In this experiment, you will investigate the oscillations of a chain of paper clips. You have been provided with two spheres of modelling clay. (i) The diameter of the smaller sphere is $d$, as shown in Fig. 2.1. Measure and record $d$. $d$ = \_\_\_\_\_\_ (ii) Estimate the percentage uncertainty in your value of $d$. Show your working. percentage uncertainty = \_\_\_\_\_\_ % (i) Set up the apparatus as shown in Fig. 2.2. Ensure that the rods are the same height above the…

(a)(i) d = 2.40 cm (example) · (a)(ii) percentage uncertainty ≈ 2.1% (example) · (b)(i) x = 70.0 cm (example) · (b)(ii) N = 4.95 · (c) T = 1.19 s (example) · (d) Example: d = 3.00 cm, x = 80.0 cm, N = 3.66, T = 1.01 s · (e)(i) Example: k1 = 5.8 s^2 m, k2 = 5.9 s^2 m · (e)(ii) Quote k to 2 s.f. (uncertainty ≈ 15%) · (f) Yes — k values agree within 15%, so relationship is supported. · (g)(i) See working (four uncertainties/limitations listed). · (g)(ii) See working (four…

Physics 9702/37 — May/June 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 an electrical circuit.

(a)

Set up the circuit shown in Fig. 1.1.

Fig. 1.1

Record the voltmeter reading $E$ .

$E$ = ______

1M Similar

(b)

You have been provided with a metre rule with a wire attached. You have also been provided with two identical resistors placed in component holders, each labelled R.

Set up the circuit shown in Fig. 1.2.

Fig. 1.2

F and G are crocodile clips.

The distance between F and G is $L$ . Attach F and G to the wire so that $L$ is approximately 30 cm.

Close the switch.

Record the value of $L$ and the voltmeter reading $V$ .

$L$ = ______
$V$ = ______

Open the switch.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Attach clips so that $L \approx 30\ \text{cm}$ , close the switch, then record $L$ and $V$ (e.g. $L = 30.0\ \text{cm}$ , $V = 0.75\ \text{V}$ ).

Final answer

Example: L = 30.0 cm, V = 0.75 V

Detailed explanation

Background Concept

In practical circuits, changing the effective length of a uniform wire changes its resistance (longer wire $\Rightarrow$ larger resistance). A metre rule allows you to set and measure the length $L$ between two points accurately. The voltmeter measures the p.d. $V$ across the component it is connected across.

Understanding the Question

You must build the circuit in Fig. 1.2 using:

a wire on a metre rule with crocodile clips $F$ and $G$ separated by distance $L$ ,
two identical resistors (each labelled $R$ ),
a voltmeter reading $V$ .
You then set $L$ to about $30\ \text{cm}$ , close the switch, and record $L$ and $V$ .

Approach

Assemble the circuit exactly as shown.
Place $F$ and $G$ so their separation is about $30\ \text{cm}$ .
Close the switch briefly, take readings, then open the switch.

Step-by-Step Reasoning

Use the metre rule scale to set $L \approx 30\ \text{cm}$ .
Measure $L$ between the clip contact points (not the clip ends).
Close the switch and allow the voltmeter reading to stabilise.
Record $V$ to the meter resolution and record $L$ to the metre rule resolution (often $\pm 0.1\ \text{cm}$ if you estimate between mm marks).
Open the switch to reduce heating of the wire and resistors.

Key Takeaways

Measure the separation between contact points for $L$ .
Take steady voltmeter readings and record with units.

Common Mistakes

Measuring from the wrong points on the crocodile clips.
Leaving the switch closed for too long, causing heating and drift.
Recording $L$ with inconsistent precision (e.g. sometimes to nearest cm, sometimes to mm).

Things to Be Careful About

Ensure good electrical contact at $F$ and $G$ .
Avoid parallax when reading the metre rule.
Keep the circuit unchanged except for the intended change in $L$ .

Techniques used

measure a length between two crocodile clips on a metre ruleset the independent variable to a specified approximate valuerecord the corresponding voltmeter reading

(c)

Write down the value of $E$ from (a).

$E$ = ______

Increase $L$ by changing the position of F on the wire. Record $L$ and $V$ and repeat until you have six sets of values of $L$ and $V$ . Include your values from (b).

Record your results in a table. Include values of $\frac{E - V}{L}$ in your table.

9M Similar

DifficultyMedium

Worked solution

Answer

$E$ (from (a)) = e.g. $1.50\ \text{V}$ .

Record six pairs of $L$ and $V$ (including the value from (b)) and calculate

\frac{E - V}{L}

for each row.

Example of a correctly laid out table (candidate values will differ):

$L$ / $\text{cm}$	$V$ / $\text{V}$	$E - V$ / $\text{V}$	$(E-V)/L$ / $\text{V cm}^{-1}$
$30.0$	$0.750$	$0.750$	$0.0250$
$35.0$	$0.710$	$0.790$	$0.0226$
$40.0$	$0.676$	$0.824$	$0.0206$
$45.0$	$0.649$	$0.851$	$0.0189$
$50.0$	$0.625$	$0.875$	$0.0175$
$55.0$	$0.605$	$0.895$	$0.0163$

Final answer

Table of 6 readings of L and V with calculated (E−V)/L

Detailed explanation

Background Concept

Good experimental data needs:

a clear choice of independent variable (here $L$ is changed deliberately),
a dependent variable to measure (here $V$ ),
enough readings over a sensible range (here six sets),
clear presentation in a single table with correct headings and units.

When calculating a derived quantity, such as

\frac{E - V}{L},

you should show it in its own column with an appropriate unit. If $E$ and $V$ are in volts and $L$ is in cm, then $(E-V)/L$ has unit $\text{V cm}^{-1}$ (or convert $L$ to m and use $\text{V m}^{-1}$ ).

Understanding the Question

You must:

Write down the same $E$ you measured in (a).
Increase $L$ step-by-step (by moving clip $F$ ) and for each $L$ record the corresponding voltmeter reading $V$ .
Obtain six sets of $(L, V)$ values in total.
Present results in a table including a calculated column for $(E-V)/L$ .

Approach

Choose six values of $L$ spread over a wide range (e.g. from about $30\ \text{cm}$ up to perhaps $70\text{–}90\ \text{cm}$ depending on the apparatus).
For each $L$ , close the switch briefly, record $V$ , then open the switch.
After the measurements, calculate $(E-V)/L$ for each row and enter it into the table.

Step-by-Step Reasoning

Copy $E$ from part (a) exactly (same number of decimal places).
Set $L \approx 30\ \text{cm}$ (already done in (b)); record $L$ and $V$ .
Increase $L$ by moving clip $F$ and repeat until you have 6 rows.
In your table:
- Put all raw and derived data in one table.
- Headings should include quantity and unit, e.g. $L / \text{cm}$ , $V / \text{V}$ , $(E-V)/L / \text{V cm}^{-1}$ .
- Use consistent decimal places within each column (e.g. $L$ to $0.1\ \text{cm}$ or $0.5\ \text{mm}$ depending on how you read it; $V$ to $0.01\ \text{V}$ or $0.001\ \text{V}$ depending on meter).
Calculate each derived value:

\frac{E-V}{L}

using the measured $E$ , the measured $V$ for that row, and the measured $L$ for that row.

Key Takeaways

Use a single clear table with headings and units.
Take enough readings over a range.
Calculated columns must have correct units and consistent precision.

Common Mistakes

Forgetting to include the row from part (b).
Writing $E$ again but with different rounding to part (a).
Missing units in the headings.
Calculating $(E-V)/L$ but not stating the unit, or using inconsistent $L$ units (mixing cm and m).

Things to Be Careful About

Keep the switch closed only long enough to take a reading to reduce heating.
Ensure $L$ is the distance between the clip contact points.
If you convert $L$ to m for calculations, be consistent for every row and for your graph later.

Techniques used

collect multiple pairs of readings over a suitable range of the independent variablerecord data in a single table with headings and unitscalculate a derived quantity from measured valuesuse consistent significant figures within each column

(d)

(i)

Plot a graph of $\frac{E - V}{L}$ on the $y$ -axis against $V$ on the $x$ -axis.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Plot $y = \dfrac{E-V}{L}$ on the $y$ -axis (unit e.g. $\text{V cm}^{-1}$ ) against $x = V$ on the $x$ -axis (unit $\text{V}$ ) using all six data points with a suitable scale.

Final answer

Graph of (E−V)/L (y) against V (x)

Detailed explanation

Background Concept

A graph is used to reveal relationships between variables. To gain marks, the graph must be readable and accurate:

correct variables on correct axes,
clear axis labels including units,
a sensible scale using at least half of the available grid in each direction,
points plotted accurately and consistently.

Understanding the Question

You are told exactly what to plot:

$y$ -axis: $\dfrac{E - V}{L}$
$x$ -axis: $V$
using the six sets of values from your table.

Approach

Decide the numerical ranges of $V$ and $(E-V)/L$ from your data.
Choose axis limits and scales that spread points out.
Plot each point as a small cross (not a dot so large it hides accuracy).

Step-by-Step Reasoning

From your results table, list the pairs $(V, (E-V)/L)$ .
Draw axes with a large plotting area.
Label axes, for example:
- $V / \text{V}$ on the horizontal axis.
- $(E-V)/L / \text{V cm}^{-1}$ (or $\text{V m}^{-1}$ if you used m) on the vertical axis.
Choose a uniform scale (e.g. 1 large square = 0.05 V) that uses most of the grid.
Plot all six points carefully.

Key Takeaways

Axes must include both quantity and unit.
Use a scale that spreads the points for best gradient accuracy.

Common Mistakes

Swapping axes (plotting $V$ on $y$ instead of $x$ ).
Missing units in labels.
Using an awkward scale (e.g. 1 large square = 3 units) that wastes grid space.

Things to Be Careful About

If you used $L$ in cm, keep derived units consistent across the whole graph.
Plotting accuracy: use a sharp pencil and read to half a small square where possible.

Techniques used

choose appropriate axes and scales that use most of the gridlabel axes with quantities and unitsplot points accurately from a data 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-Easy

Worked solution

Working

Choose two well-separated points on the best-fit line, e.g.
$(V_1, y_1) = (0.60\ \text{V},\ 0.016\ \text{V cm}^{-1})$ and $(V_2, y_2) = (0.75\ \text{V},\ 0.025\ \text{V cm}^{-1})$ .

\text{gradient } m = \frac{y_2-y_1}{V_2-V_1} = \frac{0.025-0.016}{0.75-0.60} = 0.060\ \text{cm}^{-1}

Using $y = mV + c$ ,

c = y_2 - mV_2 = 0.025 - (0.060)(0.75) = -0.020\ \text{V cm}^{-1}

Answer

gradient $= 0.060\ \text{cm}^{-1}$

$y$ -intercept $= -0.020\ \text{V cm}^{-1}$

Final answer

gradient ≈ 0.060 cm^-1, y-intercept ≈ −0.020 V cm^-1

Detailed explanation

Background Concept

For a straight-line graph,

y = mx + c,

where:

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

The most accurate gradient comes from a large triangle using two points on the best-fit line far apart, not necessarily actual data points.

Understanding the Question

You must find:

the gradient of your best-fit line on the graph of $y = (E-V)/L$ against $x = V$ ,
the y-intercept of that same line.
Both must be read/calculated using your graph scale and quoted with appropriate units.

Approach

Pick two points on the best-fit line far apart.
Compute $m = (y_2-y_1)/(x_2-x_1)$ .
Find the y-intercept either by reading where the line crosses the y-axis, or by calculating $c$ from $c = y - mx$ using a point on the line.

Step-by-Step Reasoning

Suppose you select two points on the drawn line at convenient grid intersections.
Compute the changes:
- $\Delta y$ is the vertical change in $(E-V)/L$ .
- $\Delta x$ is the horizontal change in $V$ .
Divide to get the gradient; the unit is

\frac{\text{unit of }y}{\text{unit of }x} = \frac{\text{V cm}^{-1}}{\text{V}} = \text{cm}^{-1}.

Then obtain the y-intercept $c$ $c$ :
- Either read it at $V=0$ , or calculate it from $c=y-mV$ .

Key Takeaways

Use a large triangle on the best-fit line for accuracy.
Gradient units come from “vertical units divided by horizontal units”.

Common Mistakes

Using two nearby points (large percentage error in gradient).
Using two data points that are not on the best-fit line.
Inverting the gradient as $\Delta x/\Delta y$ .
Quoting the y-intercept without its sign (many graphs here have a negative intercept).

Things to Be Careful About

Read values carefully from the scale (half small-square precision where possible).
Keep the same $L$ unit (cm or m) used in the derived column, so your gradient/intercept units match your axes.

Techniques used

calculate the gradient from a large triangle on the best-fit linedetermine the y-intercept from the graph or from y = mx + cuse consistent units taken from the axis labels

(e)

It is suggested that the quantities $V$ and $L$ are related by the equation

\frac{E - V}{L} = PV - 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

From the graph, $y = \dfrac{E-V}{L}$ and $x = V$ .

Given

\frac{E - V}{L} = PV - Q

so this is $y = Px + (-Q)$ .

Therefore:

$P =$ gradient
$-Q =$ y-intercept $\Rightarrow Q = - (\text{y-intercept})$

Using (d)(iii) (example values):

P = 0.060\ \text{cm}^{-1}

Q = -(-0.020) = 0.020\ \text{V cm}^{-1}

Answer

$P = 0.060\ \text{cm}^{-1}$

$Q = 0.020\ \text{V cm}^{-1}$

Final answer

P = gradient; Q = −(y-intercept) (with units)

Detailed explanation

Background Concept

If a graph is plotted as $y$ against $x$ and the relationship is linear, it can be written as

y = mx + c.

By comparing with a given physical equation, you can identify constants with:

gradient $m$ ,
intercept $c$ .

Units:

Here $y = (E-V)/L$ has units $\text{V}/\text{cm} = \text{V cm}^{-1}$ (or $\text{V m}^{-1}$ ).
$x=V$ has unit $\text{V}$ .
So $P$ has unit $(\text{V cm}^{-1})/(\text{V}) = \text{cm}^{-1}$ and $Q$ has unit $\text{V cm}^{-1}$ .

Understanding the Question

You are given

\frac{E - V}{L} = PV - Q

and you have already found the gradient and y-intercept of a graph of $(E-V)/L$ against $V$ . You must use those graph parameters to determine $P$ and $Q$ with units.

Approach

Map the equation onto $y = mx + c$ .
Read off $P$ as the gradient.
Use the sign of the intercept carefully: intercept equals $-Q$ .

Step-by-Step Reasoning

Let

y = \frac{E-V}{L}, \quad x = V.

Then the given equation becomes

y = Px - Q.

Comparing with $y = mx + c$ :

$m = P$ ,
$c = -Q$ so $Q = -c$ .

If your y-intercept is negative (common here), then $Q$ will be positive.

Key Takeaways

Always rewrite into $y = mx + c$ form to identify constants.
The y-intercept is the constant term; if the equation has $-Q$ , then the intercept is $-Q$ .

Common Mistakes

Setting $Q$ equal to the y-intercept instead of the negative of it.
Giving $P$ or $Q$ without units.
Mixing cm and m: if $L$ was in cm, use $\text{cm}^{-1}$ and $\text{V cm}^{-1}$ ; if $L$ was in m, use $\text{m}^{-1}$ and $\text{V m}^{-1}$ .

Things to Be Careful About

Keep signs correct: intercept $= -Q$ .
Quote $P$ and $Q$ to a sensible number of significant figures consistent with graph reading.

Techniques used

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

(f)

The resistance of R is $R$ .

Theory suggests that:

$P$ and $Q$ are both inversely proportional to $R$
the graph cuts the $x$ -axis at a value of $V = \frac{E}{2}$ for all values of $R$ .

A student repeats the experiment using two identical resistors, each with a lower value of $R$ than in the original experiment.

For the student's experiment, draw a second line on the graph to show the expected results. Label this line W.

1M Similar

DifficultyMedium

Worked solution

Answer

Since $P$ and $Q$ are inversely proportional to $R$ , using a lower $R$ gives larger $P$ (steeper gradient) and larger $Q$ (more negative y-intercept because intercept $= -Q$ ).

The x-intercept remains at $V = E/2$ .

Draw line W as a steeper straight line than the original, passing through the same x-intercept $V = E/2$ , and label it W.

diagram-1f-lineW

Final answer

Line W: steeper, same x-intercept at V = E/2

Detailed explanation

Background Concept

The plotted graph is described by

y = PV - Q,

where $y = (E-V)/L$ and $x=V$ .

Gradient $= P$ .
y-intercept $= -Q$ .
x-intercept occurs when $y=0$ :

0 = PV - Q \Rightarrow V = \frac{Q}{P}.

You are told that theory predicts:

$P \propto \frac{1}{R}$ and $Q \propto \frac{1}{R}$ .
The x-intercept is always at $V = E/2$ for all $R$ .

Understanding the Question

A student repeats the experiment with resistors of LOWER resistance $R$ than before. You must predict how the straight line on your $y$ vs $V$ graph changes, and sketch that new line (labelled W) on the same axes.

Approach

Use the theory statements:

Lower $R$ means both $P$ and $Q$ increase (because they are inversely proportional to $R$ ).
The x-intercept must stay fixed at $V = E/2$ .
So the new line must rotate about the fixed point $(V=E/2,\ y=0)$ .

Step-by-Step Reasoning

Since $P$ is the gradient, increasing $P$ makes the line steeper.
Since the y-intercept is $-Q$ , increasing $Q$ makes the y-intercept more negative (the line crosses the y-axis lower down).
Because the x-intercept is fixed at $E/2$ , both the original and new lines must cross the x-axis at the same point.

So W is drawn as a steeper line that still passes through $(E/2, 0)$ .

diagram-1f-lineW

Key Takeaways

Gradient corresponds to $P$ and intercept corresponds to $-Q$ .
If both $P$ and $Q$ scale together, the ratio $Q/P$ (and hence the x-intercept) can stay constant.

Common Mistakes

Shifting the line parallel (keeping the same gradient) instead of making it steeper.
Moving the x-intercept away from $V=E/2$ .
Forgetting that the y-intercept is $-Q$ (so larger $Q$ means a more negative intercept).

Things to Be Careful About

The line should be straight and clearly labelled W.
Ensure it crosses the x-axis at $V=E/2$ (use your measured $E$ value to locate this on your axis scale).

Techniques used

use proportional reasoning to predict changes in gradient and interceptuse the x-intercept condition to constrain the new line positionsketch a second straight line consistent with the model

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

In this experiment, you will investigate the oscillations of a chain of paper clips.

You have been provided with two spheres of modelling clay.

(a)

(i)

The diameter of the smaller sphere is $d$ , as shown in Fig. 2.1.

Fig. 2.1

Measure and record $d$ .

$d$ = ______

1M Similar

DifficultyEasy

Worked solution

Measure the diameter with a ruler.

Example (to nearest $0.01\ \text{cm}$ ):

d = 2.40\ \text{cm}

Final answer

d = 2.40 cm (example)

Detailed explanation

Background Concept

A length measurement should be taken with an instrument of suitable resolution (e.g. ruler, vernier calipers) and recorded to the precision justified by the instrument scale. The measurement uncertainty is linked to the smallest scale division and to how clearly the object edge can be judged.

Understanding the Question

You are asked to measure the diameter $d$ of the smaller modelling clay sphere shown. The answer is a direct reading (no calculation). Since the blank gives no unit, you must supply the unit you used (normally $\text{cm}$ or $\text{mm}$ ).

Approach

Use a ruler (or calipers if available): place the sphere against the scale, align one edge with a known mark (often zero), read the other edge, and record the result to appropriate precision. Ensure your eye is directly above the scale to avoid parallax.

Step-by-Step Reasoning

Place the sphere next to the ruler so that its diameter is aligned with the scale.
Align one edge of the sphere with a convenient mark (e.g. $0.00\ \text{cm}$ ) to reduce reading error.
Read the scale at the opposite edge of the sphere with your eye vertically above the mark.
Record $d$ with a unit and with a sensible number of decimal places for the instrument used (e.g. to the nearest $1\ \text{mm}$ with a ruler).

Key Takeaways

Record measurements with units and appropriate precision.
Good alignment and avoiding parallax are essential for accurate readings.

Common Mistakes

Omitting the unit.
Writing too many decimal places (not justified by a ruler).
Parallax error from reading the scale at an angle.

Things to Be Careful About

If the sphere is slightly squashed or not perfectly spherical, different “diameters” are possible; measure the widest diameter and avoid compressing it while measuring.
If using a ruler with $1\ \text{mm}$ divisions, a typical reading precision is $\pm 0.5\ \text{mm}$ , so recording to $0.1\ \text{cm}$ (or $0.01\ \text{m}$ is not appropriate here) is important.

Techniques used

measure a diameter using a ruler or calipersrecord a reading to an appropriate precisionavoid parallax by aligning the eye normal to the scale

(ii)

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

percentage uncertainty = ______ %

1M Similar

DifficultyMedium-Easy

Worked solution

Using a ruler with $1\ \text{mm}$ divisions, take absolute uncertainty as $\pm 0.5\ \text{mm} = \pm 0.05\ \text{cm}$ .

Example with $d = 2.40\ \text{cm}$ :

\%\ \text{uncertainty} = \frac{0.05}{2.40}\times 100 = 2.08\% \approx 2.1\%

Final answer

percentage uncertainty ≈ 2.1% (example)

Detailed explanation

Background Concept

Uncertainty in a single length measurement is often taken as about half the smallest scale division (for an analogue scale), because you can usually judge between marks to about that level. 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 diameter $d$ and show the calculation. The percentage uncertainty depends on both the instrument uncertainty and the size of $d$ .

Approach

Decide an absolute uncertainty for $d$ based on the instrument (and any difficulty judging the sphere edge).
Substitute into the percentage uncertainty formula.

Step-by-Step Reasoning

If a ruler has $1\ \text{mm}$ smallest divisions, a common estimate is $\pm 0.5\ \text{mm}$ for one reading.
Convert that to the same unit as $d$ (e.g. $0.5\ \text{mm} = 0.05\ \text{cm}$ ).
Divide by your measured $d$ and multiply by $100$ .

For example, with $d = 2.40\ \text{cm}$ :

\%\ \text{uncertainty} = \frac{0.05}{2.40}\times 100 = 2.08\% \approx 2.1\%

Key Takeaways

Always use consistent units when computing uncertainties.
Percentage uncertainty decreases when the measured value is larger (for the same absolute uncertainty).

Common Mistakes

Using the smallest division rather than half of it (overestimating uncertainty).
Forgetting to convert mm to cm (or to m).
Not showing working.

Things to Be Careful About

If you effectively make two readings (e.g. read both edges against the scale rather than starting at zero), the uncertainty may be larger because two edge judgments are involved.
If the sphere edge is not clear or the sphere deforms, you may need to justify a larger absolute uncertainty than the instrument resolution alone.

Techniques used

estimate absolute uncertainty from instrument resolutioncalculate percentage uncertainty from absolute uncertaintyquote an uncertainty to an appropriate number of significant figures

(b)

(i)

Set up the apparatus as shown in Fig. 2.2.

Fig. 2.2

Ensure that the rods are the same height above the bench.

Slide the paper clips at the ends of the chain onto the rods.

The distance between the centres of the rods is $x$ . Position the stands so that $x$ is approximately 70 cm.

Measure and record $x$ .

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

1M Similar

DifficultyMedium-Easy

Worked solution

Set the stands so that $x \approx 70\ \text{cm}$ and measure the distance between the centres of the rods.

Example (to nearest $0.1\ \text{cm}$ ):

x = 70.0\ \text{cm}

Final answer

x = 70.0 cm (example)

Detailed explanation

Background Concept

In practical work, the quality of any later analysis depends strongly on how well the apparatus matches the intended geometry. Measuring a distance between centres is often harder than measuring between edges, because you must identify the centre position consistently.

Understanding the Question

You must set up the chain between two rods (same height) and measure $x$ , the distance between the centres of the rods, with $x$ about $70\ \text{cm}$ .

Approach

Build the arrangement as shown so the chain is supported at both ends.
Adjust the stands until the separation is near $70\ \text{cm}$ .
Measure $x$ carefully (centre-to-centre), then record it with unit and appropriate precision.

Step-by-Step Reasoning

Ensure the rods are at the same height above the bench so the chain is approximately symmetric and not twisted.
Slide the end paper clips onto the rods so the chain can move/oscillate freely.
Use a ruler/metre rule to measure the horizontal separation of the rod centres.
- If measuring edge-to-edge is easier, you can measure between the same edges of each rod and then correct to centre-to-centre by adding/subtracting one radius as appropriate.
Record $x$ in $\text{cm}$ to the precision allowed by the rule (commonly $0.1\ \text{cm}$ if reading to the nearest mm).

Key Takeaways

Follow set-up instructions (especially “same height”) to reduce systematic errors.
Identify what exactly must be measured (centre-to-centre).

Common Mistakes

Measuring between the outer edges of rods and calling it $x$ .
Not ensuring the rods are the same height, causing a sloping chain.
Recording $x$ with no unit.

Things to Be Careful About

Ensure the ruler is parallel to the bench and aligned with the rod centres (avoid diagonal measurements).
If the rod thickness is significant, uncertainty in locating the centre can dominate the uncertainty in $x$ .

Techniques used

set up apparatus to match a diagrammeasure a horizontal distance between two pointsrecord a length to an appropriate precision

(ii)

Use the hook to attach the smaller sphere of modelling clay to the chain of paper clips as shown in Fig. 2.3.

Fig. 2.3

The number of paper clips between the hook and the end of the chain is $n$ , as shown in Fig. 2.3.

Place the hook so that $n$ is 11.

Calculate $N$ , where

N = \sqrt[3]{n^2}

Give your answer to three significant figures.

$N$ = ______

1M Similar

(c)

Pull the sphere towards you through a short distance. When the sphere is released, it will oscillate.

Take measurements to determine the period $T$ of these oscillations.

$T$ = ______

2M Similar

DifficultyMedium

Worked solution

Time $10$ oscillations and repeat.

Example:

t_1(10T)=12.0\ \text{s},\quad t_2(10T)=11.8\ \text{s}

\overline{t}(10T)=\frac{12.0+11.8}{2}=11.9\ \text{s}

T = \frac{11.9}{10} = 1.19\ \text{s}

Final answer

T = 1.19 s (example)

Detailed explanation

Background Concept

The period $T$ is the time for one complete oscillation. Measuring a single period directly with a stopwatch gives a large percentage uncertainty because reaction time (starting/stopping) is a significant fraction of $T$ . A standard improvement is to time $N$ oscillations (e.g. $N=10$ or $20$ ) and then divide by $N$ :

T = \frac{t(NT)}{N}

Repeating the timing and averaging reduces random uncertainty.

Understanding the Question

You must obtain the period $T$ of the oscillations of the modelling clay sphere attached to the paper-clip chain. You are expected to describe (by your method) and then record a value of $T$ .

Approach

Displace the sphere slightly and release to start oscillations.
Use a stopwatch to time a reasonably large number of oscillations (commonly 10 or more).
Repeat the measurement and take a mean.
Divide by the number of oscillations to obtain $T$ .

Step-by-Step Reasoning

Pull the sphere a small distance and release so it oscillates with small amplitude (more repeatable period).
Choose a reference point (e.g. when the sphere passes the centre position moving towards you).
Start timing as the sphere passes the reference point, count $10$ complete oscillations, and stop timing when it again passes the reference point in the same direction.
Repeat at least once more.
Average the measured total times and divide by $10$ to get $T$ .

Example:

T = \frac{\overline{t}(10T)}{10}

Key Takeaways

Timing multiple oscillations reduces the effect of reaction time.
Repeats and a mean improve reliability.

Common Mistakes

Timing only one oscillation (very large percentage uncertainty).
Not defining a consistent start/stop point (e.g. starting at centre but stopping at an end).
Counting half-oscillations instead of full oscillations.

Things to Be Careful About

Damping may cause amplitude to decrease; try to time early oscillations consistently.
Large amplitudes can change the period slightly; keep the initial displacement small.
Ensure the chain is oscillating in one plane and not twisting, otherwise the motion becomes irregular.

Techniques used

measure the period by timing multiple oscillationsrepeat timing measurements and calculate a meandivide total time by number of oscillations to obtain a period

(d)

Roll the two spheres into one larger sphere.

Measure and record the diameter $d$ of the larger sphere.

$d$ = ______

Repeat (b) and (c) with a value $x$ of approximately 80 cm and with the hook placed so that $n$ is 7.

$x$ = ______ $\text{cm}$
$N$ = ______
$T$ = ______

3M Similar

DifficultyMedium

Worked solution

Example measurements (student-dependent):

Larger sphere diameter:

d = 3.00\ \text{cm}

With $x \approx 80\ \text{cm}$ :

x = 80.0\ \text{cm}

With $n=7$ :

N = \sqrt[3]{n^2} = (7^2)^{1/3} = 49^{1/3} = 3.66

(to $3\ \text{s.f.}$ )

Period by timing $10$ oscillations and averaging, e.g.

\overline{t}(10T) = 10.1\ \text{s} \Rightarrow T = \frac{10.1}{10} = 1.01\ \text{s}

Final answer

Example: d = 3.00 cm, x = 80.0 cm, N = 3.66, T = 1.01 s

Detailed explanation

Background Concept

This part extends the experiment by changing the mass (by combining the clay spheres) and changing the geometry ( $x$ and $n$ ). Good practical work means keeping the measurement method consistent between the two runs so that comparisons (and later calculation of $k$ ) are meaningful.

Understanding the Question

You must:

make a larger sphere (combining the two smaller spheres), then measure its diameter $d$ ;
repeat the set-up with $x \approx 80\ \text{cm}$ ;
place the hook so $n=7$ and calculate $N = \sqrt[3]{n^2}$ ;
measure the new period $T$ using the same timing method as in (c).

Approach

Follow the same sequence each time:

Measure $d$ carefully.
Adjust stands to achieve the required $x$ and measure it.
Set $n$ by counting clips; compute $N$ .
Time multiple oscillations, repeat, average, and calculate $T$ .

Step-by-Step Reasoning

Diameter of the larger sphere: measure with ruler, record with unit and suitable precision.
Rod separation $x$ : measure centre-to-centre as before (consistency matters).
Compute $N$ for $n=7$ :

N = \sqrt[3]{7^2} = 49^{1/3} \approx 3.659... \Rightarrow 3.66\ (3\ \text{s.f.})

Period $T$ : time (e.g.) $10$ oscillations at least twice and average:

T = \frac{\overline{t}(10T)}{10}

Key Takeaways

Consistency between runs reduces systematic differences.
Derived quantities should be calculated to the requested significant figures.

Common Mistakes

Forgetting to change $x$ to approximately $80\ \text{cm}$ .
Miscounting $n$ (counting links incorrectly).
Timing method changes between runs, making $T$ values not comparable.

Things to Be Careful About

When rolling the larger sphere, it may not be perfectly spherical; measure the widest diameter without squashing it.
Keep the oscillation amplitude small in both runs so that any comparison is fair.
Record $x$ and $d$ with sensible precision (not more precise than the instrument).

Techniques used

measure a diameter and record to appropriate precisionmeasure a length x and record with unitscalculate a derived quantity from a given formuladetermine a period by timing multiple oscillations and averaging

(e)

It is suggested that the relationship between $T$ , $d$ , $N$ and $x$ is

T^2 = \frac{kdN}{x^2}

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

Worked solution

From

T^2 = \frac{kdN}{x^2}

k = \frac{T^2 x^2}{dN}

Example (convert $x,d$ to $\text{m}$ ):

First set: $T=1.19\ \text{s}$ , $x=70.0\ \text{cm}=0.700\ \text{m}$ , $d=2.40\ \text{cm}=0.0240\ \text{m}$ , $N=4.95$

k_1 = \frac{(1.19)^2(0.700)^2}{(0.0240)(4.95)} = 5.84\ \text{s}^2\ \text{m} \approx 5.8\ \text{s}^2\ \text{m}

Second set: $T=1.01\ \text{s}$ , $x=80.0\ \text{cm}=0.800\ \text{m}$ , $d=3.00\ \text{cm}=0.0300\ \text{m}$ , $N=3.66$

k_2 = \frac{(1.01)^2(0.800)^2}{(0.0300)(3.66)} = 5.94\ \text{s}^2\ \text{m} \approx 5.9\ \text{s}^2\ \text{m}

Final answer

Example: k1 = 5.8 s^2 m, k2 = 5.9 s^2 m

Detailed explanation

Background Concept

When an experiment proposes a relationship with an unknown constant $k$ , you test it by rearranging the equation to calculate $k$ from each set of measurements. If the relationship is correct and the uncertainties are reasonable, the calculated values of $k$ should agree within experimental uncertainty.

Given

T^2 = \frac{kdN}{x^2}

rearrange to

k = \frac{T^2 x^2}{dN}

Consistency of units is essential (typically SI: $x$ and $d$ in metres).

Understanding the Question

You have two runs of the experiment (small sphere with $n=11$ , and larger sphere with $n=7$ ). Using your measured values of $T$ , $d$ , $N$ and $x$ for each run, you must calculate two values of $k$ .

Approach

For each run:

Convert lengths to consistent units.
Substitute into $k = \dfrac{T^2 x^2}{dN}$ .
Compute $k$ and record it with appropriate significant figures (addressed in part (ii)).

Step-by-Step Reasoning

Start from the given model and rearrange to make $k$ the subject.
Use the measured $T$ and square it.
Convert $x$ and $d$ to metres (or keep all lengths in cm, but then $k$ will have different units; SI is usually preferred).
Calculate $x^2$ .
Divide by $dN$ .
Repeat for the second set.

If your two $k$ values are close, it suggests the proposed relationship is reasonable.

Key Takeaways

Rearranging to calculate a constant is a standard way to test a relationship.
Unit consistency matters: changing cm to m changes the numerical value (and unit) of $k$ .

Common Mistakes

Using $k = \dfrac{T^2 dN}{x^2}$ (incorrect rearrangement).
Forgetting to square $T$ or $x$ .
Mixing units (e.g. $x$ in cm and $d$ in m).

Things to Be Careful About

$N$ is dimensionless; only $T$ , $x$ , and $d$ carry units.
Because the experiment is relatively uncertain, quoting an over-precise $k$ is not justified (handled explicitly in part (ii)).

Techniques used

rearrange an equation to make the constant the subjectconvert measured quantities into consistent unitssubstitute experimental values to calculate a constant

(ii)

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

1M Similar

DifficultyMedium-Easy

Worked solution

Since the uncertainty in $k$ is of order $15\%$ , $k$ should be quoted to about $2\ \text{s.f.}$ .

(For example, $k \approx 5.8\ \text{s}^2\ \text{m}$ and $5.9\ \text{s}^2\ \text{m}$ .)

Final answer

Quote k to 2 s.f. (uncertainty ≈ 15%)

Detailed explanation

Background Concept

Significant figures should reflect the precision of the data. A useful rule is: if a quantity has a percentage uncertainty of about $10\%$ to $20\%$ , quoting more than $2\ \text{s.f.}$ is usually unjustified because the third significant figure is not meaningful.

Understanding the Question

You must explain why you gave your particular number of significant figures for $k$ . This requires linking your reported precision to the uncertainties in your measurements (and/or the stated overall uncertainty in $k$ ).

Approach

Use the dominant uncertainty argument:

either refer to the stated $15\%$ uncertainty in $k$ ,
or refer to the least precisely measured quantities (often timing and the centre-to-centre measurement of $x$ ).
Then state that $2\ \text{s.f.}$ (or at most $3\ \text{s.f.}$ if your uncertainties are smaller) is appropriate.

Step-by-Step Reasoning

A $15\%$ uncertainty means the true value could plausibly differ by roughly $\pm 0.15k$ .
For $k \approx 6$ , that is about $\pm 0.9$ .
Quoting $k$ as $5.84$ would suggest uncertainty of order $\pm 0.01$ , which is far smaller than the experimental uncertainty.
Therefore, $k$ should be quoted to about $2\ \text{s.f.}$ (e.g. $5.8$ or $5.9$ ).

Key Takeaways

Significant figures must be consistent with experimental uncertainty.
Larger uncertainties justify fewer significant figures.

Common Mistakes

Quoting many digits from a calculator without justification.
Justifying sig figs only by “what the calculator shows”.

Things to Be Careful About

If you used different units (cm vs m), your numerical value of $k$ changes, but the reasoning about significant figures does not: it still depends on the relative (percentage) uncertainty.

Techniques used

identify which measured quantities limit precisionjustify rounding based on uncertainty or least significant figureslink significant figures to percentage uncertainty

(f)

It is suggested that the percentage uncertainty in the values of $k$ is 15%.

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

1M Similar

DifficultyMedium

Worked solution

Using $15\%$ uncertainty, compare the two $k$ values.

Example:

\%\ \text{difference} = \frac{|k_1-k_2|}{\frac{1}{2}(k_1+k_2)}\times 100

With $k_1=5.8$ and $k_2=5.9$ :

\%\ \text{difference} = \frac{0.1}{5.85}\times 100 \approx 1.7\% < 15\%

So the results support the relationship.

Final answer

Yes — k values agree within 15%, so relationship is supported.

Detailed explanation

Background Concept

To decide whether results support a proposed relationship, you compare outcomes from different runs. If the model is correct, the calculated constant $k$ should be consistent. With an estimated percentage uncertainty (here $15\%$ ), you check whether the difference between values is small enough to be explained by that uncertainty.

A common comparison is percentage difference:

\%\ \text{difference} = \frac{|k_1-k_2|}{\text{mean of }k_1,k_2}\times 100

Understanding the Question

You are told the percentage uncertainty in $k$ is $15\%$ . You must use this to judge whether your two calculated values of $k$ are consistent, and therefore whether your data supports the suggested relationship.

Approach

Compute the percentage difference between $k_1$ and $k_2$ .
If the percentage difference is less than (or comparable to) $15\%$ , conclude the results support the relationship; otherwise they do not.

Step-by-Step Reasoning

Calculate the absolute difference $|k_1-k_2|$ .
Divide by the mean value $\frac{1}{2}(k_1+k_2)$ .
Convert to a percentage.
Compare to $15\%$ .

If, for example, the difference is only a few percent, it is well within $15\%$ , so the two values are consistent.

Key Takeaways

A relationship is supported when repeated/alternative runs give a consistent constant within uncertainty.
Always use the stated uncertainty to justify your conclusion.

Common Mistakes

Saying “they are close” without quantifying.
Comparing using absolute difference only (without considering the scale of $k$ ).

Things to Be Careful About

If your $k$ values differ by slightly more than $15\%$ , a cautious conclusion is that results do not clearly support (or are inconclusive) given the uncertainty estimate.
Ensure you use the same units for both $k$ values before comparing.

Techniques used

compare two experimental values using a percentage differencejudge agreement against a stated percentage uncertaintystate a conclusion about support for a proposed relationship

(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

Uncertainty in $T$ : reaction time when starting/stopping the stopwatch and judging the reference point as the sphere passes.
Uncertainty in $x$ : difficult to measure centre-to-centre of rods; rod thickness and alignment cause reading uncertainty.
Uncertainty in $d$ : modelling clay sphere not perfectly spherical / may deform when measured, so diameter varies with direction.
Limitation affecting $T$ : oscillations damped and/or motion not in a single plane (twisting of chain), making the period harder to time consistently.

Final answer

See working (four uncertainties/limitations listed).

Detailed explanation

Background Concept

An uncertainty is a quantified (or described) doubt in a measurement, often random (scatter between repeats) or systematic (a consistent bias). A limitation is a feature of the method that prevents ideal conditions (e.g. damping, poor control of variables). Good answers name the quantity affected and explain the physical reason.

Understanding the Question

You must give four sources of uncertainty or limitations for this experiment. For measurement uncertainties, you must state the quantity being measured and why it is uncertain.

Approach

Scan the experiment for:

measurements: $d$ , $x$ , $T$ , and the count $n$ ;
procedural limitations: damping, non-planar motion, inconsistent amplitude, changing support conditions.
State four distinct points, each with a clear link to the quantity and a reason.

Step-by-Step Reasoning

Examples of creditworthy points:

Timing uncertainty in $T$ : starting/stopping a stopwatch depends on human reaction time and judgement of the exact moment the sphere passes a reference point.
Distance uncertainty in $x$ : $x$ is centre-to-centre; rod thickness makes the centre hard to locate, and the ruler may not be perfectly aligned horizontally.
Diameter uncertainty in $d$ : modelling clay can deform and may not be a true sphere; different measured diameters are possible depending on orientation/pressure.
Motion limitation affecting $T$ : damping reduces amplitude and the chain may twist, so the oscillation is not perfectly simple harmonic in one plane; this reduces repeatability of timing.

Other acceptable limitations could include: rods not exactly the same height causing asymmetry; amplitude not kept small; difficulty ensuring the hook is placed consistently at the same link.

Key Takeaways

Always state the quantity and the reason.
Provide distinct, non-overlapping sources.

Common Mistakes

Writing vague phrases like “human error” with no quantity or mechanism.
Repeating the same idea (e.g. stopwatch reaction time) in different words for multiple points.
Listing improvements instead of limitations.

Things to Be Careful About

“Uncertainty in $n$ ” is usually weak because $n$ is a count; only creditable if you explain ambiguity in what counts as “between hook and end”.
Avoid claiming impossible precision (e.g. timing to $0.001\ \text{s}$ with a manual stopwatch).

Techniques used

identify random uncertainties in timing measurementsidentify systematic uncertainties in geometry measurementslink each uncertainty to a specific measured quantitystate limitations that affect repeatability or validity

(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

Measure $T$ more accurately: use a motion sensor/video analysis/light gate (or time a larger number of oscillations) to reduce reaction-time error.
Improve measurement of $x$ : mark rod centres and use a set square/tape measure aligned horizontally; keep stands fixed (e.g. clamp bases) to prevent movement.
Improve measurement of $d$ : use vernier calipers and measure diameter in several orientations, then average.
Improve the oscillations: keep amplitude small and add a fiducial marker so the same reference point is used; ensure the chain oscillates in one plane (prevent twisting).

Final answer

See working (four improvements listed).

Detailed explanation

Background Concept

Improvements should directly address identified uncertainties/limitations by either:

reducing random uncertainty (e.g. better timing method, more repeats),
reducing systematic errors (e.g. better alignment and definitions of what is measured),
improving control of variables (e.g. consistent amplitude, preventing twisting).

Understanding the Question

You must describe four improvements. They can involve better apparatus or better procedure. Each must be realistic in a school/college lab and clearly linked to how it improves the data quality.

Approach

Take the main problem areas ( $T$ , $x$ , $d$ , and the nature of the oscillation) and propose one strong improvement for each. Ensure you give four distinct ideas.

Step-by-Step Reasoning

Examples:

Reduce timing uncertainty in $T$ : use video analysis (frame-by-frame) or a motion sensor/data logger to obtain period without manual reaction time; alternatively, time $20$ oscillations and repeat several times.
Improve $x$ measurement: define the rod centre positions (e.g. mark them), measure with a tape measure kept horizontal using a set square, and secure stand bases so they do not slip during measurements.
Improve $d$ measurement: use vernier calipers and measure in several directions to account for non-sphericity; average to reduce random variation.
Improve oscillation quality: keep displacement small, use a clear fiducial marker and a fixed reference line, and prevent twisting so motion stays in one plane (more repeatable period).

Key Takeaways

Good improvements are specific and linked to a particular limitation.
Better instruments and more repeats typically reduce uncertainty.

Common Mistakes

Suggesting vague improvements like “be more careful”.
Repeating the same improvement idea (e.g. “repeat more” in multiple ways).
Suggesting unrealistic equipment without explaining how it would be used.

Things to Be Careful About

Improvements should not change the intended physics model (e.g. making large amplitudes to make timing easier can change the period).
If you propose new apparatus (e.g. motion sensor), state what it measures and how you would extract $T$ from the recorded data.

Techniques used

propose procedural changes that reduce timing uncertaintysuggest improved instruments for length measurementimprove control of variables and alignmentincrease reliability by repeating and averaging