Physics 5054/22 — May/June 2025

Walkthrough

The question asks for a distance-time graph to be plotted from the data in Table 1.1. The horizontal axis is time in seconds, and the vertical axis is distance in centimetres. We plot each pair of values from the table as a point on the grid. Because the ball is accelerating, the distance increases by larger amounts each second, so the curve must get steeper as time increases. We draw a smooth curve (not straight line segments) through the points to show the continuous motion.

Key Takeaways

• Graphs must be plotted with points within half a small square of the correct position.
• For accelerating motion, the distance-time graph is a curve with an increasing gradient.

Common Mistakes

• Drawing straight lines between the points instead of a smooth curve.
• Plotting points more than half a small square away from their correct positions.

Things to Be Careful About

• Read the scale carefully: the vertical axis is in cm and the horizontal axis is in s.
• Ensure the curve is smooth and concave down (gradient increasing), representing acceleration.

Walkthrough

Average speed is defined as the total distance travelled divided by the total time taken. From the table, at $t = 0.40\text{ s}$, the distance fallen is $78\text{ cm}$. We substitute these values into the formula to find the average speed over this interval.

Key Takeaways

• Average speed = total distance / total time.
• Ensure units are consistent; here both are in cm and s, giving cm/s.

Common Mistakes

• Using the wrong time or distance value (e.g., using $t = 0.50\text{ s}$ or $d = 123\text{ cm}$).
• Forgetting to include the unit in the final answer.

Things to Be Careful About

• The question asks for the answer in cm/s, so no unit conversion is needed.

Walkthrough

The speed of the ball at a specific instant is the instantaneous speed, which is the gradient of the distance-time graph at that point. Since the graph is a curve, we cannot simply use two points on the curve. Instead, we draw a straight line (tangent) that just touches the curve at $t = 0.40\text{ s}$. The gradient of this tangent gives the instantaneous speed at that moment.

Key Takeaways

• Instantaneous speed is the gradient of the distance-time graph at a specific time.
• For a curved graph, a tangent must be drawn at the point of interest.

Common Mistakes

• Using two points on the curve to find the gradient (this gives average speed, not instantaneous).
• Not drawing the tangent accurately.

Things to Be Careful About

• The gradient of a distance-time graph is speed, not acceleration.

Walkthrough

Two main forces act on the falling ball: its weight (downwards) and air resistance (upwards). Initially, the speed is low, so air resistance is small, and the resultant force is nearly equal to the weight, giving a large acceleration. As the ball speeds up, air resistance increases. This reduces the resultant force ($F_{\text{resultant}} = W - F_{\text{air}}$), and by Newton's second law ($F = ma$), the acceleration decreases. When the air resistance becomes large enough to equal the weight, the resultant force is zero. With no resultant force, there is no acceleration, and the ball falls at a constant speed (terminal velocity).

Key Takeaways

• Weight is constant; air resistance increases with speed.
• Resultant force determines acceleration.
• Zero resultant force means zero acceleration (terminal velocity).

Common Mistakes

• Saying 'gravity decreases' (gravity/weight is constant).
• Not mentioning that air resistance increases with speed.
• Saying 'forces cancel' without specifying which forces.

Things to Be Careful About

• Use the terms 'resultant force' and 'air resistance' or 'drag'.
• Explain the sequence: speed increases -> air resistance increases -> resultant force decreases -> acceleration decreases.

Walkthrough

When acceleration is zero, the speed is constant. On a distance-time graph, constant speed is represented by a straight line with a constant gradient (slope). The line will be straight and sloping upwards (positive gradient) since the ball is still moving downwards (increasing distance).

Key Takeaways

• Zero acceleration means constant speed.
• Constant speed on a distance-time graph is a straight line.

Common Mistakes

• Saying 'horizontal line' (this would mean zero speed, i.e., the ball has stopped).
• Saying 'curve' (this would mean changing speed/acceleration).

Things to Be Careful About

• Ensure the line is described as having a 'constant gradient' or being 'straight', not just 'straight line' without qualification.

Walkthrough

To lift the load at a constant speed, the upward force must equal the weight of the load. The work done is the force multiplied by the distance moved in the direction of the force. Here, work done = weight × height = mgh. Using $g = 9.8 \text{ N/kg}$, the weight is $50 \times 9.8 = 490 \text{ N}$. The work done is $490 \times 2.3 = 1127 \text{ J}$. Rounding to 2 significant figures gives 1100 J, which matches the mark scheme. If $g = 10 \text{ N/kg}$ is used, the work done is $50 \times 10 \times 2.3 = 1150 \text{ J}$.

Key Takeaways

• Work done against gravity equals the gain in gravitational potential energy: $W = mgh$.
• Work is measured in joules (J), mass in kg, $g$ in N/kg (or m/s²), and height in metres.

Common Mistakes

• Forgetting to multiply by $g$ and simply calculating $50 \times 2.3 = 115$.
• Using the wrong value for $g$ or omitting it entirely.

Things to Be Careful About

• The mark scheme accepts 1100 J, which is 1127 J rounded to 2 significant figures. Be consistent with significant figures based on the data given in the question.

Walkthrough

The motor has a useful output power of 600 W, which is the rate at which it does useful work on the load. Power = work done / time. Rearranging gives time = work done / power. Substituting the work done (1127 J) and the output power (600 W) gives a time of approximately 1.88 s, which rounds to 1.9 s to 2 significant figures.

Key Takeaways

• Power relates work done (or energy transferred) to time: $P = W/t$.
• Time is found by rearranging the equation to $t = W/P$.

Common Mistakes

• Using the input power (860 W) instead of the output power (600 W) to calculate the time. The output power is what is actually doing the lifting.
• Carrying forward an incorrect value from part (c)(i) without marking it as an error carried forward (ecf).

Things to Be Careful About

• Always use the useful output power when calculating the time taken to do a specific amount of useful work. The input power includes energy wasted as thermal energy, which does not contribute to lifting the load.

Walkthrough

The question gives the force $F = 1350 \text{ N}$ and the contact area $A = 0.0070 \text{ m}^2$. The pressure $P$ is defined as force per unit area, so we use the equation $P = F / A$. Substituting the values gives $P = 1350 / 0.0070 = 192857 \text{ Pa}$. Rounding to two significant figures (matching the precision of the given data) gives $1.9 \times 10^5 \text{ Pa}$.

Key Takeaways

Pressure is force distributed over an area. Large forces over small areas produce very high pressures, which is why a flat tyre (small contact area) exerts a high pressure on the road.

Common Mistakes

• Forgetting to convert units (not needed here, but common in other pressure questions).
• Giving the answer to too many significant figures (the scheme accepts $1.9 \times 10^5$, not $192857$).
• Using the wrong formula, such as $F \times A$.

Things to Be Careful About

• Always include the unit Pa (pascals) when the question asks for pressure in Pa.
• Use standard form for large numbers: $1.9 \times 10^5$ rather than $190000$.
• The area $0.0070 \text{ m}^2$ is small, so the pressure will be large; check that the order of magnitude makes sense ($10^3 / 10^{-3} = 10^6$, so $10^5$ is correct).

Walkthrough

The question states that the force exerted by the tyre on the road does not change, but the pressure inside increases. From $P = F / A$, if $F$ is constant and $P$ increases, then $A$ must decrease. As the tyre inflates, it becomes rounder and lifts off the road, reducing the contact patch.

Key Takeaways

Pressure and area are inversely proportional for a constant force. Inflating a tyre reduces the contact area with the road.

Common Mistakes

• Saying the area increases (confusing the relationship).
• Not stating that the area decreases simply; adding unnecessary details that aren't credited.

Things to Be Careful About

• The question asks to 'state' how the area changes, so a single clear sentence is sufficient.
• Do not mention the pressure outside the tyre; focus only on the contact area with the road.

Walkthrough

Momentum is a vector quantity defined as the product of an object's mass and its velocity. For a particle of air, it is the mass of the particle multiplied by its velocity.

Key Takeaways

Momentum ($p$) = mass ($m$) $\times$ velocity ($v$). It is a vector, so direction matters.

Common Mistakes

• Defining momentum as 'mass times speed' (speed is scalar, velocity is vector; momentum is a vector).
• Saying 'momentum is the force of a moving object' (confusing momentum with force or energy).

Things to Be Careful About

• Use the word 'velocity', not 'speed', in the definition.
• Do not include units in a definition unless asked.

Walkthrough

When an air particle collides with the tyre wall, it bounces back, so its velocity changes direction. This means its momentum changes. Force is defined as the rate of change of momentum ($F = \Delta p / \Delta t$). Since the particle's momentum changes, a force acts on it. By Newton's third law, the particle exerts an equal and opposite force on the tyre wall. Alternatively, you can say that momentum is transferred to the wall during the collision, and this transfer of momentum over time is a force.

Key Takeaways

Collisions involve a change in momentum. A change in momentum over time is a force. Newton's third law ensures the wall experiences this force.

Common Mistakes

• Saying 'the particle has momentum, so there is a force' (momentum alone doesn't cause force; change in momentum does).
• Forgetting to mention the change in momentum or direction.
• Not linking the force on the particle to the force on the wall (Newton's third law).

Things to Be Careful About

• The question asks to use 'ideas about momentum', so you must mention change in momentum, not just energy or speed.
• Two separate points are needed: (1) change in momentum on collision, and (2) force is related to change in momentum (or Newton's third law).

Walkthrough

Frequency is a fundamental property of any wave. It is defined as the number of complete cycles (oscillations or wavelengths) that pass a fixed point in one second. The unit is the hertz (Hz), where 1 Hz = 1 oscillation per second.

Key Takeaways

• Frequency is the rate of oscillation: number of waves per second.
• Unit is hertz (Hz).

Common Mistakes

• Saying 'number of waves per minute' or 'per hour'. The definition strictly requires 'per second'.
• Confusing frequency with wave speed. Speed is distance per second; frequency is cycles per second.

Things to Be Careful About

• The mark scheme accepts 'number of oscillations per second' or 'number of wavelength(s) passing a point per second'. Both are correct.

Walkthrough

The oscilloscope trace in Fig. 4.2 shows the waveform over a specific time interval. By counting the peaks (or complete cycles) on the grid, we see there are 2.5 complete waves in the 0.050 s time window. Frequency is defined as the number of waves per second. We calculate this by dividing the total number of waves by the total time: $f = 2.5 / 0.050 = 50$ Hz. Alternatively, we can find the time period $T$ (time for one wave) as $0.050 / 2.5 = 0.020$ s, and then use $f = 1/T = 1/0.020 = 50$ Hz.

Key Takeaways

• Reading the number of cycles from an oscilloscope trace over a known time.
• Using $f = \text{number of waves} / \text{time}$ or $f = 1/T$.

Common Mistakes

• Miscounting the number of waves (e.g., counting 3 instead of 2.5 by including the partial wave at the end incorrectly).
• Forgetting to divide by the time in seconds (e.g., $2.5 / 0.05 = 50$, but if time was in ms, conversion is needed).

Things to Be Careful About

• The mark scheme accepts 2.5 waves seen, or 1 wave taking 0.02 s. The final answer must be 50 Hz. The unit Hz is required.

Walkthrough

The amplitude of a sound wave on an oscilloscope represents the maximum displacement of the air particles, which corresponds to the energy of the wave. The energy of a sound wave determines its loudness. As the wave travels away from the source, its energy spreads out and is absorbed by the medium, causing the amplitude to decrease. A smaller amplitude means a quieter sound.

Key Takeaways

• Amplitude of a sound wave is linked to its loudness.
• Decreasing amplitude means decreasing loudness.

Common Mistakes

• Saying the pitch changes (pitch is linked to frequency, not amplitude).
• Saying the speed changes.

Things to Be Careful About

• Use words like 'quieter', 'less loud', or 'lower volume'. Avoid saying 'the sound gets smaller' without context.

Walkthrough

The original trace in Fig. 4.3 shows 2.5 waves across the grid. Half the frequency means the wave takes twice as long to complete one cycle. Therefore, the period is doubled. If the original wave took 2 horizontal grid squares for one complete cycle, the new wave must take 4 horizontal grid squares. The amplitude (height of the wave) remains the same because the tuning fork's vibration amplitude hasn't changed, only its frequency. The new trace should be drawn as a solid line, showing at least one full cycle to demonstrate the new period.

Key Takeaways

• Frequency and period are inversely proportional: $f = 1/T$. Half the frequency means double the period.
• On an oscilloscope with fixed time-base settings, a lower frequency wave is more 'stretched out' horizontally.

Common Mistakes

• Drawing a wave with half the amplitude instead of half the frequency.
• Drawing a wave with the same period but different frequency (impossible on a fixed grid).
• Not drawing at least one complete oscillation.

Things to Be Careful About

• The mark scheme requires 'half frequency shown' AND 'at least one complete oscillation seen'. Ensure the wavelength (horizontal spacing) is clearly double that of the original dashed trace.

Walkthrough

When plane wavefronts pass through a gap that is roughly the same size as the wavelength, they spread out or diffract. The diffracted wavefronts become curved (circular or semi-circular) and appear to originate from the gap. On the oscilloscope or ray diagram, we draw these as curved lines. The key requirements are: (1) the wavefronts must be curved and centered on the gap, (2) the wavelength (the perpendicular distance between adjacent wavefronts) must remain the same as on the incident side, and (3) there must be at least three wavefronts shown to represent the ongoing wave. Point P is located behind the wall, in the 'shadow' region, but because of diffraction, the sound waves spread into this region and reach P.

Key Takeaways

• Diffraction causes plane wavefronts to become curved after passing through a gap.
• The wavelength does not change during diffraction.
• The wavefronts spread into the region behind the obstacle.

Common Mistakes

• Drawing straight wavefronts after the gap (this would be no diffraction).
• Drawing wavefronts that are closer together or further apart (wavelength must be constant).
• Drawing the center of curvature away from the gap.

Things to Be Careful About

• The mark scheme awards marks for '(at least) three correctly curved wavefronts centered on gap' and 'same wavelength as on left and some curving in correct direction'. Ensure the curves are smooth and clearly centered on the door opening.

Walkthrough

The spreading out of waves as they pass through a gap or around an obstacle is called diffraction. In this case, the sound wavefronts spread out after passing through the open door, allowing the sound to reach point P which is not in a straight line from the source.

Key Takeaways

• Diffraction is the spreading of waves through a gap or around an edge.
• Significant diffraction occurs when the gap size is comparable to the wavelength.

Common Mistakes

• Saying 'refraction' (which is bending due to a change in speed/medium).
• Saying 'reflection' (bouncing back).
• Saying 'interference' (superposition of waves).

Things to Be Careful About

• The answer is simply 'diffraction'. No extra words are needed, though 'wave diffraction' is also acceptable.

Walkthrough

To find the resistance of the 9.0 cm wire, use Ohm's law in the form $R = V/I$. Pick a point on line P where the grid lines cross clearly. At $V = 6.0$ V, the current is $I = 0.8$ A. Substitute these values into the equation:

Any other valid point on line P (e.g., $V = 3.0$ V, $I = 0.4$ A) will give the same result.

Key Takeaways

• Resistance is the ratio of voltage to current ($R = V/I$), not the gradient of an I-V graph (which is $1/R$).
• Always read values from a graph where the lines cross grid intersections clearly to minimise reading errors.

Common Mistakes

• Calculating the gradient of the line ($I/V = 0.8/6.0 = 0.133$) and forgetting to invert it to get resistance.
• Reading the wrong line or misreading the axes (e.g., reading current as 0.6 A at 6.0 V).

Things to Be Careful About

• The axes are current (y-axis) against voltage (x-axis). The resistance is $V/I$, which is the reciprocal of the gradient. Ensure units are in volts and amperes to get resistance in ohms ($\Omega$).

Walkthrough

Compare lines P and Q on the current-voltage graph. At any given voltage (e.g., $V = 4.0$ V), line Q has a higher current ($0.8$ A) than line P ($\approx 0.53$ A). By Ohm's law ($R = V/I$), a higher current for the same voltage means a lower resistance. Therefore, the wire for line Q has a smaller resistance than the 9.0 cm wire.

For a uniform wire of the same material and cross-sectional area, resistance is directly proportional to length ($R \propto L$). A smaller resistance must therefore correspond to a shorter length of wire.

Key Takeaways

• On an I-V graph with current on the y-axis, a steeper line represents a lower resistance.
• Resistance of a wire is directly proportional to its length, assuming constant cross-sectional area and material.

Common Mistakes

• Stating "longer" because line Q is steeper, without linking the slope to resistance correctly.
• Forgetting to state that the resistance is smaller or that resistance is proportional to length; the mark scheme requires this reasoning to justify "shorter".

Things to Be Careful About

• Ensure the explanation explicitly links the graph observation (larger current / smaller resistance) to the physical property (shorter length). Just saying "shorter" without reasoning may not earn the mark if the question asks to explain.

Walkthrough

First, calculate the resistance of the wire for line Q using a point on the graph. At $V = 4.0$ V, $I = 0.8$ A:

We already found $R_P = 7.5 \text{ } \Omega$ for $L_P = 9.0$ cm.

Since resistance is directly proportional to length for the same wire:

Key Takeaways

• Resistance is directly proportional to length: $R = \rho L / A$, so $R_1 / L_1 = R_2 / L_2$.
• Always calculate the resistance for the unknown case before using the ratio.

Common Mistakes

• Assuming the length is proportional to the current directly, giving $9.0 \times (0.8/0.8) = 9.0$ cm.
• Using the gradient of the graph ($I/V$) directly in the ratio without inverting to get resistance.

Things to Be Careful About

• The answer must be in cm, matching the unit given in the question. Give the answer to an appropriate number of significant figures (6.0 cm, not just 6, to match the precision of 9.0 cm).

Walkthrough

In a parallel circuit, the total current supplied is equal to the sum of the currents in each individual branch. The lighting circuit has two lamps of type P and one lamp of type Q. From Table 6.1, the current for type P is 0.26 A and for type Q is 0.43 A. Therefore, the total current is $2 \times 0.26 + 0.43 = 0.95$ A. This total current flows through the lighting circuit fuse.

Key Takeaways

• In a parallel circuit, the total current is the sum of the currents in the separate branches.
• The current in the main supply line (and thus through the main fuse) is greater than the current in any single branch.

Common Mistakes

• Adding only one type P lamp current to the type Q current (giving 0.69 A) and forgetting there are two type P lamps.
• Multiplying all three currents together instead of adding them.

Things to Be Careful About

• Ensure you account for all lamps in the circuit. The diagram shows two type P lamps and one type Q lamp.

Walkthrough

First, calculate the current already used by the lamps left in the circuit. Lamp Q is removed, leaving two type P lamps. Their combined current is $2 \times 0.26 = 0.52$ A. The fuse rating is 5 A, which is the maximum current the circuit can safely carry. Subtract the used current from the maximum to find the available current: $5 - 0.52 = 4.48$ A. Divide this available current by the current drawn by one additional type P lamp (0.26 A): $4.48 / 0.26 \approx 17.23$. Because you cannot add a fraction of a lamp, you must round down to 17. If you added 18 lamps, the total current would be $0.52 + (18 \times 0.26) = 5.20$ A, which exceeds the 5 A fuse rating and would cause it to blow.

Key Takeaways

• The total current in a parallel circuit cannot exceed the rating of the main fuse.
• When calculating how many additional components can be added, always round down to the nearest whole number to ensure the total current stays within the safe limit.

Common Mistakes

• Rounding 17.23 up to 18, which would blow the fuse.
• Forgetting to subtract the current already used by the two type P lamps before calculating the remaining capacity.
• Using the total current of all three original lamps (0.95 A) instead of just the two type P lamps (0.52 A) after removing lamp Q.

Things to Be Careful About

• Always check your final answer by calculating the total current with your chosen number of additional lamps to ensure it does not exceed the fuse rating.

Walkthrough

A moment is calculated as force multiplied by the perpendicular distance from the pivot (axis of rotation) to the line of action of the force.

Here, there are two forces, each of 2.3 N, acting on opposite sides of the coil. Both forces create a turning effect in the same direction (clockwise or anti-clockwise, depending on perspective, but they add together).

• Moment from left force = $2.3 \text{ N} \times 0.15 \text{ m} = 0.345 \text{ Nm}$
• Moment from right force = $2.3 \text{ N} \times 0.15 \text{ m} = 0.345 \text{ Nm}$
• Total moment = $0.345 + 0.345 = 0.69 \text{ Nm}$

Alternatively, you can write: $\text{total moment} = 2 \times \text{force} \times \text{distance} = 2 \times 2.3 \times 0.15 = 0.69 \text{ Nm}$.

The unit of moment is the newton-metre (Nm).

Key Takeaways

• The turning effect (moment) of a force is $M = F \times d$, where $d$ is the perpendicular distance from the pivot.
• When multiple forces act to produce rotation in the same direction, their moments add together.

Common Mistakes

• Forgetting to include both forces and only calculating the moment for one side ($0.345 \text{ Nm}$). The question asks for the total moment.
• Using the full width of the coil (0.30 m) with one of the forces instead of using the distance from the axis (0.15 m) for each force.
• Forgetting the unit. The mark scheme awards a separate mark for 'Nm'.

Things to Be Careful About

• Ensure you use the correct distance: 0.15 m is the distance from the axis to each side, not the total width. The mark scheme accepts force x distance as a method mark, so showing the formula before substituting is important.

Walkthrough

The moment of a force depends on the perpendicular distance from the pivot to the line of action of the force. When the coil is horizontal (as in Fig. 7.1), the forces are vertical, and the perpendicular distance from the axis to each force is the full 0.15 m.

As the coil rotates and becomes more vertical, the sides of the coil move closer to the vertical line passing through the axis. The forces (which remain vertical, assuming a uniform horizontal magnetic field) now have a line of action that is closer to the axis. The perpendicular distance from the axis to the line of action of the force decreases.

Since $\text{moment} = F \times d_{\perp}$, and $d_{\perp}$ is decreasing while $F$ remains constant, the moment (turning effect) decreases.

When the coil is exactly vertical, the line of action of both forces passes directly through the axis, the perpendicular distance is zero, and the moment is zero. This is why a d.c. motor needs momentum to carry it past this 'dead point'.

Key Takeaways

• The turning effect is maximized when the coil is horizontal (perpendicular to the magnetic field) and zero when the coil is vertical (parallel to the magnetic field).
• It is the perpendicular distance from the pivot to the line of action of the force that matters, not the physical length of the coil arm.

Common Mistakes

• Saying 'the force becomes smaller' — the magnetic force on a current-carrying conductor in a uniform field is constant regardless of orientation (as long as the current and field are unchanged). It is the distance that changes.
• Saying 'the magnetic field becomes weaker' — the field is uniform between the pole faces.

Things to Be Careful About

• Use the precise phrase 'perpendicular distance' or 'distance between the line of action of the force and the pivot'. Simply saying 'the distance decreases' may not be precise enough if not qualified correctly. The mark scheme accepts '(perpendicular) distance between (line of action of) forces decreases' or 'coil leaves magnetic field' as alternatives.

Walkthrough

The question asks for the definition of specific heat capacity. This is a standard thermal physics definition.

Specific heat capacity ($c$) is defined as the amount of thermal energy required to raise the temperature of a unit mass (1 kg) of a substance by one unit of temperature (1 K or 1 °C). The equation linking these quantities is $E = mc\Delta\theta$, where $E$ is energy, $m$ is mass, $c$ is specific heat capacity, and $\Delta\theta$ is the temperature change.

Key Takeaways

Specific heat capacity is a material property that indicates how much energy is needed to change its temperature. A high specific heat capacity means the substance can absorb a lot of energy with only a small temperature rise.

Common Mistakes

• Forgetting to specify "per unit mass" (that makes it specific heat capacity, as opposed to just heat capacity).
• Forgetting to specify "per unit temperature rise" (that makes it specific heat capacity, as opposed to just energy).
• Saying "heat" instead of "thermal energy" or "energy" (5054 prefers "energy" or "thermal energy" over the informal "heat").

Things to Be Careful About

Ensure both conditions are met: unit mass AND unit temperature rise. The mark scheme awards one mark for the energy/temperature part (M1) and one for the unit mass part (A1).

Walkthrough

The question asks whether the coolant should have a large or small specific heat capacity and requires an explanation.

From the definition, a substance with a large specific heat capacity can absorb a large amount of thermal energy for a given temperature rise ($\Delta\theta = \frac{E}{mc}$). In a nuclear reactor, the coolant's job is to remove the large amount of thermal energy generated by fission without itself getting too hot. If the coolant had a small specific heat capacity, it would heat up very quickly, potentially reaching its boiling point and losing its effectiveness as a liquid coolant. A large specific heat capacity ensures the temperature rise is kept small and manageable for a given amount of energy absorbed.

Key Takeaways

A coolant with a high specific heat capacity is ideal for transferring large amounts of thermal energy while maintaining a relatively low and stable temperature.

Common Mistakes

• Saying "large specific heat capacity because it heats up faster" (this is the opposite; a large specific heat capacity means it heats up slower for the same energy input).
• Not providing a valid reason linked to the definition (e.g., "because it is better" without explaining the physics).
• Suggesting a small specific heat capacity and claiming it is better because it heats up faster (while this is physically true, it is a poor design choice for a reactor coolant as it risks boiling and does not absorb as much total energy before reaching a dangerous temperature).

Things to Be Careful About

The explanation must link the chosen property (large specific heat capacity) to a physical consequence (smaller temperature rise, more energy absorbed, less chance of boiling). Any one of these valid consequences is sufficient for the mark.

Walkthrough

Any object orbiting a massive body like the Sun is kept in its curved path by the gravitational attraction between the two masses. This is the universal force of gravity.

Key Takeaways

Gravity is the centripetal force that keeps planets, comets, and moons in orbit.

Common Mistakes

Saying 'centrifugal force' (which is a fictitious force in a rotating frame) or 'solar wind'. The correct term is gravitational force or gravity.

Things to Be Careful About

Accept 'gravity' or 'gravitational attraction'. Be precise and say 'gravitational force' rather than just 'force'.

Walkthrough

The gravitational force is always attractive and acts along the line connecting the two masses. Therefore, at any point on the comet's orbit, the force exerted by the Sun on the comet is directed straight towards the Sun's center. Draw an arrow at positions 1, 2, 3, and 4 pointing radially inward to the Sun.

Key Takeaways

Gravitational force is always attractive and acts along the line joining the centers of mass.

Common Mistakes

Drawing arrows tangent to the path (which would represent velocity, not force) or arrows pointing away from the Sun.

Things to Be Careful About

The arrows should be drawn by eye but must clearly point towards the Sun. They don't need to be perfectly to scale, but the direction must be correct.

Walkthrough

The gravitational force between two objects depends on the distance between them: the closer they are, the stronger the force. As the comet moves from position 1 (far away) towards position 3 (closest approach to the Sun), the distance decreases, so the gravitational force increases. After position 3, as it moves towards position 4 (moving away), the distance increases, so the force decreases.

Key Takeaways

Gravitational field strength and the resulting force increase as distance to the massive body decreases.

Common Mistakes

Saying the force is constant, or confusing the comet's speed with the force. The explanation must explicitly mention the changing distance to the Sun.

Things to Be Careful About

The description and explanation are separate marks. Give a clear description of the trend (increase then decrease) and then explain why using the relationship between force and distance.