Physics 9702/31 — May/June 2022

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

Q12MManipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and EvaluationFree sample

In this experiment, you will investigate the motion of a spring system.

You have been provided with two springs connected by string.

(a)

● Set up the apparatus as shown in Fig. 1.1.

Fig. 1.1

● The lower mass is $m$ . Arrange all of the slotted masses so that $m$ is $250\text{ g}$ and the remaining slotted masses are in the upper string loop.
● Pull the lower mass down through a short distance.
● Release the mass. The system will oscillate.
● Determine the period $T$ of the oscillations of the upper mass.

$T$ = ______ $\text{s}$

DifficultyMedium-Easy

Worked solution

Working

Set $m = 250\ \text{g}$ .

Measure time $t$ for $N$ oscillations of the upper mass (e.g. $N=10$ ), repeat and take a mean.

T = \frac{t}{N}

Example: mean $t = 12.0\ \text{s}$ for $10$ oscillations

T = \frac{12.0}{10} = 1.20\ \text{s}

Answer

$T = 1.20\ \text{s}$ (example)

Final answer

T ≈ 1.20 s (example; student-dependent)

Detailed explanation

Background Concept

The period $T$ of an oscillation is the time taken for one complete cycle of motion (e.g. from one highest point back to the next highest point).

When timing oscillations with a stopwatch, the main uncertainty is usually human reaction time (starting/stopping the watch). A key technique to reduce its effect is to time many oscillations and then divide by the number of oscillations:

T = \frac{t}{N}

where $t$ is the total time for $N$ oscillations.

Understanding the Question

You are given a two-spring system with an upper mass and a lower mass $m$ . You must:

arrange the lower mass so that $m = 250\ \text{g}$ ,
pull the lower mass down slightly and release to create oscillations,
determine the period $T$ of the oscillations of the upper mass.

Even though you pull the lower mass, the instruction is to measure the period of the upper mass oscillations.

diagram-a-setup

Approach

Set up the apparatus exactly as in Fig. 1.1, ensuring it is stable and vertical.
Set $m=250\ \text{g}$ in the lower loop.
Start oscillations with a small displacement.
Choose a clear reference point in the motion (e.g. when the upper mass passes an equilibrium position in the same direction).
Time $N$ oscillations (typically $N=10$ or more), repeat, average $t$ , then compute $T=t/N$ .

Step-by-Step Reasoning

After setting $m=250\ \text{g}$ , displace the lower mass a short distance and release gently (do not push).
Decide what counts as “one oscillation”: for example, from a highest point back to the next highest point.
Because pressing the stopwatch introduces an uncertainty (often around $\pm 0.2\ \text{s}$ for start/stop combined), timing just one period gives a large percentage uncertainty.
Instead, time (say) $10$ $10$ oscillations:
- Example: $t = 12.0\ \text{s}$ for $N=10$ .
- Then

T = \frac{t}{N} = \frac{12.0\ \text{s}}{10} = 1.20\ \text{s}.

Repeat the measurement (e.g. obtain $t_1, t_2$ ) and use the mean time to reduce random error.

Key Takeaways

Period is best obtained by timing many cycles: $T=t/N$ .
Repeat readings and average to improve reliability.
Define a consistent reference point to count oscillations accurately.

Common Mistakes

Timing only one oscillation (too much reaction-time uncertainty).
Counting oscillations inconsistently (e.g. counting half-oscillations as full).
Measuring the wrong object’s motion (timing the lower mass instead of the upper mass).
Giving $T$ without a unit.

Things to Be Careful About

Use a small amplitude so the oscillations are smooth and easier to count.
Ensure the stand is secure (G-clamp tight) to avoid movement of the support.
Avoid parallax when judging turning points; using an equilibrium crossing can be easier.
Keep significant figures sensible (typically $T$ to $0.01\ \text{s}$ or $0.1\ \text{s}$ depending on stopwatch and scatter).

Techniques used

set up the apparatus safely and as shownmeasure the time for multiple oscillations and divide by the number of oscillationsrepeat timing measurements and calculate a mean period

(b)

● Transfer some of the slotted masses from the lower string loop to the upper string loop.
● Record the value of the upper mass.

upper mass = ______

● Record the value of $m$ .

$m$ = ______

● Determine the period $T$ of the oscillations of the upper mass.

$T$ = ______ $\text{s}$

DifficultyEasy

Worked solution

Answer

Example set after transferring masses:

upper mass $= 0.200\ \text{kg}$

$m = 0.150\ \text{kg}$

Period (timed over $N$ oscillations and divided by $N$ ):

$T = 1.05\ \text{s}$ (example)

Final answer

upper mass, m and T recorded (student-dependent)

Detailed explanation

Background Concept

In practical work, “record the value of the mass” means you should write down the total load on that part of the system, with a unit, and to a precision consistent with the apparatus (slotted masses are usually known to the nearest gram or better).

The period is again best determined by timing several oscillations and dividing by the number.

Understanding the Question

You are told to change the masses by moving some slotted masses from the lower loop to the upper loop, then:

record the upper mass,
record the lower mass $m$ ,
determine the period $T$ of the upper mass oscillations.

Approach

Transfer a known amount of mass from the lower loop to the upper loop.
Add up the slotted masses to obtain the total upper mass and the new value of $m$ .
Start oscillations and time $N$ oscillations to determine $T=t/N$ .

Step-by-Step Reasoning

Suppose you move one $50\ \text{g}$ mass from the lower loop to the upper loop.
Then upper mass increases by $0.050\ \text{kg}$ and $m$ decreases by $0.050\ \text{kg}$ .
Record both totals clearly with units.
For timing: choose $N=10$ oscillations, measure total time $t$ , then

T = \frac{t}{N}.

Repeat and average if possible.

Key Takeaways

Always state masses and periods with units.
A consistent timing method improves data quality.

Common Mistakes

Writing only the mass moved, not the new total mass.
Mixing units (writing some values in $\text{g}$ and others in $\text{kg}$ without converting).
Forgetting that the period requested is for the upper mass oscillations.

Things to Be Careful About

Ensure the slotted masses sit securely in each loop (no slipping during oscillation).
Record the mass values consistently (all in $\text{kg}$ or all in $\text{g}$ ). If you will plot a graph, using $\text{kg}$ is usually safer for SI units.

Techniques used

adjust the distribution of slotted masses between the loopsrecord masses with appropriate precision and unitsmeasure period by timing multiple oscillations

(c)

Change $m$ by moving slotted masses between the two string loops and then determine $T$ .

Repeat until you have six sets of values of $m$ and $T$ . You may include your results from (a) and (b).

Record your results in a table. Include values of $\sqrt{T}$ in your table.

DifficultyMedium

Worked solution

Answer

Record six sets of $m$ and $T$ (including (a) and (b)) and calculate $\sqrt{T}$ .

Example of a correctly headed table (values are illustrative):

$m / \text{kg}$	$T / \text{s}$	$\sqrt{T} / \text{s}^{1/2}$
$0.050$	$0.33$	$0.574$
$0.100$	$0.42$	$0.648$
$0.150$	$0.53$	$0.725$
$0.200$	$0.64$	$0.800$
$0.250$	$0.77$	$0.877$
$0.300$	$0.90$	$0.949$

Final answer

Table of six (m, T) values with calculated √T (student-dependent).

Detailed explanation

Background Concept

A results table must allow someone else to see exactly what you measured and what you calculated.

Good tables in Paper 3 typically require:

one table containing all results,
clear column headings that include quantity and unit (e.g. $T / \text{s}$ ),
consistent decimal places within a column (reflecting measurement resolution),
a calculated column done correctly (here $\sqrt{T}$ ).

Since you will later plot $\sqrt{T}$ against $m$ , your table must contain both of those columns.

Understanding the Question

You must vary $m$ by moving slotted masses between loops and, each time:

measure the period $T$ of the upper mass oscillations,
repeat until you have six pairs of values $(m, T)$ ,
include $\sqrt{T}$ values in the table.

So the independent variable is $m$ (the lower mass) and the dependent measurement is $T$ , with a derived quantity $\sqrt{T}$ .

Approach

Choose at least six different values of $m$ spanning a reasonable range (not all very close together).
For each $m$ , measure $T$ by timing $N$ oscillations and dividing by $N$ .
Record $m$ and $T$ in a table with units.
Compute $\sqrt{T}$ for each row and include it as a third column with unit $\text{s}^{1/2}$ .

Step-by-Step Reasoning

Decide a set of lower masses $m$ you can make using your available slotted masses (e.g. step by $50\ \text{g}$ ).
For each $m$ $m$ :
- start oscillations with a small, similar amplitude,
- time $t$ for $N$ oscillations (often $N=10$ ),
- compute

T = \frac{t}{N}.

Then calculate the derived quantity:

\sqrt{T} = T^{1/2}.

Record all results in a single table. Headings should be in the format “quantity / unit”, for example:
- $m / \text{kg}$
- $T / \text{s}$
- $\sqrt{T} / \text{s}^{1/2}$
Precision:
- If $T$ is to $0.01\ \text{s}$ , then $\sqrt{T}$ should usually be to 3 significant figures (or consistent decimal places) so it reflects that precision.

Key Takeaways

A practical table is assessed on clarity, units, and consistency.
Derived quantities must be calculated correctly and recorded with suitable precision.
A good range and six readings improve the reliability of the graph.

Common Mistakes

Missing units in headings (e.g. writing just “ $T$ ” instead of “ $T / \text{s}$ ”).
Mixing $\text{g}$ and $\text{kg}$ within the same table.
Inconsistent decimal places in one column (suggests poor measurement discipline).
Calculating $\sqrt{T}$ incorrectly (e.g. using $\sqrt{m}$ by mistake).

Things to Be Careful About

Ensure you really change $m$ (lower loop mass), not only the upper mass.
Do not round $\sqrt{T}$ too aggressively; keep enough significant figures for graph plotting.
Keep your method of timing the same each time (same $N$ , similar amplitude), so that changes in $T$ are due to changing $m$ rather than changing technique.

Techniques used

collect multiple sets of readings over a suitable range of the independent variablecalculate a derived quantity from measured datarecord data in a table with correct headings and unitskeep consistent significant figures within each column

(d)

(i)

Plot a graph of $\sqrt{T}$ on the $y$ -axis against $m$ on the $x$ -axis.

DifficultyMedium-Easy

Worked solution

Answer

Plot $\sqrt{T}$ on the $y$ -axis against $m$ on the $x$ -axis.

Label axes: $m / \text{kg}$ and $\sqrt{T} / \text{s}^{1/2}$ .
Use a suitable scale (at least half the grid on each axis).
Plot all six points accurately.

Final answer

Graph of √T (y) against m (x) plotted with correct labels/scales.

Detailed explanation

Background Concept

A good physics graph communicates the relationship between two quantities clearly. Marks are typically awarded for:

correct choice of variables on each axis,
correct axis labels including units,
sensible scales (not cramped; not awkward values like 3 squares = 1 unit),
accurate plotting of points.

Understanding the Question

You must use your table values to plot a graph with:

$y$ -axis: $\sqrt{T}$
$x$ -axis: $m$

This is a linearisation step: if $\sqrt{T}$ is proportional to $m$ (plus a constant), the graph should be a straight line.

diagram-di-axes

Approach

Decide the range of $m$ values and $\sqrt{T}$ values from your table.
Choose scales so your data spread occupies most of the graph paper.
Label each axis with quantity and unit.
Plot each point with a small neat cross (or dot in a small circle).

Step-by-Step Reasoning

From the table, find minimum and maximum $m$ values (e.g. $0.05$ to $0.30\ \text{kg}$ ).
Set the horizontal axis to cover slightly beyond this range.
From the table, find minimum and maximum $\sqrt{T}$ values (e.g. $0.57$ to $0.95\ \text{s}^{1/2}$ ) and set the vertical axis accordingly.
Label axes clearly:
- Horizontal: $m / \text{kg}$
- Vertical: $\sqrt{T} / \text{s}^{1/2}$
Plot each pair $(m, \sqrt{T})$ carefully, checking you are using the correct row.

Key Takeaways

Always include units on axes.
Use scales that make reading gradients and intercepts accurate.
Plot all points before drawing any line.

Common Mistakes

Swapping axes (plotting $m$ on $y$ by accident).
Missing units or writing incorrect units for $\sqrt{T}$ .
Using a scale that uses only a small corner of the grid.
Plotting $T$ instead of $\sqrt{T}$ .

Things to Be Careful About

If you recorded $m$ in $\text{g}$ , your axis label must be $m / \text{g}$ and the gradient unit will change later.
Ensure your plotted points correspond to $\sqrt{T}$ values calculated from your measured $T$ , not from total times $t$ .

Techniques used

choose appropriate axes and scales that use most of the gridlabel axes with quantities and unitsplot data points accurately from a results table

(ii)

Draw the straight line of best fit.

DifficultyEasy

Worked solution

Answer

Draw a single straight line of best fit through the plotted points (balanced scatter about the line).

Final answer

Straight line of best fit drawn.

Detailed explanation

Background Concept

A line of best fit represents the overall trend of the data. For roughly linear data, the best-fit line should be straight and placed so that the points are distributed fairly evenly above and below it.

Understanding the Question

After plotting $\sqrt{T}$ against $m$ , you must draw the straight line of best fit. This line will later be used to find the gradient and the $y$ -intercept.

Approach

Use a ruler.
Do not join point-to-point.
Aim for a line that reflects the trend and balances the scatter.

Step-by-Step Reasoning

Look at the plotted points: if they follow a straight-line trend, place a ruler so the line passes through the “middle” of the cluster.
There is no need for the line to pass through every point; experimental scatter is expected.
Extend the line across most of the plotted range to help with reading gradient and intercept.

Key Takeaways

Best-fit is about trend, not connecting points.
A good best-fit line improves the accuracy of gradient/intercept.

Common Mistakes

Drawing a dot-to-dot broken line.
Forcing the line through the origin without justification.
Drawing a line that follows one outlier rather than the overall trend.

Things to Be Careful About

If one point is clearly anomalous, you still draw a best-fit line based on the main trend; do not bend the line to accommodate it.
Use a sharp pencil and a ruler to keep the line thin and accurate.

Techniques used

draw a straight line of best fit with balanced scatteravoid joining dot-to-dot between points

(iii)

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

gradient = ______
$y$ -intercept = ______

DifficultyMedium-Easy

Worked solution

Working

Using a large triangle on the best-fit line,

\text{gradient} = \frac{\Delta (\sqrt{T})}{\Delta m}

Example read from graph:

\text{gradient} = 1.50\ \text{s}^{1/2}\ \text{kg}^{-1}

$y$ -intercept (at $m=0$ ) from the line:

\text{intercept} = 0.50\ \text{s}^{1/2}

Answer

gradient $= 1.50\ \text{s}^{1/2}\ \text{kg}^{-1}$ (example)

$y$ -intercept $= 0.50\ \text{s}^{1/2}$ (example)

Final answer

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

Detailed explanation

Background Concept

For a straight-line graph of $y$ against $x$ , the gradient and intercept are defined by

y = mx + c

where:

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

Here, $y = \sqrt{T}$ and $x = m$ , so the gradient has units

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

and the intercept has units $\text{s}^{1/2}$ .

diagram-diii-triangle

Understanding the Question

You must find two numerical quantities from your best-fit line:

the gradient,
the $y$ -intercept.

These are then used in part (e) to obtain constants in the suggested equation.

Approach

Choose two well-separated points on the best-fit line (not necessarily data points).
Read their coordinates accurately.
Compute gradient as $\Delta(\sqrt{T})/\Delta m$ .
Extend the line to $m=0$ and read the intercept $\sqrt{T}$ at that point.

Step-by-Step Reasoning

Pick two points on the line far apart to reduce percentage reading error.
Suppose your chosen points are $(m_1, \sqrt{T_1})$ and $(m_2, \sqrt{T_2})$ .
Calculate differences:

\Delta m = m_2 - m_1,\qquad \Delta(\sqrt{T}) = \sqrt{T_2} - \sqrt{T_1}

Then

\text{gradient} = \frac{\Delta(\sqrt{T})}{\Delta m}.

For the intercept, set $m=0$ on the horizontal axis and read where your line crosses the vertical axis.

Key Takeaways

Always use the best-fit line for gradient, not point-to-point.
Use a large triangle for accuracy.
Quote correct units: gradient $\text{s}^{1/2}\ \text{kg}^{-1}$ , intercept $\text{s}^{1/2}$ (if $m$ is in kg).

Common Mistakes

Using $\Delta x/\Delta y$ instead of $\Delta y/\Delta x$ .
Using two adjacent points (small triangle), giving a poor gradient.
Reading the intercept from the nearest data point rather than from the line.
Forgetting units or using incorrect units (especially if $m$ was recorded in grams).

Things to Be Careful About

Ensure you are reading $\sqrt{T}$ values on the vertical axis, not $T$ .
If your line does not reach $m=0$ on the paper, extend it carefully with a ruler to read the intercept.
Keep consistent significant figures (usually 2–3 s.f. for gradient/intercept depending on scatter and scale).

Techniques used

determine gradient using a large triangle on the best-fit lineread the y-intercept from the graphcalculate and quote gradient and intercept with appropriate units

(e)

It is suggested that the quantities $T$ and $m$ are related by the equation

\sqrt{T} = Pm + Q

where $P$ and $Q$ are constants.

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

$P$ = ______
$Q$ = ______

DifficultyMedium-Easy

Worked solution

Working

Given

\sqrt{T} = Pm + Q

Comparing with $y=mx+c$ for a graph of $\sqrt{T}$ against $m$ :

P = \text{gradient},\qquad Q = y\text{-intercept}

Using (d)(iii) (example):

P = 1.50\ \text{s}^{1/2}\ \text{kg}^{-1}

Q = 0.50\ \text{s}^{1/2}

Answer

$P = 1.50\ \text{s}^{1/2}\ \text{kg}^{-1}$ (example)

$Q = 0.50\ \text{s}^{1/2}$ (example)

Final answer

P = gradient, Q = y-intercept (with appropriate units)

Detailed explanation

Background Concept

When experimental data produce a straight line, you can connect the graph to an equation by matching it to the standard straight-line form:

y = mx + c

$m$ (gradient) tells you how much $y$ changes per unit change in $x$ .
$c$ (intercept) is the value of $y$ when $x=0$ .

This is a powerful method because it lets you determine constants directly from a graph.

Understanding the Question

You are told the relationship is suggested to be

\sqrt{T} = Pm + Q

You already plotted $\sqrt{T}$ against $m$ , and in (d)(iii) you found the gradient and $y$ -intercept of that straight line. This part asks you to use those values to determine $P$ and $Q$ , including units.

Approach

Identify which variable corresponds to $y$ and which to $x$ in the straight-line form.
Match coefficients:
- $P$ corresponds to the gradient.
- $Q$ corresponds to the $y$ -intercept.
Deduce units:

[P] = \frac{[\sqrt{T}]}{[m]},\qquad [Q] = [\sqrt{T}].

Step-by-Step Reasoning

Your graph is $\sqrt{T}$ (vertical) against $m$ (horizontal), so:

\sqrt{T} = (\text{gradient})\, m + (y\text{-intercept}).

Compare with $\sqrt{T} = Pm + Q$ :

P = \text{gradient},\qquad Q = y\text{-intercept}.

Units:
- If $m$ is in $\text{kg}$ and $\sqrt{T}$ is in $\text{s}^{1/2}$ , then

P\ \text{has units}\ \frac{\text{s}^{1/2}}{\text{kg}} = \text{s}^{1/2}\ \text{kg}^{-1},

and

Q\ \text{has units}\ \text{s}^{1/2}.

Substitute your own gradient/intercept values from (d)(iii).

Key Takeaways

From a $y$ vs $x$ graph, the coefficient of $x$ is the gradient and the constant term is the intercept.
Units of constants come from the units on the axes.

Common Mistakes

Swapping $P$ and $Q$ .
Giving $P$ and $Q$ without units.
Using the gradient units incorrectly (e.g. forgetting it is per kg or per g).

Things to Be Careful About

If you plotted $m$ in grams, then $P$ would be in $\text{s}^{1/2}\ \text{g}^{-1}$ instead of $\text{s}^{1/2}\ \text{kg}^{-1}$ .
Quote $P$ and $Q$ to a sensible number of significant figures consistent with how accurately you could read the graph.

Techniques used

match a straight-line graph to the form y = mx + cidentify constants as gradient and interceptassign units to constants from the plotted variables

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