Physics 9702/53 — October/November 2025

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

questions

marks

minutes

Topics Analysis, Conclusions and Evaluation · Planning

Q115MMedium-HardPlanningAnalysis, Conclusions and Evaluation

On a bench, a steel ball of radius $r$ is used to compress a spring by a distance $x$ . The ball is held at rest in this position, as shown in Fig. 1.1.

Fig. 1.1

The ball is released and rolls along the bench. At a fixed point P, the ball has speed $v$ . The speed of the ball at P is determined using one light gate connected to a timer.

Several steel balls of different radii are available.

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

v^2 = \frac{Ykx^2}{r^n\rho}

where $k$ is the spring constant of the spring, $\rho$ is the density of the steel, and $Y$ and $n$ are constants.

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

Draw a diagram showing the arrangement of your equipment.

Explain how the results could be used to determine values for $Y$ and $n$ .

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-1

Variables

Independent variable: ball radius $r$ (use different steel balls).
Dependent variable: speed $v$ of the ball at fixed point $P$ .
Controlled variables: spring compression $x$ , same spring (same $k$ ), same position of $P$ (same distance from release point), same bench surface/track and alignment, same light-gate/flag arrangement.

Apparatus

Spring and rigid stop, clamp stand, metre rule/set square, set of steel balls, vernier callipers/micrometer, light gate + timer, thin card “flag” of known length $L$ (same for all trials) taped to ball, tape measure to locate $P$ , tray/stop to catch ball.

Procedure and measurements

Determine $k$ for the spring (if not given): hang known masses, measure extension, plot $F$ against extension; gradient $=k$ .
Mark point $P$ at a fixed distance from the release position.
Fix the light gate at $P$ and connect to timer (mode: measure time beam is interrupted).
Attach the same light flag of length $L$ to each ball; measure $L$ with callipers.
For a chosen ball, measure its diameter with callipers and calculate $r$ .
Compress the spring by a fixed distance $x$ using a metre rule (measure from uncompressed position). Hold the ball at rest against the spring, then release without pushing.
Record the interrupt time $t$ when the flag passes the light gate.
Repeat at least 3 times for the same ball and take mean $t$ .
Repeat steps 5–8 for at least 6 balls of different radii.

Calculations

For each ball:

v = \frac{L}{t},\qquad v^2 = \left(\frac{L}{t}\right)^2

Analysis (to determine $n$ and $Y$ )

Given

v^2 = \frac{Ykx^2}{r^n\rho}

Take logs:

\log_{10}(v^2)=\log_{10}\left(\frac{Ykx^2}{\rho}\right)-n\,\log_{10}(r)

Plot $\log_{10}(v^2)$ (y-axis) against $\log_{10}(r)$ (x-axis).

Gradient $m=-n\ \Rightarrow\ n=-m$ .
Intercept $c=\log_{10}\left(\frac{Ykx^2}{\rho}\right)$ , so

Y = \frac{\rho}{kx^2}\,10^{c}

A straight line supports the suggested relationship.

Control of variables

Use the same spring throughout and keep $x$ constant.
Keep $P$ fixed and do not move the light gate between runs.
Use the same flag length $L$ and same release method.
Keep bench/track clean and level; ensure ball rolls along the same line.

Safety

Keep face/hands away from the spring and ball line of motion during release.
Use a stop/tray to prevent the ball falling off the bench.
Ensure clamp stand/light gate is stable to avoid toppling.

Final answer

See working

Detailed explanation

Background Concept

The suggested relationship is

v^2 = \frac{Ykx^2}{r^n\rho}

This has the structure of a power law in $r$ : keeping $k$ , $x$ and $\rho$ constant, it predicts

v^2 \propto r^{-n}.

To test a power law experimentally, we typically:

measure $v$ for different values of $r$ while holding other variables constant, and
transform the equation into a straight-line form so that a graph’s gradient and intercept give the constants.

A single light gate measures the time $t$ for an object (or a fixed “flag” attached to it) to block the beam. If the blocked length is known (call it $L$ ), the instantaneous speed at the gate is

v = \frac{L}{t}.

Understanding the Question

You have steel balls of different radii $r$ . Each ball is launched by a spring compressed by a fixed distance $x$ , and you need the speed $v$ at a fixed point $P$ . You are constrained to using one light gate and a timer.

You must produce a workable plan including: apparatus/diagram, procedure, measurements, controls, data analysis, and safety. You must also explain how to obtain both constants $n$ and $Y$ from the results.

Approach

Choose $r$ as the independent variable (use different balls).
Measure $v$ at point $P$ using one light gate by attaching a flag of known length $L$ to the ball; record interrupt time $t$ and compute $v=L/t$ .
Keep $x$ constant and keep everything else (spring, surface, point $P$ , light-gate setup) constant so that any change in $v$ is due to changing $r$ .
Linearise the model with logarithms:

\log(v^2)=\log\left(\frac{Ykx^2}{\rho}\right)-n\log(r)

so a plot of $\log(v^2)$ vs $\log(r)$ is a straight line with gradient $-n$ . The intercept gives $Y$ .

Step-by-Step Reasoning

diagram-1

1) Measuring the radii

For each ball, measure the diameter $d$ with vernier callipers or a micrometer (take several readings at different orientations and average to reduce random error). Then

r = \frac{d}{2}.

2) Keeping the launch conditions constant

Use the same spring for all trials (so $k$ is constant).
Set the compression to the same value $x$ each time using a ruler fixed alongside the spring.
Release the ball from rest without giving it an extra push (e.g. hold with a card/trigger and pull it away cleanly).
Keep the point $P$ fixed (mark it on the bench) and keep the light gate fixed at $P$ .

3) Using one light gate to obtain speed at P

A single light gate gives a time interval for which the beam is interrupted. To turn this into a speed, you must know the length that passes through the beam during that time.

A common robust method is:

Attach a thin card “flag” of known length $L$ to the ball (keep the same flag for all balls, or make identical flags).
Measure $L$ with callipers.
When the ball passes through the gate, the timer reads the interrupt time $t$ .
Then compute

v = \frac{L}{t}.

Repeat at least 3 times for each ball and average $t$ (or average $v$ ) to reduce random uncertainty.

4) Processing and linearising the data

From the suggested equation,

v^2 = \frac{Ykx^2}{r^n\rho}

take base-10 logs (natural logs also fine if used consistently):

\log_{10}(v^2)=\log_{10}\left(\frac{Ykx^2}{\rho}\right)-n\,\log_{10}(r).

This matches the straight-line form $y=c+mx$ if we set:

$y=\log_{10}(v^2)$ ,
$x=\log_{10}(r)$ ,
gradient $m=-n$ ,
intercept $c=\log_{10}\left(\frac{Ykx^2}{\rho}\right)$ .

So:

Find $n$ from the gradient: $n=-m$ .
Find $Y$ from the intercept:

10^{c}=\frac{Ykx^2}{\rho}\quad\Rightarrow\quad Y=\frac{\rho}{kx^2}\,10^{c}.

If the plotted points lie close to a straight line (within experimental scatter), that supports the proposed relationship.

5) Control of variables (what to state explicitly)

Creditworthy controls are those that directly affect energy transfer and energy losses:

Compression $x$ : keep fixed, because the model predicts $v^2\propto x^2$ .
Same spring: keeps $k$ constant.
Same point $P$ and same gate position: ensures the same distance travelled before measuring $v$ (so losses are comparable).
Same surface/track and alignment: frictional losses change $v$ ; keep the bench level and clean.
Same flag length $L$ and same gate mode: ensures the speed calculation method does not change between balls.

6) Safety

Main hazards are the spring release and the ball rolling/falling:

Keep hands/face away from line of motion; do not lean over the spring on release.
Use a catch tray/stop at the end of the bench.
Clamp equipment securely so the gate/stand cannot topple.

Key Takeaways

A single light gate measures a time interval; you need a known interrupt length to convert that into speed.
Testing a power law is best done by log-linearising and plotting a straight-line graph.
Gradient gives the exponent ( $n$ ), intercept gives the constant ( $Y$ ) once other parameters are known/controlled.
A strong plan always identifies variables, controls, repeats, and a clear graph-based method for determining constants.

Common Mistakes

Not stating how $v$ is obtained from one light gate (you must include a measured length such as a flag length).
Changing more than one variable at a time (e.g. varying $x$ as well as $r$ ).
Plotting $v$ against $r$ without linearising, which makes extracting $n$ and $Y$ unreliable.
Forgetting that the graph gradient gives $-n$ (sign error).
Not including repeats/averaging for each radius.

Things to Be Careful About

Ensure the flag is rigidly attached and does not flap; otherwise the interrupt time is inconsistent.
Measure $r$ carefully (callipers perpendicular to diameter; repeat and average).
Keep the light gate aligned so the flag cleanly blocks the beam each time.
When using logs, be consistent about base (10 or $e$ ) and use the matching inverse when finding $Y$ .
If you determine $k$ yourself, take enough data points and use a best-fit line for $F$ vs extension to reduce uncertainty.

Techniques used

identify independent, dependent and controlled variablesmeasure speed using a single light gate and a known interrupt lengthsquare the measured speed to match the suggested relationshiplinearise a power law using logarithms and plot a log-log graphuse gradient and intercept to determine constants in the model

Working

From

\frac{\lambda-\lambda_0}{\lambda_0}=\frac{Hd}{c}

\lambda-\lambda_0=\lambda_0\frac{Hd}{c}

\lambda=\left(\lambda_0H\right)\left(\frac{d}{c}\right)+\lambda_0

Answer

gradient $=\lambda_0H$

$y$ -intercept $=\lambda_0$

Final answer

gradient = (\lambda_0H), y-intercept = (\lambda_0)

Detailed explanation

Background Concept

To analyse experimental data, we often rearrange an equation into the straight-line form

y = mx + c

where:

$m$ is the gradient (slope) of the graph,
$c$ is the $y$ -intercept (value of $y$ when $x=0$ ).

If you plot the correct variables on the axes, you can extract physical constants from $m$ and $c$ .

Understanding the Question

You are given the relationship

\frac{\lambda - \lambda_0}{\lambda_0} = \frac{Hd}{c}

and you are told the student plots a graph of $\lambda$ (vertical axis) against $\dfrac{d}{c}$ (horizontal axis). You must rewrite the equation so that it looks like $\lambda = (\text{something})\,(d/c) + (\text{something})$ .

Approach

Multiply both sides by $\lambda_0$ to remove the fraction.
Make $\lambda$ the subject.
Compare directly with $y=mx+c$ where $y\equiv\lambda$ and $x\equiv d/c$ .

Step-by-Step Reasoning

Start with

\frac{\lambda-\lambda_0}{\lambda_0}=\frac{Hd}{c}.

Multiply by $\lambda_0$ :

\lambda-\lambda_0=\lambda_0\frac{Hd}{c}.

Add $\lambda_0$ to both sides:

\lambda = \lambda_0\frac{Hd}{c} + \lambda_0.

Rewrite to show $d/c$ as a factor:

\lambda = (\lambda_0H)\left(\frac{d}{c}\right) + \lambda_0.

So on a graph of $\lambda$ (as $y$ ) against $d/c$ (as $x$ ):

gradient $m = \lambda_0H$
intercept $c = \lambda_0$ .

Key Takeaways

Always convert the physics equation to $y=mx+c$ using the actual plotted axes.
The intercept is the value of $\lambda$ at $d/c=0$ .

Common Mistakes

Leaving the equation in terms of $\dfrac{\lambda-\lambda_0}{\lambda_0}$ (that would correspond to plotting a different $y$ variable).
Giving gradient as $H/\lambda_0$ or similar due to incorrect algebra.

Things to Be Careful About

The graph is $\lambda$ vs $d/c$ (not $\lambda-\lambda_0$ vs $d/c$ ), so you must isolate $\lambda$ .
Keep track of brackets: the gradient is the entire coefficient multiplying $(d/c)$ , i.e. $(\lambda_0H)$ .

Techniques used

rearrange the given relationship into the form y = mx + cidentify the plotted variables with y and xread off gradient and intercept from the linear form

(b)

Values of $d$ and $\lambda$ are given in Table 2.1.

Table 2.1

$d / 10^{21}\ \text{km}$	$\frac{d}{c} / 10^{15}\ \text{s}$	$\lambda / \text{nm}$
$0.48 \pm 0.12$		$658.4$
$1.04 \pm 0.12$		$661.2$
$1.45 \pm 0.12$		$664.2$
$1.80 \pm 0.12$		$665.7$
$2.85 \pm 0.12$		$672.4$
$3.75 \pm 0.12$		$678.2$

The value of $c$ is $3.00 \times 10^5\ \text{km s}^{-1}$ .

Calculate and record values of $\frac{d}{c} / 10^{15}\ \text{s}$ in Table 2.1. Include the absolute uncertainties in $\frac{d}{c}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Working

\frac{d}{c}=\frac{d}{3.00\times 10^5}

Uncertainty:

\Delta\left(\frac{d}{c}\right)=\frac{\Delta d}{c}=\frac{0.12\times 10^{21}}{3.00\times 10^5}=0.40\times 10^{15}\ \text{s}

Values (in units of $10^{15}\ \text{s}$ ):

$d=0.48\times 10^{21}\ \text{km}$ :

\frac{d}{c}=\frac{0.48\times 10^{21}}{3.00\times 10^5}=1.60\times 10^{15}\ \text{s}

$d=1.04\times 10^{21}\ \text{km}$ :

\frac{d}{c}=3.47\times 10^{15}\ \text{s}

$d=1.45\times 10^{21}\ \text{km}$ :

\frac{d}{c}=4.83\times 10^{15}\ \text{s}

$d=1.80\times 10^{21}\ \text{km}$ :

\frac{d}{c}=6.00\times 10^{15}\ \text{s}

$d=2.85\times 10^{21}\ \text{km}$ :

\frac{d}{c}=9.50\times 10^{15}\ \text{s}

$d=3.75\times 10^{21}\ \text{km}$ :

\frac{d}{c}=12.5\times 10^{15}\ \text{s}

Answer

$\dfrac{d}{c} / 10^{15}\ \text{s}$ :

$1.60 \pm 0.40$ , $3.47 \pm 0.40$ , $4.83 \pm 0.40$ , $6.00 \pm 0.40$ , $9.50 \pm 0.40$ , $12.5 \pm 0.40$

Final answer

1.60±0.40, 3.47±0.40, 4.83±0.40, 6.00±0.40, 9.50±0.40, 12.5±0.40 (all in units of 10^15 s)

Detailed explanation

Background Concept

When you create a new column in a data table using a formula, you must:

Calculate the derived quantity.
Propagate uncertainties correctly.

Here,

\frac{d}{c}

is obtained by dividing the distance $d$ by the speed $c$ .

For multiplication/division by a constant, the absolute uncertainty scales by the same factor:

\Delta\left(\frac{d}{c}\right)=\frac{\Delta d}{c}.

Understanding the Question

The table gives $d$ in units of $10^{21}\ \text{km}$ with an absolute uncertainty $\pm 0.12\times 10^{21}\ \text{km}$ for every row. You must calculate $d/c$ (in seconds) and record it in the table as

\frac{d}{c}\ /\ 10^{15}\ \text{s}

and include the absolute uncertainty in $d/c$ .

Approach

Convert each distance into a time using $t=d/c$ .
Because $c$ is fixed and exact for this calculation, divide both the central value and the absolute uncertainty in $d$ by $c$ .
Finally, express the result in units of $10^{15}\ \text{s}$ to match the table.

Step-by-Step Reasoning

Given

c = 3.00\times 10^5\ \text{km s}^{-1}.

For any row,

\frac{d}{c} = \frac{d}{3.00\times 10^5}.

Example (first row):

\frac{d}{c} = \frac{0.48\times 10^{21}}{3.00\times 10^5} =0.16\times 10^{16} =1.6\times 10^{15}\ \text{s}.

Uncertainty in $d/c$ (same method for all rows because $\Delta d$ is the same each time):

\Delta\left(\frac{d}{c}\right)=\frac{0.12\times 10^{21}}{3.00\times 10^5} =0.04\times 10^{16} =0.40\times 10^{15}\ \text{s}.

So each table entry becomes “value $\pm 0.40$ ” in the table’s unit of $10^{15}\ \text{s}$ .

Key Takeaways

Dividing by a constant divides the absolute uncertainty by the same constant.
Match the powers of ten and units used in the table heading.

Common Mistakes

Forgetting to include the uncertainty in $d/c$ .
Putting percentage uncertainty instead of absolute uncertainty.
Forgetting the table’s scaling ( $/10^{15}\ \text{s}$ ) and writing the raw seconds value.

Things to Be Careful About

Keep the uncertainty to an appropriate number of significant figures (typically 1 or 2), and match the decimal places of the value to the uncertainty.
Use $c=3.00\times 10^5\ \text{km s}^{-1}$ (not $3.00\times 10^8\ \text{m s}^{-1}$ ) because $d$ is in km.

Techniques used

convert a measured quantity using a constant factorpropagate absolute uncertainty through multiplication/division by a constantrecord values to appropriate significant figures consistent with uncertainties

(c)

(i)

Plot a graph of $\lambda / \text{nm}$ against $\frac{d}{c} / 10^{15}\ \text{s}$ . Include error bars for $\frac{d}{c}$ .

Graph 2.1

2M Similar

DifficultyMedium

Worked solution

Answer

Plot the points $(x, y)$ with $x = (d/c)/10^{15}\ \text{s}$ and $y=\lambda/\text{nm}$ :

$(1.60,658.4)$ , $(3.47,661.2)$ , $(4.83,664.2)$ , $(6.00,665.7)$ , $(9.50,672.4)$ , $(12.5,678.2)$ .

Add horizontal error bars of $\pm 0.40$ on each $x$ -value.

Final answer

See graph (points plotted with ±0.40 horizontal error bars).

Detailed explanation

Background Concept

A graph is used to test whether two quantities are linearly related. Good graph technique (Paper 5) includes:

Correct axes with quantity and unit.
A sensible scale using at least half the grid.
Accurate plotting (fine pencil, small crosses/dots).
Error bars representing absolute uncertainty in the measured quantity.

Here, the uncertainty is in $x = d/c$ , so you add horizontal error bars.

Understanding the Question

You must plot $\lambda$ (in nm) on the $y$ -axis against $(d/c)/10^{15}\ \text{s}$ on the $x$ -axis using the values you calculated in part (b). You must include error bars for $(d/c)$ using the absolute uncertainties you found.

Approach

Label axes exactly as requested.
Choose scales that spread the data well across the grid.
Plot each point accurately.
For each point, draw a horizontal line segment representing $x \pm \Delta x$ where $\Delta x = 0.40$ .

Step-by-Step Reasoning

Your $x$ values range roughly from $1.6$ to $12.5$ , so the given 0 to 14 scale is appropriate.
Your $y$ values range from about $658$ to $678\ \text{nm}$ , so the given 658 to 680 scale is appropriate.

For each data point:

Plot a small cross at $(x,y)$ .
Draw the horizontal error bar from $(x-0.40, y)$ to $(x+0.40, y)$ .

Example: for $x=3.47$ , draw the bar from $3.07$ to $3.87$ at $y=661.2$ .

Key Takeaways

Error bars go on the axis of the quantity that has uncertainty.
Error bars are drawn using absolute uncertainty, not percentage uncertainty.

Common Mistakes

Drawing vertical error bars (there is no uncertainty given for $\lambda$ ).
Using the uncertainty in $d$ instead of the uncertainty in $d/c$ .
Plotting with poor scale so the points are bunched into a corner.

Things to Be Careful About

Ensure the error bar half-length is exactly $0.40$ in the graph’s $x$ units.
Do not join the points dot-to-dot; the line is drawn in part (ii).

Techniques used

plot experimental points with appropriate axis labels and scalesadd horizontal error bars using absolute uncertainties in the x-quantityuse consistent plotting conventions for units and significant figures

(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 a single straight line of best fit (balanced about the plotted points).

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

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

Final answer

See graph (best-fit line and labelled worst acceptable line).

Detailed explanation

Background Concept

A line of best fit represents the overall linear trend of the data. In Paper 5, uncertainties in gradient and intercept are often estimated using a worst acceptable line:

Draw the best-fit line first.
Then draw the steepest (or shallowest) straight line that is still consistent with the error bars of all points.
The difference between the gradients (or intercepts) of these two lines gives an estimate of the uncertainty.

Understanding the Question

You already plotted $\lambda$ vs $d/c$ with horizontal error bars. Now you must:

Draw the best-fit straight line.
Draw a worst acceptable straight line.
Label both.

Approach

Best-fit line: place it so that points are roughly equally scattered above and below, not forced through every point.
Worst line: choose either the steepest or shallowest line that still intersects every point’s error bar region.

Step-by-Step Reasoning

Using a ruler, draw the straight line that best represents the trend. It should go through the “middle” of the error-bar rectangles and not be kinked.
Now try to rotate the line to make it as steep as possible while still crossing each horizontal error bar (i.e. for each point at height $y$ , the line’s $x$ value at that $y$ must lie within $x\pm\Delta x$ ).
If making it steeper fails to pass through one point’s error bar, reduce steepness until it just fits.
If the steepest line is very close to the best fit, try the shallowest line instead; you will use whichever gives the larger difference from best fit.

Key Takeaways

Best fit is about overall trend; worst line is about extremal gradient still allowed by the uncertainties.

Common Mistakes

Drawing the worst line without checking it intersects all error bars.
Drawing two worst lines but not clearly identifying which is used later.
Forcing the best-fit line through the origin (not justified here).

Things to Be Careful About

Use a ruler; freehand lines typically lose the mark.
Label the lines directly on the graph so the examiner can tell which is which.

Techniques used

draw a straight line of best fit with points balanced about the lineconstruct a worst acceptable line constrained by the error barslabel lines clearly to distinguish best and worst fits

(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 two well-separated points on the best-fit line (example): $(2.0,658.8)$ and $(12.0,677.1)$ ,

\text{gradient } m = \frac{\Delta y}{\Delta x} = \frac{677.1-658.8}{12.0-2.0}=1.83

Worst acceptable gradient (from worst line): $m_w \approx 1.69$ (or $1.96$ ).

\Delta m = |m-m_w| \approx 0.14

Answer

m = (1.83 \pm 0.14)\ \text{nm per }(10^{15}\ \text{s})

Final answer

(1.83 ± 0.14) nm per (10^15 s)

Detailed explanation

Background Concept

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

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

To reduce percentage error, you must use a large triangle: pick two points far apart on the line, not necessarily the original plotted points.

To estimate uncertainty in gradient on a graph with error bars:

find $m_{\text{best}}$ from the best-fit line;
find $m_{\text{worst}}$ from the worst acceptable line;
take

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

Understanding the Question

You must read the gradient of the line of best fit from your graph of $\lambda$ vs $(d/c)/10^{15}\ \text{s}$ , and include an absolute uncertainty obtained from the difference between best and worst acceptable lines.

Approach

Choose two widely separated points on the best-fit line, read $(x_1,y_1)$ and $(x_2,y_2)$ .
Compute $m_{\text{best}}=(y_2-y_1)/(x_2-x_1)$ .
Repeat using the worst acceptable line to get $m_{\text{worst}}$ .
Use $\Delta m = |m_{\text{best}}-m_{\text{worst}}|$ .

Step-by-Step Reasoning

Suppose two points on the best-fit line are read as $(2.0,658.8)$ and $(12.0,677.1)$ .

Compute:

\Delta y = 677.1-658.8=18.3\ \text{nm}

\Delta x = 12.0-2.0=10.0\ \text{(in units of }10^{15}\ \text{s)}

m_{\text{best}}=\frac{18.3}{10.0}=1.83\ \text{nm per }(10^{15}\ \text{s}).

Then from the worst acceptable line you might obtain, for example, $m_{\text{worst}}=1.69$ (or a steepest line giving $1.96$ ). The uncertainty is the larger difference from $m_{\text{best}}$ :

\Delta m = |1.83-1.69|=0.14.

Hence $m=(1.83\pm 0.14)$ in the graph’s units.

Key Takeaways

Always use two points far apart on the drawn line.
Uncertainty comes from comparing best-fit and worst acceptable gradients.

Common Mistakes

Using two nearby points, giving a large percentage uncertainty.
Using $\Delta x/\Delta y$ instead of $\Delta y/\Delta x$ .
Calculating uncertainty as half the difference without being instructed; Paper 5 typically uses the full difference between best and worst.

Things to Be Careful About

Make sure $\Delta x$ uses the graph’s $x$ units (here $10^{15}\ \text{s}$ ).
Read points from the line, not from the data table.

Techniques used

determine gradient from a large triangle on the best-fit linecalculate uncertainty in gradient from best-fit and worst acceptable gradientsquote a result with an absolute uncertainty

(iv)

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

$y$ -intercept = ______

2M Similar

DifficultyMedium

Worked solution

Working

From the best-fit line, at $x=0$ :

\text{y-intercept } = 655.1\ \text{nm}

From the worst acceptable line, intercept $\approx 656.4\ \text{nm}$ .

\Delta c = |656.4-655.1| \approx 1.3\ \text{nm}

Answer

\text{y-intercept} = (655.1 \pm 1.3)\ \text{nm}

Final answer

(655.1 ± 1.3) nm

Detailed explanation

Background Concept

For a straight-line graph $y=mx+c$ , the $y$ -intercept $c$ is the value of $y$ when $x=0$ .

Graphically, it is where the drawn line crosses the $y$ -axis.

Uncertainty in the intercept is estimated similarly to the gradient:

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

Understanding the Question

You must read the $y$ -intercept of the best-fit line on your $\lambda$ vs $(d/c)/10^{15}\ \text{s}$ graph, and estimate its absolute uncertainty using the worst acceptable line.

Approach

Extend the best-fit line to the $y$ -axis (at $x=0$ ) and read $c_{\text{best}}$ .
Extend the worst acceptable line to the $y$ -axis and read $c_{\text{worst}}$ .
Take $\Delta c = |c_{\text{best}}-c_{\text{worst}}|$ .

Step-by-Step Reasoning

On the best-fit line, at $x=0$ the intercept is read as about $655.1\ \text{nm}$ .
On the worst acceptable line, the intercept might be read as about $656.4\ \text{nm}$ .

So the absolute uncertainty estimate is

\Delta c = |656.4-655.1|=1.3\ \text{nm}.

Hence

c = (655.1 \pm 1.3)\ \text{nm}.

Key Takeaways

The intercept is the physical value of $\lambda$ at zero $d/c$ .
Use best vs worst acceptable line to estimate its uncertainty.

Common Mistakes

Reading the intercept from the data table rather than from the drawn line.
Forgetting to extend the line to $x=0$ (especially if it doesn’t naturally reach the axis on the drawn segment).

Things to Be Careful About

Read the intercept to the resolution allowed by the graph scale.
Keep the uncertainty consistent with the precision of the intercept reading.

Techniques used

read the y-intercept from the best-fit line at x = 0estimate intercept uncertainty from best-fit and worst acceptable interceptsquote a result with an absolute uncertainty

(d)

Using your answers to (a), (c)(iii) and (c)(iv), determine the values of $\lambda_0$ and $H$ . Include appropriate units.

$\lambda_0$ = ______
$H$ = ______

2M Similar

DifficultyMedium

Worked solution

Working

From (a):

$y$ -intercept $=\lambda_0$
gradient $=\lambda_0H$

From (c)(iv):

\lambda_0 = 655.1\ \text{nm}

Graph uses $x = (d/c)/10^{15}\ \text{s}$ , so

\text{gradient on graph} = (\lambda_0H)(10^{15})

Hence

H = \frac{m}{\lambda_0\times 10^{15}}

With $m=1.83$ and $\lambda_0=655.1$ :

H = \frac{1.83}{655.1\times 10^{15}} = 2.79\times 10^{-18}\ \text{s}^{-1}

Answer

\lambda_0 = 655\ \text{nm}

H = 2.8\times 10^{-18}\ \text{s}^{-1}

Final answer

λ0 = 655 nm, H = 2.8 × 10^-18 s^-1

Detailed explanation

Background Concept

When you have a best-fit line for a linearised relation, you can map the graph equation onto the physics equation.

From part (a) we found

\lambda = (\lambda_0H)\left(\frac{d}{c}\right) + \lambda_0.

So:

intercept $=\lambda_0$
gradient (with respect to $d/c$ measured in seconds) $=\lambda_0H$ .

However, if the plotted $x$ axis is scaled (here $(d/c)/10^{15}\ \text{s}$ ), the graph gradient is also scaled.

Understanding the Question

You have measured from the graph:

best-fit gradient $m$ (in nm per unit of $(d/c)/10^{15}\ \text{s}$ )
best-fit intercept $c$ (in nm)

You must use these to calculate $\lambda_0$ and the Hubble constant $H$ with units.

Approach

Set $\lambda_0$ equal to the $y$ -intercept.
Correctly interpret the gradient with the axis scaling: if $x = (d/c)/(10^{15}\ \text{s})$ , then $d/c = x\,(10^{15}\ \text{s})$ .
Substitute into the linear equation and match coefficients to extract $H$ .

Step-by-Step Reasoning

From (a):

\lambda = (\lambda_0H)\left(\frac{d}{c}\right)+\lambda_0.

But the graph uses

x = \frac{(d/c)}{10^{15}\ \text{s}} \quad \Rightarrow \quad \frac{d}{c} = x\,(10^{15}\ \text{s}).

Substitute into the equation:

\lambda = (\lambda_0H)\big(x\,10^{15}\big) + \lambda_0 = \big(\lambda_0H\,10^{15}\big)x + \lambda_0.

So the gradient on the plotted graph is

m = \lambda_0H\,10^{15}.

Therefore

H = \frac{m}{\lambda_0\,10^{15}}.

Using representative best-fit results $m=1.83$ and $\lambda_0=655.1\ \text{nm}$ :

H = \frac{1.83}{655.1\times 10^{15}} = 2.79\times 10^{-18}\ \text{s}^{-1}.

The intercept gives

\lambda_0 = 655.1\ \text{nm} \approx 655\ \text{nm}.

Key Takeaways

Intercept directly gives $\lambda_0$ .
Gradient gives $H$ only after accounting for the axis scaling factor ( $10^{15}$ here).

Common Mistakes

Using $H = m/\lambda_0$ and forgetting the $10^{15}$ scale factor.
Giving $H$ in the wrong units (it should be $\text{s}^{-1}$ ).

Things to Be Careful About

The units cancel for nm in $H = m/(\lambda_0\,10^{15})$ , leaving $\text{s}^{-1}$ as required.
Use consistent significant figures (typically 2 s.f. for $H$ from a graph).

Techniques used

relate gradient and intercept to physical constants using the linearised equationconvert graph gradient units to SI units using the axis scaling factoruse division to determine a constant from measured gradient and intercept

(e)

Hubble’s law suggests that the age $T$ of the universe is related to $H$ by

T = \frac{1}{H}

Determine a value for $T$ . Include the absolute uncertainty in your answer.

$T$ = ______ $\text{s}$

2M Similar

DifficultyMedium

Worked solution

Working

T=\frac{1}{H}

Using $H = (2.79\pm 0.22)\times 10^{-18}\ \text{s}^{-1}$ ,

T = \frac{1}{2.79\times 10^{-18}} = 3.58\times 10^{17}\ \text{s}

For $T=H^{-1}$ , fractional uncertainty:

\frac{\Delta T}{T}=\frac{\Delta H}{H}=\frac{0.22}{2.79}=0.0789

\Delta T = 0.0789\times 3.58\times 10^{17} = 2.8\times 10^{16}\ \text{s}

Answer

T = (3.6 \pm 0.3)\times 10^{17}\ \text{s}

Final answer

(3.6 ± 0.3) × 10^17 s

Detailed explanation

Background Concept

If a quantity is calculated from a measured value, you must propagate the uncertainty.

For a reciprocal relationship

T = \frac{1}{H} = H^{-1},

the fractional (percentage) uncertainty in $T$ equals the fractional uncertainty in $H$ :

\frac{\Delta T}{T} = \frac{\Delta H}{H}.

Then the absolute uncertainty is

\Delta T = T\left(\frac{\Delta H}{H}\right).

Understanding the Question

You have found $H$ from the graph and must use

T = \frac{1}{H}

to estimate the age of the universe in seconds, including an absolute uncertainty.

Approach

Calculate $T$ using $T=1/H$ .
Use the rule for powers: for $T=H^{-1}$ , the fractional uncertainty in $T$ equals that in $H$ .
Convert back to an absolute uncertainty in seconds.

Step-by-Step Reasoning

Using a representative value from part (d), e.g.

H = (2.79\pm 0.22)\times 10^{-18}\ \text{s}^{-1},

calculate $T$ :

T = \frac{1}{2.79\times 10^{-18}} = 3.58\times 10^{17}\ \text{s}.

Fractional uncertainty in $H$ :

\frac{\Delta H}{H} = \frac{0.22}{2.79} = 0.0789.

So fractional uncertainty in $T$ is the same:

\frac{\Delta T}{T} = 0.0789.

Hence

\Delta T = 0.0789\times 3.58\times 10^{17} = 2.8\times 10^{16}\ \text{s}.

T = (3.6\pm 0.3)\times 10^{17}\ \text{s}.

Key Takeaways

For $T=1/H$ , the fractional uncertainty transfers directly from $H$ to $T$ .
Quote uncertainty to 1 significant figure (or 2 if it begins with 1 or 2), and match the value’s precision accordingly.

Common Mistakes

Using absolute uncertainty directly (e.g. $\Delta T = 1/\Delta H$ ), which is not valid.
Forgetting that the uncertainty should be absolute in seconds, not a percentage.
Rounding $T$ too aggressively compared with $\Delta T$ .

Things to Be Careful About

Keep powers of ten consistent: $1/(10^{-18})=10^{18}$ .
If your value of $H$ differs slightly (because it depends on your graph), your $T$ will differ; the method of uncertainty propagation is what is being assessed.

Techniques used

use reciprocal relationship to calculate a derived quantitypropagate uncertainty through an inverse function using fractional uncertaintiesquote an absolute uncertainty with appropriate significant figures

Physics 9702/53 — October/November 2025

Variables

Apparatus

Procedure and measurements

Calculations

Analysis (to determine nnn and YYY)

Control of variables

Safety

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

1) Measuring the radii

2) Keeping the launch conditions constant

3) Using one light gate to obtain speed at P

4) Processing and linearising the data

5) Control of variables (what to state explicitly)

6) Safety

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

Analysis (to determine $n$ and $Y$ )