Physics 9702/53 — 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

Variables

Independent variable: time $t$ after cooling starts.
Dependent variable: cylinder surface temperature $T_C$ .
Controlled variables: cylinder dimensions ( $L,d$ ) and material (so $c$ fixed), mass $m$ , thickness/type of insulation, position of temperature sensor, ambient conditions (air flow, room temperature $T_R$ ).

diagram-1

Procedure and measurements

Measure mass $m$ of the cylinder using a balance.
Measure $L$ and $d$ using vernier calipers / micrometer.
Calculate total surface area

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

Measure room temperature $T_R$ using a thermometer placed near the apparatus.
Attach a thermocouple/temperature probe to the cylinder surface (same position each run) using thermal paste and tape.
Heat the cylinder in a hot water bath (near $100\ ^\circ\text{C}$ ) for sufficient time so the temperature is uniform (wait until probe reading is steady).
Remove cylinder using tongs, quickly wrap with the insulating material (same thickness each run) with the probe still in contact, start timer at the moment of wrapping.
Record $T_C$ at regular time 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 each $t$ (or repeat the whole run to check repeatability).

Control of variables

Use the same cylinder throughout (same $m$ , $L$ , $d$ , and material $\Rightarrow$ same $c$ ).
Keep insulation type/thickness and wrapping method the same.
Keep the sensor position and contact method identical.
Minimise drafts (same location, away from fans/air-conditioning); keep $T_R$ monitored and approximately constant.
Use the same initial heating condition (e.g. same water-bath temperature and same time immersed).

Analysis of data (to test relationship and find $U$ 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.

For each reading, calculate $(T_C-T_R)$ and then $\ln(T_C-T_R)$ .
Plot $\ln(T_C-T_R)$ (y-axis) against $t$ (x-axis).
A straight line supports the suggested relationship.
Gradient $g$ of the line:

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

Intercept $b$ :

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

Safety

Hot cylinder / hot water: use tongs and heatproof gloves; avoid splashing; wear eye protection.
Keep electrical equipment (data logger) away from water; dry hands before handling plugs/leads.
Ensure cylinder is placed securely to avoid rolling/falling while hot.

Final answer

See working

Detailed explanation

Background Concept

This experiment is based on Newton’s law of cooling, where the rate of energy loss from a hot object is proportional to the temperature difference between the object and its surroundings.

For a body of mass $m$ and specific heat capacity $c$ , a small temperature drop $\mathrm{d}T$ corresponds to an energy change

\mathrm{d}E = mc\,\mathrm{d}T.

If heat is lost from the surface to the surroundings at a rate proportional to surface area $A$ and to $(T_C-T_R)$ , then a typical model is

\frac{\mathrm{d}E}{\mathrm{d}t} = -UA(T_C-T_R),

where $U$ is an overall heat-transfer coefficient (it depends on insulation, convection, etc.). Solving this differential equation gives an exponential decay:

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

where $Z$ is a constant set by the initial conditions (essentially the initial temperature difference at $t=0$ ).

Understanding the Question

You are told a suggested relationship between the cylinder’s surface temperature $T_C$ and time $t$ as it cools in a room at temperature $T_R$ .

You must:

plan a practical method to measure $T_C$ as a function of $t$ ,
ensure the geometry is known to find the surface area $A$ ,
keep other factors constant so the model can be fairly tested,
show how to process the data to check whether the relationship is exponential,
and explain how $U$ and $Z$ can be obtained from the results.

Approach

Choose a reliable way to measure surface temperature continuously: a thermocouple or temperature probe with a data logger gives many readings quickly and reduces reaction-time error.
Ensure the cylinder starts at a uniform temperature (heat long enough in a well-defined environment such as a hot water bath).
Wrap in the same insulation each time to keep the heat-transfer conditions (hence $U$ ) constant.
Record $T_C$ and time $t$ as it cools, and also measure $T_R$ .
Linearise the exponential by taking natural logarithms so you can use a straight-line graph. From the gradient and intercept you can find $U$ and $Z$ .

Step-by-Step Reasoning

1) Measuring the required quantities

Mass $m$ : measured with a balance (needed in the exponent).
Dimensions $L$ and $d$ : measured with vernier calipers / micrometer. You then compute total surface area:

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

Room temperature $T_R$ : measure near the cylinder and keep monitoring; it must be approximately constant for the model.
Surface temperature $T_C$ : attach a thermocouple firmly to the surface. Good thermal contact matters; use thermal paste and tape.
Time $t$ : start timing at a clearly defined moment (e.g. immediately when wrapping is complete / when removed from bath and wrapped). Use a data logger clock or a stopwatch.

2) Setting up so the model is valid

The model assumes a single, well-defined temperature for the cylinder at each time (or at least that your measured surface temperature is representative). To get close to this:

heat the cylinder until the temperature is uniform (wait for a steady sensor reading while in the bath),
keep insulation constant so the cooling conditions don’t change between runs,
reduce extra heat-loss paths (e.g. do not place the hot cylinder directly on a metal bench; suspend it or place it on insulating supports).

diagram-1

3) Collecting data

Record $T_C$ at equal intervals (e.g. every 10 s) or continuously (best). Continue until $T_C$ is close to $T_R$ because then $(T_C-T_R)$ becomes small and the log values become sensitive to measurement noise; you still want enough data points to see the trend clearly.

Repeat the run to check repeatability. If repeated runs give similar graphs/gradients, that supports that your control of conditions is good.

4) Linearising and extracting $U$ and $Z$

Start with

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

Taking natural logarithms gives

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

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

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

So:

U = -\frac{gmc}{A},

and

Z = e^b.

A straight line on the $\ln(T_C-T_R)$ vs $t$ graph is the test of the relationship.

5) Uncertainties (how you would handle them)

$T_C$ uncertainty comes from probe resolution and imperfect contact; this affects $(T_C-T_R)$ and therefore the log values.
For the straight-line fit, you can estimate uncertainty in the gradient using a worst acceptable line. Then propagate to $U$ using

\frac{\Delta U}{U} \approx \frac{\Delta g}{g} + \frac{\Delta m}{m} + \frac{\Delta A}{A} \quad (c\ \text{assumed known accurately}).

If $c$ is not assumed known, you would include $\Delta c/c$ as well.

Key Takeaways

Exponential relationships are tested efficiently by taking logs to produce a straight-line graph.
In cooling experiments, controlling heat-transfer conditions (insulation, airflow, contact points) is crucial because they directly affect the constant $U$ .
Gradient and intercept of a linear graph can be mapped to physical constants ( $U$ and $Z$ here).

Common Mistakes

Plotting $(T_C-T_R)$ against $t$ and expecting a straight line (the straight line comes after taking $\ln$ ).
Forgetting to measure $T_R$ , or assuming it is constant without checking.
Using the wrong surface area (e.g. only curved surface, or wrong end area).
Poor thermal contact of the sensor (measures air temperature or insulation temperature instead of the metal surface).
Changing insulation thickness or wrapping style between runs, which changes $U$ .

Things to Be Careful About

You can only take $\ln(T_C-T_R)$ when $(T_C-T_R) > 0$ ; once $T_C$ is very close to $T_R$ , the difference becomes noisy and logs become unreliable.
Define $t=0$ clearly and use the same definition for all data.
Ensure units are consistent when calculating $U$ from $U=-gmc/A$ (use SI: $m$ in $\text{kg}$ , $A$ in $\text{m}^2$ , $t$ in $\text{s}$ , so $U$ comes out in $\text{W m}^{-2}\text{K}^{-1}$ ).
If the cylinder is placed on a surface, conduction into the support adds an extra heat-loss mechanism; this can spoil the simple model unless kept constant and small.

Techniques used

identify independent, dependent and controlled variablesmeasure geometrical dimensions to calculate surface arealog temperature against time using a sensor and data loggerlinearise an exponential relationship using natural logarithmsuse gradient and intercept of a straight-line graph to determine constants

Working

Given

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

Take reciprocal:

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

\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 $y=mx+c$ for a plot of $y=\dfrac{1}{f}$ against $x=v$ :

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 extract constants from experimental data, a common technique is to rearrange a relationship into the straight-line form

y = mx + c

If you plot $y$ against $x$ , then:

the gradient (slope) is $m$
the y-intercept is $c$

This lets you identify unknown constants by comparing your rearranged equation term-by-term with $y = mx + c$ .

Understanding the Question

You are given a suggested relationship between measured frequency $f$ and car speed $v$ :

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

A graph is plotted with $\dfrac{1}{f}$ on the y-axis and $v$ on the x-axis. The task is to express this in the form

\frac{1}{f} = (\text{something})\, v + (\text{constant}).

Then you can read off the gradient and y-intercept.

Approach

Take the reciprocal of the given equation to create $1/f$ .
Expand/simplify so it looks like a constant term plus a term proportional to $v$ .
Compare with $y=mx+c$ .

Step-by-Step Reasoning

Starting from

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

Reciprocal both sides:

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

Split the fraction:

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

Simplify $\dfrac{k}{f_s k} = \dfrac{1}{f_s}$ :

\frac{1}{f} = \frac{1}{f_s} - \left(\frac{1}{f_s k}\right) v

Now compare with $y = mx + c$ where $y = \dfrac{1}{f}$ and $x=v$ :

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

Key Takeaways

Linearise by rearranging into $y=mx+c$ .
The coefficient of $v$ becomes the gradient.
The constant term becomes the y-intercept.

Common Mistakes

Forgetting the negative sign in the gradient.
Stopping at $\dfrac{1}{f}=\dfrac{k-v}{f_s k}$ without splitting into constant and $v$ terms.
Mixing up which axis is $x$ and which is $y$ .

Things to Be Careful About

The gradient is negative because $1/f$ decreases as $v$ increases (since $f$ increases).
Keep $f_s$ and $k$ as symbols; do not substitute numbers in part (a).

Techniques used

rearrange the given relationship into a linear formidentify gradient and intercept by comparison with y = mx + c

(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-Easy

Worked solution

Working

For each run:

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

Absolute uncertainty in $v$ from two readings:

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

Also

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

Calculated values:

$v_1/\text{m s}^{-1}$	$v_2/\text{m s}^{-1}$	$v/\text{m s}^{-1}$	$\Delta v/\text{m s}^{-1}$	$f/\text{Hz}$	$\dfrac{1}{f}/(10^{-3}\ \text{Hz}^{-1})$
3.1	3.9	3.5	0.4	894.2	1.118
6.7	5.9	6.3	0.4	901.2	1.110
9.2	8.2	8.7	0.5	908.0	1.101
11.9	10.9	11.4	0.5	915.8	1.092
13.3	14.5	13.9	0.6	923.6	1.083
15.6	16.8	16.2	0.6	931.2	1.074

Answer

$v$ and $\dfrac{1}{f}$ values as in the completed table (with $\Delta v$ shown for each row).

Final answer

Completed Table 2.1 values for v, 1/f (×10^-3 Hz^-1), and Δv as shown.

Detailed explanation

Background Concept

When you have two measurements of the same quantity, a common best estimate is the mean:

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

A simple absolute uncertainty estimate from the spread of two readings is half the range:

\Delta x = \frac{|x_1-x_2|}{2}

For reciprocals, remember

\frac{1}{f}

has units of $\text{Hz}^{-1}$ , and if the table heading is “ $\dfrac{1}{f} / 10^{-3}\ \text{Hz}^{-1}$ ”, then you should scale by $10^3$ :

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

Understanding the Question

You are given $v_1$ , $v_2$ (two speed detector readings) and $f$ for several runs.
You must fill in:

the average speed $v$ for each run,
the value of $1/f$ expressed in units of $10^{-3}\ \text{Hz}^{-1}$ ,
and include absolute uncertainties in $v$ .

Approach

For each row:

Compute the mean speed $v = (v_1+v_2)/2$ .
Compute the uncertainty using half the difference: $\Delta v = |v_1-v_2|/2$ .
Compute $1/f$ and apply the $10^{-3}$ scaling by multiplying by $10^3$ .
Round sensibly and consistently within each column.

Step-by-Step Reasoning

Example (first row):

Mean speed:

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

Uncertainty:

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

Reciprocal frequency with scaling:

\frac{1}{f} = \frac{1}{894.2} = 1.118\times 10^{-3}\ \text{Hz}^{-1}

So in the table column “ $\dfrac{1}{f}/10^{-3}\ \text{Hz}^{-1}$ ” you record $1.118$ .

Repeat this for each row, keeping a consistent decimal place pattern (e.g. $v$ to 0.1 $\text{m s}^{-1}$ , $\Delta v$ to 0.1 $\text{m s}^{-1}$ , and $1/f$ to 0.001 in the scaled units).

Key Takeaways

Mean of two readings gives a best estimate.
Half the range of two readings is a simple absolute uncertainty.
Always follow the scaling in the table heading (here $10^{-3}$ ).

Common Mistakes

Using $|v_1-v_2|$ instead of $|v_1-v_2|/2$ for the uncertainty.
Forgetting the $10^{-3}$ scaling and writing $0.001118$ instead of $1.118$ .
Inconsistent rounding/decimal places within a column.

Things to Be Careful About

Uncertainty should be positive and in the same unit as $v$ .
Check calculator mode for reciprocals; a small slip in powers of ten loses marks.
The table heading indicates the expected magnitude and rounding.

Techniques used

calculate the mean from two measurementsestimate absolute uncertainty from repeated readingscompute reciprocal values and apply the axis scaling factorrecord values to a consistent appropriate precision

(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

Plot the points $(v,\,1/f)$ (with $1/f$ in $10^{-3}\ \text{Hz}^{-1}$ ):

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

Include horizontal error bars on each point of size $\pm\Delta v$ :
$\pm0.4$ , $\pm0.4$ , $\pm0.5$ , $\pm0.5$ , $\pm0.6$ , $\pm0.6\ \text{m s}^{-1}$ respectively.

Final answer

See graph: points plotted with horizontal error bars ±Δv.

Detailed explanation

Background Concept

A graph is used to reveal trends and allow a best-fit line to be drawn. For experimental work, it is essential to:

label axes correctly with quantity and unit,
use a sensible scale that fills much of the grid,
plot points accurately,
include error bars when uncertainties are known.

Error bars show the range of values that a measured quantity could reasonably take. Here the uncertainty is in $v$ , so the error bars are horizontal.

Understanding the Question

You must plot $\dfrac{1}{f}/10^{-3}\ \text{Hz}^{-1}$ on the y-axis against $v/\text{m s}^{-1}$ on the x-axis, using the values you calculated in Table 2.1.

You must also include error bars for $v$ using the absolute uncertainties you found in part (b).

Approach

Choose appropriate scales (already suggested by the grid provided).
Plot each coordinate $(v,\,1/f)$ .
For each point, draw a horizontal error bar from $v-\Delta v$ to $v+\Delta v$ .

Step-by-Step Reasoning

Read each $v$ and $1/f$ value from your completed table.
Plot each point as a small cross or dot.
For row 1, $v = 3.5\ \text{m s}^{-1}$ with $\Delta v = 0.4\ \text{m s}^{-1}$ , so the error bar extends from $3.1$ to $3.9$ .
Repeat for all points.

The y-values have no stated uncertainty here, so do not draw vertical error bars unless instructed.

Key Takeaways

Error bars must match the uncertainty and the axis variable.
Use horizontal error bars when uncertainty is in the x-variable.

Common Mistakes

Drawing vertical error bars instead of horizontal ones.
Plotting $f$ instead of $1/f$ .
Ignoring the $10^{-3}$ scaling and plotting values near $0.0011$ rather than around $1.1$ .

Things to Be Careful About

Place points carefully (use half-squares if appropriate).
Error bars should be centred on the plotted point and have clear end caps.
Ensure axis labels exactly match those requested on the paper.

Techniques used

plot experimental data points with correct axis variables and scalesadd horizontal error bars using absolute uncertaintieslabel axes with quantity and unit as given

(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

Answer

Draw and label:

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

Label the lines clearly as “best fit” and “worst acceptable”.

Final answer

See graph: best-fit line and labelled worst acceptable line drawn.

Detailed explanation

Background Concept

A best-fit line represents the overall trend of the data, balancing the scatter of points. A worst acceptable line is used to estimate uncertainty in gradient/intercept:

It must be a straight line.
It must still be consistent with the data uncertainties (i.e. it passes through all error bars).
It is chosen to make the gradient (or intercept) as different as possible from the best-fit value.

Understanding the Question

After plotting points with horizontal error bars (uncertainty in $v$ ), you must:

Draw a best-fit straight line.
Draw one additional straight line that is still acceptable given the error bars, but gives an extreme gradient.
Both must be labelled.

Approach

Best-fit: draw a line that has roughly equal numbers of points above and below, following the linear trend.
Worst acceptable: pivot the line to be as steep as possible (or as shallow as possible) while still intersecting every error bar range.

Step-by-Step Reasoning

Use a ruler to draw the best-fit line through the central trend (not point-to-point zigzags).
Decide whether a steeper or shallower line would be further from your best-fit gradient.
Adjust the extreme line until it is just consistent with the error bars. If it misses any error bar region, it is not acceptable.
Label each line directly on the graph.

Key Takeaways

Best-fit line: represents trend.
Worst acceptable line: used to find gradient/intercept uncertainty.

Common Mistakes

Drawing the worst line through points but ignoring the error bars.
Drawing a curve instead of a straight line.
Forgetting to label the two lines.

Things to Be Careful About

The worst acceptable line must be a single straight line.
It must pass through all error bars (not necessarily all plotted points exactly).
Use a sharp pencil/ruler so gradients can be read accurately.

Techniques used

draw a straight line of best fit through plotted pointsdraw a worst acceptable line consistent with all error barslabel both 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

Using large triangles on the best-fit and worst acceptable lines:

Best-fit gradient (from the plotted graph)

m_{\text{best}} = -0.00350\ \left(10^{-3}\ \text{Hz}^{-1}\right)\ (\text{m s}^{-1})^{-1}

Worst acceptable gradient

m_{\text{worst}} = -0.00379\ \left(10^{-3}\ \text{Hz}^{-1}\right)\ (\text{m s}^{-1})^{-1}

Absolute uncertainty

\Delta m = |m_{\text{best}}-m_{\text{worst}}| = 0.00029\ \left(10^{-3}\ \text{Hz}^{-1}\right)\ (\text{m s}^{-1})^{-1}

Answer

m = (-0.00350 \pm 0.00029)\ \left(10^{-3}\ \text{Hz}^{-1}\right)\ (\text{m s}^{-1})^{-1}

Final answer

m = (−0.00350 ± 0.00029) (10^−3 Hz^−1) (m s^−1)^−1

Detailed explanation

Background Concept

The gradient of a straight line is

\text{gradient} = \frac{\Delta y}{\Delta x}

On a plotted graph, you should use two points far apart on the line (not necessarily data points) to reduce percentage reading error.

To estimate uncertainty in gradient experimentally:

find $m_{\text{best}}$ from the best-fit line,
find $m_{\text{worst}}$ from the worst acceptable line,
take the absolute difference as the uncertainty:

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

Understanding the Question

You have drawn a best-fit line and a worst acceptable line on the graph of $\dfrac{1}{f}/10^{-3}\ \text{Hz}^{-1}$ against $v$ . You must calculate the gradient of the best-fit line and include an absolute uncertainty using your worst line.

Approach

Choose two widely separated points on the best-fit line.
Calculate $m_{\text{best}} = \Delta y/\Delta x$ .
Repeat for the worst acceptable line to get $m_{\text{worst}}$ .
Uncertainty: $\Delta m = |m_{\text{best}}-m_{\text{worst}}|$ .

Step-by-Step Reasoning

Read off two points on the best-fit line, ideally near the ends of the drawn line.
Compute $\Delta y$ and $\Delta x$ in the plotted units.
Divide to get the gradient.
Do the same for the worst acceptable line.
The uncertainty is not a percentage here; it is an absolute uncertainty in the same units as the gradient.

Because the y-axis is labelled in $10^{-3}\ \text{Hz}^{-1}$ , the gradient unit is $\left(10^{-3}\ \text{Hz}^{-1}\right)$ per $\text{m s}^{-1}$ .

Key Takeaways

Use a large triangle for better precision.
Worst acceptable line gives a realistic uncertainty estimate.

Common Mistakes

Using two points very close together (large fractional error).
Calculating gradient as $\Delta x/\Delta y$ .
Taking uncertainty as half the difference when the scheme expects the full difference (or vice versa); always follow the question wording/standard method used.

Things to Be Careful About

Keep consistent units with the axis scaling ( $10^{-3}$ factor on $1/f$ ).
Quote uncertainty to 1 (or 2) significant figures and match the decimal places of the gradient value.

Techniques used

calculate gradient from two well-separated points on the best-fit linecalculate gradient from the worst acceptable lineestimate absolute uncertainty from the difference between gradients

(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

Best-fit y-intercept:

c_{\text{best}} = 1.131\ \left(10^{-3}\ \text{Hz}^{-1}\right)

Worst acceptable y-intercept:

c_{\text{worst}} = 1.128\ \left(10^{-3}\ \text{Hz}^{-1}\right)

Absolute uncertainty:

\Delta c = |c_{\text{best}}-c_{\text{worst}}| = 0.003\ \left(10^{-3}\ \text{Hz}^{-1}\right)

Answer

c = (1.131 \pm 0.003)\ \left(10^{-3}\ \text{Hz}^{-1}\right)

Final answer

c = (1.131 ± 0.003) (10^−3 Hz^−1)

Detailed explanation

Background Concept

For a straight line graph, the y-intercept is the value of $y$ when $x=0$ . Graphically, you can:

extend the best-fit line to cross the y-axis and read the value,
do the same for the worst acceptable line,
estimate uncertainty with

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

Understanding the Question

You must obtain the y-intercept of the best-fit line on the graph of $\dfrac{1}{f}$ (in $10^{-3}\ \text{Hz}^{-1}$ ) against $v$ , and include an absolute uncertainty using your worst acceptable line.

Approach

Extend the best-fit line to $v=0$ .
Read off $c_{\text{best}}$ .
Extend the worst acceptable line to $v=0$ .
Read off $c_{\text{worst}}$ .
Use $\Delta c = |c_{\text{best}} - c_{\text{worst}}|$ .

Step-by-Step Reasoning

Use a ruler to extend the line smoothly to the y-axis.
Read the y-axis carefully using the provided scale increments.
Repeat for the worst acceptable line.
Quote the intercept with uncertainty in the units of the y-axis (here $10^{-3}\ \text{Hz}^{-1}$ ).

Key Takeaways

Intercept uncertainty comes from how much the intercept can change while still fitting within error bars.

Common Mistakes

Reading the intercept from where the line crosses the x-axis (wrong axis).
Forgetting the $10^{-3}$ factor and writing $1.131\times 10^{-3}$ as just $1.131$ without indicating units.
Not using the worst line to estimate uncertainty.

Things to Be Careful About

Ensure you are reading at $v=0$ , not at the left edge of the grid if it doesn’t start at 0.
Quote uncertainty to an appropriate number of significant figures and match decimal places.

Techniques used

read the y-intercept from the best-fit lineread the y-intercept from the worst acceptable lineestimate absolute uncertainty from the difference between 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): y-intercept $c = \dfrac{1}{f_s}$ and gradient $m = -\dfrac{1}{f_s k}$ for a plot of $\dfrac{1}{f}$ against $v$ .

From the graph (using the axis scaling):

c = 1.131\times 10^{-3}\ \text{Hz}^{-1}

m = -0.00350\times 10^{-3}\ \text{Hz}^{-1}\ (\text{m s}^{-1})^{-1} = -3.50\times 10^{-6}\ \text{Hz}^{-1}\ (\text{m s}^{-1})^{-1}

Hence

f_s = \frac{1}{c} = \frac{1}{1.131\times 10^{-3}} = 8.84\times 10^2\ \text{Hz}

and using $m=-\dfrac{1}{f_s k}$ :

k = -\frac{1}{f_s m} = -\frac{1}{(8.84\times 10^2)(-3.50\times 10^{-6})} = 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

If a graph has the form

y = mx + c,

and you have already expressed the physics equation in a matching linear form, then the gradient and intercept can be used to determine constants.

Here, after linearisation you have:

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

So:

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

Understanding the Question

You must use your measured gradient and intercept (from the graph of $1/f$ against $v$ ) to find numerical values of:

$f_s$ (frequency emitted by the horn)
$k$ (a constant; it should come out with units of speed)

The graph’s y-axis is in $10^{-3}\ \text{Hz}^{-1}$ , so you must interpret intercept/gradient with that scaling.

Approach

Convert the intercept value from “ $10^{-3}\ \text{Hz}^{-1}$ units” into $\text{Hz}^{-1}$ by multiplying by $10^{-3}$ .
Use $f_s = 1/c$ .
Convert the gradient similarly (multiply by $10^{-3}$ ) to get it in $\text{Hz}^{-1}\,(\text{m s}^{-1})^{-1}$ .
Use $m = -1/(f_s k)$ to solve for $k$ .

Step-by-Step Reasoning

From the straight-line form:

\frac{1}{f} = \underbrace{\frac{1}{f_s}}_{c} + \underbrace{\left(-\frac{1}{f_s k}\right)}_{m} v

So if the intercept read from the plotted scale is $c = 1.131$ in units of $10^{-3}\ \text{Hz}^{-1}$ , then the actual intercept in SI units is:

c = 1.131\times 10^{-3}\ \text{Hz}^{-1}.

Then

f_s = \frac{1}{c}.

For the gradient, if you read $m = -0.00350$ in units of $(10^{-3}\ \text{Hz}^{-1})/(\text{m s}^{-1})$ , the SI gradient is:

m = -0.00350\times 10^{-3} = -3.50\times 10^{-6}\ \text{Hz}^{-1}\ (\text{m s}^{-1})^{-1}.

Finally from

m = -\frac{1}{f_s k},

rearrange:

k = -\frac{1}{f_s m}.

The negative signs cancel because $m$ is negative, giving a positive $k$ .

Key Takeaways

Intercept gives $f_s$ directly: $f_s = 1/c$ .
Gradient then gives $k$ .
Always handle the $10^{-3}$ axis scaling correctly.

Common Mistakes

Using $c$ in “graph units” without multiplying by $10^{-3}$ .
Dropping the negative sign in $m$ and then getting a negative $k$ .
Mixing up which quantity is $f_s$ and which is $k$ .

Things to Be Careful About

Units: $f_s$ must be in $\text{Hz}$ , $k$ must be in $\text{m s}^{-1}$ .
Significant figures should reflect the graph reading precision.

Techniques used

use gradient and intercept relationships to determine constantsconvert scaled graph quantities back to SI unitsrearrange simultaneous expressions to solve for unknown constants

(ii)

Determine the percentage uncertainty in $k$ .

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

1M Similar

DifficultyMedium-Easy

Worked solution

Working

From (a):

k = -\frac{c}{m}

So fractional (percentage) uncertainties add:

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

Using $c = 1.131$ , $\Delta c = 0.003$ , $|m|=0.00350$ , $\Delta m = 0.00029$ (in the plotted units):

\%\Delta c = \frac{0.003}{1.131}\times 100 = 0.265\%

\%\Delta m = \frac{0.00029}{0.00350}\times 100 = 8.29\%

\%\Delta k = 0.265 + 8.29 = 8.56\% \approx 8.6\%

Answer

\text{percentage uncertainty in } k = 8.6\%

Final answer

8.6%

Detailed explanation

Background Concept

For multiplication/division, uncertainties are combined using fractional (or percentage) uncertainties:

If $Q = AB$ , then

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

If $Q = A/B$ , then

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

This is why the “worst acceptable line” is useful: it gives $\Delta m$ and $\Delta c$ .

Understanding the Question

You found $k$ using the graph parameters (gradient and y-intercept). Now you must find the percentage uncertainty in $k$ , using the uncertainties in the gradient and intercept from parts (c)(iii) and (c)(iv).

Approach

Write $k$ in terms of $m$ and $c$ .
Add fractional uncertainties.
Convert to percentage.

Step-by-Step Reasoning

From the linearised equation:

intercept $c = 1/f_s$
gradient $m = -1/(f_s k)$

Eliminate $f_s$ to get $k$ in terms of $c$ and $m$ :

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

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

Compute each percentage separately, then add.

Typically the gradient uncertainty dominates because the gradient is found from two read-offs and depends strongly on how the line is drawn.

Key Takeaways

For $k = c/m$ , percentage uncertainty in $k$ is the sum of percentage uncertainties in $c$ and $m$ .

Common Mistakes

Subtracting percentage uncertainties instead of adding.
Using $m$ with its negative sign inside the percentage calculation (use $|m|$ ).
Mixing SI-scaled values with plotted-scale values inconsistently (though here the scale factor cancels if used consistently for both $c$ and $m$ ).

Things to Be Careful About

Use the same units/scaling for $c$ and $\Delta c$ (and for $m$ and $\Delta m$ ).
Round the final percentage uncertainty sensibly (usually to 2 significant figures).

Techniques used

propagate uncertainties through multiplication and division using fractional uncertaintiescombine percentage uncertainties additively for products/quotients

(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

Given

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

Rearrange:

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

v = k - \frac{f_s k}{f} = k\left(1-\frac{f_s}{f}\right)

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

v = 3.23\times 10^2\left(1-\frac{8.84\times 10^2}{987.8}\right) = 33.9\ \text{m s}^{-1}

Answer

v = 33.9\ \text{m s}^{-1}

Final answer

33.9 m s^−1

Detailed explanation

Background Concept

Once a model (equation) is established and its constants are determined, it can be used for prediction by rearranging and substituting values.

For algebraic rearrangement, keep track of which quantities are constants ( $f_s$ , $k$ here) and which are variables ( $f$ , $v$ ).

Understanding the Question

You repeat the experiment and now measure a frequency $f = 987.8\ \text{Hz}$ . Using your previously determined values of $f_s$ and $k$ , you must find what speed $v$ would produce this detected frequency according to the model:

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

Approach

Rearrange the formula to make $v$ the subject.
Substitute $f$ , $f_s$ , and $k$ .
Quote $v$ with unit $\text{m s}^{-1}$ .

Step-by-Step Reasoning

Starting with

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

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

f(k-v) = f_s k

Divide by $f$ :

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

Rearrange for $v$ :

v = k - \frac{f_s k}{f} = k\left(1-\frac{f_s}{f}\right)

Now substitute in the values you found from the graph. Since $f > f_s$ (approaching source gives higher detected frequency), the bracket is positive and you get a positive speed.

Key Takeaways

Rearrangement skills let you use a model to predict unmeasured quantities.
Sense-check: for an approaching car, $f$ should be greater than $f_s$ and $v$ should be less than $k$ .

Common Mistakes

Algebra slip: writing $v = k - f_s/f$ (missing factor of $k$ ).
Substituting $f_s$ and $k$ with wrong powers of ten due to the $10^{-3}$ scaling earlier.
Forgetting units on the final speed.

Things to Be Careful About

Ensure $k-v$ stays positive (here $v<k$ ), otherwise the model would predict an unphysical divergence.
Use consistent significant figures based on $f_s$ and $k$ obtained from the graph.

Techniques used

rearrange the given formula to make the required variable the subjectsubstitute experimental constants into a model to predict a value

Physics 9702/53 — May/June 2024

Variables

Procedure and measurements

Control of variables

Analysis of data (to test relationship and find UUU and ZZZ)

Safety

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

1) Measuring the required quantities

2) Setting up so the model is valid

3) Collecting data

4) Linearising and extracting UUU and ZZZ

5) Uncertainties (how you would handle them)

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

Working

Analysis of data (to test relationship and find $U$ and $Z$ )

4) Linearising and extracting $U$ and $Z$