In this experiment, you will investigate the rolling of a plastic bottle. (i) You are provided with a plastic bottle with a cap, as shown in Fig. 2.1. The diameter of the base of the bottle is $d$. Measure and record $d$. $d$ = \_\_\_\_\_\_ (ii) Estimate the percentage uncertainty in your value of $d$. Show your working. percentage uncertainty = \_\_\_\_\_\_ $\%$ (b) Set up the apparatus as shown in Fig. 2.2. Adjust the two metre rules so that the rules are approximately…

(a)(i) Example: d = 6.50 cm · (a)(ii) 0.8% · (b) Example: L = 80.0 cm · (c)(i) Example: h = 10.0 cm, t = 2.30 s · (c)(ii) 0.303 m s^-2 · (c)(iii) a quoted to 3 s.f., limited by L and t. · (d) Example: h = 5.0 cm, t = 2.70 s, a = 0.219 m s^-2 · (e) Example: k ≈ 16.6 and 17.0 · (f) Yes, k values agree within 15% so relationship is supported. · (g)(i) Four limitations listed (timing, L definition, h reading, repeatability due to alignment/friction). · (g)(ii) Four improvements…

Physics 9702/35 — October/November 2025

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

questions

marks

120

minutes

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

Q1MediumManipulation, Measurement and ObservationAnalysis, Conclusions and EvaluationPresentation of Data and Observations

In this experiment, you will investigate the equilibrium of a wooden rod.

Some of the apparatus has been set up for you.

(a)

(i)

Fig. 1.1 shows the rod with two eyes.

Fig. 1.1

The distance between the two eyes on the rod is $S$ .

Measure and record $S$ .

$S$ = ______

Complete the set-up of the apparatus as shown in Fig. 1.2.

Fig. 1.2

P and Q are masses.

The distance between the centre of mass P and the centre of the right-hand eye is $p$ , as shown in Fig. 1.2.

The distance between the centre of mass Q and the centre of the right-hand eye is $q$ , as shown in Fig. 1.2.

The angle between the string and the rod is $\theta$ .

Use some of the adhesive putty to attach Q to the rod so that $q$ is approximately $26\text{ cm}$ .

Use some of the adhesive putty to attach P to the rod. Adjust the position of P and the position of the stand with the pulley so that the rod is parallel to the bench and $\theta$ is approximately $45^{\circ}$ .

Do not move the stands for the remainder of the experiment.

Measure and record $\theta$ , $p$ and $q$ .

$\theta$ = ______ $^{\circ}$
$p$ = ______
$q$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

(Representative example values, measured with a ruler to nearest $0.1\ \text{cm}$ and protractor to nearest $1^{\circ}$ .)

$S = 30.0\ \text{cm}$

$\theta = 45^{\circ}$

$p = 18.4\ \text{cm}$

$q = 26.0\ \text{cm}$

Final answer

Example: S = 30.0 cm, θ = 45°, p = 18.4 cm, q = 26.0 cm

Detailed explanation

Background Concept

In static equilibrium, an object has no linear acceleration and no angular acceleration. In practical work, “the rod is parallel to the bench” is a convenient observable condition indicating the rod is not rotating and is in a steady equilibrium position.

The key experimental skills here are:

measuring lengths using a ruler (choosing a sensible reference point, reading at eye level to reduce parallax),
measuring an angle with a protractor,
setting up apparatus to match a diagram, and
recording values with appropriate precision and units.

Understanding the Question

You are asked to:

measure the separation $S$ between the two eyes on the rod (Fig. 1.1),
complete the apparatus setup (Fig. 1.2),
set $q \approx 26\ \text{cm}$ , then adjust $P$ and the pulley stand until the rod is horizontal and $\theta \approx 45^{\circ}$ ,
measure and record $\theta$ , $p$ and $q$ .

The note “Do not move the stands for the remainder of the experiment” matters because later readings assume the same geometry (especially $\theta$ ).

Approach

Use the ruler to measure $S$ between the centres of the eyes.
Place $Q$ so that $q$ is close to $26\ \text{cm}$ .
Adjust $P$ and the pulley stand until the rod is horizontal and $\theta$ is close to $45^{\circ}$ .
Measure $p$ and $q$ from the centre of each mass to the centre of the right-hand eye.
Measure $\theta$ between the string and the rod.
Record all readings with correct units and sensible precision.

Step-by-Step Reasoning

Measuring $S$

Align the ruler along the rod.
Identify the centres of the two eyes.
Measure the distance between these two centres.
Record $S$ to the nearest mm (i.e. $0.1\ \text{cm}$ ) if using a standard ruler.

Setting $q \approx 26\ \text{cm}$

Measure from the centre of mass $Q$ to the centre of the right-hand eye.
Reposition $Q$ (using adhesive putty) until the measurement is close to $26\ \text{cm}$ .

Making the rod horizontal and $\theta \approx 45^{\circ}$

Move $P$ along the rod to change its turning effect.
Move the pulley stand (initially) to adjust the direction of the string and hence $\theta$ .
When the rod is parallel to the bench, stop and do not move the stands afterwards.

Measuring $p$ , $q$ , and $\theta$

$p$ : distance from centre of mass $P$ to the right-hand eye centre.
$q$ : distance from centre of mass $Q$ to the right-hand eye centre.
$\theta$ : angle between the string and the rod, measured with a protractor.

Key Takeaways

Record experimental measurements with correct precision and units.
Use the diagram definitions carefully: all distances are from the right-hand eye.
Achieving the stated condition (rod horizontal) is part of the skill being assessed.

Common Mistakes

Measuring $p$ or $q$ from the wrong eye (left-hand instead of right-hand).
Measuring to the edge of a mass rather than to its centre.
Recording angles without the degree symbol or without appropriate precision.
Moving the stands after the instruction not to (changes $\theta$ and ruins later comparisons).

Things to Be Careful About

Parallax: read the ruler scale at eye level.
Reference points: use the centres of the eyes and centres of masses.
Units: if your ruler reads in cm, write cm explicitly.
Consistency: use the same precision (e.g. all lengths to $0.1\ \text{cm}$ ) across the experiment.

Techniques used

measure lengths with a ruler and record to appropriate precisionmeasure angles with a protractor and record to appropriate precisionadjust positions to satisfy a stated equilibrium conditionrecord readings with units and consistent significant figures

(ii)

The moment of the force about the eye due to mass P is $T_P$ .

The moment of the force about the eye due to mass Q is $T_Q$ .

The values of $T_P$ and $T_Q$ are given by:

T_P = 5Wp \quad \text{and} \quad T_Q = 7Wq

where $W$ has the value $0.0981\text{ N}$ .

Calculate $T_P$ and $T_Q$ .

$T_P$ = ______
$T_Q$ = ______

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Using example values $p = 18.4\ \text{cm} = 0.184\ \text{m}$ and $q = 26.0\ \text{cm} = 0.260\ \text{m}$ :

T_P = 5Wp = 5(0.0981)(0.184) = 9.03 \times 10^{-2}\ \text{N m}

T_Q = 7Wq = 7(0.0981)(0.260) = 1.79 \times 10^{-1}\ \text{N m}

Answer

$T_P = 9.03 \times 10^{-2}\ \text{N m}$

$T_Q = 1.79 \times 10^{-1}\ \text{N m}$

Final answer

Example: TP = 9.03×10^-2 N m, TQ = 1.79×10^-1 N m

Detailed explanation

Background Concept

A moment (turning effect) about a point is

\text{moment} = \text{force} \times \text{perpendicular distance}

Its unit is $\text{N m}$ . In this question you are given the expressions for the moments due to the weights of masses $P$ and $Q$ about a particular eye:

T_P = 5Wp, \quad T_Q = 7Wq

Here $W = 0.0981\ \text{N}$ is a weight (force). For the moment to come out in $\text{N m}$ , the distances $p$ and $q$ must be in metres.

Understanding the Question

You have measured $p$ and $q$ (likely in cm). You are asked to calculate numerical values of $T_P$ and $T_Q$ using the given formulae and the given value of $W$ .

Approach

Convert $p$ and $q$ into metres.
Substitute into $T_P = 5Wp$ and $T_Q = 7Wq$ .
Quote answers with unit $\text{N m}$ and sensible significant figures.

Step-by-Step Reasoning

Using the representative readings (your readings will differ):

Convert to SI units: $p = 18.4\ \text{cm} = 0.184\ \text{m}, \quad q = 26.0\ \text{cm} = 0.260\ \text{m}$
Substitute into $T_P$ : $T_P = 5(0.0981)(0.184) = 0.0903\ \text{N m}$
Substitute into $T_Q$ : $T_Q = 7(0.0981)(0.260) = 0.179\ \text{N m}$

Key Takeaways

Moments are measured in $\text{N m}$ , so distances must be in metres.
A correct unit conversion is often the main mark in a short calculation like this.

Common Mistakes

Using $p$ and $q$ in cm without converting to m (gives answers too large by a factor of 100).
Omitting units or writing $\text{N cm}$ .
Rounding too aggressively (e.g. 1 s.f.) so later graphing becomes poor.

Things to Be Careful About

Keep enough significant figures for plotting (typically 3 s.f. is safe).
Ensure you use the correct multiplier (5 for $T_P$ , 7 for $T_Q$ ).

Techniques used

convert measured lengths into SI unitssubstitute values into given expressionscalculate moments and include correct units

(b)

Change the position of Q and adjust the position of P until the rod is again parallel to the bench. Measure $p$ and $q$ . Repeat until you have six sets of values of $p$ and $q$ .

Record your results in a table. Include values of $T_P$ and $T_Q$ in your table.

8M Similar

DifficultyMedium

Worked solution

Answer

Record six sets of readings of $p$ and $q$ (to nearest $0.1\ \text{cm}$ ), and calculate $T_P$ and $T_Q$ for each set using

T_P = 5Wp, \quad T_Q = 7Wq

with $p$ and $q$ in metres and $W = 0.0981\ \text{N}$ .

(Example of a correctly formatted table.)

set	$p / \text{cm}$	$q / \text{cm}$	$T_P / \text{N m}$	$T_Q / \text{N m}$
1	12.0	20.0	0.0589	0.137
2	15.0	23.0	0.0736	0.158
3	18.0	26.0	0.0883	0.179
4	21.0	29.0	0.103	0.199
5	24.0	32.0	0.118	0.220
6	27.0	35.0	0.132	0.240

Final answer

See working (table of six sets of p, q with calculated TP and TQ)

Detailed explanation

Background Concept

In a practical investigation, marks for a results table usually come from presentation quality and correct processing, not from any single “correct” numerical value.

A good table:

contains all raw and derived quantities needed later,
has clear headings with quantity symbol / unit (e.g. $p / \text{cm}$ ),
uses consistent precision within a column,
includes enough rows (here six sets) and a reasonable spread of the independent variable.

Understanding the Question

You must:

move mass $Q$ to a new position,
adjust mass $P$ until the rod is again horizontal,
measure $p$ and $q$ for that equilibrium,
repeat until you have six pairs $(p,q)$ ,
calculate and include $T_P$ and $T_Q$ in the table.

Approach

Decide on a range of positions for $Q$ so that $q$ changes by a noticeable amount between trials.
For each trial: set $Q$ (choose $q$ ), then slide $P$ until the rod is horizontal, then measure $p$ and $q$ .
Immediately compute $T_P$ and $T_Q$ so you can spot anomalies while you still have the apparatus set up.

Step-by-Step Reasoning

Collecting data

Choose 6 different positions for $Q$ (e.g. evenly spaced along the rod) to give a good range of $q$ .
Each time you move $Q$ , re-adjust $P$ until the rod is parallel to the bench (equilibrium condition).
Measure $p$ and $q$ from the centre of each mass to the centre of the right-hand eye.

Tabulating raw data

Record $p$ and $q$ in cm to $0.1\ \text{cm}$ if using a mm scale.
Keep the same number of decimal places down each column.

Calculating $T_P$ and $T_Q$

Convert $p$ and $q$ to metres before substitution.
Use $T_P = 5Wp, \quad T_Q = 7Wq$ with $W = 0.0981\ \text{N}$ .
Record moments in $\text{N m}$ , typically to 3 s.f. (or a consistent decimal place suitable for your data size).

Key Takeaways

Six well-spread data points improve the reliability of the later graph.
Correct headings (symbol and unit) and consistent precision are often the easiest marks to secure.
Always include derived quantities needed for later parts (here $T_P$ and $T_Q$ ).

Common Mistakes

Missing units in headings (e.g. writing just $p$ instead of $p / \text{cm}$ ).
Mixing precisions (e.g. 12.0, 12.35, 12 in the same column).
Forgetting to include $T_P$ and $T_Q$ columns.
Calculating $T_P$ and $T_Q$ using cm instead of m.

Things to Be Careful About

Don’t move the stands after initial set-up (your $\theta$ should remain constant for the investigation).
Check that the rod is truly horizontal before reading $p$ and $q$ ; small tilts can shift the equilibrium point.
If one reading looks inconsistent with the trend, repeat that trial rather than leaving an outlier unchecked.

Techniques used

vary one position and re-establish equilibrium before taking readingstake repeated sets of measurements over a suitable rangecalculate derived quantities and record them in the same tableuse correct column headings with quantity and unitmaintain consistent decimal places within each column

(c)

(i)

Plot a graph of $T_Q$ on the $y$ -axis against $T_P$ on the $x$ -axis.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Plot $T_Q$ on the $y$ -axis against $T_P$ on the $x$ -axis.

Label axes $T_P / \text{N m}$ (horizontal) and $T_Q / \text{N m}$ (vertical).
Use a suitable scale (at least half the grid in each direction).
Plot all six points accurately.

Final answer

Graph of TQ (y) against TP (x) plotted with labelled axes and suitable scales

Detailed explanation

Background Concept

A graph is used to reveal relationships between variables. Good graphing practice in Cambridge practical papers includes:

axes labelled as “quantity / unit”,
a sensible linear scale that uses most of the grid,
points plotted with small, neat crosses,
and later (in part (ii)) a best-fit line.

Understanding the Question

You have a table containing values of $T_P$ and $T_Q$ for six equilibrium positions. You must plot $T_Q$ on the vertical axis and $T_P$ on the horizontal axis.

Approach

Decide the min and max values of $T_P$ and $T_Q$ from your table.
Choose axis scales that comfortably include all values and use at least half the graph paper.
Label axes with units.
Plot each pair $(T_P, T_Q)$ .

Step-by-Step Reasoning

Read the range of $T_P$ and $T_Q$ from your table.
On the x-axis, mark a linear scale for $T_P$ that covers the full range.
On the y-axis, mark a linear scale for $T_Q$ that covers the full range.
Label:
- x-axis: $T_P / \text{N m}$
- y-axis: $T_Q / \text{N m}$
Plot each of the six points using a sharp pencil; use small crosses (not large blobs).

Key Takeaways

Correct axis choice and labelling is assessed.
Scale choice affects how well you can later draw a best-fit line and read gradient/intercept.

Common Mistakes

Swapping axes (plotting $T_P$ on $y$ and $T_Q$ on $x$ ).
Missing units on axes.
Using an awkward scale (e.g. 1 big square = 3 units) that makes plotting inaccurate.

Things to Be Careful About

Do not force the axes to start at zero unless it helps; choose what gives best spread and accuracy.
Ensure you plot the processed values ( $T_P$ , $T_Q$ ), not the raw distances $p$ , $q$ .

Techniques used

choose suitable axis scales that use most of the graph paperlabel axes with quantity and unitplot experimental points accurately

(ii)

Draw the straight line of best fit.

1M Similar

(iii)

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

gradient = ______
$y$ -intercept = ______

2M Similar

DifficultyMedium

Worked solution

Working

Using two well-separated points on the best-fit line (example):

\text{gradient} = \frac{\Delta T_Q}{\Delta T_P} = \frac{0.240 - 0.137}{0.132 - 0.0589} = 1.41

$y$ -intercept read from graph (example):

\text{intercept} \approx 5.5 \times 10^{-2}\ \text{N m}

Answer

gradient $= 1.41$

$y$ -intercept $= 5.5 \times 10^{-2}\ \text{N m}$

Final answer

Example: gradient = 1.41, y-intercept = 5.5×10^-2 N m

Detailed explanation

Background Concept

For a straight-line graph of $y$ against $x$ :

y = mx + c

the gradient is

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

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

In this question, $y$ is $T_Q$ and $x$ is $T_P$ , so the gradient has no units (because it is $\text{(N m)}/\text{(N m)}$ ), while the intercept has units of $T_Q$ , i.e. $\text{N m}$ .

Understanding the Question

You have already plotted $T_Q$ vs $T_P$ and drawn a best-fit straight line. Now you must extract two numerical features of that line:

its gradient,
and its y-intercept.

These values are then used in part (d).

Approach

Choose two points on the best-fit line that are far apart (to reduce percentage reading error).
Read their coordinates $(T_P, T_Q)$ .
Compute gradient $= \Delta T_Q / \Delta T_P$ .
Find y-intercept either by reading where the line crosses the y-axis or by using $c = y - mx$ with a point on the line.

Step-by-Step Reasoning

Pick two points on the line
Choose points near the ends of your drawn line (not necessarily data points), e.g.

Point 1: $(T_{P1}, T_{Q1})$
Point 2: $(T_{P2}, T_{Q2})$

Calculate the gradient

\text{gradient} = \frac{T_{Q2} - T_{Q1}}{T_{P2} - T_{P1}}

Using widely spaced points makes $\Delta T_P$ and $\Delta T_Q$ larger, so the same small ruler-reading error has a smaller effect on the final gradient.

Find the y-intercept

Either read the intersection of the line with the y-axis directly.
Or use

\text{intercept} = T_Q - (\text{gradient})\,T_P

with a point $(T_P, T_Q)$ on the best-fit line.

Key Takeaways

Gradient comes from the best-fit line, not from joining adjacent points.
Use a large triangle to reduce uncertainty.
Units: gradient is dimensionless here; y-intercept has units $\text{N m}$ .

Common Mistakes

Using two adjacent plotted points (gives a very uncertain gradient).
Calculating gradient as $\Delta x/\Delta y$ instead of $\Delta y/\Delta x$ .
Reading coordinates from the plotted points rather than from the best-fit line.
Forgetting the y-intercept unit.

Things to Be Careful About

Write enough significant figures consistent with graph-reading precision (often 2–3 s.f.).
Ensure you use the same scale units as your axes when reading coordinates.
If the line does not reach the y-axis, extend it carefully with a ruler before reading the intercept.

Techniques used

determine the gradient from a large triangle on the best-fit lineread the y-intercept from the best-fit linecalculate gradient using $\Delta y / \Delta x$

(d)

(i)

It is suggested that the quantities $T_Q$ and $T_P$ are related by the equation

T_Q = A + BT_P

where $A$ and $B$ are constants.

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

$A$ = ______
$B$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Given

T_Q = A + BT_P

Comparing with $y = c + mx$ for a graph of $T_Q$ (y) against $T_P$ (x):

A = \text{y-intercept}, \quad B = \text{gradient}

Answer

$A = 5.5 \times 10^{-2}\ \text{N m}$

$B = 1.41$ (no unit)

Final answer

Example: A = 5.5×10^-2 N m, B = 1.41

Detailed explanation

Background Concept

If experimental data give a straight-line relationship, it can be written in the linear form

y = mx + c

where:

$m$ is the gradient,
$c$ is the y-intercept.

Here the suggested theory is

T_Q = A + BT_P

which is already in the same structure, with $T_Q$ playing the role of $y$ and $T_P$ playing the role of $x$ .

Understanding the Question

You have already found from the graph:

the gradient,
the y-intercept.

Now you must convert those graph features into the constants $A$ and $B$ and state appropriate units.

Approach

Identify $A$ as the y-intercept because it is the value of $T_Q$ when $T_P = 0$ .
Identify $B$ as the gradient because it multiplies $T_P$ .
Assign units: $A$ must have the same units as $T_Q$ ; $B$ is a ratio of $T_Q$ to $T_P$ .

Step-by-Step Reasoning

Compare:

T_Q = A + BT_P

with

T_Q = (\text{gradient})\,T_P + (\text{y-intercept}).

Therefore:

$A =$ y-intercept.
$B =$ gradient.

Units:

$T_Q$ is a moment, so unit is $\text{N m}$ , hence $A$ is in $\text{N m}$ .
$B$ is

\frac{T_Q}{T_P} = \frac{\text{N m}}{\text{N m}} = 1

so $B$ is dimensionless (no unit).

Key Takeaways

Constants in a linear model come directly from intercept and gradient.
Always check units: the additive constant has the same unit as the dependent variable.

Common Mistakes

Swapping $A$ and $B$ .
Giving units to $B$ when both axes have the same unit.
Using raw data points instead of the best-fit line values.

Things to Be Careful About

Quote $A$ and $B$ to a reasonable number of significant figures consistent with your graph readings.
Ensure you use your (c)(iii) values, not recalculated values from a single data point.

Techniques used

match a straight-line graph to the form $y = mx + c$identify constants from gradient and interceptstate appropriate units for constants

(ii)

Theory suggests that

R = F \sin\theta - \frac{2A}{S}

where $R$ is the weight of the rod and $F$ has the value $2.94\text{ N}$ .

Use your answers in (a)(i) and (d)(i) to determine a value for $R$ . Give an appropriate unit.

$R$ = ______

1M Similar

DifficultyMedium

Worked solution

Working

Using $R = F \sin\theta - \dfrac{2A}{S}$ , with example values $F = 2.94\ \text{N}$ , $\theta = 45^{\circ}$ , $A = 5.5 \times 10^{-2}\ \text{N m}$ and $S = 30.0\ \text{cm} = 0.300\ \text{m}$ :

R = 2.94\sin45^{\circ} - \frac{2(5.5 \times 10^{-2})}{0.300}

R = 2.08 - 0.367 = 1.71\ \text{N}

Answer

$R = 1.71\ \text{N}$

Final answer

Example: R = 1.71 N

Detailed explanation

Background Concept

The given theoretical relationship combines two ideas:

resolving a force into a component using $F\sin\theta$ ,
using a graph-derived constant $A$ (which has units of moment, $\text{N m}$ ) together with a length $S$ to form a force term $\dfrac{2A}{S}$ .

Dimensional check:

\frac{A}{S} \sim \frac{\text{N m}}{\text{m}} = \text{N}

so the equation produces a force (weight) as required.

Understanding the Question

You are told:

R = F\sin\theta - \frac{2A}{S}

with $F = 2.94\ \text{N}$ . You must use:

$\theta$ and $S$ from part (a)(i),
$A$ from part (d)(i),

to calculate $R$ , the weight of the rod.

Approach

Convert $S$ into metres because $A$ is in $\text{N m}$ .
Compute $F\sin\theta$ .
Compute $\dfrac{2A}{S}$ .
Subtract to obtain $R$ and give the unit $\text{N}$ .

Step-by-Step Reasoning

Using representative values (your values will be based on your own measurements and graph):

Convert $S$ :

S = 30.0\ \text{cm} = 0.300\ \text{m}

Calculate the upward/vertical component term:

F\sin\theta = 2.94\sin45^{\circ} \approx 2.94(0.707) \approx 2.08\ \text{N}

Calculate the term involving $A$ :

\frac{2A}{S} = \frac{2(5.5\times 10^{-2})}{0.300} = \frac{0.110}{0.300} \approx 0.367\ \text{N}

Substitute into the formula:

R = 2.08 - 0.367 = 1.71\ \text{N}

Key Takeaways

Always make units consistent: moments in $\text{N m}$ require lengths in metres when used in calculations.
A constant obtained from a graph can be substituted into further theoretical expressions.
Trigonometric components like $F\sin\theta$ are common when forces act at angles.

Common Mistakes

Using $S$ in cm in the term $2A/S$ (gives an answer wrong by a factor of 100).
Using $\cos\theta$ instead of $\sin\theta$ .
Forgetting the unit for $R$ (should be N, since it is a weight).

Things to Be Careful About

Ensure your calculator is in degree mode when using $45^{\circ}$ .
The final significant figures should reflect the precision of $A$ (from a graph) and $\theta$ (from a protractor); 2–3 s.f. is usually appropriate.

Techniques used

substitute experimental values into a derived equationconvert lengths to SI units to match moment unitscalculate a force using trigonometric resolution

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

In this experiment, you will investigate the rolling of a plastic bottle.

(a)

(i)

You are provided with a plastic bottle with a cap, as shown in Fig. 2.1.

Fig. 2.1

The diameter of the base of the bottle is $d$ .

Measure and record $d$ .

$d$ = ______

1M Similar

DifficultyEasy

Worked solution

Answer

Measure the base diameter with a ruler (or calipers) and record to the nearest $1\ \text{mm}$ .

Example:

d = 6.50\ \text{cm}

Final answer

Example: d = 6.50 cm

Detailed explanation

Background Concept

In practical work, a measurement must be recorded with a precision that matches the instrument used. A metre rule typically has $1\ \text{mm}$ smallest divisions, so a length read from it is usually recorded to the nearest $1\ \text{mm}$ (i.e. $0.1\ \text{cm}$ ).

Understanding the Question

You are given a bottle and asked to measure the diameter $d$ of its circular base. This is a straightforward length measurement.

Approach

Place the bottle so you can view the base clearly.
Use a ruler (or calipers if available) across the widest part of the base.
Read the scale with your eye directly above the mark to minimise parallax.
Record $d$ to the correct resolution.

Step-by-Step Reasoning

Align the zero of the ruler with one edge of the base.
Read the position of the opposite edge.
The diameter is the difference between these two readings.
Record in $\text{cm}$ to $0.1\ \text{cm}$ (or in $\text{mm}$ to the nearest $1\ \text{mm}$ ).

A typical bottle might give, for example, $d = 6.50\ \text{cm}$ .

Key Takeaways

Match the recorded decimal places to the measuring instrument.
Avoid parallax by reading the scale straight-on.

Common Mistakes

Recording too many decimal places (e.g. $6.503\ \text{cm}$ from a ruler).
Measuring a chord that is not the true diameter (not across the widest point).
Parallax error from viewing the scale at an angle.

Things to Be Careful About

Ensure the ruler is actually across the centre of the circular base.
If the bottle base is not perfectly circular, note that the “diameter” may vary slightly depending on orientation; take the largest consistent value if instructed to measure the diameter.

Techniques used

use a ruler or calipers to measure a diametertake a reading perpendicular to the scale to reduce parallaxrecord a measurement to the instrument resolution

(ii)

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

percentage uncertainty = ______ $\%$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Using a ruler with $1\ \text{mm}$ divisions:

\Delta d = \pm 0.5\ \text{mm} = \pm 0.05\ \text{cm}

Example with $d = 6.50\ \text{cm}$ :

\%\text{ uncertainty} = \frac{0.05}{6.50} \times 100 = 0.77\% \approx 0.8\%

Answer

0.8\%

Final answer

0.8%

Detailed explanation

Background Concept

For a direct reading on an analogue scale (like a ruler), a common estimate of absolute uncertainty is half the smallest scale division.

Percentage uncertainty is defined by

\%\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 working. The key is identifying the likely absolute uncertainty in $d$ from your measuring instrument.

Approach

Decide the instrument used (usually a ruler with $1\ \text{mm}$ resolution).
Convert the reading uncertainty into the same units as $d$ .
Substitute into the percentage uncertainty formula.

Step-by-Step Reasoning

If the smallest division is $1\ \text{mm}$ , the reading uncertainty is typically $\pm 0.5\ \text{mm}$ .
Convert to the unit used for $d$ .
Divide by $d$ and multiply by $100$ .

Example:

$d = 6.50\ \text{cm}$
$\Delta d = 0.05\ \text{cm}$

\%\text{ uncertainty} = \frac{0.05}{6.50} \times 100 = 0.77\%

So a sensible quoted value is $0.8\%$ .

Key Takeaways

Use half a scale division as the absolute uncertainty for a single analogue reading.
Always convert to consistent units before calculating a percentage.

Common Mistakes

Using $\pm 1\ \text{mm}$ instead of $\pm 0.5\ \text{mm}$ for a single reading.
Mixing units (e.g. using $0.5\ \text{mm}$ with $d$ in $\text{cm}$ without conversion).
Writing an uncertainty with too many significant figures.

Things to Be Careful About

If you measured $d$ by taking two ruler readings (one at each edge) and subtracting, some schemes allow a larger uncertainty (two readings). If not specified, the standard half-division approach is usually accepted in Paper 3, but be consistent with how you measured it.

Techniques used

estimate absolute uncertainty from instrument resolutioncalculate percentage uncertaintypresent uncertainty to a sensible number of significant figures

(b)

Set up the apparatus as shown in Fig. 2.2.

Fig. 2.2

Adjust the two metre rules so that the rules are approximately parallel to each other, as shown in Fig. 2.3.

Fig. 2.3

Pour all the water from the beaker into the bottle.

Place the bottle on the two rules as shown in Fig. 2.4.

Fig. 2.4

Release the bottle. Adjust the rules so that the bottle rolls to the end of the rules.

The distance that the bottle rolls on the rules is $L$ , as shown in Fig. 2.4.

Measure and record $L$ .

$L$ = ______

1M Similar

(c)

(i)

Stand the bottle upright on the bench.

The height of the water in the bottle is $h$ , as shown in Fig. 2.5.

Fig. 2.5

Measure and record $h$ .

$h$ = ______

The time for the bottle to roll distance $L$ on the rules is $t$ .

Take measurements to determine $t$ .

$t$ = ______

3M Similar

DifficultyMedium

Worked solution

Working

Measure $h$ to the nearest $1\ \text{mm}$ .

Example:

h = 10.0\ \text{cm}

Determine $t$ by timing the roll over distance $L$ several times and taking a mean.

Example timings: $2.31\ \text{s}$ , $2.29\ \text{s}$ , $2.30\ \text{s}$

t = \frac{2.31 + 2.29 + 2.30}{3} = 2.30\ \text{s}

Answer

h = 10.0\ \text{cm},\qquad t = 2.30\ \text{s}

Final answer

Example: h = 10.0 cm, t = 2.30 s

Detailed explanation

Background Concept

Two key practical skills are being assessed:

Measuring a height $h$ of a liquid column: the main uncertainty is parallax and judging the level/meniscus.
Measuring a time interval $t$ : the main uncertainty is reaction time. Repeating readings and taking a mean improves reliability and helps identify anomalies.

Understanding the Question

You must:

Stand the bottle upright and measure the water height $h$ (as in Fig. 2.5).
Measure the time $t$ taken for the bottle to roll the previously-measured distance $L$ along the rules.

Both $h$ and $t$ will be used later to calculate acceleration $a$ and then the constant $k$ .

Approach

For $h$ : use a ruler alongside the bottle, read the level at eye height.
For $t$ : use a stopwatch, but reduce random error by repeating the timing several times and averaging.

Step-by-Step Reasoning

Measuring $h$

Put the bottle on a flat bench.
Place a ruler next to the bottle with the zero at the base level.
Look horizontally at the water level (eye at the same height as the surface) to avoid parallax.
Record to $\pm 1\ \text{mm}$ (so usually as $x.x\ \text{cm}$ ).

Measuring $t$

Mark the start position and the end position (so you start/stop the timing at the same points each run).
Release the bottle without pushing.
Start the stopwatch as the bottle begins rolling and stop it when it reaches the end point.
Repeat at least 3 times and calculate the mean time.

Example:

$h = 10.0\ \text{cm}$
times $2.31\ \text{s}$ , $2.29\ \text{s}$ , $2.30\ \text{s}$ , mean $t = 2.30\ \text{s}$ .

Key Takeaways

Repeat timings to improve reliability.
Parallax is a major issue for reading liquid levels.

Common Mistakes

Only taking one timing and not averaging.
Starting timing before releasing or after the bottle has already moved.
Measuring $h$ with the ruler not aligned to the base level.

Things to Be Careful About

Water can slosh if the bottle is moved; allow it to settle before measuring $h$ .
Use the same start/end reference each time; otherwise the time corresponds to a different distance.
If one timing is clearly anomalous, repeat rather than averaging it in without question.

Techniques used

measure a liquid height with a ruler and avoid parallaxtime motion with a stopwatch using repeat readingscalculate a mean time from repeated trials

(ii)

A value for the acceleration $a$ of the bottle is given by

a = \frac{2L}{t^2}

Calculate $a$ .

$a$ = ______

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Using $a = \dfrac{2L}{t^2}$ .

Example with $L = 80.0\ \text{cm} = 0.800\ \text{m}$ and $t = 2.30\ \text{s}$ :

a = \frac{2(0.800)}{(2.30)^2} = \frac{1.60}{5.29} = 0.303\ \text{m s}^{-2}

Answer

a = 0.303\ \text{m s}^{-2}

Final answer

0.303 m s^-2

Detailed explanation

Background Concept

If an object starts from rest and moves with constant acceleration $a$ over distance $L$ in time $t$ , then

L = \frac{1}{2}at^2

Rearranging gives

a = \frac{2L}{t^2}

In this experiment, the given formula is used as a working model for the bottle’s motion along the rules.

Understanding the Question

You are given the equation for $a$ in terms of your measured $L$ and $t$ and asked to calculate $a$ . The key practical point is using consistent units (preferably SI).

Approach

Convert $L$ into metres.
Substitute $L$ and $t$ into $a = 2L/t^2$ .
Calculate and present the result with appropriate significant figures and unit $\text{m s}^{-2}$ .

Step-by-Step Reasoning

Using the example values:

$L = 80.0\ \text{cm} = 0.800\ \text{m}$
$t = 2.30\ \text{s}$

Compute $t^2$ :

t^2 = (2.30)^2 = 5.29\ \text{s}^2

Then

a = \frac{2(0.800)}{5.29} = 0.303\ \text{m s}^{-2}

Key Takeaways

Always convert to SI before substituting into kinematics-style equations.
Keep the unit of acceleration as $\text{m s}^{-2}$ .

Common Mistakes

Using $L$ in $\text{cm}$ and still writing the unit as $\text{m s}^{-2}$ .
Forgetting to square the time.
Writing $a = 2t/L^2$ or other rearrangement errors.

Things to Be Careful About

Ensure your calculator brackets include the whole $t^2$ in the denominator.
If your $t$ has been averaged, use the averaged value consistently.

Techniques used

substitute measured values into a given formulaconvert centimetres to metres for consistencycalculate an acceleration from distance and time

(iii)

Justify the number of significant figures that you have given for your value of $a$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

$a$ is calculated from $L$ and $t$ .

$L$ is recorded to $3\ \text{s.f.}$ (e.g. $0.800\ \text{m}$ ) and $t$ to $3\ \text{s.f.}$ (e.g. $2.30\ \text{s}$ ), so $a$ is given to $3\ \text{s.f.}$ .

Final answer

a quoted to 3 s.f., limited by L and t.

Detailed explanation

Background Concept

A calculated quantity should not be quoted to more significant figures than are justified by the precision of the measurements used to calculate it. For multiplication/division (and powers), the significant figures in the result are limited by the least precise input (in significant-figure terms).

Understanding the Question

You must justify the number of significant figures used for $a$ in part (c)(ii). That means referring to how precisely $L$ and $t$ were measured/recorded.

Approach

Identify the significant figures of $L$ and $t$ as recorded.
State that since $a = 2L/t^2$ depends on both, $a$ should be quoted to the same (or fewer) significant figures as the least precise of $L$ and $t$ .

Step-by-Step Reasoning

Example:

If $L = 0.800\ \text{m}$ , that is $3\ \text{s.f.}$ .
If $t = 2.30\ \text{s}$ (mean time), that is $3\ \text{s.f.}$ .

So $a$ should be given to $3\ \text{s.f.}$ (e.g. $0.303\ \text{m s}^{-2}$ ).

If your time is only recorded to $2\ \text{s.f.}$ (e.g. $2.3\ \text{s}$ ), then $a$ should be quoted to $2\ \text{s.f.}$ .

Key Takeaways

Quote calculated results to a precision justified by raw measurements.
Time measurements often limit precision in mechanics experiments.

Common Mistakes

Giving $a$ to many decimal places just because the calculator displays them.
Justifying based on the number of decimal places instead of significant figures.
Ignoring that $t$ is squared (it can strongly affect the uncertainty), though the sig-fig rule still follows the recorded precision.

Things to Be Careful About

Be consistent: if you record $t$ as $2.30\ \text{s}$ , don’t later treat it as if it were $2.3\ \text{s}$ .
If $t$ is a mean, you should still record it to a sensible precision based on stopwatch resolution and spread of repeats (don’t overstate precision).

Techniques used

identify the limiting measurement for significant figuresmatch calculated significant figures to raw data precisionstate a clear justification linked to measurement resolution

(d)

Pour approximately half the water from the bottle into the beaker.

Repeat (c)(i) and (c)(ii).

$h$ = ______
$t$ = ______
$a$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

After pouring out about half the water, repeat the measurements and calculation.

Example:

h = 5.0\ \text{cm}

Mean time (example):

t = 2.70\ \text{s}

With $L = 0.800\ \text{m}$ :

a = \frac{2(0.800)}{(2.70)^2} = \frac{1.60}{7.29} = 0.219\ \text{m s}^{-2}

Answer

h = 5.0\ \text{cm},\quad t = 2.70\ \text{s},\quad a = 0.219\ \text{m s}^{-2}

Final answer

Example: h = 5.0 cm, t = 2.70 s, a = 0.219 m s^-2

Detailed explanation

Background Concept

To test a suggested relationship involving $a$ and $h$ , you need more than one data set. Changing the amount of water changes $h$ ; repeating the procedure gives a second value of $a$ to compare.

Understanding the Question

You must remove about half the water, then repeat the measurements of $h$ and $t$ and recalculate $a$ using the same $L$ (unless $L$ was changed).

Approach

Keep the apparatus arrangement the same as before (rules separation, slope, and $L$ ).
Measure the new water height $h$ .
Time the roll several times again and compute the mean $t$ .
Calculate $a$ from $a = 2L/t^2$ .

Step-by-Step Reasoning

Example (illustrative):

New height: $h = 5.0\ \text{cm}$ .
Mean time: $t = 2.70\ \text{s}$ .
Use the same $L$ as before, converted into metres.

Then

a = \frac{2L}{t^2}

Substituting $L = 0.800\ \text{m}$ and $t = 2.70\ \text{s}$ :

a = \frac{1.60}{7.29} = 0.219\ \text{m s}^{-2}

Key Takeaways

Repeating the experiment with a changed condition provides additional data.
Keep all other variables as constant as possible when testing a relationship.

Common Mistakes

Changing the spacing/angle of the rules unintentionally between runs (this changes the motion for reasons unrelated to $h$ ).
Not re-measuring $h$ after pouring out water.
Using a single timing for the second run.

Things to Be Careful About

Ensure “approximately half” is not wildly different each attempt; the important thing is that $h$ is clearly different and measured accurately.
Let the water settle before measuring $h$ and before releasing the bottle.

Techniques used

repeat measurements after changing one conditionrecord new values of height and time to consistent precisionrecalculate acceleration using the same formula

(e)

It is suggested that the relationship between $a$ and $h$ is

a = \frac{kd\sqrt{\pi h}}{2}

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

Working

From

a = \frac{kd\sqrt{\pi h}}{2}

k = \frac{2a}{d\sqrt{\pi h}}

Example using $d = 6.50\ \text{cm} = 0.0650\ \text{m}$ .

First set: $h = 10.0\ \text{cm} = 0.100\ \text{m}$ , $a = 0.303\ \text{m s}^{-2}$

k_1 = \frac{2(0.303)}{0.0650\sqrt{\pi(0.100)}} = 16.6

Second set: $h = 5.0\ \text{cm} = 0.050\ \text{m}$ , $a = 0.219\ \text{m s}^{-2}$

k_2 = \frac{2(0.219)}{0.0650\sqrt{\pi(0.050)}} = 17.0

Answer

First value of $k = 16.6$

Second value of $k = 17.0$

Final answer

Example: k ≈ 16.6 and 17.0

Detailed explanation

Background Concept

When a relationship contains an unknown constant (here $k$ ), you can use experimental measurements to evaluate it by rearranging the equation.

Given

a = \frac{kd\sqrt{\pi h}}{2}

solve for $k$ :

k = \frac{2a}{d\sqrt{\pi h}}

To test whether $k$ is constant, calculate it for different values of $h$ and see if the results agree within uncertainty.

Understanding the Question

You have two sets of measured data (one with more water, one with about half). Using each set, calculate a value of $k$ . If the suggested model is correct, the two $k$ values should be similar.

Approach

Rearrange the given equation to make $k$ the subject.
Convert $d$ and $h$ to metres and use $a$ in $\text{m s}^{-2}$ .
Calculate $k$ twice (once per data set).

Step-by-Step Reasoning

Using the example measurements:

$d = 0.0650\ \text{m}$ .

Data set 1

$h_1 = 0.100\ \text{m}$ , $a_1 = 0.303\ \text{m s}^{-2}$ .

Compute the square-root factor:

\sqrt{\pi h_1} = \sqrt{\pi(0.100)} = \sqrt{0.314} = 0.560

Then

k_1 = \frac{2a_1}{d\sqrt{\pi h_1}} = \frac{0.606}{0.0650 \times 0.560} = 16.6

Data set 2

$h_2 = 0.050\ \text{m}$ , $a_2 = 0.219\ \text{m s}^{-2}$ .

\sqrt{\pi h_2} = \sqrt{\pi(0.050)} = \sqrt{0.157} = 0.396

k_2 = \frac{2a_2}{d\sqrt{\pi h_2}} = \frac{0.438}{0.0650 \times 0.396} = 17.0

The two values are close, suggesting $k$ is approximately constant.

Key Takeaways

Rearranging a model lets you compute a constant from experimental results.
Using consistent SI units avoids hidden scaling errors.

Common Mistakes

Forgetting the factor of $2$ in the rearrangement.
Using $h$ in $\text{cm}$ inside $\sqrt{\pi h}$ while using $d$ in $\text{m}$ .
Calculating only one value of $k$ instead of two.

Things to Be Careful About

Keep enough significant figures in intermediate steps (especially square roots) to avoid rounding too early.
If you include units for $k$ , be consistent; however, many mark schemes for Paper 3 mainly credit correct numerical evaluation and consistency between the two values.

Techniques used

rearrange an equation to make a constant the subjectsubstitute experimental values into an expressioncalculate and compare two values of a derived constant

(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-Easy

Worked solution

Working

Example values: $k_1 = 16.6$ , $k_2 = 17.0$ .

Percentage difference (using mean):

\frac{|17.0-16.6|}{(17.0+16.6)/2} \times 100 = 2.4\%

Since $2.4\% < 15\%$ , the values agree within uncertainty.

Answer

Yes. The two values of $k$ are consistent within $15\%$ , so the results support the relationship.

Final answer

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

Detailed explanation

Background Concept

To judge whether results support a suggested relationship, you compare measured/derived values against the experimental uncertainty.

If two values are supposed to be the same (here $k$ should be constant), they are considered consistent if their difference is not larger than the stated uncertainty level.

Understanding the Question

You are told the percentage uncertainty in $k$ is $15\%$ . Using that, you must decide whether your two calculated values of $k$ support the model in (e). This is essentially a consistency check.

Approach

Calculate how different the two $k$ values are (a percentage difference is a clear method).
Compare this difference to $15\%$ .
If the difference is less than (or comparable to) $15\%$ , conclude the results support the relationship.

Step-by-Step Reasoning

Using illustrative values $k_1 = 16.6$ and $k_2 = 17.0$ :

Difference:

\Delta k = |k_2-k_1| = 0.4

A sensible percentage difference is relative to the mean:

\%\text{ difference} = \frac{\Delta k}{(k_1+k_2)/2} \times 100

\%\text{ difference} = \frac{0.4}{16.8} \times 100 = 2.4\%

Because $2.4\%$ is much less than $15\%$ , the two $k$ values agree within the stated uncertainty. Therefore the results support the relationship.

Key Takeaways

Use uncertainty to decide whether differences are meaningful.
A relationship is supported if derived constants are consistent within experimental uncertainty.

Common Mistakes

Stating “they are close” with no reference to the $15\%$ uncertainty.
Using an incorrect percentage calculation (e.g. dividing by the larger value without explanation).
Concluding the relationship is wrong just because the values are not identical (experimental data never match perfectly).

Things to Be Careful About

Make sure you are comparing like with like (both $k$ values calculated using consistent units and methods).
If your two values differ by more than $15\%$ , you should conclude the results do not support the model (or are inconclusive), not try to force agreement.

Techniques used

compare two experimental values using a percentage differenceapply an uncertainty threshold to judge consistencystate a conclusion about support for a model

(g)

(i)

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

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

4M Similar

DifficultyMedium

Worked solution

Answer

Timing $t$ : stopwatch reaction time when starting/stopping; hard to judge the exact instant the bottle starts moving and reaches the end.
Distance $L$ : difficulty defining the exact start/end position of the bottle (which part of the bottle is the reference), so $L$ may not be measured consistently.
Height $h$ : parallax / unclear water level (meniscus) when reading $h$ on the bottle, especially if the bottle is not perfectly vertical.
Motion not perfectly repeatable: rules may not be exactly parallel / surface friction varies, so the bottle may wobble or rub a rule, changing the acceleration between runs.

Final answer

Four limitations listed (timing, L definition, h reading, repeatability due to alignment/friction).

Detailed explanation

Background Concept

Experimental uncertainty comes from:

Instrument limits (resolution of ruler/stopwatch)
Human factors (reaction time, judgement)
Systematic effects (misalignment, friction, changing conditions)

Good evaluation statements identify the measured quantity and explain why it is uncertain or why the method is limited.

Understanding the Question

You must describe four sources of uncertainty or limitations. For any measurement uncertainty you mention, you must name the quantity (e.g. $t$ , $h$ , $L$ ) and give a reason.

Approach

Choose distinct points covering different parts of the experiment:

timing the motion
measuring distances/heights
repeatability of the run (track alignment, friction)
any physical effect that violates assumptions (e.g. sloshing water)

State each one clearly as a separate bullet.

Step-by-Step Reasoning

Examples of creditworthy limitations:

Timing uncertainty ( $t$ ): manual stopwatch timing introduces reaction time (random error) and ambiguity of when the bottle starts/finishes.
Distance uncertainty ( $L$ ): the bottle is not a point object; if you don’t always measure from the same reference point on the bottle, $L$ changes. Also, the end position may not be sharply defined.
Water height uncertainty ( $h$ ): the water surface may not be perfectly level, and the meniscus is hard to judge through curved plastic; parallax occurs if not viewed at eye level.
Non-repeatable rolling: if the rules are not exactly parallel or have uneven surfaces, the bottle can drift and rub, changing resistive forces; this affects the acceleration and therefore the derived $a$ and $k$ .

Other acceptable limitations (if kept distinct) could include:

water moving inside the bottle during rolling (changing mass distribution)
small changes in slope/track geometry between runs
bottle slipping rather than rolling without slipping

Key Takeaways

Link uncertainties to specific measured quantities.
Give physical reasons, not just “human error”.

Common Mistakes

Writing vague statements like “parallax error” without saying which reading (e.g. $h$ or $L$ ).
Repeating the same idea four times (e.g. “reaction time” for everything).
Listing improvements instead of limitations.

Things to Be Careful About

Make sure you give four distinct sources.
For measurement uncertainties, always include quantity + reason (e.g. “ $t$ because reaction time”).

Techniques used

identify random and systematic uncertainty sourceslink each uncertainty to a 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

Worked solution

Answer

Measure $t$ using light gates / motion sensor / video analysis instead of a stopwatch to remove reaction time.
Use a fixed release mechanism (e.g. a clamp/stop at the start) so the bottle is released from the same position without an initial push.
Fix the rules with clamps/spacers so they are truly parallel and do not move; keep the separation constant for all runs.
Increase reliability by repeating each timing several times (and/or using a longer $L$ ) and taking a mean to reduce random error.

Final answer

Four improvements listed (automatic timing, controlled release, fixed parallel rules, repeats/longer L).

Detailed explanation

Background Concept

Improvements should directly reduce uncertainties or remove limitations. Strong improvement statements:

specify what to change
explain what uncertainty it reduces

Common categories: better instruments, better control of variables, more repeats, and improved geometry/alignment.

Understanding the Question

You must describe four improvements (apparatus and/or procedure) to make the experiment more accurate/reliable.

Approach

Take the limitations from (g)(i) and propose one improvement for each. Ensure all four are different.

Step-by-Step Reasoning

Examples:

Replace stopwatch timing with light gates or video tracking. This removes reaction time and gives a clearer definition of start/finish time.
Standardise the release: use a rigid stop at the start and lift it away quickly, or a simple gate mechanism. This avoids giving the bottle an extra push and keeps the start position fixed.
Maintain track geometry: clamp the metre rules and use spacers to keep them parallel and at constant separation; this improves repeatability and reduces rubbing/wobbling.
More data / better averaging: repeat each run many times and calculate a mean (and possibly exclude anomalies). Alternatively, increase $L$ so that the same reaction time produces a smaller percentage error in $t$ .

Other possible improvements (if needed):

Use a set square/vertical reference to measure $h$ more accurately.
Mark the start/end points on the rules for consistent $L$ .
Use a smoother track surface to reduce variable friction.

Key Takeaways

Improvements should clearly target the dominant uncertainties (often timing and repeatability).
Standardising start conditions improves reliability.

Common Mistakes

Suggesting vague improvements like “be more careful”.
Giving an improvement that does not address any stated limitation.
Repeating the same idea in different words (e.g. “do more repeats” four times).

Things to Be Careful About

Provide exactly four distinct improvements.
Include enough detail so the improvement is actually implementable in a school laboratory.

Techniques used

propose modifications that reduce a stated uncertaintyselect appropriate alternative instrumentsincrease reliability by repeating and standardising procedure