In this experiment, you will investigate the oscillations of a rod. (i) The length of the rod is $L$, as shown in Fig. 2.1. Measure and record $L$. $L$ = \_\_\_\_\_\_ (ii) The mass of the rod is $M$. Measure and record $M$. $M$ = \_\_\_\_\_\_ (iii) Calculate $S$, where $$ S = \frac{ML^2}{12} $$ $S$ = \_\_\_\_\_\_ (iv) Justify the number of significant figures that you have given for your value of $S$. (i) ● Wrap one end of the copper wire tightly three times around the…

(a)(i) L = 60.0 cm · (a)(ii) M = 120.0 g · (a)(iii) S = 3.60 × 10^-3 kg m^2 · (a)(iv) S is quoted to 3 significant figures (limited by L). · (b)(i) m = 50.0 g, a = 1.50 cm · (b)(ii) 3.3 % · (c) T = 1.25 s · (d) m = 10.0 g, a = 4.50 cm, T = 1.26 s · (e) k1 = 4.33×10^2 s^2 kg^-1 m^-2; k2 = 4.39×10^2 s^2 kg^-1 m^-2 · (f) Yes; k values agree within 10%. · (g)(i) Four limitations listed (a measurement uncertainty, timing uncertainty, slippage, asymmetry/level error). · (g)(ii)…

Physics 9702/31 — May/June 2024

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

questions

marks

120

minutes

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

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

In this experiment, you will investigate an electrical circuit.

You have been provided with a metre rule with a wire attached. You have also been provided with six identical resistors. Four of the resistors are connected in series and attached to a wooden block. The other resistors are labelled X and Z.

(a)

● Set up the circuit shown in Fig. 1.1.

Fig. 1.1

● E, F, G and H are crocodile clips.

$n$ resistors on the wooden block are connected in parallel with X. Connect F so that $n = 2$ , as shown in Fig. 1.1.

● The distance between G and H is $y$ . Attach H to the wire so that $y$ is approximately $50\text{ cm}$ .

● Close the switch.

● Record $n$ , $y$ and the ammeter reading $I$ .

$n$ = ______
$y$ = ______
$I$ = ______

● Open the switch.

2M Similar

DifficultyEasy

Worked solution

Answer

(Example of acceptable recorded readings)

$n = 2$

$y = 48.0\ \text{cm}$

$I = 0.130\ \text{A}$

Final answer

Example: n = 2, y = 48.0 cm, I = 0.130 A

Detailed explanation

Background Concept

In this practical you are adjusting a circuit and taking direct measurements:

$n$ is the effective number of identical resistors on the block that are connected in parallel with resistor $X$ (set by the position of crocodile clip $F$ ).
$y$ is the length of resistance wire between clips $G$ and $H$ , measured on the metre rule.
$I$ is the circuit current measured by the ammeter in series.

Changing $n$ changes the effective resistance of one part of the circuit; changing $y$ changes the resistance of the wire section (longer wire (\Rightarrow) larger resistance). Both changes affect the total circuit resistance and therefore the current.

Understanding the Question

You are told exactly how to set up the circuit (Fig. 1.1), then to:

fix $n = 2$ using clip $F$ ,
set $y \approx 50\ \text{cm}$ by placing $H$ about half-way along the wire from $G$ ,
close the switch and record $n$ , $y$ and the ammeter reading $I$ .

The marks are for taking sensible readings and recording them clearly, with appropriate precision.

Approach

Build the circuit exactly as in the diagram.
Ensure the ammeter is in series and reads steady.
Measure $y$ directly on the metre rule between the positions of $G$ and $H$ .
Record values neatly (and with consistent significant figures / decimal places for measured quantities).

Step-by-Step Reasoning

Place clip $F$ at the position that makes $n = 2$ (as indicated in Fig. 1.1: two resistors on the block are in parallel with $X$ ).
Attach clip $H$ so that the distance from $G$ to $H$ along the wire is about $50\ \text{cm}$ .
Close the switch and wait briefly for the current to stabilise.
Read the ammeter to its available resolution (typically to 2–3 s.f., depending on the meter).
Record:
- $n$ as an exact value (here, 2)
- $y$ to the nearest mm (or nearest 0.1 cm) if using a metre rule scale
- $I$ to the meter’s resolution.

Key Takeaways

Follow circuit diagrams accurately.
Length measurements should match the instrument resolution.
Ammeter readings should be recorded with sensible significant figures.

Common Mistakes

Measuring $y$ from the wrong reference point (not from $G$ to $H$ ).
Recording $y$ with unrealistic precision (e.g. many decimal places from a metre rule).
Forgetting to open the switch between adjustments, causing heating of the wire and drifting readings.

Things to Be Careful About

Ensure crocodile clips make good contact; poor contact changes resistance and makes $I$ unstable.
Avoid parallax error when reading $y$ on the metre rule.
Keep $y$ close to $50\ \text{cm}$ for this first reading as instructed; do not optimise it yet.

Techniques used

assemble the circuit according to the diagramset a specified configuration using crocodile clipsmeasure a length using a metre rulerecord an ammeter reading to appropriate precision

(b)

● Connect Z as shown in Fig. 1.2.

Fig. 1.2

When Z is connected in parallel with the first of the resistors on the block, the total value of $n$ is reduced by 0.5.

For the arrangement in Fig. 1.2, the value of $n$ is 1.5.

● Close the switch.

● Change the position of H on the wire until the value of $I$ is as close as possible to your value in (a).

● Record $n$ and $y$ .

$n$ = ______
$y$ = ______

● Open the switch.

● Disconnect Z.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

(Example of acceptable recorded readings)

$n = 1.5$

$y = 50.8\ \text{cm}$

Final answer

Example: n = 1.5, y = 50.8 cm

Detailed explanation

Background Concept

If you want the current $I$ to be the same, then the total circuit resistance must be the same (since the supply p.d. is fixed). In this experiment you change part of the circuit (by adding resistor $Z$ in parallel with the first resistor on the block), which changes the effective resistance of that section.

To restore the original current, you compensate by changing $y$ , which changes the resistance of the wire section.

Understanding the Question

You are instructed to connect resistor $Z$ as in Fig. 1.2. This makes the effective value of $n$ smaller by 0.5 (because one identical resistor in parallel with another identical resistor halves that resistor’s resistance, i.e. it behaves like “half a resistor” in series-count terms).

For this arrangement, $n = 1.5$ . You then move clip $H$ until the ammeter reading $I$ is as close as possible to your value in (a), and record $n$ and $y$ .

Approach

Add $Z$ exactly as shown.
Close the switch.
Slide $H$ along the wire, watching the ammeter, until $I$ matches the part (a) current.
Record the final $y$ when the match is best.

Step-by-Step Reasoning

Connect $Z$ in parallel with the first resistor on the block (between the appropriate nodes shown).
Set $n = 1.5$ (given for this configuration).
Close the switch and note whether $I$ is higher or lower than in (a).
Move $H$ $H$ slightly:
- Increasing $y$ increases the wire resistance, tending to reduce $I$ .
- Decreasing $y$ decreases the wire resistance, tending to increase $I$ .
Iterate until the ammeter reading is as close as possible to the part (a) value.
Measure and record $y$ at this position.

Key Takeaways

Matching a target current is done by adjusting the variable resistance (here, wire length).
When a circuit is altered, you can compensate by changing another resistance to keep $I$ constant.

Common Mistakes

Recording $y$ before the current is matched.
Forgetting that changing $y$ affects resistance (and therefore current).
Not disconnecting $Z afterwards (the instruction says to disconnect it for later parts).

Things to Be Careful About

Heating of the wire can change its resistance; keep the switch closed only while adjusting.
Make sure the “best match” is judged with a steady reading (wait for fluctuations to settle).

Techniques used

modify the circuit by adding a component in paralleladjust a variable resistance length to match a target currentrecord a measured length with appropriate precision

(c)

Vary $n$ by changing the position of F and connecting and disconnecting Z.

For each value of $n$ , change the position of H until the value of $I$ is as close as possible to your value in (a).

Repeat until you have six sets of values of $n$ and $y$ . Include your values from (a) and (b).

Record your results in a table. Include values of $\frac{n}{n+1}$ to two significant figures in your table.

8M Similar

DifficultyMedium

Worked solution

Answer

(Example of a correctly formatted table with six sets of values)

Target current: $I \approx 0.130\ \text{A}$ (from part (a))

$n$	$y / \text{cm}$	$\dfrac{n}{n+1}$ (2 s.f.)
0.5	62.0	0.33
1.0	55.0	0.50
1.5	50.8	0.60
2.0	48.0	0.67
2.5	46.0	0.71
3.0	44.5	0.75

Final answer

See table (six sets of n, y and n/(n+1) to 2 s.f.).

Detailed explanation

Background Concept

A good results table in Paper 3 must:

contain all results in one clear table,
have correct column headings with quantity and unit (units only for quantities that have units),
use consistent raw-data precision (e.g. $y$ to nearest 0.1 cm),
include calculated quantities to the required significant figures.

Here, you are asked to calculate

\frac{n}{n+1}

for each row and give it to two significant figures.

Understanding the Question

You must obtain six pairs of values $(n, y)$ such that, for each chosen $n$ , you adjust $y$ until the current $I$ matches the value found in part (a). You then tabulate the results, including your part (a) and (b) readings.

So:

Independent variable: $n$ (you choose by moving $F$ and adding/removing $Z$ ).
Dependent variable you record: $y$ (the length required to keep $I$ at the target).
Control/target: $I$ kept the same as in (a).

Approach

Pick a sensible spread of $n$ values (not all close together).
For each $n$ , move $H$ until the ammeter matches the target current.
Record $y$ each time.
Calculate $n/(n+1)$ for each row and round to 2 s.f.

Step-by-Step Reasoning

Decide on six $n$ settings. Typical available values are $0.5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0$ depending on clip positions and whether $Z$ is connected.
For each $n$ $n$ :
- connect/disconnect $Z$ as required,
- set clip $F$ to the desired tap,
- close the switch,
- slide $H$ until $I$ matches the part (a) value,
- measure $y$ from $G$ to $H$ and record it.
Calculate the derived column. Example for $n=2.0$ :

\frac{n}{n+1} = \frac{2.0}{3.0} = 0.666\ldots \approx 0.67\ (2\ \text{s.f.})

Ensure table conventions:
- $n$ has no unit.
- $y$ has a unit (cm or m, but be consistent).
- $n/(n+1)$ has no unit.
- Use consistent decimal places within a column where appropriate.

Key Takeaways

Collect a range of data points (six sets) for a reliable graph.
Derived quantities must be calculated correctly and rounded as instructed.
Tables must have clear headings and units.

Common Mistakes

Forgetting to include the derived column or not using 2 s.f.
Mixing units (some $y$ in cm, others in m) or not stating the unit in the heading.
Using inconsistent precision (e.g. some $y$ values to 0.1 cm and others to 1 cm).
Not actually matching $I$ to the target current before recording $y$ .

Things to Be Careful About

Write $\dfrac{n}{n+1}$ values to 2 s.f. exactly as requested (e.g. $0.67$ , not $0.6667$ ).
Keep the switch closed only as long as needed to avoid heating (which changes resistances and shifts $I$ ).

Techniques used

choose a suitable range of the independent variableadjust the wire length to keep a dependent reading constantcompute a derived quantity and round to stated significant figurestabulate results with correct headings and units

(d)

(i)

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

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Graph plotted with:

$x$ -axis: $\dfrac{n}{n+1}$ (no unit)
$y$ -axis: $y / \text{cm}$
suitable scales using at least half the grid
all six points plotted accurately.

Final answer

Graph of y (y-axis) against n/(n+1) (x-axis) plotted.

Detailed explanation

Background Concept

A good experimental graph should:

have axes labelled with quantity and unit (unit omitted if dimensionless),
use a sensible linear scale (no awkward jumps),
use at least half the graph paper in both directions,
show plotted points clearly (small crosses or dots with circles),
plot the dependent variable on the $y$ -axis and independent on the $x$ -axis.

Understanding the Question

You are told exactly what to plot:

vertical axis: $y$ ,
horizontal axis: $\dfrac{n}{n+1}$ .

So you must transfer your table values onto graph paper with correct presentation.

Approach

Put the calculated values of $\dfrac{n}{n+1}$ on the $x$ -axis.
Put the measured $y$ on the $y$ -axis (use the same unit as in your table).
Pick scales that spread the data across the page.

Step-by-Step Reasoning

Determine the range of your data:
- $\dfrac{n}{n+1}$ will lie between 0 and 1 (often around 0.3 to 0.8 in this experiment).
- $y$ will be around tens of cm.
Choose axis limits slightly beyond your smallest and largest values, so all points fit comfortably.
Mark equal intervals and write numbers at regular spacing.
Plot each point $(x, y)$ from the table carefully.

Key Takeaways

Correct axis labelling and sensible scales are essential for graph marks.
Accuracy in plotting affects your later gradient/intercept.

Common Mistakes

Swapping axes (plotting $y$ on the $x$ -axis).
Missing units on the $y$ axis label.
Choosing a scale that uses only a small corner of the grid.
Plotting $\dfrac{n}{n+1}$ values with too much rounding beyond what you calculated.

Things to Be Careful About

$\dfrac{n}{n+1}$ is dimensionless, so do not write a unit for that axis.
Plot using your recorded $y$ precision; do not invent extra decimal places.

Techniques used

choose appropriate axis scales to use most of the gridlabel axes with correct quantities and unitsplot 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

Using two points on the best-fit line, e.g. $(0.33,\ 62.0\ \text{cm})$ and $(0.75,\ 44.5\ \text{cm})$ :

\text{gradient } m = \frac{44.5 - 62.0}{0.75 - 0.33} = \frac{-17.5}{0.42} = -41.7\ \text{cm}

y = mx + c\ \Rightarrow\ c = 62.0 - (-41.7)(0.33) = 75.8\ \text{cm}

Answer

gradient $= -41.7\ \text{cm}$

$y$ -intercept $= 75.8\ \text{cm}$

Final answer

gradient = −41.7 cm, y-intercept = 75.8 cm

Detailed explanation

Background Concept

For a straight-line graph,

y = mx + c

$m$ is the gradient (slope):

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

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

The best practice is to use two points on the drawn best-fit line that are far apart to reduce percentage reading error.

Understanding the Question

You have a graph of $y$ (vertical) against $x = \dfrac{n}{n+1}$ (horizontal). You must find:

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

These will be used in the next part to find constants $P$ and $Q$ .

Approach

Pick two widely separated points that lie on the best-fit line (not necessarily actual plotted data points).
Compute the gradient using $\Delta y / \Delta x$ .
Find the intercept by either reading where the line crosses the $y$ -axis (if within range) or by substituting one point into $y = mx + c$ .

Step-by-Step Reasoning

Choose two points on the line far apart in $x$ to minimise uncertainty.
Read their coordinates carefully using the graph scales.
Compute changes:
- $\Delta y$ is the change in the vertical coordinate.
- $\Delta x$ is the change in the horizontal coordinate.
Calculate

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

Keep units: since $x$ is dimensionless, the gradient has the same unit as $y$ (e.g. cm).
5) Determine $c$ :

If the line crosses the $y$ -axis on your paper, read it directly.
Otherwise use $c = y - mx$ with a point on the line.

Key Takeaways

Always use points on the best-fit line, and make them far apart.
Gradient uses $\Delta y / \Delta x$ (not the other way around).
Units: gradient has units of $y$ when $x$ is dimensionless.

Common Mistakes

Using two actual plotted points instead of points on the best-fit line (can give a poorer estimate).
Calculating $\Delta x / \Delta y$ by accident.
Forgetting units for gradient and intercept.
Reading intercept incorrectly if the line does not extend to $x=0$ .

Things to Be Careful About

Use a large triangle (wide separation) for the gradient calculation.
Keep enough significant figures during calculation, then round appropriately at the end.

Techniques used

choose two well-separated points on a best-fit linecalculate gradient using $\Delta y / \Delta x$determine y-intercept by reading from the graph or substituting into $y = mx + c$

(e)

It is suggested that the quantities $y$ and $n$ are related by the equation

y = - \frac{Pn}{n+1} + 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

y = -\frac{Pn}{n+1} + Q

Let $x = \dfrac{n}{n+1}$ . Then

y = -Px + Q

So gradient $m = -P$ and intercept $c = Q$ .

With $m = -41.7\ \text{cm}$ and $c = 75.8\ \text{cm}$ :

P = -m = 41.7\ \text{cm}

Q = c = 75.8\ \text{cm}

Answer

$P = 41.7\ \text{cm}$

$Q = 75.8\ \text{cm}$

Final answer

P = 41.7 cm, Q = 75.8 cm

Detailed explanation

Background Concept

A key practical skill is relating a straight-line graph to an equation.
If you plot $y$ against $x$ and obtain a straight line,

y = mx + c

then the gradient is $m$ and the intercept is $c$ .

Here the suggested relationship is

y = -\frac{Pn}{n+1} + Q

which becomes linear if you define

x = \frac{n}{n+1}.

Understanding the Question

You already found the gradient and intercept of the graph of $y$ against $\dfrac{n}{n+1}$ . You now use them to determine constants $P$ and $Q$ , including their units.

Approach

Rewrite the given equation in the form $y = mx + c$ where $x = \dfrac{n}{n+1}$ .
Compare term-by-term to identify $m$ and $c$ .
Use your measured graph values for $m$ and $c$ to compute $P$ and $Q$ .

Step-by-Step Reasoning

Set $x = \dfrac{n}{n+1}$ . Then

y = -Px + Q.

Compare with $y = mx + c$ :

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

Units:

$x$ is dimensionless.
Therefore $P$ must have the same unit as $y$ .
$Q$ is an intercept on the $y$ -axis, so it also has the same unit as $y$ .
If $y$ was measured in cm, then $P$ and $Q$ are in cm; if $y$ was measured in m, then $P$ and $Q$ are in m.

Key Takeaways

Linearising by defining $x = \dfrac{n}{n+1}$ makes the equation match $y = mx + c$ .
$P$ comes from the negative of the gradient; $Q$ is the intercept.

Common Mistakes

Forgetting the minus sign and writing $P = m$ .
Giving $P$ and $Q$ in different units.
Adding inappropriate units to $\dfrac{n}{n+1}$ (it is unitless).

Things to Be Careful About

Keep sign conventions consistent: if your gradient is negative, $P$ should come out positive.
Use the same unit system as your graph when stating $P$ and $Q$ .

Techniques used

match an experimental straight-line form to a given equationidentify constants from gradient and interceptassign units by comparison with measured quantities

(f)

Theory suggests that

\frac{P}{Q} = \frac{X}{C}

where the resistance $X$ of resistor X is $12\ \Omega$ and $C$ is the resistance of the whole circuit.

Use your values in (e) to determine a value for $C$ .

$C$ = ______ $\Omega$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Given

\frac{P}{Q} = \frac{X}{C}

C = \frac{XQ}{P}

With $X = 12\ \Omega$ , $P = 41.7\ \text{cm}$ , $Q = 75.8\ \text{cm}$ :

C = \frac{12 \times 75.8}{41.7} = 21.8\ \Omega

Answer

$C = 21.8\ \Omega$

Final answer

21.8 Ω

Detailed explanation

Background Concept

A theoretical link is provided:

\frac{P}{Q} = \frac{X}{C}

where $X$ is known and $C$ is the total circuit resistance to be determined.

Because $P$ and $Q$ have the same units, $P/Q$ is dimensionless, matching $X/C$ (also dimensionless). This is a good quick unit-check.

Understanding the Question

You have found $P$ and $Q$ from your graph, and you are told $X = 12\ \Omega$ . You must calculate $C$ .

Approach

Rearrange the equation to make $C$ the subject, then substitute values.

Step-by-Step Reasoning

Start with

\frac{P}{Q} = \frac{X}{C}.

Cross-multiply:

PC = XQ.

Solve for $C$ :

C = \frac{XQ}{P}.

Substitute your values. Since $Q/P$ is a ratio, it does not matter whether $P$ and $Q$ are in cm or m, as long as they are in the same unit.
Quote $C$ in ohms.

Key Takeaways

Rearrangement and substitution from experimental constants is a standard final step.
Ratios like $Q/P$ are unitless if both quantities share the same unit.

Common Mistakes

Inverting the ratio and using $C = XP/Q$ .
Mixing units for $P$ and $Q$ (e.g. $P$ in cm and $Q$ in m) which would make $Q/P$ incorrect.
Forgetting to state the unit $\Omega$ for $C$ .

Things to Be Careful About

Ensure you use your own experimental values of $P$ and $Q$ from (e); the numerical value of $C$ is therefore student-dependent.
Round sensibly (typically 2–3 significant figures, consistent with the precision of the graph-derived values).

Techniques used

rearrange a proportionality equation to solve for an unknownsubstitute experimental constants into a theoretical relationshipuse consistent units and ratios

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

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

(a)

(i)

The length of the rod is $L$ , as shown in Fig. 2.1.

Fig. 2.1

Measure and record $L$ .

$L$ = ______

1M Similar

(ii)

The mass of the rod is $M$ .

Measure and record $M$ .

$M$ = ______

1M Similar

(iii)

Calculate $S$ , where

S = \frac{ML^2}{12}

$S$ = ______

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Convert to SI:

M = 0.120\ \text{kg},\qquad L = 0.600\ \text{m}

S = \frac{ML^2}{12} = \frac{0.120\times (0.600)^2}{12}

S = 3.60\times 10^{-3}\ \text{kg m}^2

Answer

$S = 3.60\times 10^{-3}\ \text{kg m}^2$

Final answer

S = 3.60 × 10^-3 kg m^2

Detailed explanation

Background Concept

When you calculate a new quantity from measured values, you must:

use consistent units (SI is safest),
substitute into the given equation correctly,
and include the correct derived unit.

Here,

S = \frac{ML^2}{12}

If $M$ is in $\text{kg}$ and $L$ is in $\text{m}$ , then $S$ has units:

\text{kg}\times \text{m}^2 = \text{kg m}^2

Understanding the Question

You measured $M$ (mass of rod) and $L$ (length of rod). You must calculate $S$ using the given formula.

Approach

Convert $M$ and $L$ into SI units.
Calculate $L^2$ .
Multiply by $M$ .
Divide by $12$ .
Quote $S$ with unit $\text{kg m}^2$ .

Step-by-Step Reasoning

Using the example readings:

Convert mass: $120.0\ \text{g} = 0.1200\ \text{kg}$ (often rounded to $0.120\ \text{kg}$ to match sig figs).
Convert length: $60.0\ \text{cm} = 0.600\ \text{m}$ .
Square the length: $L^2 = (0.600)^2 = 0.360\ \text{m}^2$ .
Multiply by $M$ : $ML^2 = 0.120\times 0.360 = 0.0432\ \text{kg m}^2$ .
Divide by $12$ : $S = 0.0432/12 = 0.00360\ \text{kg m}^2$ .
Write in standard form if preferred: $3.60\times 10^{-3}\ \text{kg m}^2$ .

Key Takeaways

Always use SI units before substituting.
Check units to confirm the calculation is consistent.

Common Mistakes

Using $M$ in grams and $L$ in metres (or vice versa) without correcting units.
Forgetting to square $L$ .
Missing the unit on $S$ .

Things to Be Careful About

Keep enough guard digits during working, then round at the end.
Ensure your final sig figs match what is justified by your measurements (part (iv)).

Techniques used

convert measured quantities to SI unitssubstitute into a given formulacalculate and quote a derived quantity with correct units

(iv)

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

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

$S$ is calculated from $S = \dfrac{ML^2}{12}$ .

$L$ was recorded to $3\ \text{s.f.}$ and $M$ was recorded to $4\ \text{s.f.}$ , so the limiting value is $L$ (3 s.f.).

Therefore $S$ is given to $3\ \text{s.f.}$ .

Final answer

S is quoted to 3 significant figures (limited by L).

Detailed explanation

Background Concept

Significant figures (s.f.) reflect the precision of measurements. For quantities calculated by multiplication/division, the result should be quoted to the same number of significant figures as the least precise factor.

S = \frac{ML^2}{12}

then $S$ depends on $M$ and $L$ (and $12$ is an exact number, not a measurement).

Understanding the Question

You must explain why you wrote your value of $S$ to a particular number of significant figures.

Approach

Identify how many significant figures were used for $M$ and $L$ .
Decide which is the limiting measurement (fewest s.f.).
Quote $S$ to that number of s.f. (since $12$ is exact).

Step-by-Step Reasoning

Suppose $L$ was measured as $60.0\ \text{cm}$ : this is $3\ \text{s.f.}$ .
Suppose $M$ was measured as $120.0\ \text{g}$ : this is $4\ \text{s.f.}$ .
In $ML^2$ , the precision is limited by $L$ because it has fewer s.f.
Dividing by $12$ does not change the appropriate s.f. because $12$ is exact.
Therefore, quote $S$ to $3\ \text{s.f.}$ (e.g. $3.60\times 10^{-3}\ \text{kg m}^2$ ).

Key Takeaways

For multiplication/division: final s.f. = smallest s.f. among measured inputs.
Exact constants (like $12$ ) do not limit s.f.

Common Mistakes

Quoting $S$ to the same number of decimal places as $M$ or $L$ .
Over-rounding midway through the calculation and losing accuracy.

Things to Be Careful About

If $L$ is squared, its percentage uncertainty doubles, but the significant-figure rule still comes from the original recorded precision of $L$ .
Keep working to extra digits, then round once at the end.

Techniques used

identify the limiting measurement in significant figuresapply significant-figure rules to multiplication and divisionjustify rounding based on instrument resolution

(b)

(i)

● Wrap one end of the copper wire tightly three times around the centre of the rod, as shown in Fig. 2.2.

Fig. 2.2

● Slide a $50\text{ g}$ slotted mass onto each end of the rod.

● Record the mass $m$ on one end of the rod.

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

● Adjust the positions of the masses so that they are equally spaced from the centre of the rod and their centres are approximately $3\text{ cm}$ apart, as shown in Fig. 2.3. You may need to use some of the adhesive putty to keep the masses in position.

Fig. 2.3

● Set up the apparatus as shown in Fig. 2.4.

Fig. 2.4

● Make a hook in the wire and place the hook on the rubber band.

● The distance between the centre of each mass and the wire is $a$ .

Adjust the position of the masses until the rod is parallel to the bench and each mass is the same distance $a$ from the wire.

● Measure and record $a$ .

$a$ = ______

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

$m = 50.0\ \text{g}$

$a = 1.50\ \text{cm}$

Final answer

m = 50.0 g, a = 1.50 cm

Detailed explanation

Background Concept

In practical work, distances like $a$ must be defined clearly (here: from the centre of each mass to the wire). To reduce systematic error:

keep the rod horizontal,
ensure both masses are the same distance from the centre,
measure to the centre of the masses (not the edge).

Understanding the Question

You must:

record the mass $m$ on one end of the rod (a single slotted mass),
adjust the masses so the rod is horizontal and symmetric,
measure and record $a$ , the distance from the wire (at the rod centre) to the centre of one mass.

Approach

Use the label on the slotted mass set or a balance to determine $m$ .
Use a ruler to measure from the wire position to the centre of a mass.
Record $a$ with appropriate precision (e.g. nearest $1\ \text{mm}$ ).

Step-by-Step Reasoning

Confirm each end has the same slotted mass and record one mass as $m$ .
Adjust both masses so their centres are the same distance from the wire and the rod is parallel to the bench.
Identify the centre of each mass (by geometry of the cylinder) and measure from the wire position (centre of rod) to that centre.
Record the value with unit.

(Example readings recorded: $m = 50.0\ \text{g}$ and $a = 1.50\ \text{cm}$ .)

Key Takeaways

For distances involving objects, measure between well-defined reference points (centre-to-centre here).
Symmetry (same $a$ on both sides) is essential to match the intended model.

Common Mistakes

Measuring to the near edge of the mass instead of its centre.
Measuring $a$ on only one side without ensuring the other side matches.
Writing $m$ without units, or mixing $\text{g}$ and $\text{kg}$ later.

Things to Be Careful About

Parallax when reading the ruler.
The wire at the centre may have thickness; choose a consistent reference line (e.g. the centre of the wire).
Ensure the masses do not slide after measuring (use putty if needed).

Techniques used

set up apparatus symmetrically according to a diagrammeasure a distance between two reference pointsrecord measured quantities with units and appropriate precision

(ii)

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

percentage uncertainty = ______ $\%$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

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

\Delta a \approx \pm 0.5\ \text{mm} = \pm 0.05\ \text{cm}

\%\ \text{uncertainty in } a = \frac{0.05}{1.50}\times 100 = 3.3\%

Answer

percentage uncertainty $= 3.3\%$

Final answer

3.3 %

Detailed explanation

Background Concept

Percentage uncertainty shows the fractional uncertainty of a measurement:

\%\ \text{uncertainty} = \frac{\Delta x}{x}\times 100

where $x$ is the measured value and $\Delta x$ is the absolute uncertainty.

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

Understanding the Question

You measured $a$ in part (b)(i). Now you must estimate the percentage uncertainty in $a$ and show the working.

Approach

Choose a sensible absolute uncertainty for $a$ based on the ruler resolution.
Substitute into the percentage uncertainty formula.
Quote the result as a percentage.

Step-by-Step Reasoning

If the ruler has $1\ \text{mm}$ marks, the reading uncertainty is about half a division:
$\Delta a \approx \pm 0.5\ \text{mm}$ .
Convert to the same units as $a$ . If $a = 1.50\ \text{cm}$ , then $0.5\ \text{mm} = 0.05\ \text{cm}$ .
Compute:

\frac{0.05}{1.50}\times 100 = 3.3\%

Key Takeaways

Always express $\Delta a$ and $a$ in the same units before dividing.
For rulers, $\pm 0.5\ \text{mm}$ is a typical absolute uncertainty per reading.

Common Mistakes

Using $\pm 1\ \text{mm}$ instead of $\pm 0.5\ \text{mm}$ without justification.
Mixing units (e.g. dividing $0.5\ \text{mm}$ by $1.50\ \text{cm}$ directly).
Forgetting to multiply by $100$ .

Things to Be Careful About

If $a$ is determined from two readings (e.g. position difference), the absolute uncertainty can be larger (sum of uncertainties). Decide based on how you actually measured $a$ .

Techniques used

estimate absolute uncertainty from instrument resolutioncalculate percentage uncertaintyshow uncertainty working clearly

(c)

● Rotate the rod horizontally through $90^\circ$ .

● Release the rod. The rod will oscillate.

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

$T$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Time $10$ oscillations twice:

t_1 = 12.4\ \text{s},\qquad t_2 = 12.6\ \text{s}

Mean time for $10$ oscillations:

\bar{t} = \frac{12.4 + 12.6}{2} = 12.5\ \text{s}

Period:

T = \frac{\bar{t}}{10} = \frac{12.5}{10} = 1.25\ \text{s}

Answer

$T = 1.25\ \text{s}$

Final answer

T = 1.25 s

Detailed explanation

Background Concept

The period $T$ is the time for one complete oscillation. Timing a single oscillation gives a large percentage uncertainty because reaction time is comparable to the measured time. A standard technique is:

time $N$ oscillations (with $N \ge 5$ or $10$ ),
then divide by $N$ to get $T$ ,
and repeat to find a mean.

Understanding the Question

You must take measurements to determine the oscillation period $T$ after rotating the rod through $90^\circ$ and releasing it.

Approach

Choose a fixed reference point in the motion (e.g. when the rod passes through its central equilibrium position in the same direction each time).
Use a stopwatch to time $N$ oscillations.
Repeat the timing.
Average the total times and divide by $N$ .

Step-by-Step Reasoning

Start the stopwatch as the rod passes the reference point.
Count $10$ full oscillations (returning to the same position and direction).
Stop the watch after the 10th oscillation.
Repeat for a second value to reduce random error.
Average the two times for 10 oscillations, then divide by 10.

The example calculation shows how this produces $T = 1.25\ \text{s}$ .

Key Takeaways

Timing many oscillations reduces percentage uncertainty.
Repeats and averaging improve reliability.

Common Mistakes

Timing half-oscillations by accident.
Starting/stopping at different points in the cycle.
Timing too few oscillations (large random uncertainty).

Things to Be Careful About

Damping may reduce amplitude; still time using the same reference point.
Ensure the rod is oscillating freely (masses not slipping, wire not rubbing).

Techniques used

time multiple oscillations and divide to obtain the periodrepeat timing measurements and take a meanuse consistent reference points when timing oscillations

(d)

● Remove the hook from the rubber band.

● Remove the $50\text{ g}$ masses from the rod.

● Place the $10\text{ g}$ masses on the rod so that their centres are approximately $9\text{ cm}$ apart.

● Record $m$ .

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

● Place the hook on the rubber band.

● Adjust the position of the masses until the rod is parallel to the bench and each mass is the same distance $a$ from the wire.

● Measure and record $a$ .

$a$ = ______

● Repeat (c).

$T$ = ______

2M Similar

(e)

It is suggested that the relationship between $T$ , $S$ , $a$ and $m$ is

T^2 = k(S + a^2m)

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

T^2 = k(S + a^2m)

k = \frac{T^2}{S+a^2m}

Using $S = 3.60\times 10^{-3}\ \text{kg m}^2$ .

First set: $T = 1.25\ \text{s}$ , $a = 0.0150\ \text{m}$ , $m = 0.0500\ \text{kg}$

k = \frac{(1.25)^2}{3.60\times 10^{-3} + (0.0150)^2(0.0500)} = 4.33\times 10^{2}\ \text{s}^2\ \text{kg}^{-1}\ \text{m}^{-2}

Second set: $T = 1.26\ \text{s}$ , $a = 0.0450\ \text{m}$ , $m = 0.0100\ \text{kg}$

k = \frac{(1.26)^2}{3.60\times 10^{-3} + (0.0450)^2(0.0100)} = 4.39\times 10^{2}\ \text{s}^2\ \text{kg}^{-1}\ \text{m}^{-2}

Answer

first value of $k = 4.33\times 10^{2}\ \text{s}^2\ \text{kg}^{-1}\ \text{m}^{-2}$

second value of $k = 4.39\times 10^{2}\ \text{s}^2\ \text{kg}^{-1}\ \text{m}^{-2}$

Final answer

k1 = 4.33×10^2 s^2 kg^-1 m^-2; k2 = 4.39×10^2 s^2 kg^-1 m^-2

Detailed explanation

Background Concept

If a model suggests

T^2 = k(S + a^2m)

then for each set of measurements you can compute

k = \frac{T^2}{S+a^2m}

If the relationship is correct and the experiment is done well, $k$ should be (approximately) the same for different trials.

Units check:

$S$ has units $\text{kg m}^2$ .
$a^2m$ has units $\text{m}^2\times \text{kg} = \text{kg m}^2$ .
So $(S+a^2m)$ is in $\text{kg m}^2$ and $k$ has units:

\frac{\text{s}^2}{\text{kg m}^2} = \text{s}^2\ \text{kg}^{-1}\ \text{m}^{-2}

Understanding the Question

You have two sets of data (from the $50\ \text{g}$ and the $10\ \text{g}$ configurations). You must calculate two values of $k$ using your measured $T$ , $a$ , and known $S$ .

Approach

For each trial:

convert $m$ to $\text{kg}$ and $a$ to $\text{m}$ ,
compute $T^2$ ,
compute $a^2m$ ,
add $S + a^2m$ ,
compute $k$ from $k = T^2/(S+a^2m)$ .

Step-by-Step Reasoning

First trial (example numbers):

Convert: $a = 1.50\ \text{cm} = 0.0150\ \text{m}$ , $m = 50.0\ \text{g} = 0.0500\ \text{kg}$ .
$T^2 = (1.25\ \text{s})^2 = 1.5625\ \text{s}^2$ .
$a^2m = (0.0150)^2(0.0500) = 1.125\times 10^{-5}\ \text{kg m}^2$ .
Denominator: $S+a^2m \approx 3.60\times 10^{-3} + 1.125\times 10^{-5}$ .
Divide to find $k$ .

Second trial is identical method with the second set of $a,m,T$ .

Key Takeaways

Always convert to SI before using a relationship with squared terms.
If a constant is truly constant, computed values from different runs should agree within uncertainty.

Common Mistakes

Using $a$ in $\text{cm}$ inside $a^2m$ (gives a $10^4$ error because $\text{cm}^2$ vs $\text{m}^2$ ).
Using $m$ in grams without converting to kilograms.
Squaring $a$ incorrectly or forgetting to square it.

Things to Be Careful About

Keep enough significant figures during intermediate steps.
$S$ may dominate over $a^2m$ depending on the values; rounding too aggressively can distort $k$ .

Techniques used

rearrange an equation to solve for a constantsubstitute experimental data into a modeluse consistent SI units across derived quantities

(f)

It is suggested that the percentage uncertainty in the values of $k$ is $10\%$ .

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

1M Similar

DifficultyMedium-Easy

Worked solution

Working

\%\ \text{difference} = \frac{|439-433|}{\frac{439+433}{2}}\times 100 \approx 1.4\%

Since $1.4\% < 10\%$ , the two values of $k$ agree within the stated uncertainty.

Answer

Yes, the results support the relationship because the values of $k$ are consistent within $10\%$ .

Final answer

Yes; k values agree within 10%.

Detailed explanation

Background Concept

To decide whether results support a relationship that predicts a constant $k$ , you compare values of $k$ from different trials.

A common method is percentage difference:

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

If this percentage difference is less than (or comparable to) the stated percentage uncertainty, the results are considered consistent.

Understanding the Question

You are told the percentage uncertainty in $k$ is $10\%$ . You have two calculated values of $k$ . You must say whether the results support the suggested relationship.

Approach

Compute the percentage difference between the two values of $k$ .
Compare it with $10\%$ .
State a clear conclusion.

Step-by-Step Reasoning

Take the two calculated values, e.g. $k_1 = 4.33\times 10^2$ and $k_2 = 4.39\times 10^2$ .
Find the absolute difference: $|k_1-k_2|$ .
Divide by the mean of the two values and multiply by $100$ .
If the result is less than $10\%$ , then the discrepancy is small enough to be explained by experimental uncertainty.

Key Takeaways

“Supports the relationship” means the data are consistent with it within uncertainty.
Use a quantitative comparison (percentage difference) rather than a vague statement.

Common Mistakes

Comparing by subtraction only without converting to a percentage.
Using $10\%$ as an absolute uncertainty instead of a percentage.
Claiming the relationship is proved; in practical work you only say “supported” or “consistent within uncertainty”.

Things to Be Careful About

If one value is close to the other but both have large uncertainties, agreement is expected; the uncertainty criterion is the key.
Use sensible rounding so that the comparison isn’t distorted by premature rounding.

Techniques used

compare two experimental values using a percentage differenceapply an uncertainty criterion to judge agreementdraw a conclusion consistent with uncertainty bounds

(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

Measuring $a$ : difficult to judge the centre of each mass and the reference line of the wire, so $a$ has parallax/centre-position uncertainty.
Timing $T$ : stopwatch reaction time when starting/stopping and difficulty deciding the exact instant the rod passes a reference position.
Mass position: masses may slip/move on the rod during oscillations (putty deformation), changing $a$ during the run.
Rod not exactly horizontal / unequal $a$ on each side: if the rod is not level or masses not symmetrically placed, the motion differs from the assumed arrangement and affects $T$ .

Final answer

Four limitations listed (a measurement uncertainty, timing uncertainty, slippage, asymmetry/level error).

Detailed explanation

Background Concept

In oscillation experiments, uncertainty comes from:

measurement resolution and judgement (lengths like $a$ ),
human reaction time (timing $T$ ),
changes during the oscillation (slipping masses, damping),
and departures from the idealised setup assumed in the formula.

Good exam answers name the quantity affected and the reason the measurement/procedure is uncertain.

Understanding the Question

You must give four sources of uncertainty/limitations. For measurement uncertainties, you must state (i) what you are measuring and (ii) why it is uncertain.

Approach

Think through the steps that generate your data ( $a$ , $T$ , and the stability of the arrangement). For each, ask: “What makes this hard to measure accurately or keep constant?” Then write each point explicitly.

Step-by-Step Reasoning

Creditable limitations include:

Uncertainty in $a$ : you must measure from the wire to the centre of the mass. The centre is not a sharp point, and the wire has thickness; plus parallax can occur when reading the ruler.
Uncertainty in $T$ from reaction time: start/stop timing depends on the observer’s judgement. This is especially significant if you time only a few oscillations.
Difficulty identifying a consistent timing reference: torsional oscillations can have small amplitude changes, and deciding the “same point in the cycle” each time can vary.
Masses may move: if the masses slide, or the putty deforms, then $a$ changes during oscillation, so the run does not correspond to one fixed value of $a$ .
Rod not perfectly horizontal / symmetry not perfect: if the rod tilts or the masses are not exactly equal distances from the wire, the oscillation is not the intended symmetric system.

Any four well-explained points like these score.

Key Takeaways

Link each limitation to a specific measured quantity or experimental assumption.
Give a physical reason, not just “human error”.

Common Mistakes

Writing vague statements like “parallax” without saying what was being measured.
Giving improvements instead of limitations (that belongs in (g)(ii)).
Repeating the same idea in different words (e.g. “reaction time” and “human error” as two separate points).

Things to Be Careful About

Make sure each of the four points is distinct.
If you mention an uncertainty, make it clear whether it is random (varies between repeats) or systematic (biases all readings).

Techniques used

identify random and systematic uncertainty sources in an oscillation experimentlink 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

Measure $a$ more precisely by marking the mass centres and using vernier calipers (or a set square/rule at eye level) to reduce centre/parallax uncertainty.
Reduce timing uncertainty by timing a larger number of oscillations (e.g. $20$ ) and repeating several times, then averaging.
Use video analysis/light sensor with a data logger to determine $T$ automatically (removes reaction time).
Prevent masses moving by clamping them firmly (e.g. with a screw clamp/tape) so that $a$ stays constant during oscillations.

Final answer

Four improvements given (better a measurement, more oscillations & repeats, automatic timing, secure masses).

Detailed explanation

Background Concept

An improvement should directly reduce a specific uncertainty or correct a limitation. The strongest answers:

identify what is improved (e.g. $a$ or $T$ ),
name the instrument/procedure,
and explain how it reduces uncertainty.

Understanding the Question

You must suggest four improvements to the experiment (apparatus or procedure changes). These should be practical and aimed at improving accuracy/precision.

Approach

Take the limitations from (g)(i) and propose fixes:

for $a$ measurement issues: better tools and clearer reference points,
for timing: increase total timed interval, repeat, or automate,
for instability: make the setup more secure and controlled.

Step-by-Step Reasoning

Examples of strong improvements:

Improve $a$ measurement: mark the rod centre and mass centres; use vernier calipers or careful alignment tools to reduce judgement and parallax.
Improve $T$ measurement precision: time more oscillations (larger $N$ ) so reaction time becomes a smaller fraction of the total time; repeat and average.
Automate timing: use a motion sensor/photogate/video tracking to identify crossings of a reference position and compute $T$ objectively.
Stop masses sliding: use stronger fixing (clamps, tape, or a mechanical stop) so $a$ remains constant during oscillation.

Other valid improvements could include shielding from drafts, ensuring the support is rigid, or using a pointer and scale to define the equilibrium position more clearly.

Key Takeaways

Improvements should be specific and linked to an error source.
Better instruments and better procedures both count.

Common Mistakes

Saying “use more accurate equipment” without naming what or how.
Suggesting changing the physics model (irrelevant) rather than improving measurement.
Repeating the same improvement (e.g. “repeat readings” stated in multiple ways).

Things to Be Careful About

Improvements must be realistic with typical lab equipment.
Avoid suggesting changes that alter the defined variables (e.g. changing the wire material) unless justified as controlling a variable.

Techniques used

propose procedural improvements that reduce specific uncertaintiesmatch each improvement to a stated limitationsuggest suitable alternative apparatus for more precise measurement