Physics 9702/51 — May/June 2024

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

Fig. 1.1

Fig. 1.1 shows a small solid metal cylinder of mass $m$ , length $L$ and diameter $d$ .

The cylinder is heated to a uniform temperature. The cylinder is then removed from the heat source and the cylinder is wrapped in an insulating material.

The temperature of the room is $T_R$ . At time $t$ after the cylinder starts to cool, the surface temperature of the cylinder is $T_C$ .

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

(T_C - T_R) = Ze^{-\frac{UAt}{mc}}

where $A$ is the total surface area of the cylinder, $c$ is the specific heat capacity of the metal, and $U$ and $Z$ are constants.

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

Draw a diagram showing the arrangement of your equipment.

Explain how the results could be used to determine values for $U$ and $Z$ .

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

Answer

Variables

Independent variable: time $t$ since the cylinder is removed from the heater and insulated.
Dependent variable: cylinder surface temperature $T_C$ .
Controlled variables: same cylinder ( $m, L, d$ fixed so $A$ fixed), same insulation material and thickness, same room temperature $T_R$ (avoid draughts / same location), same initial temperature (heat to the same high temperature each run), same position of temperature sensor on cylinder and same contact pressure.

Apparatus

Metal cylinder, heat source (hot water bath or electric heater), insulation (lagging), thermocouple/temperature probe + data logger, stopwatch (if not logging time automatically), top-pan balance, vernier calipers/micrometer, retort stand/tape to hold probe in contact, ruler.

diagram-1

Procedure and measurements

Measure mass $m$ using a balance.
Measure length $L$ and diameter $d$ (several readings, take mean). Calculate total surface area

A = \pi dL + 2\pi\left(\frac{d}{2}\right)^2 = \pi dL + \frac{\pi d^2}{2}.

Measure room temperature $T_R$ using the same probe.
Heat the cylinder for long enough so its temperature is uniform (e.g. leave in a stirred boiling water bath; ensure full immersion but dry surface before insulation).
Quickly remove cylinder (tongs/heatproof gloves), wrap immediately in the same insulation each time, leaving a small hole for the thermocouple junction to contact the metal surface.
Start timing at $t=0$ as cooling begins. Record $T_C$ at regular intervals (or continuously with a data logger) until $T_C$ is close to $T_R$ .
Repeat the cooling run and average $T_C$ values at the same times (or repeat full run to check repeatability).

Analysis (to obtain $U$ and $Z$ )

Given

(T_C-T_R)=Ze^{-\frac{UAt}{mc}}.

Take natural logs:

\ln(T_C-T_R)=\ln Z-\frac{UA}{mc}t.

Calculate $(T_C-T_R)$ and then $\ln(T_C-T_R)$ for each time.
Plot $y=\ln(T_C-T_R)$ against $x=t$ .
Gradient $g$ of best-fit line:

g=-\frac{UA}{mc} \Rightarrow U=-\frac{gmc}{A}.

Intercept $b$ :

b=\ln Z \Rightarrow Z=e^b.

(Use worst acceptable line to estimate uncertainty in gradient and hence in $U$ .)

Safety

Hot metal cylinder: use tongs/heatproof gloves; keep cylinder on a heatproof mat.
Hot water bath/heater: avoid splashes, keep electrics dry, switch off heater when not needed.
Ensure insulation and stand are stable to prevent rolling/falling cylinder.

Final answer

See working

Detailed explanation

Background Concept

The suggested relationship

(T_C - T_R) = Ze^{-\frac{UAt}{mc}}

has the form of exponential cooling. The quantity $(T_C-T_R)$ is the temperature excess above room temperature. Exponential decay means that equal fractions (not equal amounts) of the excess temperature are lost in equal time intervals.

To test an exponential relationship in the lab, it is standard to linearise it using logarithms:

\ln(T_C-T_R)=\ln Z-\frac{UA}{mc}t.

This is now in the straight-line form $y = b + gx$ with:

$y = \ln(T_C-T_R)$
$x = t$
gradient $g = -\dfrac{UA}{mc}$
intercept $b = \ln Z$

So a straight line on a graph of $\ln(T_C-T_R)$ vs $t$ supports the model, and the gradient/intercept allow $U$ and $Z$ to be found.

Understanding the Question

You have a metal cylinder with known mass $m$ and dimensions $L$ and $d$ , heated to a uniform temperature and then insulated while cooling in a room at temperature $T_R$ .

You must plan an experiment that:

Measures how surface temperature $T_C$ changes with time $t$ .
Controls relevant variables so the only intended change is with time.
Uses the data to test whether $(T_C-T_R)$ follows the given exponential form.
Extracts numerical values for the constants $U$ and $Z$ .

Because the equation includes $A$ , $m$ , and $c$ , your plan must make clear how $A$ and $m$ are measured (and that $c$ is known from data tables for the metal, or is provided/assumed known).

Approach

Decide variables: vary/record $t$ ; measure $T_C$ ; measure $T_R$ .
Set up a temperature probe that stays in good thermal contact with the cylinder surface while the cylinder is wrapped in insulation.
Ensure a uniform initial temperature by heating in a well-controlled way (e.g. stirred water bath).
Record $T_C$ at many times during cooling (data logger is ideal for many points).
Compute $(T_C-T_R)$ , then $\ln(T_C-T_R)$ , and plot vs $t$ .
Use gradient/intercept to determine $U$ and $Z$ .

Step-by-Step Reasoning

diagram-1

1) Measurements needed before cooling

Measure $m$ with a balance.
Measure $L$ and $d$ accurately with vernier calipers (or micrometer for $d$ ). Repeat readings along the cylinder to reduce random error.
Calculate surface area $A$ of the whole cylinder:

A = \text{curved area} + \text{two end faces} = \pi dL + 2\pi\left(\frac{d}{2}\right)^2 = \pi dL + \frac{\pi d^2}{2}.

This matters because $A$ appears directly in the exponent.

2) Producing a uniform starting temperature

A uniform temperature is important: if different parts start at different temperatures, the early-time cooling will not match a single-temperature model.

Put the cylinder in a stirred hot water bath for sufficient time.
If using boiling water, the cylinder approaches a known temperature close to $100\ ^\circ\text{C}$ (depending on pressure), which makes repeat runs consistent.

3) Recording temperature during cooling

Measure $T_R$ using the same probe, away from the hot bath.
Remove the cylinder, dry it quickly (so evaporation does not add extra cooling), and immediately wrap in the same insulation each time.
Ensure the thermocouple junction is pressed against the metal surface at a fixed position (e.g. mid-length), then cover it with insulation so it is not directly cooled by air.
Start timing at the moment cooling begins ( $t=0$ ) and record $T_C$ at fixed intervals (e.g. every 10 s) or continuously.

The key is to get many readings, because testing an exponential depends on the overall shape; a data logger makes this easier and reduces reading/response-time errors.

4) Testing the model and finding $U$ and $Z$

From

(T_C-T_R)=Ze^{-\frac{UAt}{mc}},

take natural logs:

\ln(T_C-T_R)=\ln Z-\frac{UA}{mc}t.

For each reading:

Compute $(T_C-T_R)$ .
Compute $\ln(T_C-T_R)$ .
Plot $\ln(T_C-T_R)$ (vertical) against $t$ (horizontal).

If the points lie close to a straight line, the relationship is supported.

Find gradient $g$ of best-fit line using a large triangle:

g = \frac{\Delta\ln(T_C-T_R)}{\Delta t}.

Then:

g=-\frac{UA}{mc} \Rightarrow U=-\frac{gmc}{A}.

Read the y-intercept $b$ at $t=0$ ; then:

b=\ln Z \Rightarrow Z=e^b.

5) Uncertainty (evaluation)

Use a worst acceptable line to estimate uncertainty in gradient: compare gradients of best line and worst line to find $\Delta g$ .
Propagate to $U$ (if $m$ and $A$ uncertainties are small compared with gradient uncertainty, the dominant contribution is from $g$ ):

\frac{\Delta U}{U} \approx \frac{\Delta g}{g} + \frac{\Delta m}{m} + \frac{\Delta A}{A}.

This is the usual Paper 5 expectation for evaluating constants from a straight-line graph.

Key Takeaways

Exponentials are tested by taking logs to produce a straight-line graph.
A good plan identifies variables, shows how to measure each quantity, and explains how to control conditions.
For $y=\ln(T_C-T_R)$ vs $t$ , the gradient gives $U$ and the intercept gives $Z$ .
Accurate and frequent temperature measurements (preferably logged) are essential for a convincing test.

Common Mistakes

Plotting $(T_C-T_R)$ vs $t$ directly and expecting a straight line (it should be exponential, not linear).
Forgetting to subtract $T_R$ before taking the logarithm (must use $\ln(T_C-T_R)$ , not $\ln T_C$ ).
Allowing the probe to measure the insulation temperature rather than the metal surface (poor contact gives wrong $T_C$ ).
Not controlling the insulation thickness/coverage between runs.
Using too few readings, making it impossible to judge whether the graph is truly straight.

Things to Be Careful About

Ensure $(T_C-T_R)>0$ before taking $\ln$ ; stop before the cylinder reaches room temperature.
Use consistent units (time in $\text{s}$ ; $L,d$ in $\text{m}$ if calculating $A$ in $\text{m}^2$ ).
Decide how $c$ is obtained (from data book / given / known material); state it clearly.
Start timing consistently (define $t=0$ clearly) and minimise delay between removing from heater and wrapping.
Avoid draughts and changing room conditions; convection changes the effective heat transfer and hence affects the fitted $U$ .

Techniques used

identify independent, dependent and controlled variablesmeasure geometry to calculate surface arealog-linearise an exponential relationshipplot a straight-line graph and obtain gradient and interceptdetermine constants from gradient and intercept with units

Working

f = \frac{f_s k}{k - v}

\frac{1}{f} = \frac{k-v}{f_s k} = \frac{1}{f_s} - \frac{v}{f_s k}

Answer

gradient $= -\dfrac{1}{f_s k}$

$y$ -intercept $= \dfrac{1}{f_s}$

Final answer

gradient = -1/(f_s k), y-intercept = 1/f_s

Detailed explanation

Background Concept

To analyse experimental data, we often turn a relationship into the straight-line form

y = mx + c

where $m$ is the gradient and $c$ is the $y$ -intercept.

Here the suggested model is

f = \frac{f_s k}{k - v}

where $f$ varies with $v$ . A convenient choice is to plot $\frac{1}{f}$ against $v$ , because taking the reciprocal often turns a “variable in the denominator” into a linear expression.

Understanding the Question

You are told that a graph of $\frac{1}{f}$ (vertical axis) against $v$ (horizontal axis) is plotted. The question asks for the expressions (in terms of $f_s$ and $k$ ) for:

the gradient of this graph
the $y$ -intercept of this graph.

Approach

Rearrange the given equation to make $\frac{1}{f}$ the subject.
Put it into the form $\frac{1}{f} = (\text{constant}) + (\text{constant})\,v$ .
Compare with $y = mx + c$ to read off gradient and intercept.

Step-by-Step Reasoning

Start with

f = \frac{f_s k}{k - v}

Take the reciprocal:

\frac{1}{f} = \frac{k - v}{f_s k}

Split the fraction into two terms:

\frac{1}{f} = \frac{k}{f_s k} - \frac{v}{f_s k} = \frac{1}{f_s} - \frac{1}{f_s k}v

Comparing with $y = mx + c$ (where $y=\frac{1}{f}$ and $x=v$ ):

gradient $m = -\frac{1}{f_s k}$
intercept $c = \frac{1}{f_s}$

Key Takeaways

Linearising an equation is done by rearranging into $y = mx + c$ .
The coefficient of $v$ gives the gradient; the constant term gives the intercept.

Common Mistakes

Forgetting the negative sign in the gradient.
Treating $k-v$ as $v-k$ (which would flip the sign).
Stating gradient as $-\frac{1}{k}$ (missing the $f_s$ factor).

Things to Be Careful About

The graph described is $\frac{1}{f}$ vs $v$ , not $f$ vs $v$ .
Keep algebra symbolic; do not substitute numbers here.

Techniques used

rearrange the given relationship into straight-line formidentify gradient and intercept by comparing with a linear equation

(b)

Values of $v_1$ , $v_2$ and $f$ are given in Table 2.1.

Table 2.1

$v_1 / \text{m s}^{-1}$	$v_2 / \text{m s}^{-1}$	$f / \text{Hz}$
3.1	3.9	894.2
6.7	5.9	901.2
9.2	8.2	908.0
11.9	10.9	915.8
13.3	14.5	923.6
15.6	16.8	931.2

Calculate and record values of $v / \text{m s}^{-1}$ and $\frac{1}{f} / 10^{-3} \text{ Hz}^{-1}$ in Table 2.1.

Include the absolute uncertainties in $v$ .

2M Similar

DifficultyMedium

Worked solution

Working

Average speed:

v = \frac{v_1+v_2}{2}

Absolute uncertainty:

\Delta v = \frac{|v_1-v_2|}{2}

And

\frac{1}{f}\ /(10^{-3}\ \text{Hz}^{-1}) = \frac{1000}{f}

Answer

Row 1: $v = 3.5 \pm 0.4\ \text{m s}^{-1}$ , $\frac{1}{f} = 1.118\ (10^{-3}\ \text{Hz}^{-1})$
Row 2: $v = 6.3 \pm 0.4\ \text{m s}^{-1}$ , $\frac{1}{f} = 1.110\ (10^{-3}\ \text{Hz}^{-1})$
Row 3: $v = 8.7 \pm 0.5\ \text{m s}^{-1}$ , $\frac{1}{f} = 1.101\ (10^{-3}\ \text{Hz}^{-1})$
Row 4: $v = 11.4 \pm 0.5\ \text{m s}^{-1}$ , $\frac{1}{f} = 1.092\ (10^{-3}\ \text{Hz}^{-1})$
Row 5: $v = 13.9 \pm 0.6\ \text{m s}^{-1}$ , $\frac{1}{f} = 1.083\ (10^{-3}\ \text{Hz}^{-1})$
Row 6: $v = 16.2 \pm 0.6\ \text{m s}^{-1}$ , $\frac{1}{f} = 1.074\ (10^{-3}\ \text{Hz}^{-1})$

Final answer

v values and 1/f values completed with uncertainties (see working)

Detailed explanation

Background Concept

When two measurements of the same quantity are taken, a common estimate of the best value is the mean:

\text{mean} = \frac{\text{reading 1} + \text{reading 2}}{2}

A simple absolute uncertainty estimate from two readings is half the range (half the difference):

\Delta v = \frac{|v_1 - v_2|}{2}

For the reciprocal, if $f$ is measured in $\text{Hz}$ , then $\frac{1}{f}$ has unit $\text{Hz}^{-1}$ (equivalently $\text{s}$ ). If the table wants $\frac{1}{f}$ in units of $10^{-3}\ \text{Hz}^{-1}$ , you calculate

\frac{1}{f}\ /(10^{-3}\ \text{Hz}^{-1}) = \frac{1/f}{10^{-3}} = 1000\left(\frac{1}{f}\right)=\frac{1000}{f}.

Understanding the Question

You must fill the missing columns:

$v$ calculated from $v_1$ and $v_2$
$\frac{1}{f}$ expressed in $10^{-3}\ \text{Hz}^{-1}$
And you must include the absolute uncertainties in $v$ (these will become the horizontal error bars later).

Approach

For each row:

Compute $v = (v_1+v_2)/2$ .
Compute $\Delta v = |v_1-v_2|/2$ .
Compute $1000/f$ to get the plotted value of $\frac{1}{f}$ in the requested scale.
Round consistently (typically to the same decimal places as the table/grid implies).

Step-by-Step Reasoning

Example (Row 1):

v=\frac{3.1+3.9}{2}=3.5\ \text{m s}^{-1}

\Delta v=\frac{|3.1-3.9|}{2}=0.4\ \text{m s}^{-1}

\frac{1}{f}\ /(10^{-3}\ \text{Hz}^{-1})=\frac{1000}{894.2}=1.118\ (\text{to }3\ \text{d.p.})

Repeat exactly the same process for each row.

Key Takeaways

Mean value: average the two readings.
Absolute uncertainty from repeats: half the difference is a standard Paper 5 method.
Converting to a scaled axis label like $10^{-3}\ \text{Hz}^{-1}$ is equivalent to multiplying by $1000$ .

Common Mistakes

Using $|v_1-v_2|$ instead of $|v_1-v_2|/2$ for the uncertainty.
Forgetting to convert $1/f$ into the $10^{-3}$ scale (writing $0.001118$ instead of $1.118$ ).
Inconsistent rounding within a column.

Things to Be Careful About

Keep units with $v$ and $\Delta v$ .
The uncertainty is absolute (same unit as $v$ ), not a percentage.
The $\frac{1}{f}$ values are plotted numbers; the axis label carries the $10^{-3}\ \text{Hz}^{-1}$ factor.

Techniques used

calculate a mean from repeated measurementscalculate absolute uncertainty from repeated readingscompute reciprocal values with consistent roundingrecord results in a correctly formatted table

(c)

(i)

Plot a graph of $\frac{1}{f} / 10^{-3} \text{ Hz}^{-1}$ against $v / \text{m s}^{-1}$ . Include error bars for $v$ .

Graph 2.1

2M Similar

DifficultyMedium

Worked solution

Answer

Points plotted on Graph 2.1:

(3.5,1.118),\ (6.3,1.110),\ (8.7,1.101),\ (11.4,1.092),\ (13.9,1.083),\ (16.2,1.074)

with horizontal error bars $\pm 0.4,\ \pm 0.4,\ \pm 0.5,\ \pm 0.5,\ \pm 0.6,\ \pm 0.6\ \text{m s}^{-1}$ respectively.

Final answer

See plotted graph (points with horizontal error bars in v)

Detailed explanation

Background Concept

A good graph in Paper 5 must:

have correctly labelled axes with units
use a sensible scale (typically using at least half the grid)
show points as small crosses/dots
show error bars where requested (here, uncertainties in $v$ give horizontal error bars).

Understanding the Question

You are given pre-drawn axes for $\frac{1}{f}/10^{-3}\ \text{Hz}^{-1}$ (vertical) against $v/\text{m s}^{-1}$ (horizontal). You must plot the six data points from Table 2.1 and include error bars for $v$ using the absolute uncertainties you calculated in (b).

Approach

Transfer each pair $(v,\ 1000/f)$ onto the grid.
For each plotted point, draw a horizontal error bar from $(v-\Delta v)$ to $(v+\Delta v)$ at the same $y$ value.

Step-by-Step Reasoning

Using the completed table:

Plot $(3.5,1.118)$ and draw an error bar from $3.5-0.4=3.1$ to $3.5+0.4=3.9$ .
Plot $(6.3,1.110)$ with error bar $5.9$ to $6.7$ .
Plot $(8.7,1.101)$ with error bar $8.2$ to $9.2$ .
Plot $(11.4,1.092)$ with error bar $10.9$ to $11.9$ .
Plot $(13.9,1.083)$ with error bar $13.3$ to $14.5$ .
Plot $(16.2,1.074)$ with error bar $15.6$ to $16.8$ .

Key Takeaways

Uncertainty in the $x$ -quantity produces horizontal error bars.
Error bars should be centred on the plotted mean value.

Common Mistakes

Drawing vertical error bars instead of horizontal ones.
Using full error bar length $\Delta v$ on each side (should be $\pm\Delta v$ about the mean).
Plotting $1/f$ as $0.001118$ instead of $1.118$ on the given scaled axis.

Things to Be Careful About

Make plotted points small and precise; thick blobs lose accuracy.
Ensure the error bars have end-caps so the examiner can see their extent.
Do not join points dot-to-dot; a best-fit line is drawn in (ii).

Techniques used

choose suitable axis scales to use most of the gridplot data points accurately from a tableadd horizontal error bars using absolute uncertainties

(ii)

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

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

A straight line of best fit is drawn through the points.

A worst acceptable straight line is drawn (most different gradient) that still passes through all horizontal error bars.

Both lines are labelled on the graph.

Final answer

See graph (best-fit and worst-acceptable lines drawn and labelled)

Detailed explanation

Background Concept

A best-fit line represents the overall trend of the data, balancing points above and below the line. A worst acceptable line is a straight line that is still consistent with the uncertainties (it must pass through all relevant error bars) but has the most different gradient from the best-fit line. This is used to estimate uncertainty in gradient and intercept.

Understanding the Question

After plotting points with horizontal error bars, you must:

draw the best-fit straight line
draw a worst acceptable straight line
label both.

Approach

Best-fit line: use a ruler to draw a single straight line that best represents the trend (do not force it through every point).
Worst acceptable line: pivot the line to make it as steep or as shallow as possible while still intersecting every horizontal error bar.
Label the two lines clearly (e.g. “best fit” and “worst”).

Step-by-Step Reasoning

Place the ruler so the line passes centrally among the points (roughly equal scatter about it).
For the worst acceptable line, focus on the extreme points (lowest $v$ and highest $v$ ), and use the ends of their horizontal error bars to maximise or minimise the gradient while remaining consistent.
Write labels directly next to each line.

Key Takeaways

Worst acceptable lines must be consistent with error bars, not just the plotted points.
You must label lines for the examiner to know which is which.

Common Mistakes

Drawing a curve instead of a straight line.
Drawing a “worst” line that does not pass through all error bars.
Forgetting to label the two lines.

Things to Be Careful About

Use a sharp pencil and a ruler.
The worst acceptable line should be clearly different from the best-fit line (otherwise your uncertainty will be unrealistically small).

Techniques used

draw a straight line of best fit through the trenddraw a worst acceptable line consistent with all error barslabel lines clearly on the 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

From the best-fit line (two well-separated points):

m_{\text{best}}=\frac{1.074-1.118}{16.2-3.5}=-3.50\times 10^{-3}

Worst acceptable line (using ends of error bars):

m_{\text{worst}}=\frac{1.074-1.118}{15.6-3.9}=-3.79\times 10^{-3}

Absolute uncertainty:

\Delta m = |m_{\text{best}}-m_{\text{worst}}|=0.30\times 10^{-3}

Answer

\text{gradient} = (-3.50 \pm 0.30)\times 10^{-3}

Final answer

(-3.50 ± 0.30) × 10^-3

Detailed explanation

Background Concept

The gradient of a straight-line graph is

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

To reduce percentage reading error, choose two points far apart on the line (a large triangle).

To estimate uncertainty in the gradient experimentally, Paper 5 commonly uses:

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

where $m_{\text{worst}}$ is the gradient of a worst acceptable line that still passes through all error bars.

Understanding the Question

You must find the gradient of the best-fit line on your graph of $\frac{1}{f}/10^{-3}\ \text{Hz}^{-1}$ against $v/\text{m s}^{-1}$ , and include the absolute uncertainty. The uncertainty comes from comparing the best-fit line to a worst acceptable straight line.

Approach

Read two well-separated coordinates on the best-fit line and calculate $m_{\text{best}}$ using $\Delta y/\Delta x$ .
Draw a worst acceptable line using the horizontal error bars in $v$ , then repeat to get $m_{\text{worst}}$ .
Use $\Delta m = |m_{\text{best}}-m_{\text{worst}}|$ .

Step-by-Step Reasoning

diagram-2c-iii-1

Choose two points on the best-fit line, ideally near the ends of the plotted range to maximise $\Delta x$ . For example, near $(v,y)=(3.5,1.118)$ and $(16.2,1.074)$ .
Compute

m_{\text{best}}=\frac{1.074-1.118}{16.2-3.5}

which is negative because the graph slopes down.

For the worst acceptable line, use the ends of the horizontal error bars to make the line as steep as possible (or as shallow as possible), while still being consistent. Using the high- $v$ point at its smallest $v$ and the low- $v$ point at its largest $v$ gives a steeper (more negative) slope.
Calculate $m_{\text{worst}}$ from two points on that worst line.
The absolute uncertainty is the difference in the gradients.

Key Takeaways

Always use a large triangle for gradient.
Worst acceptable line must still pass through all error bars.
Uncertainty is found from best vs worst, not from scatter alone.

Common Mistakes

Using two neighbouring points (gives a large fractional error).
Calculating $\Delta x/\Delta y$ instead of $\Delta y/\Delta x$ .
Choosing a “worst” line that does not intersect all error bars.

Things to Be Careful About

Keep consistent units: here $y$ is plotted in $10^{-3}\ \text{Hz}^{-1}$ units, so your gradient matches that plotted scale.
Quote the uncertainty to 1 significant figure (often) and the value to match.
The sign matters: the gradient should be negative.

Techniques used

determine gradient from two well-separated points on a straight lineuse error bars to construct a worst acceptable lineestimate absolute uncertainty from best vs worst values

(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

For best-fit line:

c_{\text{best}} = y - mx = 1.118 - (-3.50\times 10^{-3})(3.5)=1.131

For worst acceptable line:

c_{\text{worst}} = 1.118 - (-3.79\times 10^{-3})(3.9)=1.133

\Delta c = |c_{\text{best}}-c_{\text{worst}}|=0.0026\approx 0.003

Answer

\text{$y$-intercept} = 1.131 \pm 0.003

Final answer

1.131 ± 0.003

Detailed explanation

Background Concept

For a straight line

y = mx + c

$c$ is the $y$ -intercept (value of $y$ when $x=0$ ). On a graph, you can find $c$ either by extending the line to cross the $y$ -axis or by calculating it using a known point $(x,y)$ on the line:

c = y - mx.

Uncertainty in intercept is found using the same best-vs-worst method:

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

Understanding the Question

You must determine the $y$ -intercept of the best-fit line on your graph of $\frac{1}{f}/10^{-3}\ \text{Hz}^{-1}$ vs $v$ , and include an absolute uncertainty using your worst acceptable line.

Approach

Find $c_{\text{best}}$ by reading the intercept from the graph or using $c=y-mx$ .
Find $c_{\text{worst}}$ in the same way from the worst acceptable line.
Compute $\Delta c = |c_{\text{best}}-c_{\text{worst}}|$ .

Step-by-Step Reasoning

Using a point on the best-fit line (well read from the line), substitute into $c=y-mx$ .
Repeat with the worst acceptable line. The worst line should have a different gradient, so it will generally cross the $y$ -axis at a different point.
The difference gives the absolute uncertainty.

Key Takeaways

The intercept can be computed from a point and the gradient; you do not have to read it only at $x=0$ .
Worst acceptable line provides a straightforward uncertainty estimate.

Common Mistakes

Using a point not on the line (using a plotted point that is off the line).
Forgetting the sign of $m$ when calculating $c=y-mx$ .
Quoting $\Delta c$ as a percentage instead of an absolute uncertainty.

Things to Be Careful About

Use consistent units: the intercept here is in the plotted units of $\frac{1}{f}/10^{-3}\ \text{Hz}^{-1}$ .
Quote intercept and uncertainty to consistent decimal places.

Techniques used

read y-intercept from a straight-line graphcalculate intercept from y = mx + c using a point on the lineestimate uncertainty from best vs worst line intercepts

(d)

(i)

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

$f_s$ = ______
$k$ = ______

2M Similar

DifficultyMedium

Worked solution

Working

From (a), for a plot of $1/f$ against $v$ :

\text{intercept} = \frac{1}{f_s},\qquad \text{gradient} = -\frac{1}{f_s k}

But the plotted quantity is $\left(\frac{1}{f}\right)/(10^{-3}\ \text{Hz}^{-1}) = \frac{1000}{f}$ , so

c = \frac{1000}{f_s}

and

m = -\frac{1000}{f_s k}

Hence

f_s = \frac{1000}{c} = \frac{1000}{1.131} = 8.84\times 10^2\ \text{Hz}

and

k = -\frac{c}{m} = -\frac{1.131}{-3.50\times 10^{-3}} = 3.23\times 10^2\ \text{m s}^{-1}

Answer

f_s = 8.84\times 10^2\ \text{Hz}

k = 3.23\times 10^2\ \text{m s}^{-1}

Final answer

f_s = 8.84×10^2 Hz, k = 3.23×10^2 m s^-1

Detailed explanation

Background Concept

Once data are linearised, the gradient and intercept are linked to physical constants. If

\frac{1}{f} = \frac{1}{f_s} - \frac{1}{f_s k}v,

then comparing with $y=mx+c$ gives

c=\frac{1}{f_s},\qquad m=-\frac{1}{f_s k}.

From these,

f_s = \frac{1}{c},\qquad k = -\frac{c}{m}.

However, if the graph uses a scaled axis (here $\frac{1}{f}/10^{-3}\ \text{Hz}^{-1}$ ), you must include the scale factor.

Understanding the Question

You must use your measured gradient and intercept (from the graph in part (c)) together with your expressions from (a) to determine numerical values of $f_s$ and $k$ , with units.

Approach

Write the straight-line equation in terms of the plotted variables.
Relate the plotted intercept $c$ and gradient $m$ to $f_s$ and $k$ .
Substitute your numerical values of $c$ and $m$ .
Quote with correct units: $f_s$ in $\text{Hz}$ and $k$ in $\text{m s}^{-1}$ .

Step-by-Step Reasoning

Because the vertical axis is $\left(\frac{1}{f}\right)/(10^{-3}\ \text{Hz}^{-1})$ , the plotted $y$ value is

y_{\text{plot}}=\frac{1/f}{10^{-3}}=1000\left(\frac{1}{f}\right).

So if the true line is $\frac{1}{f} = -\frac{1}{f_s k}v + \frac{1}{f_s}$ , the plotted line is

y_{\text{plot}} = 1000\left(-\frac{1}{f_s k}v + \frac{1}{f_s}\right) = \left(-\frac{1000}{f_s k}\right)v + \frac{1000}{f_s}.

Thus

c_{\text{plot}}=\frac{1000}{f_s}\ \Rightarrow\ f_s=\frac{1000}{c_{\text{plot}}}

and

m_{\text{plot}}=-\frac{1000}{f_s k},\quad \text{so } k=-\frac{c_{\text{plot}}}{m_{\text{plot}}}.

Then substitute your numerical intercept and gradient.

Key Takeaways

Always account for axis scaling factors like “ $/10^{-3}$ ”.
The speed-like constant $k$ comes from the ratio $-c/m$ .

Common Mistakes

Using $f_s=1/c$ instead of $f_s=1000/c$ (forgetting the $10^{-3}$ scale).
Forgetting the negative sign when using $k=-c/m$ .
Omitting units.

Things to Be Careful About

Use consistent significant figures (typically 3 s.f. for constants from a graph).
$k$ should be of order $3\times 10^2\ \text{m s}^{-1}$ ; a value off by a factor of $1000$ indicates a scaling error.

Techniques used

use gradient and intercept to determine model constantshandle axis scaling factors when converting intercepts and gradientscheck units for derived constants

(ii)

Determine the percentage uncertainty in $k$ .

percentage uncertainty in $k$ = ______ $\%$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

k=-\frac{c}{m}

\frac{\Delta k}{k}=\frac{\Delta c}{c}+\frac{\Delta m}{|m|}

Using $c = 1.131 \pm 0.003$ and $m = (-3.50\pm 0.30)\times 10^{-3}$ :

\%\ \text{uncertainty in }k = \left(\frac{0.003}{1.131}+\frac{0.30\times 10^{-3}}{3.50\times 10^{-3}}\right)\times 100 = 8.9\%

Answer

8.9\%

Final answer

8.9%

Detailed explanation

Background Concept

When a quantity is found from multiplication/division, fractional uncertainties add. For

Q=\frac{A}{B},

the fractional uncertainty is

\frac{\Delta Q}{Q}=\frac{\Delta A}{A}+\frac{\Delta B}{B}.

To convert to percentage uncertainty, multiply the fractional uncertainty by $100$ .

Understanding the Question

You have already determined $k$ using the gradient and intercept of your graph. Now you must find the percentage uncertainty in $k$ , using the uncertainties you found for the gradient and the intercept.

Approach

Express $k$ in terms of the measured graph quantities $c$ and $m$ .
Write down the fractional uncertainty rule for a quotient.
Substitute the absolute uncertainties and compute the percentage.

Step-by-Step Reasoning

From part (d)(i),

k=-\frac{c}{m}.

Treating $c$ and $m$ as measured quantities with uncertainties,

\frac{\Delta k}{k}=\frac{\Delta c}{c}+\frac{\Delta m}{|m|}

(use $|m|$ because uncertainty is a magnitude).

Substitute your values of $c,\Delta c,m,\Delta m$ and multiply by $100$ to get the percentage.

Key Takeaways

For products/quotients, add fractional (percentage) uncertainties.
Use absolute uncertainty divided by the best estimate.

Common Mistakes

Subtracting uncertainties instead of adding them.
Using $m$ (negative) rather than $|m|$ in the denominator.
Mixing up $0.30\times 10^{-3}$ with $0.30$ .

Things to Be Careful About

Keep consistent powers of ten when dividing $\Delta m$ by $m$ .
Quote the final percentage uncertainty to 2 significant figures typically.

Techniques used

propagate fractional uncertainties through divisionadd fractional uncertainties for multiplied or divided quantitiesconvert fractional uncertainty to percentage uncertainty

(e)

The experiment is repeated. Determine the speed $v$ that gives a value of $f$ of $987.8 \text{ Hz}$ .

$v$ = ______ $\text{m s}^{-1}$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

f = \frac{f_s k}{k-v} \Rightarrow f(k-v)=f_s k

fk-fv=f_s k \Rightarrow v = k\frac{f-f_s}{f}

Using $f=987.8\ \text{Hz}$ , $f_s=8.84\times 10^2\ \text{Hz}$ , $k=3.23\times 10^2\ \text{m s}^{-1}$ :

v = 3.23\times 10^2\times \frac{987.8-884}{987.8}=33.8\ \text{m s}^{-1}

Answer

v=33.8\ \text{m s}^{-1}

Final answer

33.8 m s^-1

Detailed explanation

Background Concept

Once you have found model constants, you can use the model equation to predict values. Here the model is

f = \frac{f_s k}{k-v}

If you need $v$ , you rearrange algebraically to make $v$ the subject, then substitute numerical values.

Understanding the Question

In a repeat of the experiment, the microphone reads $f=987.8\ \text{Hz}$ . Using the $f_s$ and $k$ you determined from the graph, you must calculate the corresponding car speed $v$ .

Approach

Rearrange the equation to isolate $v$ .
Substitute $f$ , $f_s$ and $k$ .
Give the result with unit $\text{m s}^{-1}$ .

Step-by-Step Reasoning

Start with

f = \frac{f_s k}{k-v}.

Multiply both sides by $(k-v)$ :

f(k-v)=f_s k.

Expand and collect terms in $v$ :

fk - fv = f_s k \Rightarrow fv = fk - f_s k = k(f-f_s).

v = k\frac{f-f_s}{f}.

Substitute the values you found in part (d) together with $f=987.8\ \text{Hz}$ to get $v$ .

Key Takeaways

Rearranging before substituting reduces algebra mistakes.
The result should be physically sensible: $v$ must be less than $k$ because the original formula has $k-v$ in the denominator.

Common Mistakes

Rearranging to $v=k\frac{f_s-f}{f}$ (wrong sign, giving negative speed).
Using $f_s=1/c$ instead of $1000/c$ from the scaled intercept (carrying forward a previous scaling error).
Forgetting units.

Things to Be Careful About

Check that $f>f_s$ for an approaching source (so $v$ comes out positive).
Ensure consistent significant figures based on your graph-derived constants.

Techniques used

rearrange the model equation to make a required variable the subjectsubstitute experimentally determined constants into the modelcalculate a predicted value from the model

Physics 9702/51 — May/June 2024

Answer

Variables

Apparatus

Procedure and measurements

Analysis (to obtain UUU and ZZZ)

Safety

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

1) Measurements needed before cooling

2) Producing a uniform starting temperature

3) Recording temperature during cooling

4) Testing the model and finding UUU and ZZZ

5) Uncertainty (evaluation)

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

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

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Analysis (to obtain $U$ and $Z$ )

4) Testing the model and finding $U$ and $Z$