Physics 5054/41 — October/November 2025

Walkthrough

The measuring cylinder has major markings every 10 cm³ and minor markings every 1 cm³. The bottom of the meniscus (the curved surface of the liquid) is read at eye level. In the magnified inset, the meniscus bottom rests at the second small division above the 50 cm³ mark, giving a reading of 52 cm³. Since the scale allows reading to the nearest 1 cm³, we record it as 52.0 cm³ to show the precision of the instrument.

Key Takeaways

• Always read the bottom of the meniscus for water and aqueous solutions.
• Record the reading to the precision of the smallest division on the instrument.

Common Mistakes

• Reading the top of the meniscus instead of the bottom.
• Failing to include the trailing zero to indicate the precision of the measuring cylinder (e.g., writing 52 instead of 52.0).

Things to Be Careful About

• Ensure the eye is level with the meniscus to avoid parallax error, though in a printed diagram we simply read the value shown. The mark scheme accepts 52 or 52.0.

Walkthrough

The voltmeter scale runs from 0 to 5 V with major markings every 1 V and minor markings every 0.1 V. The needle points to the fourth small division past 3, giving 3.4 V. We record this as 3.40 V to show precision to two decimal places.
The ammeter scale runs from 0 to 1.5 A with major markings every 0.5 A and minor markings every 0.05 A. The needle points exactly halfway between 1.4 and 1.5, giving 1.45 A.

Key Takeaways

• Read analogue meters to the nearest small division.
• Include trailing zeros if they indicate the precision of the scale.

Common Mistakes

• Misreading the ammeter scale by confusing the 0.5 A and 1.0 A major markings.
• Not recording the voltmeter reading to the correct number of decimal places.

Things to Be Careful About

• The mark scheme accepts 3.4 or 3.40 for the voltage, and 1.45 for the current. Ensure you read the correct meter for V and I.

Walkthrough

The thermometer has major markings every 10 °C and minor markings every 1 °C. The liquid column level rests one small division below the 20 °C mark in the magnified view, giving a reading of 21 °C. We record this as 21.0 °C to indicate the precision of the thermometer.

Key Takeaways

• Thermometers are read to the nearest degree or half-degree depending on the scale.
• Recording with a trailing zero shows awareness of instrument precision.

Common Mistakes

• Reading the scale from top to bottom instead of bottom to top.
• Miscounting the small divisions between major markings.

Things to Be Careful About

• The mark scheme accepts 21 or 21.0. Ensure you read the bottom of the liquid column if it has a meniscus, though liquid-in-glass thermometers typically have a flat top.

Walkthrough

The temperature change is the difference between the final temperature and the initial temperature. Using the recorded values: $\theta_F = 24.5 ^\circ\text{C}$ and $\theta_0 = 21.0 ^\circ\text{C}$. Substituting into the given equation gives $\Delta\theta = 24.5 - 21.0 = 3.5 ^\circ\text{C}$.

Key Takeaways

• Temperature change is always final minus initial.
• Ensure both temperatures are in the same units before subtracting.

Common Mistakes

• Subtracting in the wrong order (initial minus final), which would give a negative value.
• Forgetting to include the unit in the final answer.

Things to Be Careful About

• Use the value of $\theta_0$ you recorded in part (c)(i), not a different value from the image description if it differs from the mark scheme.

Walkthrough

When the heater is switched off, the water near the coil is still hotter than the water further away. Stirring the water continues to mix the hot and cold regions, allowing the temperature to become uniform throughout the beaker. This ensures the thermometer measures the true average final temperature of the water, which is necessary for an accurate calculation of the energy gained.

Key Takeaways

• Stirring promotes convection and ensures thermal equilibrium within a liquid.
• Without stirring, temperature gradients would lead to an inaccurate final temperature reading.

Common Mistakes

• Saying 'to cool the water down' or 'to mix the water' without explaining the purpose (uniform temperature or complete heat absorption).

Things to Be Careful About

• The mark scheme specifically looks for 'uniform temperature' or 'all heat absorbed'. Avoid vague answers like 'to make it mixed'.

Walkthrough

The electrical energy supplied by the heater is calculated using the formula $Q_H = I \times V \times t$. The current $I = 1.45 \text{ A}$, the p.d. $V = 3.40 \text{ V}$, and the time $t = 300 \text{ s}$ (5 minutes). Substituting these values gives $Q_H = 1.45 \times 3.40 \times 300 = 1479 \text{ J}$. The mark scheme accepts values around 1480 J depending on rounding during intermediate steps.

Key Takeaways

• Electrical energy can be calculated as the product of current, p.d., and time.
• Time must be in seconds for the result to be in joules.

Common Mistakes

• Forgetting to convert minutes to seconds (using 5 instead of 300).
• Multiplying the wrong values or using incorrect readings from the meters.

Things to Be Careful About

• The mark scheme allows 1480 or 1500 J if error carried forward is applied from earlier parts. Use the exact values recorded in parts (b)(ii).

Walkthrough

The thermal energy gained by the water is calculated using the formula $Q_W = W \times 4.2 \times \Delta\theta$. Here, $W$ is the mass of the water in grams (since the specific heat capacity of water is given as $4.2 \text{ J/(g}^\circ\text{C)}$ implicitly by the formula using volume in cm³). $W = 52.0 \text{ g}$ (since $1 \text{ cm}^3$ of water has a mass of $1 \text{ g}$), and $\Delta\theta = 3.5 ^\circ\text{C}$. Substituting gives $Q_W = 52.0 \times 4.2 \times 3.5 = 764.4 \text{ J}$. The mark scheme accepts 764 J.

Key Takeaways

• The formula $Q = mc\Delta\theta$ is used, where $c = 4.2 \text{ J/(g}^\circ\text{C)}$ for water when mass is in grams.
• The volume in cm³ is numerically equal to the mass in grams for water.

Common Mistakes

• Converting the mass to kilograms and then not adjusting the specific heat capacity (which would require $c = 4200 \text{ J/(kg}^\circ\text{C)}$).
• Using the wrong value for $W$ or $\Delta\theta$ from previous parts.

Things to Be Careful About

• The equation provided in the question uses $W$ directly in cm³, implying the specific heat capacity is $4.2 \text{ J/(cm}^3\cdot^\circ\text{C)}$ or that 1 cm³ = 1 g. Do not convert to kg unless you also change 4.2 to 4200.

Walkthrough

Efficiency is the ratio of useful energy output to total energy input. Here, the useful energy is the thermal energy gained by the water ($Q_W = 764.4 \text{ J}$), and the total energy input is the electrical energy supplied by the heater ($Q_H = 1479 \text{ J}$). Dividing these gives $\text{efficiency} = 764.4 / 1479 \approx 0.517$. This can also be expressed as $51.7\%$. The mark scheme expects a value between 0.51 and 0.52.

Key Takeaways

• Efficiency is always a ratio of output to input and is dimensionless.
• It can be expressed as a decimal or a percentage.

Common Mistakes

• Inverting the ratio (calculating $Q_H / Q_W$ instead of $Q_W / Q_H$).
• Forgetting to use the values calculated in the previous parts.

Things to Be Careful About

• The mark scheme allows error carried forward from previous parts. If a student used slightly different values for $Q_W$ and $Q_H$, an efficiency between 0.51 and 0.52 is still accepted.

Walkthrough

In optics experiments, the normal is defined as an imaginary reference line drawn perpendicular (at $90^\circ$) to the refracting surface at the point of incidence.

To construct the normal:

1. Place the centre mark of a protractor at point $M$ on the line $XY$.
2. Align the base line of the protractor precisely along the line $XY$.
3. Mark the $90^\circ$ mark on the paper.
4. Using a sharp pencil and ruler, draw a continuous straight line (or dashed line) passing through point $M$ and the $90^\circ$ mark, extending it well above the block into the air and below the top surface inside the block.

Key Takeaways

• A normal is always at right angles ($90^\circ$) to the boundary surface.
• Extending the normal both above and below the surface allows angles of incidence and refraction to be measured directly.

Common Mistakes

• Drawing a line that is not perpendicular to $XY$.
• Forgetting to extend the line below $XY$ into the block.

Things to Be Careful About

• Use a sharp pencil and ensure exact alignment with point $M$ to maintain accuracy within $\pm 1^\circ$.

Walkthrough

The angle of incidence $i$ is measured between the incident ray and the normal, NOT between the ray and the surface of the block.

1. Place the protractor at point $M$, aligned along the normal line constructed in (a)(i).
2. Measure an angle of $40^\circ$ to the left of the normal, above the line $XY$.
3. Draw a straight line from $M$ through this $40^\circ$ mark towards the top left.
4. Label the far end of this line as $L$.

Key Takeaways

• The angle of incidence is always measured relative to the normal line, not the surface boundary.

Common Mistakes

• Measuring $40^\circ$ from the horizontal surface $XY$ instead of from the vertical normal.
• Drawing the ray to the right of the normal instead of the left.

Things to Be Careful About

• Ensure the line is drawn cleanly to within the mark scheme tolerance of $\pm 1^\circ$.

Walkthrough

The two crosses represent points along the path of the emerging ray of light after it leaves the block.

1. Lay a ruler across both crosses marked below the block.
2. Draw a single straight line through both crosses.
3. Extend this line upwards until it intersects the lower horizontal boundary of the block.
4. Label this point of intersection on the bottom edge as point $P$.

Key Takeaways

• Light travels in straight lines in a uniform medium (air), so the path outside the block is a straight line through the marked points.

Common Mistakes

• Stopping the line at the upper cross rather than extending it to meet the bottom edge of the block.

Things to Be Careful About

• Use a sharp pencil to ensure the line passes precisely through the centre of each cross.

Walkthrough

Point $M$ is the point where the incident ray enters the block, and point $P$ is the point where the ray emerges from the block. Since the block is made of a uniform transparent material, light travels in a straight line between $M$ and $P$.

1. Place the ruler between point $M$ on the top boundary and point $P$ on the bottom boundary.
2. Draw a straight line inside the block connecting $M$ to $P$.

Key Takeaways

• The line $MP$ represents the refracted ray inside the transparent block.

Common Mistakes

• Drawing a curved line or missing either point $M$ or $P$.

Things to Be Careful About

• Ensure the line starts exactly at $M$ and terminates cleanly at $P$.

Walkthrough

The angle of refraction $r$ is defined as the angle between the normal line (extended downwards into the block from $M$) and the refracted ray $MP$.

1. Place the centre point of the protractor at point $M$.
2. Align the base line with the normal line extending downwards from $M$.
3. Read the angle to the refracted line $MP$.
4. The measured value is $25^\circ$ (acceptable range: $24^\circ$ to $26^\circ$).

Key Takeaways

• Angle of refraction $r$ is always measured between the normal and the refracted ray inside the medium.

Common Mistakes

• Measuring the angle between the line $MP$ and the surface $XY$ instead of the normal.

Things to Be Careful About

• Read the scale carefully to the nearest degree.

Walkthrough

To find the mass of the 5 marbles alone, subtract the mass of the empty dish (tare) from the combined mass of the dish and the 5 marbles:

1. From Fig. 3.1, the reading on the top pan balance for the empty dish is $4.2\ \text{g}$.
2. From Fig. 3.2, the reading with the 5 marbles added to the dish is $29.6\ \text{g}$.
3. Calculate the difference:

Key Takeaways

• When measuring mass using a container, the mass of the object is found by subtracting the mass of the empty container from the total mass.
• Maintain consistent precision matching the instrument readings (here to 1 decimal place).

Common Mistakes

• Forgetting to subtract the dish mass and giving $29.6\ \text{g}$ as the answer.
• Adding the two readings instead of subtracting them.

Things to Be Careful About

• Ensure the unit matches the blank (the blank provides $\text{g}$, so the numerical answer is 25.4).

Walkthrough

To calculate the average mass of a single marble, divide the total mass of the 5 marbles found in part (a)(i) by 5:

This can be rounded to 2 significant figures as $5.1\ \text{g}$ or given to 3 significant figures as $5.08\ \text{g}$. Both are accepted by the mark scheme.

Key Takeaways

• Measuring multiple identical items and dividing by the count improves accuracy by reducing the relative effect of reading uncertainty.

Common Mistakes

• Dividing the total mass of the dish and marbles ($29.6\ \text{g}$) instead of the net mass of the marbles ($25.4\ \text{g}$).
• Arithmetic errors when dividing by 5.

Things to Be Careful About

• Always carry forward your answer from the previous step correctly.

Walkthrough

The average time $t_{av}$ is the arithmetic mean of the three repeated measurements $t_1$, $t_2$, and $t_3$:

1. Sum the three values:

1. Divide by the total number of readings (3):

1. Express to a sensible precision consistent with the data (2 decimal places):

Key Takeaways

• Taking repeated measurements and calculating a mean reduces the effect of random timing errors.
• The calculated mean should generally be quoted to the same number of decimal places as the original readings.

Common Mistakes

• Leaving an unrounded recurring decimal like $2.21333...\ \text{s}$.
• Forgetting one of the three readings when summing.

Things to Be Careful About

• Check calculator inputs carefully to avoid typo errors.

Walkthrough

1. First, fill in the first row for $h = 4.0\ \text{cm}$ using the values given in part (b)(i):

   • $t_1 = 2.16\ \text{s}$
   • $t_2 = 2.23\ \text{s}$
   • $t_3 = 2.25\ \text{s}$
   • $t_{av} = 2.21\ \text{s}$

2. Compute $t_{av}$ for each subsequent row by taking the mean of $t_1, t_2, t_3$:

   • For $h = 6.0\ \text{cm}$: $\frac{2.04 + 1.94 + 1.91}{3} = \frac{5.89}{3} = 1.9633... \rightarrow 1.96\ \text{s}$
   • For $h = 8.0\ \text{cm}$: $\frac{1.75 + 1.82 + 1.70}{3} = \frac{5.27}{3} = 1.7566... \rightarrow 1.76\ \text{s}$
   • For $h = 10.0\ \text{cm}$: $\frac{1.46 + 1.42 + 1.38}{3} = \frac{4.26}{3} = 1.42\ \text{s}$
   • For $h = 12.0\ \text{cm}$: $\frac{1.36 + 1.23 + 1.31}{3} = \frac{3.90}{3} = 1.30\ \text{s}$

3. Ensure all values in the $t_{av}$ column are recorded to a consistent number of decimal places (2 decimal places, including trailing zero in $1.30$).

Key Takeaways

• Data in any given column of a table must have a consistent number of decimal places.
• A zero at the end of a decimal (like 1.30) must not be omitted when keeping 2 decimal places.

Common Mistakes

• Omitting the trailing zero and writing 1.3 instead of 1.30.
• Forgetting to copy the readings for $h = 4.0\ \text{cm}$ into the first row.

Things to Be Careful About

• Rounding each average correctly: $1.963...$ rounds to $1.96$, and $1.756...$ rounds to $1.76$.

Walkthrough

To score full marks on the graph:

1. Axes and Labels (Mark 1):

   • Label the horizontal axis with quantity and unit: $h / \text{cm}$.
   • Label the vertical axis with quantity and unit: $t_{av} / \text{s}$.

2. Scales (Mark 2):

   • Choose scales that are easy to use (multiples of 1, 2, or 5 per major division) and occupy more than half the grid in both directions.
   • For example:
     • $x$-axis: $2\ \text{cm}$ per major grid square (ranging from $0$ to $14\ \text{cm}$, or $4$ to $14\ \text{cm}$).
     • $y$-axis: $0.2\ \text{s}$ per major grid square (ranging from $1.0\ \text{s}$ to $2.4\ \text{s}$).
   • Avoid awkward scales such as multiples of 3 or 7.

3. Plotting (Mark 3):

   • Plot each of the five coordinates with a small neat cross $(\times)$ or dot in a circle $(\odot)$ to within $\pm 0.5$ small square:
     • $(4.0, 2.21)$
     • $(6.0, 1.96)$
     • $(8.0, 1.76)$
     • $(10.0, 1.42)$
     • $(12.0, 1.30)$

4. Line of Best Fit (Mark 4):

   • Use a sharp pencil and a clear ruler to draw a single, thin, continuous straight line.
   • Ensure the points are evenly distributed above and below the line with no kinks or thick 'hairy' lines.

Key Takeaways

• Good graph practice requires: clear axis labels with units, linear sensible scales covering $\ge 50\%$ of the grid, accurate point plotting, and a well-balanced line of best fit.

Common Mistakes

• Swapping the $x$ and $y$ axes.
• Choosing awkward scales where each small square represents an inconvenient fraction.
• Drawing a 'connect-the-dots' zigzag line instead of a single best-fit line.
• Drawing a thick or double line.

Things to Be Careful About

• Points must be plotted to within half a small grid square.

Walkthrough

Looking at the data in Table 3.1 or the plotted graph:

• When the ramp height $h$ increases from $4.0\ \text{cm}$ to $12.0\ \text{cm}$, the average time $t_{av}$ taken for the marble to roll down decreases from $2.21\ \text{s}$ to $1.30\ \text{s}$.
• Therefore, the relationship is: as $h$ increases, $t_{av}$ decreases.

Key Takeaways

• To describe a relationship between two variables, state clearly how one changes as the other increases.

Common Mistakes

• Stating that $h$ and $t_{av}$ are 'inversely proportional' without proving that $h \times t_{av} = \text{constant}$ (which is not true here, as the line does not represent an inverse proportion with a zero intercept).

Things to Be Careful About

• Use clear comparative language: 'as $h$ increases, $t_{av}$ decreases'.

Walkthrough

1. Locate $h = 7.0\ \text{cm}$ on the horizontal axis ($x$-axis).
2. Draw a dashed construction line vertically up from $7.0\ \text{cm}$ until it hits the straight line of best fit.
3. Draw a horizontal dashed line from that point to the vertical axis ($y$-axis).
4. Read the corresponding value of $t_{av}$ on the vertical scale. For a well-drawn line of best fit, this reading is approximately $1.86\ \text{s}$ (accepted range typically $1.84\text{--}1.88\ \text{s}$).

Both the visible construction lines on the graph and the correctly read value earn 1 mark each.

Key Takeaways

• 'Show on the graph how you find...' requires clear construction lines (usually dashed with arrows) from the axis to the line and across to the other axis.

Common Mistakes

• Forgetting to draw the construction lines on the graph (costs 1 mark).
• Calculating a value mathematically instead of reading it directly off the drawn line of best fit.

Things to Be Careful About

• Read the scale carefully to half a small square precision.

Walkthrough

1. Calculate the speed:
   Use the formula provided:

   $$v = \frac{0.90}{t_{av}}$$

   Substitute the value of $t_{av}$ obtained from part (c)(i) (e.g., $1.86\ \text{s}$):

   $$v = \frac{0.90}{1.86} \approx 0.484\ \text{m/s}$$

   Acceptable values in the mark scheme range from $0.48$ to $0.49$ inclusive (or ecf from the candidate's graph reading).

2. Identify the unit:
   The numerator is distance in metres ($0.90\ \text{m}$, representing $90.0\ \text{cm}$) and the denominator is time in seconds ($\text{s}$).
   Therefore, the unit is metres per second, written as $\text{m/s}$.

Key Takeaways

• Speed is calculated as distance divided by time: $v = \frac{d}{t}$.
• When distance is in metres and time is in seconds, the unit of speed is $\text{m/s}$.

Common Mistakes

• Giving the wrong unit, e.g., $\text{cm/s}$ (since $0.90$ is in metres, not centimetres) or omitting the unit entirely.
• Rounding incorrectly to 1 significant figure ($0.5$).

Things to Be Careful About

• Ensure the unit blank is explicitly filled with $\text{m/s}$.