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) d = 6.50 cm · (a)(ii) 0.8% · (b) L = 85.0 cm · (c)(i) h = 19.0 cm, t = 1.90 s · (c)(ii) 0.471 m s^-2 · (c)(iii) 3 s.f. (e.g. a = 0.471 m s^-2) · (d) h = 9.5 cm, t = 2.30 s, a = 0.321 m s^-2 · (e) k1 = 18.8 m^-1/2 s^-2, k2 = 18.1 m^-1/2 s^-2 · (f) Yes, the k values agree within 15% so the relationship is supported. · (g)(i) See working (four sources listed). · (g)(ii) See working (four improvements listed).

Physics 9702/37 — October/November 2025

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

questions

marks

120

minutes

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

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

In this experiment, you will investigate 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

(Example set of readings; your values depend on your apparatus.)

$S = 50.0\ \text{cm}$

$\theta = 45\ ^\circ$

$p = 18.0\ \text{cm}$

$q = 26.0\ \text{cm}$

Final answer

Student-dependent (e.g. S = 50.0 cm, θ = 45°, p = 18.0 cm, q = 26.0 cm)

Detailed explanation

Background Concept

In practical mechanics questions, marks are often awarded for good measurement technique and appropriate recording. A length should be measured between the correct reference points (often centres of holes/eyes or centres of masses), and recorded with a sensible precision (typically to the nearest millimetre, i.e. $0.1\ \text{cm}$ , for a standard ruler). An angle measured with a protractor is usually recorded to the nearest degree.

In equilibrium set-ups, the instruction “rod parallel to the bench” is important: it fixes the geometry so that the angle $\theta$ is well-defined and the distances $p$ and $q$ correspond to the intended horizontal lever arms along the rod.

Understanding the Question

You are asked to:

measure $S$ , the separation of the two eyes on the rod;
complete the set-up and adjust it until the rod is horizontal (parallel to the bench) and $\theta \approx 45^\circ$ ;
measure and record $\theta$ , and the distances $p$ and $q$ from the right-hand eye to the centre of mass of masses $P$ and $Q$ .

These measured values will be used later to calculate moments and then plot a graph.

diagram-a-i-setup

Approach

Measure $S$ directly on the rod between the same points indicated by the diagram (centre-to-centre of the eyes).
Attach $Q$ so that $q \approx 26\ \text{cm}$ (this is an approximate placement step).
Adjust $P$ and the pulley stand until the rod is horizontal and the string makes about $45^\circ$ to the rod.
Once the rod is horizontal and the stands are fixed (do not move them afterwards), take careful measurements of $\theta$ , $p$ , and $q$ .

Step-by-Step Reasoning

Measuring $S$ :
- Place a ruler along the rod.
- Measure from the centre of the left eye to the centre of the right eye.
- Record $S$ to the nearest $0.1\ \text{cm}$ (or to the smallest division you can reliably read).
Setting $q \approx 26\ \text{cm}$ :
- Attach $Q$ with adhesive putty.
- Measure from the centre of the right-hand eye to the centre of the mass $Q$ .
- Adjust until the reading is close to $26\ \text{cm}$ .
Making the rod parallel to the bench:
- Adjust the position of $P$ and the pulley stand until the rod is horizontal.
- A simple check is to view the rod against the bench edge or use a set square/spirit level if available.
Measuring $\theta$ :
- Place the protractor with its baseline aligned with the rod (horizontal direction).
- Read the angle between the string and the rod.
- Record to the nearest degree.
Measuring $p$ and $q$ :
- Measure along the rod from the centre of the right-hand eye to the centre of mass of $P$ (for $p$ ) and to the centre of mass of $Q$ (for $q$ ).
- Record each to the nearest $0.1\ \text{cm}$ .

Key Takeaways

Identify the correct reference points (centre of eye, centre of mass).
Record measurements with appropriate precision.
Ensure the stated condition (rod horizontal, stands fixed) before recording values.

Common Mistakes

Measuring from the edge of an eye or mass instead of its centre.
Recording lengths with inconsistent precision (e.g. some to $0.1\ \text{cm}$ , others to $1\ \text{cm}$ ).
Reading the protractor from the wrong scale (inner vs outer) or not aligning the baseline with the rod.
Parallax error when reading the ruler/protractor.

Things to Be Careful About

Once the instruction says “Do not move the stands”, moving them changes the geometry and makes later comparisons invalid.
Ensure $p$ and $q$ are measured from the same right-hand eye each time.
Check the rod is genuinely parallel to the bench before taking readings; small tilts can change the effective geometry and the repeatability of results.

Techniques used

measure lengths with a ruler and quote to appropriate precisionmeasure an angle with a protractor and quote to appropriate precisionadjust the apparatus until a stated geometric condition is satisfiedavoid parallax when taking scale readings

(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 the example readings from (a)(i): $p = 18.0\ \text{cm} = 0.180\ \text{m}$ , $q = 26.0\ \text{cm} = 0.260\ \text{m}$ )

T_P = 5Wp = 5(0.0981)(0.180) = 8.83 \times 10^{-2}\ \text{N m}

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

Answer

$T_P = 8.83 \times 10^{-2}\ \text{N m}$

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

Final answer

Student-dependent (e.g. TP = 8.83×10^-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 from pivot}

In this experiment, the given expressions already combine the force due to each mass with the relevant distance along the rod (treated as the perpendicular lever arm about the eye), so you only need to substitute values.

The constant $W = 0.0981\ \text{N}$ is the weight corresponding to a $10\ \text{g}$ mass (since $0.010\ \text{kg} \times 9.81\ \text{m s}^{-2} = 0.0981\ \text{N}$ ). The factors $5$ and $7$ therefore represent weights of $50\ \text{g}$ and $70\ \text{g}$ respectively.

Moments are measured in $\text{N m}$ , so distances must be in metres.

Understanding the Question

You are told that

T_P = 5Wp, \qquad T_Q = 7Wq,

with $W = 0.0981\ \text{N}$ . You must calculate $T_P$ and $T_Q$ using your measured values of $p$ and $q$ from (a)(i), and give answers in appropriate units.

Approach

Convert $p$ and $q$ into metres.
Substitute into the given formulae.
Quote $T_P$ and $T_Q$ in $\text{N m}$ to a sensible number of significant figures (often 3 s.f. if your lengths are measured to $0.1\ \text{cm}$ ).

Step-by-Step Reasoning

Suppose you measured (example) $p = 18.0\ \text{cm}$ and $q = 26.0\ \text{cm}$ .
Convert:

p = 18.0\ \text{cm} = 0.180\ \text{m}, \qquad q = 26.0\ \text{cm} = 0.260\ \text{m}.

Calculate $T_P$ :

T_P = 5Wp = 5(0.0981)(0.180).

Multiplying gives $T_P = 0.08829\ \text{N m}$ , which rounds to $8.83 \times 10^{-2}\ \text{N m}$ .

Calculate $T_Q$ :

T_Q = 7Wq = 7(0.0981)(0.260) = 0.178542\ \text{N m} \approx 1.79 \times 10^{-1}\ \text{N m}.

Your numerical answers will differ if your $p$ and $q$ differ.

Key Takeaways

Always convert centimetres to metres before calculating moments in $\text{N m}$ .
Use the given formulae directly; the factors $5$ and $7$ are part of the force term.

Common Mistakes

Leaving $p$ and $q$ in $\text{cm}$ , giving moments 100 times too large.
Giving units as $\text{N cm}$ instead of $\text{N m}$ .
Rounding too aggressively (e.g. 1 s.f.) so later graph work is degraded.

Things to Be Careful About

Keep consistent significant figures: if $p$ is to 3 s.f., $T_P$ should usually be to 3 s.f.
Use $\text{N m}$ (not $\text{Nm}$ without a space is fine, but $\text{N m}$ is clearer).
Ensure you substitute the correct $p$ into $T_P$ and the correct $q$ into $T_Q$ .

Techniques used

convert measured lengths into SI unitssubstitute values into given expressionscalculate a moment as force multiplied by perpendicular distancequote answers with appropriate significant figures and 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 $p$ and $q$ (with the rod horizontal each time) in one table, and calculate $T_P$ and $T_Q$ for each set using

T_P = 5Wp, \qquad T_Q = 7Wq \qquad (W = 0.0981\ \text{N})

A suitable table format is:

$p / \text{cm}$	$q / \text{cm}$	$T_P / \text{N m}$	$T_Q / \text{N m}$
15.0	22.0	0.0736	0.151
16.5	24.0	0.0809	0.165
18.0	26.0	0.0883	0.179
19.5	28.0	0.0957	0.192
21.0	30.0	0.103	0.206
22.5	32.0	0.110	0.220

(Values shown are an example; your readings will differ.)

Final answer

Completed results table with six sets of p, q and calculated TP, TQ (student-dependent).

Detailed explanation

Background Concept

In Paper 3 practical questions, marks for a results table typically come from:

taking enough readings (here, six sets);
choosing a suitable range of the variable you change (here by moving $Q$ to change $q$ );
recording raw measurements ( $p$ , $q$ ) and including calculated quantities ( $T_P$ , $T_Q$ );
correct table layout: clear headings with quantity and unit, consistent precision, and values recorded sensibly.

Understanding the Question

You must repeatedly:

change the position of $Q$ (so $q$ changes),
adjust the position of $P$ until the rod is again parallel to the bench,
measure $p$ and $q$ ,
calculate the corresponding moments $T_P$ and $T_Q$ ,
record six complete sets in a table.

So your table must contain both measured distances and calculated moments.

Approach

Decide a range for $q$ (e.g. move $Q$ in steps of a few cm so the points are well spread).
Each time, re-level the rod by moving $P$ .
Measure $p$ and $q$ carefully from the same reference (the right-hand eye) to the centres of the masses.
Convert to metres when calculating $T_P$ and $T_Q$ so the units come out as $\text{N m}$ .
Present all data in a single clear table with appropriate headings.

Step-by-Step Reasoning

Choosing the range: If your $q$ values are too close together, your graph later will be clustered and the gradient/intercept will be uncertain. Aim for a spread (for example, $q$ covering roughly $20$ to $32\ \text{cm}$ depending on rod length).
Taking each reading:
- Place $Q$ at a new position.
- Adjust $P$ until the rod is horizontal again.
- Measure $p$ and $q$ to the nearest $0.1\ \text{cm}$ .
Calculating moments: For each row, use

T_P = 5W p, \qquad T_Q = 7W q,

with $p$ and $q$ in metres.

Example for one row (illustrative): if $p = 18.0\ \text{cm} = 0.180\ \text{m}$ ,

T_P = 5(0.0981)(0.180) = 0.0883\ \text{N m}.

Table conventions that gain credit:
- Each column has a heading like $p / \text{cm}$ , not just “ $p$ ”.
- The derived columns have correct unit $\text{N m}$ .
- Values in a column use consistent decimal places (e.g. $p$ and $q$ each to 0.1 cm; moments to 3 s.f.).

Key Takeaways

In practical papers, presentation is assessed: a clear, correctly headed table is part of the physics skill.
A wide range of readings improves the reliability of a graph and the constants derived from it.

Common Mistakes

Fewer than six sets of readings.
Omitting $T_P$ and $T_Q$ columns.
Missing units in headings, or mixing units within a column.
Using inconsistent precision (e.g. some $p$ values to $0.1\ \text{cm}$ and others to $1\ \text{cm}$ ).
Calculating moments using $p$ and $q$ in centimetres (gives values 100 times too large).

Things to Be Careful About

Keep the stands fixed after the initial set-up, as instructed, so the geometry (especially $\theta$ ) stays constant.
Ensure the rod is horizontal for every set before measuring; otherwise the relationship you graph later may not be linear.
Record measured values immediately to avoid transcription errors; calculate $T_P$ and $T_Q$ carefully (a calculator slip affects the graph).

Techniques used

vary the independent variable across a suitable rangemeasure and record repeated sets of readings systematicallycalculate derived quantities from raw measurementsrecord results in a single table with correct headings and unitsuse 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 $T_Q$ ( $\text{N m}$ ) on the $y$ -axis against $T_P$ ( $\text{N m}$ ) on the $x$ -axis using a suitable scale (using at least half the grid in each direction). Plot all six points accurately.

Final answer

Graph of TQ (y) against TP (x) plotted (student-dependent).

Detailed explanation

Background Concept

A graph is used to test whether two quantities are linearly related and to allow constants to be determined from the gradient and intercept. To gain plotting marks, you typically need:

correctly labelled axes with units,
sensible scales (not cramped, not using awkward increments like 3 per big square),
accurate plotting of each point.

Understanding the Question

You have calculated $T_P$ and $T_Q$ for six equilibria. You must plot $T_Q$ on the vertical axis and $T_P$ on the horizontal axis.

This choice of axes is important because later parts ask for gradient and intercept of $T_Q$ vs $T_P$ .

diagram-c-i-graph

Approach

Put $T_P$ on the $x$ -axis and $T_Q$ on the $y$ -axis.
Use the minimum and maximum values of your data to decide scales so that the points occupy a large portion of the grid.
Plot each point as a small cross; if a plotting method is specified by your centre, follow it.

Step-by-Step Reasoning

From your table, identify the range of $T_P$ and $T_Q$ .
Choose axis limits slightly beyond the smallest and largest values.
Mark equal intervals with an easy scale (e.g. $0.02\ \text{N m}$ per large square).
Label axes:
- $x$ -axis: $T_P / \text{N m}$
- $y$ -axis: $T_Q / \text{N m}$
Plot each of the six data pairs $(T_P, T_Q)$ .
Check that each point is in the correct position by re-reading the coordinates.

Key Takeaways

Correct axis choice and units are essential; you lose marks even if your points are correct.
Good scaling reduces uncertainty in the gradient and intercept.

Common Mistakes

Swapping axes (plotting $T_P$ on $y$ and $T_Q$ on $x$ ).
Missing units in axis labels.
Using a scale that wastes space (points clustered in a corner) or an awkward scale.
Plotting blobs/dots too large to judge accuracy.

Things to Be Careful About

Use consistent units ( $\text{N m}$ ) for both axes.
Plot the calculated moments, not the raw distances $p$ and $q$ .
Ensure the origin does not have to be included unless it helps; choose what best fits your data range (unless your teacher instructs otherwise).

Techniques used

choose appropriate axis scales that use most of the graph gridlabel axes with quantities and unitsplot experimental points accuratelyuse a sharp pencil and small, clear plot symbols

(ii)

Draw the straight line of best fit.

1M Similar

DifficultyEasy

Worked solution

Answer

Draw one straight line of best fit through the plotted points (not point-to-point), with the line balanced so that points are roughly evenly scattered about it.

Final answer

Straight line of best fit drawn.

Detailed explanation

Background Concept

A best-fit line represents the overall trend in data when a linear relationship is expected. Because experimental results scatter, the line should not be forced through every point; instead, it should pass through the middle of the cluster.

Understanding the Question

After plotting $T_Q$ against $T_P$ , you must draw the best-fit straight line. This line will be used in the next part to determine gradient and intercept.

Approach

Use a ruler to draw a single straight line.
Position it so that the spread of points above and below the line is roughly balanced.
Do not join points in a zig-zag.

Step-by-Step Reasoning

Visually judge the trend: your points should lie approximately along a straight line.
Place the ruler so the line passes close to as many points as possible.
Adjust slightly so that the deviations (vertical distances) are not all on one side.
Draw the line across the full range of your plotted data (not just between the first and last point).

Key Takeaways

Best-fit means “overall trend”, not “through all points”.
A good best-fit line improves the reliability of the gradient and intercept.

Common Mistakes

Drawing a line through the first and last points regardless of the middle points.
Drawing a line that only covers a short section of the data.
Joining dots point-to-point.

Things to Be Careful About

If there is an outlier, do not force the line to go through it; keep the line representative of the majority of points.
Use a sharp pencil and a ruler for a thin, clear line.

Techniques used

draw a single straight line of best fit through the trend of pointsbalance the line so residuals are roughly even on both sidesignore minor scatter rather than joining point-to-point

(iii)

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

gradient = ______
$y$ -intercept = ______

2M Similar

DifficultyMedium

Worked solution

Working

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

\text{gradient} = \frac{\Delta T_Q}{\Delta T_P}

Read the $y$ -intercept where $T_P = 0$ .

Answer

(Example only.)

gradient $= 1.00$

$y$ -intercept $= 0.100\ \text{N m}$

Final answer

Student-dependent (gradient and y-intercept from best-fit line).

Detailed explanation

Background Concept

For a straight-line graph of the form

y = mx + c,

the gradient is $m = \Delta y / \Delta x$ (using two points on the line),
the y-intercept is $c$ , the value of $y$ when $x = 0$ .

On a $T_Q$ vs $T_P$ graph, the gradient is dimensionless because both axes have the same units ( $\text{N m}$ ), while the intercept has units of $T_Q$ , i.e. $\text{N m}$ .

Understanding the Question

You must obtain numerical values for:

gradient of your best-fit line,
y-intercept of your best-fit line.

These will be used directly in part (d).

Approach

Use a large triangle on the best-fit line: pick two points far apart to reduce percentage reading uncertainty.
Compute gradient as $\Delta T_Q / \Delta T_P$ .
Find the intercept by reading where the line crosses the $T_Q$ axis (at $T_P = 0$ ), extending the line if necessary.

Step-by-Step Reasoning

Identify two points on the line (not necessarily measured points) that lie exactly on grid intersections if possible.
Read their coordinates $(T_{P1}, T_{Q1})$ and $(T_{P2}, T_{Q2})$ .
Calculate changes:

\Delta T_Q = T_{Q2} - T_{Q1}, \qquad \Delta T_P = T_{P2} - T_{P1}.

Gradient:

\text{gradient} = \frac{\Delta T_Q}{\Delta T_P}.

For the y-intercept, set $T_P = 0$ and read the corresponding $T_Q$ where the line crosses the y-axis.
Quote the gradient to 2–3 s.f. and the intercept to a similar precision to your plotted values.

Key Takeaways

Use a large triangle: the bigger the $\Delta x$ and $\Delta y$ , the smaller the percentage error from reading.
Gradient is $\Delta y / \Delta x$ , not $\Delta x / \Delta y$ .

Common Mistakes

Using two data points rather than two points on the best-fit line.
Using a small triangle (large uncertainty in gradient).
Calculating gradient as $T_Q / T_P$ from one point (wrong unless line passes through origin).
Forgetting units for the intercept.

Things to Be Careful About

Keep the order consistent so that both $\Delta T_Q$ and $\Delta T_P$ are positive (or both negative); the gradient sign must match the line slope.
If the y-axis does not start at zero, read the intercept carefully using the axis scale.
Remember: intercept has units ( $\text{N m}$ ), gradient here is unitless because it is $(\text{N m})/(\text{N m})$ .

Techniques used

determine gradient from a large triangle on the best-fit linecalculate gradient as change in y divided by change in xread y-intercept by extrapolating the best-fit line to the y-axisquote gradient and intercept with appropriate units

(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

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

A = \text{$y$-intercept}, \qquad B = \text{gradient}.

Answer

$A =$ (your $y$ -intercept) in $\text{N m}$

$B =$ (your gradient), unitless

(Example only: $A = 0.100\ \text{N m}$ , $B = 1.00$ .)

Final answer

A = y-intercept (N m), B = gradient (unitless) (student-dependent).

Detailed explanation

Background Concept

If experimental data follows

T_Q = A + BT_P,

and you plot $T_Q$ against $T_P$ , you are plotting a straight line with:

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

Units:

$T_Q$ and $T_P$ are moments, so both are in $\text{N m}$ .
$A$ must have the same unit as $T_Q$ , so $\text{N m}$ .
$B$ multiplies $T_P$ to give $T_Q$ , so $B$ has units $(\text{N m})/(\text{N m}) = 1$ , i.e. no unit.

Understanding the Question

You have already found the gradient and y-intercept of the graph in (c)(iii). This part asks you to translate those into the constants $A$ and $B$ in the suggested equation.

Approach

Recognise the mapping to $y = mx + c$ .
Read off $A$ from the intercept and $B$ from the gradient.
State correct units.

Step-by-Step Reasoning

From the graph:

$y$ corresponds to $T_Q$ .
$x$ corresponds to $T_P$ .

So comparing

T_Q = A + BT_P

with

y = c + mx,

gives:

A = c = \text{y-intercept}, \qquad B = m = \text{gradient}.

Then assign units:

$A$ in $\text{N m}$ .
$B$ dimensionless.

Key Takeaways

A straight-line equation can be read directly from the graph: intercept and slope become constants.
Units can be checked by dimensional consistency.

Common Mistakes

Swapping $A$ and $B$ .
Giving a unit for $B$ (it is unitless here).
Using a gradient/intercept from a wrongly labelled graph (e.g. axes swapped earlier).

Things to Be Careful About

If you plotted $T_P$ vs $T_Q$ by mistake, then your gradient/intercept would not correspond to $A$ and $B$ as stated.
Quote $A$ and $B$ to a sensible number of significant figures consistent with (c)(iii).

Techniques used

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

(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

Use

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

(Example using $F = 2.94\ \text{N}$ , $\theta = 45^\circ$ , $A = 0.100\ \text{N m}$ , $S = 50.0\ \text{cm} = 0.500\ \text{m}$ .)

R = 2.94\sin45^\circ - \frac{2(0.100)}{0.500}

R = 2.08 - 0.400 = 1.68\ \text{N}

Answer

$R = 1.68\ \text{N}$ (example only; use your values)

Final answer

Student-dependent (R in N).

Detailed explanation

Background Concept

This part uses substitution into a provided theoretical relationship:

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

Key points:

$F\sin\theta$ is a component of a force (units: N).
$A$ has units $\text{N m}$ and $S$ has units m, so $(2A/S)$ has units N.
Therefore $R$ comes out in newtons, consistent with being a weight.

Understanding the Question

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

your measured $\theta$ from (a)(i),
your measured $S$ from (a)(i),
your $A$ from (d)(i),

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

Approach

Convert $S$ to metres.
Calculate $\sin\theta$ (ensure calculator is in degrees).
Compute $F\sin\theta$ .
Compute $2A/S$ .
Subtract to get $R$ and give the unit N.

Step-by-Step Reasoning

Using example values to demonstrate the method:

Suppose $\theta = 45^\circ$ and $S = 50.0\ \text{cm}$ .

Convert $S$ :

S = 50.0\ \text{cm} = 0.500\ \text{m}.

Suppose from the graph you found $A = 0.100\ \text{N m}$ .
Compute the force component:

F\sin\theta = 2.94\sin45^\circ = 2.94(0.7071) \approx 2.08\ \text{N}.

Compute the second term:

\frac{2A}{S} = \frac{2(0.100)}{0.500} = 0.400\ \text{N}.

Then

R = 2.08 - 0.400 = 1.68\ \text{N}.

Your final value depends on your measured $\theta$ , $S$ , and your experimental intercept $A$ .

Key Takeaways

Check units: $A$ must be in $\text{N m}$ and $S$ in m so that $2A/S$ is in N.
The final answer is a weight, so it should be in newtons.

Common Mistakes

Using $S$ in centimetres in the formula (makes $2A/S$ 100 times too small).
Calculator in radians instead of degrees when evaluating $\sin\theta$ .
Using $B$ instead of $A$ in the formula.
Forgetting the unit N for $R$ .

Things to Be Careful About

Make sure your $\theta$ is the angle between the string and the rod (as defined), not the angle to the vertical.
If your $A$ is negative from the graph, keep the sign consistent in $-2A/S$ (subtracting a negative becomes adding).
Quote $R$ to an appropriate number of significant figures, usually limited by $A$ , $S$ , and $\theta$ precision.

Techniques used

substitute measured and derived quantities into a given formulause trigonometry to evaluate a sine termconvert a measured length to SI units before substitutioncheck units for consistency to obtain the unit of the final quantity

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

Measured diameter (example):

d = 6.50\ \text{cm}

Final answer

d = 6.50 cm

Detailed explanation

Background Concept

A diameter is a straight-line distance across a circular cross-section through the centre. When measuring a length with a ruler, the key ideas are:

choose an instrument with suitable resolution (smallest scale division)
avoid parallax (read the scale with your eye directly above the mark)
record to a precision consistent with the instrument (typically to the nearest mm for a ruler).

Understanding the Question

You are given a plastic bottle (Fig. 2.1) and asked to measure the diameter $d$ of its base. This is a straightforward measurement to be written in the answer space.

Approach

Place the bottle so you can measure across the widest part of the base. Use a ruler (or vernier calipers if available). Read the value carefully and record it with appropriate units and precision.

Step-by-Step Reasoning

Place the base of the bottle next to the ruler, aligning one edge with the zero (or note the reading at one edge if zero is damaged).
Measure across to the opposite edge of the base, ensuring the ruler passes through the centre (maximum width).
Read the scale at eye level.
Record $d$ to the nearest mm (or to the caliper resolution).

A typical bottle might give $d \approx 6.50\ \text{cm}$ (example only).

Key Takeaways

A measurement mark is earned by a sensible value recorded with correct unit and appropriate precision.
Parallax and zero-error are the common issues to control.

Common Mistakes

Recording without a unit.
Measuring not through the centre (giving a chord smaller than the diameter).
Reading the ruler at an angle (parallax).

Things to Be Careful About

If you do not start at zero, use $d = \text{reading}_2 - \text{reading}_1$ .
Ensure the bottle is not tilted and the ruler is not slipping around the curved surface.

Techniques used

measure a diameter with a suitable length scaletake readings at eye level to reduce parallaxrecord a measurement to an appropriate precision

(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 mm divisions, take absolute uncertainty as $\pm 0.5\ \text{mm} = \pm 0.05\ \text{cm}$ .

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

Answer

\text{percentage uncertainty} \approx 0.8\%

Final answer

0.8%

Detailed explanation

Background Concept

Uncertainty for a single reading is typically related to the instrument resolution:

For a ruler with smallest division 1 mm, a common estimate is $\pm 0.5\ \text{mm}$ .
Percentage uncertainty is

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

Understanding the Question

You have measured the diameter $d$ and must estimate the percentage uncertainty in that measurement, showing working.

Approach

Decide a sensible absolute uncertainty in $d$ from the measuring instrument.
Divide by the measured $d$ .
Multiply by 100 to convert to a percentage.

Step-by-Step Reasoning

If you used a ruler (1 mm scale), the reading uncertainty is usually $\pm 0.5\ \text{mm}$ because you are judging between the smallest marks.
Convert that to the same unit as $d$ .
Substitute into the percentage formula.

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

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

(The units cancel, as they should for a percentage.)

Key Takeaways

Always use a consistent unit before dividing.
State your chosen absolute uncertainty clearly.

Common Mistakes

Using $\pm 1\ \text{mm}$ automatically instead of $\pm 0.5\ \text{mm}$ for a single ruler reading.
Forgetting to multiply by 100.
Mixing units (e.g. mm in numerator, cm in denominator).

Things to Be Careful About

If you measured $d$ by taking two ruler readings (one at each edge), the uncertainty can be larger because it involves two readings; many students still use $\pm 0.5\ \text{mm}$ if they aligned one edge with zero, but if you subtracted two readings you should consider the combined effect.
Quote the final percentage to a sensible number of significant figures (usually 1 s.f. is fine for an uncertainty).

Techniques used

estimate absolute uncertainty from instrument resolutioncalculate percentage uncertaintyshow substitution and final percentage

(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

DifficultyEasy

Worked solution

Answer

Measured rolling distance (example):

L = 85.0\ \text{cm}

Final answer

L = 85.0 cm

Detailed explanation

Background Concept

In practical work, a distance like $L$ should be measured between clearly defined reference points, using an appropriate instrument (here a metre rule). Good practice includes:

aligning the zero with the start point (or subtracting two readings)
measuring along the direction of motion
recording to the instrument precision.

Understanding the Question

After setting up the two metre rules as a track and adjusting them so the bottle rolls to the end, you must measure and record the distance $L$ that the bottle rolls along the rules (as shown in Fig. 2.4).

Approach

Mark or identify the bottle’s start position and the end of the rules, then measure the separation along the rules with a ruler/metre rule and record it with units.

Step-by-Step Reasoning

Place the bottle at the chosen start position on the rules.
Ensure the bottle rolls to the end after adjusting the rules.
Measure from the start position to the end of the rules (or the point defined in the diagram) along the direction of roll.
Record $L$ to the nearest mm (0.1 cm) if using a metre rule.

Example: $L = 85.0\ \text{cm}$ .

Key Takeaways

Define start and end points consistently.
Record with appropriate precision and units.

Common Mistakes

Measuring the horizontal projection instead of the distance along the rules.
Changing the start point between trials without noticing.
Recording an over-precise value (e.g. many decimal places) when using a ruler.

Things to Be Careful About

If the bottle does not always stop at the same point, define the endpoint as the end of the rules rather than the bottle’s final resting position.
Avoid parallax when reading the scale.

Techniques used

set up apparatus to obtain a usable range of motionidentify the measured distance along the trackrecord a length to appropriate precision

(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-Easy

Worked solution

Working

Measured water height (example):

h = 19.0\ \text{cm}

Time for bottle to roll distance $L$ (repeat and mean):

t_1 = 1.92\ \text{s},\quad t_2 = 1.88\ \text{s},\quad t_3 = 1.90\ \text{s}

\bar{t} = \frac{1.92+1.88+1.90}{3} = 1.90\ \text{s}

Answer

t = 1.90\ \text{s}

Final answer

h = 19.0 cm, t = 1.90 s

Detailed explanation

Background Concept

Two measurements are required here:

The height $h$ of the water column in the upright bottle. This is a length reading; for liquids you should read the meniscus and avoid parallax.
The time $t$ taken for the bottle to roll the distance $L$ . Human reaction time makes single stopwatch timings unreliable, so repeating and averaging improves data quality.

Understanding the Question

With the bottle upright, measure the water height $h$ (Fig. 2.5). Then, for the same set-up used earlier (distance $L$ ), determine the time $t$ taken for the bottle to roll that distance.

Approach

Measure $h$ with a ruler placed next to the bottle, reading the water level carefully.
For $t$ , use a stopwatch and repeat the roll several times, keeping the same start position and distance $L$ , then calculate a mean time.

Step-by-Step Reasoning

Measuring $h$
- Stand the bottle upright on a level surface.
- Place a ruler next to the bottle and read the height of the water surface above the base.
- Read at eye level to avoid parallax.
Measuring $t$
- Put the bottle on the rules at the defined start point.
- Start the stopwatch at the instant you release the bottle (try not to push it).
- Stop the watch when the bottle reaches the end point corresponding to $L$ .
- Repeat at least three times and compute the mean.

Example:

\bar{t} = \frac{t_1+t_2+t_3}{3}.

Key Takeaways

Repeats and a mean reduce random timing scatter.
Keep the definition of start/end consistent between runs.

Common Mistakes

Only one timing reading (large random error).
Starting/stopping the stopwatch late due to unclear start/end points.
Measuring $h$ from the bench instead of from the base of the bottle.

Things to Be Careful About

If the water surface is moving, wait for it to settle before measuring $h$ .
Use the same $L$ and same start position for every timing so that $t$ corresponds to the same motion each time.

Techniques used

measure a liquid height with a ruler and read the meniscustime a motion using repeated trials and calculate a meankeep the measured distance constant while timing

(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

Convert to SI units:

L = 85.0\ \text{cm} = 0.850\ \text{m},\quad t = 1.90\ \text{s}

a = \frac{2L}{t^2} = \frac{2(0.850)}{(1.90)^2} = 0.471\ \text{m s}^{-2}

Answer

a = 0.471\ \text{m s}^{-2}

Final answer

0.471 m s^-2

Detailed explanation

Background Concept

If an object starts from rest and accelerates uniformly along a straight line, then

s = \frac{1}{2}at^2.

Rearranging gives

a = \frac{2s}{t^2}.

Here $s$ is the distance travelled (given as $L$ ) and $t$ is the time taken.

Understanding the Question

You are given the formula

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

and must calculate $a$ using your measured $L$ and $t$ .

Approach

Convert $L$ to metres so that $a$ comes out in $\text{m s}^{-2}$ .
Substitute values into the equation.
Evaluate carefully, especially the square of $t$ .

Step-by-Step Reasoning

Convert $L$ from cm to m:

85.0\ \text{cm} = 0.850\ \text{m}.

Square the time:

t^2 = (1.90\ \text{s})^2 = 3.61\ \text{s}^2.

Substitute into the formula:

a = \frac{2(0.850)}{3.61} = 0.471\ \text{m s}^{-2}.

Key Takeaways

Always put lengths into metres before calculating acceleration.
Check that the final unit is $\text{m s}^{-2}$ .

Common Mistakes

Using $L$ in cm but still writing $\text{m s}^{-2}$ .
Forgetting to square $t$ .
Rounding too early (carry extra digits, round at the end).

Things to Be Careful About

This method assumes approximately uniform acceleration from rest; in real rolling motion the acceleration may vary, but the question instructs you to use the formula as defined.

Techniques used

convert measured quantities to SI unitssubstitute into a provided formulacalculate 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

$a$ depends on $L$ and $t^2$ . Here $t = 1.90\ \text{s}$ (3 s.f.) and $L = 0.850\ \text{m}$ (3 s.f.), so $a$ should be given to 3 s.f.

a = 0.471\ \text{m s}^{-2}\ (3\ \text{s.f.})

Final answer

3 s.f. (e.g. a = 0.471 m s^-2)

Detailed explanation

Background Concept

Significant figures (s.f.) communicate the precision of a measured or calculated value. For a calculated quantity, the final answer should not imply greater precision than the least precise measurement used to calculate it.

A common practical rule used at A Level:

Quote the calculated result to the same number of significant figures as the least precise input quantity (or sometimes to the limiting uncertainty).

Understanding the Question

You have calculated acceleration $a$ using

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

The question asks you to justify the number of significant figures in your quoted value of $a$ .

Approach

Identify the significant figures (or practical precision) of $L$ and $t$ .
Since $t$ is squared, it still limits precision because it is measured with the same stopwatch resolution and reaction-time scatter.
Quote $a$ to a sensible number of s.f. consistent with the least precise measurement.

Step-by-Step Reasoning

If you recorded, for example:

$L = 0.850\ \text{m}$ (3 s.f., measured to about $\pm 0.001\ \text{m}$ )
$t = 1.90\ \text{s}$ (3 s.f., but in practice limited by stopwatch resolution and reaction time)

then quoting $a$ to 3 s.f. is appropriate, e.g. $0.471\ \text{m s}^{-2}$ . Quoting 4 or 5 s.f. would suggest unrealistic precision.

Key Takeaways

Derived results should reflect measurement precision.
Over-precise answers can lose marks in practical papers.

Common Mistakes

Writing many digits from a calculator (e.g. $0.470914\ldots$ ).
Quoting $a$ to fewer s.f. than the data supports without reason.

Things to Be Careful About

If your timing readings show noticeable scatter, that may justify fewer s.f. even if you wrote $t$ to 3 s.f.
Always keep units with the final value.

Techniques used

identify the least precise measured quantitymatch significant figures of a derived quantity to the input datajustify rounding using measurement precision

(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, measured (example):

h = 9.5\ \text{cm}

Repeated timings and mean:

t_1 = 2.29\ \text{s},\quad t_2 = 2.32\ \text{s},\quad t_3 = 2.28\ \text{s}

t = \frac{2.29+2.32+2.28}{3} = 2.30\ \text{s}

With $L = 0.850\ \text{m}$ :

a = \frac{2L}{t^2} = \frac{2(0.850)}{(2.30)^2} = 0.321\ \text{m s}^{-2}

Answer

h = 9.5\ \text{cm},\quad t = 2.30\ \text{s},\quad a = 0.321\ \text{m s}^{-2}

Final answer

h = 9.5 cm, t = 2.30 s, a = 0.321 m s^-2

Detailed explanation

Background Concept

Repeating an experiment with a changed condition (here, a smaller water height $h$ ) gives another data point. To make a fair comparison, you should keep other factors as constant as possible (same $L$ , same track set-up, same release technique) and use repeats to reduce random error.

Understanding the Question

You must pour out about half the water and then repeat the measurements and calculation from parts (c)(i) and (c)(ii): measure $h$ , determine $t$ for the same distance $L$ , and calculate $a$ .

Approach

Change only the water amount.
Measure the new $h$ .
Repeat the timing procedure several times and take the mean to find $t$ .
Use the same $L$ and compute $a = 2L/t^2$ .

Step-by-Step Reasoning

After removing about half the water, stand bottle upright and measure $h$ as before.
Place bottle on rules at the same start point and time the roll over distance $L$ .
Repeat timings (at least 3) and calculate the mean time.
Substitute $L$ and mean $t$ into the acceleration formula.

Example calculation:

If $t = 2.30\ \text{s}$ ,

a = \frac{2(0.850)}{(2.30)^2} = 0.321\ \text{m s}^{-2}.

Key Takeaways

Consistency is crucial: keep $L$ and the track unchanged while changing $h$ .
Averaging repeated times improves reliability.

Common Mistakes

Changing the start position, so $L$ is no longer the same.
Not re-measuring $h$ after pouring out water.
Using a single timing reading.

Things to Be Careful About

Water can slosh; let it settle before timing and before measuring $h$ .
The rules may shift when you adjust them; ensure the set-up is stable before collecting timings.

Techniques used

repeat measurements after changing one variabletime the motion and calculate a meancalculate acceleration using the same method as before

(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}}

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

First set ( $h = 19.0\ \text{cm} = 0.190\ \text{m}$ , $a = 0.471\ \text{m s}^{-2}$ ):

k_1 = \frac{2(0.471)}{0.0650\sqrt{\pi(0.190)}} = 18.8\ \text{m}^{-1/2}\ \text{s}^{-2}

Second set ( $h = 9.5\ \text{cm} = 0.095\ \text{m}$ , $a = 0.321\ \text{m s}^{-2}$ ):

k_2 = \frac{2(0.321)}{0.0650\sqrt{\pi(0.095)}} = 18.1\ \text{m}^{-1/2}\ \text{s}^{-2}

Answer

\text{first value of } k = 18.8\ \text{m}^{-1/2}\ \text{s}^{-2}

\text{second value of } k = 18.1\ \text{m}^{-1/2}\ \text{s}^{-2}

Final answer

k1 = 18.8 m^-1/2 s^-2, k2 = 18.1 m^-1/2 s^-2

Detailed explanation

Background Concept

When a relationship includes an unknown constant $k$ , you can test consistency by rearranging the equation to make $k$ the subject and calculating it from each set of measurements. If the model is correct and uncertainties are reasonable, the calculated values of $k$ should be approximately the same.

Here,

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

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

Understanding the Question

You have two sets of data (with different water heights $h$ ) and corresponding accelerations $a$ . Using your measured bottle diameter $d$ , you must calculate two values of $k$ using the suggested relationship.

Approach

Rearrange to $k = 2a/(d\sqrt{\pi h})$ .
Convert $d$ and $h$ to metres (SI), so the calculation is consistent.
Substitute values for each dataset to obtain $k_1$ and $k_2$ .

Step-by-Step Reasoning

Rearrange the equation:

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

Convert lengths to metres (important because $\sqrt{h}$ appears):

$d$ in m
$h$ in m

For each dataset:

compute $\pi h$
take the square root
multiply by $d$
divide $2a$ by that result

Compare the two $k$ values: if close, it supports the proposed model.

Key Takeaways

Rearrangement and consistent units are essential.
Calculating $k$ twice is a consistency check on the suggested relationship.

Common Mistakes

Leaving $d$ and $h$ in cm, which changes the numerical value (and units) of $k$ .
Forgetting the factor of 2.
Calculating $\sqrt{\pi}\,h$ instead of $\sqrt{\pi h}$ .

Things to Be Careful About

Use brackets: $\sqrt{\pi h}$ means the square root of the product.
Because $\sqrt{h}$ is involved, a factor of 100 in $h$ becomes a factor of 10 in $\sqrt{h}$ , so unit errors can be subtle but large.

Techniques used

rearrange an equation to make the constant the subjectsubstitute experimental values consistently in SI unitscalculate and report two values for a 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

Using $k_1 = 18.8$ and $k_2 = 18.1$ :

\%\ \text{difference} = \frac{|18.8-18.1|}{\frac{18.8+18.1}{2}}\times 100 = 3.8\%

Since $3.8\% < 15\%$ , the values agree within the stated uncertainty.

Answer

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

Final answer

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

Detailed explanation

Background Concept

To decide whether results support a model, you compare calculated constants (here $k$ ) from different trials. If differences between them are no larger than the expected experimental uncertainty, then the data are consistent with the model.

A common way is percentage difference between two values:

\%\ \text{difference} = \frac{|x_1-x_2|}{\text{mean of }x_1,x_2} \times 100.

Alternatively, you can check overlap of uncertainty ranges (e.g. $k \pm 15\%$ ).

Understanding the Question

You are told that the percentage uncertainty in $k$ is $15\%$ . Using that, you must say whether your two calculated $k$ values support the suggested relationship.

Approach

Compute how different $k_1$ and $k_2$ are (percentage difference).
Compare that with $15\%$ .
If the difference is smaller, conclude that they agree within uncertainty and thus support the relationship.

Step-by-Step Reasoning

With example values $k_1 = 18.8$ and $k_2 = 18.1$ :

\text{mean }k = \frac{18.8+18.1}{2} = 18.45.

\%\ \text{difference} = \frac{0.7}{18.45}\times 100 \approx 3.8\%.

Because $3.8\%$ is less than $15\%$ , the two values are consistent with each other given the stated uncertainty, so the relationship is supported.

Key Takeaways

“Support” means “consistent within uncertainty”, not “exactly equal”.
Use a quantitative comparison, not just “they look close”.

Common Mistakes

Comparing the absolute difference (e.g. $0.7$ ) without converting to a percentage.
Using $15\%$ incorrectly (e.g. subtracting 15 instead of 15%).
Saying the relationship is proven (experiments can support, not prove).

Things to Be Careful About

If your two $k$ values differ by more than $15\%$ , the correct conclusion is that the results do not support the relationship (or that uncertainty is underestimated / systematic error exists).

Techniques used

compare two experimental values using percentage differenceapply an uncertainty criterion to judge agreementstate a conclusion linked to the calculated values

(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-Easy

Worked solution

Answer

Time $t$ : stopwatch reaction time when starting/stopping; difficult to judge exact release and exact end point.
Time $t$ : bottle may not be released identically each time (small push / different initial conditions), so timings vary.
Distance $L$ : uncertainty in defining the start position of the bottle and the instant it reaches the end; bottle edge is curved so the reference point is unclear.
Height $h$ : water level (meniscus) is difficult to read accurately; parallax and water surface may not be level/steady.

(Any other valid limitations: rules not perfectly parallel / changing friction; water sloshing during motion.)

Final answer

See working (four sources listed).

Detailed explanation

Background Concept

Uncertainties and limitations are factors that cause measured values to vary (random error) or to be consistently offset (systematic error). High-mark practical answers:

name the quantity affected (e.g. $t$ , $h$ , $L$ , $d$ )
give a clear reason based on the method/apparatus
avoid vague statements like “human error” without explanation.

Understanding the Question

You must describe four sources of uncertainty or limitations in this rolling-bottle experiment. For measurement uncertainties, you must state the quantity measured and why it is uncertain.

Approach

Look at each measured quantity ( $d$ , $L$ , $h$ , $t$ ) and the procedure steps (release, rolling, judging the end point). Identify what makes each measurement hard or variable, then express four distinct points.

Step-by-Step Reasoning

Examples of creditworthy points:

Timing $t$ : reaction time
- Starting the stopwatch exactly at release and stopping it exactly at the end point depends on human judgment, giving random uncertainty.
Timing $t$ : inconsistent release
- A slight push, different angle, or different contact with the rules changes the motion and hence $t$ .
Distance $L$ : defining start/end
- The bottle is curved, so the reference point for where it “starts” and when it “reaches the end” can be ambiguous, affecting $L$ .
Height $h$ : meniscus/parallax and stability
- The water surface may be curved and may slosh; reading the meniscus with a ruler can introduce parallax error.

Other valid limitations (if needed):

Track/rolling conditions: rules may not remain exactly parallel or may flex; friction may change along the track.
Water movement: water sloshing changes the distribution of mass during rolling, so acceleration may not be constant.

Key Takeaways

Always tie uncertainty to a named measurement.
Give a mechanism: why does this step introduce scatter or bias?

Common Mistakes

Writing “human error” without specifying what and why.
Listing the same idea twice in different words (e.g. “reaction time” and “slow stopwatch”).
Giving improvements instead of limitations (those belong in part (g)(ii)).

Things to Be Careful About

Make sure your four points are distinct.
Prefer practical-specific statements (start/end judgment, sloshing, rule alignment) over generic claims.

Techniques used

identify random and systematic uncertainty sourceslink each 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 $t$ using light gates / motion sensor / video analysis instead of a stopwatch to remove reaction-time uncertainty.
Use a mechanical release (e.g. gate or clamp) so the bottle is released without a push and from the same position each time.
Fix the metre rules with spacers/clamps to keep them parallel and prevent movement; mark a single clear start line.
Increase reliability by repeating many trials for each $h$ and taking a mean (and possibly use a longer $L$ to reduce fractional timing uncertainty).

(Other valid: use vernier calipers for $d$ ; use a set square to read $h$ .)

Final answer

See working (four improvements listed).

Detailed explanation

Background Concept

Improvements are changes that reduce uncertainty or remove limitations. Strong answers:

are specific (name apparatus/procedure)
explain what uncertainty is reduced
are practical and realistic in a school laboratory.

Understanding the Question

You must describe four improvements to the rolling-bottle experiment. You may suggest different apparatus or procedures.

Approach

Take the limitations from (g)(i) and propose one improvement for each. Focus on the largest uncertainty sources: timing, consistency of release, track alignment, and repeatability.

Step-by-Step Reasoning

Examples of clear improvement statements:

Replace stopwatch timing
- Use two light gates a known distance apart (or a data logger/video). This gives objective timing and reduces reaction-time error.
Standardise the release
- Use a mechanical gate so the bottle starts from rest at exactly the same position every time, avoiding accidental pushes.
Stabilise and align the track
- Clamp the rules and use spacers to keep a fixed separation and parallel alignment; this reduces run-to-run changes in friction/contact.
Improve reliability with repeats (and/or longer distance)
- Take more repeated timings and calculate a mean; if possible increase $L$ so the time is longer, reducing the percentage timing uncertainty.

Other good improvements:

Use vernier calipers for $d$ .
Use a set square against the bottle to read $h$ without parallax.

Key Takeaways

Improvements should directly address the biggest uncertainties.
Mentioning appropriate apparatus (light gates, video, clamps) is often rewarded.

Common Mistakes

Repeating the limitation instead of improving it.
Suggesting unrealistic changes (e.g. “use a frictionless surface”).
Giving vague statements like “be more careful”.

Things to Be Careful About

Ensure each improvement is distinct.
If you suggest repeats, explicitly say “calculate the mean” to show how it improves data quality.

Techniques used

propose apparatus changes that reduce a named uncertaintypropose procedural changes to improve repeatabilitylink each improvement to a specific limitation