Physics 5054/21 — May/June 2024

Walkthrough

On a distance$-$time graph:

• The gradient (slope) of the line equals the speed of the object: $\text{speed} = \frac{\Delta d}{\Delta t}$.
• If the graph is curved upwards (the slope becomes steeper), the speed is increasing, which means the ball is accelerating. On Fig. 1.2, this occurs from $t = 0\text{ s}$ to about $0.8\text{ s}$ (as the ball rolls down from A to B) and again from about $t = 1.6\text{ s}$ to $2.0\text{ s}$ (as it rolls down from C to D). Point $\text{P}$ can be placed anywhere in either of these regions.
• If the graph is a straight line with a constant gradient, the speed is constant. From $t = 2.0\text{ s}$ to $2.5\text{ s}$, the graph is a straight line, corresponding to the flat section DE of the track. Point $\text{Q}$ must be placed in this region.

Key Takeaways

• Speed is given by the gradient of a distance$-$time graph.
• A changing gradient indicates acceleration (speeding up or slowing down).
• A straight, non-horizontal line indicates constant (uniform) speed.

Common Mistakes

• Confusing distance$-$time graphs with speed$-$time graphs (where a horizontal line means constant speed and a straight sloped line means constant acceleration).
• Placing $\text{Q}$ on a horizontal part or at the inflection point instead of the straight section.

Things to Be Careful About

• Check the axes carefully: this is a distance$-$time graph, not a speed$-$time graph.

Walkthrough

To find the speed at a specific instant on a curved distance$-$time graph, we find the gradient of the curve at that point by drawing a tangent:

1. Place a ruler along the curve at point $\text{X}$ ($t = 1.0\text{ s}$) so that it touches the curve only at $\text{X}$ without crossing it.
2. Draw a long, straight tangent line extending across the grid.
3. Choose two convenient points on the tangent line that lie on grid intersections, spaced well apart for accuracy.
4. Calculate the gradient using:

1. Any correctly calculated gradient from a properly drawn tangent within the range $1.9\text{ m/s}$ to $2.4\text{ m/s}$ is accepted.

Key Takeaways

• Instantaneous speed on a distance$-$time graph is found by drawing a tangent at that point and determining its gradient.
• Use a large triangle (large $\Delta t$) to minimise reading uncertainty.

Common Mistakes

• Calculating speed by simply dividing the coordinates of point $\text{X}$ directly ($1.7 / 1.0 = 1.7\text{ m/s}$) instead of drawing a tangent and finding its gradient.
• Drawing a secant (a line cutting across the curve) instead of a true tangent.

Things to Be Careful About

• Ensure the tangent line is long enough to read values accurately from the grid.
• Always include the correct unit ($\text{m/s}$ is printed, so the numerical value must match).

Walkthrough

To explain how a gas exerts pressure according to the kinetic particle model:

1. Steam consists of water molecules in rapid, random motion.
2. As these particles collide with the piston face, their momentum changes as they bounce off.
3. By Newton's second and third laws, each collision exerts a small force on the surface of the piston.
4. Millions of such collisions occur every second across the piston's surface area. The total average force divided by the area of the piston gives the macroscopic pressure ($P = F / A$).

Key Takeaways

• Gas pressure arises from continuous collisions of particles with the container walls/surfaces.
• Force is the rate of change of momentum during collisions, and pressure is force per unit area.

Common Mistakes

• Saying particles "push" without mentioning collisions/hitting.
• Forgetting to link the collision forces to the area to explain pressure.

Things to Be Careful About

• Ensure both essential ideas are stated: (1) particle collisions with the piston, and (2) force exerted per collision / over an area / momentum change.

Walkthrough

1. Temperature is proportional to the average kinetic energy of the gas particles. When the temperature of the steam increases, the particles gain kinetic energy and move at higher average speeds.
2. Because the particles are faster:
   • They hit the piston more frequently (more collisions per second).
   • Each individual collision involves a larger change in momentum, producing a larger force per collision.
3. Both factors increase the total force exerted per unit area, resulting in an increased pressure.

Key Takeaways

• Higher temperature $\rightarrow$ higher average kinetic energy $\rightarrow$ higher particle speeds.
• Increased speed leads to both more frequent collisions and harder/more forceful collisions.

Common Mistakes

• Stating only that particles move faster without explaining how that alters collisions.
• Stating that particles expand instead of moving faster.

Things to Be Careful About

• Mention at least two distinct points: particle speed/kinetic energy increasing, and collision frequency/force per collision increasing.

Walkthrough

1. State the equation for momentum:

1. Substitute the given values ($m = 3.0 \times 10^4\ \text{kg}$ and $v = 28\ \text{m/s}$):

Key Takeaways

• Momentum is the product of mass and velocity ($p = mv$).

Common Mistakes

• Squaring the velocity by confusion with kinetic energy formula $\frac{1}{2}mv^2$.

Things to Be Careful About

• Express the answer clearly in standard form or full notation with appropriate units.

Walkthrough

There are two equivalent ways to solve this:

Method 1: Using Impulse and Momentum

1. Resultant force is the rate of change of momentum:

1. The aircraft starts from rest, so initial momentum $p_{\text{initial}} = 0$, giving $\Delta p = 8.4 \times 10^5\ \text{kg m/s}$.
2. Rearrange for time $t$:

Method 2: Using $F = ma$ and $a = \frac{v - u}{t}$

1. Acceleration $a = \frac{F}{m} = \frac{6.0 \times 10^5}{3.0 \times 10^4} = 20\ \text{m/s}^2$.
2. Time $t = \frac{v - u}{a} = \frac{28 - 0}{20} = 1.4\ \text{s}$.

Key Takeaways

• Impulse equals change in momentum: $F \times t = \Delta p$.
• Both momentum methods and equations of motion ($F = ma$) lead directly to the same result.

Common Mistakes

• Using the steam force without subtracting atmospheric pressure (error carried forward from part (b) is allowed if clearly shown).
• Forgetting that the aircraft starts from rest ($u = 0$).

Things to Be Careful About

• Keep significant figures appropriate (2 s.f.: 1.4 s).

Walkthrough

The melting point of ice is 0 °C. To convert a temperature from degrees Celsius to kelvin, add 273 (or 273.15, but 273 is standard at O Level).

Key Takeaways

The Kelvin scale is an absolute temperature scale. 0 K is absolute zero. The size of one kelvin is the same as one degree Celsius. To convert from °C to K, add 273.

Common Mistakes

Forgetting to add 273, or subtracting 273 instead. Writing 'degrees kelvin' (K) instead of just 'kelvin' (K). The unit is K, not °K.

Things to Be Careful About

The mark scheme accepts 273. Using 273.15 is technically more accurate but 273 is the expected value at this level. Ensure the unit is K, not °C.

Walkthrough

During melting, the substance changes state from solid to liquid. The thermal energy supplied is not used to increase the kinetic energy of the particles (which would raise the temperature). Instead, it is used as latent heat to overcome or weaken the intermolecular forces holding the particles in their fixed positions in the solid lattice. This increases the potential energy of the particles. Because the average kinetic energy of the particles does not change, the temperature remains constant until all the ice has melted.

Key Takeaways

During a change of state, temperature remains constant. The energy supplied is latent heat, which increases the potential energy of the particles by weakening or breaking the forces between them.

Common Mistakes

Saying 'the energy is used to melt the ice' without explaining what that means at the particle level. Saying 'temperature is constant because the ice is melting' is circular reasoning. Forgetting to mention that kinetic energy (and thus temperature) stays constant.

Things to Be Careful About

Use the term 'latent heat' or 'potential energy'. Do not say 'heat is used to break bonds' if the bonds are intermolecular forces, not chemical bonds. The mark scheme accepts 'weaken forces between particles' or 'more potential energy'.

Walkthrough

The energy required to warm the ice is calculated using the specific heat capacity equation $Q = mc\Delta T$, where $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature.
Given:
$m = 1.5 \text{ kg}$
$c = 2100 \text{ J/(kg }\degree\text{C)}$
$\Delta T = 0 - (-15) = 15 \degree\text{C}$
Substituting these values:

Rounding to 2 significant figures (as the mass is given to 2 s.f.), the energy is 47000 J.

Key Takeaways

The equation $Q = mc\Delta T$ calculates the thermal energy needed to change the temperature of a substance without changing its state. Ensure $\Delta T$ is the final temperature minus the initial temperature.

Common Mistakes

Forgetting to subtract a negative initial temperature (e.g., using $0 - 15 = -15$ instead of $0 - (-15) = 15$). Using the wrong value for specific heat capacity (the value for water is 4200, but the question specifies ice).

Things to Be Careful About

The mark scheme accepts 47000 or 47250. Using 2 significant figures is appropriate here because the mass (1.5 kg) and the temperature change (15 °C) are given to 2 s.f. Always include the unit J in the final answer if asked, though the blank here implies the unit is already there.

Walkthrough

Power is the rate of energy transfer, given by $P = E/t$. From part (iii), the energy transferred to warm the ice is $E = 47250 \text{ J}$. From Fig. 3.2, the time taken for this warming process (from -15 °C to 0 °C) is 2 minutes. Convert this time to seconds:

Now calculate the power:

Rounding to 2 significant figures gives 390 W.

Key Takeaways

Power is energy divided by time. Always ensure time is in seconds when calculating power in watts. Reading values from a graph requires careful attention to the axis scales and units.

Common Mistakes

Forgetting to convert minutes to seconds (using $t = 2$ gives $P = 23625 \text{ W}$, which is wrong). Using the energy from the wrong part of the graph. Not rounding to the correct number of significant figures.

Things to Be Careful About

The mark scheme accepts 390 or 394. Using 2 significant figures is consistent with the data given. Ensure you use the energy calculated in part (iii), not a different value. The graph clearly shows the warming phase ends at $t = 2 \text{ minutes}$.

Walkthrough

The gradient of a temperature-time graph (with constant power input) is inversely proportional to the specific heat capacity of the substance, since $P = mc\Delta T/t$, so $\Delta T/t = P/(mc)$. A smaller gradient means a larger specific heat capacity. Water has a specific heat capacity of 4200 J/(kg °C), which is roughly double that of ice (2100 J/(kg °C)). Therefore, for the same mass and the same power input, the temperature of liquid water rises more slowly than that of ice.

Key Takeaways

The gradient of a heating curve is steeper for substances with a lower specific heat capacity. Water has an unusually high specific heat capacity compared to many other substances, including its solid form, ice.

Common Mistakes

Saying 'water is denser than ice' or 'water has stronger bonds'. While water does have stronger hydrogen bonds, the direct reason for the smaller gradient is the higher specific heat capacity. The mark scheme specifically looks for 'higher specific heat capacity'.

Things to Be Careful About

The question asks for 'one reason'. Stating 'higher specific heat capacity' is sufficient and is the most direct answer. Avoid over-explaining or giving multiple reasons if only one is required, though the mark scheme accepts 'stronger bonds' as an alternative explanation for the higher specific heat capacity.

Walkthrough

Frequency is defined as the number of complete wave cycles that pass a fixed point in one second. This is a direct recall definition from the wave properties section of the syllabus.

Key Takeaways

Frequency ($f$) is measured in hertz (Hz) and is the reciprocal of the time period ($T$). It describes how often the wave oscillates.

Common Mistakes

Candidates sometimes say 'the number of waves per minute' or omit 'per second'. The definition must specify a time interval of one second.

Things to Be Careful About

Ensure the answer refers to 'complete waves' or 'cycles' passing a 'fixed point' or 'a point'.

Walkthrough

The wave speed equation links speed ($v$), frequency ($f$) and wavelength ($\lambda$): $v = f \lambda$. The frequency is given as $30\text{ kHz}$, which must be converted to standard units (Hz) by multiplying by $1000$, giving $30\,000\text{ Hz}$. The wavelength is $0.011\text{ m}$. Substituting these values gives $v = 30\,000 \times 0.011 = 330\text{ m/s}$.

Key Takeaways

Always convert frequency to hertz (Hz) and wavelength to metres (m) before using $v = f \lambda$ to get the speed in $\text{m/s}$.

Common Mistakes

Forgetting to convert $30\text{ kHz}$ to $30\,000\text{ Hz}$ and calculating $30 \times 0.011 = 0.33$, which is wrong by a factor of $1000$.

Things to Be Careful About

The question asks for the answer in $\text{m/s}$, so the unit must be included in the final answer. The mark scheme awards a C1 for the correct equation in any form and an A1 for the correct numerical answer.

Walkthrough

Ultrasound is used in many applications because it can travel through solids and liquids and can be reflected by boundaries between different materials. Common uses include medical imaging (scanning fetuses or internal organs), sonar for finding the depth of the sea or locating fish, ultrasonic cleaning of delicate objects, and non-destructive testing of materials.

Key Takeaways

Ultrasound has frequencies above the upper limit of human hearing ($20\text{ kHz}$) and is useful for imaging and cleaning due to its short wavelength and high energy.

Common Mistakes

Saying 'hearing' or 'music', which are uses of audible sound, not ultrasound.

Things to Be Careful About

Any valid use scores. Medical scanning, sonar, and cleaning are the most common accepted answers.

Walkthrough

The key difference between the two wave types lies in the direction of particle oscillation relative to the direction of energy transfer (wave travel). For longitudinal waves (like sound), particles move parallel to the wave direction, creating compressions and rarefactions. For transverse waves (like light or water waves), particles move perpendicular to the wave direction, creating crests and troughs. A labelled diagram clearly showing these oscillation directions earns the third mark.

Key Takeaways

Longitudinal = parallel oscillation; Transverse = perpendicular oscillation. Remember that the wave itself travels forward, but the particles only oscillate about a fixed position.

Common Mistakes

Saying 'particles move with the wave' — particles do not travel with the wave; they oscillate about a mean position. Confusing the direction of particle motion with the direction of wave travel.

Things to Be Careful About

The mark scheme specifically requires reference to 'the movement of the particles'. Simply saying 'sound is longitudinal and light is transverse' does not answer the question. A diagram is optional but highly recommended to secure the third mark.

Walkthrough

Earthquakes generate seismic waves, which include P-waves (primary waves, which are longitudinal) and S-waves (secondary waves, which are transverse). Lightning is another valid answer, as it produces thunder (sound waves, which are longitudinal) and light (electromagnetic waves, which are transverse).

Key Takeaways

Some natural events generate multiple types of waves. Seismic waves from earthquakes are a classic example of both longitudinal and transverse mechanical waves being produced.

Common Mistakes

Saying 'a speaker' (only produces longitudinal sound waves) or 'a string' (only produces transverse waves). The source must generate BOTH types.

Things to Be Careful About

Ensure the example genuinely produces both types. Earthquakes and lightning are the standard accepted answers in the mark scheme.

Walkthrough

The graph plots current (y-axis) against voltage (x-axis). For an ohmic conductor at constant temperature, this would be a straight line through the origin ($I = V/R$). However, for a filament lamp, as the voltage increases, the current increases, causing the filament to heat up. The temperature rise causes the resistance of the tungsten filament to increase. Since $I = V/R$, a increasing $R$ means the current increases less than proportionally with voltage. On an I-V graph, the gradient is $1/R$. As $R$ increases, the gradient decreases. Therefore, the curve starts steep and bends to become flatter, approaching the voltage axis.

Key Takeaways

The current-voltage graph for a filament lamp is non-linear because resistance is not constant; it increases with temperature.

Common Mistakes

• Drawing a straight line (this would be for an ohmic conductor like a fixed resistor at constant temperature).
• Bending the curve the wrong way (towards the current axis), which would imply resistance is decreasing.

Things to Be Careful About

Check which quantity is on the y-axis and which is on the x-axis. Here, current is on the y-axis, so a decreasing gradient means the curve flattens out. If voltage were on the y-axis, the curve would bend upwards.

Walkthrough

The shape of the graph is determined by how resistance changes. In a filament lamp, the filament is made of tungsten. As current flows, electrical energy is converted to thermal energy, heating the filament. The resistance of a metal conductor increases as its temperature increases. Therefore, at higher voltages and currents, the filament is hotter and has a higher resistance. This higher resistance limits the current more than it would at lower temperatures, causing the graph to curve.

Key Takeaways

Resistance in metallic conductors is not always constant; it depends on temperature. Filament lamps are non-ohmic devices.

Common Mistakes

• Saying 'resistance increases because voltage increases' without mentioning temperature. The direct cause is temperature.
• Saying 'current decreases' (current still increases, just at a slower rate).

Things to Be Careful About

The mark scheme awards one mark for 'temperature increases' and one mark for 'resistance increases'. Both links in the chain must be stated clearly.

Walkthrough

The fuse must be rated slightly higher than the normal operating current of the circuit so that it does not blow during normal use, but low enough to blow quickly if a fault causes a dangerous current.

Normal operating current = lamp current + heater current = $1.5 \text{ A} + 10 \text{ A} = 11.5 \text{ A}$.

Standard fuse ratings are typically 3 A, 5 A, 13 A, 20 A, 30 A, etc. A 13 A fuse is the smallest standard rating above 11.5 A, making it suitable. Values up to 19 A are also accepted by the mark scheme as long as they are not so large that they fail to protect the circuit.

Key Takeaways

Fuse ratings must be chosen to be above the normal working current but below the current that would damage the wiring or cause a fire.

Common Mistakes

• Suggesting a fuse rating lower than 11.5 A (e.g., 10 A), which would blow during normal operation.
• Suggesting a very large rating like 30 A, which would not protect the circuit.

Things to Be Careful About

The question asks for 'a suitable fuse rating'. Any integer between 12 A and 19 A is acceptable. 13 A is the most common practical choice for a 240 V domestic circuit of this size.

Walkthrough

A fuse is a safety device. Its purpose is to melt and break the circuit when the current exceeds a safe value (a fault current). If a fuse with a very large rating (e.g., 30 A) is used in a circuit that normally draws 11.5 A, a fault might cause a current of 20 A. This is dangerous and could overheat the cables, but the 30 A fuse would not melt, so the circuit would remain live and the danger would persist.

Key Takeaways

Fuses protect circuits by melting when current is too high. A correctly rated fuse ensures protection without nuisance blowing.

Common Mistakes

• Saying 'the fuse would not work' (it does work, just too late).
• Not mentioning that the circuit needs to be shut down or that overheating/danger is the consequence.

Things to Be Careful About

Focus on the consequence of the fuse not blowing: the current remains large/dangerous, and the circuit does not shut down quickly enough.

Walkthrough

In a mains supply, the live wire carries the high alternating voltage (240 V in the UK), while the neutral wire is at approximately 0 V. If a fault occurs and the fuse melts in the live wire, the connection to the high voltage is broken. The rest of the circuit (the lamp, the heaters, the wiring) is no longer connected to the live supply and is safe to touch.

If the fuse were placed in the neutral wire, melting it would break the circuit and stop the current from flowing, so the appliances would stop working. However, the live wire would still be connected to the appliances, meaning they would still be at 240 V. If a person touched the appliance internals, they could receive a fatal electric shock.

Key Takeaways

Safety devices like fuses must be placed in the live wire to ensure the circuit is de-energised when they operate.

Common Mistakes

• Saying 'the earth wire carries the current' (earth is a safety wire, not a current-carrying wire under normal conditions).
• Not explaining that cutting the live wire makes the rest of the circuit safe (not live).

Things to Be Careful About

Be precise with terminology: use 'live', 'voltage', 'electrocution', or 'dangerous' rather than vague terms like 'electricity' or 'power'.

Walkthrough

The force on a current-carrying conductor in a magnetic field is proportional to the magnetic flux density. Using a stronger magnet increases the magnetic field strength, which increases the force on each side of the coil. This larger force creates a larger turning effect (moment) about the axis, so the coil accelerates more and the motor turns faster.

Key Takeaways

A stronger magnetic field increases the force on a current-carrying conductor, leading to a larger turning effect in a motor.

Common Mistakes

Students often write 'the motor works better' without specifying what happens (e.g., turns faster, stronger force). 'Larger current' is incorrect because the question asks about changing the magnet, not the power supply.

Things to Be Careful About

Accept any valid statement linking the stronger magnet to a larger force, larger moment, or faster rotation. Avoid vague answers like 'more powerful' without context.

Walkthrough

Magnetic field direction: Magnetic field lines outside a magnet always point from the North pole to the South pole. In Fig. 6.1, the N pole is on the left and the S pole is on the right, so draw arrows pointing from left to right across the gap between the magnets.

Current direction: Look at the battery symbol. The longer vertical line is the positive terminal, and the shorter, thicker line is the negative terminal. Current flows from positive to negative in the external circuit. Follow the wire from the positive terminal to the right carbon brush. The current goes up into the right half of the split-ring commutator, then into the right side of the coil. Following the coil around, the current goes along the right side (away from you), across the back, along the left side (towards you), down the left side of the commutator, and back to the negative terminal. Mark arrows on the visible front-facing parts of the coil to show this direction.

Key Takeaways

Magnetic field lines go from N to S. Current flows from the positive (longer line) to the negative (shorter line) terminal of a battery in the external circuit.

Common Mistakes

Drawing the magnetic field from S to N. Getting the battery polarity wrong (the longer line is positive). Forgetting to show the current direction on the coil sides themselves, not just the wires.

Things to Be Careful About

The question asks to mark and label on Fig. 6.1. Ensure arrows are clearly drawn on the coil and in the magnetic field gap. Labels like 'N to S' and 'current' are helpful but the arrows themselves carry the marks.

Walkthrough

A current-carrying conductor in a magnetic field experiences a force. This is the motor effect. In the motor, the two long sides of the rectangular coil are perpendicular to the magnetic field, so they experience forces.

To find the direction of the force, use Fleming's left-hand rule (or the right-hand grip rule to find the coil's poles). With the current flowing towards the viewer on the left side and the field pointing left to right, the force on the left side is upwards. With the current flowing away from the viewer on the right side, the force on the right side is downwards.

Because these two equal and opposite forces act on different sides of the coil (not along the same line), they form a couple and produce a turning effect (moment) about the central axis. This is why the coil rotates.

Alternatively, you can explain it using magnetic poles: the current in the coil makes it act like a bar magnet. The face towards the top becomes a North pole. This North pole is repelled by the permanent magnet's North pole and attracted to its South pole, causing the coil to turn.

Key Takeaways

Forces on opposite sides of a current-carrying coil in a magnetic field are in opposite directions, creating a turning effect (couple) that causes rotation.

Common Mistakes

Saying 'the coil is attracted to the magnet' without specifying why it rotates. Forgetting to mention that the forces are in opposite directions on opposite sides. Using 'heat' instead of 'thermal energy' (not applicable here, but a common theme). Saying 'the magnetic field pushes the current' instead of 'the magnetic field exerts a force on the current-carrying conductor'.

Things to Be Careful About

The mark scheme awards one mark for stating that a force is caused (or the coil becomes a magnet), and one mark for explaining the direction of the forces (up on one side, down on the other) or the resulting magnetic poles. Both parts are needed for full marks. Ensure you use the word 'force' or 'turning effect' / 'couple'.

Walkthrough

First, determine the direction of the magnetic field around each individual wire using the right-hand grip rule. Point your right thumb in the direction of the current, and your fingers curl in the direction of the magnetic field lines.

• Left wire: Current is downwards into the card (represented by the cross, like the tail feathers of an arrow). Point your right thumb into the card. Your fingers curl clockwise. Draw concentric circles with clockwise arrows around the left wire.
• Right wire: Current is upwards out of the card (represented by the dot, like the tip of an arrow). Point your right thumb out of the card. Your fingers curl anti-clockwise. Draw concentric circles with anti-clockwise arrows around the right wire.

Next, consider the superposition of these two fields. Between the two wires, the clockwise field from the left wire points downwards, and the anti-clockwise field from the right wire also points downwards. The fields add together, making the magnetic field stronger between the wires. Draw the field lines more closely spaced in this region to indicate the stronger field. Outside the wires, the fields partially cancel and splay outwards.

Key Takeaways

The right-hand grip rule gives the field direction around a straight current-carrying wire. For two parallel wires with opposite currents, the fields between them add together, creating a stronger field.

Common Mistakes

Drawing the field lines as straight lines instead of circles. Getting the direction of the circles wrong (mixing up clockwise and anti-clockwise). Drawing the field lines crossing each other (field lines never cross). Forgetting to show the direction of the field with arrows. Drawing the field lines equally spaced everywhere, missing the stronger field between the wires.

Things to Be Careful About

The question asks to sketch the pattern on Fig. 6.3. Ensure the circles are clearly drawn around the existing cross and dot symbols. The key is provided, so use it correctly: cross is into the page, dot is out of the page. The mark scheme awards one mark for circles close to and around each wire, one for the correct overall shape (stronger between wires), and one for correct direction on at least one line with none wrong. Ensure at least one arrow direction is clearly correct and no arrows are drawn in the wrong direction.

Walkthrough

This question asks for the mechanism behind the force between two current-carrying wires. It is a two-step explanation.

Step 1: The current flowing through the first wire generates a magnetic field around it (Oersted's discovery). This magnetic field extends outwards and is present at the location of the second wire.

Step 2: The second wire also carries a current. When a current-carrying conductor is placed in an external magnetic field, it experiences a force (this is the motor effect, described by $F = BIL$). The magnetic field from the first wire acts on the current in the second wire, producing a force. By Newton's third law, the second wire's field acts on the first wire with an equal and opposite force.

Since the currents are in opposite directions, the forces are repulsive. You can also explain this using field superposition: between the wires, the magnetic fields from both wires point in the same direction and add together, creating a stronger field. Outside the wires, they point in opposite directions and cancel. The wires are pushed away from the region of stronger field towards the region of weaker field, resulting in repulsion.

Key Takeaways

A current-carrying wire produces a magnetic field. Another current-carrying wire in that field experiences a force. This is the fundamental interaction between parallel conductors.

Common Mistakes

Saying 'the wires have magnetic fields so they repel' without explaining the mechanism. Forgetting to mention that the second wire is in a magnetic field. Saying 'the currents repel' (currents are not objects that can repel; the wires do). Using 'heat' or 'thermal energy' incorrectly (not applicable here).

Things to Be Careful About

The mark scheme awards one mark for stating that the current in one wire causes a magnetic field at the other (or that the field between them is stronger), and one mark for stating that a current experiences a force in a magnetic field (or that wires move away from the strongest field). Both parts are needed. Ensure you use the phrase 'magnetic field' and 'force on a current-carrying conductor' or equivalent.

Walkthrough

The proton number is the bottom number in the nuclide notation $^{A}_{Z}\text{X}$. Reading from Table 7.1:

• thorium-236: $Z = 90$
• protactinium-236: $Z = 91$
• uranium-235: $Z = 92$
• uranium-238: $Z = 92$
  The two nuclei with the same number of protons (92) are uranium-235 and uranium-238.

Key Takeaways

Nuclide notation is $^{A}_{Z}\text{X}$, where $A$ is the nucleon number (top) and $Z$ is the proton number (bottom). Elements with the same proton number are the same element, regardless of the number of neutrons.

Common Mistakes

Confusing the top number (nucleon number) with the bottom number (proton number). Forgetting that the element name is determined solely by the proton number.

Things to Be Careful About

Ensure you are reading the correct subscript for the proton number. In this case, both uranium isotopes have $Z = 92$.

Walkthrough

Nucleons are the total number of protons and neutrons in the nucleus. To find the number of neutrons, subtract the proton number ($Z$, the bottom number) from the nucleon number ($A$, the top number).

Calculating for each:

• $^{236}_{90}\text{Th}$: $236 - 90 = 146$ neutrons
• $^{236}_{91}\text{Pa}$: $236 - 91 = 145$ neutrons
• $^{235}_{92}\text{U}$: $235 - 92 = 143$ neutrons
• $^{238}_{92}\text{U}$: $238 - 92 = 146$ neutrons

Thorium-236 and uranium-238 both have 146 neutrons.

Key Takeaways

Isotopes have the same number of protons but different numbers of neutrons. Nuclei of different elements can have the same number of neutrons (these are called isotones).

Common Mistakes

Assuming that nuclei with the same nucleon number have the same number of neutrons. Forgetting to subtract the proton number from the nucleon number.

Things to Be Careful About

Always perform the subtraction explicitly. A power-of-ten error or misreading the top/bottom numbers will lead to the wrong neutron count.

Walkthrough

The nucleon number (also called the mass number) is the top number in the nuclide notation $^{A}_{Z}\text{X}$. Looking at Table 7.1:

• thorium-236: $A = 236$
• protactinium-236: $A = 236$
• uranium-235: $A = 235$
• uranium-238: $A = 238$

Thorium-236 and protactinium-236 both have a nucleon number of 236.

Key Takeaways

Nuclei with the same nucleon number but different proton numbers are called isobars. They have the same total number of protons and neutrons combined, but a different ratio of the two.

Common Mistakes

Confusing nucleon number with neutron number or proton number. Remember that the nucleon number is simply the top number in the symbol.

Things to Be Careful About

The nucleon number is an integer and represents the total count of protons and neutrons. Do not confuse it with the atomic mass in atomic mass units, which can be a decimal.

Walkthrough

When using a Geiger-Müller tube to measure radiation from a source, the counter will always record some counts even when no source is present. This is because radiation is constantly coming from natural sources in the environment, such as cosmic rays from space, radioactive isotopes in rocks and soil (like radon gas), and even from building materials and food.

To get an accurate measurement of the radiation from the source alone, this background count rate must be subtracted from the total count rate measured when the source is present.

Key Takeaways

Background radiation is unavoidable and must always be accounted for in radioactivity experiments. The corrected count rate from the source is: $\text{corrected count rate} = \text{total count rate} - \text{background count rate}$.

Common Mistakes

Saying 'background radiation is from man-made sources' (this is incorrect; it is primarily natural). Saying 'background radiation is zero' (it is never zero). Forgetting to subtract the background when calculating the source's activity.

Things to Be Careful About

The mark scheme accepts 'natural' or 'present without a source'. Do not just say 'radiation from the ground' as it is too specific; cosmic rays and internal body radiation also contribute.

Walkthrough

The counter records a total of 420 counts per minute when the source is present. This total includes both the radiation from the source and the background radiation. To find the count rate due to the source alone, subtract the background count rate (20 counts per minute, measured earlier with no source).

Key Takeaways

Always subtract the background count rate to isolate the activity of the radioactive source. This corrected value is what you use for all subsequent half-life calculations.

Common Mistakes

Forgetting to subtract the background radiation and using 420 in the half-life calculation. This will lead to an incorrect final answer.

Things to Be Careful About

Ensure the units are consistent. Both the background and total count rates are in 'counts per minute', so no unit conversion is needed.

Walkthrough

First, determine how many half-lives have passed in 90 minutes:

Next, halve the corrected count rate (400 counts per minute) for each half-life:

• After 1 half-life (45 minutes): $400 / 2 = 200$ counts per minute
• After 2 half-lives (90 minutes): $200 / 2 = 100$ counts per minute

Finally, the question asks for the 'reading on the counter'. The counter will still detect background radiation, so add the background count rate (20 counts per minute) back to the corrected source count rate:

Key Takeaways

Half-life calculations must be done on the corrected count rate (source only). However, the final answer to 'what will the counter read' must include the background radiation.

Common Mistakes

Forgetting to add the background radiation back at the end. If you stop at 100, you lose the final mark. Halving the uncorrected total (420 -> 210 -> 105) is also incorrect because the background does not decay.

Things to Be Careful About

Read the question carefully: 'reading on the counter' means the total count rate, not the corrected count rate from the source. Always track whether you are working with total or corrected values.

Walkthrough

We cannot predict exactly when a specific nucleus will decay. Radioactive decay is a random, spontaneous process at the level of individual atoms. While the half-life gives us a reliable average for a large number of atoms, the actual count rate measured by a Geiger-Müller tube will show statistical fluctuations (Poisson statistics).

Therefore, the calculated value of 120 counts per minute is an expected average, and the actual reading on the counter at exactly 90 minutes could be slightly higher or lower (e.g., 118 or 123).

Key Takeaways

Half-life is a statistical average that applies to large numbers of nuclei. Individual decay events are random and unpredictable.

Common Mistakes

Saying 'because the half-life changes' (half-life is constant for a given isotope). Saying 'because of human error' (this is not a valid physics reason for the inherent uncertainty in decay counts).

Things to Be Careful About

Use the word 'random' or 'fluctuates'. 'Unpredictable' is also acceptable for individual decays, but 'random' is the precise term used in the syllabus.

Walkthrough

The gravitational field strength $g$ at the surface of a planet is given by $g = GM/R^2$, where $M$ is the mass and $R$ is the radius.

Density $\rho = M / (\frac{4}{3}\pi R^3)$, so $M = \rho \cdot \frac{4}{3}\pi R^3$.

Substituting this into the equation for $g$:

Since the average densities ($\rho$) of Mercury and Venus are similar, $g$ is directly proportional to the radius $R$. Mercury has a smaller radius, so it has a smaller mass, which results in a weaker gravitational field strength.

Key Takeaways

• Gravitational field strength depends on both mass and radius.
• For planets with similar densities, the smaller planet will have a weaker surface gravity.

Common Mistakes

• Stating 'smaller size' without linking it to mass.
• Saying 'less gravity' without explaining why.

Things to Be Careful About

• The question asks to 'suggest why', so a clear causal link between mass and gravitational field strength is required.