5054/42 May/June 2025 — Physics Solved Past Paper

Q110MMediumExperimental ContextsObservations and MeasurementsAnalysis, Conclusions and Evaluation

A student investigates the refraction of a ray of light passing through a transparent block and determines the refractive index $n$ of the block.

The student's ray-trace sheet is shown full size in Fig. 1.1.

The outline of the transparent block is shown by the rectangle ABCD. A line SR that meets side AD of the block is also marked.

Fig. 1.1

(a)

On Fig. 1.1, draw a normal to the block at the point R. Extend your normal 6 cm above side AD and below side BC.

Label the point where the normal crosses side BC of the block with the letter T.

Measure the angle $\theta$ between SR and the normal.

$\theta$ = ______ $^\circ$

1M Similar

(b)

The student:

positions an illuminated slit on the ray-trace sheet so that a ray of light passes along the line SR towards R
marks with small crosses (x) two points on the ray that leaves side BC of the block
removes the transparent block.

On Fig. 1.1, draw a straight line through the two crosses to meet side BC of the block.

Label the point where this line meets side BC with the letter E.

Label the other end of this line as F.

Draw a straight line from E to R.

This shows the path of the ray of light through the block.

1M Similar

(c)

2M

(i)

Measure the length $a$ of line ET on Fig. 1.1.

$a$ = ______ $\text{cm}$

1M Similar

(ii)

Measure the length $b$ of line ER on Fig. 1.1.

$b$ = ______ $\text{cm}$

1M Similar

(d)

On Fig. 1.1, extend the line FE into the block, until it meets the line RT.

Label the point where FE meets RT with the letter G.

Measure the length $c$ of line EG on Fig. 1.1.

$c$ = ______ $\text{cm}$

1M Similar

(e)

2M

(i)

Use your values from (c)(i) and (c)(ii) to calculate a first value $n_1$ for the refractive index of the block.

Use the equation shown.

n_1 = \frac{b}{2a}

$n_1$ = ______

1M Similar

(ii)

Use your values from (c)(ii) and (d) to calculate a second value $n_2$ for the refractive index of the block.

Use the equation shown.

n_2 = \frac{b}{c}

$n_2$ = ______

1M Similar

(f)

Two quantities can be considered to be the same within the limits of experimental accuracy if their values are within 10% of each other.

Compare your value $n_1$ for the refractive index calculated in (e)(i) with the value $n_2$ calculated in (e)(ii).

State if your two values can be considered to be the same.

Support your statement with a calculation.

calculation

statement ______

2M Similar

(g)

One source of inaccuracy in this experiment is careless measurement.

Suggest another source of inaccuracy in this experiment.

1M Similar

Q210MMediumObservations and MeasurementsExperimental ContextsUse of Techniques, Apparatus and MaterialsAnalysis, Conclusions and Evaluation

A student investigates the resistance of a thermistor at different temperatures.

The student constructs the circuit shown in Fig. 2.1.

Fig. 2.1

The thermistor shown in Fig. 2.1 is placed in an empty beaker and is at room temperature.

(a)

The student measures the room temperature $\theta_R$ .

The thermometer is shown in Fig. 2.2.

Fig. 2.2

Record the reading of $\theta_R$ on the answer line.

$\theta_R$ = ______ $^\circ\text{C}$

1M Similar

(b)

2M

(i)

The student:

closes the switch
records the potential difference across the thermistor $V_{XY}$ while the thermistor is at room temperature $\theta_R$ in the top row of Table 2.1
opens the switch.

The voltmeter is shown in Fig. 2.3.

Fig. 2.3

Record the potential difference $V_{XY}$ while the thermistor is at room temperature $\theta_R$ in Table 2.1.

Table 2.1

	$V_{XY} / \text{V}$	$V_{YZ} / \text{V}$	$I / \text{A}$
thermistor at room temperature $\theta_R$	______	1.4	______
thermistor at temperature of hot water $\theta_H$	1.2	3.3	______

1M Similar

(ii)

The student:

disconnects the voltmeter from points X and Y
reconnects the voltmeter across the 220 $\Omega$ resistor between points Y and Z
closes the switch and records the potential difference across the 220 $\Omega$ resistor $V_{YZ}$ while the thermistor is at room temperature $\theta_R$ in the top row of Table 2.1.

Re-draw the circuit shown in Fig. 2.1 on page 5 to show the voltmeter connected to measure the potential difference across the 220 $\Omega$ resistor.

1M Similar

(c)

The student:

pours some hot water into the beaker containing the thermistor
places the thermometer into the hot water and stirs the water gently
waits for 30 s
measures the temperature of the hot water $\theta_H$
measures new values for potential differences $V_{XY}$ and $V_{YZ}$ while the thermistor is at the temperature of the hot water $\theta_H$ .

The temperature of the hot water $\theta_H$ is $83^\circ\text{C}$ .

The student's measurements are recorded in the bottom row of Table 2.1.

2M

(i)

Explain why the student waits for 30 s before measuring the temperature of the hot water.

1M Similar

(ii)

Explain why the student stirs the water before reading the temperature of the hot water from the thermometer.

1M Similar

(d)

Examine the data in Table 2.1.

Compare the reading for the potential difference across the thermistor $V_{XY}$ at room temperature $\theta_R$ with the reading for the potential difference across the thermistor $V_{XY}$ at the temperature of the hot water $\theta_H$ .

Suggest what causes the difference in the readings.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The potential difference across the thermistor decreases from 3.1 V at room temperature to 1.2 V at the higher temperature. Since the current in the series circuit increases (as shown by the higher $V_{YZ}$ ), the resistance of the thermistor must decrease. The resistance of the thermistor decreases as temperature increases.

Final answer

The resistance of the thermistor decreases

Detailed explanation

Walkthrough

In the series circuit, the total potential difference from the power source is constant. The sum of the potential differences across the thermistor ( $V_{XY}$ ) and the $220\ \Omega$ resistor ( $V_{YZ}$ ) equals the supply voltage. At room temperature, $V_{XY} = 3.1\ \text{V}$ and $V_{YZ} = 1.4\ \text{V}$ . At the higher temperature, $V_{XY} = 1.2\ \text{V}$ and $V_{YZ} = 3.3\ \text{V}$ . The decrease in $V_{XY}$ and the increase in $V_{YZ}$ indicate that the current in the circuit has increased. By Ohm's law ( $R = V/I$ ), if the potential difference across a component decreases while the current through it increases, its resistance must have decreased. This is the characteristic behavior of an NTC (Negative Temperature Coefficient) thermistor.

Key Takeaways

In a series circuit with a constant supply voltage, a decrease in potential difference across one component (and a corresponding increase across another) indicates a decrease in resistance of the first component, assuming the current has increased.

Common Mistakes

Saying 'the voltage decreases so the resistance decreases' without mentioning current. Resistance is not simply proportional to voltage; it depends on both voltage and current ( $R = V/I$ ). You must consider the change in current as well.

Things to Be Careful About

Ensure you state that the resistance of the thermistor decreases, not just the potential difference. The question asks for the cause of the difference in readings, which is the change in resistance due to temperature.

Techniques used

compare potential difference readings and link to thermistor resistance

(e)

The current $I$ in the circuit is calculated using the equation:

I = \frac{V_{YZ}}{R}

where $R = 220\ \Omega$ .

Use the measurements recorded in Table 2.1 to calculate the current $I$ at room temperature $\theta_R$ and at the temperature of the hot water $\theta_H$ .

Record your answers in Table 2.1.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

At room temperature $\theta_R$ :

I = \frac{V_{YZ}}{R} = \frac{1.4}{220} = 0.00636\ \text{A}

At the temperature of hot water $\theta_H$ :

I = \frac{V_{YZ}}{R} = \frac{3.3}{220} = 0.015\ \text{A}

Answer

	$V_{XY} / \text{V}$	$V_{YZ} / \text{V}$	$I / \text{A}$
thermistor at room temperature $\theta_R$	3.1	1.4	0.00636
thermistor at temperature of hot water $\theta_H$	1.2	3.3	0.015

Final answer

0.00636 A and 0.015 A

Detailed explanation

Walkthrough

The current $I$ is the same everywhere in a series circuit. The $220\ \Omega$ resistor is a standard resistor with a known resistance, so the current can be calculated using the potential difference across it ( $V_{YZ}$ ) and Ohm's law: $I = V_{YZ} / R$ .

For room temperature:

I = \frac{1.4\ \text{V}}{220\ \Omega} = 0.0063636...\ \text{A} \approx 0.00636\ \text{A}

For hot water temperature:

I = \frac{3.3\ \text{V}}{220\ \Omega} = 0.015\ \text{A}

These values are recorded in the $I / \text{A}$ column of Table 2.1.

Key Takeaways

In a series circuit, the current is uniform. Using a known resistor and its potential difference allows you to calculate the circuit current, which is then used to find the resistance of an unknown component.

Common Mistakes

Using the wrong potential difference (e.g., using $V_{XY}$ instead of $V_{YZ}$ ) in the calculation. Forgetting to divide by 220, or using the wrong unit for resistance.

Things to Be Careful About

Give your answers to 2 or 3 significant figures. 0.00636 A is appropriate; 0.0064 A may also be accepted depending on the scheme's tolerance, but 0.00636 is more precise. 0.015 A is exact to 2 significant figures.

Techniques used

calculate current using Ohm's law

(f)

The resistance $R_T$ of the thermistor is calculated using the equation:

R_T = \frac{V_{XY}}{I}

Use the data in Table 2.1 on page 6 to calculate $R_T$ at room temperature $\theta_R$ and at the temperature of the hot water $\theta_H$ .

$R_T$ at room temperature $\theta_R$ = ______ $\Omega$
$R_T$ at temperature of hot water $\theta_H$ = ______ $\Omega$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

At room temperature $\theta_R$ :

R_T = \frac{V_{XY}}{I} = \frac{3.1}{0.00636} = 487.4\ \Omega \approx 487\ \Omega

At the temperature of hot water $\theta_H$ :

R_T = \frac{V_{XY}}{I} = \frac{1.2}{0.015} = 80\ \Omega

Answer

$R_T$ at room temperature $\theta_R$ = 487 $\Omega$ (any answer between 480 and 520 is acceptable)
$R_T$ at temperature of hot water $\theta_H$ = 80 $\Omega$

Final answer

487 Ω and 80 Ω

Detailed explanation

Walkthrough

The resistance of the thermistor $R_T$ is found using Ohm's law rearranged as $R_T = V_{XY} / I$ , where $V_{XY}$ is the potential difference across the thermistor and $I$ is the current calculated in part (e).

For room temperature:

R_T = \frac{3.1\ \text{V}}{0.00636\ \text{A}} = 487.4\ \Omega

Rounding to 3 significant figures gives 487 $\Omega$ . The mark scheme accepts a range from 480 to 520 $\Omega$ to allow for rounding differences in the current calculation (e.g., if a student used 0.0064 A, they would get $3.1 / 0.0064 = 484.375\ \Omega$ ).

For hot water temperature:

R_T = \frac{1.2\ \text{V}}{0.015\ \text{A}} = 80\ \Omega

Key Takeaways

The resistance of a component can be calculated from the potential difference across it and the current through it. For a thermistor, this resistance will change significantly with temperature.

Common Mistakes

Using the wrong potential difference (e.g., $V_{YZ}$ instead of $V_{XY}$ ). Forgetting to use the current calculated in part (e) and instead trying to calculate resistance from the total voltage and total resistance.

Things to Be Careful About

Pay attention to the range of acceptable answers for the room temperature resistance. If you used a rounded value for the current (e.g., 0.006 A), your resistance will be off. Use at least 3 significant figures for intermediate calculations to avoid rounding errors.

Techniques used

calculate resistance using Ohm's law

(g)

Calculate $\alpha$ , the average change in the resistance of the thermistor per degree Celsius, for the thermistor as the temperature of the thermistor rises from room temperature $\theta_R$ to the temperature of the hot water $\theta_H$ .

Use the equation shown.

\alpha = \frac{\text{change in resistance of thermistor}}{\text{change in temperature}}

Give your answer to 2 significant figures.

$\alpha$ = ______ $\Omega\text{ / }^\circ\text{C}$

2M Similar

DifficultyMedium

Worked solution

Working

Change in resistance of thermistor:

\Delta R_T = 487 - 80 = 407\ \Omega

Change in temperature:

\Delta \theta = 83 - 21 = 62\ ^\circ\text{C}

Average change in resistance per degree Celsius ( $\alpha$ ):

\alpha = \frac{\Delta R_T}{\Delta \theta} = \frac{407}{62} = 6.5645...\ \Omega/\ ^\circ\text{C}

Rounding to 2 significant figures:

\alpha = 6.6\ \Omega/\ ^\circ\text{C}

Answer

$\alpha = 6.6\ \Omega\text{ / }^\circ\text{C}$

Final answer

6.6

Detailed explanation

Walkthrough

The question asks for the average change in resistance per degree Celsius, $\alpha$ . This is calculated as the total change in resistance divided by the total change in temperature.

Change in resistance:

\Delta R_T = R_T(\theta_R) - R_T(\theta_H) = 487 - 80 = 407\ \Omega

(Note: Using the unrounded value 487.4 gives $\Delta R_T = 407.4\ \Omega$ )

Change in temperature:

\Delta \theta = \theta_H - \theta_R = 83 - 21 = 62\ ^\circ\text{C}

Calculate $\alpha$ :

\alpha = \frac{407}{62} = 6.5645...\ \Omega/\ ^\circ\text{C}

The question specifically asks for the answer to 2 significant figures. Rounding 6.5645... to 2 significant figures gives 6.6.

Key Takeaways

When calculating a rate of change (like $\alpha$ ), always find the difference in the dependent variable and divide by the difference in the independent variable. Pay close attention to the required number of significant figures in the final answer.

Common Mistakes

Forgetting to calculate the change in resistance and temperature, and instead dividing the final resistance by the final temperature. Failing to round to 2 significant figures as explicitly requested.

Things to Be Careful About

The mark scheme awards one mark for the correct substitution $(487 - 80) / (83 - 21)$ and one mark for the final answer to 2 significant figures. If you use 480 or 520 for the room temperature resistance, your final answer will be slightly different (e.g., $(480-80)/62 = 6.45 \rightarrow 6.5$ or $(520-80)/62 = 7.1$ ), but the scheme typically allows a range. Stick to your calculated value of 487 $\Omega$ for consistency. Ensure your final answer is exactly 2 significant figures: 6.6, not 6.56 or 7.

Techniques used

calculate the rate of change of resistance with temperature

Q314MMediumObservations and MeasurementsExperimental ContextsUse of Techniques, Apparatus and MaterialsAnalysis, Conclusions and Evaluation

A student investigates the stretching of a spring.

The student suspends the spring from the clamp of a retort stand as shown in Fig. 3.1.

Fig. 3.1

(a)

The student takes readings from the metre rule to determine the length of the spring.

3M

(i)

On Fig. 3.2, take readings from the metre rule level with the top of the spring and level with the bottom of the spring.

Do not include the loops at the top and the bottom of the spring in your measurements.

Record your readings to the nearest 0.1 cm.

Fig. 3.2

reading level with top of spring = ______ $\text{cm}$
reading level with bottom of spring = ______ $\text{cm}$

2M Similar

(ii)

Draw on Fig. 3.2 to show how the student uses a set-square to take a reading from the metre rule level with the bottom of the spring.

1M Similar

(b)

2M

(i)

Calculate the length $l$ of the coiled part of the spring. Use the equation shown.

l = \text{reading level with bottom of spring} - \text{reading level with top of spring}

Show your working.

Record $l$ in Table 3.1 on page 12 for a load $L = 0.0\ \text{N}$ .

1M Similar

(ii)

The student:

places a load $L = 1.0\ \text{N}$ on the spring
takes readings to determine the new length $l$ of the spring.

The top of the spring does not move but the bottom of the spring moves downwards.

The student determines that the new reading on the rule level with the bottom of spring is 45.2 cm.

Calculate the new length $l$ of the spring.

Record your answer for the new length $l$ for a load $L = 1.0\ \text{N}$ in Table 3.1 on page 12.

1M Similar

(c)

The student repeats the procedure in (b)(ii) for loads $L = 2.0\ \text{N}$ , $3.0\ \text{N}$ , $4.0\ \text{N}$ and $5.0\ \text{N}$ . The results are shown in Table 3.1.

Table 3.1

$\text{load } L / \text{N}$	0	1.0	2.0	3.0	4.0	5.0
$\text{length } l / \text{cm}$	______	______	10.0	14.0	18.2	21.9

On the grid provided in Fig. 3.3 on page 13, plot a graph of $l$ on the $y$ -axis against $L$ on the $x$ -axis.

Start from the origin (0, 0).

Draw the straight line of best fit.

Fig. 3.3

4M Similar

DifficultyMedium

Worked solution

Answer

Axes and Labels:
- Horizontal axis: $\text{load } L / \text{N}$ (or $L\text{ (N)}$ ), scaled from $0$ to at least $5.0\text{ N}$ (e.g. $2\text{ cm} = 1.0\text{ N}$ ).
- Vertical axis: $\text{length } l / \text{cm}$ (or $l\text{ (cm)}$ ), scaled from $0$ to at least $24.0\text{ cm}$ (e.g. $2\text{ cm} = 2.0\text{ cm}$ or $2\text{ cm} = 4.0\text{ cm}$ ).
Scales: Linear, sensible, non-awkward intervals occupying more than half the grid.
Plotting: All 6 points from Table 3.1 plotted accurately to within half a small square:
- $(0, 2.1)$
- $(1.0, 6.2)$
- $(2.0, 10.0)$
- $(3.0, 14.0)$
- $(4.0, 18.2)$
- $(5.0, 21.9)$
Line of best fit: A single, thin, clear straight line drawn with a ruler passing evenly through the points.

Final answer

Graph plotted with correctly labelled axes, linear scales, accurate points, and a straight line of best fit

Detailed explanation

Walkthrough

Axes setup:
- The question specifies $l$ on the $y$ -axis and $L$ on the $x$ -axis.
- Label the $x$ -axis: $\text{load } L / \text{N}$ .
- Label the $y$ -axis: $\text{length } l / \text{cm}$ .
Choosing scales:
- Must start from the origin $(0, 0)$ as instructed.
- $x$ -axis range is $0$ to $5.0\text{ N}$ . A convenient scale is $2\text{ cm} = 1.0\text{ N}$ (or $10\text{ small squares} = 1.0\text{ N}$ ).
- $y$ -axis range is $0$ to $21.9\text{ cm}$ (up to $25\text{ cm}$ ). A convenient scale is $2\text{ cm} = 2.0\text{ cm}$ or $2\text{ cm} = 4.0\text{ cm}$ , ensuring the plot covers more than half the grid area.
Plotting the points:
- Plot each of the pairs: $(0, 2.1)$ , $(1.0, 6.2)$ , $(2.0, 10.0)$ , $(3.0, 14.0)$ , $(4.0, 18.2)$ , $(5.0, 21.9)$ using neat small crosses ( $+$ or $\times$ ) or encircled dots within $\pm 0.5$ small square.
Line of best fit:
- Use a transparent ruler to draw a straight line that balances the points on either side. Note that this line has a $y$ -intercept around $2.1\text{ cm}$ and does NOT pass through $(0,0)$ .

Key Takeaways

In physics graph plotting, axes must be labelled with quantity and unit in the format $\text{quantity} / \text{unit}$ .
Scales must be linear (e.g. steps of $1, 2, 5$ ) and never awkward multiples like $3$ or $7$ .
A best-fit line represents the underlying trend; do not join points dot-to-dot.

Common Mistakes

Forcing the line of best fit to go through $(0, 0)$ when the $y$ -intercept is the unstretched length $l_0 \approx 2.1\text{ cm}$ .
Inverting the axes ( $L$ on $y$ instead of $x$ ).
Drawing a thick, fuzzy, or multiple-stroke line.

Things to Be Careful About

The plotted points must cover more than half the available grid area in both directions.

Techniques used

choose linear graph scales occupying more than half the gridplot experimental points accuratelydraw a straight line of best fit

(d)

Use the data in Table 3.1 and the graph in Fig. 3.3 to determine the extension of the spring when a load of 3.5 N is added to the spring.

Show your working.

extension = ______ $\text{cm}$

2M Similar

(e)

A student suggests that the stretched length $l$ of the spring is proportional to load $L$ .

State if the data in Table 3.1 supports this suggestion. Justify your statement using the data in Table 3.1 or the graph in Fig. 3.3.

statement ______

justification ______

1M Similar

(f)

Line of sight (parallax) errors can occur when readings are taken from the metre rule.

State one practical technique, other than using a set square, that ensures accurate readings are taken from a metre rule.

1M Similar

(g)

The student repeats the procedure described in (c) two more times and averages the readings before calculating the spring lengths $l$ for each load $L$ .

Explain why the student does this.

1M Similar

Q46MMedium-EasyPlanning Experiments and Investigations

As a metal ball falls through a liquid, it experiences a frictional force from the liquid that opposes the motion of the metal ball.

Plan an experiment to determine the relationship between the density of a liquid contained in a measuring cylinder and the average speed of a metal ball falling through the liquid from the surface of the liquid to the bottom of the cylinder.

The average speed of the ball is calculated using the equation:

\text{average speed} = \frac{\text{distance travelled}}{\text{time taken}}

The arrangement of the apparatus is shown in Fig. 4.1.

Fig. 4.1

The apparatus available includes:

a measuring cylinder
a metal ball
a selection of different liquids whose densities are known.

You may also use any other apparatus commonly found in a school laboratory.

In your plan, include:

any other apparatus needed
a brief description of the method, including what you will measure and how you make sure that your measurements are accurate
the variables you will control
a results table to record your measurements (you are not required to enter any readings in the table)
how you will process your results to draw a conclusion.

DifficultyMedium-Easy

Worked solution

Answer

Additional apparatus needed:

ruler (or measuring tape)
stopwatch (or digital timer)

Method:

Measure the height (distance) of the liquid in the measuring cylinder using the ruler.
Drop the metal ball from rest at the surface of the liquid.
Use the stopwatch to measure the time taken for the ball to fall to the bottom.
Calculate the average speed using $\text{average speed} = \frac{\text{distance}}{\text{time}}$ .
Repeat the experiment for each of the other liquids with different known densities.

Control variables:

Keep the height (or volume) of the liquid the same for all trials.
Use the same metal ball (same mass, volume, and shape) for all trials.

Results table:

Density of liquid / $\text{kg/m}^3$	Time taken / s	Average speed / $\text{m/s}$

Processing results and conclusion:

Calculate the average speed for each liquid from the measured time and distance.
Plot a graph of density of liquid (x-axis) against average speed (y-axis).
Draw a line of best fit and observe whether the graph is horizontal (no effect) or has a gradient (density affects speed). Alternatively, compare the calculated average speeds for different densities to see if there is a clear relationship.

Final answer

See working

Detailed explanation

Walkthrough

1. Additional apparatus:
The experiment requires measuring two quantities: the distance the ball falls and the time it takes. The distance is the height of the liquid in the measuring cylinder, which requires a ruler or measuring tape. The time requires a stopwatch or digital timer.

2. Method:
The candidate must describe the physical actions. First, measure the liquid height with the ruler. Then, drop the ball from rest at the surface and start the stopwatch simultaneously. Stop the timer when the ball reaches the bottom. Calculate the speed using the given formula. Finally, the method must include repeating the procedure for each of the different liquids to gather a range of data points.

3. Control variables:
To ensure that only the density of the liquid affects the ball's speed, all other factors must be kept constant. The height of the liquid must be the same for every trial so the distance is constant. The metal ball must be the same each time, meaning its mass, volume, and shape are unchanged. Changing the ball would change its mass and surface area, altering the frictional and gravitational forces.

4. Results table:
The table must record the independent variable (density of the liquid), the raw measured dependent variable (time taken), and the calculated derived variable (average speed). Crucially, units must be included in the column headings (e.g., $\text{kg/m}^3$ , s, $\text{m/s}$ ). Blank rows are provided for the candidate to fill in their readings.

5. Processing and conclusion:
After calculating the average speed for each trial, the standard way to analyse the relationship between two variables is to plot a graph of density against average speed. Drawing a line of best fit allows the candidate to see the trend: a horizontal line means density has no effect, while a sloped line indicates a relationship. Alternatively, simply comparing the numerical values of the average speeds for different densities to identify a trend is an acceptable conclusion method.

Key Takeaways

A planning question requires a structured response covering apparatus, method, control variables, a results table, and data processing.
The results table must include units in the column headings; this is a separate mark.
Control variables must be specific to the experiment. "Same apparatus" is too vague; specify "same ball" and "same liquid height".
Data processing should involve either plotting a graph or explicitly comparing calculated values to draw a conclusion about the relationship.

Common Mistakes

Forgetting units in the table: The mark scheme awards a mark specifically for units in the table headings. Writing just "Time" and "Density" without units loses the mark.
Vague control variables: Writing "keep the same conditions" or "use the same cylinder" does not score. The candidate must specify what is kept the same (e.g., liquid height, ball mass/shape).
Missing the repeat for other liquids: The method must explicitly state that the procedure is repeated for the different liquids, not just that the ball is timed once.
Confusing independent and dependent variables: Density is the independent variable (it is chosen/known), time is the raw dependent variable (measured), and speed is the derived dependent variable (calculated). The table and graph should reflect this.

Things to Be Careful About

Precision of apparatus: A standard school ruler is precise to $1\text{ mm}$ , and a stopwatch to $0.01\text{ s}$ . The method should imply reading to the appropriate precision (e.g., reading the meniscus at eye level if measuring liquid volume, though here we measure height from the top/bottom markings).
Zero errors: If using a digital stopwatch, ensure it is reset to zero before each timing. If using a ruler, check for a zero error at the bottom of the cylinder.
Repeatability: While the mark scheme focuses on repeating for different liquids, mentioning taking repeats for the same liquid and averaging the time is a good practice that demonstrates an understanding of accuracy, even if not explicitly marked here.

Techniques used

identify additional apparatus needed for measuring distance and timedescribe a method to measure falling time and calculate average speedidentify control variables to ensure a fair testdesign a results table with appropriate columns and unitsplan data processing using a graph to find a relationship

Physics 5054/42 — May/June 2025

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Things to Be Careful About

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Working