Physics 5054/21 — October/November 2024

Walkthrough

1. The force exerted by a spring within its limit of proportionality is given by Hooke's law: $F = kx$.
2. Substitute the spring constant $k = 14\text{ N/m}$ and the extension $x = 0.035\text{ m}$.
3. $F = 14 \times 0.035 = 0.49\text{ N}$.

Key Takeaways

• Hooke's law states that the force exerted by a spring is directly proportional to its extension ($F = kx$).
• The spring constant $k$ is in $\text{N/m}$ and extension $x$ must be in metres for the force to be in Newtons.

Common Mistakes

• Using an incorrect formula, such as $F = k / x$.
• Confusing extension with the total length of the spring.

Things to Be Careful About

• The extension is already given in metres ($0.035\text{ m}$), so no conversion is required. The force is constant because the extension is kept constant.

Walkthrough

1. Work done is defined as force multiplied by the distance moved in the direction of the force: $W = Fx$.
2. Use the constant force from part (i): $F = 0.49\text{ N}$.
3. The distance moved is $x = 0.86\text{ m}$.
4. $W = 0.49 \times 0.86 = 0.4214\text{ J}$. Rounding to two significant figures gives $0.42\text{ J}$.

Key Takeaways

• Work done $W$ (in joules) equals force $F$ (in newtons) times distance $x$ (in metres).
• When a constant force is applied over a distance, the work done is simply the product of the two.

Common Mistakes

• Using the wrong formula for work done, such as $W = F / x$.
• Carrying forward an incorrect force value from part (i) without recalculating.

Things to Be Careful About

• The force is constant because the extension is kept constant at $0.035\text{ m}$. This justifies using $W = Fx$ directly. Round the final answer to an appropriate number of significant figures (two, matching the given data).

Walkthrough

1. Power is defined as the rate of doing work, or work done per unit time: $P = W / t$.
2. As the trolley accelerates, its speed increases.
3. Because the speed is higher, the distance moved in any given time interval (such as one second) is greater.
4. Since the force is constant and the distance moved in that time is greater, the work done $W = Fx$ in that time interval increases.
5. Therefore, the work done per unit time (power) increases.

Key Takeaways

• Power is the rate of energy transfer or work done: $P = W / t$.
• An increase in speed means more distance is covered in the same amount of time, leading to more work done per second.

Common Mistakes

• Stating "power increases because speed increases" without explaining the link to work done or distance.
• Confusing power with force or work done.

Things to Be Careful About

• The mark scheme specifically looks for the link between speed, distance moved in a given time, and work done in that time. Ensure all three steps are clearly stated.

Walkthrough

The ball is 40 m above the ground, so it has gravitational potential energy because of its position in the Earth's gravitational field. The equation is

where $m$ is the mass in kg, $g$ is the gravitational field strength and $h$ is the height in m. Substituting the values gives

The ground is taken as the zero level of gravitational potential energy, so this is the initial GPE of the ball.

Key Takeaways

• Gravitational potential energy is stored energy due to height in a gravitational field.
• The formula $E_p = mgh$ uses mass in kg, height in m and $g$ in N/kg.
• The unit of energy is the joule (J).

Common Mistakes

• Using the weight ($0.20 \times 9.8 = 1.96$ N) as the energy instead of multiplying by the height.
• Forgetting to multiply by $g$.
• Using the height in cm without converting to m.

Things to Be Careful About

• Use $g = 9.8\text{ N/kg}$ as used in the mark scheme.
• The unit is joules, not newtons.
• The answer 78.4 J is acceptable; the mark scheme also accepts 78 J.

Walkthrough

Air resistance is ignored, so no energy is lost to the surroundings. The initial gravitational potential energy is all converted to kinetic energy just before the ball hits the ground:

So

The mass cancels, giving

This is the speed immediately before impact.

Key Takeaways

• Energy is conserved when there is no friction or air resistance.
• The kinetic energy just before impact equals the initial gravitational potential energy.
• Because the mass cancels, the impact speed does not depend on the mass of the ball.

Common Mistakes

• Forgetting to take the square root, giving 784 m/s.
• Using $v = \sqrt{gh}$ instead of $v = \sqrt{2gh}$.
• Thinking the mass must be known to find the speed.

Things to Be Careful About

• The question says 'ignore air resistance'; if air resistance were included, the speed would be lower.
• Speed is a scalar, so no direction is needed.
• The answer is 28 m/s.

Walkthrough

When the ball stops, its kinetic energy is transferred to internal energy of the ball. The maximum temperature rise happens if all 78.4 J of kinetic energy is used to warm the ball. The specific heat capacity equation is

where $\Delta E$ is the thermal energy supplied, $m$ is the mass, $c$ is the specific heat capacity and $\Delta\theta$ is the temperature rise. Rearranging gives

This is the maximum possible temperature increase because it assumes no energy is transferred elsewhere.

Key Takeaways

• Specific heat capacity tells how much energy is needed to raise the temperature of 1 kg of a substance by 1 °C.
• The equation $\Delta E = mc\Delta\theta$ links thermal energy, mass, specific heat capacity and temperature change.
• 'Maximum' means all the kinetic energy goes into warming the ball.

Common Mistakes

• Using the mass incorrectly (e.g. dividing by 0.20 instead of multiplying).
• Forgetting to use the energy from part (a)(i).
• Giving the answer without units or with wrong units.

Things to Be Careful About

• Use $\Delta E = 78.4\text{ J}$ from part (a)(i).
• The unit of specific heat capacity is $\text{J/(kg }^\circ\text{C)}$, so the temperature rise comes out in °C.
• Show the substitution clearly to earn the method mark.

Walkthrough

The calculated value in (b)(i) assumes all the kinetic energy becomes internal energy of the ball. In reality, some energy is transferred to the ground when the ball hits it, and some may be used to deform the modelling clay. So less energy is available to raise the ball's temperature, and the actual rise is smaller. Any one valid reason is enough.

Key Takeaways

• Energy is never destroyed; it is transferred between stores.
• 'Maximum possible' means assuming no losses.
• The temperature rise is smaller when some energy goes to the surroundings or is used to change the shape of the ball.

Common Mistakes

• Saying 'energy is lost' without saying where it goes.
• Saying 'heat escapes' without identifying the transfer.
• Giving a reason that contradicts the assumption in part (a)(ii), unless it is clearly about the real situation.

Things to Be Careful About

• The mark scheme accepts: energy transferred to the ground, work done deforming the clay, or air resistance reducing the kinetic energy.
• Use 'thermal energy' rather than 'heat' for precision.
• Only one reason is needed for the mark.

Walkthrough

Temperature is a measure of the average kinetic energy of the particles in a substance. As the temperature of the gas increases, the average kinetic energy of the gas particles increases. Since kinetic energy is proportional to the square of the speed ($E_k = \frac{1}{2}mv^2$), an increase in average kinetic energy means an increase in the average speed of the particles.

Key Takeaways

• Temperature is directly related to the average kinetic energy of particles.
• Higher temperature means higher average particle speed.

Common Mistakes

• Saying 'particles get bigger' or 'particles move further apart' as the temperature increases (these describe volume expansion, not the direct effect on particle motion at constant volume).
• Forgetting the word 'average' when describing particle speed, as individual particle speeds vary widely.

Things to Be Careful About

• The question asks to 'state', so only one concise sentence is needed. Do not over-explain.

Walkthrough

Pressure is the force exerted per unit area on the walls of a container. In a gas, this force arises from particles colliding with the walls. As the temperature increases, the average speed of the particles increases (from part (b)(i)). This has two effects:

1. Each particle hits the wall with a greater speed, so it undergoes a greater change in momentum (greater impulse) per collision, exerting a greater force.
2. The particles are moving faster, so they travel to the walls and back more quickly, meaning they collide with the walls more frequently.
   Both effects combine to increase the total force on the walls, and thus the pressure.

Key Takeaways

• Gas pressure is caused by particle collisions with the container walls.
• Increasing temperature increases both the force per collision and the frequency of collisions.

Common Mistakes

• Saying 'particles get bigger' or 'particles expand'.
• Not mentioning that the particles collide with the walls at all.
• Saying 'particles push on the walls' without referring to collisions or momentum change.

Things to Be Careful About

• The question asks to explain 'in terms of the particles', so you must refer to particle motion and collisions.
• All three points (collision, greater force/impulse, more often/harder) are needed for full marks.

Walkthrough

The piston is released and moves to the right, increasing the volume of the gas. The temperature is kept constant at $100\ ^\circ\text{C}$, so the average speed of the particles remains the same. Because the gas now occupies a larger volume, the particles are more spread out (further apart). This means it takes longer for a particle to travel from one side of the cylinder to the other, so the particles collide with the walls less frequently. Fewer collisions per unit time means a smaller total force on the walls, and thus a lower pressure.

Key Takeaways

• At constant temperature, increasing volume decreases pressure because collision frequency with the walls decreases.
• Particle speed (and thus force per collision) remains unchanged if temperature is constant.

Common Mistakes

• Saying 'particles move slower' (temperature is constant, so average speed is constant).
• Saying 'particles are further apart so they don't hit the walls' (they still hit, just less often).

Things to Be Careful About

• The question asks to explain in terms of the particles, so refer to particle spacing and collision frequency.
• Only one mark is awarded here, so a single clear statement is sufficient.

Walkthrough

The piston moves to the right because initially the pressure of the hot gas inside is greater than the atmospheric pressure outside, creating a resultant force to the right. As the piston moves, the volume increases and the gas pressure decreases (from part (c)(i)). The piston will continue to move until the gas pressure inside equals the atmospheric pressure outside. At this point, the force exerted by the gas on the right face of the piston equals the force exerted by the atmosphere on the left face. With no resultant force acting on the piston, it stops moving (Newton's first law).

Key Takeaways

• An object stops accelerating when the resultant force on it is zero.
• For a piston in a cylinder open to the atmosphere, equilibrium is reached when internal gas pressure equals external atmospheric pressure.

Common Mistakes

• Saying 'the pressure runs out' or 'the gas stops expanding'.
• Not mentioning that the pressures become equal or that the forces balance.
• Saying 'friction stops it' without acknowledging the pressure balance (though friction/air resistance is an acceptable alternative for the second mark if pressures are not equal, the primary expected answer is pressure equality).

Things to Be Careful About

• Two marks are available: one for the pressure equality, one for the resultant force being zero (or an alternative like friction). Ensure both concepts are covered clearly.

Walkthrough

When white light is dispersed by a prism, it separates into the colours of the visible spectrum. The standard sequence from the least deviated (red) to the most deviated (violet) is red, orange, yellow, green, blue, indigo, and violet. The question asks for the five colours between red and violet, which are orange, yellow, green, blue, and indigo. Listing all seven in order from red to violet fully satisfies the requirement.

Key Takeaways

• The visible spectrum consists of seven colours: red, orange, yellow, green, blue, indigo, violet.
• Red is at the long-wavelength end (least refracted) and violet is at the short-wavelength end (most refracted).

Common Mistakes

• Forgetting 'indigo' or mixing up the order of blue and indigo.
• Listing the colours in reverse order (violet to red) when the question specifies red to violet.

Things to Be Careful About

• Ensure the list is in the correct order from red to violet as requested. Any four in order, or all five in any order, will score, but providing the full seven in order is the safest approach.

Walkthrough

As you move from red light to violet light in the visible spectrum, the wavelength gets shorter. Since the speed of light in a vacuum (or air) is constant for all electromagnetic waves, the wave equation $v = f \lambda$ means that as wavelength $\lambda$ decreases, frequency $f$ must increase. Therefore, going from red to violet, the frequency increases and the wavelength decreases.

Key Takeaways

• Red light has the longest wavelength and lowest frequency in the visible spectrum.
• Violet light has the shortest wavelength and highest frequency in the visible spectrum.
• Frequency and wavelength are inversely proportional for waves traveling at the same speed.

Common Mistakes

• Stating that frequency decreases or wavelength increases (the opposite of the truth).
• Confusing the trend with amplitude or loudness/brightness.

Things to Be Careful About

• The question asks for the change 'going from red light to violet light', so the direction of the change must be clearly stated (increases/decreases), not just the relative values.

Walkthrough

1. Recall the wave equation $v = f\lambda$, where $v$ is the wave speed, $f$ is the frequency and $\lambda$ is the wavelength.
2. Rearrange to make $\lambda$ the subject: $\lambda = v/f$.
3. The frequency is given in MHz — convert to Hz: $8.4\ \text{MHz} = 8.4 \times 10^6\ \text{Hz}$.
4. Substitute the speed $1500\ \text{m/s}$ and the frequency: $\lambda = 1500 / (8.4 \times 10^6)$.
5. Calculate: $\lambda = 1.79 \times 10^{-4}\ \text{m}$, which rounds to $1.8 \times 10^{-4}\ \text{m}$.

The C1 mark is for writing the correct equation or the substitution; a second C1 is for the correct order of magnitude ($1.8 \times 10^N$); the A1 is for the final value.

Key Takeaways

The wave equation $v = f\lambda$ links speed, frequency and wavelength.
Always convert megahertz (MHz) to hertz (Hz) before substituting.
Small wavelengths like this are the reason ultrasound can resolve fine detail in medical imaging.

Common Mistakes

Forgetting to convert MHz to Hz — using 8.4 directly gives $\lambda = 178.6\ \text{m}$, which is physically absurd for ultrasound in tissue.
Using $\lambda = vf$ instead of $v/f$.
Quoting the answer without the unit — the answer blank requires 'm'.

Things to Be Careful About

The speed is in m/s and the frequency in Hz, so the wavelength automatically comes out in metres.
Write the answer in standard form as $1.8 \times 10^{-4}\ \text{m}$ — a decimal like 0.00018 m is correct but less tidy; the mark scheme expects the standard form.

Walkthrough

When a wave crosses a boundary into a different medium, its frequency is fixed by the source — the transmitter keeps emitting at 8.4 MHz, so the frequency cannot change. This is the key physical idea and earns the first B1.

The wave equation $v = f\lambda$ then tells you what happens to the wavelength: if $v$ increases and $f$ stays the same, then $\lambda$ must increase. The wave is travelling faster in bone, so its wavelength is longer. This earns the second B1.

Key Takeaways

Frequency is set by the source and never changes when a wave moves between media.
Speed and wavelength change together according to $v = f\lambda$; a greater speed with constant frequency means a longer wavelength.

Common Mistakes

Saying 'frequency increases' because the speed increased — a classic error; frequency is source-determined.
Saying 'wavelength decreases' — that is what would happen if the wave slowed down, not if it speeded up.
Answering in terms of sound changing pitch — the frequency here is fixed.

Things to Be Careful About

The question states the speed in bone is 'greater than 1500 m/s', so the wavelength must increase.
The word 'state' means a short decision is enough — no calculation or explanation is required for the marks, but the reasoning above is what justifies the answers.

Walkthrough

When a magnet is placed in an external magnetic field, its poles experience forces. The North pole of a magnet is always pulled in the direction of the external magnetic field lines. The South pole is always pulled in the opposite direction to the external magnetic field lines.
In Fig. 6.2, the external magnetic field due to the electromagnet is directed from left to right. Therefore:

• The N pole (bottom-left) experiences a horizontal force to the right.
• The S pole (top-right) experiences a horizontal force to the left.
  Draw a horizontal arrow pointing right at the N pole and a horizontal arrow pointing left at the S pole. Both arrows should be parallel to the field lines.

Key Takeaways

In a uniform external magnetic field, the N pole of a magnet is pushed along the field lines and the S pole is pushed against them. This creates a turning effect (couple) if the magnet is not already aligned with the field.

Common Mistakes

Drawing the force on the N pole to the left or the S pole to the right. Drawing the forces at an angle instead of horizontally parallel to the field lines. Drawing forces of different lengths (they should be equal in magnitude for a uniform field).

Things to Be Careful About

The forces are horizontal and parallel to the electromagnet's field lines, not along the length of the bar magnet. The question asks for horizontal forces, so ensure the arrows are strictly left/right.

Walkthrough

1. Forces produce a turning effect: From part (c)(i), there is a force to the right on the N pole and a force to the left on the S pole. These forces are equal in magnitude (uniform field) and opposite in direction, but they do not act along the same line of action because the magnet is tilted. Two equal and opposite forces whose lines of action do not coincide form a couple. A couple produces a pure turning effect (or moment / torque).
2. Direction of rotation: The N pole is pulled to the right and the S pole is pulled to the left. Looking at the diagram, this will cause the magnet to rotate anticlockwise.
3. Final position: The magnet will continue to rotate until it aligns with the external magnetic field. When the magnet is horizontal (parallel to the field lines), the force on the N pole (right) and the force on the S pole (left) will act along the same line through the pivot. At this point, the turning effect is zero, and the magnet is in equilibrium.

Key Takeaways

A magnet in an external magnetic field experiences a couple if it is not aligned with the field. This couple rotates the magnet until it is parallel to the field lines, where the forces act along the same line and the net moment is zero.

Common Mistakes

Saying the magnet 'moves to the right' or 'translates'. The resultant force is zero (equal and opposite forces), so there is no translation, only rotation. Forgetting to mention that the forces form a couple or produce a moment. Not stating the final position (parallel to the field lines). Giving the wrong direction of rotation (clockwise instead of anticlockwise).

Things to Be Careful About

Use precise physics terminology: 'couple', 'moment', 'turning effect', or 'torque'. Explain why there is a turning effect (forces are not along the same line). Describe the motion ('rotates anticlockwise') and the final state ('parallel to the field lines').

Walkthrough

The question provides the power transferred by the lamp ($P = 24\text{ W}$) and the voltage across it ($V = 12\text{ V}$). The relevant equation linking these to current is $P = IV$. Rearranging for current gives $I = P/V$. Substituting the given values yields $I = 24 / 12 = 2.0\text{ A}$. The mark scheme awards one mark for the correct equation or substitution ($P/V$ or $24/12$) and one mark for the final numerical answer with the correct unit (A).

Key Takeaways

The electrical power dissipated or transferred by a component is the product of the current through it and the potential difference across it ($P = IV$). If any two of these three quantities are known, the third can be calculated directly.

Common Mistakes

Candidates sometimes try to find resistance first using an incorrect formula, or they forget to include the unit 'A' in the final answer. The mark scheme explicitly requires the unit for the answer mark.

Things to Be Careful About

Always check that units are consistent: power in watts (W) and voltage in volts (V) will naturally yield current in amperes (A). Do not drop the trailing zero if it is significant to the precision of the data, though '2 A' is often accepted, '2.0 A' matches the mark scheme's '2.0 (A)'.

Walkthrough

Now that the current is known to be $I = 2.0\text{ A}$ and the voltage is $V = 12\text{ V}$, use Ohm's law to find the resistance: $R = V/I$. Substituting the values gives $R = 12 / 2.0 = 6.0\text{ }\Omega$. The mark scheme awards one mark for the correct equation or substitution ($V/I$ or $12/2.0$) and one mark for the final answer with the correct unit (Ω).

Key Takeaways

Ohm's law states that resistance is the ratio of potential difference to current ($R = V/I$). This can be used to find the resistance of a component at a specific operating point.

Common Mistakes

Using the wrong rearrangement, such as $R = IV$ or $R = I/V$. Forgetting the unit 'Ω' is a common error that costs the answer mark.

Things to Be Careful About

Use the current value calculated in part (i). If a candidate made an error in part (i) but used their incorrect current value correctly here, they would still earn the method and answer marks (error carried forward). Ensure the unit is written as 'Ω' or 'ohms'.

Walkthrough

To obtain a range of current and voltage readings for the lamp, the voltage across it must be varied. This is achieved by adding a variable resistor in series with the battery and the lamp. Alternatively, a potential divider circuit could be used. To measure the electrical characteristics, an ammeter is placed in series with the circuit to measure the current flowing through the lamp, and a voltmeter is placed in parallel (across) the lamp to measure the potential difference directly across it.

Key Takeaways

When measuring the I-V characteristics of a component, a variable resistor is used in series to change the current and voltage. The ammeter always goes in series with the component, and the voltmeter always goes in parallel across the component.

Common Mistakes

Placing the voltmeter in series (which would block the current due to its high resistance) or placing the ammeter in parallel (which would short-circuit the component). Forgetting to include a variable resistor or potential divider means the voltage cannot be varied to give a range of readings.

Things to Be Careful About

The mark scheme accepts either a variable resistor in series or a correctly connected potential divider. Ensure the ammeter and voltmeter symbols are correct (circle with 'A' or 'V'). The voltmeter must be connected across the lamp only, not across the whole circuit.

Walkthrough

A filament lamp is a non-ohmic conductor. As the voltage increases, the current increases, but not in a directly proportional way. The graph of current against voltage starts at the origin. Initially, the gradient is steep (low resistance), but as the voltage and current increase, the filament heats up and its resistance increases. This means the current increases more slowly for each additional volt, so the gradient of the curve decreases. The curve bends towards the voltage axis (it is concave down).

Key Takeaways

The I-V graph for a filament lamp is a curve, not a straight line, because its resistance changes with temperature. The decreasing gradient on an I-V graph indicates increasing resistance.

Common Mistakes

Drawing a straight line through the origin (which represents an ohmic conductor like a fixed resistor at constant temperature). Drawing a curve that bends the other way (towards the current axis), which would imply decreasing resistance. Starting the curve away from the origin.

Things to Be Careful About

The gradient of an I-V graph is $1/R$. A decreasing gradient means $1/R$ is decreasing, so $R$ is increasing. Ensure the curve starts exactly at (0,0) and that the gradient is always positive (current never decreases as voltage increases). The curve should pass through or near the point (12, 2.0) based on the data in part (a).

Walkthrough

From the current-voltage graph, the gradient decreases as voltage increases. Since resistance $R = V/I$ (the reciprocal of the gradient on an I-V graph), a decreasing gradient means the resistance is increasing. The physical reason for this is that as current flows through the filament, electrical energy is converted into thermal energy, heating the filament. At the particle level, the metal ions in the lattice vibrate more vigorously as their thermal energy increases. These more vigorous vibrations cause the moving electrons to collide more frequently with the lattice ions, impeding their flow and thus increasing the resistance.

Key Takeaways

The resistance of a metal conductor increases with temperature. At the microscopic level, this is due to increased thermal vibrations of the positive ions in the lattice, which scatter the conduction electrons more effectively.

Common Mistakes

Saying 'resistance increases because voltage increases' (this is circular and doesn't explain the mechanism). Saying 'electrons slow down' (current is the rate of flow of charge; the drift velocity might change, but the fundamental reason is increased collisions). Using the word 'heat' instead of 'thermal energy' or 'temperature' in some contexts, though 'filament gets hot' is generally acceptable.

Things to Be Careful About

The mark scheme requires two distinct points: the statement that resistance increases, and the explanation involving temperature and particle vibration. Both must be present to earn full marks. Ensure the explanation links the increased temperature to increased vibration and then to the impediment of electron flow.

Walkthrough

The equation must conserve both the total nucleon number (top number) and the total proton number (bottom number).

On the left, hydrogen-3 has a nucleon number of 3 and a proton number of 1.

A beta particle is an electron emitted from the nucleus. It is written as $^{0}_{-1}\beta$: it contains no nucleons and its charge is $-1$.

For Q, the top number must be $3 - 0 = 3$. The bottom number must satisfy $x + (-1) = 1$, so $x = 2$. Therefore Q is $^{3}_{2}\text{Q}$.

There are three separate marks: one for the beta particle, one for the top number of Q, and one for the bottom number of Q.

Key Takeaways

• In any radioactive decay equation, the total top numbers must balance and the total bottom numbers must balance.
• A beta particle has nucleon number 0 and proton number $-1$.
• The bottom number is the proton number, which identifies the element.

Common Mistakes

• Writing the beta particle as $^{1}_{-1}\beta$ or $^{0}_{1}\beta$.
• Forgetting that the bottom number of the beta particle is $-1$.
• Putting 3 and 1 on Q without correcting for the beta particle's charge.

Things to Be Careful About

• Use conservation of numbers, not memory, to fill each blank.
• The top number of Q stays 3 because the beta particle contributes 0 nucleons.
• The bottom number of Q becomes 2, which will be used in the next part.

Walkthrough

From the balanced equation, Q has a proton number of 2. Every atom with 2 protons is an atom of the element helium. Its nucleon number is 3, so Q is the isotope helium-3.

Key Takeaways

• The proton number defines the element.
• A nucleus with 2 protons must be helium, whichever isotope it is.

Common Mistakes

• Saying hydrogen because the original nucleus was hydrogen; beta decay changes the proton number.
• Giving the nuclide notation instead of the element name.

Things to Be Careful About

• Make sure you have the correct proton number from part b(i) before naming the element.
• Write 'helium', not 'Q'. Q is only a symbol used in the equation.

Walkthrough

Half-life is a time. The definition needs two ideas: first, that it is the time taken for something to halve; second, that the 'something' is the number of radioactive nuclei or the corrected count rate. Corrected count rate means the measured count rate with background radiation subtracted.

The mark scheme gives one mark for the general idea of 'time for something to halve' and one mark for specifying the number of nuclei or corrected count rate.

Key Takeaways

• Half-life is a time, measured in seconds, hours or years.
• It can be defined using the number of radioactive nuclei or the corrected count rate.
• It does not depend on the initial amount of the sample.

Common Mistakes

• Saying 'time for the mass to halve'; the mass of the sample is not the quantity used in the definition.
• Omitting the time part and only saying 'the count rate halves'.

Things to Be Careful About

• If using count rate, say 'corrected count rate' to show you are allowing for background radiation.
• Include both the 'time' and the 'halving' idea to score both marks.

Walkthrough

The count rates in the table go down, up, then down again: 98, 89, 93, 85. This is not a steady fall, so it cannot be explained simply by the sample running out.

Radioactive decay is random and spontaneous. We cannot predict which nucleus will decay next, and each nucleus decays independently. In one day a larger number of nuclei may happen to decay, and the next day a smaller number may happen to decay. This causes the count rate to fluctuate, increasing on some days and decreasing on others.

Key Takeaways

• Radioactive decay is random and spontaneous.
• Count rates from a radioactive source are not smooth; they fluctuate.
• A single measurement of count rate is not a precise value.

Common Mistakes

• Saying the count rate decreases because the sample is running out; this would only explain a steady decrease, not an increase.
• Blaming background radiation alone; the credited idea here is the random nature of the decay.

Things to Be Careful About

• Use the word 'random' or 'spontaneous' to score the mark.
• Explain that the number of decays in successive equal time intervals is not exactly the same.

Walkthrough

The question asks for two things: the name of the nuclear reaction powering a star, and a description of what happens during it. Inside the core of a main-sequence star, conditions are extremely hot and dense. Hydrogen nuclei (protons) collide at high speed and fuse (join) together to form helium nuclei. This process is called nuclear fusion.

The mark scheme gives one mark for the name “fusion”, one for saying that hydrogen nuclei join/fuse, and one for saying that helium is produced. A full answer therefore is: “Nuclear fusion — hydrogen nuclei fuse together to produce helium.”

Key Takeaways

• A stable star produces its energy by nuclear fusion in its core.
• Fusion joins small nuclei (hydrogen) into a bigger nucleus (helium).

Common Mistakes

• Writing “nuclear fission” — this is the splitting of heavy nuclei and is wrong here.
• Naming the reaction but not describing it — the description carries two of the three marks.
• Forgetting to say that helium is the product.

Things to Be Careful About

• Say “fusion” and then cover both halves of the description: hydrogen nuclei join, and they produce helium. Each half is a separate mark.

Walkthrough

This is an “explain” question, and the mark scheme wants three linked physical ideas, each worth one mark:

1. The fusion reaction transfers energy thermally — it releases energy as heat. (B1)
2. Because the core is so hot, it produces a high pressure / outward force that pushes outwards. (B1)
3. This outward pressure balances the inward gravitational force, preventing the star from collapsing under its own gravity. (B1)

The full chain is: fusion → energy → heat → high outward pressure → gravity pulling in → balanced → no collapse. A star is stable precisely because these two forces balance.

Key Takeaways

• A star is stable when the outward pressure from fusion energy exactly balances the inward pull of gravity.
• This is an equilibrium — both forces must be mentioned to show the balance.

Common Mistakes

• Only saying “the star gets hot” without linking temperature to pressure.
• Only mentioning gravity without the balancing outward force.
• Saying “gravity prevents collapse” on its own, without stating that the outward pressure balances it.

Things to Be Careful About

• The mark scheme wants the word “pressure” or “outward force” from the high temperature, and the word “balanced” for the gravitational force.
• Keep the chain in order: energy → pressure → balance.

Walkthrough

A red supergiant has run out of hydrogen fuel in its core. Without hydrogen, fusion cannot continue, so no more energy is released and the outward pressure drops. The star can no longer hold itself up against gravity. The mark scheme simply wants the reason: not enough hydrogen (left) / runs out of hydrogen.

Key Takeaways

• Fusion needs hydrogen fuel; when the fuel runs out, the balancing outward pressure stops too.

Common Mistakes

• Giving a vague answer like “fusion stops” without saying why — the hydrogen has run out.

Things to Be Careful About

• This is a one-mark question: say clearly that the hydrogen fuel has run out.

Walkthrough

When a red supergiant can no longer support itself, it collapses and then explodes in a supernova. This colossal explosion:

• produces heavy elements (elements heavier than iron are made in supernovae),
• throws out material that forms a nebula (a cloud of gas and dust),
• and leaves behind a neutron star or a black hole in the centre.

Any three of the five listed points earn the three marks. A full answer: “It explodes as a supernova, producing heavy elements and forming a nebula, leaving behind a neutron star or a black hole.”

Key Takeaways

• Massive stars end their lives in a supernova explosion.
• The leftovers are a neutron star (for lighter massive stars) or a black hole (for heavier ones).

Common Mistakes

• Giving the low-mass star ending (white dwarf — the Sun’s fate), which is wrong for a massive star.
• Listing only two points when three are needed for the three marks.

Things to Be Careful About

• Pick three of the five points exactly as in the scheme: explodes, supernova, heavy elements, nebula, neutron star / black hole.
• Either “neutron star” or “black hole” is acceptable as the remnant — the scheme allows both.