In this experiment, you will investigate the movement of a ball. (a) You are provided with two table tennis balls A and B, each attached to a string. The mass of A and its string is $m$. Use the balance to measure $m$. $m$ = ______ $\text{g}$ (i) ● Attach the clamp to the stand using the boss. ● Place both strings between the two wooden strips and secure the wooden strips in the clamp, as shown in Fig. 2.1. ● Adjust the apparatus so that the bottoms of the two wooden strips…

(a) m = 3.80 g (example; student-dependent) · (b)(i) b = 3.5 cm, d = 7.5 cm, \sqrt{d-b} = 2.00 cm^{1/2} (example; student-dependent) · (b)(ii) Quote \sqrt{d-b} to sig. figs. consistent with b and d (e.g. 2 s.f. if b and d are 0.1 cm). · (b)(iii) H = 6.5 cm (example; student-dependent) · (b)(iv) 8% (example, using \Delta H = 0.5 cm and H = 6.5 cm) · (b)(v) M = 3.80 g, \sqrt{H-b} = 1.73 cm^{1/2} (example; student-dependent) · (c) H = 9.8 cm, M = 13.8 g, \sqrt{H-b} = 2.51…

Physics 9702/33 — October/November 2024

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 determine the resistivity of a metal.

(a)

● Set up the circuit shown in Fig. 1.1.

Fig. 1.1

● The distance between the two crocodile clips is $L$ . The reading on the voltmeter is $V$ .

Adjust the position of crocodile clip F so that $L$ is approximately $45\text{ cm}$ .

● Close the switch.

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

$L$ = ______ $\text{cm}$
$V$ = ______ $\text{V}$

● Open the switch.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Example of acceptable readings (student-dependent):

$L = 45.0\ \text{cm}$

$V = 0.56\ \text{V}$

Final answer

Example: L = 45.0 cm, V = 0.56 V

Detailed explanation

Background Concept

The experiment measures the potential difference $V$ across a chosen length $L$ of a metal wire. A voltmeter must be connected in parallel with the section of wire being measured. The length $L$ is set by moving a crocodile clip along the wire on a metre rule.

Understanding the Question

You are instructed to set the crocodile clip so that the distance between the fixed end and clip F is about $45\ \text{cm}$ , then close the switch and record:

the length $L$ (from the metre rule), and
the voltmeter reading $V$ across that length.

Approach

Set up the circuit exactly as shown, ensuring the voltmeter is across the wire segment.
Move clip F until $L \approx 45\ \text{cm}$ .
Close the switch briefly, wait for a steady reading, then record $L$ and $V$ .
Open the switch to reduce heating of the wire.

Step-by-Step Reasoning

Measure $L$ along the metre rule between the fixed contact and the position of clip F.
Record $L$ to the nearest mm (typical: $0.1\ \text{cm}$ ) because a metre rule has 1 mm divisions.
With the switch closed, read $V$ directly from the voltmeter scale/display and record to the voltmeter resolution.
Open the switch after taking the reading to minimise temperature rise (resistance increases with temperature, affecting $V$ ).

Key Takeaways

The voltmeter must be in parallel with the wire section whose p.d. is measured.
Length readings should match the precision of the ruler scale.
Avoid heating by keeping the switch closed only when taking readings.

Common Mistakes

Measuring $L$ from the wrong end of the wire or not from the fixed contact.
Leaving the switch closed for a long time so the wire warms up and readings drift.
Recording $L$ without a unit or with unrealistic precision (e.g. $45.000\ \text{cm}$ ).

Things to Be Careful About

Avoid parallax when reading the metre rule.
Ensure crocodile clips make good electrical contact; a loose contact gives unstable $V$ .
Record the actual $L$ you set (it only needs to be approximately $45\ \text{cm}$ , not exactly).

Techniques used

adjust the position of a sliding contact to set a required lengthread a length from a metre rule with appropriate precisionmeasure a potential difference using a voltmeter connected in parallelrecord readings with consistent units and appropriate significant figures

(b)

Vary $L$ in the range $15.0\text{ cm} \leq L \leq 95.0\text{ cm}$ by adjusting the position of F. For each position of F, measure $L$ and $V$ . Repeat until you have six sets of values of $L$ and $V$ .

Record your results in a table. Include values of $\frac{1}{L}$ and $\frac{1}{V}$ in your table.

8M Similar

DifficultyMedium

Worked solution

Answer

A single table with headings including units, and six sets of readings spanning $15.0$ to $95.0\ \text{cm}$ , plus calculated $1/L$ and $1/V$ .

Example (student-dependent):

$L\ /\ \text{cm}$	$V\ /\ \text{V}$	$\left(\frac{1}{L}\right)\ /\ \text{cm}^{-1}$	$\left(\frac{1}{V}\right)\ /\ \text{V}^{-1}$
$15.0$	$0.25$	$0.0667$	$4.00$
$30.0$	$0.43$	$0.0333$	$2.34$
$45.0$	$0.56$	$0.0222$	$1.78$
$60.0$	$0.67$	$0.0167$	$1.50$
$75.0$	$0.75$	$0.0133$	$1.34$
$95.0$	$0.84$	$0.0105$	$1.20$

Final answer

Table of six L and V values (15.0–95.0 cm) with calculated 1/L and 1/V, correctly headed with units.

Detailed explanation

Background Concept

To test a relationship experimentally, you need:

a suitable spread of the independent variable (here $L$ ),
repeated measurements of the dependent variable (here $V$ ), and
clear recording of raw and calculated values.

Because later parts ask for a graph of $1/V$ against $1/L$ , you must calculate the reciprocal columns carefully and present them with correct units.

Understanding the Question

You must vary the clip position so that $L$ covers the range $15.0\ \text{cm}$ to $95.0\ \text{cm}$ and record six pairs $(L, V)$ . You must then add two derived columns $1/L$ and $1/V$ .

Approach

Choose six values of $L$ spaced across the range (e.g. roughly evenly).
For each $L$ , close the switch briefly, read $V$ , then open the switch.
Record all values in one table with units in headings.
Compute $1/L$ and $1/V$ for each row, keeping sensible significant figures.

Step-by-Step Reasoning

Select $L$ values across the whole interval: using only a narrow range makes the graph unreliable.
For each $L$ $L$ :
- set the clip position,
- measure $L$ (typically to $0.1\ \text{cm}$ ),
- close the switch, allow the reading to settle, record $V$ , then open the switch.
Construct a table:
- each heading should show quantity and unit, e.g. $L/\text{cm}$ , $V/\text{V}$ ,
- derived headings must include units: $(1/L)/\text{cm}^{-1}$ and $(1/V)/\text{V}^{-1}$ ,
- keep consistent decimal places in a column (especially for $L$ ).
Compute reciprocals using a calculator:
- e.g. if $L = 45.0\ \text{cm}$ , then $1/L = 1/45.0 = 0.0222\ \text{cm}^{-1}$ ,
- if $V = 0.56\ \text{V}$ , then $1/V = 1/0.56 = 1.79\ \text{V}^{-1}$ (to 3 s.f. if appropriate).

Key Takeaways

Use a wide range and enough readings (six) for a reliable straight-line test.
Put units in the table headings, not repeatedly in the body.
Derived quantities need correct units and sensible rounding.

Common Mistakes

Fewer than six sets of readings.
Using $L$ values outside the required range.
Missing units or incorrect units for $1/L$ and $1/V$ .
Inconsistent rounding (e.g. mixed 2 d.p. and 3 d.p. within one column).

Things to Be Careful About

If the wire heats up, $V$ for the same $L$ may drift; minimise by opening the switch between readings.
Ensure you calculate $1/L$ using the recorded $L$ value (not the intended value).
Do not write $1/L$ as $\text{cm}$ ; it must be $\text{cm}^{-1}$ .

Techniques used

vary the independent variable over a stated rangecollect multiple sets of paired readingscalculate reciprocal quantities for linearisationpresent readings in a single table with correct headings and unitsuse consistent decimal places within each column

(c)

(i)

Plot a graph of $\frac{1}{V}$ on the $y$ -axis against $\frac{1}{L}$ on the $x$ -axis.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Plot $y = 1/V\ (\text{V}^{-1})$ against $x = 1/L\ (\text{cm}^{-1})$ using a suitable scale (at least half the grid in each direction) and plot all six points accurately as small crosses.

graph-axes

Final answer

Graph of 1/V (V^-1) vs 1/L (cm^-1) plotted with suitable scales and all points.

Detailed explanation

Background Concept

A graph is used to test whether two variables have a linear relationship. When the question specifies plotting $1/V$ against $1/L$ , it is telling you to use a linearised form so the data should lie close to a straight line.

Good graph technique is assessed by:

correct axis labels (quantity and unit),
sensible scales,
accurate plotting.

Understanding the Question

You have a table of six values of $L$ and $V$ and their reciprocals. You must plot:

vertical axis: $1/V$ in $\text{V}^{-1}$ ,
horizontal axis: $1/L$ in $\text{cm}^{-1}$ .

Approach

Decide the ranges of $1/L$ and $1/V$ from your table.
Choose scales so the plotted points fill most of the graph paper.
Label axes with both quantity and unit.
Plot each point as a small cross centered on the coordinate.

Step-by-Step Reasoning

From typical data, $1/L$ might range from about $0.010\ \text{cm}^{-1}$ to $0.067\ \text{cm}^{-1}$ .
$1/V$ might range from about $1.2\ \text{V}^{-1}$ to $4.0\ \text{V}^{-1}$ .
Pick scales such as:
- $x$ -axis: 0.010 to 0.070 with convenient divisions,
- $y$ -axis: 1.0 to 4.2 with convenient divisions.
Plot all points carefully; accuracy matters because the next part uses the best-fit line to find gradient/intercept.

graph-axes

Key Takeaways

Axes must be labelled as $1/L\ /\ \text{cm}^{-1}$ and $1/V\ /\ \text{V}^{-1}$ .
Use a scale that spreads points out for a more accurate gradient.

Common Mistakes

Swapping axes (plotting $1/L$ on $y$ ).
Missing units or writing units incorrectly (e.g. $\text{V}$ instead of $\text{V}^{-1}$ ).
Using a cramped scale so all points bunch together.

Things to Be Careful About

Do not force the graph to go through the origin unless the data demands it.
Plotting should be to within about half a small square.
Use crosses, not large dots, so the best-fit line can be judged fairly.

Techniques used

choose appropriate axis scales that use most of the gridlabel axes with quantity and unitplot data points accurately using a sharp crossuse calculated reciprocal values as plotted variables

(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

Using two well-separated points on the best-fit line,

\text{gradient} = \frac{\Delta (1/V)}{\Delta (1/L)}

Example values from a suitable best-fit line:

\text{gradient} \approx 50\ \text{cm V}^{-1}

$y$ -intercept at $1/L = 0$ :

\text{$y$-intercept} \approx 0.67\ \text{V}^{-1}

Answer

$\text{gradient} = 50\ \text{cm V}^{-1}$

$y$ -intercept $= 0.67\ \text{V}^{-1}$

Final answer

gradient = 50 cm V^-1, y-intercept = 0.67 V^-1

Detailed explanation

Background Concept

For a straight-line graph,

y = mx + c

the gradient $m$ is

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

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

Here, $y$ is $1/V$ and $x$ is $1/L$ , so:

gradient has units $\text{(V}^{-1})/(\text{cm}^{-1}) = \text{cm V}^{-1}$ ,
intercept has units $\text{V}^{-1}$ .

Understanding the Question

You must obtain two numerical values from your best-fit line:

the gradient,
the $y$ -intercept.

These will be used directly in part (d).

Approach

Choose two points on the drawn best-fit line that are far apart (to reduce percentage reading error).
Read their coordinates $(x_1, y_1)$ and $(x_2, y_2)$ .
Compute gradient $m = (y_2 - y_1)/(x_2 - x_1)$ .
Find the intercept by reading where the line crosses the $y$ -axis (at $x=0$ ).

Step-by-Step Reasoning

Pick points on the line (not the experimental crosses). A large separation in $x$ makes $\Delta x$ large and reduces the effect of small reading errors.
Calculate:

\text{gradient} = \frac{\Delta (1/V)}{\Delta (1/L)}

For the intercept, extend the line to meet the $y$ -axis at $1/L = 0$ and read $1/V$ there.
Quote both with appropriate units: $\text{cm V}^{-1}$ for gradient and $\text{V}^{-1}$ for intercept.

Key Takeaways

Always use two points on the best-fit line, not two measured points.
Gradient is $\Delta y/\Delta x$ , never $\Delta x/\Delta y$ .
Units come from the axes.

Common Mistakes

Using points that are too close together, giving an inaccurate gradient.
Forgetting that the $x$ -axis is $1/L$ , not $L$ .
Writing the intercept with the wrong unit (e.g. $\text{V}$ instead of $\text{V}^{-1}$ ).

Things to Be Careful About

Read coordinates carefully from the scale you chose.
If your graph uses $L$ in cm, your gradient will be in $\text{cm V}^{-1}$ (conversion to SI may be needed later).

Techniques used

read two well-separated points on the best-fit linecalculate gradient using \Delta y / \Delta xdetermine the y-intercept by extrapolation to x = 0quote gradient and intercept with appropriate units

(d)

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

\frac{1}{V} = \frac{J}{L} + W

where $J$ and $W$ are constants.

Using your answers in (c)(iii), determine the values of $J$ and $W$ . Give appropriate units.

$J$ = ______
$W$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Given

\frac{1}{V} = \frac{J}{L} + W

Plot is $y = 1/V$ against $x = 1/L$ , so

y = Jx + W

Hence

J = \text{gradient}, \quad W = y\text{-intercept}

Answer

$J = 50\ \text{cm V}^{-1}$

$W = 0.67\ \text{V}^{-1}$

Final answer

J = 50 cm V^-1, W = 0.67 V^-1

Detailed explanation

Background Concept

If experimental data fits a straight line, you can compare the equation to the standard linear form:

y = mx + c

Here,

\frac{1}{V} = \frac{J}{L} + W

If you define $y = 1/V$ and $x = 1/L$ , then

y = Jx + W

So the gradient is $J$ and the intercept is $W$ .

Understanding the Question

You have already found the gradient and $y$ -intercept from your graph of $1/V$ against $1/L$ . Now you must use those to state numerical values for the constants $J$ and $W$ , including units.

Approach

Identify $y$ and $x$ from the axes.
Compare the given equation with $y = mx + c$ .
Assign $J$ to the gradient and $W$ to the intercept.
Use axis units to give units for $J$ and $W$ .

Step-by-Step Reasoning

Since the graph is $1/V$ (units $\text{V}^{-1}$ ) vs $1/L$ (units $\text{cm}^{-1}$ ), the gradient unit is:

\frac{\text{V}^{-1}}{\text{cm}^{-1}} = \text{cm V}^{-1}

The intercept is read from the $y$ -axis, so it has units $\text{V}^{-1}$ .
Therefore, whatever numbers you obtained in (c)(iii) are directly $J$ and $W$ .

Key Takeaways

Linearisation means constants can often be read directly as gradient/intercept.
Units for gradient/intercept come directly from the graph axes.

Common Mistakes

Using $L$ instead of $1/L$ when matching to $y = mx + c$ .
Swapping $J$ and $W$ .
Giving $J$ the wrong unit (e.g. $\text{V}^{-1}$ instead of $\text{cm V}^{-1}$ ).

Things to Be Careful About

If you later need SI units, convert $J$ from $\text{cm V}^{-1}$ to $\text{m V}^{-1}$ by dividing by $100$ .

Techniques used

match a linear equation to y = mx + cidentify gradient and intercept from the graphassign units to constants from dimensional consistency

(e)

(i)

Use a micrometer to measure the diameter $d$ of the wire.

$d$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Example (student-dependent), micrometer reading to $0.01\ \text{mm}$ with repeats:

$d = 0.32\ \text{mm}$

micrometer

Final answer

Example: d = 0.32 mm

Detailed explanation

Background Concept

The wire diameter $d$ is needed to find its cross-sectional area $A$ :

A = \frac{\pi d^2}{4}

Because $A$ depends on $d^2$ , any percentage uncertainty in $d$ is doubled in $d^2$ , so careful measurement of $d$ is important.

A micrometer screw gauge typically reads to $0.01\ \text{mm}$ .

Understanding the Question

You must measure and record the diameter of the wire using a micrometer. Two marks typically require:

correct precision (e.g. $0.01\ \text{mm}$ ), and
good technique such as repeats and/or zero-error check.

Approach

Check the micrometer for zero error.
Measure the wire diameter by gently closing the jaws using the ratchet.
Repeat at several positions and average.
Record $d$ with unit.

Step-by-Step Reasoning

Close the micrometer gently until the ratchet clicks; this applies a consistent force.
If the micrometer reads a non-zero value when fully closed, apply a zero correction.
Measure the diameter at different points along the wire and at different orientations (the wire may not be perfectly circular).
Average the readings and record as $d$ .

micrometer

Key Takeaways

Use the ratchet to avoid compressing the wire.
Repeat measurements and average for reliability.
Record to the instrument resolution and include the unit.

Common Mistakes

Not using the ratchet, leading to inconsistent pressure.
Forgetting to correct for zero error.
Recording $d$ with too many decimal places or without units.

Things to Be Careful About

Ensure the wire is perpendicular to the anvil/spindle faces.
Do not measure over a kinked or flattened section of wire.
Keep track of units: $0.32\ \text{mm} = 3.2 \times 10^{-4}\ \text{m}$ .

Techniques used

use a micrometer screw gauge correctly (ratchet, alignment)check and correct for zero errortake repeated diameter measurements at different points and averageconvert between mm and m correctly

(ii)

Theory suggests that

J = \frac{\pi d^2 WR}{4\rho}

where $R$ is $33\ \Omega$ and $\rho$ is the resistivity of the metal of the wire.

Using your answers in (d) and (e)(i), determine a value for $\rho$ .

$\rho$ = ______ $\Omega\text{m}$

1M Similar

DifficultyMedium

Worked solution

Working

Given

J = \frac{\pi d^2 W R}{4\rho}

Rearrange:

\rho = \frac{\pi d^2 W R}{4J}

Convert $J$ to SI: $J = 50\ \text{cm V}^{-1} = 0.50\ \text{m V}^{-1}$ .

Convert $d$ : $d = 0.32\ \text{mm} = 3.2 \times 10^{-4}\ \text{m}$ .

Substitute $W = 0.67\ \text{V}^{-1}$ and $R = 33\ \Omega$ :

\rho = \frac{\pi (3.2 \times 10^{-4})^2 (0.67)(33)}{4(0.50)}

\rho = 3.6 \times 10^{-6}\ \Omega\text{ m}

Answer

$\rho = 3.6 \times 10^{-6}\ \Omega\text{ m}$

Final answer

3.6 × 10^-6 Ω m

Detailed explanation

Background Concept

Resistivity $\rho$ is a material property linking resistance to dimensions:

R = \rho\frac{L}{A}

This practical uses a graph to obtain constants ( $J$ and $W$ ), then combines them with the wire diameter to determine $\rho$ via the given theoretical relationship:

J = \frac{\pi d^2 W R}{4\rho}

Understanding the Question

You are given:

$R = 33\ \Omega$ ,
your measured $d$ from (e)(i),
your $J$ and $W$ from (d).

You must calculate $\rho$ in $\Omega\text{ m}$ . This requires SI units, so any values in cm or mm must be converted to m.

Approach

Rearrange the given equation to make $\rho$ the subject.
Convert $d$ to metres and ensure $J$ corresponds to $L$ in metres.
Substitute values with units.
Quote $\rho$ with appropriate significant figures and unit $\Omega\text{ m}$ .

Step-by-Step Reasoning

Start with:

J = \frac{\pi d^2 W R}{4\rho}

Multiply both sides by $\rho$ and divide by $J$ :

\rho = \frac{\pi d^2 W R}{4J}

Unit consistency:

If your graph used $L$ in cm, then $J$ came out in $\text{cm V}^{-1}$ . The theoretical equation expects SI (metres), so convert:

J(\text{m V}^{-1}) = \frac{J(\text{cm V}^{-1})}{100}

Convert diameter from mm to m:

d(\text{m}) = d(\text{mm}) \times 10^{-3}

Then substitute:

compute $d^2$ carefully (it strongly affects the answer),
keep track of powers of ten,
ensure final unit is $\Omega\text{ m}$ .

Key Takeaways

Practical calculations often depend critically on unit conversions.
When a graph uses non-SI units, the gradient/intercept may need conversion before use in theory equations.
Squared quantities (like $d^2$ ) amplify measurement uncertainty.

Common Mistakes

Using $J$ in $\text{cm V}^{-1}$ directly without converting to $\text{m V}^{-1}$ .
Using $d$ in mm instead of m.
Rearranging incorrectly (e.g. putting $J$ in the numerator).
Forgetting the unit $\Omega\text{ m}$ .

Things to Be Careful About

Convert $J$ only if the theoretical equation assumes $L$ in metres (it does here).
Keep significant figures consistent with your measured values (often 2 or 3 s.f.).
Write the unit as $\Omega\text{ m}$ (not $\Omega/\text{m}$ ).

Techniques used

rearrange an equation to make the required quantity the subjectconvert measured quantities into SI unitssubstitute values consistently with unitscalculate and present an answer in standard form where appropriate

Q2MediumManipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and Evaluation

In this experiment, you will investigate the movement of a ball.

(a)

You are provided with two table tennis balls A and B, each attached to a string.

The mass of A and its string is $m$ .

Use the balance to measure $m$ .

$m$ = ______ $\text{g}$

1M Similar

DifficultyEasy

Worked solution

Measure the mass of ball A plus its string using the balance and record it to the balance resolution.

Example (typical):

m = 3.80\ \text{g}

Final answer

m = 3.80 g (example; student-dependent)

Detailed explanation

Background Concept

A balance measures mass by comparing the weight of an object with an internal calibrated reference. The key exam skills are:

zeroing (tare) the balance so the reading is not offset,
placing the whole object to be measured on the pan,
recording the mass with the correct unit and to the correct resolution (number of decimal places shown).

Understanding the Question

You are told that the mass of ball A together with its string is $m$ . You must use the provided balance to obtain a value of $m$ and write it in grams.

Approach

Ensure the balance reads $0.00\ \text{g}$ (or $0.0\ \text{g}$ depending on its resolution) with an empty pan.
Place ball A and its string on the pan without anything else touching.
Wait for the reading to stabilise.
Record the value with unit $\text{g}$ and with the same decimal places as the balance display.

Step-by-Step Reasoning

If the balance displays to $0.01\ \text{g}$ , record to $0.01\ \text{g}$ (e.g. $3.80\ \text{g}$ ).
If it displays to $0.1\ \text{g}$ , record to $0.1\ \text{g}$ (e.g. $3.8\ \text{g}$ ).
Do not round to fewer decimal places than the balance provides.

Key Takeaways

Always tare/zero first.
Record mass with the correct unit and matching the instrument resolution.

Common Mistakes

Forgetting to zero the balance.
Measuring only the ball and not the string.
Recording in $\text{kg}$ when the answer line asks for $\text{g}$ .

Things to Be Careful About

Ensure the string is fully supported by the balance pan and not touching the bench.
Avoid draughts or movement that makes the reading fluctuate.

Techniques used

zero the balance before usemeasure mass using a top-pan balancerecord a reading to the balance resolution

(b)

(i)

● Attach the clamp to the stand using the boss.

● Place both strings between the two wooden strips and secure the wooden strips in the clamp, as shown in Fig. 2.1.

Fig. 2.1

● Adjust the apparatus so that the bottoms of the two wooden strips are approximately $50\text{ cm}$ above the bench and the bottoms of the two balls are approximately $1.5\text{ cm}$ above the bench.

● Displace A and adjust the position of the block so that, when they are touching, the centre of A is level with the top of the block, as shown in Fig 2.2.

Fig. 2.2

● When A is in this position, the distance between the centre of B and the bench is $b$ , as shown in Fig. 2.2.

The distance between the top of the block and the bench is $d$ .

Measure and record $b$ and $d$ .

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

● Calculate $\sqrt{d-b}$ . Give an appropriate unit.

$\sqrt{d-b}$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Example measurements (typical):

b = 3.5\ \text{cm}

d = 7.5\ \text{cm}

\sqrt{d-b} = \sqrt{7.5-3.5} = \sqrt{4.0} = 2.00\ \text{cm}^{1/2}

Final answer

b = 3.5 cm, d = 7.5 cm, \sqrt{d-b} = 2.00 cm^{1/2} (example; student-dependent)

Detailed explanation

Background Concept

When you measure lengths with a ruler/metre rule, the key limitation is reading uncertainty (set by the smallest scale division) and parallax error if your eye is not directly in line with the mark.

Here, $b$ and $d$ are vertical distances from the bench to specific points. The quantity $d-b$ is a length, so it has unit $\text{cm}$ (if $b$ and $d$ are in $\text{cm}$ ). Taking a square root gives a derived unit:

[d-b] = \text{cm} \quad \Rightarrow \quad \left[\sqrt{d-b}\right] = \sqrt{\text{cm}} = \text{cm}^{1/2}.

Understanding the Question

You set up the apparatus so that when ball A touches the block, the centre of A is level with the top of the block.

$b$ is the height of the centre of ball B above the bench in this configuration.
$d$ is the height of the top of the block above the bench.
You must measure $b$ and $d$ , then calculate $\sqrt{d-b}$ and include an appropriate unit.

Approach

Use a ruler/metre rule to measure heights from the bench to the relevant point.
Record $b$ and $d$ in $\text{cm}$ to sensible precision (typically $0.1\ \text{cm}$ with a mm scale).
Compute $d-b$ .
Take the square root and write the correct unit $\text{cm}^{1/2}$ .

Step-by-Step Reasoning

diagram-b-i-1

Measuring $d$ : hold a metre rule vertically with its zero on the bench beside the block; read the scale at the top surface of the block.
Measuring $b$ : with B hanging, identify the centre of the ball (half its diameter above its lowest point). Measure its height above the bench. A common practical method is to measure the height of the bottom of B above the bench and add the radius of the ball.
Subtraction: compute $d-b$ using the same unit (both in cm).
Square root: apply $\sqrt{\ }$ to $d-b$ .
Unit: because you took a square root of a length, write $\text{cm}^{1/2}$ .

Key Takeaways

Keep units consistent before subtracting.
A square root changes units: $\sqrt{\text{cm}} = \text{cm}^{1/2}$ .

Common Mistakes

Using different units for $b$ and $d$ before subtracting.
Forgetting the unit or writing the unit as $\text{cm}$ instead of $\text{cm}^{1/2}$ .
Measuring to the wrong point (e.g. top of the ball rather than centre).

Things to Be Careful About

Parallax: your eye must be level with the mark being read.
Ensure the bench is the reference level for both $b$ and $d$ .
If you use the ball diameter to locate the centre, measure the diameter once and keep that value consistent.

Techniques used

take ruler readings with eye level to avoid parallaxsubtract two measured values to obtain a differenceapply a square root to a measured differenceassign units to a derived quantity

(ii)

Justify the number of significant figures that you have given for your value of $\sqrt{d-b}$ .

1M Similar

DifficultyMedium-Easy

Worked solution

If $b$ and $d$ are measured to the nearest $0.1\ \text{cm}$ , then $d-b$ is only known to about $\pm 0.1\ \text{cm}$ (or worse if both readings contribute). Therefore $\sqrt{d-b}$ should be quoted to an appropriate number of significant figures consistent with this (e.g. $2\ \text{s.f.}$ ).

Final answer

Quote \sqrt{d-b} to sig. figs. consistent with b and d (e.g. 2 s.f. if b and d are 0.1 cm).

Detailed explanation

Background Concept

Significant figures in a calculated quantity should reflect the uncertainty/precision of the measurements used.

A metre rule typically allows readings to $\pm 0.05\ \text{cm}$ (nearest mm) or you record to $0.1\ \text{cm}$ .
When you subtract two similar numbers, the absolute uncertainty in the difference can be relatively large compared with the difference.
Taking a square root does not improve measurement precision; the result should not contain unjustified extra digits.

Understanding the Question

You calculated $\sqrt{d-b}$ in (b)(i). This part asks you to explain why you wrote it to the number of significant figures you chose.

Approach

State the resolution of the ruler readings for $b$ and $d$ .
Explain that $d-b$ inherits uncertainty from both measurements.
Conclude that $\sqrt{d-b}$ should be rounded so it does not claim more precision than the measurements allow.

Step-by-Step Reasoning

Suppose $b$ and $d$ are each recorded to $0.1\ \text{cm}$ . That implies each has an uncertainty on the order of $\pm 0.05\ \text{cm}$ .
The difference $d-b$ combines both uncertainties, so its uncertainty is roughly the sum of the absolute uncertainties (order-of-magnitude reasoning is sufficient in Paper 3).
Therefore $d-b$ might be uncertain by about $\pm 0.1\ \text{cm}$ (or more), so giving $\sqrt{d-b}$ to many decimal places would be unjustified.
A sensible choice is usually 2 or 3 significant figures depending on the size of $d-b$ and the ruler precision.

Key Takeaways

Your calculated value must not imply higher precision than the measurements.
Subtraction can reduce the meaningful precision.

Common Mistakes

Quoting $\sqrt{d-b}$ to 4 or 5 significant figures because a calculator shows them.
Claiming “more s.f. is better”; it is not if it is not supported by measurement precision.

Things to Be Careful About

Be explicit: mention the measurement precision of $b$ and $d$ (e.g. nearest $0.1\ \text{cm}$ ) and connect it to the rounding of $\sqrt{d-b}$ .

Techniques used

link significant figures to instrument resolutionpropagate the effect of subtraction on precisionround a calculated quantity appropriately

(iii)

● Release A so that it moves towards B.

● When A hits B, it causes B to move.

At the maximum height of ball B, the vertical distance between the centre of B and the bench is $H$ , as shown in Fig. 2.3.

Fig. 2.3

Determine $H$ .

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

2M Similar

DifficultyMedium

Worked solution

Determine the maximum vertical height of the centre of ball B above the bench.

Example (typical):

H = 6.5\ \text{cm}

Final answer

H = 6.5 cm (example; student-dependent)

Detailed explanation

Background Concept

A pendulum bob rises to a maximum height (turning point) where its speed is momentarily zero. In experiments, the turning point is difficult to judge because it occurs briefly and the bob may oscillate.

A good measurement of a maximum height typically needs:

a fixed scale (metre rule) referenced to the bench,
a method to avoid parallax,
repeats and averaging.

Understanding the Question

After releasing A, it collides with B and B swings upward. You must find $H$ , the vertical distance from the bench to the centre of B at its maximum height.

Approach

Place a vertical scale beside the path of ball B (metre rule clamped upright).
Use a marker/pointer aligned with the centre of B to identify its highest position.
Repeat several times and take a consistent reading for $H$ .

Step-by-Step Reasoning

diagram-b-iii-1

Fix a metre rule vertically with its zero on the bench.
Watch ball B and identify the highest point of its centre (a piece of card behind the ball can help you see the centre).
Read the height at the turning point; to reduce random error, repeat the release several times and take the mean of consistent readings.

Key Takeaways

Turning-point measurements are often the largest source of uncertainty.
Repeating and averaging improves reliability.

Common Mistakes

Measuring to the bottom or top of the ball instead of the centre.
Reading the scale after the ball has already started descending.
Holding the ruler by hand so it moves between trials.

Things to Be Careful About

Keep the metre rule fixed and vertical.
Align your eye with the centre of the ball to minimise parallax.
Ensure each trial uses the same release position of A so that collisions are comparable.

Techniques used

observe a maximum height in a swinging motionmeasure a vertical height from a reference levelrepeat measurements and take a mean

(iv)

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

percentage uncertainty = ______ %

1M Similar

DifficultyMedium-Easy

Worked solution

Example (typical): take absolute uncertainty in $H$ as $\pm 0.5\ \text{cm}$ (difficulty judging maximum height).

\%\ \text{uncertainty} = \frac{0.5}{6.5}\times 100 = 7.7\% \approx 8\%

Final answer

8% (example, using \Delta H = 0.5 cm and H = 6.5 cm)

Detailed explanation

Background Concept

Percentage uncertainty expresses how large an absolute uncertainty is compared with the measured value:

\%\ \text{uncertainty in } H = \frac{\Delta H}{H} \times 100.

Here, $\Delta H$ should reflect the limiting factor in measuring $H$ . For a turning point, the dominant uncertainty is often not the ruler resolution but the difficulty of judging the exact highest position.

Understanding the Question

You measured $H$ in (b)(iii). You must estimate the percentage uncertainty in $H$ and show the calculation.

Approach

Decide a reasonable absolute uncertainty $\Delta H$ $Δ H$ .
- If the metre rule reads to $1\ \text{mm}$ , the resolution might suggest $\pm 0.05\ \text{cm}$ .
- But because $H$ is a turning-point reading, a larger uncertainty (e.g. $\pm 0.2\ \text{cm}$ to $\pm 0.5\ \text{cm}$ ) is often more realistic.
Substitute into $\Delta H/H \times 100$ .

Step-by-Step Reasoning

Choose an uncertainty that matches the real limitation (turning point). For example, if you can only judge the maximum to about half a centimetre, then $\Delta H = 0.5\ \text{cm}$ .
Compute:

\frac{\Delta H}{H} \times 100 = \frac{0.5}{6.5}\times 100 = 7.7\%.

Round appropriately (often to 1 s.f. or 2 s.f.): $\approx 8\%$ .

Key Takeaways

Always use $\% = (\text{absolute}/\text{measured})\times 100$ .
For maximum-height measurements, the dominant uncertainty is often observational.

Common Mistakes

Using the ruler resolution only (e.g. $\pm 0.1\ \text{cm}$ ) when the turning point dominates.
Using $H-b$ in the denominator when the question asks uncertainty in $H$ .
Forgetting to multiply by 100.

Things to Be Careful About

State clearly what absolute uncertainty you assumed for $H$ and why.
Use consistent units (both $H$ and $\Delta H$ in cm).

Techniques used

estimate an absolute uncertainty from measurement methodcalculate percentage uncertaintypresent uncertainty to appropriate significant figures

(v)

With this arrangement of the apparatus, the total mass of A is equal to $m$ measured in (a).

● Record the total mass $M$ of A.

$M$ = ______ $\text{g}$

● Calculate $\sqrt{H-b}$ .

$\sqrt{H-b}$ = ______

1M Similar

(c)

● Use the adhesive putty to attach the $10.0\text{ g}$ slotted mass to A.

The total mass $M$ of A is now equal to the sum of $m$ and the mass of the slotted mass.

● Displace A and adjust the position of the block so that, when they are touching, the centre of A is level with the top of the block, as shown in Fig. 2.2.

● Repeat (b)(iii) and (b)(v).

$H$ = ______ $\text{cm}$
$M$ = ______ $\text{g}$
$\sqrt{H-b}$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Attach the $10.0\ \text{g}$ mass to A, repeat the measurement of $H$ and record the new total mass $M=m+10.0\ \text{g}$ .

Example (typical):

H = 9.8\ \text{cm}

M = 3.80 + 10.0 = 13.8\ \text{g}

\sqrt{H-b} = \sqrt{9.8-3.5} = \sqrt{6.3} = 2.51\ \text{cm}^{1/2}

Final answer

H = 9.8 cm, M = 13.8 g, \sqrt{H-b} = 2.51 cm^{1/2} (example; student-dependent)

Detailed explanation

Background Concept

Changing the mass of the moving ball A changes the collision outcome, so you take a second set of readings. In practical work, you keep the geometry the same (same $b$ and same $d$ procedure) and change only one variable (the mass of A).

The new total mass is:

M = m + 10.0\ \text{g}.

Understanding the Question

You must attach a known $10.0\ \text{g}$ mass to A, then repeat what you did earlier to obtain:

a new maximum height $H$ for ball B,
the new total mass $M$ ,
the new value of $\sqrt{H-b}$ .

Approach

Secure the slotted mass firmly using putty so it does not slip during motion.
Re-adjust the block so that when A touches it, the centre of A is level with the top of the block (same condition as before).
Release A, measure $H$ at the maximum height of B.
Calculate $M$ and then $\sqrt{H-b}$ .

Step-by-Step Reasoning

Attaching the mass changes $M$ but does not change the measured $m$ of the ball + string.
You must re-set the block position because the string may stretch/shift slightly when the mass is added, changing the geometry.
Once you have a consistent $H$ value, substitute into $\sqrt{H-b}$ and include $\text{cm}^{1/2}$ .

Key Takeaways

Change one factor (mass) and repeat measurements consistently.
Re-adjusting the apparatus to the stated condition is essential for fair comparison.

Common Mistakes

Forgetting to add $10.0\ \text{g}$ to $m$ when calculating $M$ .
Allowing the attached mass to be off-centre so A twists, giving inconsistent collisions.
Not re-checking the block alignment condition (centre of A level with block top).

Things to Be Careful About

Ensure the putty holds the mass securely throughout the swing.
Record $M$ to appropriate precision (often 0.1 g when adding 10.0 g).
Use the same $b$ value unless you have re-measured it after adjusting the setup.

Techniques used

modify apparatus by adding a known massrepeat measurements under a new conditioncalculate a derived quantity from measured values

(d)

It is suggested that the relationship between $H$ , $b$ , $m$ , $M$ and $d$ is

\frac{1}{\sqrt{H-b}} = k + \frac{m}{2M\sqrt{d-b}}

where $k$ is a constant.

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

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

1M Similar

DifficultyMedium

Worked solution

From

\frac{1}{\sqrt{H-b}} = k + \frac{m}{2M\sqrt{d-b}}

k = \frac{1}{\sqrt{H-b}} - \frac{m}{2M\sqrt{d-b}}

Using example data set 1 ( $M=m=3.80\ \text{g}$ , $\sqrt{H-b}=1.73\ \text{cm}^{1/2}$ , $\sqrt{d-b}=2.00\ \text{cm}^{1/2}$ ):

k_1 = \frac{1}{1.73} - \frac{3.80}{2\times 3.80\times 2.00} = 0.578 - 0.250 = 0.328\ \text{cm}^{-1/2}

Using example data set 2 ( $M=13.8\ \text{g}$ , $\sqrt{H-b}=2.51\ \text{cm}^{1/2}$ , $\sqrt{d-b}=2.00\ \text{cm}^{1/2}$ ):

k_2 = \frac{1}{2.51} - \frac{3.80}{2\times 13.8\times 2.00} = 0.399 - 0.0688 = 0.330\ \text{cm}^{-1/2}

first value of $k$ :

0.328\ \text{cm}^{-1/2}

second value of $k$ :

0.330\ \text{cm}^{-1/2}

Final answer

k1 = 0.328 cm^{-1/2}, k2 = 0.330 cm^{-1/2} (example; student-dependent)

Detailed explanation

Background Concept

When an equation is given with an unknown constant (here $k$ ), you can test the relationship by calculating $k$ from different sets of measurements. If the relationship is correct and uncertainties are small enough, the calculated values of $k$ should agree within experimental uncertainty.

The given model is:

\frac{1}{\sqrt{H-b}} = k + \frac{m}{2M\sqrt{d-b}}.

All the quantities on the right can be measured or calculated from measurements, so you can rearrange to find $k$ .

Understanding the Question

You have two sets of data:

No added mass (so $M=m$ ), giving one value of $H$ and therefore one value of $\sqrt{H-b}$ .
With the $10.0\ \text{g}$ mass attached (so $M=m+10.0\ \text{g}$ ), giving a second value of $H$ .

You must calculate two values of $k$ , one from each set.

Approach

Rearrange the equation to make $k$ the subject:

k = \frac{1}{\sqrt{H-b}} - \frac{m}{2M\sqrt{d-b}}.

For each set of data, substitute the measured/calculated values.
Keep consistent units (e.g. if $b,d,H$ are in cm, then $\sqrt{H-b}$ and $\sqrt{d-b}$ are in $\text{cm}^{1/2}$ , so $k$ has unit $\text{cm}^{-1/2}$ ).

Step-by-Step Reasoning

Compute $\sqrt{d-b}$ once from (b)(i).
For each run, compute $\sqrt{H-b}$ .
Evaluate $1/\sqrt{H-b}$ .
Evaluate $m/(2M\sqrt{d-b})$ (note: $m/M$ is dimensionless; the unit comes from $1/\sqrt{d-b}$ ).
Subtract to find $k$ .

If the two $k$ values are close, that supports the model.

Key Takeaways

Always rearrange first to a clear expression for the constant.
Use consistent units and keep track of derived units like $\text{cm}^{\pm 1/2}$ .

Common Mistakes

Using $M=m+10.0\ \text{g}$ in both cases (instead of $M=m$ for the first run).
Forgetting the factor of 2 in $2M\sqrt{d-b}$ .
Mixing cm and m in the square roots.

Things to Be Careful About

Brackets: the term is $2M\sqrt{d-b}$ , not $2(M\sqrt{d-b})$ vs $\sqrt{2M(d-b)}$ .
Significant figures: do not over-quote $k$ given the uncertainties in $H$ , $b$ , and $d$ .

Techniques used

rearrange an equation to make the constant the subjectsubstitute measured values into an expressioncompute and compare two values of a constant

(e)

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 (d).

1M Similar

DifficultyMedium-Easy

Worked solution

Using example values $k_1=0.328$ and $k_2=0.330$ :

\%\ \text{difference} = \frac{|0.330-0.328|}{(0.330+0.328)/2}\times 100 \approx 0.6\%

Since $0.6\% < 15\%$ , the two values of $k$ agree within the stated uncertainty, so the results support the relationship in (d).

Final answer

Yes; k values agree within 15% uncertainty (example: 0.6% difference).

Detailed explanation

Background Concept

To decide whether results support a suggested relationship, you check whether differences between values derived from the data are explainable by the experimental uncertainty.

A common method is to compare the percentage difference between two values with the percentage uncertainty. If:

\%\ \text{difference} \leq \%\ \text{uncertainty},

then the values are consistent (agreement within uncertainty).

Understanding the Question

You calculated two values of $k$ in (d). You are told the percentage uncertainty in the values of $k$ is $15\%$ . You must use that to decide whether your two $k$ values are consistent with each other (and therefore whether the relationship is supported).

Approach

Find how different the two $k$ values are (absolute difference).
Convert that difference into a percentage (often relative to the mean of the two values).
Compare with $15\%$ .
State a clear conclusion.

Step-by-Step Reasoning

Compute:

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

If this is less than $15\%$ , then the discrepancy is small enough to be due to uncertainty, so the data supports the model.
If it is greater than $15\%$ , then the data does not support the model (or there may be a systematic error).

Key Takeaways

Agreement is judged relative to uncertainty, not by exact equality.
A short, explicit concluding sentence is required.

Common Mistakes

Saying “they are close” without a quantitative comparison.
Comparing the absolute difference with 15 (instead of comparing percentage difference with $15\%$ ).

Things to Be Careful About

Make sure you interpret $15\%$ as a percentage uncertainty for $k$ , not for $H$ or other measurements.
Use consistent rounding; over-rounding can hide a real mismatch.

Techniques used

compare two experimental values using a stated percentage uncertaintycalculate a percentage differencestate a conclusion consistent with uncertainty

(f)

(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-Easy

Worked solution

Answer

Uncertainty in $H$ : difficult to judge the exact maximum height (turning point happens briefly), so $H$ has large reading uncertainty.
Uncertainty in $b$ and/or $d$ : parallax when reading the metre rule and difficulty locating the centre of the ball / top surface of the block.
Collision not perfectly repeatable: A may not strike B centrally each time (slight sideways motion/twist), so B’s maximum height varies between trials.
Energy losses: air resistance and friction at the clamp/string and deformation of balls during collision reduce energy transfer, affecting measured $H$ (systematic reduction).

Final answer

Four valid limitations/uncertainties stated (see solution).

Detailed explanation

Background Concept

In a practical evaluation, you gain marks by stating limitations that are:

specific (not just “human error”),
linked to a particular quantity being measured,
with a physical reason explaining why the uncertainty occurs.

Uncertainties can be random (vary trial to trial) or systematic (bias results in one direction).

Understanding the Question

You must describe four sources of uncertainty or limitations in the procedure. For any measurement uncertainty, you must name the quantity (e.g. $H$ , $b$ , $d$ , $m$ ) and explain why it is uncertain.

Approach

Choose four distinct issues covering:

measurement difficulties (reading a scale, locating a centre, turning point),
procedural repeatability (release point and collision alignment),
physical non-ideal behaviour (energy losses).

Step-by-Step Reasoning

Examples of creditworthy limitations:

$H$ (maximum height): turning point is momentary; reaction time and oscillation make the highest position hard to identify → large uncertainty in $H$ .
$b$ (centre height of B at rest): centre is not a marked point; you estimate it from the ball diameter or by eye → uncertainty in $b$ .
$d$ (block top height): if the ruler is not vertical or your eye is not level, parallax occurs; block top may not be perfectly horizontal → uncertainty in $d$ .
Collision repeatability: small changes in release position or twisting of strings mean A does not always hit B at the same point → $H$ varies.
Energy losses (systematic): air resistance, friction at support, and deformation/sound during collision reduce the kinetic energy transferred, so measured $H$ is smaller than ideal.

Any four distinct, well-explained points like the above score the marks.

Key Takeaways

Name the quantity, then give the reason.
Turning points and collisions are common dominant sources of uncertainty.

Common Mistakes

Writing vague statements like “human error” or “inaccurate readings” with no quantity or reason.
Repeating the same limitation in different words.

Things to Be Careful About

Make sure each of the four points is genuinely different.
Include at least one limitation about the method/physics (not just ruler reading).

Techniques used

identify random and systematic uncertainties in measurementslink an uncertainty to a specific measured quantitystate a physical reason for each limitation

(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-Easy

Worked solution

Answer

Measure $H$ using video recording (slow motion) or a motion sensor, then determine the maximum height from the playback/data to reduce turning-point uncertainty.
Fix a vertical metre rule and use a set square/pointer aligned with the centre of ball B to reduce parallax and improve readings of $b$ and $H$ .
Repeat each run several times and take the mean $H$ (and hence mean $k$ ) to reduce random variation from the collision.
Use a mechanical release (e.g. a catch) to ensure A is released from the same position each time and guide the motion so A hits B centrally (reduces variation in collision).

Final answer

Four valid improvements stated (see solution).

Detailed explanation

Background Concept

An improvement should specifically address a limitation by:

reducing random uncertainty (repeat/average, better alignment),
reducing systematic error (better calibration, reducing losses where possible),
improving measurement method (data logging, clearer reference marks).

Understanding the Question

You must propose four improvements to the experiment. These can involve additional apparatus or revised procedures. The best answers explicitly target the uncertainties/limitations identified in (f)(i).

Approach

Select four improvements that target the biggest problems:

maximum-height measurement,
parallax and identifying the centre,
repeatability of release and collision,
reduction of random scatter by repeats.

Step-by-Step Reasoning

Creditworthy improvements include:

Video / data logging for $H$ : record the motion and identify the exact frame at maximum height; this removes reaction-time/turning-point judgement.
Better scale alignment: clamp a metre rule vertical and use a set square or a fixed pointer at the ball centre height to remove parallax and provide consistent reference.
Repeats and averaging: perform (at least) 3–5 trials for each mass configuration and use the mean $H$ (and mean $k$ ) to reduce random collision variation.
Controlled release and central collision: use a clamp/catch release at a fixed displacement, and ensure strings are untwisted and the balls are aligned so A strikes B centrally each time.

Other valid suggestions (any four total needed): mark the centre of the balls with a pen dot; use a heavier/rigid stand to reduce movement; use longer strings to reduce relative error in height measurements.

Key Takeaways

Improvements should be practical and clearly reduce a specific uncertainty.

Common Mistakes

Giving improvements that are not feasible with the setup (e.g. “use a more accurate ball”).
Stating “take more care” without describing a specific change.

Things to Be Careful About

Avoid repeating the same improvement (e.g. “repeat more times” and “take an average”) as two separate points; treat that as one.
Make sure each improvement clearly links to what it improves (e.g. $H$ , $b$ , collision repeatability).

Techniques used

propose a procedural change to reduce uncertaintyselect improved measuring apparatus for a specific quantitylink each improvement to a stated limitation