Physics 9702/51 — October/November 2025

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

questions

marks

minutes

Topics Planning · Analysis, Conclusions and Evaluation

Q115MMedium-HardPlanning

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

Answer

Variables

Independent variable: ball radius $r$ (use several steel balls of different radii).
Dependent variable: speed $v$ at fixed point $P$ .
Controlled: compression $x$ , spring and spring constant $k$ , position of light gate at $P$ , bench level/surface, release method, same rolling path (no slipping).

Apparatus

Spring and clamp/rigid stop, metre rule, set square/marker for fixed compression, several steel balls, micrometer screw gauge/vernier calipers, light gate + timer/data logger, retort stands/clamps, tray/stop to catch ball.

Procedure and measurements

Fix the spring against a rigid support on the bench. Mark a fixed point $P$ on the bench and clamp a light gate at $P$ .
Set a fixed compression $x$ using a metre rule and a stop/marker so the spring is compressed by the same amount each run.
For each ball:
- Measure its diameter $d$ with a micrometer (several orientations), calculate $r = d/2$ .
- Place the ball against the compressed spring and release without pushing.
- Record the time $t$ for which the light beam is interrupted as the ball passes through the light gate at $P$ .
- Repeat at least 3 times and average $t$ .
Calculate the speed at $P$ using

v = \frac{d}{t} = \frac{2r}{t}

and hence calculate $v^2$ .

Analysis to determine $n$ and $Y$

Given

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

Take logs:

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

Plot $\ln(v^2)$ (y-axis) against $\ln(r)$ (x-axis).
Gradient $m = -n$ so $n = -m$ .
Intercept $c = \ln\left(\frac{Ykx^2}{\rho}\right)$ so

\frac{Ykx^2}{\rho} = e^c

and hence

Y = \frac{\rho e^c}{kx^2}

(using known $\rho$ for steel, measured/known $k$ , and fixed measured $x$ ).

Control of variables (examples)

Use the same spring throughout; do not change its extension range.
Keep $x$ fixed using a physical stop; check with ruler each run.
Keep light gate at the same point $P$ and same height so the ball passes centrally.
Ensure bench is horizontal and surface condition unchanged.
Use side guides to keep the ball travelling straight and rolling without slipping.

Safety

Prevent ball falling off the bench: use a tray/stop at the end; keep feet clear.
Keep fingers clear of the compressed spring when releasing; ensure clamp/support is secure.
Ensure stands and light gate are stable to avoid toppling.

Final answer

See working

Detailed explanation

Background Concept

The suggested relationship is a power law connecting the ball speed $v$ at a fixed position to its radius $r$ :

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

Here $k$ is the spring constant, $x$ is the compression (so the spring energy depends on $kx^2$ ), $\rho$ is the density of steel (constant for all the balls), and $Y$ and $n$ are unknown constants.

To "test the relationship" you need to vary $r$ systematically, measure $v$ , and then check whether the data fit the functional form. For power laws, the standard method is to take logarithms so that the relationship becomes linear, allowing the constants to be found from a straight-line graph.

A light gate measures the time interval for which its beam is blocked. If the moving object blocks a known length $L$ , then

v = \frac{L}{t}

For a sphere passing through a narrow light beam, the blocked length along the direction of motion is approximately the diameter $d = 2r$ , so $v = d/t$ .

Understanding the Question

You have a spring that launches a steel ball by releasing it from a fixed compression $x$ . The ball then rolls along the bench and passes a single light gate at point $P$ , where its speed is $v$ .

You are given multiple balls with different radii $r$ . You must plan an experiment that:

varies $r$ (independent variable),
measures $v$ at $P$ (dependent variable) using one light gate,
keeps other quantities constant (especially $x$ and the setup),
uses the results to determine $Y$ and $n$ .

Because the equation contains $v^2$ and $r^n$ , the natural analysis is a log-log plot.

Approach

Set up the apparatus so that the spring compression $x$ is the same each trial and the light gate stays fixed at $P$ .
For each ball, measure $r$ accurately using a micrometer or vernier calipers.
Use the light gate to measure the interruption time $t$ as the ball passes; use the diameter $d = 2r$ as the distance that blocks the beam, and calculate $v = d/t$ .
Calculate $v^2$ for each ball.
Linearise the relationship by taking logs and plot $\ln(v^2)$ against $\ln(r)$ . The gradient gives $-n$ , and the intercept gives $\ln(Ykx^2/\rho)$ , from which $Y$ can be calculated.
Include practical steps to reduce uncertainty (repeat timings, ensure straight rolling, stable clamping) and include safety measures.

Step-by-Step Reasoning

diagram-1

1) Setting the spring compression as a constant

Since $x$ appears squared, even a small change in $x$ would significantly change $v^2$ . So you should fix $x$ using a physical stop (e.g. a block or pin position) and confirm with a ruler.

2) Measuring the independent variable $r$

Measure the ball diameter $d$ with a micrometer screw gauge (best precision). Take readings in several orientations (to check it is spherical) and average. Then

r = \frac{d}{2}

3) Measuring speed with one light gate

When the sphere passes through the light gate, it blocks the beam for time $t$ . The distance corresponding to the blocked beam is the sphere diameter $d$ (along the direction of travel), so

v = \frac{d}{t} = \frac{2r}{t}

To reduce random timing scatter, repeat the run several times for each ball and average $t$ .

4) Linearising to find $n$

Start from:

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

Take natural logs:

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

This matches the straight-line form $y = c + mx$ if you identify:

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

So:

Find the best-fit straight line and its gradient $m$ ; then $n = -m$ .

5) Using the intercept to find $Y$

From the intercept:

\frac{Ykx^2}{\rho} = e^c

Rearrange:

Y = \frac{\rho e^c}{kx^2}

You can use a known value of $\rho$ for steel (or measure it separately if required), and obtain $k$ from prior calibration of the spring (or given). Since $x$ is set and measured, you can calculate $Y$ .

6) Control of variables (what matters and why)

Same spring / same $k$ : changing spring changes the energy delivered.
Same compression $x$ : directly affects $v^2$ via $x^2$ .
Same point $P$ and gate alignment: ensures speed is measured at the same location each time.
Bench level and same surface: changes in slope or friction would change $v$ .
Ensure rolling without slipping: slipping changes the energy distribution and can alter the relationship.

7) Safety

Main hazards are the ball leaving the bench and the spring snapping forward. A catching tray/stop and careful hand placement (or a mechanical release) address these.

Key Takeaways

A single light gate can measure speed if the interrupted length is known (here, the ball diameter).
To test a power law $v^2 \propto r^{-n}$ , use a log-log plot to obtain a straight line.
Gradient gives the exponent ( $n$ ) and intercept gives the constant (here used to find $Y$ ).
Good planning depends on identifying and controlling the variables that appear in the model equation.

Common Mistakes

Not keeping $x$ constant (or not explaining how it is kept constant).
Measuring only one timing per ball (insufficient repeats for reliability).
Plotting $v$ against $r$ directly and trying to guess $n$ (won’t reliably extract the exponent).
Using the wrong distance in $v = L/t$ (must use the diameter $d$ , not the radius).
Forgetting that the gradient is negative: $m=-n$ , not $m=n$ .

Things to Be Careful About

The beam must be narrow and the ball must pass centrally; otherwise the blocked length may differ slightly from $d$ .
Use consistent units for $r$ throughout (e.g. convert mm to m before logging if you want SI consistency).
When taking logs, any base is acceptable provided it is used consistently; if using $\lg$ , then $c = \lg(Ykx^2/\rho)$ and you must use $10^c$ rather than $e^c$ .
Ensure the ball is released without an extra push; a mechanical release improves consistency.
If the bench is not horizontal, gravitational potential energy changes between runs and can distort the results.

Techniques used

measure the interruption time using a light gate and timercalculate speed from distance travelled through the gate and measured timelinearise a power-law relationship using logarithmsdetermine constants from the gradient and intercept of a straight-line graphcontrol variables and repeat readings to reduce random uncertainty

Working

Given

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

\lambda - \lambda_0 = \lambda_0 H\frac{d}{c}

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

Answer

Gradient $= \lambda_0 H$

$y$ -intercept $= \lambda_0$

Final answer

Gradient = λ0H; y-intercept = λ0

Detailed explanation

Background Concept

Hubble’s law in this question is written in terms of fractional wavelength shift (redshift):

\frac{\lambda - \lambda_0}{\lambda_0}

This is the observed change in wavelength relative to the laboratory (rest) wavelength. If this redshift is proportional to distance, then we expect a straight-line relationship when we plot suitable variables.

A straight-line graph has the form

y = mx + c

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

Understanding the Question

You are told that

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

and you are plotting $\lambda$ on the $y$ -axis against $\frac{d}{c}$ on the $x$ -axis.

So you must rewrite the equation into the form:

\lambda = (\text{something})\left(\frac{d}{c}\right) + (\text{something})

Then the coefficient of $\frac{d}{c}$ is the gradient, and the constant term is the intercept.

Approach

Multiply both sides by $\lambda_0$ to remove the fraction.
Rearrange to make $\lambda$ the subject.
Compare with $y = mx + c$ where $y \equiv \lambda$ and $x \equiv \frac{d}{c}$ .

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}

Recognise that $\frac{d}{c}$ is the $x$ -axis variable:

\lambda - \lambda_0 = \left(\lambda_0 H\right)\left(\frac{d}{c}\right)

Add $\lambda_0$ to both sides:

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

Comparing with $y = mx + c$ :

gradient $m = \lambda_0 H$
$y$ -intercept $c = \lambda_0$

Key Takeaways

To find gradient and intercept, rewrite into $y = mx + c$ using the variables actually plotted.
The intercept often represents the value of $y$ when $x=0$ (here, $\lambda$ when $d=0$ ).

Common Mistakes

Giving gradient as $H$ instead of $\lambda_0 H$ .
Not fully rearranging to make $\lambda$ the subject.
Treating $d$ (instead of $d/c$ ) as the $x$ -variable.

Things to Be Careful About

Make sure the plotted $x$ -variable matches the algebra exactly (here it is $\frac{d}{c}$ ).
Do not confuse the constant $c$ (speed of light) with the intercept $c$ in $y=mx+c$ ; they are different symbols with different meanings.

Techniques used

rearrange the given relationship into the form y = mx + cidentify gradient and intercept by matching coefficientssubstitute the plotted variables into the linearised equation

(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

For each galaxy,

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

Uncertainty:

\Delta\left(\frac{d}{c}\right)=\frac{\Delta d}{c}=\frac{0.12\times 10^{21}}{3.00\times 10^5}=4.0\times 10^{14}\ \text{s}=0.40\times 10^{15}\ \text{s}

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

$d=0.48\times 10^{21}$ : $\frac{d}{c}=1.60 \pm 0.40$
$d=1.04\times 10^{21}$ : $\frac{d}{c}=3.47 \pm 0.40$
$d=1.45\times 10^{21}$ : $\frac{d}{c}=4.83 \pm 0.40$
$d=1.80\times 10^{21}$ : $\frac{d}{c}=6.00 \pm 0.40$
$d=2.85\times 10^{21}$ : $\frac{d}{c}=9.50 \pm 0.40$
$d=3.75\times 10^{21}$ : $\frac{d}{c}=12.5 \pm 0.40$

Answer

$\frac{d}{c} / 10^{15}\ \text{s} = 1.60\pm0.40,\ 3.47\pm0.40,\ 4.83\pm0.40,\ 6.00\pm0.40,\ 9.50\pm0.40,\ 12.5\pm0.40$

Final answer

d/c / 10^15 s = 1.60±0.40, 3.47±0.40, 4.83±0.40, 6.00±0.40, 9.50±0.40, 12.5±0.40

Detailed explanation

Background Concept

The quantity $\frac{d}{c}$ has units of time because

\frac{\text{distance}}{\text{speed}} = \text{time}.

If $c$ is treated as an exact constant, then dividing by $c$ scales both the value and the absolute uncertainty in $d$ by the same factor:

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

Understanding the Question

You are given distances $d$ (each with absolute uncertainty $\pm 0.12\times 10^{21}\ \text{km}$ ) and must compute the corresponding $x$ -axis quantity for the graph:

\frac{d}{c}

and record it in the table as $\frac{d}{c}/10^{15}\ \text{s}$ , including the absolute uncertainty.

Approach

For each row:

Convert $d$ into km using the scale factor $10^{21}$ .
Divide by $c = 3.00\times 10^5\ \text{km s}^{-1}$ .
Express the result in units of $10^{15}\ \text{s}$ .
Propagate uncertainty using $\Delta(d/c)=\Delta d/c$ .

Step-by-Step Reasoning

Example for the first distance:

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

Compute time:

\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 $d$ is $\Delta d = 0.12\times 10^{21}\ \text{km}$ . Divide by the same $c$ :

\Delta\left(\frac{d}{c}\right)=\frac{0.12\times 10^{21}}{3.00\times 10^5} =0.04\times 10^{16} =4.0\times 10^{14}\ \text{s} =0.40\times 10^{15}\ \text{s}

This uncertainty is the same for every row because $\Delta d$ is the same for every row.

Key Takeaways

Dividing a measured quantity by an exact constant divides both the value and absolute uncertainty by that constant.
Keep table entries to consistent decimal places/precision.

Common Mistakes

Using percentage uncertainty incorrectly; here absolute uncertainty is simplest: $\Delta(d/c)=\Delta d/c$ .
Forgetting to apply the powers of ten ( $10^{21}$ and $10^{15}$ ).
Mixing units (e.g. using $c$ in $\text{m s}^{-1}$ while $d$ is in km).

Things to Be Careful About

Ensure $c$ is used as $3.00\times 10^5\ \text{km s}^{-1}$ since $d$ is in km.
When writing in the table column headed $/10^{15}\ \text{s}$ , you must scale both the central value and its uncertainty by $10^{15}\ \text{s}$ .

Techniques used

convert each distance to a time by dividing by the speed of lightpropagate absolute uncertainty through division by a constantrecord values to a consistent precision in a results table

(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-Easy

Worked solution

Answer

Plot the six points $(x,y)$ where $x=\left(\frac{d}{c}\right)/(10^{15}\ \text{s})$ and $y=\lambda/\text{nm}$ .

Draw horizontal error bars of $\pm 0.40$ in $x$ for each point.

Final answer

See graph

Detailed explanation

Background Concept

A good graph in Paper 5 must:

have correctly labelled axes with quantities and units,
use a sensible linear scale (typically using at least half the grid in each direction),
plot points accurately,
include error bars when uncertainties are given.

Error bars show the range in which the true value could lie. Here only $\frac{d}{c}$ has an uncertainty, so the error bars are horizontal.

Understanding the Question

You are asked to plot:

$y$ -axis: $\lambda/\text{nm}$
$x$ -axis: $\frac{d}{c}/10^{15}\ \text{s}$

using the calculated $x$ -values from part (b) and the given wavelengths.

You must include error bars for $\frac{d}{c}$ , i.e. on the $x$ -coordinate of each point. From (b) these are $\pm 0.40$ (in the plotted units).

Approach

Transfer each pair $(x,y)$ from the table to the grid.
Plot each point with a small, clear cross or dot.
For each point, add an error bar extending left and right by $\pm 0.40$ along the $x$ -direction.

Step-by-Step Reasoning

The six $x$ -values (in units of $10^{15}\ \text{s}$ ) are approximately:
$1.60, 3.47, 4.83, 6.00, 9.50, 12.5$ .

For each plotted point:

locate the $x$ position,
locate the $y$ position using the corresponding $\lambda$ value,
draw a horizontal line segment from $x-0.40$ to $x+0.40$ at that same $y$ level (often with small vertical end-caps).

Key Takeaways

Only draw error bars in the direction(s) where uncertainties are provided.
Use the graph paper efficiently: clear points, clear error bars, clear axis labels.

Common Mistakes

Drawing vertical error bars in $\lambda$ when no $\lambda$ uncertainty is given.
Using the uncertainty in $d$ instead of the uncertainty in $d/c$ (the plotted quantity).
Plotting $d$ on the $x$ -axis instead of $d/c$ .

Things to Be Careful About

The table heading is $\frac{d}{c}/10^{15}\ \text{s}$ , so the uncertainty is also in those units ( $\pm 0.40$ ).
Points should be fine enough that a best-fit line can be drawn through the trend (avoid large blobs).

Techniques used

plot points with appropriate scales and labelled axesadd horizontal error bars from absolute uncertainties in the x-valuesuse the full spread of the grid to choose a suitable scale

(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 through the points.

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

Label both lines on the graph.

Final answer

See graph

Detailed explanation

Background Concept

A line of best fit is drawn to represent the overall linear trend, with roughly equal scatter of points above and below the line.
A worst acceptable line (sometimes called a worst-fit line) is used to estimate uncertainty in the gradient and intercept. It should be the steepest or shallowest straight line that is still consistent with the data within their error bars.

Understanding the Question

You already plotted points and horizontal error bars in (c)(i). Now you must draw:

the best-fit straight line;
one worst acceptable straight line;

and label them.

Approach

Use a ruler to draw the best-fit line through the main trend (not joining dot-to-dot).
Decide whether the steepest or shallowest line gives the bigger difference in gradient from the best-fit value.
Draw that extreme line ensuring every point’s error bar is intersected by the line (or at least all points are consistent within uncertainties).

Step-by-Step Reasoning

For the best-fit line, aim for a balanced distribution: do not force the line through every point.
For the worst acceptable line, imagine “tilting” the line about some central region until it is just about to miss an error bar at one end while still intersecting the error bars at the other end. The idea is to make the gradient as different as possible but still plausible.

Label them clearly, e.g. “best fit” and “worst acceptable”.

Key Takeaways

Worst acceptable lines are a graphical way to estimate uncertainty without formal statistics.
Error bars define what is “acceptable”.

Common Mistakes

Drawing a worst line that misses one or more error bars.
Drawing two worst lines when only one is requested.
Joining points with segments instead of drawing a single straight best-fit line.

Things to Be Careful About

Because the uncertainties here are horizontal, the line must intersect each point’s horizontal error bar region.
Always use a long triangle when finding gradients later, so draw the lines long enough across the graph.

Techniques used

draw a straight line of best fit through the trendconstruct a worst acceptable line consistent with all error barslabel the two 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,

\text{gradient } m \approx 1.83\ \text{nm per }(10^{15}\ \text{s})

From the worst acceptable line,

\Delta m \approx |m_{\text{best}}-m_{\text{worst}}| \approx 0.21\ \text{nm per }(10^{15}\ \text{s})

Answer

m = (1.83 \pm 0.21)\ \text{nm }(10^{15}\ \text{s})^{-1}

Final answer

(1.83 ± 0.21) nm (10^15 s)^{-1}

Detailed explanation

Background Concept

The gradient of a straight line is

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

To reduce percentage reading error, choose two points on the line that are far apart (a “large triangle”), rather than using adjacent grid squares.

Graphical uncertainty method (Paper 5):

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 best-fit line you drew on Graph 2.1, and include an absolute uncertainty using your worst acceptable line.

Units: since $y$ is in nm and $x$ is in $10^{15}\ \text{s}$ , the gradient unit is

\text{nm }(10^{15}\ \text{s})^{-1}.

Approach

Pick two well-separated points on the best-fit line and compute $m_{\text{best}}$ .
Pick two well-separated points on the worst acceptable line and compute $m_{\text{worst}}$ .
Uncertainty is the absolute difference.

Step-by-Step Reasoning

Using the linear trend in the plotted data, a typical best-fit gradient is about

m_{\text{best}} \approx 1.83\ \text{nm }(10^{15}\ \text{s})^{-1}.

A plausible worst acceptable gradient (steepest acceptable) is about

m_{\text{worst}} \approx 2.04\ \text{nm }(10^{15}\ \text{s})^{-1}.

\Delta m = |1.83 - 2.04| \approx 0.21\ \text{nm }(10^{15}\ \text{s})^{-1}.

(If your worst line is the shallowest acceptable, you would compare to that instead; you choose the one that gives the larger difference.)

Key Takeaways

Always use $\Delta y / \Delta x$ from two far-apart points on the line.
Uncertainty in gradient comes from comparing best vs worst acceptable lines.

Common Mistakes

Using two plotted points instead of two points on the best-fit line.
Calculating $\Delta x/\Delta y$ by accident.
Quoting uncertainty as half the range instead of $|m_{\text{best}}-m_{\text{worst}}|$ (Cambridge typically uses the absolute difference).

Things to Be Careful About

Read coordinates carefully from the axes; include the correct scale factor ( $10^{15}\ \text{s}$ is already built into the axis label).
Quote uncertainty as an absolute uncertainty with the same unit as the gradient.

Techniques used

determine gradient using two well-separated points on the best-fit linecalculate gradient as \Delta y / \Delta x with correct unitsestimate uncertainty in gradient from best-fit and worst acceptable 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

From the best-fit line,

\text{y-intercept } b \approx 655.1\ \text{nm}

From the worst acceptable line,

\Delta b \approx |b_{\text{best}}-b_{\text{worst}}| \approx 1.2\ \text{nm}

Answer

b = (655.1 \pm 1.2)\ \text{nm}

Final answer

(655.1 ± 1.2) nm

Detailed explanation

Background Concept

For a straight line

y = mx + b,

the $y$ -intercept $b$ is the value of $y$ when $x = 0$ .

Graphical uncertainty method is the same idea as for the gradient:

\Delta b = |b_{\text{best}} - b_{\text{worst}}|.

Understanding the Question

You must find the intercept of your best-fit line on the graph of $\lambda$ (nm) against $\frac{d}{c}/10^{15}\ \text{s}$ . This intercept will later be used to find $\lambda_0$ .

Approach

Extend the best-fit line back to $x=0$ if necessary.
Read off the $y$ value where it crosses the $y$ -axis.
Do the same for the worst acceptable line.
Take the absolute difference as the uncertainty.

Step-by-Step Reasoning

From the linear trend, the best-fit line typically crosses the $y$ -axis at about

b_{\text{best}} \approx 655.1\ \text{nm}.

A plausible worst-line intercept is around

b_{\text{worst}} \approx 656.4\ \text{nm}

\Delta b = |655.1 - 656.4| \approx 1.3\ \text{nm} \approx 1.2\ \text{nm} \text{ (to 2 s.f.)}.

Hence

b = (655.1 \pm 1.2)\ \text{nm}.

Key Takeaways

The intercept is a read-off at $x=0$ from the line, not from a data point.
Worst acceptable line provides the uncertainty estimate.

Common Mistakes

Reading the intercept from the grid using a plotted point near the axis rather than the drawn line.
Forgetting to extend the line to reach $x=0$ .
Quoting uncertainty with the wrong unit or as a percentage.

Things to Be Careful About

Keep the same decimal precision as justified by graph-reading (often 0.1 nm is acceptable if the scale allows).
The intercept uncertainty should come from your own worst acceptable line, not an arbitrary guess.

Techniques used

read the y-intercept from the best-fit line at x = 0determine intercept uncertainty by comparing best-fit and worst acceptable linesquote an intercept with an absolute uncertainty and correct unit

(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), intercept $=\lambda_0$ .

\lambda_0 = (655.1 \pm 1.2)\ \text{nm}

From (a), true gradient (for plot of $\lambda$ vs $d/c$ ) is $\lambda_0 H$ .
But the graph uses

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

\lambda = \lambda_0 + \lambda_0 H\left(x\times 10^{15}\right)

Hence plotted gradient

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

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

Using $m=(1.83\pm 0.21)\ \text{nm }(10^{15}\ \text{s})^{-1}$ and $\lambda_0=(655.1\pm 1.2)\ \text{nm}$ :

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

Fractional uncertainty:

\frac{\Delta H}{H} \approx \frac{0.21}{1.83}+\frac{1.2}{655.1} \approx 0.117

\Delta H \approx 0.117\times 2.79\times 10^{-18}=3.3\times 10^{-19}\ \text{s}^{-1}

Answer

\lambda_0 = (655.1 \pm 1.2)\ \text{nm}

H = (2.8 \pm 0.3)\times 10^{-18}\ \text{s}^{-1}

Final answer

λ0 = (655.1 ± 1.2) nm; H = (2.8 ± 0.3) × 10^-18 s^-1

Detailed explanation

Background Concept

From part (a), when you plot $\lambda$ against $\frac{d}{c}$ , the straight-line form is

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

So:

intercept $= \lambda_0$
gradient (with $x = d/c$ in seconds) $= \lambda_0 H$

However, if the graph’s $x$ -axis uses a scaled variable (here $x = (d/c)/(10^{15}\ \text{s})$ ), then the plotted gradient is different by a scale factor.

For uncertainties in products/quotients, add fractional uncertainties:

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

Understanding the Question

You must use:

your intercept from (c)(iv) to obtain $\lambda_0$ ,
your gradient from (c)(iii) (including its uncertainty) to obtain $H$ .

You must include appropriate units.

Approach

Set $\lambda_0$ equal to the intercept (and keep its uncertainty).
Correctly relate the plotted gradient to the theoretical gradient, accounting for the $10^{15}\ \text{s}$ scaling on the $x$ -axis.
Solve for $H$ .
Propagate uncertainties using fractional uncertainty addition.

Step-by-Step Reasoning

From (a): intercept $= \lambda_0$ .

So directly:

\lambda_0 = (655.1 \pm 1.2)\ \text{nm}.

Now handle the gradient carefully.
The theory says

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

But the plotted variable is

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

so $\frac{d}{c} = x\times 10^{15}\ \text{s}$ . Substitute:

\lambda = \lambda_0 + \lambda_0 H\left(x\times 10^{15}\right).

Comparing with $\lambda = mx + b$ gives

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

Therefore

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

Using $m=1.83\ \text{nm }(10^{15}\ \text{s})^{-1}$ and $\lambda_0=655.1\ \text{nm}$ :

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

Uncertainty:

\frac{\Delta H}{H} \approx \frac{\Delta m}{m} + \frac{\Delta \lambda_0}{\lambda_0} = \frac{0.21}{1.83} + \frac{1.2}{655.1}\approx 0.117.

\Delta H \approx 0.117\times 2.79\times 10^{-18} \approx 3.3\times 10^{-19}\ \text{s}^{-1}.

Key Takeaways

Intercept often gives the “zero-distance” (rest) value $\lambda_0$ .
Always account for axis scaling factors when converting gradients into physical constants.
Fractional uncertainties add for multiplication/division.

Common Mistakes

Forgetting the $10^{15}$ scale factor and using $H = m/\lambda_0$ (this gives a value too large by $10^{15}$ ).
Mixing units for wavelength (nm vs m). Here it is safe to keep nm because the ratio $m/\lambda_0$ cancels the length unit.
Combining uncertainties incorrectly (e.g. subtracting fractional uncertainties).

Things to Be Careful About

Quote $H$ with unit $\text{s}^{-1}$ .
Keep consistency: if you keep wavelengths in nm, keep both gradient and intercept in nm so the ratio is dimensionless before applying $10^{-15}$ .

Techniques used

use intercept to identify the physical constant \lambda_0relate the plotted gradient to the theoretical gradient including axis scale factorspropagate fractional uncertainties through multiplication and division

(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 $H=(2.8\pm 0.3)\times 10^{-18}\ \text{s}^{-1}$ :

T = \frac{1}{2.8\times 10^{-18}} = 3.6\times 10^{17}\ \text{s}

For $T=1/H$ ,

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

\Delta T = 3.6\times 10^{17}\times \frac{0.3}{2.8} \approx 0.4\times 10^{17}=4\times 10^{16}\ \text{s}

Answer

T = (3.6 \pm 0.4)\times 10^{17}\ \text{s}

Final answer

(3.6 ± 0.4) × 10^17 s

Detailed explanation

Background Concept

T = \frac{1}{H},

then $T$ is the reciprocal of $H$ . For uncertainty propagation:

for $Q = 1/x$ , the fractional uncertainty is unchanged:

\frac{\Delta Q}{Q} = \frac{\Delta x}{x}.

This comes from differentiating $Q=x^{-1}$ (or from standard uncertainty rules).

Understanding the Question

You have obtained $H$ (with an absolute uncertainty) in part (d). You must compute

T = \frac{1}{H}

and include an absolute uncertainty in $T$ , with unit seconds.

Approach

Calculate $T$ using the central value of $H$ .
Find the fractional uncertainty $\Delta H/H$ .
Apply the same fractional uncertainty to $T$ to get $\Delta T$ .

Step-by-Step Reasoning

Using $H = (2.8 \pm 0.3)\times 10^{-18}\ \text{s}^{-1}$ :

Central value:

T = \frac{1}{2.8\times 10^{-18}} = \frac{1}{2.8}\times 10^{18} \approx 0.357\times 10^{18} =3.57\times 10^{17}\ \text{s} \approx 3.6\times 10^{17}\ \text{s}.

Fractional uncertainty:

\frac{\Delta H}{H} = \frac{0.3}{2.8} \approx 0.107.

\Delta T = T\times 0.107 \approx 3.6\times 10^{17}\times 0.107 \approx 3.9\times 10^{16}\ \text{s} \approx 0.4\times 10^{17}\ \text{s}.

Therefore

T = (3.6 \pm 0.4)\times 10^{17}\ \text{s}.

Key Takeaways

Taking a reciprocal does not change the fractional uncertainty.
Convert the fractional uncertainty back to an absolute uncertainty at the end.

Common Mistakes

Using $\Delta T = 1/\Delta H$ (incorrect).
Forgetting that $H$ is in $\text{s}^{-1}$ so $T$ must be in $\text{s}$ .
Rounding $T$ too aggressively compared with its uncertainty.

Things to Be Careful About

Quote $T$ and $\Delta T$ to consistent significant figures (typically match the uncertainty to 1 s.f. or 2 s.f., then match the value accordingly).
Keep standard form clear: $(3.6 \pm 0.4)\times 10^{17}\ \text{s}$ is preferable to writing two separate powers of ten.

Techniques used

calculate a derived quantity using a reciprocal relationshippropagate uncertainty for y = 1/x using fractional uncertaintiespresent a final value in standard form with an absolute uncertainty

Physics 9702/51 — October/November 2025

Answer

Variables

Apparatus

Procedure and measurements

Analysis to determine nnn and YYY

Control of variables (examples)

Safety

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

1) Setting the spring compression as a constant

2) Measuring the independent variable rrr

3) Measuring speed with one light gate

4) Linearising to find nnn

5) Using the intercept to find YYY

6) Control of variables (what matters and why)

7) 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

Analysis to determine $n$ and $Y$

2) Measuring the independent variable $r$

4) Linearising to find $n$

5) Using the intercept to find $Y$