Physics 9702/52 — May/June 2024

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

questions

marks

minutes

Topics Planning · Analysis, Conclusions and Evaluation

Q115MMedium-HardPlanning

A spring is attached to a strong cylindrical magnet of length $L$ and cross-sectional area $A$ . The magnet is placed on thin card on top of a magnetic sheet on the bench, as shown in Fig. 1.1.

Fig. 1.1

The thickness of the card is $t$ . The magnetic flux density at one of the poles of the magnet is $B$ .

A force is applied upwards to the spring. The extension of the spring when the magnet just leaves the card is $s$ .

It is suggested that $s$ is related to $t$ by the relationship

ks = \frac{ALBZ}{t}

where $k$ is the spring constant of the spring and $Z$ is a constant.

Plan a laboratory experiment to test the relationship between $s$ and $t$ .

Draw a diagram showing the arrangement of your equipment.

Explain how the results could be used to determine a value for $Z$ .

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.

DifficultyMedium-Hard

Worked solution

Diagram

diagram-1

Variables

Independent variable: total card thickness $t$ .
Dependent variable: spring extension $s$ when the magnet just leaves the card.
Controlled variables: same magnet (fixed $A$ , $L$ , pole), same spring (fixed $k$ ), same magnetic sheet and its position, same orientation and alignment of magnet on the card, same contact area and card material (use identical cards stacked), same temperature.

Apparatus

Retort stand and clamp, spring, cylindrical magnet, magnetic sheet, set of identical thin cards (or card shims) to vary $t$ , micrometer screw gauge (or digital calipers) for $t$ , metre rule / mm scale fixed next to spring with a pointer attached to the spring or magnet, set square to read the pointer, (optional) camera/phone to judge detachment.

Procedure and measurements

Set up the magnet attached to the bottom of the vertical spring. Place the magnetic sheet on the bench and place card(s) on top. Put the magnet upright on the card, centred at the same marked position each time.
Measure the total thickness $t$ of the card stack using a micrometer. Take several readings at different points and use the mean.
With no upward pull, record the spring length $l_0$ (or pointer reading) with the magnet resting on the card.
Increase the upward pull slowly and steadily (e.g. raise the top clamp in small steps). Identify the instant the magnet just leaves the card (first visible gap / card can be slid freely / video frame where separation begins).
Record the spring length $l$ (or pointer reading) at that instant and calculate the extension

s = l - l_0.

Repeat steps 3–5 at least three times for the same $t$ and calculate the mean $s$ .
Change $t$ by adding/removing identical cards to obtain at least 6 values of $t$ over a suitable range, and repeat.

Determining $k$

Determine $k$ separately by hanging known masses from the spring and measuring extension $x$ :

F = mg = kx \Rightarrow k = \frac{mg}{x}

(using a graph of $F$ vs $x$ ; gradient $=k$ ).

Analysis of data and determination of $Z$

Given

ks = \frac{ALBZ}{t}

rearrange to

s = \left(\frac{ALBZ}{k}\right)\left(\frac{1}{t}\right).

Calculate $1/t$ for each card thickness.
Plot a graph of $s$ (y-axis) against $1/t$ (x-axis).
A straight line through the origin supports the relationship.
Gradient $m$ is

m = \frac{ALBZ}{k}

Z = \frac{mk}{ALB}.

Safety

Strong magnet: keep fingers clear to avoid pinching; keep away from phones/computers/credit cards and anyone with a pacemaker.
Secure the retort stand so it cannot topple; do not overload the spring; keep feet clear if masses are used to determine $k$ .

Final answer

See working

Detailed explanation

Background Concept

The suggested relationship is

ks = \frac{ALBZ}{t}.

Here $k$ is the spring constant (units $\text{N m}^{-1}$ ), so $ks$ is a force. The right-hand side must also have units of force, and the relationship predicts an inverse proportionality between the extension at lift-off ( $s$ ) and the card thickness ( $t$ ).

To test a relationship experimentally, you vary one variable (independent), measure the response (dependent), keep all other relevant quantities constant (controlled), and then analyse the data in a way that would produce a straight line if the relationship is correct.

A key Paper 5 skill is linearisation. If

s \propto \frac{1}{t}

then plotting $s$ against $1/t$ should give a straight line through the origin. The gradient of that line contains the constant(s) you want, allowing you to determine $Z$ .

Understanding the Question

You are given an apparatus: a spring attached to a cylindrical magnet resting on card on a magnetic sheet. You pull upwards until the magnet just detaches, and you measure the spring extension $s$ at that instant.

You must plan an experiment to check whether changing the card thickness $t$ changes $s$ according to

ks = \frac{ALBZ}{t}.

You must also explain how to use your results to find $Z$ . That means your plan must include (i) how to vary and measure $t$ , (ii) how to measure $s$ reliably at the moment of detachment, (iii) how to measure or know $k$ , and (iv) what graph to plot so that $Z$ can be obtained from a gradient.

Approach

Choose $t$ as the independent variable: use stacks of identical cards so only thickness changes.
Measure $t$ with a micrometer because it is a small thickness.
For each $t$ , slowly increase the extension until the magnet just lifts; record $s$ using a fixed scale and pointer to avoid parallax.
Repeat for each $t$ to reduce random uncertainty.
Determine $k$ separately using weights and a force-extension graph.
Linearise the relationship by plotting $s$ vs $1/t$ and use the gradient to calculate $Z$ .

Step-by-Step Reasoning

diagram-1

1) Setting up and choosing variables

Independent variable: $t$ (easy to vary by adding cards).
Dependent variable: $s$ at the instant of detachment.
Controlled variables are the quantities appearing in the formula that must not change: $A$ , $L$ , $B$ , $k$ , and also practical factors that could change the magnetic interaction (same sheet, same orientation, same position).

2) Measuring thickness $t$ well

A single card may be too thin to measure precisely. Stacking cards increases $t$ so the percentage uncertainty in $t$ is smaller.
Use a micrometer to measure total thickness. Take several readings at different points because card can be slightly uneven; average them.

3) Measuring the extension $s$ consistently

You need a clear definition of “just leaves the card”. Good operational definitions include:
- the first moment you can see a gap,
- the card can be slid with negligible friction,
- a video frame where separation begins.
Record a reference reading $l_0$ with the magnet resting on the card (no upward pull). Then at lift-off record $l$ . The extension is

s = l - l_0.

Using a pointer and a fixed vertical scale reduces parallax; reading with a set square helps keep the eye level correct.

4) Repeats

Detachment is a threshold event and can vary slightly due to small vibrations or how steadily you pull.
Repeat at least three times for each $t$ and take the mean $s$ .

5) Measuring the spring constant $k$
You cannot assume $k$ unless it is given, so determine it:

Hang known masses $m$ , measure extension $x$ .
Compute force $F = mg$ .
Plot $F$ against $x$ ; the gradient is $k$ because

F = kx.

A graph is better than a single pair of readings because it averages out random error and reveals any non-linearity.

6) Linearising and extracting $Z$
Start with

ks = \frac{ALBZ}{t}.

Divide both sides by $k$ and rearrange to match $y = mx$ :

s = \left(\frac{ALBZ}{k}\right)\left(\frac{1}{t}\right).

So if you plot $s$ (y-axis) against $1/t$ (x-axis):

You expect a straight line.
The line should pass through the origin (within experimental uncertainty) because when $1/t = 0$ (very large $t$ ), the model predicts $s \to 0$ .
The gradient $m$ is

m = \frac{ALBZ}{k}.

Hence

Z = \frac{mk}{ALB}.

If $A$ , $L$ , and $B$ are known (or measured separately), then once you have $m$ and $k$ you can calculate $Z$ .

Key Takeaways

A good plan clearly identifies independent, dependent, and controlled variables.
For an inverse relationship, plotting against the reciprocal of the independent variable is a standard linearisation.
Threshold measurements (like “just detaches”) require a clear, repeatable criterion and repeats.
Constants in a model are commonly obtained from the gradient of a straight-line graph.

Common Mistakes

Plotting $s$ against $t$ and expecting a straight line (the model predicts $s \propto 1/t$ , not $s \propto t$ ).
Not stating how $k$ is obtained, even though $k$ is required to find $Z$ .
Varying more than one factor at a time (e.g. changing card material as well as thickness, which could change separation, friction, or magnetic properties).
Measuring $t$ with a ruler (insufficient resolution for thin card).
No clear definition of “just leaves the card”, leading to inconsistent readings.

Things to Be Careful About

Ensure the magnet is always oriented the same way (same pole toward the sheet) because $B$ at the pole could differ if orientation changes.
Keep the contact area and alignment consistent: always place the magnet at the same marked position on the card and keep it vertical.
Pull up slowly to reduce overshoot; vibrations can cause premature detachment.
Use a sufficiently wide range of $t$ (and therefore $1/t$ ) to make the graph gradient more reliable.
Quote $t$ and $s$ to sensible precision (micrometer resolution for $t$ ; mm scale for $s$ ) and use mean values from repeats.
Handle strong magnets safely and keep them away from sensitive devices and medical implants.

Techniques used

identify independent, dependent and controlled variablesdesign a method to detect the instant of detachment consistentlymeasure small lengths using a micrometer and avoid parallax on scaleslinearise the suggested relationship to obtain a straight-line graphdetermine a constant from the gradient of a graph with repeat readings

Answer

gradient (= n)

(y)-intercept (= \lg!\left(\frac{2}{C}\right))

Final answer

gradient = n; y-intercept = lg(2/C)

Detailed explanation

Background Concept

A power-law relationship can be tested by taking logarithms. If

T = kL^n

then taking base-10 logs gives

\lg T = \lg k + n\,\lg L

which is a straight-line form (y = mx + c) with (y = \lg T), (x = \lg L), gradient (m = n) and intercept (c = \lg k).

Understanding the Question

You are told

T = \frac{2L^n}{C}

and that a graph of (\lg T) (vertical axis) against (\lg L) (horizontal axis) is plotted. The question asks what the gradient and the y-intercept represent in terms of the constants (n) and (C).

Approach

Take (\lg) of both sides.
Rearrange into (\lg T = (\text{something}),\lg L + (\text{constant})).
Compare with (y = mx + c).

Step-by-Step Reasoning

Start with

T = \frac{2L^n}{C}

Take (\lg) of both sides:

\lg T = \lg\!\left(\frac{2L^n}{C}\right)

Use log rules (\lg(ab)=\lg a+\lg b) and (\lg(L^n)=n\lg L):

\lg T = \lg\!\left(\frac{2}{C}\right) + \lg(L^n) = \lg\!\left(\frac{2}{C}\right) + n\lg L

So for a plot of (\lg T) against (\lg L):

gradient (= n)
y-intercept (= \lg(2/C))

Key Takeaways

Logarithms turn (T \propto L^n) into a straight line.
The power (n) becomes the gradient on a log-log plot.
The constant factor becomes the y-intercept (as a log).

Common Mistakes

Writing the intercept as (\lg(2C)) instead of (\lg(2/C)).
Forgetting that (\lg(L^n)=n\lg L).

Things to Be Careful About

The base of the log is (10) because (\lg) is used.
Keep the linear form clearly as (\lg T = n\lg L + \lg(2/C)) before reading off (m) and (c).

Techniques used

take logarithms to linearise a power-law relationshipmatch a linearised equation to the form y = mx + cidentify gradient and y-intercept from the linear form

(b)

Values of $L$ and $T$ are given in Table 2.1.

Table 2.1

$L / \text{cm}$	$T / 10^{-5}\ \text{s}$	$\lg(L / \text{cm})$	$\lg(T / 10^{-5}\ \text{s})$
54	$24 \pm 1$
70	$32 \pm 1$
86	$39 \pm 1$
108	$49 \pm 2$
140	$64 \pm 2$
167	$74 \pm 2$

Calculate and record values of $\lg(L / \text{cm})$ and $\lg(T / 10^{-5}\ \text{s})$ in Table 2.1.
Include the absolute uncertainties in $\lg T$ .

2M Similar

DifficultyMedium

Worked solution

For (\lg T),

\Delta(\lg T)=\frac{\Delta T}{T\ln 10}=0.434\,\frac{\Delta T}{T}

Completed Table 2.1 (logs to 3 d.p.):

(L/\text{cm})	(T/10^{-5},\text{s})	(\lg(L/\text{cm}))	(\lg(T/10^{-5},\text{s}))
54	(24\pm1)	(1.732)	(1.380\pm0.018)
70	(32\pm1)	(1.845)	(1.505\pm0.014)
86	(39\pm1)	(1.935)	(1.591\pm0.011)
108	(49\pm2)	(2.033)	(1.690\pm0.018)
140	(64\pm2)	(2.146)	(1.806\pm0.014)
167	(74\pm2)	(2.223)	(1.869\pm0.012)

Final answer

lgL: 1.732, 1.845, 1.935, 2.033, 2.146, 2.223; lgT: 1.380±0.018, 1.505±0.014, 1.591±0.011, 1.690±0.018, 1.806±0.014, 1.869±0.012

Detailed explanation

Background Concept

When plotting a log-log graph, you need (\lg) values. If a measured quantity (T) has an absolute uncertainty (\Delta T), the uncertainty in (\lg T) is found from differentiation:

\lg T = \frac{\ln T}{\ln 10}

\Delta(\lg T) \approx \left|\frac{\mathrm{d}(\lg T)}{\mathrm{d}T}\right|\Delta T = \frac{1}{T\ln 10}\,\Delta T =0.434\,\frac{\Delta T}{T}

This gives an absolute uncertainty in the plotted (y)-value, used for error bars.

Understanding the Question

You are given values of (L) (in cm) and (T) (in units of (10^{-5},\text{s})) with uncertainties in (T). You must:

calculate (\lg(L/\text{cm})) for each row,
calculate (\lg(T/10^{-5},\text{s})) for each row,
include absolute uncertainties in (\lg T) (i.e. in the log values of (T)).

Approach

Use a calculator to find each (\lg) value.
For each (T\pm \Delta T), compute (\Delta(\lg T)=0.434(\Delta T/T)).
Record logs to a consistent number of decimal places (typically 3 d.p. for graphing).

Step-by-Step Reasoning

Example for the first row:

(T = 24) (in units of (10^{-5},\text{s})), (\Delta T = 1).

Log value:

\lg T = \lg(24)=1.380\,(3\ \text{d.p.})

Uncertainty in log:

\Delta(\lg T)=0.434\frac{1}{24}=0.0181\approx 0.018

So record (1.380\pm0.018).

Repeat the same method for each row; the fractional uncertainty (\Delta T/T) is what matters.

Key Takeaways

Log values are plotted, so you must also convert uncertainties into log-space.
For (\lg), use (\Delta(\lg T)=\Delta T/(T\ln 10)).
Consistent decimal places help accurate graph reading.

Common Mistakes

Using (\Delta(\lg T)=\lg(\Delta T)) (incorrect).
Forgetting the factor (\ln 10) (or (0.434)).
Writing too few decimal places, making graphing and gradients inaccurate.

Things to Be Careful About

The table uses (T/10^{-5},\text{s}), so you take (\lg) of the numbers shown (24, 32, ...).
Uncertainty in (\lg T) should be absolute (e.g. (\pm 0.014)), not a percentage.
Keep rounding sensible: log values typically to 3 d.p.; log-uncertainties often to 2–3 s.f.

Techniques used

calculate base-10 logarithms of measured quantitiespropagate an absolute uncertainty through a logarithm using differentiationrecord values to consistent decimal places in a results table

(c)

(i)

Plot a graph of $\lg(T / 10^{-5}\ \text{s})$ against $\lg(L / \text{cm})$ . Include error bars for $\lg T$ .

Graph

2M Similar

DifficultyMedium

Worked solution

Plot (y=\lg(T/10^{-5},\text{s})) against (x=\lg(L/\text{cm})) with vertical error bars (\pm\Delta(\lg T)).

Points to plot ((x, y)):

((1.732,\ 1.380)) with (\pm 0.018)
((1.845,\ 1.505)) with (\pm 0.014)
((1.935,\ 1.591)) with (\pm 0.011)
((2.033,\ 1.690)) with (\pm 0.018)
((2.146,\ 1.806)) with (\pm 0.014)
((2.223,\ 1.869)) with (\pm 0.012)

Final answer

See plotted points with error bars (coordinates given in working).

Detailed explanation

Background Concept

A graph is used to test whether a relationship is linear and to extract constants from the gradient and intercept. When uncertainties are provided for the dependent variable, they must be shown as error bars so that a realistic uncertainty in the gradient/intercept can be estimated.

Understanding the Question

You must plot (\lg(T/10^{-5},\text{s})) (y-axis) against (\lg(L/\text{cm})) (x-axis), using the values you calculated in part (b). You must also include error bars for (\lg T), using the absolute uncertainties (\Delta(\lg T)).

Approach

Choose sensible scales so the plotted points occupy at least half the grid in each direction.
Label axes exactly as stated (including the quantity inside the log and the units).
Plot each point accurately.
Draw a vertical error bar through each point of total height (2\Delta(\lg T)).

Step-by-Step Reasoning

Put (x = \lg(L/\text{cm})) on the horizontal axis and (y = \lg(T/10^{-5},\text{s})) on the vertical axis.
For each row, locate (x) and (y) on the grid and mark a small cross.
Add the error bar: from (y-\Delta(\lg T)) to (y+\Delta(\lg T)) at that same (x).

For example, the first row is (x=1.732), (y=1.380) and (\Delta y=0.018), so the error bar runs from (1.362) to (1.398).

Key Takeaways

Correct axis labels and scales are essential marks in Paper 5.
Error bars must match the uncertainties in the plotted quantity (here, (\lg T)).

Common Mistakes

Plotting (T) against (L) instead of their logarithms.
Using error bars based on (\Delta T) rather than (\Delta(\lg T)).
Omitting units inside the (\lg) label or writing ambiguous labels.

Things to Be Careful About

Error bars are vertical only here because only (T) uncertainties are given.
Ensure plotting precision: each point should be within about half a small square of the correct location.
Use consistent rounding from the table (don’t mix 2 d.p. and 3 d.p. values).

Techniques used

plot points on log-log axes using calculated coordinateslabel axes with quantity and unit in log formadd vertical error bars using absolute uncertainties in the y-values

(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 and label:

a straight line of best fit through the points,
a worst acceptable straight line (steepest or shallowest) that still passes within all error bars.

(For example, best-fit may be represented by (y \approx 1.00x - 0.35) and a worst acceptable line by (y \approx 1.06x - 0.47).)

Final answer

Best-fit line and a labelled worst acceptable line drawn.

Detailed explanation

Background Concept

A best-fit line represents the most likely linear relationship. To estimate the uncertainty in gradient (and intercept) from experimental scatter and error bars, Paper 5 commonly uses a “worst acceptable line”: the steepest or shallowest straight line that is still consistent with all the plotted error bars.

Understanding the Question

After plotting the data with error bars, you must draw:

a straight line of best fit (most balanced among the points), and
one worst acceptable line (either maximum gradient or minimum gradient) that still fits the data within the error bars.

Approach

Best-fit line: draw a single straight line so that points are roughly equally distributed above and below the line (do not join dots).
Worst acceptable line: pivot the line to make it as steep or as shallow as possible while still passing through (or at least intersecting) every vertical error bar.
Label each line clearly (e.g. “best fit” and “worst”).

Step-by-Step Reasoning

Place a ruler and draw a straight best-fit line through the central trend.
For the worst line, choose steepest OR shallowest:
- Steepest line: try to go through the bottom of the first error bar and the top of the last error bar (or equivalent extremes), then check all intermediate error bars are still intersected.
- Shallowest line: try to go through the top of the first error bar and bottom of the last error bar, then check again.
Once you have an acceptable extreme, label it “worst acceptable line”.

Key Takeaways

Best-fit gives the best estimate of the relationship.
Worst acceptable line provides an uncertainty estimate in gradient/intercept.

Common Mistakes

Drawing the worst line that misses one or more error bars (not acceptable).
Drawing two worst lines when only one is requested, or failing to label them.
Forcing the best-fit line through the origin without justification.

Things to Be Careful About

The worst acceptable line must be straight and must be plausible by the error bars, not by the points alone.
Choose the worst line that gives the greatest difference in gradient from the best-fit line, because that maximises the uncertainty estimate (as required by the method).

Techniques used

draw a straight line of best fit through scattered dataconstruct a worst acceptable line consistent with all error barslabel lines clearly for later gradient/intercept extraction

(iii)

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

gradient = ______

2M Similar

DifficultyMedium-Hard

Worked solution

Using the best-fit line (large triangle):

\text{gradient } m = \frac{\Delta y}{\Delta x} \approx 1.00

Worst acceptable line gives (e.g.)

m_{\text{worst}} \approx 1.06

Absolute uncertainty:

\Delta m = \left|m_{\text{worst}}-m\right| \approx 0.06

Answer

(\text{gradient} = 1.00 \pm 0.06)

Final answer

1.00 ± 0.06

Detailed explanation

Background Concept

For a straight line graph of (y) against (x), the gradient is

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

To estimate uncertainty in (m) from a graph with error bars:

find (m) from the best-fit line,
find (m_{\text{worst}}) from the worst acceptable line,
take

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

This is the standard Paper 5 “best vs worst” method.

Understanding the Question

You must read the gradient of your best-fit line and include an absolute uncertainty. The uncertainty comes from comparing with the gradient of your worst acceptable line (drawn in part (ii)).

Approach

On the best-fit line, choose two well-separated points (to reduce reading error).
Compute (m=\Delta y/\Delta x).
Repeat on the worst acceptable line to get (m_{\text{worst}}).
Quote (m \pm \Delta m).

Step-by-Step Reasoning

If you use endpoints far apart (large triangle), small reading errors in (x) and (y) have less effect on (m).
Suppose the best-fit line gives (m\approx 1.00).
A steepest acceptable line (still within all error bars) might give (m_{\text{worst}}\approx 1.06) (alternatively the shallowest might be (\approx 0.94); you choose the one that is furthest from the best-fit gradient).
Then

\Delta m = |1.06-1.00|=0.06

So you report (1.00\pm 0.06).

Key Takeaways

Gradient uncertainty from graphs is obtained by comparing best-fit and worst acceptable gradients.
Use a large triangle for reliable gradients.

Common Mistakes

Calculating (\Delta x/\Delta y) instead of (\Delta y/\Delta x).
Using two points that are too close together on the line.
Taking uncertainty as half the difference between two worst lines (not the method when one worst line is drawn).

Things to Be Careful About

Keep enough significant figures: typically 2–3 s.f. for gradients read from graphs.
The worst line must be consistent with the error bars; otherwise the uncertainty estimate is not valid.

Techniques used

calculate gradient using a large triangle on the best-fit linecalculate gradient of a worst acceptable linefind absolute uncertainty from the difference between best and worst gradients

(iv)

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

$y$ -intercept = ______

2M Similar

DifficultyMedium-Hard

Worked solution

Best-fit line gives y-intercept

c \approx -0.35

Worst acceptable line gives (e.g.)

c_{\text{worst}} \approx -0.47

Absolute uncertainty:

\Delta c = \left|c_{\text{worst}}-c\right| \approx 0.13

Answer

(y\text{-intercept} = -0.35 \pm 0.13)

Final answer

-0.35 ± 0.13

Detailed explanation

Background Concept

For a straight line

y = mx + c

the y-intercept is the value of (y) when (x=0). On a graph, you can obtain (c) by extending the line to the y-axis, or by substituting a point on the line into (c=y-mx).

Uncertainty in (c) from a graph with error bars is commonly estimated using best-fit vs worst acceptable line:

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

Understanding the Question

You must find the y-intercept of the best-fit line, and include an absolute uncertainty, using your worst acceptable line as the comparison.

Approach

Determine (c) from the best-fit line (extend to y-axis or compute (y-mx)).
Determine (c_{\text{worst}}) from the worst acceptable line the same way.
Find (\Delta c = |c_{\text{worst}}-c|).

Step-by-Step Reasoning

From the best-fit line, you may read an intercept around (-0.35).
Using the worst acceptable line (for example the steepest one), the intercept might be around (-0.47).
Then

\Delta c = |-0.47-(-0.35)| = 0.12 \approx 0.13

So report (c=-0.35\pm0.13).

Key Takeaways

The intercept is part of the linear model and is found from the line, not from any single data point.
The uncertainty comes from how far the worst acceptable intercept differs from the best-fit intercept.

Common Mistakes

Taking the intercept from the plotted points instead of from the line.
Forgetting that the intercept may lie off the plotted grid (needs careful extension).
Using a worst line that is not acceptable (misses some error bars), giving a meaningless uncertainty.

Things to Be Careful About

If extending lines beyond the grid, use a ruler and keep the extension straight.
Quote intercept uncertainty to a sensible precision (usually 1–2 s.f.).

Techniques used

read the y-intercept of the best-fit lineread the y-intercept of a worst acceptable lineestimate intercept uncertainty from the best-vs-worst difference

(d)

Using your answers to (a), (c)(iii) and (c)(iv), determine the values of $n$ and $C$ . Include the absolute uncertainties in your values. You need not be concerned with units.

$n$ = ______
$C$ = ______

3M Similar

DifficultyMedium-Hard

Worked solution

From (a): gradient (=n), intercept (=\lg(2/C)).

Using (c)(iii):

n = 1.00 \pm 0.06

Using (c)(iv), with (c=-0.35\pm0.13):

10^{c}=\frac{2}{C} \quad\Rightarrow\quad C=\frac{2}{10^{c}}

Best value:

C = \frac{2}{10^{-0.35}} = 4.4

Uncertainty:

\frac{\Delta C}{C}=\ln(10)\,\Delta c = 2.303(0.13)=0.30

\Delta C = 0.30\times 4.4 \approx 1.3

Answer

(n = 1.00\pm0.06)

(C = 4.4\pm1.3)

Final answer

n = 1.00 ± 0.06; C = 4.4 ± 1.3

Detailed explanation

Background Concept

From the linearised relationship

\lg T = n\,\lg L + \lg\!\left(\frac{2}{C}\right)

we identify:

gradient (m = n)
intercept (c = \lg(2/C))

To recover (C) from the intercept:

10^{c} = \frac{2}{C} \quad \Rightarrow \quad C = \frac{2}{10^{c}}

If (C\propto 10^{-c}), then uncertainties propagate via

\ln C = \ln 2 - c\ln 10 \;\Rightarrow\; \frac{\Delta C}{C} = (\ln 10)\,\Delta c

Understanding the Question

You are told to use your answers from (a), (c)(iii) and (c)(iv) to find (n) and (C), including absolute uncertainties. So:

(n) comes directly from the graph gradient.
(C) comes from the intercept using (c=\lg(2/C)).

Approach

Set (n) equal to the gradient (including its uncertainty).
Rearrange (c=\lg(2/C)) to find (C).
Convert intercept uncertainty (\Delta c) into (\Delta C) using fractional uncertainty (\Delta C/C = (\ln 10)\Delta c).

Step-by-Step Reasoning

From the graph work, suppose

m = 1.00\pm0.06

Then

n = 1.00\pm0.06

Intercept is

c = -0.35\pm0.13

Use

C=\frac{2}{10^{c}}=2\times 10^{-c}

Best value:

C = \frac{2}{10^{-0.35}} = 2\times 10^{0.35} \approx 2\times 2.24 \approx 4.4

Uncertainty:

\frac{\Delta C}{C} = (\ln 10)\Delta c = 2.303\times 0.13 \approx 0.30

\Delta C \approx 0.30\times 4.4 \approx 1.3

Hence (C=4.4\pm 1.3).

Key Takeaways

On a log-log plot, the gradient gives the power (n).
The intercept gives the multiplicative constant, but because it is a log, recovering (C) involves powers of 10.
Intercept uncertainty produces a relatively large percentage uncertainty in (C).

Common Mistakes

Using (C=2\times 10^{c}) instead of (C=2/10^{c}).
Treating (\Delta c) as a percentage uncertainty directly (it is not).
Forgetting to include uncertainties in the final (n) and (C).

Things to Be Careful About

Be consistent with the definition of the intercept you used: it must match the linearised equation from part (a).
When propagating uncertainty from (c) to (C), use (\Delta C/C = (\ln 10)\Delta c), not (\Delta C/C = \Delta c).
Quote uncertainties to appropriate significant figures (usually 1–2 s.f.), and match the value’s decimal place accordingly.

Techniques used

use gradient-intercept results to determine constants in a linearised modelrearrange an intercept expression to solve for a constantpropagate uncertainty from a logarithmic intercept into a multiplicative constant

(e)

The experiment is repeated. Determine the length $L$ of the rod that gives a value of $T$ of $0.10\ \text{ms}$ .

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

1M Similar

DifficultyMedium-Easy

Worked solution

Use

T=\frac{2L^{n}}{C}

With (T=0.10,\text{ms}=1.0\times10^{-4},\text{s}=10\times 10^{-5},\text{s}), so (T/10^{-5},\text{s}=10).

Using (n\approx 1.00) and (C\approx 4.4):

10=\frac{2L}{4.4}

L=\frac{4.4\times 10}{2}=22\ \text{cm}

Answer

(L = 22\ \text{cm})

Final answer

22 cm

Detailed explanation

Background Concept

Once constants in a model are found, you can use the model to predict values. Here the suggested relationship is

T=\frac{2L^n}{C}

If (n\approx 1), this is approximately a direct proportionality between (T) and (L).

Understanding the Question

You repeat the experiment and want the rod length (L) that produces a period (T=0.10,\text{ms}). You should use the values of (n) and (C) obtained in part (d). The table/graph worked with (T) measured in units of (10^{-5},\text{s}) and (L) in cm, so your (T) should be expressed consistently.

Approach

Convert (0.10,\text{ms}) into (10^{-5},\text{s}) units.
Substitute (T), (n), (C) into (T=2L^n/C).
Rearrange to solve for (L).

Step-by-Step Reasoning

Convert the target period:

0.10\,\text{ms}=0.10\times 10^{-3}\,\text{s}=1.0\times 10^{-4}\,\text{s}

In units of (10^{-5},\text{s}):

\frac{T}{10^{-5}\,\text{s}}=\frac{1.0\times 10^{-4}}{1.0\times 10^{-5}}=10

Use (n\approx 1.00) and (C\approx 4.4):

10=\frac{2L^{1.00}}{4.4}

Rearrange:

L=\frac{4.4\times 10}{2}=22\ \text{cm}

Key Takeaways

Always convert the target value into the same units/scale used when determining constants.
Rearranging a proportionality is straightforward once (n) and (C) are known.

Common Mistakes

Using (T=0.10) directly without converting units (ms to (10^{-5},\text{s})).
Squaring or taking logs unnecessarily instead of direct rearrangement.

Things to Be Careful About

The predicted (L) here is an extrapolation beyond the measured range (data started at (54,\text{cm})), so experimentally it may be less reliable.
The question does not ask for an uncertainty in (L), so only the best estimate is required.

Techniques used

convert a target period into the same units used in the modelrearrange a power-law relationship to solve for lengthsubstitute determined constants to predict a new value

Physics 9702/52 — May/June 2024

Diagram

Variables

Apparatus

Procedure and measurements

Determining kkk

Analysis of data and determination of ZZZ

Safety

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

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

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

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

Determining $k$

Analysis of data and determination of $Z$