Physics 5054/22 — May/June 2024

Walkthrough

A speed-time graph plots speed on the vertical axis and time on the horizontal axis. The gradient of this graph represents acceleration. If the line is straight, the gradient is constant, meaning the acceleration is uniform. In Fig. 1.2, the line for trolley B from $t = 0.30\text{ s}$ to $t = 0.50\text{ s}$ is a straight diagonal line, proving the acceleration is uniform.

Key Takeaways

• The gradient of a speed-time graph equals acceleration.
• A straight line on a speed-time graph indicates uniform (constant) acceleration.

Common Mistakes

• Saying 'the speed is constant' (that would be a horizontal line).
• Not referring specifically to the line being straight or the gradient being constant.

Things to Be Careful About

• Ensure you are looking at the correct trolley's line. Trolley A's line is also straight during the collision, but the question asks about trolley B.

Walkthrough

The magnitude of acceleration is given by the steepness (gradient) of the line on a speed-time graph. Looking at Fig. 1.2 between $t = 0.30\text{ s}$ and $t = 0.50\text{ s}$, the line for trolley B rises from $0\text{ m/s}$ to $0.40\text{ m/s}$, while the line for trolley A falls from $0.70\text{ m/s}$ to $0.40\text{ m/s}$. The line for trolley B is visually steeper, indicating a larger magnitude of acceleration.

Key Takeaways

• Steeper gradient on a speed-time graph means larger acceleration.
• You can compare accelerations by comparing the steepness of the lines or by comparing the change in speed over the same time interval.

Common Mistakes

• Saying 'trolley B has a larger speed' (speed and acceleration are different).
• Not mentioning 'steeper' or 'larger change in speed'.

Things to Be Careful About

• The question asks for the magnitude (size) of the acceleration, so direction does not matter. Just compare how steep the lines are.

Walkthrough

To find the acceleration at $t = 0.40\text{ s}$, we use the fact that the acceleration is uniform during the collision. We can calculate the gradient of the straight line for trolley B between $t = 0.30\text{ s}$ and $t = 0.50\text{ s}$. The initial speed $u = 0\text{ m/s}$ at $t = 0.30\text{ s}$, and the final speed $v = 0.40\text{ m/s}$ at $t = 0.50\text{ s}$. The time interval $\Delta t = 0.50 - 0.30 = 0.20\text{ s}$. Substituting into $a = \Delta v / \Delta t$ gives $0.40 / 0.20 = 2.0\text{ m/s}^2$.

Key Takeaways

• Acceleration can be calculated from the gradient of a speed-time graph: $a = \Delta v / \Delta t$.
• For uniform acceleration, any two points on the straight line section can be used.

Common Mistakes

• Using the wrong time interval (e.g., $0.50 - 0 = 0.50\text{ s}$ instead of $0.20\text{ s}$).
• Forgetting to divide by the time interval.

Things to Be Careful About

• Ensure units are correct: speed in $\text{m/s}$ and time in $\text{s}$, giving acceleration in $\text{m/s}^2$. Give the answer to 2 significant figures as indicated by the graph readings.

Walkthrough

First, calculate the volume of water in the pool. The pool is a rectangular prism, so its volume is length × width × depth:

$$V = 4.0 \times 2.1 \times 0.80 = 6.72 \text{ m}^3$$

Next, use the density formula $\rho = m/V$, rearranged to $m = \rho V$, to find the mass:

$$m = 1000 \times 6.72 = 6720 \text{ kg}$$

The question asks to show the mass is approximately $6700 \text{ kg}$, which matches $6720 \text{ kg}$ rounded to two significant figures.

Key Takeaways

• The volume of a rectangular container is found by multiplying its three dimensions.
• Mass can be calculated from density and volume using $m = \rho V$.

Common Mistakes

• Forgetting to include the depth ($0.80 \text{ m}$) when calculating volume, using only length and width.
• Using the wrong units for density or dimensions.

Things to Be Careful About

• Ensure all dimensions are in metres before calculating volume, as density is given in $\text{kg/m}^3$.
• The question asks to show the mass is approximately $6700 \text{ kg}$, so rounding $6720 \text{ kg}$ to two significant figures is expected.

Walkthrough

Pressure is a fundamental quantity in physics defined as the normal force applied per unit area over which that force is distributed. The mathematical definition is $P = F/A$.

Key Takeaways

• Pressure is not just force; it depends on the area over which the force is spread.
• The standard SI unit for pressure is the pascal (Pa), which is equivalent to $\text{N/m}^2$.

Common Mistakes

• Defining pressure as simply "force" without mentioning area.
• Confusing pressure with force or stress.

Things to Be Careful About

• Use the exact phrasing "force per unit area" as this is the standard definition credited in mark schemes.

Walkthrough

The pressure exerted by a liquid at a given depth depends on the depth, the density of the liquid, and the gravitational field strength. The formula is $P = \rho g h$.

Substitute the given values:

$$P = 1000 \times 9.8 \times 0.80 = 7840 \text{ Pa}$$

Alternatively, you could calculate the weight of the water ($W = mg = 6720 \times 9.8 = 65856 \text{ N}$) and divide by the area of the base ($A = 4.0 \times 2.1 = 8.4 \text{ m}^2$):

$$P = \frac{F}{A} = \frac{65856}{8.4} = 7840 \text{ Pa}$$

Both methods yield the same result. The mark scheme accepts $7800 \text{ Pa}$ (rounded to 2 s.f.) or $7840 \text{ Pa}$.

Key Takeaways

• Liquid pressure depends only on depth, density, and $g$, not on the shape or total volume of the container.
• $P = \rho g h$ is derived from $P = F/A$ where $F = mg = \rho V g = \rho (A h) g$.

Common Mistakes

• Using the total mass of the water instead of density in the formula.
• Forgetting to convert units or using the wrong value for $g$ (e.g., using $10 \text{ N/kg}$ gives $8000 \text{ Pa}$, which may not match the expected answer if $9.8$ is required).

Things to Be Careful About

• The mark scheme specifically lists $7800 \text{ Pa}$ as the accepted answer, implying $g = 9.8 \text{ N/kg}$ and rounding to 2 significant figures. Always check which value of $g$ is expected in your exam series.

Walkthrough

First, identify the known quantities:

• Energy transferred, $E = 5.1 \times 10^8 \text{ J}$
• Mass of water, $m = 6720 \text{ kg}$ (from part a(i))
• Specific heat capacity of water, $c = 4200 \text{ J/(kg }^\circ\text{C)}$
• Initial temperature, $T_{\text{initial}} = 10 \text{ }^\circ\text{C}$

Use the specific heat capacity equation $E = mc\Delta T$ to find the temperature rise $\Delta T$:

$$\Delta T = \frac{E}{mc} = \frac{5.1 \times 10^8}{6720 \times 4200}$$

Calculate the denominator:

$$6720 \times 4200 = 2.8224 \times 10^7$$

Now divide:

$$\Delta T = \frac{5.1 \times 10^8}{2.8224 \times 10^7} \approx 18.07 \text{ }^\circ\text{C}$$

Round to a sensible number of significant figures (2 s.f. based on the given data): $\Delta T = 18 \text{ }^\circ\text{C}$.

Finally, add this temperature change to the initial temperature to find the final temperature:

$$T_{\text{final}} = 10 + 18 = 28 \text{ }^\circ\text{C}$$

Key Takeaways

• The specific heat capacity equation $E = mc\Delta T$ can be rearranged to find any one of the four variables.
• Always remember to add the temperature change to the initial temperature to get the final temperature; a common error is to stop at $\Delta T$.

Common Mistakes

• Using $6700 \text{ kg}$ instead of the more precise $6720 \text{ kg}$ from part (a), leading to a slightly different $\Delta T$.
• Forgetting to add the initial $10 \text{ }^\circ\text{C}$ and answering $18 \text{ }^\circ\text{C}$ instead of $28 \text{ }^\circ\text{C}$.
• Arithmetic errors when dividing large numbers like $5.1 \times 10^8$.

Things to Be Careful About

• Carry forward the unrounded mass value ($6720 \text{ kg}$) from part (a) to avoid compounding rounding errors.
• Pay attention to significant figures; the energy is given to 2 s.f. ($5.1 \times 10^8$), so the final answer should logically be to 2 s.f. ($28 \text{ }^\circ\text{C}$).

Walkthrough

To explain evaporation and cooling in terms of the particle model, we need to describe what happens to the particles during the process:

1. Particle energy distribution: In any liquid, the particles have a range of kinetic energies. Some are moving fast (high energy) and some are moving slow (low energy).
2. Escape: Evaporation happens at the surface. Only the particles with the highest kinetic energy (the fastest-moving ones) have enough energy to break the intermolecular forces of attraction holding them in the liquid and escape into the air as gas.
3. Cooling effect: When these high-energy particles leave, the average kinetic energy of the particles left behind in the liquid is lower. Since temperature is directly proportional to the average kinetic energy of the particles, the temperature of the liquid drops. This is why evaporation causes cooling.

Key Takeaways

• Evaporation is a surface phenomenon that depends on the energy distribution of particles.
• Cooling occurs because the highest-energy particles are removed, lowering the average energy of the remaining liquid.

Common Mistakes

• Saying "the particles lose energy" without specifying that it is the fastest or most energetic particles that escape.
• Using the word "heat" instead of "kinetic energy" or "thermal energy" when describing what is lost.
• Confusing evaporation with boiling (boiling happens throughout the liquid at a fixed temperature, evaporation happens at the surface at any temperature).

Things to Be Careful About

• The question specifically asks for the explanation "in terms of the movement of particles". You must mention particle speed or energy.
• Both marks are awarded for separate points: (1) fastest/most energetic particles escape, and (2) leaving slower/less energetic particles. Ensure both are stated clearly.

Walkthrough

The rate of evaporation from a liquid surface depends on several environmental factors. To increase evaporation, you can:

1. Increase the temperature: Higher temperatures mean the particles in the liquid have more average kinetic energy, so more particles will have the minimum energy required to escape.
2. Increase wind speed: Wind blows away the water vapour molecules that have just evaporated from the surface. This keeps the concentration of water vapour near the surface low, allowing more particles to escape (the rate of condensation back into the liquid is reduced).
3. Decrease humidity (lower water vapour concentration in the air): Similar to wind, drier air means a lower concentration of water vapour, increasing the net rate of evaporation.
4. Increase the surface area: Though not strictly an "environmental factor" of the air, a larger surface area allows more particles to escape at once.

The mark scheme accepts any two from: higher temperature, wind, low humidity.

Key Takeaways

• Evaporation rate is influenced by temperature, air movement (wind), and humidity.
• All these factors affect the balance between the rate of evaporation and the rate of condensation at the surface.

Common Mistakes

• Giving factors that affect boiling (like atmospheric pressure) instead of evaporation.
• Saying "increase the pressure" which would actually decrease evaporation.

Things to Be Careful About

• The question asks for changes to environmental factors in the surroundings of the pool. Factors like "using a heater" are methods, not the environmental factor itself; the factor is "higher temperature".
• Only two factors are required; providing more will not gain extra marks but may introduce errors.

Walkthrough

The question asks to complete an energy transfer diagram showing the useful flow of energy from the Sun to the phone battery. Energy is stored in different ways depending on the object. The Sun produces energy through nuclear fusion in its core, so the energy store in the Sun is nuclear. The energy is transferred by light (electromagnetic radiation) to the solar panel, which converts it into electrical energy. This electrical energy is then transferred to the phone battery, where it is stored as chemical energy ready to be used later. We simply fill in these two missing store names in the diagram.

Key Takeaways

Energy is stored in specific ways: nuclear in stars, chemical in batteries. Energy transfers between stores via light, electrical currents, heating, or mechanical work. Always name the store, not the transfer mechanism.

Common Mistakes

• Writing 'solar' or 'light' for the energy store in the Sun. 'Solar' is not a standard energy store in the 5054 syllabus; the correct store is nuclear.
• Writing 'electricity' or 'electrical' for the energy store in the battery. The battery stores energy chemically; the electrical current is the transfer mechanism, not the store.

Things to Be Careful About

The diagram distinguishes between energy stores (inside the boxes) and energy transfers (on the arrows). 'Light from Sun' and 'electrical current' are transfers; 'nuclear' and 'chemical' are stores. Do not mix them up.

Walkthrough

The solar-powered charger converts light energy from the Sun into electrical energy. The rate at which it does this (its power output) depends on the intensity of the light hitting the panel. Clouds block and scatter sunlight, so less light energy reaches the solar panel. With less input energy, the charger produces a smaller electrical current. Since current is the rate of flow of charge (and thus the rate at which energy is transferred to the battery), a smaller current means the battery takes longer to charge.

Key Takeaways

The power output of a solar panel depends on the intensity of incident light. Reduced light input directly reduces the electrical current produced, slowing down the charging process.

Common Mistakes

• Saying 'there is no energy on a cloudy day'. There is still some energy, just less. Use 'less light' or 'reduced light', not 'no light'.
• Forgetting to link the reduced light to the reduced current. The mark scheme requires the chain: less light -> less current -> slower charging.

Things to Be Careful About

Be precise with wording. 'Less light energy' or 'less light reaching the panel' is acceptable. Simply saying 'it is cloudy' is not enough; you must explain the physical consequence (less energy/current).

Walkthrough

The light ray enters the glass at point P normally, meaning it is perpendicular to the top surface. The angle of incidence is defined as the angle between the incident ray and the normal to the surface at the point of incidence. Since the ray and the normal are the same line, the angle between them is 0 degrees.

Key Takeaways

The angle of incidence is always measured between the ray and the normal, never between the ray and the surface. When a ray enters a medium normally (perpendicularly), the angle of incidence is 0 and the ray does not bend.

Common Mistakes

Writing 90 degrees by measuring the angle between the ray and the surface instead of the normal. Remember: angle of incidence = angle with the normal.

Things to Be Careful About

Always draw the normal first as a dashed line perpendicular to the surface. Measure the angle from this normal.

Walkthrough

At point R, the light ray QR is travelling horizontally and hits the right-hand sloped face of the prism. To find the angle of incidence, we must first draw the normal: a dashed line perpendicular to the surface at point R, pointing outwards. The angle of incidence is the angle between this normal and the incoming ray QR. This angle must be clearly marked with an arc and labelled 'angle of incidence' or '$i$'.

Key Takeaways

The normal is always drawn perpendicular to the surface at the point where the ray hits. The angle of incidence is always between the normal and the incident ray.

Common Mistakes

Drawing the normal parallel to the face instead of perpendicular. Labeling the angle between the ray and the surface as the angle of incidence.

Things to Be Careful About

Use a ruler and set square to ensure the normal is exactly perpendicular to the sloped face. The angle of incidence at R is actually 45 degrees, but the question only asks you to draw and label it, not calculate the value.

Walkthrough

For total internal reflection to occur at point Q (or R), two conditions must be met simultaneously. First, the light must be travelling from a medium with a higher refractive index (more optically dense, like glass) into a medium with a lower refractive index (less optically dense, like air). Second, the angle at which the light hits the boundary (the angle of incidence) must be greater than the critical angle for that pair of media. If both conditions are met, all the light is reflected back into the denser medium and none is refracted out.

Key Takeaways

Total internal reflection requires: (1) travel from higher to lower refractive index, and (2) angle of incidence > critical angle. Both conditions are necessary.

Common Mistakes

Stating only one condition. Saying 'the angle is large' without referencing the critical angle. Saying 'from glass to air' is acceptable, but the general principle 'from more dense to less dense' is preferred.

Things to Be Careful About

Ensure you mention the direction of travel. If light travels from air into glass, total internal reflection cannot happen regardless of the angle.

Walkthrough

Optical fibres can transmit information using pulses of visible light, but infrared radiation is also widely used because it experiences less attenuation (signal loss) in glass fibres over long distances.

Key Takeaways

Infrared radiation is commonly used in optical fibre communications alongside visible light.

Common Mistakes

Saying 'ultraviolet' or 'microwaves'. Microwaves are used in wireless communications (satellites, mobile phones), not typically in glass fibres.

Things to Be Careful About

The question asks for a type of electromagnetic radiation, so ensure you name a region of the spectrum (infrared), not a device or application.

Walkthrough

Glass fibres (optical fibres) offer several advantages over copper wires for transmitting internet data:

• Bandwidth/Speed: Light has a much higher frequency than electrical signals, allowing more data to be transmitted per second (greater bandwidth).
• Interference: Light signals are not affected by electromagnetic interference from nearby power cables or other signals, unlike electrical signals in copper wires.
• Attenuation: Light signals lose less energy over distance, so they can travel longer distances before needing to be boosted.
• Physical properties: Optical fibres are thinner, lighter, and use less raw material than thick copper cables.
• Security: It is harder to tap into an optical fibre without breaking the signal, making it more secure.

Any two of these points earn full marks.

Key Takeaways

Optical fibres provide higher bandwidth, less interference, less signal loss over distance, and are physically lighter and thinner than copper wires.

Common Mistakes

Saying 'light is faster than electricity' as a general statement without context; the speed in the medium is actually slower, but the data rate (bandwidth) is higher. Saying 'it is cheaper' is not always true and is not a standard mark scheme point.

Things to Be Careful About

Be specific. 'Less interference' is better than 'better quality'. 'Less attenuation' is better than 'goes further' (though 'longer transmission distances' is also accepted).

Walkthrough

When two different materials are rubbed together, electrons can be transferred from one material to the other. The material that gains electrons becomes negatively charged, and the material that loses electrons becomes positively charged. Since the balloon is stated to have a negative charge, it must have gained electrons. Therefore, the woollen cloth must have lost electrons. Losing negatively charged electrons leaves the woollen cloth with a net positive charge.

Key Takeaways

Frictional charging involves the transfer of electrons, not protons. The object gaining electrons becomes negative; the object losing electrons becomes positive.

Common Mistakes

• Stating that 'protons' are transferred. Protons are bound in the nucleus and do not move during ordinary frictional charging.
• Saying the cloth becomes 'neutral' or 'uncharged'. Losing electrons creates a positive charge.

Things to Be Careful About

Ensure the blanks are filled in the correct boxes. The top blank in both boxes is 'electrons'. The bottom blank on the cloth box is 'positive'.

Walkthrough

For a static charge to dissipate, the charged particles (electrons) must be able to move away from the object. In conductors (like metals), electrons are free to move throughout the material and can easily flow to the ground or through the air if ionised. Rubber is a classic insulator; its electrons are tightly bound to their atoms and cannot move freely through the material. Additionally, the air inside and around the balloon is also an insulator under normal conditions, meaning there are very few free ions to carry charge away. (Background radiation causes slight ionisation in air, but it is insufficient to discharge the balloon quickly.) Because neither the rubber nor the air allows charge to flow, the negative charge remains trapped on the balloon for a long time.

Key Takeaways

Insulators hold static charge because charge carriers cannot move through them. Conductors allow charge to flow and dissipate.

Common Mistakes

• Saying 'heat' or 'current' cannot flow. We are talking about static charge/electrons.
• Forgetting to mention that the material is an insulator. Just saying 'electrons can't move' is not enough; you must identify the material property.

Things to Be Careful About

The mark scheme accepts 'little ionisation in air' as an alternative or additional point. Ensure you mention that the charge/electrons cannot move through the material.

Walkthrough

Fig. 6.1 shows a simple series circuit containing a battery, a fixed resistor, and an LDR. In a series circuit, there is only one path for the current to flow. Therefore, the current is identical through every component in the loop. The question states the current in the LDR is 0.050 A, so the current in the fixed resistor must also be 0.050 A.

Key Takeaways

In a series circuit, the current is the same through all components. This is a fundamental rule of series circuits and is often used to find currents in parts of the circuit where direct measurement isn't given.

Common Mistakes

• Trying to calculate the current using Ohm's law without first knowing the voltage across the fixed resistor. While possible later, the series rule is the direct and intended method.
• Assuming the current splits or changes between components. Current does not 'get used up' in a circuit.

Things to Be Careful About

The question asks for the current in Amperes, and the blank is followed by the unit A. Simply provide the numerical value 0.050. Do not add extra units or change the number of significant figures unnecessarily.

Walkthrough

To find the e.m.f. of the cell, we need to find the total voltage supplied to the series circuit. In a series circuit, the total e.m.f. is equal to the sum of the potential differences across each component.

1. Voltage across the fixed resistor: We know the current I = 0.050 A and the resistance R = 240 $\Omega$. Using $V = IR$:
   $V_{\text{fixed}} = 0.050 \times 240 = 12\text{ V}$
2. Voltage across the LDR: We are given the current I = 0.050 A. We use Fig. 6.2 (the line for 'lamp switched on') to find the corresponding voltage. Locate 0.050 A on the vertical axis, move across to the 'lamp on' line, and read down to the horizontal axis. The voltage is 4 V.
3. Total e.m.f.: Add the two voltages:
   $\text{e.m.f.} = 12\text{ V} + 4\text{ V} = 16\text{ V}$

Alternative method: Calculate the total resistance. $R_{\text{LDR}} = 4 / 0.050 = 80\ \Omega$. Total $R = 240 + 80 = 320\ \Omega$. Then $\text{e.m.f.} = I \times R_{\text{total}} = 0.050 \times 320 = 16\text{ V}$.

Key Takeaways

In a series circuit, the total e.m.f. equals the sum of the p.d. across each component. You can find individual p.d.s using $V = IR$ for known resistors and reading values from characteristic graphs (like the LDR's I-V graph) for unknown resistors.

Common Mistakes

• Forgetting to add the two voltages together and only calculating the p.d. across the fixed resistor (12 V).
• Reading the wrong line from Fig. 6.2 (using the 'lamp off' line instead of 'lamp on').
• Using the wrong current value. The current is 0.050 A for the entire series circuit.

Things to Be Careful About

• Ensure you read the graph correctly. At 0.050 A on the 'lamp on' line, the voltage is exactly 4 V (midway between 0.04 and 0.06 on the y-axis corresponds to 4 V on the x-axis).
• The e.m.f. is the total voltage, so it must be the sum of the p.d.s across all components in the series loop.

Walkthrough

The needle of a plotting compass is a small magnet. Its N pole points in the direction of the magnetic field at that point. In Fig. 7.1, the compass is placed to the left of the bar magnet, and its N pole is pointing to the left (away from the bar magnet). This means the magnetic field lines at the compass are pointing away from the bar magnet. Since magnetic field lines emerge from the N pole of a magnet and enter the S pole, the left end of the bar magnet must be the N pole, and the right end must be the S pole.

Key Takeaways

A plotting compass N pole always points along the magnetic field line in the direction from N to S outside the magnet.

Common Mistakes

Students often assume the compass N pole is attracted to the magnet's N pole. Remember that like poles repel; the compass N pole is repelled by the magnet's N pole (or attracted to its S pole).

Things to Be Careful About

Always label the poles directly on the diagram where the question asks. Do not just write 'N' and 'S' in your working; the mark is awarded for the correct placement on the figure.

Walkthrough

To plot a magnetic field line, you trace the direction of the field step by step. Starting at point P, you place the compass down. The needle aligns with the local magnetic field. You mark the tip of the N pole (or one end). Then you move the compass so that the opposite end is now at the mark you just made. The needle will point in a new direction; you mark this new tip. Repeating this process traces out the path of the field line. Finally, you connect the dots with a smooth curve to represent the continuous field line.

Key Takeaways

Plotting a field line is a sequential process: place, mark, move, mark, join. Each step relies on the previous mark to determine the new position.

Common Mistakes

• Forgetting to move the compass to the mark made by the previous step.
• Drawing straight lines between marks instead of smooth curves.
• Not joining the marks at the end to form a continuous line.

Things to Be Careful About

Ensure you mark the paper clearly. The marks must be joined with a smooth curve, not a jagged line, to represent the continuous nature of a magnetic field line.

Walkthrough

The magnetic field direction at any point is defined as the direction that the N pole of a small compass would point when placed at that point. Therefore, simply placing the compass on P and reading the orientation of the N pole gives the field direction.

Key Takeaways

The N pole of a plotting compass always indicates the direction of the magnetic field line at that location.

Common Mistakes

• Saying the S pole points in the direction of the field.
• Forgetting to state that the compass must be placed on P first.

Things to Be Careful About

Be precise with wording: it is the 'N pole' or 'shaded end' that points in the direction of the field, not the 'S pole' or 'white end'.

Walkthrough

Current is only produced when there is a change in the magnetic environment of the coil. Moving the magnet changes the magnetic field passing through the coil (magnetic flux). According to the principle of electromagnetic induction, a changing magnetic flux through a conductor induces an e.m.f. Since the coil is connected in a closed circuit, this induced e.m.f. causes a current to flow.

Key Takeaways

Relative motion between a magnet and a coil changes the magnetic flux, inducing an e.m.f. and current.

Common Mistakes

• Saying 'the magnet creates a current' without mentioning induction or e.m.f.
• Stating that the magnetic field 'is' in the coil rather than 'changes' or 'cuts' the coil.

Things to Be Careful About

Use precise terminology: 'cuts the coil' or 'changes the magnetic flux' are both acceptable, but 'induces an e.m.f.' is the key principle that must be stated for the second mark.

Walkthrough

Frequency ($f$) is the number of complete cycles per second. A frequency of $0.50\text{ Hz}$ means 0.5 cycles per second. The period ($T$) is the time for one complete cycle, given by $T = 1/f = 1/0.50 = 2\text{ s}$. Therefore, one complete in-and-out movement must take 2 seconds.

Key Takeaways

Frequency and period are reciprocals: $T = 1/f$.

Common Mistakes

• Confusing frequency with period (saying 'once every 0.5 s').
• Not specifying that the movement is 'in and out' (a full cycle).

Things to Be Careful About

Ensure the units are correct. $0.50\text{ Hz}$ is half a cycle per second, so one cycle takes 2 seconds.

Walkthrough

An alternating current (a.c.) constantly reverses direction. A centre-zero ammeter is designed to show positive current in one direction and negative current in the opposite direction. If the current were direct current (d.c.), the needle would deflect to only one side of zero and stay there. The fact that the reading is on both sides of zero proves the current is alternating.

Key Takeaways

A centre-zero ammeter deflects in opposite directions for current flowing in opposite directions, making it ideal for detecting a.c.

Common Mistakes

• Saying the ammeter 'moves back and forth' without specifying it goes to both sides of zero.
• Not mentioning that d.c. would only show one side.

Things to Be Careful About

Be precise: the reading must be on 'both sides of zero' or 'positive and negative'.

Walkthrough

The magnitude of the induced e.m.f. depends on the rate of change of magnetic flux (or the rate at which field lines are cut). If the frequency of the magnet's movement is increased, the magnet moves faster through the coil. This means the magnetic flux changes more rapidly, cutting the coil's field lines at a higher rate. A faster rate of change of flux induces a larger e.m.f. according to Faraday's law (qualitatively), resulting in a larger current.

Key Takeaways

The size of the induced e.m.f. is proportional to the rate of change of magnetic flux.

Common Mistakes

• Saying 'more field lines are cut' instead of 'cut faster' or 'rate of cutting is higher'.
• Not linking the larger e.m.f. to a larger current (though the question only asks for the magnitude of the a.c., which implies current or e.m.f.).

Things to Be Careful About

Use the phrase 'rate of change' or 'rate of cutting'. Simply saying 'the field changes more' is not precise enough; it must be the rate of change that increases.

Walkthrough

Our solar system, including the Sun, is located in a spiral galaxy called the Milky Way. This is a standard astronomical fact covered in the Earth and the Solar System and Stars and the Universe topics.

Key Takeaways

Students should know the name of our home galaxy and that it contains hundreds of billions of stars.

Common Mistakes

Students may confuse the Milky Way with the Andromeda galaxy or other nearby galaxies. The answer must be specifically "the Milky Way".

Things to Be Careful About

Ensure the name is spelled correctly and capitalised as "The Milky Way" or "Milky Way".

Walkthrough

A light-year is a unit of distance, not time. It is defined as the distance that light travels in a vacuum in one Julian year (365.25 days). Since the speed of light is constant, this provides a convenient way to express vast astronomical distances.

Key Takeaways

Students must distinguish between a light-year (distance) and a year (time). The definition is straightforward: distance = speed of light × time (one year).

Common Mistakes

A very common mistake is to say a light-year is a unit of time. Students must explicitly state it is a distance.

Things to Be Careful About

Ensure the answer specifies "distance" and not "time". The mark scheme accepts "distance travelled by light in one year".

Walkthrough

To find the speed of the Sun, we need the total distance it travels in one orbit and the time it takes. The orbit is circular, so the distance is the circumference $2\pi R$. The radius $R$ is given in light-years, so we must convert it to metres using the speed of light and the number of seconds in a year.

1. Calculate the length of one light-year in metres: $c \times 1\text{ year} = 3.0 \times 10^8 \times 3.2 \times 10^7 = 9.6 \times 10^{15}\text{ m}$.
2. Calculate the orbital radius $R$: $26\,000 \times 9.6 \times 10^{15} = 2.496 \times 10^{20}\text{ m}$.
3. Calculate the orbital circumference: $2\pi R = 2\pi \times 2.496 \times 10^{20} \approx 1.568 \times 10^{21}\text{ m}$.
4. Calculate the speed: $v = \frac{\text{distance}}{\text{time}} = \frac{1.568 \times 10^{21}}{7.3 \times 10^{15}} \approx 214\,837\text{ m/s}$.
5. Round to an appropriate number of significant figures (two, based on the given data): $210\,000\text{ m/s}$ or $2.1 \times 10^5\text{ m/s}$.

Key Takeaways

When dealing with astronomical scales, unit conversion is essential. Always convert given units (like light-years) into standard SI units (metres) before applying kinematic equations. The circular orbit means the distance is $2\pi R$.

Common Mistakes

• Forgetting to convert light-years to metres and using $26\,000$ directly as the radius in metres.
• Using the diameter instead of the radius when calculating the circumference ($2\pi R$, not $\pi R^2$).
• Dividing the radius by the time instead of the circumference by the time.
• Significant figures: the final answer should be given to 2 or 3 significant figures. $210\,000$ or $2.1 \times 10^5$ is acceptable.

Things to Be Careful About

• Ensure all powers of ten are handled correctly during multiplication and division.
• The mark scheme accepts the unrounded value $214\,800$ or similar, but $210\,000$ is the expected 2-sig-fig answer. Writing $2.1 \times 10^5$ is also correct and often clearer for large numbers.
• Do not confuse the time for one orbit (period $T$) with the time it takes light to travel one light-year.

Walkthrough

A cloud chamber works because a charged particle such as an alpha particle ionises the air as it passes through. If the chamber is filled with alcohol vapour and then cooled, the vapour becomes supersaturated. The ions produced by the alpha particle act as nuclei for condensation, so a visible line of droplets forms along the track.

The labelled drawing should show:

• a transparent window or side for viewing,
• a small radioactive source inside,
• a layer of alcohol,
• a cooling method, such as dry ice or a piston,
• a dashed track of condensation droplets.

Key Takeaways

• Cloud chambers detect alpha particles by the tracks of condensation they produce.
• The apparatus needs a vapour source, a way of cooling, and a transparent window.

Common Mistakes

• Missing the labels on the drawing; each label is part of the mark.
• Forgetting to show the source of alcohol vapour or the cooling method.
• Not showing a transparent window for viewing.

Things to Be Careful About

If you choose a spark counter instead, you would need to show a fine wire, a plate or grille, and a high-voltage supply. The drawing must be clear and fully labelled to score all three marks.

Walkthrough

In a cloud chamber, the tracks appear at unpredictable times. If the emission were regular, the tracks would be equally spaced. The fact that the intervals between tracks vary is direct evidence that the emission is random.

Key Takeaways

• Random means the time between emissions is not fixed.
• The variation in track spacing is the observable evidence.

Common Mistakes

• Saying 'the tracks go in different directions' is acceptable, but the clearest answer is that the time between tracks varies.
• Do not say that the tracks appear at regular intervals.

Things to Be Careful About

If you described a spark counter, say that the time between sparks varies. The same idea applies.

Walkthrough

The count falls from 120 tracks per minute to 15 tracks per minute. Each half-life halves the count:

120 → 60 → 30 → 15

That is three halvings, so 3 half-lives have passed in 6.0 hours.

Therefore one half-life is 6.0 ÷ 3 = 2.0 hours.

No correction for background is needed because the question says no tracks are produced without the source.

Key Takeaways

• Half-life is the time for the number of radioactive nuclei to halve.
• Work out the number of half-lives by successive halving.

Common Mistakes

• Dividing 6.0 by 120 and then by 15, which gives the wrong answer.
• Forgetting the unit 'hours' in the final answer.

Things to Be Careful About

Count the number of halvings carefully. Here 120 to 15 is 3 halvings, not 2. The answer must be 2.0 hours, with the unit included.