In this experiment, you will investigate the deformation of paper cylinders. (a) You have been provided with two pieces of paper. The width of a piece of paper is the length $w$ of the shorter side, as shown in Fig. 2.1. • Select the smaller piece of paper. • Measure and record $w$. $w$ = \_\_\_\_\_\_ $\text{cm}$ • Roll the paper into a cylinder and use two paper clips to hold the paper in place, as shown in Fig. 2.2. • The diameter of the cylinder is $d$, as shown in Fig.…

(a) w = 14.8 cm, d = 7.0 cm (example) · (b)(i) y = 32.0 cm (example) · (b)(ii) 0.31 % · (c)(i) p = 28.4 cm (example) · (c)(ii) 3.6 cm · (d) w = 21.0 cm, d = 7.0 cm, y = 32.0 cm, p = 27.0 cm, (y − p) = 5.0 cm (example) · (e)(i) k ≈ 4.1 and 4.2 (example) · (e)(ii) k to 2 s.f. · (f) Supports relationship (values consistent within 10%) · (g)(i) Four valid uncertainties/limitations listed · (g)(ii) Four valid improvements listed

Physics 9702/31 — October/November 2025

Cambridge AS Level · Advanced Practical Skills · worked solutions for every part, with the mark scheme

questions

marks

120

minutes

Topics Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation

Q1MediumManipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and Evaluation

In this experiment, you will investigate oscillations.

You have been provided with a double spring and two single springs.

(a)

(i)

• Slide the double spring and the two single springs onto the longer wooden rod.
• Set up the apparatus as shown in Fig. 1.1.

Fig. 1.1

• Use the bosses to fix the wooden rod approximately $55\text{ cm}$ above the bench and parallel to the bench.
• Hang a total mass of $270\text{ g}$ from the double spring.
• The mass hanging from the double spring is $m$ .

Record $m$ .

$m$ = ______

1M Similar

(ii)

• Gently pull the mass down through a short distance. When released, the mass will oscillate.
• Take measurements to determine the period $T_1$ of the oscillations.

$T_1$ = ______

• Remove the mass from the double spring.

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Time $N=10$ oscillations and divide by $10$ .

Example:

t_{10} = 12.5\ \text{s}

T_1 = \frac{t_{10}}{10} = \frac{12.5}{10} = 1.25\ \text{s}

Answer

$T_1 = 1.25\ \text{s}$

Final answer

T1 ≈ 1.25 s

Detailed explanation

Background Concept

The period $T$ of an oscillation is the time for one complete cycle. For a mass oscillating vertically on a spring, each cycle is short enough that reaction time becomes a significant fraction of the time if you try to time just one oscillation.

A standard practical method is therefore:

time $N$ oscillations (with $N \ge 5$ , typically $10$ or more), measuring a total time $t_N$
calculate the period using

T = \frac{t_N}{N}

Repeating the measurement and averaging reduces random error.

Understanding the Question

You are asked to “take measurements to determine the period $T_1$ ”. This means you must decide a sensible timing method and then produce a value for $T_1$ in seconds.

Approach

Displace the mass slightly and release so the motion is steady and approximately simple harmonic.
Choose a fixed reference point (e.g. equilibrium position) and always count oscillations at the same point in the cycle.
Measure the time for $N$ complete oscillations.
Divide by $N$ to obtain $T_1$ .
Repeat and average.

Step-by-Step Reasoning

Start oscillations with a small amplitude (large amplitudes can make the motion less regular).
Decide $N=10$ oscillations (large enough to reduce percentage timing uncertainty).
Use a stopwatch to measure $t_{10}$ , the time for 10 complete oscillations.
Calculate:

T_1 = \frac{t_{10}}{10}

Repeat timing (e.g. three values of $t_{10}$ ) and use the mean $\bar t_{10}$ to get the best estimate:

T_1 = \frac{\bar t_{10}}{10}

(Any sensible repeat scheme would earn credit in the practical.)

Key Takeaways

Always time multiple oscillations: $T=t_N/N$ .
Repeat and average to improve reliability.
Use a consistent reference point to count cycles.

Common Mistakes

Timing one oscillation only (too much reaction-time error).
Counting half-oscillations incorrectly.
Starting/stopping the stopwatch at different points in the cycle each time.

Things to Be Careful About

Make sure you time complete oscillations (same position and same direction of motion).
Keep amplitude small and avoid sideways swinging.
Quote $T_1$ to a sensible precision consistent with stopwatch resolution and repeats.

Techniques used

time multiple oscillations and divide to obtain the periodrepeat timing measurements and calculate a mean valueuse a consistent reference point to reduce reaction-time error

(b)

• Using the shorter wooden rod, hang mass $m$ as shown in Fig. 1.2.

Fig. 1.2

• Gently place the hook of the mass hanger near the centre of the shorter rod.
• Carefully adjust the position of the mass hanger until the shorter rod is approximately parallel to the bench, as shown in Fig. 1.2.
• Gently pull the mass down through a short distance. When released, the mass will oscillate.
• Take measurements to determine the period $T_2$ of the oscillations.

$T_2$ = ______

• Remove the mass.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Time $N=10$ oscillations and divide by $10$ .

Example:

t_{10} = 10.7\ \text{s}

T_2 = \frac{t_{10}}{10} = \frac{10.7}{10} = 1.07\ \text{s}

Answer

$T_2 = 1.07\ \text{s}$

Final answer

T2 ≈ 1.07 s

Detailed explanation

Background Concept

For oscillations you determine the period the same way regardless of the exact spring arrangement:

T = \frac{t_N}{N}

The additional practical skill here is setting up the system so it starts from a well-defined equilibrium (rod approximately horizontal), which makes the oscillations repeatable.

Understanding the Question

You must:

attach mass $m$ to the shorter rod (suspended by the two single springs)
slide the mass hanger hook until the rod is approximately parallel to the bench (equilibrium)
then measure the oscillation period $T_2$ .

Approach

Adjust the hanger position so the rod is horizontal (net turning effect about the support is zero).
Displace the rod/mass slightly and release.
Measure time for $N$ oscillations and divide by $N$ .
Repeat and average.

Step-by-Step Reasoning

With the mass not at the centre, the rod would tilt because the weight produces a turning effect; sliding the hanger until the rod is horizontal ensures a steady equilibrium position.
Start oscillations with a small amplitude.
Measure $t_{10}$ for 10 oscillations, then:

T_2 = \frac{t_{10}}{10}

Repeat timing to reduce random error.

Key Takeaways

Achieving a good equilibrium position improves repeatability.
Timing many oscillations reduces percentage uncertainty.

Common Mistakes

Not making the rod horizontal before timing (motion becomes irregular).
Allowing sideways swinging/twisting as well as vertical oscillations.
Counting oscillations inconsistently.

Things to Be Careful About

Keep oscillations small.
Ensure the support rod and springs are not rubbing against stands/bosses.
Start timing after the motion becomes steady (not during an initial wobble).

Techniques used

adjust the position of the load to achieve horizontal equilibriumtime multiple oscillations and divide to obtain the periodrepeat measurements and calculate a mean period

(c)

Vary $m$ . For each value of $m$ , determine $T_1$ and $T_2$ . Repeat until you have five sets of values of $m$ , $T_1$ and $T_2$ . Do not use values of $m$ less than $200\text{ g}$ .

Record your results in a table. Include values of $\sqrt{T_1}$ and $\sqrt{T_2}$ in your table.

8M Similar

DifficultyMedium

Worked solution

Answer

Record at least five sets with $m \ge 200\ \text{g}$ in one table, including calculated $\sqrt{T_1}$ and $\sqrt{T_2}$ (units $\text{s}^{1/2}$ ).

Example of a correctly set out table:

$m / \text{kg}$	$T_1 / \text{s}$	$T_2 / \text{s}$	$\sqrt{T_1} / \text{s}^{1/2}$	$\sqrt{T_2} / \text{s}^{1/2}$
0.200	1.00	0.903	1.000	0.950
0.240	1.10	0.988	1.049	0.994
0.280	1.20	1.07	1.095	1.036
0.320	1.30	1.16	1.140	1.076
0.360	1.40	1.24	1.183	1.115

Final answer

Single clear table with m, T1, T2, √T1, √T2 (≥5 sets; m ≥ 200 g).

Detailed explanation

Background Concept

In Paper 3, marks are awarded for the quality of your data and how you present it.

A good results table should:

be a single table containing all raw and calculated data
have headings that include the quantity and unit (e.g. $T_1/\text{s}$ )
use consistent decimal places within each column
include a suitable range and at least five values of the independent variable

Here, you must also calculate $\sqrt{T_1}$ and $\sqrt{T_2}$ . Since $T$ is in seconds, $\sqrt{T}$ has unit $\text{s}^{1/2}$ .

Understanding the Question

You must vary the mass $m$ (but not below $200\ \text{g}$ ), and for each mass measure both periods $T_1$ and $T_2$ . You then need five sets of $(m, T_1, T_2)$ and you must include $\sqrt{T_1}$ and $\sqrt{T_2}$ in your table.

Approach

Choose five (or more) masses with a sensible spacing, all $\ge 200\ \text{g}$ .
For each $m$ $m$ :
- measure $T_1$ by timing $N$ oscillations and dividing by $N$ (repeat and average)
- measure $T_2$ similarly
Calculate $\sqrt{T_1}$ and $\sqrt{T_2}$ for each row.
Present all values in one well-formatted table.

Step-by-Step Reasoning

Select masses, e.g. $0.200, 0.240, 0.280, 0.320, 0.360\ \text{kg}$ (any sensible set is fine).
For each mass, you could do:
- time $10$ oscillations twice or three times
- average $t_{10}$
- compute $T = \bar t_{10}/10$
After obtaining $T_1$ and $T_2$ , calculate:

\sqrt{T_1} \quad \text{and} \quad \sqrt{T_2}

For example, if $T_1 = 1.20\ \text{s}$ then:

\sqrt{T_1} = \sqrt{1.20} = 1.095\ \text{s}^{1/2}

Record the calculated values to a consistent precision (often 3 s.f. is reasonable for square roots).

Key Takeaways

Collect enough data: at least five sets and a good range.
Period measurements should be averaged and based on multiple oscillations.
A good table has units in headings and consistent formatting.

Common Mistakes

Using masses below $200\ \text{g}$ .
Splitting results into multiple tables.
Missing units in headings.
Inconsistent decimal places (e.g. mixing $1.2$ and $1.23$ in the same column with no reason).
Calculating $\sqrt{T}$ but forgetting the unit $\text{s}^{1/2}$ .

Things to Be Careful About

When you take square roots, your significant figures should reflect the precision of $T$ .
Keep the oscillations small and consistent between trials.
Check you are not mixing seconds and milliseconds in calculations.

Techniques used

vary the independent variable over a suitable range with at least five setsmeasure periods by timing multiple oscillations and averagingcalculate derived quantities and record them with consistent significant figuresconstruct a single results table with quantity and unit in each heading

(d)

(i)

Plot a graph of $\sqrt{T_2}$ on the $y$ -axis against $\sqrt{T_1}$ on the $x$ -axis.

3M Similar

DifficultyMedium

Worked solution

Answer

Plot $\sqrt{T_2}$ on the $y$ -axis against $\sqrt{T_1}$ on the $x$ -axis.

Axis labels:

$x$ -axis: $\sqrt{T_1} / \text{s}^{1/2}$
$y$ -axis: $\sqrt{T_2} / \text{s}^{1/2}$

Use a suitable linear scale (at least half the grid) and plot all points with small crosses.

Final answer

Graph of √T2 (y) against √T1 (x) with correct labels/units and accurately plotted points.

Detailed explanation

Background Concept

Graph marks in Paper 3 come from clear communication:

correct choice of variables on each axis
correct units in axis labels
a sensible scale (not cramped and not awkward like 3 squares = 1 unit)
accurate plotting (small, neat crosses or dots)

Here you are plotting transformed variables: $\sqrt{T_1}$ and $\sqrt{T_2}$ . Both have units $\text{s}^{1/2}$ .

Understanding the Question

You must use your table values of $\sqrt{T_1}$ and $\sqrt{T_2}$ and plot a graph with:

horizontal axis: $\sqrt{T_1}$
vertical axis: $\sqrt{T_2}$

Approach

Decide the range of your $\sqrt{T_1}$ and $\sqrt{T_2}$ values.
Choose scales so the plotted data spans a large area of the grid.
Label axes as “ $\sqrt{T_1} / \text{s}^{1/2}$ ” and “ $\sqrt{T_2} / \text{s}^{1/2}$ ”.
Plot each pair $(\sqrt{T_1},\sqrt{T_2})$ as a small cross.

Step-by-Step Reasoning

Suppose your $\sqrt{T_1}$ values range from about $1.00$ to $1.18\ \text{s}^{1/2}$ , then set an $x$ -axis range such as $0.95$ to $1.20\ \text{s}^{1/2}$ .
Similarly choose a $y$ -axis range to fit all $\sqrt{T_2}$ values.
Mark each axis with evenly spaced numbers.
Plot points carefully by reading across from the $x$ value then up to the $y$ value.

Key Takeaways

Labels must include both the quantity and the unit.
Scale choice matters: large spread improves gradient accuracy.

Common Mistakes

Swapping axes (plotting $\sqrt{T_1}$ on $y$ ).
Missing units or writing just “ $\sqrt{T_1}$ ” without $\text{s}^{1/2}$ .
Using a scale that uses only a small part of the grid.

Things to Be Careful About

Both axes are square-root time, so both units are $\text{s}^{1/2}$ .
Plotting accuracy: avoid large blobs; use small crosses so the best-fit line can be judged.

Techniques used

choose suitable axis scales to use most of the graph paperlabel axes with the correct variables and unitsplot points accurately from a results table

(ii)

Draw the straight line of best fit.

1M Similar

(iii)

Determine the gradient and $y$ -intercept of this line.

gradient = ______
$y$ -intercept = ______

2M Similar

DifficultyMedium

Worked solution

Working

Choose two well-separated points on the best-fit line.

Example points on line: $(x_1,y_1)=(1.00,0.95)$ and $(x_2,y_2)=(1.18,1.11)$ (units $\text{s}^{1/2}$ ).

\text{gradient} = \frac{\Delta y}{\Delta x} = \frac{1.11-0.95}{1.18-1.00} = \frac{0.16}{0.18} = 0.89

$y$ -intercept from the line at $x=0$ :

y\text{-intercept} = 0.05\ \text{s}^{1/2}

Answer

gradient $= 0.89$

$y$ -intercept $= 0.05\ \text{s}^{1/2}$

Final answer

gradient ≈ 0.89; y-intercept ≈ 0.05 s^{1/2}

Detailed explanation

Background Concept

For a straight-line graph, the gradient and intercept summarise the relationship.

If you plot $y$ against $x$ and obtain a straight line, then:

y = mx + c

where:

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

From a plotted best-fit line:

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

Using a large triangle (widely separated points) reduces percentage reading error.

Understanding the Question

You have a graph of $\sqrt{T_2}$ (vertical) against $\sqrt{T_1}$ (horizontal). You must determine:

the gradient of your best-fit line
the $y$ -intercept of your best-fit line

Units:

both $x$ and $y$ are in $\text{s}^{1/2}$
so the gradient is dimensionless (because units cancel)
the $y$ -intercept has unit $\text{s}^{1/2}$

Approach

Draw a large triangle on the best-fit line and read two points on the line (not necessarily data points).
Compute gradient using $\Delta y/\Delta x$ .
Extend the line (if needed) to read the intercept at $x=0$ .

Step-by-Step Reasoning

diagram-diii-gradient

Pick two points far apart on the line, e.g. where the line crosses convenient grid intersections.
Read coordinates $(x_1,y_1)$ and $(x_2,y_2)$ .
Calculate changes:

\Delta y = y_2-y_1

\Delta x = x_2-x_1

Gradient:

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

Intercept: find where the line crosses the $y$ -axis (at $x=0$ ), and read off $c$ .

Notes that examiners expect:

a triangle large enough to use much of the graph
correct use of $\Delta y/\Delta x$ (not inverted)
intercept read from the best-fit line, not from a single point.

Key Takeaways

Use a large triangle for accurate gradients.
Gradient uses $\Delta y/\Delta x$ .
Intercept is from the line at $x=0$ .

Common Mistakes

Using two adjacent points (small triangle) giving a poor gradient.
Calculating $\Delta x/\Delta y$ by mistake.
Using two plotted points that are not on the best-fit line.
Giving a unit for the gradient here (it is dimensionless).

Things to Be Careful About

Read values to the precision allowed by the graph scale.
If the line does not reach the $y$ -axis on the drawn region, extend it carefully with a ruler before reading the intercept.
Keep consistent units: both axes are $\text{s}^{1/2}$ , so the intercept must be $\text{s}^{1/2}$ .

Techniques used

use a large triangle on the best-fit line to calculate gradientread the y-intercept from the best-fit lineapply gradient formula \u0394y/\u0394x with correct units

(e)

It is suggested that the quantities $\sqrt{T_2}$ and $\sqrt{T_1}$ are related by the equation

\sqrt{T_2} = P \sqrt{T_1} + Q

where $P$ and $Q$ are constants.

Using your answers in (d)(iii), determine the values of $P$ and $Q$ . Give appropriate units.

$P$ = ______
$Q$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Given

\sqrt{T_2} = P\sqrt{T_1} + Q

Comparing with $y=mx+c$ for the graph of $y=\sqrt{T_2}$ against $x=\sqrt{T_1}$ :

P = \text{gradient}

Q = y\text{-intercept}

Units: $P$ is dimensionless and $Q$ has unit $\text{s}^{1/2}$ .

Answer

$P = 0.89$ (no unit)

$Q = 0.05\ \text{s}^{1/2}$

Final answer

P ≈ 0.89 (dimensionless), Q ≈ 0.05 s^{1/2}

Detailed explanation

Background Concept

When experimental data give a straight-line graph, you can identify constants by comparing the line equation with:

y = mx + c

Here:

$m$ (gradient) multiplies $x$
$c$ is the value of $y$ when $x=0$

Unit check:

If $y$ and $x$ have the same units, then the gradient $m$ is dimensionless.
The intercept $c$ has the same units as $y$ .

Understanding the Question

You are told that:

\sqrt{T_2} = P\sqrt{T_1} + Q

and you have already found the gradient and $y$ -intercept from your graph of $\sqrt{T_2}$ (y-axis) vs $\sqrt{T_1}$ (x-axis). You must use those to state $P$ and $Q$ , including units.

Approach

Match each symbol to the straight-line form:

$P$ corresponds to the gradient
$Q$ corresponds to the intercept
Then assign units from the plotted quantities.

Step-by-Step Reasoning

Let:

$y = \sqrt{T_2}$
$x = \sqrt{T_1}$

Then the suggested equation becomes:

y = Px + Q

Comparing with $y = mx + c$ :

P = m

Q = c

Units:

both $\sqrt{T_2}$ and $\sqrt{T_1}$ have unit $\text{s}^{1/2}$ , so $P$ is dimensionless.
$Q$ has unit $\text{s}^{1/2}$ .

Key Takeaways

The constant multiplying the x-variable is the gradient.
The constant added is the intercept.
Use units of the axes to infer units of constants.

Common Mistakes

Giving $P$ the unit $\text{s}^{1/2}$ (it should have no unit here).
Writing $Q$ with unit $\text{s}$ instead of $\text{s}^{1/2}$ .
Swapping $P$ and $Q$ .

Things to Be Careful About

Make sure you are using the gradient/intercept from the correct graph ( $\sqrt{T_2}$ vs $\sqrt{T_1}$ , not $T_2$ vs $T_1$ ).
State units explicitly for $Q$ and explicitly state “no unit” (or omit units) for $P$ .

Techniques used

match a straight-line equation to y = mx + cidentify constants as the gradient and interceptdeduce units from the units of plotted variables

Q2MediumManipulation, Measurement and ObservationAnalysis, Conclusions and EvaluationPresentation of Data and Observations

In this experiment, you will investigate the deformation of paper cylinders.

(a)

You have been provided with two pieces of paper.

The width of a piece of paper is the length $w$ of the shorter side, as shown in Fig. 2.1.

Fig. 2.1

• Select the smaller piece of paper.
• Measure and record $w$ .

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

• Roll the paper into a cylinder and use two paper clips to hold the paper in place, as shown in Fig. 2.2.

Fig. 2.2

• The diameter of the cylinder is $d$ , as shown in Fig. 2.2.

Adjust the paper and paper clips until $d$ is as close as possible to $7.0\text{ cm}$ .

• Measure and record $d$ .

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

2M Similar

DifficultyEasy

Worked solution

Answer

(Example readings, measured with a ruler to the nearest $0.1\,\text{cm}$ )

$w = 14.8\,\text{cm}$

Adjusted cylinder so $d \approx 7.0\,\text{cm}$ , then measured:

$d = 7.0\,\text{cm}$

Final answer

w = 14.8 cm, d = 7.0 cm (example)

Detailed explanation

Background Concept

In Paper 3 practical questions, marks are often awarded for making and recording measurements correctly. A length measured using a ruler should normally be recorded to the nearest millimetre (i.e. nearest $0.1\,\text{cm}$ ), and written with a unit.

Here you measure:

$w$ : the width (shorter side) of the chosen sheet of paper.
$d$ : the cylinder diameter after adjusting it to be as close as possible to $7.0\,\text{cm}$ .

Understanding the Question

You are told to:

Choose the smaller sheet.
Measure and record its width $w$ .
Roll it into a cylinder, secure with paper clips, adjust until the diameter is close to $7.0\,\text{cm}$ .
Measure and record the cylinder diameter $d$ .

The key skill is careful measurement and clear recording.

Approach

Use a ruler to measure $w$ directly across the shorter edge.
Form the cylinder and adjust overlap until the diameter is about $7.0\,\text{cm}$ .
Measure $d$ across the widest part of the circular cross-section (straight across the centre), and record to the same precision.

Step-by-Step Reasoning

Place the ruler along the short side of the smaller sheet and read $w$ with your eye directly above the scale to avoid parallax.
Roll the sheet and hold it with two paper clips so the cylinder keeps its shape.
Adjust the overlap until the cylinder is close to the target diameter ( $7.0\,\text{cm}$ ). The target value is given to $0.1\,\text{cm}$ , so recording $d$ to $0.1\,\text{cm}$ is appropriate.
Measure the diameter $d$ across the cylinder through the centre (not along the curved surface) and record.

Key Takeaways

Record length measurements with appropriate precision (typically $0.1\,\text{cm}$ ) and units.
For a diameter, measure straight across the circle through the centre.

Common Mistakes

Recording $w$ or $d$ with no unit.
Recording with inconsistent precision (e.g. $7\,\text{cm}$ instead of $7.0\,\text{cm}$ when using a mm ruler).
Measuring around the curved surface (circumference) instead of the diameter.
Parallax error from viewing the scale at an angle.

Things to Be Careful About

Ensure the ruler is aligned with the edge being measured (not skewed).
Make sure the cylinder cross-section is as circular as possible when measuring $d$ .
Take the diameter at the middle of the cylinder (where it is most uniform).

Techniques used

measure a length with a ruler to an appropriate precisionadjust the apparatus to match a target dimensionrecord readings with units and consistent decimal places

(b)

(i)

• Set up the apparatus as shown in Fig. 2.3.

Fig. 2.3

• Slide the loop of the upper spring onto the wooden rod.
• Hang a mass of $200\text{ g}$ from the lower spring.
• Adjust the height of the boss until the bottom of the mass is approximately $10\text{ cm}$ above the bench.
• Place the paper cylinder so that the middle of the cylinder is under the mass, as shown in Fig. 2.3.
• The length of the springs is $y$ , as shown in Fig. 2.3.

Measure and record $y$ .

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

1M Similar

DifficultyEasy

Worked solution

Answer

(Example reading)

$y = 32.0\,\text{cm}$

Final answer

y = 32.0 cm (example)

Detailed explanation

Background Concept

When measuring the length of an extended spring (or springs in series), the main issues are reading the scale accurately and defining clearly what points you measure between.

The experiment defines $y$ as the length of the springs as shown in the diagram, so you should measure the same endpoints consistently each time.

Understanding the Question

You are instructed to:

Assemble the springs and mass as in Fig. 2.3.
Set the boss height so the bottom of the mass is about $10\,\text{cm}$ above the bench.
Place the paper cylinder under the mass.
Measure and record the spring length $y$ .

This value will later be compared with a new length $p$ when the cylinder supports part of the mass.

Approach

After setting the geometry, use a ruler to measure the total spring length indicated as $y$ .
Record the reading to a sensible precision (typically nearest $0.1\,\text{cm}$ with a mm ruler).

Step-by-Step Reasoning

Assemble the two springs in series and hang the $200\,\text{g}$ mass.
Adjust the boss so the mass is safely above the bench (about $10\,\text{cm}$ ) to allow later movement.
Wait for any oscillations to die down so the spring length is steady.
Hold a ruler close to the springs and read the length $y$ between the points indicated in the figure.
Record $y$ with unit.

Key Takeaways

In experiments, the reliability of later calculations depends on careful and consistent length measurements.

Common Mistakes

Measuring from different reference points each time.
Reading while the mass is still bouncing.
Estimating to an over-precise value not supported by the ruler scale.

Things to Be Careful About

Keep the ruler parallel to the spring length you are measuring.
Read at eye level to reduce parallax.
Make sure the cylinder does not touch the mass at this stage (it should be under the mass, not supporting it).

Techniques used

set up apparatus following a diagrammeasure a length with a ruler to an appropriate precisionavoid parallax when reading a scale

(ii)

Estimate the percentage uncertainty in your value of $y$ . Show your working.

percentage uncertainty = ______ $\%$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Ruler resolution $= 1\,\text{mm} \Rightarrow \Delta y = \pm 0.1\,\text{cm}$ .

\%\text{ uncertainty in }y = \frac{0.1}{32.0} \times 100 = 0.3125\% \approx 0.31\%

Answer

$0.31\%$

Final answer

0.31 %

Detailed explanation

Background Concept

An uncertainty is an estimate of the range within which the true value probably lies.

For a ruler:

The smallest division is typically $1\,\text{mm}$ .
A common estimate for the absolute uncertainty in a single ruler reading is $\pm 1\,\text{mm}$ (or sometimes $\pm 0.5\,\text{mm}$ depending on the mark scheme and how the reading is taken).

Percentage uncertainty is:

\%\text{ uncertainty} = \frac{\text{absolute uncertainty}}{\text{measured value}} \times 100

Understanding the Question

You must estimate the percentage uncertainty in your measured $y$ and show working. So you need:

an absolute uncertainty $\Delta y$ (from the ruler), and
your measured value of $y$ .

Approach

Decide $\Delta y$ based on the ruler resolution.
Substitute into $\frac{\Delta y}{y} \times 100$ .
Round sensibly.

Step-by-Step Reasoning

If the ruler has $1\,\text{mm}$ divisions, take $\Delta y = \pm 1\,\text{mm} = \pm 0.1\,\text{cm}$ .
Divide by the measured value (example $y = 32.0\,\text{cm}$ ) and multiply by $100$ :

\frac{0.1}{32.0} \times 100 = 0.3125\%

Quoting $0.31\%$ is appropriate; the uncertainty estimate is itself not exact, so giving many digits would be misleading.

Key Takeaways

Convert instrument resolution into an absolute uncertainty.
Percentage uncertainty is a ratio, then multiplied by $100$ .

Common Mistakes

Using $0.01\,\text{cm}$ for a mm ruler (wrong conversion).
Forgetting to multiply by $100$ .
Writing the uncertainty as a decimal (e.g. $0.0031$ ) but labelling it as a percentage.

Things to Be Careful About

If $y$ is found from two ruler readings (difference of two positions), the absolute uncertainty is usually larger (often the sum of the two reading uncertainties). Here the question implies a direct measurement of $y$ as a length, so a single-reading estimate is typically used.
Keep units consistent (cm with cm) before forming the ratio.

Techniques used

estimate an absolute uncertainty from instrument resolutioncalculate percentage uncertainty using $\text{percentage} = \frac{\Delta x}{x} \times 100$quote uncertainty to an appropriate number of significant figures

(c)

(i)

• By adjusting the height of the boss, lower the mass to squash the middle of the paper cylinder until the bottom of the mass is $2.5\text{ cm}$ above the bench, as shown in Fig. 2.4.

Fig. 2.4

• The length of the springs is $p$ , as shown in Fig. 2.4.

Measure and record $p$ .

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

1M Similar

(ii)

Calculate $(y - p)$ .

$(y - p)$ = ______ $\text{cm}$

1M Similar

(d)

Using the larger sheet of paper, repeat (a), (b)(i), (c)(i) and (c)(ii).

$w$ = ______ $\text{cm}$
$d$ = ______ $\text{cm}$
$y$ = ______ $\text{cm}$
$p$ = ______ $\text{cm}$
$(y - p)$ = ______ $\text{cm}$

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

(Example readings for larger sheet)

$w = 21.0\,\text{cm}$

$d = 7.0\,\text{cm}$

$y = 32.0\,\text{cm}$

$p = 27.0\,\text{cm}$

(y-p) = 32.0 - 27.0 = 5.0\,\text{cm}

Final answer

w = 21.0 cm, d = 7.0 cm, y = 32.0 cm, p = 27.0 cm, (y − p) = 5.0 cm (example)

Detailed explanation

Background Concept

Repeating measurements for a second condition (here, using a larger sheet) is how practical experiments test relationships: you need at least two data sets to compare.

The key is to keep the method the same so any changes in results are due to the changed variable (the paper width $w$ ), not due to a different measuring technique.

Understanding the Question

You must repeat parts (a), (b)(i), (c)(i) and (c)(ii) with the larger paper:

measure $w$ for the larger sheet,
make a cylinder with $d \approx 7.0\,\text{cm}$ and measure $d$ ,
measure $y$ in the initial position,
squash to $2.5\,\text{cm}$ above the bench and measure $p$ ,
calculate $(y-p)$ .

Approach

Use the same ruler and the same precision.
Adjust the cylinder to the same target diameter $d$ (as close to $7.0\,\text{cm}$ as possible) so the only deliberate difference is $w$ .
Measure $y$ and $p$ using the same reference points as before.

Step-by-Step Reasoning

Measure the width $w$ of the larger sheet to $0.1\,\text{cm}$ .
Roll into a cylinder, clip it, adjust overlap until the diameter is close to $7.0\,\text{cm}$ , then measure $d$ .
Set the apparatus with the mass about $10\,\text{cm}$ above the bench and measure $y$ .
Lower the mass until it is $2.5\,\text{cm}$ above the bench (cylinder squashed) and measure $p$ .
Compute $(y-p)$ and record with the correct unit.

Key Takeaways

Repeats should be carried out with consistent method and precision.
Derived quantities like $(y-p)$ should match the precision of the original measurements.

Common Mistakes

Forgetting to re-adjust the cylinder diameter to $7.0\,\text{cm}$ .
Recording inconsistent decimal places across the two trials.
Not recalculating $(y-p)$ for the larger sheet.

Things to Be Careful About

Ensure the mass contacts the middle of the cylinder in both trials; otherwise the deformation (and $p$ ) may not be comparable.
Keep the same mass ( $200\,\text{g}$ ) for both trials.

Techniques used

repeat a measurement procedure for a second conditionrecord multiple measured quantities with consistent precisionadjust the cylinder diameter to a target value before measuring

(e)

It is suggested that the relationship between $w$ , $y$ and $p$ is

w = k(y - p)

where $k$ is a constant.

(i)

Using your data, calculate two values of $k$ .

first value of $k$ = ______
second value of $k$ = ______

1M Similar

DifficultyMedium-Easy

Worked solution

Working

From $w = k(y-p)$ ,

k = \frac{w}{(y-p)}

Small sheet:

k_1 = \frac{14.8}{3.6} = 4.11 \approx 4.1

Large sheet:

k_2 = \frac{21.0}{5.0} = 4.2

Answer

first value of $k = 4.1$

second value of $k = 4.2$

Final answer

k ≈ 4.1 and 4.2 (example)

Detailed explanation

Background Concept

If two quantities are related by

w = k(y-p)

then the constant of proportionality is found by rearranging:

k = \frac{w}{(y-p)}

In a practical, you often calculate $k$ more than once (from different trials). If the relationship is correct and random uncertainties are not too large, the values of $k$ should be similar.

Understanding the Question

Using your two sets of measurements (small sheet and large sheet), you must calculate two values of $k$ .

You already have, for each sheet:

$w$ (paper width)
$(y-p)$ (change in spring length)

Approach

Rearrange to $k = \frac{w}{(y-p)}$ .
Substitute the measured values from each trial separately.
Quote each value of $k$ to a sensible number of significant figures (this is discussed in part (e)(ii)).

Step-by-Step Reasoning

Start with the given relationship:

w = k(y-p)

Divide both sides by $(y-p)$ :

k = \frac{w}{(y-p)}

Use the small-sheet readings to compute $k_1$ .
Use the large-sheet readings to compute $k_2$ .
Compare whether $k_1$ and $k_2$ are close.

Key Takeaways

Rearranging a linear proportionality lets you calculate a constant from measured values.
Two trials give two estimates to check consistency.

Common Mistakes

Using $k = \frac{(y-p)}{w}$ (inverted).
Forgetting that $(y-p)$ must be calculated for each trial before using it.
Rounding too aggressively (losing meaningful comparison) or giving too many digits (false precision).

Things to Be Careful About

If $(y-p)$ is small, its percentage uncertainty can be large; this strongly affects $k$ .
Ensure you use consistent units (here cm cancels, so $k$ is dimensionless).

Techniques used

rearrange a proportionality to compute a constantsubstitute experimental values into an equationcalculate and compare two estimates of a constant

(ii)

Justify the number of significant figures that you have given for your values of $k$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

$k$ is calculated from $k = \dfrac{w}{(y-p)}$ .

$w$ is measured to $\pm 0.1\,\text{cm}$ , and $y$ and $p$ are each measured to $\pm 0.1\,\text{cm}$ , so

\Delta (y-p) \approx \pm (0.1+0.1)=\pm 0.2\,\text{cm}

For example, if $(y-p)=3.6\,\text{cm}$ then percentage uncertainty is

\frac{0.2}{3.6}\times 100 \approx 6\%

So quoting $k$ to $2\,$ s.f. (e.g. $4.1$ , $4.2$ ) is justified.

Final answer

k to 2 s.f.

Detailed explanation

Background Concept

Significant figures (s.f.) should reflect the precision of the data. A calculated result cannot be more precise than the measurements used to obtain it.

A key point here is that $(y-p)$ is a difference of two measured lengths. When you subtract, the absolute uncertainties add (worst-case estimate):

\Delta (y-p) \approx \Delta y + \Delta p

Then for a quotient like $k = \frac{w}{(y-p)}$ , the fractional (percentage) uncertainty in $k$ is approximately the sum of the fractional uncertainties in $w$ and $(y-p)$ .

Understanding the Question

You are not asked to calculate the uncertainty in $k$ here; you are asked to justify the number of significant figures used for $k$ .

So you should explain which measurement limits the precision, and show that the uncertainty is at the level of a few percent (so 2 s.f. is appropriate).

Approach

Identify measurement precision: ruler readings for $w$ , $y$ , $p$ .
Find the absolute uncertainty in $(y-p)$ by adding the absolute uncertainties in $y$ and $p$ .
Turn that into a percentage uncertainty for a typical $(y-p)$ .
Decide s.f. from that scale of uncertainty.

Step-by-Step Reasoning

Suppose the ruler reads to $1\,\text{mm}$ , so each length has about $\pm 0.1\,\text{cm}$ uncertainty.
Since $(y-p)$ uses two measurements, a worst-case absolute uncertainty is:

\Delta (y-p) \approx 0.1 + 0.1 = 0.2\,\text{cm}

For a typical value $(y-p) \approx 3.6\,\text{cm}$ , this is about:

\frac{0.2}{3.6} \times 100 \approx 5.6\% \approx 6\%

That means $k$ is uncertain by several percent, so quoting $k$ to $2\,$ s.f. is sensible. Writing $4.11$ would imply about $\pm 0.01$ precision (about $0.2\%$ ), which is far better than the measurements justify.

Key Takeaways

Differences amplify uncertainty: absolute uncertainties add.
Significant figures should match the realistic uncertainty (often a few percent in practical work).

Common Mistakes

Choosing s.f. based only on $w$ and ignoring that $(y-p)$ is a subtraction.
Giving too many digits because a calculator shows them.
Rounding inconsistently between the two calculated $k$ values.

Things to Be Careful About

If your measured $(y-p)$ is smaller than in the example, the percentage uncertainty becomes even larger, making fewer s.f. appropriate.
The mark scheme typically accepts sensible justification; it does not require an exact uncertainty propagation, but it must be logically linked to measurement precision.

Techniques used

identify the least precise measured quantity controlling a calculated resultpropagate absolute uncertainties through subtractionchoose significant figures consistent with an estimated percentage uncertainty

(f)

It is suggested that the percentage uncertainty in the values of $k$ is $10\%$ .

Using this uncertainty, explain whether your results support the relationship in (e).

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Using $k_1 = 4.1$ and $k_2 = 4.2$ .

Mean

\bar{k} = \frac{4.1+4.2}{2} = 4.15

Percentage difference

\frac{|4.2-4.1|}{4.15} \times 100 \approx 2.4\%

Since $2.4\% < 10\%$ , the two values of $k$ agree within the stated uncertainty.

Answer

Yes, the results support the relationship because the values of $k$ are consistent within $10\%$ .

Final answer

Supports relationship (values consistent within 10%)

Detailed explanation

Background Concept

Experimental results rarely match perfectly because of uncertainty. A common way to test whether two values are consistent is to see whether their difference is small compared with the allowed uncertainty.

If the percentage uncertainty in $k$ is $10\%$ , then values that differ by only a few percent are considered consistent, supporting the proposed relationship.

Understanding the Question

You have two calculated values of $k$ (from the two cylinders). You are told to assume the percentage uncertainty in $k$ is $10\%$ and then decide whether your results support $w = k(y-p)$ .

Approach

Compare $k_1$ and $k_2$ .
Compute a percentage difference (often difference divided by mean, expressed as a percentage).
If this percentage difference is less than (or comparable to) $10\%$ , conclude the results support the relationship.

Step-by-Step Reasoning

Find the absolute difference $|k_2-k_1|$ .
Convert it to a percentage of a representative value (the mean is commonly used):

\%\text{ difference} = \frac{|k_2-k_1|}{\bar{k}} \times 100

Compare with $10\%$ .
Make a clear statement: either “consistent within uncertainty” (supports) or “not consistent” (does not support).

Key Takeaways

A relationship is supported when repeated estimates of a constant agree within experimental uncertainty.

Common Mistakes

Saying “supports” just because the numbers look similar, without referencing the $10\%$ .
Dividing by the wrong reference (e.g. dividing by the smaller value without stating what you are doing).
Forgetting to express the comparison as a percentage.

Things to Be Careful About

If your two $k$ values differ by around $10\%$ , the conclusion may be borderline; you should then say something like “difference is comparable to uncertainty, so results are consistent (or only just consistent)”.
Use your own measured $k_1$ and $k_2$ in the exam; the numbers here are examples.

Techniques used

compare two experimental estimates using a stated percentage uncertaintycalculate a percentage differencedecide whether results are consistent within uncertainty

(g)

(i)

Describe four sources of uncertainty or limitations of the procedure for this experiment.

For any uncertainties in measurement that you describe, you should state the quantity being measured and a reason for the uncertainty.

4M Similar

DifficultyMedium

Worked solution

Answer

Uncertainty in measuring $y$ and $p$ : difficult to judge the exact endpoints of the spring length (spring loops/thickness) and possible parallax when reading the ruler.
Uncertainty in setting the bottom of the mass to $10\,\text{cm}$ and $2.5\,\text{cm}$ above the bench: the reference point (bottom of mass) is not sharp and reading the vertical height can have parallax.
Cylinder not perfectly circular / diameter not exactly $7.0\,\text{cm}$ : overlap and paper clips can distort the shape, changing how the cylinder deforms.
Load may not be applied centrally: if the mass is not directly above the middle of the cylinder, the cylinder squashes unevenly and the support force (hence $p$ and $y-p$ ) changes.

Final answer

Four valid uncertainties/limitations listed

Detailed explanation

Background Concept

In practical work, limitations come from:

instrument resolution (e.g. mm ruler),
reading technique (parallax, unclear reference points),
difficulty controlling variables (e.g. keeping diameter fixed),
physical behaviour of the system (paper may deform non-uniformly).

Good answers:

identify the quantity affected,
explain the reason for uncertainty,
explain the consequence (how it could change results).

Understanding the Question

You must give four sources of uncertainty or limitations in this procedure. For any measurement uncertainty, you must name the measured quantity and why it is uncertain.

The main measurements are $w$ , $d$ , $y$ , $p$ and the set heights (10 cm and 2.5 cm above the bench).

Approach

Think through each stage:

Making the cylinder and measuring $w$ and $d$ .
Measuring $y$ and $p$ (spring lengths).
Ensuring the geometry is correct (mass height, central loading).

Select four distinct points (not repeats of the same idea) and write each as: quantity + reason + effect.

Step-by-Step Reasoning

Examples of creditworthy limitations:

Spring length readings ( $y$ , $p$ ): endpoints are not perfectly defined (loops, thickness of coils); ruler may not be exactly alongside the spring; eye-level error gives parallax. This changes the calculated $(y-p)$ .
Setting height of the mass above the bench (10 cm, 2.5 cm): difficult to judge the exact bottom point of the mass and read the height accurately. If the mass is not at the intended height, the amount of squashing changes, altering $p$ .
Cylinder diameter control ( $d$ ): paper clips and overlap cause the cylinder to be slightly oval, and $d$ may vary with orientation. Since deformation depends on geometry, this affects how much load is transferred from springs to cylinder.
Central loading and alignment: if the mass is not directly above the centre of the cylinder, the cylinder deforms asymmetrically; friction/tilting changes the support force and hence the spring length $p$ .

Key Takeaways

Limitations should be specific and linked to quantities you measure.
Geometry/alignment issues are often major sources of scatter in mechanics-based practicals.

Common Mistakes

Writing vague statements like “human error” without naming a quantity.
Repeating the same idea four times (e.g. parallax for every point) without variety.
Giving improvements instead of limitations (those belong in part (g)(ii)).

Things to Be Careful About

Try to include both measurement uncertainties (ruler readings) and procedural/physical limitations (cylinder shape, alignment).
Make sure each of the four points is clearly distinct and explained.

Techniques used

identify random and systematic uncertainties in measurementslink each limitation to a specific measured quantityexplain why each limitation affects the result

(ii)

Describe four improvements that could be made to this experiment. You may suggest the use of other apparatus or different procedures.

4M Similar

DifficultyMedium

Worked solution

Answer

Clamp a ruler beside the springs and use a fixed reference pointer (or set square) to read $y$ and $p$ at eye level, reducing parallax and improving repeatability.
Use a set square / metre rule fixed vertically to set the mass heights (10 cm and 2.5 cm above bench) more accurately.
Improve cylinder geometry: mark a fixed overlap line and use more clips (or tape) to keep the cylinder truly circular and keep $d$ close to $7.0\,\text{cm}$ ; check $d$ in two perpendicular directions.
Ensure central loading: add a guide (e.g. a vertical rod/frame) so the mass always contacts the centre of the cylinder; repeat and average readings to reduce random scatter.

Final answer

Four valid improvements listed

Detailed explanation

Background Concept

Improvements aim to reduce uncertainty and improve reliability by:

using better measurement techniques (fixed scales, eye-level readings),
improving control of variables (constant diameter and alignment),
increasing repeatability (repeats and averaging).

Understanding the Question

You must give four improvements. These can include using different apparatus or changing procedure.

A strong set of improvements should clearly address the limitations identified in part (g)(i).

Approach

For each improvement:

state what you would change,
state what uncertainty/limitation it reduces and how.

Step-by-Step Reasoning

Examples of good improvements:

Reduce parallax / unclear endpoints for $y$ and $p$ : Clamp a ruler next to the springs and attach a small pointer at the endpoint so readings are consistent; read at eye level.
Set heights more accurately: Use a vertical metre rule fixed to the stand and a set square to check the bottom of the mass is at exactly $10\,\text{cm}$ and $2.5\,\text{cm}$ .
Control cylinder diameter more reliably: Mark the overlap position on the paper, use tape rather than clips, and measure $d$ in two directions to ensure the cross-section is circular.
Improve alignment / central loading and reliability: Use a guide so the mass is always above the centre of the cylinder; take multiple trials for each sheet and average $(y-p)$ to reduce random uncertainty.

Key Takeaways

Improvements should be practical, specific, and clearly linked to reduced uncertainty.
Repeats and better alignment often give big gains in reliability.

Common Mistakes

Giving the limitation again without proposing a change.
Suggesting unrealistic equipment without explaining how it helps.
Writing fewer than four distinct improvements.

Things to Be Careful About

Ensure each improvement is a real change to the method or apparatus.
Where you suggest repeats, state what you would repeat (e.g. measure $p$ several times) and that you would average the results.

Techniques used

propose procedural changes to reduce uncertaintymatch each improvement to a stated limitationsuggest more suitable measuring instruments