calculating power \(P = 0.35 \times 12 = 4.2\text{ W}\) first). 1 mark for correct final answer with unit: \(1260\text{ J}\) (or \(1.26\text{ kJ}\)).

First, find the frequency of the wave. Frequency is the number of complete oscillations per second: \(f = \frac{15}{10\text{ s}} = 1.5\text{ Hz}\). Next, use the wave speed equation: \(v = f \times \lambda\). Substitute the frequency and the given wavelength: \(v = 1.5\text{ Hz} \times 0.60\text{ m} = 0.90\text{ m/s}\).

1 mark for calculating the frequency: \(f = 1.5\text{ Hz}\). 1 mark for selecting and using the wave equation: \(v = f \times \lambda\). 1 mark for correct speed with unit: \(0.90\text{ m/s}\) (or \(0.9\text{ m/s}\)).

First, calculate the pressure exerted by the seawater column using: \(p = \rho \times g \times h\). Substitute the given values: \(p_{\text{water}} = 1025\text{ kg/m}^3 \times 10\text{ m/s}^2 \times 18\text{ m} = 184,500\text{ Pa} = 184.5\text{ kPa}\). Total pressure is the sum of atmospheric pressure and the liquid pressure: \(p_{\text{total}} = 101\text{ kPa} + 184.5\text{ kPa} = 285.5\text{ kPa}\) (or \(285,500\text{ Pa}\)).

1 mark for calculating water pressure: \(184,500\text{ Pa}\) (or \(184.5\text{ kPa}\)). 1 mark for adding atmospheric pressure to water pressure in consistent units. 1 mark for correct final answer with unit: \(285.5\text{ kPa}\) (or \(285,500\text{ Pa}\) or \(2.9 \times 10^5\text{ Pa}\)).

First, calculate the useful work done in lifting the mass: \(W = m \times g \times h = 40\text{ kg} \times 10\text{ N/kg} \times 6.0\text{ m} = 2400\text{ J}\). Next, calculate the useful power output of the motor: \(P_{\text{out}} = \frac{W}{t} = \frac{2400\text{ J}}{8.0\text{ s}} = 300\text{ W}\). Since efficiency is \(75\%\) (or \(0.75\)), calculate the total power input: \(P_{\text{in}} = \frac{P_{\text{out}}}{\text{efficiency}} = \frac{300\text{ W}}{0.75} = 400\text{ W}\).

1 mark for calculating useful work done (\(2400\text{ J}\)) or useful power output (\(300\text{ W}\)). 1 mark for correct use of efficiency equation: \(P_{\text{in}} = \frac{P_{\text{out}}}{0.75}\). 1 mark for correct final answer with unit: \(400\text{ W}\).

First, find the energy required to melt the ice block: \(Q_1 = m \times L = 0.15\text{ kg} \times 3.3 \times 10^5\text{ J/kg} = 49,500\text{ J}\). Second, find the energy required to heat the resulting water from \(0^\circ\text{C}\) to \(12^\circ\text{C}\): \(Q_2 = m \times c \times \Delta T = 0.15\text{ kg} \times 4200\text{ J/kg}^\circ\text{C} \times 12^\circ\text{C} = 7560\text{ J}\). Total energy required: \(Q_{\text{total}} = Q_1 + Q_2 = 49,500\text{ J} + 7560\text{ J} = 57,060\text{ J}\) (or \(57\text{ kJ}\)).

1 mark for calculating the melting energy: \(49,500\text{ J}\). 1 mark for calculating the heating energy: \(7560\text{ J}\). 1 mark for correct total energy with unit: \(57,060\text{ J}\) (or \(57\text{ kJ}\) or \(5.7 \times 10^4\text{ J}\)).

By conservation of momentum: \(\text{Total momentum before collision} = \text{Total momentum after collision}\). Initial momentum: \(p_{\text{initial}} = m_1 u_1 + m_2 u_2 = (2.0 \times 4.5) + (1.0 \times 0) = 9.0\text{ kg m/s}\). Total mass after collision: \(m_{\text{total}} = 2.0 + 1.0 = 3.0\text{ kg}\). Let \(v\) be the common velocity: \(m_{\text{total}} \times v = 9.0\), so \(3.0 \times v = 9.0\), giving \(v = 3.0\text{ m/s}\).

1 mark for calculating the initial momentum: \(9.0\text{ kg m/s}\). 1 mark for equating initial and final momentum: \(3.0 \times v = 9.0\) (or rearranged: \(v = \frac{9.0}{3.0}\)). 1 mark for correct final velocity with unit: \(3.0\text{ m/s}\) (or \(3\text{ m/s}\)).

First, convert the temperatures to Kelvin: \(T_1 = 17 + 273 = 290\text{ K}\) and \(T_2 = 75 + 273 = 348\text{ K}\). Since volume and mass are constant, use the pressure law: \(\frac{p_1}{T_1} = \frac{p_2}{T_2}\). Rearrange to solve for \(p_2\): \(p_2 = \frac{p_1 \times T_2}{T_1} = \frac{1.5 \times 10^5\text{ Pa} \times 348\text{ K}}{290\text{ K}} = 1.8 \times 10^5\text{ Pa}\).

1 mark for converting both temperatures to Kelvin: \(290\text{ K}\) and \(348\text{ K}\). 1 mark for using the correct relationship: \(p_2 = \frac{p_1 \times T_2}{T_1}\). 1 mark for correct final pressure with unit: \(1.8 \times 10^5\text{ Pa}\) (or \(180\text{ kPa}\)).

Step 1: Convert the initial temperature to Kelvin: \(T_1 = 15 + 273 = 288\text{ K}\). Step 2: Use the pressure law for constant volume, \(P_1/T_1 = P_2/T_2\), rearranged to find \(T_2\): \(T_2 = T_1 \times (P_2 / P_1) = 288 \times (1.45 \times 10^5 / 1.10 \times 10^5) = 379.6\text{ K}\). Step 3: Convert the final temperature back to Celsius: \(t_2 = 379.6 - 273 = 106.6\text{ }^\circ\text{C}\). This rounds to \(107\text{ }^\circ\text{C}\).

MP1: Conversion of initial temperature to Kelvin (\(288\text{ K}\)) (1 mark). MP2: Recall and substitution into pressure law to find Kelvin temperature (\(380\text{ K}\) or \(379.6\text{ K}\)) (1 mark). MP3: Final conversion to Celsius (\(107\text{ }^\circ\text{C}\), accept range \(106.6\text{ }^\circ\text{C}\) to \(107\text{ }^\circ\text{C}\)) (1 mark).

Step 1: Calculate the useful work done on the crate using \(W = m \times g \times h = 450 \times 10 \times 12 = 54,000\text{ J}\). Step 2: Calculate the useful power output using \(P_{\text{out}} = W / t = 54,000 / 15 = 3600\text{ W}\). Step 3: Calculate the electrical power input using the efficiency formula: \(P_{\text{in}} = P_{\text{out}} / \text{efficiency} = 3600 / 0.75 = 4800\text{ W}\). (Alternatively: Total energy input = \(54,000 / 0.75 = 72,000\text{ J}\), so power input = \(72,000 / 15 = 4800\text{ W}\).)

MP1: Calculation of useful work done (\(54,000\text{ J}\)) or useful power output (\(3600\text{ W}\)) (1 mark). MP2: Correct use of efficiency equation to find input energy (\(72,000\text{ J}\)) or input power (1 mark). MP3: Correct calculation of final power input with unit (\(4800\text{ W}\) or \(4.8\text{ kW}\)) (1 mark).

Step 1: Calculate the power of the heating element using \(P = I^2 R = (6.0)^2 \times 8.5 = 36 \times 8.5 = 306\text{ W}\). Step 2: Convert the time from minutes to seconds: \(t = 5.0 \times 60 = 300\text{ s}\). Step 3: Calculate the electrical energy transferred using \(E = P \times t = 306 \times 300 = 91,800\text{ J}\). This rounds to \(92,000\text{ J}\) or \(92\text{ kJ}\) (accept \(91,800\text{ J}\)).

MP1: Calculation of power (\(306\text{ W}\)) or voltage (\(51\text{ V}\)) (1 mark). MP2: Conversion of time to seconds (\(300\text{ s}\)) (1 mark). MP3: Calculation of electrical energy with unit (\(91,800\text{ J}\) or \(92,000\text{ J}\) or \(92\text{ kJ}\)) (1 mark).

Step 1: Convert the frequency to hertz: \(f = 42\text{ kHz} = 42,000\text{ Hz}\). Step 2: Convert the wavelength to meters: \(\lambda = 3.6\text{ cm} = 0.036\text{ m}\). Step 3: Use the wave equation \(v = f \times \lambda\) to find speed: \(v = 42,000 \times 0.036 = 1512\text{ m/s}\). This rounds to \(1500\text{ m/s}\) to 2 significant figures (accept \(1512\text{ m/s}\)).

MP1: Conversion of frequency to Hz (\(42,000\text{ Hz}\)) OR wavelength to meters (\(0.036\text{ m}\)) (1 mark). MP2: Correct substitution into the wave speed equation \(v = f \lambda\) using converted values (1 mark). MP3: Correct calculation of speed with unit (\(1512\text{ m/s}\) or \(1500\text{ m/s}\)) (1 mark).

1. Place the dry stone on an electronic balance to measure and record its mass, \(m\).
2. Pour water into a measuring cylinder and record the initial volume, \(V_1\).
3. Lower the stone carefully into the measuring cylinder using a thin thread, ensuring it is fully submerged and no water splashes out.
4. Record the new volume of the water, \(V_2\). Calculate the volume of the stone as \(V = V_2 - V_1\).
5. Use the formula \(\text{density} = \frac{\text{mass}}{\text{volume}}\) to calculate the density of the stone.

Award 1 mark for each of the following points, up to a maximum of 4 marks:
- MP1: Measure the mass of the stone using a balance / electronic scale.
- MP2: Use a measuring cylinder containing water to find the volume of the stone by displacement (recording initial and final volumes, or using a eureka / displacement can to collect and measure displaced water).
- MP3: Calculate the volume of the stone by subtracting the initial volume from the final volume (\(V = V_2 - V_1\)).
- MP4: State and use the formula \(\text{density} = \frac{\text{mass}}{\text{volume}}\).

[Accept alternative detail for accuracy, such as reading the volume at eye level to avoid parallax error, ensuring no air bubbles adhere to the stone, or using a thin thread that does not displace significant water.]

1. Place the rectangular glass block on a piece of paper and trace around it with a pencil.
2. Shine a single ray of light from a ray box into one of the long sides of the block at an angle.
3. Draw crosses along the incident and emergent rays to mark their paths. Remove the block, join the crosses, and draw the refracted ray inside the outline of the block.
4. Draw a normal line at the point of entry. Use a protractor to measure the angle of incidence, \(i\), and the angle of refraction, \(r\).
5. Repeat the measurement for several different angles of incidence.
6. Plot a graph of \(\sin(i)\) on the y-axis against \(\sin(r)\) on the x-axis. Draw a line of best fit; the gradient of this line equals the refractive index, \(n\), of the glass.

Award 1 mark for each of the following points, up to a maximum of 4 marks:
- MP1: Trace the outline of the block on paper, shine a ray of light into it, and mark the entry and exit paths of the light ray to draw the refracted ray.
- MP2: Draw a normal at the boundary and measure the angle of incidence, \(i\), and the angle of refraction, \(r\), using a protractor.
- MP3: Repeat the experiment for a range of different angles of incidence.
- MP4: Plot a graph of \(\sin(i)\) against \(\sin(r)\) and state that the gradient of the line of best fit is equal to the refractive index, \(n\) (or calculate \(n = \frac{\sin(i)}{\sin(r)}\) for each set of angles and calculate a mean).

When water waves travel from deep water to shallow water, their wave speed decreases due to increased friction with the seabed.

The frequency of the wave is determined solely by the source generating the waves and does not change when the wave enters a new medium (it remains constant).

Using the wave equation:
\(v = f\lambda\)

Since frequency \(f\) is constant, a decrease in wave speed \(v\) must result in a proportional decrease in wavelength \(\lambda\).

Therefore, row A is the correct description.

1 mark for the correct option (A).

The relationship between pressure, density, gravitational field strength, and depth is given by:
\(p = \rho g h\)

First, find the change in depth (\(\Delta h\)):
\(\Delta h = 15.0\text{ m} - 5.0\text{ m} = 10.0\text{ m}\)

Now calculate the change in pressure (\(\Delta p\)):
\(\Delta p = \rho g \Delta h\)
\(\Delta p = 1000\text{ kg/m}^3 \times 10\text{ N/kg} \times 10.0\text{ m} = 100\,000\text{ Pa}\)

Convert Pa to kPa:
\(\Delta p = 100\text{ kPa}\)

Therefore, the correct option is B.

1 mark for the correct option (B).

When the negatively charged balloon is brought near the wall, it repels the negatively charged electrons near the surface of the wall deeper into the wall. This leaves a net positive induced charge on the surface of the wall closest to the balloon. Because opposite charges attract, an electrostatic force of attraction is created between the negative balloon and the positive surface of the wall, causing it to cling.

MP1: State that negative charges or electrons in the wall are repelled away from the surface. (1) MP2: State that this leaves a positive induced charge on the surface of the wall, leading to electrostatic attraction. (1)

Increasing the temperature of the gas increases the average kinetic energy of its molecules, meaning they move at higher average speeds. As a result, the molecules collide with the container walls more frequently and with greater force per collision (greater change in momentum). This increases the overall average force exerted on the walls, and since pressure is force per unit area, the pressure increases.

MP1: Identify that higher temperature increases the average kinetic energy or speed of the gas molecules. (1) MP2: Explain that this leads to more frequent collisions with the walls OR greater force / change in momentum per collision. (1)

A voltmeter is designed to measure the potential difference (voltage drop) between two distinct points in a circuit, so it must span across the component (in parallel). Additionally, voltmeters are engineered to have an extremely high resistance so they do not draw significant current from the circuit. If connected in series, the voltmeter's high resistance would drastically reduce the current flowing through the lamp and the entire circuit.

MP1: Explain that a voltmeter measures potential difference across a component (requiring connection to both sides). (1) MP2: Explain that voltmeters have very high resistance, so a series connection would severely reduce the circuit current. (1)

Alpha particles have a large mass and a strong positive charge (+2e), which makes them highly effective at stripping electrons from atoms, resulting in high ionizing power. Each time an alpha particle ionizes an atom, it transfers some of its kinetic energy to that atom. Because they cause a high density of ionizations over a short distance, they lose all of their kinetic energy very rapidly and come to a stop, which directly results in their very short range in air.

MP1: State that alpha particles lose energy each time they cause an ionization. (1) MP2: Explain that because they ionize strongly, they lose their energy very quickly/over a short distance, resulting in low penetration. (1)

A step-up transformer increases the voltage of the electrical supply before transmission. For a given electrical power (\(P = IV\)), increasing the voltage results in a proportionally lower current flowing through the transmission cables. Since power loss in wires is given by \(P = I^2 R\), a lower current significantly reduces the energy wasted as heat in the resistance of the cables, making transmission much more efficient.

MP1: Explain that increasing the voltage decreases the transmission current. (1) MP2: Explain that a lower current reduces the power/energy lost as heat in the cables (accept reference to \(P = I^2 R\)). (1)

In a longitudinal wave, the oscillations or vibrations of the particles in the medium occur back and forth along the same line as the wave is travelling. This means the direction of particle vibration is parallel to the direction of wave propagation and energy transfer, creating alternating regions of compressions and rarefactions.

MP1: Mention the vibration or oscillation of particles/medium. (1) MP2: State that this vibration is parallel to (or in the same direction as) the direction of wave travel / energy transfer. (1)

Nuclear fusion requires two positively charged atomic nuclei to come close enough together to join. Because both nuclei are positively charged, they experience a strong electrostatic force of repulsion as they approach each other. High temperatures give the nuclei extremely high kinetic energies (moving very fast), and high pressures force them close together, allowing them to overcome this electrostatic repulsion and come close enough for the strong nuclear force to bind them.

MP1: State that the colliding nuclei are both positively charged and repel each other. (1) MP2: Explain that high temperature/kinetic energy (or high pressure) is needed to overcome this electrostatic repulsion. (1)

The observed red-shift of light from distant galaxies means that the wavelengths of the light have been stretched as they travel to Earth, shifting towards the red end of the spectrum. This indicates that these galaxies are moving away (receding) from us. Since almost all distant galaxies show red-shift, and more distant ones show greater red-shift, it demonstrates that space itself is expanding in all directions, supporting the theory of an expanding Universe.

MP1: State that red-shift shows that distant galaxies are moving away (receding) from Earth. (1) MP2: Explain that the movement of galaxies away from each other indicates that the Universe/space is expanding. (1)

1. When the magnet is moved in and out, it cuts the magnetic field lines of the coil (changing the magnetic flux linkage), which induces a voltage across the coil. 2. Because the magnet changes direction of motion repeatedly, the direction of the induced voltage and current also reverses repeatedly, resulting in an alternating current.

M1: For stating that moving the magnet cuts magnetic field lines / changes magnetic flux to induce a voltage/current [1 mark]. M2: For stating that the direction of the induced voltage/current reverses when the direction of the magnet's motion reverses [1 mark].

1. When the charged balloon is brought near the neutral wall, it induces a charge on the wall's surface by repelling charges of the same sign deeper into the wall and attracting opposite charges to the surface. 2. This separation of charge results in an electrostatic force of attraction between the balloon and the wall that is strong enough to support the balloon's weight.

M1: For explaining that the charged balloon induces a charge on the wall (by attracting opposite charges / repelling like charges) [1 mark]. M2: For stating that this results in an electrostatic force of attraction between the balloon and the wall [1 mark].

1. An increase in temperature increases the average kinetic energy of the gas particles, meaning they move at higher speeds. 2. As a result, the particles collide with the internal walls of the canister more frequently and with a larger force (or larger change in momentum per collision), which increases the pressure.

M1: For stating that raising the temperature increases the kinetic energy / speed of the gas particles [1 mark]. M2: For stating that particles collide with the walls more frequently OR with greater force / momentum change [1 mark].

1. Nuclei are positively charged and therefore experience a strong electrostatic force of repulsion when they are brought close together. 2. Extremely high temperatures mean the nuclei have very high kinetic energy, which allows them to overcome this electrostatic repulsion and get close enough for the strong nuclear force to bind them together.

M1: For identifying that nuclei have positive charges and repel each other electrostatically [1 mark]. M2: For explaining that high temperatures provide the high kinetic energy needed to overcome this electrostatic repulsion [1 mark].

1. During a crash, the seatbelt is designed to stretch slightly, which increases the time interval \(t\) over which the passenger's momentum changes to zero. 2. Since force is equal to the rate of change of momentum (\(F = \frac{\Delta p}{t}\)), a longer collision time results in a significantly reduced average force acting on the passenger, minimizing physical trauma.

M1: For stating that the seatbelt stretches to increase the time taken for the momentum to change (or time taken to stop) [1 mark]. M2: For explaining that this reduces the average force on the passenger because force is the rate of change of momentum [1 mark].

1. Glass has a higher refractive index (optical density) than air, causing the speed of the light wave to decrease as it enters the glass. 2. Because the light ray enters at an angle, different parts of the wavefront cross the boundary and slow down at different times, which causes the wavefronts to pivot and bend the path of the ray towards the normal.

M1: For stating that light slows down as it enters the glass due to its higher optical density / refractive index [1 mark]. M2: For explaining that the change in speed of the wavefronts at an angle causes the change in direction towards the normal [1 mark].

1. If an electrical fault causes the live wire to contact the metal casing, the casing becomes live. The earth wire provides a low-resistance path for the current to flow to the ground. 2. This prevents a user touching the casing from receiving a dangerous electric shock, and the low-resistance path allows a large current to flow, which quickly melts the fuse and cuts off the electricity supply.

M1: For stating that the earth wire provides a safe, low-resistance path to ground if the metal casing becomes live [1 mark]. M2: For explaining that this prevents the user from receiving an electric shock / causes a large current to flow which blows the fuse [1 mark].

1. Determine the total orbital radius \(r\): \(r = 6400\text{ km} + 400\text{ km} = 6800\text{ km}\). 2. Convert orbital period \(T\) to seconds: \(T = 90 \times 60 = 5400\text{ s}\). 3. Use orbital speed formula: \(v = \frac{2 \pi r}{T} = \frac{2 \times \pi \times 6800}{5400} \approx 7.91\text{ km/s}\).

1 mark for calculating total orbital radius as \(6800\text{ km}\) or period as \(5400\text{ s}\). 1 mark for correct substitution into orbital speed equation. 1 mark for final speed of \(7.91\text{ km/s}\) (allow \(7.9\) to \(7.91\)).

1. Change in momentum \(\Delta p = m(v - u)\). Taking the rebound direction as negative: \(u = 32\text{ m/s}\) and \(v = -24\text{ m/s}\). Therefore, \(\Delta p = 0.058 \times (-24 - 32) = -3.248\text{ kg m/s}\). The magnitude of change in momentum is \(3.248\text{ kg m/s}\). 2. Average force \(F = \frac{\Delta p}{\Delta t}\) where \(t = 8.0\text{ ms} = 0.008\text{ s}\). 3. Calculate force: \(F = \frac{3.248}{0.008} = 406\text{ N}\).

1 mark for calculating change in momentum of \(3.248\text{ kg m/s}\). 1 mark for dividing by contact time in seconds (\(0.008\text{ s}\)). 1 mark for correct final answer of \(406\text{ N}\).

1. Recall the relationship: \(V = \frac{E}{Q}\). 2. Substitute the values: \(V = \frac{4.5 \times 10^8\text{ J}}{15\text{ C}} = 3.0 \times 10^7\text{ V}\). 3. Convert to megavolts: \(3.0 \times 10^7\text{ V} = 30\text{ MV}\).

1 mark for rearranging or selecting formula \(V = E / Q\). 1 mark for correct substitution to find potential difference in volts (\(3.0 \times 10^7\text{ V}\)). 1 mark for correct conversion to megavolts (\(30\text{ MV}\)).

1. Convert initial temperature to Kelvin: \(T_1 = 27 + 273 = 300\text{ K}\). 2. Use the pressure law for constant volume: \(\frac{p_1}{T_1} = \frac{p_2}{T_2}\). Rearranging gives \(T_2 = T_1 \times \frac{p_2}{p_1} = 300 \times \frac{180}{120} = 450\text{ K}\). 3. Convert back to degrees Celsius: \(T_2 = 450 - 273 = 177^\circ\text{C}\).

1 mark for converting Celsius temperature to Kelvin (\(300\text{ K}\)). 1 mark for correct substitution and calculation of Kelvin temperature (\(450\text{ K}\)). 1 mark for correct conversion back to Celsius (\(177^\circ\text{C}\)).

1. For a 100% efficient transformer: \(V_p I_p = V_s I_s\). 2. Rearrange the formula to solve for primary current: \(I_p = \frac{V_s I_s}{V_p}\). 3. Substitute values: \(I_p = \frac{12 \times 2.5}{230} = \frac{30}{230} \approx 0.1304\text{ A}\). Rounding to 2 significant figures gives \(0.13\text{ A}\).

1 mark for recall/use of the equation \(V_p I_p = V_s I_s\). 1 mark for substituting values correctly into the equation. 1 mark for the correct final answer of \(0.13\text{ A}\).

1. Calculate the energy supplied: \(t = 12 \times 60 = 720\text{ s}\), so \(Q = P \times t = 50\text{ W} \times 720\text{ s} = 36000\text{ J}\). 2. Use specific latent heat formula: \(Q = m L\). 3. Solve for \(L\): \(L = \frac{Q}{m} = \frac{36000\text{ J}}{0.25\text{ kg}} = 144000\text{ J/kg}\).

1 mark for converting time to seconds (\(720\text{ s}\)). 1 mark for calculating energy supplied (\(36000\text{ J}\)). 1 mark for final correct calculation of specific latent heat of fusion (\(144000\text{ J/kg}\) or \(1.44 \times 10^5\text{ J/kg}\)).

1. Calculate the pressure due to the seawater column: \(p_{\text{liquid}} = \rho g h = 1025 \times 10 \times 120 = 1230000\text{ Pa} = 1230\text{ kPa}\). 2. Add atmospheric pressure to find total pressure: \(p_{\text{total}} = p_{\text{liquid}} + p_{\text{atmospheric}}\). 3. Calculate total pressure: \(p_{\text{total}} = 1230\text{ kPa} + 101\text{ kPa} = 1331\text{ kPa}\).

1 mark for correct calculation of hydrostatic pressure (\(1230000\text{ Pa}\) or \(1230\text{ kPa}\)). 1 mark for adding the atmospheric pressure. 1 mark for correct final answer (\(1331\text{ kPa}\)).

1. Calculate the change in wavelength: \(\Delta \lambda = 658.3\text{ nm} - 656.3\text{ nm} = 2.0\text{ nm}\). 2. Use the redshift equation: \(\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}\) where \(\lambda_0 = 656.3\text{ nm}\). 3. Rearrange and solve for recessional velocity: \(v = c \times \frac{\Delta \lambda}{\lambda_0} = 3.0 \times 10^8 \times \frac{2.0}{656.3} \approx 914216\text{ m/s}\). Convert to km/s: \(v \approx 914\text{ km/s}\).

1 mark for calculating wavelength change of \(2.0\text{ nm}\). 1 mark for correctly substituting into redshift formula. 1 mark for final correct answer of \(914\text{ km/s}\) (accept \(910\) to \(915\text{ km/s}\)).

1. Weigh the dry cork on a digital mass balance to find its mass \(m\). 2. Fill a measuring cylinder with a known volume of water. Attach a heavy metal sinker to a thread and submerge it completely in the water. Record this initial volume reading as \(V_1\). 3. Tie the cork to the sinker using the thread and submerge both the sinker and the cork completely in the measuring cylinder. Record the new volume reading as \(V_2\). 4. Calculate the volume of the cork as \(V = V_2 - V_1\). 5. Calculate the density of the cork using the equation \(\text{density} = \frac{\text{mass}}{\text{volume}}\). To ensure accuracy, read the measuring cylinder at eye level to avoid parallax error and measure from the bottom of the meniscus.

1 mark: Measure the mass of the dry cork using a mass balance. 1 mark: Submerge a heavy sinker in a measuring cylinder of water to record the initial volume \(V_1\), then submerge both sinker and cork together to record the final volume \(V_2\). 1 mark: Explain how to find the cork's volume by subtraction: \(V = V_2 - V_1\). 1 mark: Mention a relevant experimental precaution to improve accuracy, such as reading the measuring cylinder at eye level / avoiding parallax error / ensuring no air bubbles are trapped. 0.5 marks: Use the formula \(\text{density} = \text{mass} / \text{volume}\) to calculate the final value.

1. Set up two identical funnels containing crushed melting ice, placing each over a separate beaker on a mass balance. 2. Place an electrical heater inside the ice of the experimental funnel, and leave the second funnel without a heater as a control. 3. Connect the heater to a joulemeter and turn it on for a timed interval, recording the total electrical energy \(E\) supplied. 4. Measure the mass of water collected from the experimental funnel, \(m_{\text{experimental}}\), and the control funnel, \(m_{\text{control}}\), during this same time period. 5. Determine the mass of ice melted solely by the heater: \(m = m_{\text{experimental}} - m_{\text{control}}\). 6. Use the formula \(L_f = \frac{E}{m}\) to calculate the specific latent heat of fusion.

1 mark: Use a control funnel with melting ice but no heater alongside an experimental funnel with a heater. 1 mark: Measure the electrical energy input \(E\) using a joulemeter (or by recording voltage, current, and time and using \(E = VIt\)). 1 mark: Measure the mass of water collected from both funnels over the same time interval using a balance. 1 mark: Calculate the mass of ice melted solely by the heater by subtracting the control mass from the experimental mass: \(m = m_{\text{experimental}} - m_{\text{control}}\). 0.5 marks: State the formula used to find the specific latent heat of fusion: \(L_f = E / m\).

1. Setup: Clamp a vertical plastic guide tube directly above a coil of wire with a known number of turns \(N\). 2. Measurement apparatus: Connect the coil to a cathode-ray oscilloscope (CRO) or a fast-sampling voltage data logger to accurately record the rapid transient peak voltage. 3. Variables to control: Drop the bar magnet from a fixed, measured height above the coil each time, maintaining the same orientation of the magnet's poles and the same magnet strength. 4. Procedure: Drop the magnet, record the maximum (peak) induced voltage \(V_{\text{max}}\). Repeat this process 3 times and find the average peak voltage. Repeat the entire experiment using coils with different numbers of turns, keeping all other factors constant. 5. Analysis: Plot a graph of average maximum induced e.m.f. on the y-axis against the number of turns \(N\) on the x-axis. A straight line starting at the origin shows direct proportionality.

1 mark: Identify the independent variable (number of turns of wire) and the dependent variable (maximum induced voltage/e.m.f.). 1 mark: Identify at least two control variables, specifically the drop height of the magnet and the magnetic field strength/mass of the magnet. 1 mark: Connect the coil to an oscilloscope or data logger to capture the peak voltage (do NOT accept standard analog/digital voltmeters due to speed of drop). 1 mark: Describe repeating drops for each coil to calculate average peak values and plotting a graph of maximum voltage against number of turns. 0.5 marks: Correctly describe the expected trend (direct proportionality, straight line through the origin).