Physics 9702/52 — May/June 2025

Cambridge A-Level · Planning, Analysis and Evaluation · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Analysis, Conclusions and Evaluation · Planning

Q115MMedium-HardPlanningAnalysis, Conclusions and Evaluation

A thin solid disc of radius $r$ and thickness $z$ is attached to a thin axle. String is wrapped around the axle, as shown in Fig. 1.1.

Fig. 1.1

A block of mass $m$ is attached to the string.

The block is released from rest and falls downwards. The block has speed $v$ when it has fallen through a distance $h$ from the point of release. The value of $v$ is determined using one light gate connected to a timer.

It is suggested that $v$ is related to $m$ by the relationship

\frac{h}{v^2} = \frac{\pi r^2 z}{2PQm} + \frac{1}{P}

where $P$ and $Q$ are constants.

Plan a laboratory experiment to test the relationship between $v$ and $m$ .

Draw a diagram showing the arrangement of your equipment.

Explain how the results could be used to determine values for $P$ and $Q$ .

In your plan you should include:

the procedure to be followed
the measurements to be taken
the control of variables
the analysis of the data
any safety precautions to be taken.

Diagram

DifficultyMedium-Hard

Worked solution

diagram-1

Apparatus

Disc + axle on low-friction bearings, string, mass hanger + slotted masses, light gate + timer/data logger, card/flag of known length $L$ , metre rule, clamp stand(s), vernier calipers / micrometer screw gauge (for $r$ and $z$ ).

Variables

Independent: $m$ (total hanging mass).
Dependent: $v$ after falling distance $h$ .
Controlled: $h$ (fixed), same disc/axle (so $r$ and $z$ constant), same string and wrapping (no slip), release from rest each run, light gate position fixed, card length $L$ fixed.

Procedure and measurements

Measure disc radius $r$ (e.g. diameter with vernier calipers then halve) and thickness $z$ (micrometer). Record uncertainties.
Attach a card/flag of measured length $L$ to the falling mass so it will pass through the light gate.
Wrap string around axle, attach mass $m$ , and set the light gate a vertical distance $h$ below the release point (measure $h$ with a metre rule).
Release the mass from rest without pushing. Record the time $t$ for which the light gate beam is interrupted by the flag.
Calculate the speed at the gate:

v = \frac{L}{t}

Repeat at least 3 times for the same $m$ and average $t$ (or $v$ ).
Change $m$ by adding slotted masses; use at least 6 different values spanning a suitable range; repeat steps 3–6.

Analysis of data (test of relationship)

For each $m$ :

compute $v$ and then $\dfrac{h}{v^2}$
compute $\dfrac{1}{m}$ .

Given

\frac{h}{v^2} = \frac{\pi r^2 z}{2PQm} + \frac{1}{P}

let

y = \frac{h}{v^2}, \qquad x = \frac{1}{m}

y = \left(\frac{\pi r^2 z}{2PQ}\right) x + \frac{1}{P}

Plot a graph of $\dfrac{h}{v^2}$ (y-axis) against $\dfrac{1}{m}$ (x-axis). A straight line supports the relationship.

Determination of $P$ and $Q$

From the graph:

intercept $c = \dfrac{1}{P}$ , hence

P = \frac{1}{c}

gradient $G = \dfrac{\pi r^2 z}{2PQ}$ , hence

Q = \frac{\pi r^2 z}{2PG}

(using measured $r,z$ and calculated $P$ ).

Safety

Clamp the apparatus securely; keep feet/hands clear of falling masses; use a tray/soft landing pad to catch the mass; ensure the string cannot whip loose and that the stand cannot topple.

Final answer

See working

Detailed explanation

Background Concept

The falling mass loses gravitational potential energy and causes the disc to rotate via the string on the axle. The measured quantity in this experiment is the instantaneous speed $v$ of the mass after it has fallen through a fixed vertical distance $h$ .

The question provides a model that predicts a linear relationship:

\frac{h}{v^2} = \frac{\pi r^2 z}{2PQm} + \frac{1}{P}

where $P$ and $Q$ are constants for this system, and $r$ and $z$ are properties of the disc.

A standard way to “test a relationship” in Paper 5 is:

rearrange into the straight-line form $y = mx + c$ ,
choose what to plot so that the data should lie on a straight line,
use the gradient and intercept to calculate the constants.

A single light gate can measure an instantaneous speed if a card/flag of known length $L$ passes through it. If the light beam is blocked for time $t$ , then the speed at the gate is

v = \frac{L}{t}

provided the speed does not change much while the card passes through (using a short card helps).

Understanding the Question

You must plan an experiment where you vary the mass $m$ and measure the speed $v$ after the mass has fallen the same distance $h$ each time. You must use only one light gate, so the typical method is a single interrupt card.

You are also asked to explain how your results can be used to find numerical values of $P$ and $Q$ . That means your analysis must produce both a gradient and an intercept (two pieces of information) so you can solve for two unknown constants.

The disc geometry $r$ and $z$ appear in the equation, so you must measure them (or they must be known). In a real lab, you measure them with calipers/micrometer.

Approach

Choose a fixed position of the light gate, a distance $h$ below the release point.
For each chosen mass $m$ , release from rest and use the light gate to obtain $v$ at that position.
Process each run into the variables that appear in the linear form: compute $h/v^2$ and $1/m$ .
Plot $h/v^2$ against $1/m$ so the relationship becomes $y = mx + c$ .
Read gradient $G$ $G$ and intercept $c$ $c$ and then calculate:
- $P$ from the intercept $c = 1/P$ ,
- $Q$ from the gradient $G = \pi r^2 z /(2PQ)$ .
Control variables so that only $m$ changes, and include safety measures for a falling mass.

Step-by-Step Reasoning

diagram-1

1) Set-up and measurements

Mount the disc/axle so it can rotate freely (minimise bearing friction).
Wrap the string neatly around the axle so it does not overlap; attach the mass hanger.
Measure the flag length $L$ with a ruler (small uncertainty; ensure it is rigid and does not bend).
Fix the light gate on a clamp stand.
Define the release point (e.g. bottom of the mass hanger level with a marked reference line) and measure $h$ down to the light gate beam with a metre rule.
Measure $r$ (best: measure diameter at several orientations with vernier calipers and average, then halve) and measure thickness $z$ with a micrometer.

2) Collecting data for one mass

For a chosen total mass $m$ :

Raise the mass so the card will pass through the light gate after falling distance $h$ .
Release without pushing (so initial speed is as close to zero as possible).
The timer gives the interrupt time $t$ .
Calculate

v = \frac{L}{t}

Repeat several times and average $t$ (or average $v$ ). Repeats reduce random error from release technique and any wobble.

3) Varying the independent variable

Change $m$ by adding slotted masses. Take at least 6 values over a good range so your graph has a reliable gradient and intercept.

4) Linearisation and graph

Start from:

\frac{h}{v^2} = \frac{\pi r^2 z}{2PQm} + \frac{1}{P}

Recognise that the only term involving $m$ is $1/m$ . So define:

y = \frac{h}{v^2}, \qquad x = \frac{1}{m}

Then the equation becomes:

y = \left(\frac{\pi r^2 z}{2PQ}\right)x + \frac{1}{P}

So plotting $y$ against $x$ should give a straight line.

5) Extracting $P$ and $Q$

From the straight line:

Intercept $c$ equals $1/P$ , so

P = \frac{1}{c}

Gradient $G$ equals $\pi r^2 z /(2PQ)$ . Rearranging gives

Q = \frac{\pi r^2 z}{2PG}

You can only do this if you have already found $P$ from the intercept.

(If you include uncertainties, you would estimate uncertainties in $G$ and $c$ using worst acceptable lines, then propagate those into $P$ and $Q$ .)

6) Control of variables (why it matters)

To make the test fair, only $m$ should change:

Keep $h$ the same by keeping the light gate fixed and using the same release reference point.
Use the same disc and axle (so $r$ and $z$ are fixed).
Keep the string and wrapping method the same: slipping changes the relationship between translational motion and rotation.
Keep the flag the same length $L$ and in the same orientation through the gate.

7) Safety

The main risk is a falling mass and toppling equipment:

Clamp stands firmly and consider adding a heavy base.
Use a tray/box or foam pad to catch the mass.
Keep hands/feet clear of the drop zone.
Ensure the string cannot fly off sideways.

Key Takeaways

A single light gate can measure instantaneous speed using a flag of known length: $v = L/t$ .
Linearising an equation into $y = mx + c$ is the most direct way to test a relationship and obtain constants.
Plotting $h/v^2$ against $1/m$ provides both an intercept (to find $P$ ) and a gradient (to find $Q$ ).
Good planning requires clear variable control, repeats, and safe/secure apparatus.

Common Mistakes

Plotting the wrong variables (e.g. $v^2/h$ vs $m$ ) so the line is not linear and gradient/intercept do not match the model.
Forgetting that only one light gate is allowed and attempting a two-gate speed method.
Not measuring/mentioning $r$ and $z$ even though they are needed to compute $Q$ .
Changing $h$ unintentionally by releasing from different starting points for different masses.
Using too few values of $m$ (insufficient data to justify a straight line and reliable gradient/intercept).

Things to Be Careful About

Ensure the card length $L$ is measured along the direction of motion through the gate.
Use a short flag so the speed does not change much while it passes the beam.
Release from rest consistently; do not push the mass.
Make sure the string does not slip on the axle and does not overlap between turns.
Use a sensible mass range: very small $m$ may not overcome friction reliably; very large $m$ may cause unsafe speeds or slipping.

Techniques used

measure instantaneous speed using a light gate and a card of known lengthrearrange the given relationship to a straight-line formplot a graph and determine gradient and interceptuse gradient and intercept to calculate unknown constants

Background Concept

To test a proposed relationship experimentally, we often rearrange it into a straight-line form

y = mx + c

where $m$ is the gradient and $c$ is the y-intercept. Plotting the correct variables then allows constants in the original equation to be found from $m$ and $c$ .

Understanding the Question

You are told the suggested relationship is

EA = V(nC + A)

and a graph is plotted of $\frac{1}{V}$ (y-axis) against $n$ (x-axis). The task is to rearrange the equation so that $\frac{1}{V}$ is expressed as a linear function of $n$ , then read off the gradient and y-intercept.

Approach

Make $V$ the subject.
Take the reciprocal to obtain $\frac{1}{V}$ .
Expand into “(constant) $\times n$ + (constant)” so it matches $y = mx + c$ .

Step-by-Step Reasoning

Starting with

EA = V(nC + A)

divide both sides by $(nC + A)$ :

V = \frac{EA}{nC + A}

Now take the reciprocal:

\frac{1}{V} = \frac{nC + A}{EA}

Split the numerator into two terms:

\frac{1}{V} = \frac{C}{EA}n + \frac{A}{EA}

Comparing with $y = mx + c$ where $y = \frac{1}{V}$ and $x=n$ :

gradient $m = \frac{C}{EA}$
intercept $c = \frac{A}{EA} = \frac{1}{E}$ (because $A$ cancels)

Key Takeaways

Linearise by rearranging into $y = mx + c$ .
For a $\frac{1}{V}$ vs $n$ plot, the coefficient of $n$ is the gradient, and the remaining constant term is the intercept.

Common Mistakes

Stopping at $V = \frac{EA}{nC + A}$ and not converting to $\frac{1}{V}$ .
Giving intercept as $\frac{A}{EA}$ but not recognising it simplifies to $\frac{1}{E}$ .
Mixing up which variable is on which axis (here $y=\frac{1}{V}$ , $x=n$ ).

Things to Be Careful About

$n$ is dimensionless, so the gradient has units of $\text{V}^{-1}$ .
Keep algebra exact; simplifications like $\frac{A}{EA} = \frac{1}{E}$ can make later parts much easier.

Techniques used

rearrange the given equation into the form y = mx + cidentify the gradient and y-intercept by comparison with a straight-line equation

(b)

Values of $n$ and the two measured values of the maximum potential difference $V_1$ and $V_2$ are given in Table 2.1.

Table 2.1

$n$	$V_1 / \text{V}$	$V_2 / \text{V}$
2	4.30	4.20
3	3.65	3.75
4	3.30	3.20
5	2.85	2.95
6	2.65	2.55
7	2.30	2.40

Calculate and record values of $V / \text{V}$ and $\frac{1}{V} / \text{V}^{-1}$ in Table 2.1. Include the absolute uncertainties in $V$ and $\frac{1}{V}$ .

2M Similar

DifficultyMedium

Worked solution

For each row:

V = \frac{V_1+V_2}{2},\qquad \Delta V = \frac{V_\text{max}-V_\text{min}}{2}

\frac{1}{V} = V^{-1},\qquad \Delta\left(\frac{1}{V}\right)=\frac{\Delta V}{V^2}

Completed values (with absolute uncertainties):

$n$	$V/\text{V}$	$\Delta V/\text{V}$	$\frac{1}{V}/\text{V}^{-1}$	$\Delta\left(\frac{1}{V}\right)/\text{V}^{-1}$
2	$4.25$	$0.05$	$0.235$	$0.003$
3	$3.70$	$0.05$	$0.270$	$0.004$
4	$3.25$	$0.05$	$0.308$	$0.005$
5	$2.90$	$0.05$	$0.345$	$0.006$
6	$2.60$	$0.05$	$0.385$	$0.007$
7	$2.35$	$0.05$	$0.426$	$0.009$

Final answer

See completed table in working

Detailed explanation

Background Concept

When a quantity is measured more than once, a common estimate of the absolute uncertainty from repeats is half the range:

\Delta V = \frac{V_{\max}-V_{\min}}{2}

If you then calculate a derived quantity, you must propagate uncertainty. For a reciprocal,

y = \frac{1}{V} = V^{-1}

the approximate absolute uncertainty is found using differentiation:

\Delta y \approx \left|\frac{\mathrm{d}}{\mathrm{d}V}(V^{-1})\right|\Delta V = \frac{\Delta V}{V^2}

Understanding the Question

You are given two readings $V_1$ and $V_2$ for each $n$ . You must:

Find the mean maximum p.d. $V$ .
Find $\frac{1}{V}$ .
Include absolute uncertainties in both $V$ and $\frac{1}{V}$ in the table.

Approach

Compute the mean: $V = (V_1+V_2)/2$ .
Use half-range for uncertainty in $V$ .
Compute $1/V$ .
Propagate uncertainty for the reciprocal using $\Delta(1/V)=\Delta V/V^2$ .

Step-by-Step Reasoning

Example for $n=2$ :

Mean:

V = \frac{4.30+4.20}{2} = 4.25\ \text{V}

Uncertainty in $V$ (half range):

\Delta V = \frac{4.30-4.20}{2} = 0.05\ \text{V}

Reciprocal:

\frac{1}{V} = \frac{1}{4.25} = 0.235\ \text{V}^{-1}

Uncertainty in reciprocal:

\Delta\left(\frac{1}{V}\right) = \frac{0.05}{(4.25)^2} \approx 2.8\times 10^{-3}\ \text{V}^{-1}\ \approx 0.003\ \text{V}^{-1}

Repeat the same steps for all $n$ .

Key Takeaways

Mean from repeats: $\text{mean} = (V_1+V_2)/2$ .
Repeat uncertainty estimate: half the range.
Reciprocal uncertainty: $\Delta(1/V)=\Delta V/V^2$ .

Common Mistakes

Using the full range instead of half the range for $\Delta V$ .
Calculating percentage uncertainty when the question asks for absolute uncertainty.
Forgetting units for $1/V$ (it is $\text{V}^{-1}$ ).
Rounding $1/V$ too aggressively so the plotted points lose precision.

Things to Be Careful About

Keep consistent decimal places within a column.
Uncertainties should usually be to 1 (or sometimes 2) significant figures, and the quoted value should match that precision.
Use the mean $V$ in $\Delta(1/V)=\Delta V/V^2$ , not one of the individual readings.

Techniques used

calculate a mean from repeated readingsestimate absolute uncertainty from repeat spreadpropagate uncertainty through a reciprocal calculationrecord calculated quantities with consistent significant figures

(c)

(i)

Plot a graph of $\frac{1}{V} / \text{V}^{-1}$ against $n$ . Include error bars for $\frac{1}{V}$ .

Graph 2.1

2M Similar

DifficultyMedium-Easy

Worked solution

Plot the points $(n,\,1/V)$ from Table 2.1 and draw vertical error bars of size $\pm\Delta(1/V)$ for each point.

Final answer

See plotted graph with error bars

Detailed explanation

Background Concept

A graph is used to test a linear relationship and to extract constants from its gradient and intercept. Error bars show the uncertainty in measured (or derived) values; here only $1/V$ has uncertainty bars because $n$ is exact.

Understanding the Question

You must plot $\frac{1}{V}$ on the y-axis against $n$ on the x-axis, using your calculated values, and include error bars for $\frac{1}{V}$ using the absolute uncertainties you found in part (b).

Approach

Label axes with quantity and unit: $n$ (no unit) and $1/V\ (\text{V}^{-1})$ .
Plot each point accurately.
For each point, draw a vertical error bar from $\left(\frac{1}{V}-\Delta\left(\frac{1}{V}\right)\right)$ to $\left(\frac{1}{V}+\Delta\left(\frac{1}{V}\right)\right)$ .

Step-by-Step Reasoning

Choose a scale that uses at least half of the provided grid in both directions.
Plot the six points (for $n=2$ to $7$ ) at their calculated y-values.
For each plotted point, use your uncertainty to mark the top and bottom of the error bar in y.
Draw neat, centred error bars (usually with small horizontal end-caps).

Key Takeaways

Error bars represent the uncertainty range for the plotted quantity.
Here, $n$ is a count, so it is treated as exact (no x-error bars).

Common Mistakes

Omitting units in the axis label for $1/V$ .
Drawing error bars of the wrong size (e.g. using $\Delta V$ instead of $\Delta(1/V)$ ).
Plotting $V$ instead of $1/V$ .

Things to Be Careful About

Plotting precision: use at least half a small square when reading positions.
Ensure error bars are vertical (since they apply to $1/V$ only).

Techniques used

plot data points with appropriate axis scaling and labelsadd vertical error bars using absolute uncertaintiesuse a sensible plotting precision for coordinates

(ii)

Draw the straight line of best fit and a worst acceptable straight line on your graph. Label both lines.

2M Similar

DifficultyMedium

Worked solution

Draw a straight line of best fit through the plotted points.

Draw a worst acceptable straight line (steepest or shallowest) that still passes through all the error bars.

Label the two lines clearly (e.g. “best fit” and “worst acceptable”).

Final answer

See labelled best-fit and worst-acceptable lines

Detailed explanation

Background Concept

A line of best fit represents the overall trend of the data (balancing points above and below). A worst acceptable line is drawn to estimate uncertainty in gradient/intercept: it is the steepest or shallowest line that is still consistent with the data within their error bars.

c-ii-best-worst-line

Understanding the Question

You have already plotted points of $1/V$ vs $n$ with vertical error bars. Now you must add:

one best-fit straight line,
one worst acceptable straight line,
and label them.

Approach

Best-fit line: aim for roughly equal scatter about the line across the whole range.
Worst acceptable line: pivot to make it as steep (or as shallow) as possible while still intersecting every vertical error bar somewhere.
Label both; this is essential so the examiner knows which line you used for each calculation.

Step-by-Step Reasoning

Use a ruler to draw one straight line that represents the trend: do not join dot-to-dot.
Check visually that the line passes near the middle of the error bars overall.
Draw a second line:
- try the steepest possible line that still crosses all error bars,
- and/or the shallowest possible line that still crosses all error bars.
Choose the line (steepest or shallowest) that gives the largest difference in gradient (and/or intercept) compared with your best-fit line; this is typically used in the next parts.
Write labels directly on the graph near each line.

Key Takeaways

Worst acceptable line is an uncertainty tool, not a “second best-fit”.
It must be consistent with the error bars, not necessarily with the central points.

Common Mistakes

Drawing the worst line through the plotted points rather than through the error bars.
Drawing a line that misses one or more error bars (then it is not acceptable).
Forgetting to label which line is which.

Things to Be Careful About

Use the full span of the graph when judging acceptability.
Keep the lines straight and drawn with a ruler; small wobbles can change gradient readings.

Techniques used

draw a straight line of best fit through scattered dataconstruct a worst acceptable line consistent with all error barslabel multiple lines clearly on a graph

(iii)

Determine the gradient of the line of best fit. Include the absolute uncertainty in your answer.

gradient = ______

2M Similar

DifficultyMedium

Worked solution

Working

Using a large triangle on the best-fit line,

\text{gradient} = \frac{\Delta(1/V)}{\Delta n} \approx 3.80\times 10^{-2}\ \text{V}^{-1}

Worst acceptable gradient gives

\Delta(\text{gradient}) = |m_{\text{best}}-m_{\text{worst}}| \approx 2.4\times 10^{-3}\ \text{V}^{-1}

Answer

\text{gradient} = (3.80 \pm 0.24)\times 10^{-2}\ \text{V}^{-1}

Final answer

(3.80 ± 0.24) × 10^-2 V^-1

Detailed explanation

Background Concept

For a straight-line graph of $y$ against $x$ , the gradient is

m = \frac{\Delta y}{\Delta x}

To reduce percentage reading error, you should use a large triangle (points far apart on the line).

To estimate the uncertainty in gradient with error bars, Paper 5 typically uses a “best line” and a “worst acceptable line”. A common method is:

\Delta m = |m_{\text{best}} - m_{\text{worst}}|

where $m_{\text{worst}}$ is the steepest or shallowest acceptable line (whichever is further from $m_{\text{best}}$ ).

Understanding the Question

You must find the gradient of the best-fit line on your $1/V$ vs $n$ graph, and include an absolute uncertainty based on your worst acceptable line.

Approach

Choose two widely separated points on the best-fit line (not necessarily the plotted data points).
Calculate $m_{\text{best}} = \Delta(1/V)/\Delta n$ .
Repeat using the worst acceptable line to get $m_{\text{worst}}$ .
Take the absolute difference to get the uncertainty.

Step-by-Step Reasoning

Because $n$ is dimensionless, the unit of gradient is simply the unit of $1/V$ , i.e. $\text{V}^{-1}$ .
Using widely spaced points keeps fractional reading uncertainty low.
The worst acceptable line is constrained by the error bars; it represents an extreme but still plausible gradient.

A representative set of values gives:

m_{\text{best}} \approx 0.0380\ \text{V}^{-1}

and a worst acceptable gradient around

m_{\text{worst}} \approx 0.0356\ \text{V}^{-1}\ \text{or}\ 0.0404\ \text{V}^{-1}

\Delta m \approx 0.0024\ \text{V}^{-1}

Key Takeaways

Use a large triangle for the gradient.
Uncertainty in gradient comes from comparing best and worst acceptable lines.

Common Mistakes

Using $\frac{\Delta n}{\Delta(1/V)}$ instead of $\frac{\Delta(1/V)}{\Delta n}$ .
Using two points very close together, giving a poor gradient.
Taking half the difference between gradients when the scheme expects $|m_{\text{best}}-m_{\text{worst}}|$ .

Things to Be Careful About

Read values from the line, not from the individual plotted points.
Quote the gradient to a sensible number of significant figures consistent with the graph precision.

Techniques used

calculate a gradient using a large triangle on a straight-line graphdetermine gradient uncertainty by comparing best-fit and worst acceptable linesattach correct units to a graph-derived gradient

(iv)

Determine the $y$ -intercept of the line of best fit. Include the absolute uncertainty in your answer.

$y$ -intercept = ______

2M Similar

DifficultyMedium

Worked solution

Working

From the best-fit line,

\text{y-intercept} \approx 0.157\ \text{V}^{-1}

From the worst acceptable line,

\Delta(\text{y-intercept}) = |c_{\text{best}}-c_{\text{worst}}| \approx 0.010\ \text{V}^{-1}

Answer

\text{y-intercept} = (0.157 \pm 0.010)\ \text{V}^{-1}

Final answer

(0.157 ± 0.010) V^-1

Detailed explanation

Background Concept

y = mx + c

$c$ is the y-intercept, i.e. the value of $y$ when $x=0$ . On a graph, it is where the line crosses the y-axis.

The uncertainty in $c$ is found by comparing the intercept from the best-fit line with the intercept from the worst acceptable line:

\Delta c = |c_{\text{best}} - c_{\text{worst}}|

Understanding the Question

You must determine the y-intercept of your best-fit line on the $1/V$ vs $n$ graph, and give an absolute uncertainty using the worst acceptable line.

Approach

Extend the best-fit line to the y-axis ( $n=0$ ) and read off $c_{\text{best}}$ .
Do the same for the worst acceptable line to get $c_{\text{worst}}$ .
Take the absolute difference.

Step-by-Step Reasoning

Since the y-axis is $1/V$ , the intercept has units $\text{V}^{-1}$ .
Even though your data starts at $n=2$ , you can still find the intercept by extending the straight line to $n=0$ .

A representative reading gives

c_{\text{best}} \approx 0.157\ \text{V}^{-1}

and a worst acceptable intercept around

c_{\text{worst}} \approx 0.167\ \text{V}^{-1}\ \text{or}\ 0.152\ \text{V}^{-1}

\Delta c \approx 0.010\ \text{V}^{-1}

Key Takeaways

The intercept is read at $n=0$ by extending the straight line.
Use the worst acceptable line to estimate an intercept uncertainty.

Common Mistakes

Reading the intercept at $n=2$ instead of at $n=0$ .
Forgetting to extend the line accurately to the axis.
Quoting an intercept uncertainty that is unrealistically small compared with the graph scale.

Things to Be Careful About

Use the same method for both best and worst lines.
Ensure your uncertainty is an absolute uncertainty (not a percentage here).

Techniques used

read the y-intercept of a best-fit straight linedetermine y-intercept uncertainty by comparing best-fit and worst acceptable linesquote an intercept with appropriate unit and precision

(d)

(i)

Using your answers to (a), (c)(iii) and (c)(iv), determine the values of $E$ and $C$ . Include appropriate units.

Data: $A = (2.2 \pm 0.2)\ \text{mF}$

$E$ = ______
$C$ = ______

2M Similar

DifficultyMedium

Worked solution

From (a):

\frac{1}{V} = \frac{C}{EA}n + \frac{1}{E}

So y-intercept $c = 1/E$ :

E = \frac{1}{c} = \frac{1}{0.157} = 6.37\ \text{V}

Gradient $m = C/(EA)$ , hence

C = mEA = mA\left(\frac{1}{c}\right)= \frac{mA}{c}

Using $A=2.2\ \text{mF}$ , $m=3.80\times 10^{-2}\ \text{V}^{-1}$ , $c=0.157\ \text{V}^{-1}$ :

C = \frac{(3.80\times 10^{-2})(2.2\ \text{mF})}{0.157} = 0.53\ \text{mF}

Answer

E = 6.37\ \text{V}

C = 0.53\ \text{mF}

Final answer

E = 6.37 V, C = 0.53 mF

Detailed explanation

Background Concept

Once a relationship has been linearised to

\frac{1}{V} = mn + c

the graph directly gives $m$ (gradient) and $c$ (intercept). By comparing this with the theoretical rearrangement, you can solve for the physical constants.

Understanding the Question

You have already found in (a) that

\frac{1}{V} = \frac{C}{EA}n + \frac{1}{E}

You also determined from the graph:

best-fit gradient $m$ (with units $\text{V}^{-1}$ ),
best-fit intercept $c$ (with units $\text{V}^{-1}$ ).

You are given $A = (2.2\pm 0.2)\ \text{mF}$ and must calculate $E$ and $C$ .

Approach

Use the intercept: since $c = 1/E$ , invert it to get $E$ .
Use the gradient: $m = C/(EA)$ , rearrange to $C = mEA$ .
Substitute $E = 1/c$ so that you only need $m$ , $c$ , and $A$ .

Step-by-Step Reasoning

From

\frac{1}{V} = \frac{C}{EA}n + \frac{1}{E}

identify

c = \frac{1}{E}\Rightarrow E = \frac{1}{c}

So if $c\approx 0.157\ \text{V}^{-1}$ ,

E \approx \frac{1}{0.157} = 6.37\ \text{V}

Now use

m = \frac{C}{EA}\Rightarrow C = mEA

and substitute $E=1/c$ :

C = mA\left(\frac{1}{c}\right)=\frac{mA}{c}

Putting $A$ in mF will give $C$ in mF (since $m/c$ is dimensionless):

C \approx \frac{(3.80\times 10^{-2})(2.2)}{0.157}\ \text{mF} \approx 0.53\ \text{mF}

Key Takeaways

Intercept gives $E$ directly here because it simplifies to $1/E$ .
Gradient together with $E$ and $A$ gives $C$ .
Always carry units through to confirm the result makes sense.

Common Mistakes

Using $c = A/(EA)$ but forgetting it simplifies, then making algebra errors.
Mixing up gradient and intercept when substituting.
Converting $A$ to farads but leaving $C$ in mF without stating units clearly.

Things to Be Careful About

$c$ is in $\text{V}^{-1}$ , so $1/c$ is in V.
$n$ has no unit, so both gradient and intercept have the same unit $\text{V}^{-1}$ .
Use the best-fit values (not the worst-line values) for the central estimates of $E$ and $C$ .

Techniques used

use gradient and intercept to solve for constantssubstitute numerical values with correct unitscheck dimensional consistency of derived constants

(ii)

Determine the percentage uncertainty in your value of $C$ .

percentage uncertainty = ______ $\%$

1M Similar

DifficultyMedium-Easy

Worked solution

Using

C = \frac{mA}{c}

\frac{\Delta C}{C} = \frac{\Delta m}{m} + \frac{\Delta A}{A} + \frac{\Delta c}{c}

\frac{\Delta m}{m} = \frac{0.0024}{0.0380}=0.063

\frac{\Delta A}{A} = \frac{0.2}{2.2}=0.091

\frac{\Delta c}{c} = \frac{0.010}{0.157}=0.064

\%\text{ uncertainty in }C = (0.063+0.091+0.064)\times 100 \approx 22\%

Answer

22\%

Final answer

22%

Detailed explanation

Background Concept

For multiplication and division, fractional uncertainties add:

If $Q = ab$ , then $\frac{\Delta Q}{Q} = \frac{\Delta a}{a} + \frac{\Delta b}{b}$ .
If $Q = \frac{a}{b}$ , then $\frac{\Delta Q}{Q} = \frac{\Delta a}{a} + \frac{\Delta b}{b}$ .

This is the standard Paper 5 method for percentage uncertainty.

Understanding the Question

You found $C$ using $m$ (gradient), $c$ (intercept), and $A$ . You are asked for the percentage uncertainty in $C$ , so you must combine the uncertainties in these inputs.

Approach

Use the expression you used to calculate $C$ :

C = \frac{mA}{c}

Add fractional uncertainties from $m$ , $A$ , and $c$ .
Multiply by $100$ to convert to a percentage.

Step-by-Step Reasoning

Start from

C = \frac{mA}{c}

\frac{\Delta C}{C} = \frac{\Delta m}{m} + \frac{\Delta A}{A} + \frac{\Delta c}{c}

$A = 2.2\pm 0.2\ \text{mF}$ gives $\Delta A/A = 0.2/2.2$ .
$\Delta m$ and $\Delta c$ come from your best-fit vs worst acceptable line method.

Add the three contributions and convert to a percentage.

Key Takeaways

Always pick the correct form of $C$ before combining uncertainties.
For products/quotients, add fractional uncertainties.

Common Mistakes

Subtracting fractional uncertainties because one quantity is in the denominator (you still add them).
Using percentage uncertainty in $m$ but absolute uncertainty in $A$ (must be consistent: fractional then convert).
Forgetting to multiply by $100$ .

Things to Be Careful About

Use the absolute uncertainties you actually obtained from the graph.
Quote the final percentage uncertainty to a sensible whole number (typically).

Techniques used

combine fractional uncertainties for products and quotientsconvert fractional uncertainty to percentage uncertaintyuse absolute uncertainties from graph parameters

(e)

The experiment is repeated with 10 capacitors, each of capacitance $C$ , connected in parallel between P and Q. Determine the maximum potential difference $V$ between P and Q.

$V$ = ______ $\text{V}$

1M Similar

DifficultyMedium-Easy

Worked solution

Using

V = \frac{EA}{nC + A}

with $n=10$ , $E=6.37\ \text{V}$ , $A=2.2\ \text{mF}$ , $C=0.53\ \text{mF}$ :

V = \frac{(6.37)(2.2)}{10(0.53)+2.2} = 1.86\ \text{V}

Answer

V = 1.9\ \text{V}

Final answer

1.9 V

Detailed explanation

Background Concept

Once you have determined the constants in an empirical/theoretical relationship, you can use the relationship to predict the dependent variable for new values of the independent variable.

Here the model is

EA = V(nC + A)

so for a given $n$ you can calculate $V$ .

Understanding the Question

The experiment is repeated with $n=10$ capacitors in parallel. Using your calculated values of $E$ and $C$ (and given $A$ ), you must predict the maximum potential difference $V$ between P and Q.

Approach

Rearrange to make $V$ the subject:

V = \frac{EA}{nC + A}

Then substitute $n=10$ and your values for $E$ , $A$ , and $C$ .

Step-by-Step Reasoning

Start from

EA = V(nC + A)

Divide by $(nC+A)$ :

V = \frac{EA}{nC + A}

Substitute values. If you keep $A$ and $C$ both in mF, the mF cancels in numerator/denominator, giving volts as required.
Evaluate and round sensibly (typically to 2 s.f. for a 1-mark numerical answer).

Key Takeaways

Use the derived relationship to extrapolate/interpolate to a new $n$ .
Keep capacitances in consistent units so they cancel cleanly.

Common Mistakes

Substituting $n=10$ into the graph equation for $1/V$ but forgetting to invert to get $V$ .
Mixing units (e.g. putting $A$ in F and $C$ in mF).
Arithmetic error in the denominator $nC + A$ .

Things to Be Careful About

$nC + A$ must be evaluated before dividing.
Quote $V$ with unit V and appropriate significant figures.

Techniques used

substitute derived constants into the given relationshiprearrange to make the required variable the subjectevaluate an expression and quote an appropriate number of significant figures

Physics 9702/52 — May/June 2025

Apparatus

Variables

Procedure and measurements

Analysis of data (test of relationship)

Determination of PPP and QQQ

Safety

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

1) Set-up and measurements

2) Collecting data for one mass

3) Varying the independent variable

4) Linearisation and graph

5) Extracting PPP and QQQ

6) Control of variables (why it matters)

7) Safety

Key Takeaways

Common Mistakes

Things to Be Careful About

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Determination of $P$ and $Q$

5) Extracting $P$ and $Q$