Physics 5054/31 — October/November 2025

Walkthrough

With the switch closed, you read the ammeter and the voltmeter. The mark scheme expects the current recorded to 0.01 A or better (e.g. 1.30 A) and the p.d. recorded to 0.1 V or better (e.g. 3.0 V). These are candidate-dependent readings — typical values for this experiment are around $I = 1.30\ \text{A} \pm 0.50\ \text{A}$ and $V = 3.0\ \text{V} \pm 0.5\ \text{V}$.

The precision is the key point: the reading must reflect the scale divisions on the instrument.

Key Takeaways

• An ammeter reading is recorded to 0.01 A or better.
• A voltmeter reading is recorded to 0.1 V or better.

Common Mistakes

• Recording the current as 1.3 A instead of 1.30 A — the trailing zero shows the precision.
• Mixing up the units (A for current, V for p.d.).

Things to Be Careful About

• Read the instruments without parallax error.
• Record both readings at the same time, immediately after closing the switch.

Walkthrough

After 5 minutes of heating (with gentle stirring) you open the switch, stir for a further minute, and record the final temperature $\theta_F$. The final temperature must be greater than or equal to the initial temperature (the water has been heated). Then you calculate the temperature change:

For example, if $\theta_0 = 25.0\ ^\circ\text{C}$ and $\theta_F = 29.5\ ^\circ\text{C}$, then $\Delta\theta = 4.5\ ^\circ\text{C}$.

The mark is for recording $\theta_F$ (with $\theta_F \geq \theta_0$) and recording $\Delta\theta$.

Key Takeaways

• Temperature change is final minus initial.
• The final temperature must be at least the initial temperature — the heater adds energy.

Common Mistakes

• Recording $\theta_F < \theta_0$ — physically impossible since the heater warms the water.
• Forgetting to calculate $\Delta\theta$.

Things to Be Careful About

• Units: °C.
• Record $\theta_F$ to the same precision as $\theta_0$.

Walkthrough

The water is stirred for a further minute after the switch is opened so that the temperature is uniform throughout the beaker. Without stirring, the water near the coil would be hotter than the water at the edges, and the thermometer reading would not represent the average temperature of all the water. Stirring also ensures that all the heat from the coil is transferred to the water (rather than some staying in the coil or in a local hot spot).

The mark scheme accepts either: '(to ensure the water is at) a uniform temperature' or '(to ensure that) all the heat from the coil is absorbed by the water'.

Key Takeaways

• Stirring equalises the temperature so the reading is representative.
• This is a technique for accurate observations.

Common Mistakes

• Saying 'to cool it down' — stirring does not cool the water.
• Saying 'to make the reading more accurate' without saying why (uniform temperature).

Things to Be Careful About

• The mark is for the idea of a uniform temperature, so state it explicitly.

Walkthrough

The energy supplied by the heater is calculated from the electrical power and the time:

where $I$ is the current, $V$ is the p.d. across the heater, and $t = 300\ \text{s}$ (5 minutes). Using the example values $I = 1.30\ \text{A}$ and $V = 3.0\ \text{V}$:

The mark is for the correct calculation from your own values of $I$ and $V$. Show the substitution so the working is visible.

Key Takeaways

• Electrical energy = power × time = $I \times V \times t$.
• Convert minutes to seconds before substituting (5 min = 300 s).

Common Mistakes

• Using $t = 5$ instead of $t = 300\ \text{s}$.
• Not showing the substitution.

Things to Be Careful About

• The answer is in joules (J).
• Use the candidate's own $I$ and $V$ values.

Walkthrough

The energy gained by the water is:

Here 50 is the mass of water in grams (50 cm³ of water has a mass of 50 g), 4.2 is the specific heat capacity of water in J/(g °C), and $\Delta\theta$ is the temperature rise. Using the example $\Delta\theta = 4.5\ ^\circ\text{C}$:

The mark scheme also requires $Q_W \leq Q_H$ — the water cannot gain more energy than the heater supplies, so if your numbers give $Q_W > Q_H$ something is wrong.

Key Takeaways

• $Q_W = mc\Delta\theta$ with $m = 50\ \text{g}$ and $c = 4.2\ \text{J/(g}\,^\circ\text{C)}$.
• The energy gained by the water must be less than the energy supplied by the heater.

Common Mistakes

• Getting $Q_W > Q_H$ — physically impossible; check the arithmetic.
• Forgetting the unit (J).

Things to Be Careful About

• The mass is in grams here (50 g), not kg — the equation is set up for grams.
• Use the candidate's own $\Delta\theta$.

Walkthrough

Efficiency is the useful energy output divided by the energy input:

Using the example values:

The efficiency is a ratio with no unit. It can also be expressed as a percentage (81%).

Key Takeaways

• Efficiency = useful output energy / total input energy.
• Efficiency is dimensionless (or given as a percentage).

Common Mistakes

• Inverting the ratio ($Q_H / Q_W$) — efficiency must be less than 1.
• Giving the answer as a percentage without the % sign, or as a decimal without explanation.

Things to Be Careful About

• Use the candidate's own $Q_W$ and $Q_H$ values.
• The efficiency must be between 0 and 1 (or 0% and 100%).

Walkthrough

Using a protractor, align the baseline along the line $XY$ with the centre mark at point $M$. Make a mark at $90^\circ$. Using a ruler, draw a continuous straight line through point $M$ that extends $8\text{ cm}$ above $XY$ and $8\text{ cm}$ below $XY$. This line is the normal to the boundary.

Key Takeaways

• A normal line is always perpendicular ($90^\circ$) to the boundary surface.
• Exact lengths specified in experimental instructions must be followed carefully using a ruler.

Common Mistakes

• Drawing the normal not strictly at $90^\circ$ to the line $XY$.
• Forgetting to extend the line both above and below line $XY$.

Things to Be Careful About

• Ensure the line is drawn thinly and sharply with a sharp pencil.

Walkthrough

Place the protractor on the normal line with its centre at point $M$. Measure an angle of $40^\circ$ to the left of the normal (in the upper half above $XY$). Mark this angle and draw a straight line from $M$ through the mark. Label the top-left end of this line $L$. This represents the incident ray with an angle of incidence $i = 40^\circ$.

Key Takeaways

• The angle of incidence is always measured between the incident ray and the normal, NOT between the incident ray and the surface.

Common Mistakes

• Measuring $40^\circ$ from the line $XY$ instead of from the normal.
• Drawing the line to the right of the normal instead of to the left.

Things to Be Careful About

• Keep pencil lines sharp and align the protractor accurately with the normal line.

Walkthrough

Place the transparent rectangular block onto the paper so that its top long edge aligns exactly with the line $XY$ and its top left corner touches point $X$. Hold the block firmly so it does not slip, and carefully draw around its other three edges with a pencil.

Key Takeaways

• Accurately outlining an optical component is essential so that if it shifts, it can be repositioned correctly.

Common Mistakes

• Allowing the block to shift while tracing around it.

Things to Be Careful About

• Ensure the pencil is held vertically against the edges of the block to trace its exact boundaries.

Walkthrough

Align the light box slit so a narrow ray shines along the incident line $LM$ into the block at $M$. The ray refracts inside the block and emerges from the opposite face below $XY$. Observe the emergent ray on the paper. Mark two precise crosses along the middle of this beam, ensuring they are spaced well apart (at least $5\text{ cm}$ apart) to minimise alignment errors when joining them later.

Key Takeaways

• Spacing marking points / pins as far apart as possible significantly improves the angular accuracy of the drawn ray line.

Common Mistakes

• Placing the two crosses too close together (less than $5\text{ cm}$ apart), which increases angular inaccuracy.
• Marking the crosses off-centre of the light beam.

Things to Be Careful About

• Keep the room reasonably dark or shield the light to see the centre of the beam clearly.

Walkthrough

Remove the transparent block. Use a ruler to draw a straight line connecting both crosses and extend this line back to intersect the lower horizontal line of the rectangular outline. Label this intersection point $P$. Point $P$ represents the point where the light ray emerged from the block.

Key Takeaways

• Extending the emergent ray backwards gives the exact point of exit from the optical medium.

Common Mistakes

• Forgetting to extend the line all the way to the boundary of the block.
• Forgetting to label the intersection point $P$.

Things to Be Careful About

• Ensure the straight line passes through the centres of both crosses precisely.

Walkthrough

Using a straight ruler, draw a single line from point $M$ (where the incident ray enters the block) to point $P$ (where the ray leaves the block). This line represents the path of the refracted ray inside the transparent block.

Key Takeaways

• Light travels in straight lines within a uniform, homogeneous medium such as glass or Perspex.

Common Mistakes

• Drawing a curved or wobbly line instead of a clean, straight ruler line.

Things to Be Careful About

• Line $MP$ must start exactly at $M$ and end exactly at $P$.

Walkthrough

Place the centre of the protractor at point $M$, aligning the baseline of the protractor with the normal line extending below line $XY$. Read the angle between this normal and the line $MP$. For a typical glass or acrylic block with $n \approx 1.5$ and an angle of incidence $i = 40^\circ$, the angle of refraction $r$ is approximately $25^\circ$ (values in the range $23^\circ$ to $27^\circ$ are accepted).

Key Takeaways

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

Common Mistakes

• Measuring the angle between line $MP$ and the surface $XY$ instead of the normal.
• Misreading the inner/outer scale of the protractor.

Things to Be Careful About

• Ensure the protractor is centred accurately at point $M$.

Walkthrough

The top pan balance is tared (zeroed) with the empty dish placed on it. Adding 5 identical marbles allows the candidate to measure their combined mass directly. A typical set of 5 glass marbles has a mass in the range of $23.0\text{ g}$ to $27.0\text{ g}$ ($25.0 \pm 2.0\text{ g}$).

Key Takeaways

• Taring the balance subtracts the mass of the container (the dish), giving directly the mass of the marbles alone.

Common Mistakes

• Forgetting to tare the balance before adding the marbles, which results in including the mass of the dish.

Things to Be Careful About

• Ensure the reading is recorded to the precision shown on the balance (typically 1 or 2 decimal places).

Walkthrough

To find the average mass of a single marble, divide the measured mass of the 5 marbles from part (a)(i) by 5:

For a measured value of $25.0\text{ g}$, the average mass is $5.0\text{ g}$.

Key Takeaways

• Measuring multiple items together and dividing by the count reduces the relative uncertainty of a single measurement.

Common Mistakes

• Arithmetic errors in dividing by 5.

Things to Be Careful About

• Keep significant figures consistent with the mass recorded in (a)(i).

Walkthrough

1. The marble is released from rest with its right-hand edge at the $90.0\text{ cm}$ mark on the track set at height $h = 4.0\text{ cm}$.
2. The stopwatch is started on release and stopped when the marble hits the stopper at the bottom ($0\text{ cm}$ mark).
3. Three separate trials are timed to obtain $t_1$, $t_2$, and $t_3$.
4. The average time is calculated using:

Key Takeaways

• Repeating measurements and calculating the mean minimizes the effect of random timing and human reaction-time errors.

Common Mistakes

• Rounding the average incorrectly or omitting one of the three readings when calculating the mean.

Things to Be Careful About

• Ensure all individual times and the average are recorded to a consistent precision (typically 2 decimal places for digital stopwatches, or 1 decimal place if taking reaction time into account).

Walkthrough

1. For each height $h = 6.0\text{ cm}, 8.0\text{ cm}, 10.0\text{ cm}, 12.0\text{ cm}$, the marble is released from the $90.0\text{ cm}$ mark three times.
2. All three trials ($t_1, t_2, t_3$) are entered in Table 3.1.
3. The average time $t_{av}$ is computed for each row: $\frac{t_1 + t_2 + t_3}{3}$.
4. As the height $h$ increases, the slope becomes steeper, so the acceleration of the marble increases and the transit time decreases down the table.

Key Takeaways

• Increasing ramp height increases the component of gravitational force along the slope, resulting in greater acceleration and shorter rolling times.

Common Mistakes

• Inconsistent decimal places down a single column.
• Arithmetic mistakes when calculating row averages.

Things to Be Careful About

• Ensure the trend shows $t_{av}$ decreasing steadily as $h$ increases.

Walkthrough

1. Label Axes: Label the $y$-axis with '$t_{av} / \text{s}$' and the $x$-axis with '$h / \text{cm}$'.
2. Choose Scales: Select linear, easy-to-read scale divisions (e.g., 2 cm per large block on the x-axis, 0.2 s per large block on the y-axis). The data points must occupy at least half of the grid in both directions.
3. Plot Points: Mark each pair of $(h, t_{av})$ values with a small, sharp cross ($+$ or $\times$) or a small dot with a circle around it. Points must be plotted to within half a small grid square.
4. Line of Best Fit: Draw a thin, smooth best-fit line (or smooth curve) showing the downward trend, balancing points evenly on either side of the line without forcing it through (0,0).

Key Takeaways

• Good graph-plotting technique requires clear axis labels with units, non-awkward linear scales covering $\ge 50\%$ of the grid, precise plotting, and a smooth, unforced line of best fit.

Common Mistakes

• Using awkward scales such as multiples of 3 or 7, which makes plotting difficult and loses scale marks.
• Drawing thick, 'fuzzy', or multi-stroke lines.

Things to Be Careful About

• The question explicitly notes: 'You do not need to start your axes at (0, 0)', allowing candidates to choose a false origin for the y-axis to spread the points across the grid.

Walkthrough

From the experimental data and the plotted graph, as the ramp height $h$ increases, the time taken $t_{av}$ for the marble to roll down the slope becomes shorter. Therefore, as $h$ increases, $t_{av}$ decreases.

Key Takeaways

• Stating a relationship requires describing how changes in the independent variable ($h$) cause changes in the dependent variable ($t_{av}$).

Common Mistakes

• Stating 'they are inversely proportional' without mathematical justification (such as testing $h \times t_{av} = \text{constant}$). Simply stating 'as $h$ increases, $t_{av}$ decreases' is safe and fully credited.

Things to Be Careful About

• Ensure both variables ($h$ and $t_{av}$) are mentioned explicitly.

Walkthrough

1. Locate $h = 7.0\text{ cm}$ on the horizontal axis of Fig. 3.2.
2. Draw a dashed vertical construction line up from $7.0\text{ cm}$ until it intersects the line of best fit.
3. From this intersection point, draw a dashed horizontal line across to the vertical ($t_{av}$) axis.
4. Read off the value of $t_{av}$ from the scale (e.g. $2.14\text{ s}$).

Key Takeaways

• When asked to 'show on the graph how you find the value', visible construction/tie lines must be drawn on the grid.

Common Mistakes

• Forgetting to draw the construction lines on the graph grid, which loses 1 mark.

Things to Be Careful About

• Read the scale division carefully to ensure the interpolated time matches the scale precision.

Walkthrough

1. The distance travelled by the marble along the track is $d = 90.0\text{ cm}$ (or $0.900\text{ m}$).
2. Using the equation provided:

1. Substitute $d$ and the value of $t_{av}$ obtained from (c)(i):
   • If using $d = 90.0\text{ cm}$: $v = \frac{90.0}{2.14} = 42.1\text{ cm/s}$.
   • If using $d = 0.900\text{ m}$: $v = \frac{0.900}{2.14} = 0.421\text{ m/s}$.
2. State the numerical value and the matching unit ($\text{cm/s}$ or $\text{m/s}$).

Key Takeaways

• Average speed is the total distance divided by the total time taken ($v = d/t$).
• The unit must match the unit used for distance and time.

Common Mistakes

• Giving the unit as $\text{m/s}$ while using $d = 90\text{ cm}$ without converting to metres.
• Omitting the unit on the unit blank line.

Things to Be Careful About

• Round the final speed to 2 or 3 significant figures.