Physics 9702/33 — May/June 2025

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

questions

marks

120

minutes

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

Q1Medium-HardManipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and Evaluation

In this experiment, you will investigate an electrical circuit.

(a)

• Set up the circuit shown in Fig. 1.1.

Fig. 1.1

• Record the voltmeter reading $E$ .

$E$ = ______ $\text{V}$

1M Similar

(b)

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

• Set up the circuit shown in Fig. 1.2.

Fig. 1.2

• F and G are crocodile clips.

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

• Close the switch.

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

$L$ = ______
$V$ = ______

• Open the switch.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Example readings (to appropriate precision):

L = 0.317\ \text{m}

V = 1.04\ \text{V}

Final answer

L = 0.317 m, V = 1.04 V (example)

Detailed explanation

Background Concept

A metre rule allows you to measure a distance along the wire between two crocodile clips. The length $L$ is the separation between the contact points of clips F and G. A voltmeter measures the p.d. $V$ across the component(s) to which it is connected in parallel.

Understanding the Question

You must build the circuit in Fig. 1.2, set $L$ to about $30\ \text{cm}$ , then close the switch and record $L$ and the voltmeter reading $V$ . Finally, you open the switch.

Approach

Attach clips F and G so that their separation is approximately $0.30\ \text{m}$ .
Close the switch only while taking readings (to reduce heating changes).
Read $L$ from the metre rule and read $V$ from the voltmeter.

Step-by-Step Reasoning

Place F and G on the wire and measure $L$ as the difference between their positions on the metre rule.
Close the switch and wait briefly for the reading to settle.
Record $V$ from the voltmeter display.
Open the switch after recording to limit temperature rise in the wire/resistors.

Key Takeaways

Measure lengths between the actual contact points.
Take electrical readings with the switch closed only when needed.

Common Mistakes

Measuring $L$ from the wrong reference points (e.g. clip ends rather than contact points).
Leaving the switch closed continuously (heating changes resistance and affects readings).

Things to Be Careful About

Record $L$ to a sensible precision (typically $\pm 1\ \text{mm}$ , i.e. $0.001\ \text{m}$ ) and $V$ to the voltmeter resolution.
Ensure clips make firm contact with the wire (oxidation/loose contact increases resistance).

Techniques used

set a length using a metre rule and crocodile clipsmeasure length between two points on a wiremeasure potential difference with a voltmeter

(c)

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

$E$ = ______

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

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

9M Similar

DifficultyMedium-Hard

Worked solution

Answer

Use $E$ from (a). Example:

E = 1.50\ \text{V}

Record at least six sets of $L$ and $V$ (including (b)) and calculate

\frac{E - V}{L}

Example table (values illustrative):

$L / \text{m}$	$V / \text{V}$	$(E-V)/L\ /\ \text{V m}^{-1}$
0.317	1.04	1.45
0.400	1.00	1.25
0.482	0.97	1.10
0.589	0.94	0.951
0.682	0.92	0.850
0.800	0.90	0.750

Final answer

Table of six (L, V) readings with calculated (E−V)/L (student-dependent; example shown)

Detailed explanation

Background Concept

In practical work you often measure two quantities and then calculate a third “derived” quantity from them. Here the derived quantity is

\frac{E - V}{L}

where $E$ and $V$ are in volts and $L$ is a length along the wire. If $L$ is recorded in metres, then $(E-V)/L$ has unit $\text{V m}^{-1}$ .

Good tables:

include all readings in one place,
have clear column headings with quantity symbols and units,
show consistent precision (same decimal places) within a column.

Understanding the Question

You must:

Write down the value of $E$ from part (a).
Increase $L$ several times, each time recording $L$ and the corresponding voltmeter reading $V$ .
Obtain six sets of $(L, V)$ in total (including your first set from (b)).
Add a calculated column for $(E-V)/L$ .

Approach

Choose a sensible range of $L$ values (e.g. from about $0.30\ \text{m}$ up to a larger value, without exceeding the wire length).
For each $L$ $L$ :
- close the switch briefly,
- record $V$ once stable,
- open the switch.
After taking measurements, calculate $(E-V)/L$ for each row.

Step-by-Step Reasoning

Record $E$ once (from part (a)); use this same $E$ value for all rows.
Generate six data points:
- Start with $L \approx 0.30\ \text{m}$ .
- Move clip F to increase $L$ in suitable steps (to produce a good spread of values).
Fill the table:
- Column 1: $L$ with unit (preferably m). Keep a consistent number of decimal places (e.g. 0.001 m).
- Column 2: $V$ with unit (V). Keep consistent decimal places (e.g. 0.01 V).
Calculate derived column for each row:

\frac{E - V}{L}

Example calculation using one row (illustrative numbers):

\frac{E - V}{L} = \frac{1.50 - 1.04}{0.317} = \frac{0.46}{0.317} = 1.45\ \text{V m}^{-1}

Key Takeaways

Use a single, clear results table with headings and units.
Take enough readings (six) with a good range.
Show derived quantities calculated correctly and to sensible significant figures.

Common Mistakes

Forgetting to include the calculated $(E-V)/L$ column.
Missing units in headings, or writing units in the body of the table instead of the heading.
Inconsistent precision (e.g. mixing 0.3 m and 0.317 m in the same column).
Not including the data point from part (b) among the six.

Things to Be Careful About

Use the same $E$ value for every calculation.
Keep the switch closed only while reading $V$ to reduce heating.
Avoid parallax error when reading the metre rule: read with your eye directly above the scale mark at the clip contact point.

Techniques used

collect a suitable range of values of the independent variablerecord results in a table with headings and unitscalculate a derived quantity from measured valueskeep significant figures consistent within each column

(d)

(i)

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

3M Similar

DifficultyMedium

Worked solution

Answer

Plot a graph of $y = \dfrac{E-V}{L}$ (in $\text{V m}^{-1}$ ) on the $y$ -axis against $x = V$ (in V) on the $x$ -axis.

diagram-d-i-graph

Final answer

Graph of (E−V)/L against V (see plotting instructions)

Detailed explanation

Background Concept

A graph is used to reveal whether two quantities are related linearly and to allow constants to be found from the gradient and intercept. Good graph technique includes:

clear axis labels with units,
a sensible scale using at least half the grid in each direction,
accurate plotting with small, neat crosses,
plotting the correct variables on the correct axes.

Understanding the Question

You must plot $\dfrac{E-V}{L}$ on the vertical axis and $V$ on the horizontal axis using your six results from (c).

Approach

Compute $y = (E-V)/L$ for each reading.
Set up axes: horizontal axis $V$ (V), vertical axis $(E-V)/L$ ( $\text{V m}^{-1}$ if $L$ in m).
Choose scales that spread the points well.
Plot all six points.

Step-by-Step Reasoning

Decide the range of $V$ values (min to max) and the range of $(E-V)/L$ .
Draw axes and label them fully:
- $x$ -axis: $V / \text{V}$
- $y$ -axis: $(E-V)/L\ /\ \text{V m}^{-1}$
Pick scales (e.g. 1 large square = 0.05 V on $x$ , 1 large square = 0.1 $\text{V m}^{-1}$ on $y$ ) so points occupy much of the grid.
Plot each pair $(V, (E-V)/L)$ carefully.

diagram-d-i-graph

Key Takeaways

Always plot the variables exactly as specified.
Labels must include both symbol and unit.

Common Mistakes

Swapping axes (plotting $V$ on $y$ and $(E-V)/L$ on $x$ ).
Missing units in axis labels.
Choosing a scale that compresses points into a small area.

Things to Be Careful About

If you used $L$ in cm, then $(E-V)/L$ is in $\text{V cm}^{-1}$ ; be consistent and label accordingly.
Use a sharp pencil and plot small crosses; large blobs make best-fit difficult.

Techniques used

select appropriate axes and scales for a graphlabel axes with quantities and unitsplot data points accurately from a table

(ii)

Draw the straight line of best fit.

1M Similar

(iii)

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

gradient = ______
$y$ -intercept = ______

2M Similar

DifficultyMedium

Worked solution

Working

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

(V_1, y_1) = (0.90\ \text{V},\ 0.750\ \text{V m}^{-1}),\quad (V_2, y_2) = (1.04\ \text{V},\ 1.45\ \text{V m}^{-1})

\text{gradient} = \frac{\Delta y}{\Delta x} = \frac{1.45 - 0.750}{1.04 - 0.90} = \frac{0.700}{0.14} = 5.0\ \text{m}^{-1}

Using $y = mx + c$ with $(V, y) = (1.00\ \text{V},\ 1.25\ \text{V m}^{-1})$ :

c = y - mV = 1.25 - 5.0(1.00) = -3.75\ \text{V m}^{-1}

Answer

\text{gradient} = 5.0\ \text{m}^{-1}

y\text{-intercept} = -3.75\ \text{V m}^{-1}

Final answer

gradient = 5.0 m^-1, y-intercept = -3.75 V m^-1 (example)

Detailed explanation

Background Concept

For a straight-line graph, the equation is

y = mx + c

where:

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

Using two points that are far apart reduces the percentage uncertainty from reading the graph.

Understanding the Question

You have plotted $y = (E-V)/L$ against $x = V$ and drawn a best-fit line. You must now find:

the gradient of the best-fit line,
the $y$ -intercept of the best-fit line.

Approach

Pick two points on the drawn best-fit line (not necessarily your plotted data points), far apart.
Read their coordinates carefully.
Compute $m = (y_2-y_1)/(x_2-x_1)$ .
Find $c$ either by reading where the line crosses the $y$ -axis, or by substituting one point into $c = y - mx$ .

Step-by-Step Reasoning

Choose two points far apart on the best-fit line to form a large triangle.
Read values of $V$ from the $x$ -axis and $y=(E-V)/L$ from the $y$ -axis.
Calculate gradient:

\text{gradient} = \frac{\Delta y}{\Delta x}

Units: $y$ has units $\text{V m}^{-1}$ and $x$ has units V, so the gradient has units $\text{m}^{-1}$ .
4. Find intercept:

Either read directly at $V=0$ (where the line crosses the $y$ -axis),
Or calculate using $c = y - mV$ with any point on the line.

diagram-d-iii-gradient

Key Takeaways

Use two points on the line, not two experimental points close together.
Gradient is always $\Delta y/\Delta x$ .

Common Mistakes

Using $\Delta x/\Delta y$ (inverts the gradient).
Using two plotted points that are too close (large fractional uncertainty).
Forgetting units for gradient and intercept.

Things to Be Careful About

Read the coordinates from the line, not from the nearest plotted crosses if they are off the line.
Keep consistent units: if you plotted $y$ in $\text{V cm}^{-1}$ , the gradient unit becomes $\text{cm}^{-1}$ and intercept becomes $\text{V cm}^{-1}$ .

Techniques used

determine the gradient using two well-separated points on the best-fit linecalculate gradient as \u0394y/\u0394x with unitsdetermine the y-intercept from the best-fit line

(e)

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

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

where $P$ and $Q$ are constants.

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

$P$ = ______
$Q$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Given

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

This matches $y = mx + c$ with $y = \dfrac{E-V}{L}$ and $x = V$ .

P = \text{gradient}

and

-Q = y\text{-intercept}\ \Rightarrow\ Q = -\,(y\text{-intercept})

Using (d)(iii) (example):

P = 5.0\ \text{m}^{-1}

Q = -(-3.75\ \text{V m}^{-1}) = 3.75\ \text{V m}^{-1}

Answer

P = 5.0\ \text{m}^{-1}

Q = 3.75\ \text{V m}^{-1}

Final answer

P = gradient, Q = −(y-intercept) with units (example: P = 5.0 m^-1, Q = 3.75 V m^-1)

Detailed explanation

Background Concept

If a graph plots $y$ against $x$ and the relationship can be written as

y = mx + c

then:

gradient $m$ is the coefficient of $x$ ,
intercept $c$ is the constant term.

Units come from the fact that $m = y/x$ and $c$ has the same units as $y$ .

Understanding the Question

You are given the suggested relationship

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

You already found the gradient and $y$ -intercept from your graph of $\dfrac{E-V}{L}$ (on the $y$ -axis) against $V$ (on the $x$ -axis). You must now identify $P$ and $Q$ (with units).

Approach

Compare the given equation with $y = mx + c$ .
Read off which constant corresponds to the gradient and which corresponds to the intercept.
Use the plotted units to determine units of $P$ and $Q$ .

Step-by-Step Reasoning

Let

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

Then the given equation becomes

y = Px - Q

So the gradient is $P$ and the intercept is $-Q$ .

Therefore $P = \text{gradient}$ .
If the $y$ -intercept is $c$ , then $c = -Q \Rightarrow Q = -c$ .

Units:

If $L$ is in m, then $y = (E-V)/L$ is in $\text{V m}^{-1}$ .
Since $x$ is in V,

[P] = \frac{\text{V m}^{-1}}{\text{V}} = \text{m}^{-1}

[Q] = [y] = \text{V m}^{-1}

Key Takeaways

Gradient gives the coefficient of $V$ .
Intercept sign matters: here the intercept is $-Q$ , not $Q$ .
Units follow from the axes.

Common Mistakes

Taking $Q$ equal to the $y$ -intercept instead of the negative of it.
Giving incorrect units (e.g. writing $Q$ in V instead of $\text{V m}^{-1}$ ).

Things to Be Careful About

If you used $L$ in cm on your table/graph, your $P$ and $Q$ units will be $\text{cm}^{-1}$ and $\text{V cm}^{-1}$ respectively; do not mix cm-based values with m-based units.

Techniques used

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

(f)

The resistance of R is $R$ .

Theory suggests that:

• $P$ and $Q$ are both inversely proportional to $R$

• the graph cuts the $x$ -axis at a value of $V = \frac{E}{2}$ for all values of $R$ .

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

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

1M Similar

DifficultyMedium

Worked solution

Working

The graph is

y = PV - Q

Lower $R$ gives larger $P$ and larger $Q$ (both $\propto 1/R$ ).

The $x$ -intercept is where $y=0$ :

0 = PV - Q \Rightarrow V = \frac{Q}{P} = \frac{E}{2}\ \text{(unchanged)}

So the new line passes through $V = E/2$ on the $x$ -axis, but has a larger gradient (steeper). Since $y$ -intercept $= -Q$ , it is more negative.

Answer

diagram-f-lineW

Final answer

Line W is steeper and pivots about the same x-intercept at V = E/2; y-intercept is more negative.

Detailed explanation

Background Concept

For a straight line

y = PV - Q

gradient $= P$ ,
$y$ -intercept $= -Q$ ,
$x$ -intercept occurs when $y=0$ :

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

If both $P$ and $Q$ change by the same factor, the ratio $Q/P$ stays the same, so the $x$ -intercept stays fixed.

Understanding the Question

You are told the theory predictions:

$P$ and $Q$ are both inversely proportional to $R$ ,
the graph cuts the $x$ -axis at $V = E/2$ for all values of $R$ .

A student repeats with smaller resistors (so smaller $R$ ). You must draw the expected new line on the same axes and label it W.

Approach

Work out what happens to the gradient and intercept when $R$ decreases.
Use the given fixed $x$ -intercept at $V=E/2$ .
Sketch a second straight line consistent with these constraints.

Step-by-Step Reasoning

Since $P \propto 1/R$ , decreasing $R$ increases $P$ , so the line becomes steeper.
Since $Q \propto 1/R$ , decreasing $R$ increases $Q$ .
The $y$ -intercept is $-Q$ , so increasing $Q$ makes the $y$ -intercept more negative (the line crosses the $y$ -axis lower down).
The $x$ -intercept is

V = \frac{Q}{P}

If both $Q$ and $P$ increase in the same inverse proportion to $R$ , the ratio $Q/P$ stays constant, so the line must still cross the $x$ -axis at

V = \frac{E}{2}.

So the new line W is best thought of as a line that pivots about the fixed point $(V=E/2,\ y=0)$ and becomes steeper.

diagram-f-lineW

Key Takeaways

Changing a parameter can change both gradient and intercept.
A fixed $x$ -intercept provides a strong constraint for sketching the new line.

Common Mistakes

Drawing the new line parallel to the old one (wrong because gradient changes).
Shifting the $x$ -intercept away from $V=E/2$ (explicitly stated to stay the same).
Making the $y$ -intercept less negative when $Q$ increases (sign error: intercept is $-Q$ ).

Things to Be Careful About

Ensure line W is clearly labelled.
The line should remain straight (same functional form) and should cross the $x$ -axis at the same $V=E/2$ point as the original line.

Techniques used

use proportional reasoning to predict how gradient and intercept changeidentify a fixed x-intercept from a linear equationsketch a second straight line consistent with changed parameters

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

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

You have been provided with two spheres of modelling clay.

(a)

(i)

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

Fig. 2.1

Measure and record $d$ .

$d$ = ______

1M Similar

(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 (1 mm resolution), take absolute uncertainty in $d$ as $\pm 0.05\ \text{cm}$ .

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

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

Answer

\text{percentage uncertainty} = 2.1\%

Final answer

2.1%

Detailed explanation

Background Concept

Uncertainty is an estimate of the likely range of values for a measured quantity.

For a ruler:

smallest division is usually 1 mm.
a single scale reading is often taken as $\pm 0.5\ \text{mm}$ .

Percentage uncertainty is

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

Understanding the Question

You must estimate the percentage uncertainty in your measured diameter $d$ and show working.

Approach

Decide the absolute uncertainty in $d$ based on the ruler resolution.
Divide by your measured $d$ .
Multiply by 100 to get a percentage.

Step-by-Step Reasoning

If the ruler has 1 mm markings, a reasonable uncertainty in a single reading is $\pm 0.5\ \text{mm} = \pm 0.05\ \text{cm}$ .
For example, if $d = 2.40\ \text{cm}$ :

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

(If you instead took $\pm 0.1\ \text{cm}$ you would get a larger percentage; the key is that your value must match your stated absolute uncertainty.)

Key Takeaways

Percentage uncertainty comes from absolute uncertainty divided by the measured value.
Always show the calculation.

Common Mistakes

Using $\pm 1\ \text{mm}$ instead of $\pm 0.5\ \text{mm}$ without justification.
Forgetting to multiply by 100.
Mixing mm and cm in the same calculation.

Things to Be Careful About

If you measure $d$ by subtracting two ruler readings, some students argue for a larger uncertainty (two readings). If you do this, be consistent and show it clearly.
Quote the final percentage to a sensible number of significant figures (usually 2 s.f.).

Techniques used

estimate absolute uncertainty from instrument resolutioncalculate percentage uncertainty using a ratio

(b)

(i)

• Set up the apparatus as shown in Fig. 2.2.

Fig. 2.2

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

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

• The distance between the centres of the rods is $x$ .
Position the stands so that $x$ is approximately $70\text{ cm}$ .

• Measure and record $x$ .

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

1M Similar

(ii)

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

Fig. 2.3

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

Place the hook so that $n$ is 11.

• Calculate $N$ , where

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

Give your answer to three significant figures.

$N$ = ______

1M Similar

(c)

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

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

$T$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Time $10$ oscillations twice and average.

Example:

t_1 = 12.1\ \text{s},\quad t_2 = 11.9\ \text{s}

\bar{t} = \frac{12.1+11.9}{2} = 12.0\ \text{s}

T = \frac{\bar{t}}{10} = 1.20\ \text{s}

Answer

T = 1.20\ \text{s}

Final answer

T = 1.20 s

Detailed explanation

Background Concept

The period $T$ is the time for one complete oscillation. A stopwatch has reaction-time uncertainty (often ~0.2 s for a single start/stop), so timing just one oscillation gives a large percentage uncertainty.

A standard method is:

time $N$ oscillations (e.g. $N=10$ or $20$ ),
then divide by $N$ :

T = \frac{t}{N}

Repeating and averaging reduces random uncertainty.

Understanding the Question

You must “take measurements to determine the period $T$ ” of the oscillating clay sphere on the paper-clip chain.

Approach

Choose a clear reference point (e.g. centre position).
Time a large number of oscillations.
Repeat the timing.
Average and divide by the number of oscillations to obtain $T$ .

Step-by-Step Reasoning

Displace the sphere a small distance and release so it oscillates with small amplitude.
Start the stopwatch as the sphere passes the reference point in a chosen direction.
Count 10 complete oscillations and stop the watch when it returns to the same point moving in the same direction.
Repeat the timing.
Example:

\bar{t} = 12.0\ \text{s for 10 oscillations} \Rightarrow T = 12.0/10 = 1.20\ \text{s}

Key Takeaways

Time many oscillations to reduce percentage uncertainty.
Repeat and average to reduce random error.

Common Mistakes

Timing only one oscillation.
Not using the same reference point/direction for start and stop.
Counting oscillations incorrectly (e.g. counting half-oscillations).

Things to Be Careful About

Keep amplitude small; large amplitudes can change the period and cause the motion to become irregular.
If the oscillations decay (damping), take timings early and consistently.

Techniques used

time multiple oscillations and divide to obtain the periodrepeat timings and calculate a meanuse a consistent reference point for timing

(d)

• Roll the two spheres into one larger sphere.

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

$d$ = ______

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

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

3M Similar

DifficultyMedium-Easy

Worked solution

Working

Example readings after combining spheres:

d = 3.02\ \text{cm},\quad x = 80.0\ \text{cm}

With $n=7$ :

N = \sqrt[3]{n^2} = \sqrt[3]{7^2} = \sqrt[3]{49} = 3.659\ldots = 3.66

Timing $10$ oscillations (example $t=10.2\ \text{s}$ ):

T = \frac{10.2}{10} = 1.02\ \text{s}

Answer

d = 3.02\ \text{cm}

x = 80.0\ \text{cm}

N = 3.66

T = 1.02\ \text{s}

Final answer

d = 3.02 cm, x = 80.0 cm, N = 3.66, T = 1.02 s

Detailed explanation

Background Concept

This part repeats the same measurement and timing skills:

measure lengths ( $d$ , $x$ ) to the instrument resolution,
compute a derived quantity $N$ from the given formula,
measure period $T$ by timing multiple oscillations and dividing.

Understanding the Question

You must:

Combine the two clay spheres into one larger sphere and measure its diameter $d$ .
Re-set the apparatus with $x \approx 80\ \text{cm}$ .
Place the hook so $n=7$ and calculate $N=\sqrt[3]{n^2}$ .
Measure the period $T$ of oscillation again.

Approach

Follow the same methods used earlier:

diameter with ruler (or calipers if available),
$x$ measured centre-to-centre,
compute $N$ and round to 3 s.f. if needed,
time 10–20 oscillations and divide to get $T$ .

Step-by-Step Reasoning

Roll the clay into a sphere; because clay deforms, rotate it and check several diameters, then take a representative value.
Measure $x$ as the centre-to-centre separation.
With $n=7$ :

N = \sqrt[3]{7^2} = \sqrt[3]{49} = 3.659\ldots \approx 3.66

For the period, choose a number of oscillations such as 10, measure time $t$ , then

T = \frac{t}{10}

Repeat and average if possible.

Key Takeaways

Consistency: use the same measurement definitions (centre-to-centre, full oscillations) each time.
Derived quantities must be rounded appropriately.

Common Mistakes

Forgetting to change $n$ to 7.
Using $N=n^{1/3}$ instead of $N=n^{2/3}$ .
Measuring $x$ between rod edges rather than centres.

Things to Be Careful About

The larger sphere may be less spherical; this increases uncertainty in $d$ .
Ensure rods remain at equal height; otherwise the motion can become irregular and affect $T$ .

Techniques used

measure a diameter and a separation distance with appropriate precisionevaluate a derived quantity from a given formulatime multiple oscillations and divide to obtain the period

(e)

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

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

where $k$ is a constant.

(i)

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

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

1M Similar

DifficultyMedium

Worked solution

Working

From

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

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

Using SI units.

Run 1 (example): $T=1.20\ \text{s}$ , $x=0.700\ \text{m}$ , $d=0.0240\ \text{m}$ , $N=4.95$

k_1 = \frac{(1.20)^2(0.700)^2}{(0.0240)(4.95)} = 5.94\ \text{s}^2\ \text{m}

Run 2 (example): $T=1.02\ \text{s}$ , $x=0.800\ \text{m}$ , $d=0.0302\ \text{m}$ , $N=3.66$

k_2 = \frac{(1.02)^2(0.800)^2}{(0.0302)(3.66)} = 6.03\ \text{s}^2\ \text{m}

Answer

first value of $k$ :

5.94\ \text{s}^2\ \text{m}

second value of $k$ :

6.03\ \text{s}^2\ \text{m}

Final answer

k1 = 5.94 s^2 m, k2 = 6.03 s^2 m

Detailed explanation

Background Concept

If a relationship is correct, a constant calculated from different experimental runs should agree within experimental uncertainty.

From

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

we rearrange to calculate $k$ :

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

You must use consistent units for $x$ and $d$ (both in m, or both in cm). The numerical value (and unit) of $k$ depends on that choice.

Understanding the Question

You have two sets of data (small sphere run and large sphere run). You must calculate $k$ for each run to obtain two values, then you can later judge agreement.

Approach

Make $k$ the subject.
For each run, substitute $T$ , $x$ , $d$ , $N$ .
Compute $k$ and record to a sensible number of significant figures.

Step-by-Step Reasoning

Rearrangement:

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

Substitute Run 1 values (example): square $T$ and $x$ carefully, then divide by $dN$ .
Repeat for Run 2.

A good check is that if the relationship holds, the two $k$ values should be similar (not identical).

Key Takeaways

Rearranging is essential: isolate the constant.
Consistent units are critical when calculating derived constants.

Common Mistakes

Forgetting to square $x$ or $T$ .
Mixing cm and m for $x$ and $d$ .
Using rounded $N$ too aggressively (can shift $k$ ).

Things to Be Careful About

Quote $k$ to appropriate significant figures based on your least precise measurement.
Include the unit of $k$ if you used SI units (here $\text{s}^2\ \text{m}$ ).

Techniques used

rearrange an equation to make a constant the subjectsubstitute measured values consistently with unitscalculate and report two derived values for comparison

(ii)

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

1M Similar

(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 example values $k_1=5.94$ and $k_2=6.03$ :

\%\ \text{difference} = \frac{|6.03-5.94|}{(6.03+5.94)/2}\times 100 = 1.50\%

Since $1.50\% < 15\%$ , the values agree within the uncertainty.

Answer

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

Final answer

Yes, consistent within 15%.

Detailed explanation

Background Concept

To test whether two results agree within an uncertainty, you compare their difference with the allowed uncertainty range.

If the percentage uncertainty in $k$ is $15\%$ , then two values of $k$ should typically differ by no more than about $15\%$ (order-of-magnitude check) to be considered consistent.

A common quantitative check is percentage difference:

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

Understanding the Question

You are told to assume a $15\%$ uncertainty in $k$ and decide whether your experimental results support the proposed relationship.

Approach

Compute how far apart the two $k$ values are (as a percentage).
Compare this with $15\%$ .
State a clear conclusion (supports / does not support).

Step-by-Step Reasoning

Using example values:

\%\ \text{difference} = 1.50\%

Because $1.50\%$ is much less than $15\%$ , the two values are consistent within the stated uncertainty.
Therefore the experimental results support the relationship.

Key Takeaways

“Support” means agreement within uncertainty, not perfect equality.

Common Mistakes

Comparing absolute difference without considering the size of the values.
Using $15\%$ of the larger value but not stating what you did.
Giving a conclusion without referencing the uncertainty.

Things to Be Careful About

If your two $k$ values differ by more than $15\%$ , you should say results do not support (within that uncertainty) and could mention experimental limitations as a possible reason.

Techniques used

compare two experimental values using a percentage differencejudge consistency using a stated uncertainty bound

(g)

(i)

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

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

4M Similar

DifficultyMedium

Worked solution

Answer

Uncertainty in $T$ : stopwatch reaction time and difficulty judging the instant the sphere passes a fixed reference point.
Uncertainty in $T$ : oscillations may not stay in one vertical plane (twisting of the chain), so the motion is not perfectly periodic.
Uncertainty in $x$ : difficult to measure centre-to-centre of the rods accurately; parallax and uncertainty in locating the rod centres.
Uncertainty in $d$ : modelling clay is not a perfect rigid sphere and can deform, so the measured diameter depends on orientation/pressure.

Final answer

Four limitations/uncertainties listed (T reaction time; non-planar oscillation; x centre-to-centre; d deformation).

Detailed explanation

Background Concept

A good evaluation point:

names the quantity affected (e.g. $T$ , $x$ , $d$ ),
states why it is uncertain (instrument limits, human judgement, motion not ideal),
indicates whether it is random (varies trial-to-trial) or systematic (bias).

Understanding the Question

You must give four sources of uncertainty or limitations in this procedure. For measurement uncertainties, you must state the quantity and the reason.

Approach

Look at each measured or controlled quantity in the experiment ( $d$ , $x$ , $n$ / $N$ , $T$ , and the oscillation conditions) and identify what makes it hard to measure reliably or what makes the model imperfect.

Step-by-Step Reasoning

Four strong, creditworthy examples are:

Period $T$ (reaction time / judgement): starting and stopping a stopwatch depends on human reaction time, and it is hard to decide the exact moment the sphere passes the reference position.
Period $T$ (non-ideal motion): the chain can twist so the sphere oscillates in an ellipse or with coupled modes rather than a single simple oscillation, so the period is not constant.
Distance $x$ (definition of centres): $x$ is defined as centre-to-centre of rods; the centre is not a sharp point on the apparatus, giving reading uncertainty and possible parallax.
Diameter $d$ (deformation): modelling clay is compressible and may not form a perfect sphere; pressing it against a ruler can change its size slightly, and different orientations give different diameters.

Other acceptable limitations (if distinct) could include damping (amplitude decays, making later oscillations harder to time) or inconsistency in release (not releasing from same displacement).

Key Takeaways

Evaluation marks come from specific, physics-based limitations tied to the measurements.

Common Mistakes

Vague statements like “human error” without naming the quantity or cause.
Repeating the same point (e.g. reaction time) in different words.
Stating an improvement instead of a limitation (belongs in part (g)(ii)).

Things to Be Careful About

Ensure each of the four points is genuinely different.
For measurement uncertainty, always include both: the measured quantity and why it is uncertain.

Techniques used

identify random and systematic uncertainties in measurementslink 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

Worked solution

Answer

Improve $T$ measurement: use a video camera / motion sensor and determine $T$ from playback or data logging instead of a stopwatch.
Improve $T$ measurement: time a larger number of oscillations (e.g. 20) and repeat several times, then average.
Improve $x$ measurement: mark the rod centres and measure $x$ with a set square/metre rule aligned horizontally to reduce parallax and improve centre-to-centre accuracy.
Improve $d$ measurement: use vernier calipers and measure several diameters at different orientations, then take the mean.

Final answer

Four improvements listed (data-logged T; more oscillations and repeats; better x measurement; better d measurement).

Detailed explanation

Background Concept

Improvements should be specific and should reduce the uncertainties/limitations identified. Good improvements either:

reduce random uncertainty (repeat/average, better timing method),
reduce systematic error (better definition of measurement points),
make the motion closer to the ideal model (reduce twisting/damping effects).

Understanding the Question

You must describe four improvements (apparatus or procedure changes) that would make the experiment more reliable/accurate.

Approach

For each main limitation (timing, geometry measurement, non-ideal sphere), propose a concrete change and briefly state how it helps.

Step-by-Step Reasoning

Replace stopwatch timing of $T$ : use video analysis or a motion sensor/data logger to find the period from recorded position-time data; this removes reaction-time error.
Reduce random uncertainty in $T$ : time 20 oscillations rather than 10 and repeat multiple trials; average the calculated periods.
Measure $x$ more accurately: mark the centres of rods and use a rigid metre rule aligned at the same height; using a set square helps align the rule and reduces parallax.
Measure $d$ more accurately: use vernier calipers and measure several diameters at different orientations; average to reduce the effect of non-sphericity.

Other possible improvements include ensuring oscillations stay in one plane (guides or preventing twist) and ensuring the stands/rods are level using a spirit level.

Key Takeaways

Improvements must be practical and linked to identifiable uncertainties.

Common Mistakes

Saying “use more accurate equipment” without naming it.
Giving improvements that do not address a real limitation in the experiment.
Repeating the same improvement (e.g. “repeat readings”) for multiple marks without adding new detail.

Things to Be Careful About

Keep improvements distinct.
Make sure each improvement is described clearly enough that an experimenter could implement it.

Techniques used

propose apparatus changes to reduce measurement uncertaintymodify procedure to improve repeatabilitylink each improvement to a corresponding limitation