Physics 9702/36 — October/November 2025

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

questions

marks

120

minutes

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

Q120MMediumManipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and Evaluation

In this experiment, you will investigate the phase difference between the oscillations of two mass–spring systems.

(a)

• Assemble the apparatus as shown in Fig. 1.1.

Fig. 1.1

• Mass A and mass B are each $200\text{ g}$ .
• Add a mass $z$ of $40\text{ g}$ to mass B.
Record the value of $z$ .

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

• $M$ is given by $M = 200\text{ g} + z$ .
Calculate $M$ .

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

• Pull both A and B down a short distance and release them together. Observe the oscillations. A and B initially oscillate in phase (both moving up and down together), then their oscillations go out of phase and then become in phase again.
• The time from A and B oscillating in phase to the next time they oscillate in phase is $P$ .
Measure and record $P$ .

$P$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Recorded added mass:

z = 40\ \text{g}

M = 200\ \text{g} + z = 200\ \text{g} + 40\ \text{g} = 240\ \text{g}

Measured time between successive in-phase conditions (example):

P = 18.5\ \text{s}

Answer

z = 40\ \text{g}

M = 240\ \text{g}

P = 18.5\ \text{s}\ \text{(example)}

Final answer

z = 40 g, M = 240 g, P = 18.5 s (example)

Detailed explanation

Background Concept

Two oscillators (mass–spring systems) with slightly different periods will drift in and out of phase. When they start in phase, the phase difference increases with time because one oscillator completes cycles slightly faster than the other. After some time, they return to being in phase again.

The quantity $P$ defined here is the time from one “in phase” condition to the next “in phase” condition (often called a beat period in this context).

Understanding the Question

You are told:

mass A is $200\ \text{g}$ ,
mass B is $200\ \text{g}$ plus an extra mass $z$ (initially $40\ \text{g}$ ),
$M = 200\ \text{g} + z$ .

You must:

record $z$ from the balance, 2) calculate $M$ , and 3) observe the motion and measure the time $P$ between successive times when A and B are in phase.

Approach

Read $z$ directly from the mass pieces / balance and record to the balance resolution.
Add $200\ \text{g}$ to obtain $M$ .
For $P$ , use the definition in the question: start timing when A and B are clearly in phase (moving together), stop when they are next clearly in phase again.

Step-by-Step Reasoning

Added mass is specified as $40\ \text{g}$ , so record:

z = 40\ \text{g}.

Use the given relationship:

M = 200\ \text{g} + z = 200\ \text{g} + 40\ \text{g} = 240\ \text{g}.

To measure $P$ reliably:

Pull both masses down by a small similar distance and release together.
Watch for the condition “in phase” (both moving up together and down together). Choose a clear reference, e.g. when both pass through equilibrium moving upward.
Start the stopwatch at one in-phase moment and stop at the next in-phase moment.
Record $P$ to the stopwatch resolution (commonly $0.1\ \text{s}$ ).

(Your numerical value of $P$ depends on your apparatus and readings, so any sensible measured value with correct precision is acceptable.)

Key Takeaways

Record directly measured values to the instrument’s resolution.
Use the provided formula for a derived quantity ( $M$ ).
Apply the operational definition of $P$ (in-phase to next in-phase) to decide when to start/stop timing.

Common Mistakes

Forgetting to include units (g for $z$ and $M$ , s for $P$ ).
Using the wrong timing interval (e.g. timing from in phase to out of phase).
Recording $P$ with inappropriate precision (e.g. many decimal places not justified by a handheld stopwatch).

Things to Be Careful About

Keep oscillations small so the motion is close to SHM and easier to judge.
Ensure both are released at the same time; otherwise the initial phase is not well-defined.
Decide a consistent “in phase” reference event (e.g. both at lowest point, or both crossing equilibrium in the same direction).

Techniques used

measure mass using a balance to an appropriate resolutioncalculate a derived quantity by additionmeasure a time interval using a stopwatch with a clear start and end condition

(b)

Change $z$ and determine $P$ . Repeat until you have six sets of values of $z$ and $P$ .
Record your results in a table. Include values of $M$ , $\frac{1}{\sqrt{M}}$ and $\frac{1}{P}$ in your table.

10M Similar

DifficultyMedium

Worked solution

Answer

Obtain six different values of $z$ and for each value:

calculate $M = (200 + z)\ \text{g}$ ,
measure $P$ (repeat and average),
calculate $\dfrac{1}{\sqrt{M}}$ and $\dfrac{1}{P}$ .

Record all results in one table with headings including units, e.g.

$z / \text{g}$
$P / \text{s}$
$M / \text{g}$
$1/\sqrt{M} / \text{g}^{-1/2}$
$1/P / \text{s}^{-1}$

(Values are student-dependent.)

Final answer

Single clear results table with six sets of z and P, plus computed M, 1/√M and 1/P (with units).

Detailed explanation

Background Concept

A good practical investigation needs:

enough data points to reveal a trend (here, six sets),
a suitable range of the independent variable ( $z$ hence $M$ ),
repeat readings to reduce random uncertainty,
correct processing of data into forms that allow a straight-line graph.

Calculated quantities should be presented with consistent significant figures and with units.

Understanding the Question

You must change the added mass $z$ and determine the corresponding time $P$ (time between successive in-phase conditions). You then need a table containing not only $z$ and $P$ but also:

M = 200\ \text{g} + z,

\frac{1}{\sqrt{M}},\quad \frac{1}{P}.

Six sets means six different values of $z$ with a matching measured $P$ for each.

Approach

Choose six different values of $z$ (spread out, not all close together).
For each $z$ :

assemble the mass on B,
start both oscillations together,
measure $P$ (repeat at least twice and take a mean).

Fill the table row-by-row:

compute $M$ by addition,
compute $1/\sqrt{M}$ using a calculator,
compute $1/P$ .

Ensure headings show quantity and unit, and ensure consistent dp / sf within each column.

Step-by-Step Reasoning

Choosing $z$ values: Use values that make $M$ clearly different each time (for example, increments of $20\ \text{g}$ or $40\ \text{g}$ , depending on available masses). This improves the spread on the later graph.
Measuring $P$ well:
- Keep amplitude small.
- Start timing at a clearly defined in-phase event (e.g. both at the lowest point, or both crossing equilibrium upward).
- Stop timing at the next time the same event happens with both in phase.
- Repeat the measurement of $P$ and average to reduce random reaction-time effects.
Tabulation conventions:
- Put all data in a single table.
- Include units in the heading (not in every cell).
- Keep decimal places consistent for each calculated column.
Derived columns:
- For each row, compute $M = 200 + z$ in grams.
- Compute $1/\sqrt{M}$ ; if $M$ is in g, then the unit is $\text{g}^{-1/2}$ .
- Compute $1/P$ ; if $P$ is in s, then unit is $\text{s}^{-1}$ .

Key Takeaways

Six points and a good range are essential for a reliable straight-line test.
Repeats improve precision; means are more reliable than single timings.
Correct headings and units are part of the assessment in Paper 3.

Common Mistakes

Forgetting to include either $M$ , $1/\sqrt{M}$ , or $1/P$ in the table.
Mixing units (some $M$ in g, some in kg) within the same data set.
Inconsistent significant figures (e.g. wildly different dp in the same column).
Too small a range of $z$ , producing a cramped graph with large percentage uncertainties in gradient.

Things to Be Careful About

If you decide to convert $M$ to kg, do it for every row and update units: $1/\sqrt{M}$ would then be $\text{kg}^{-1/2}$ .
Don’t round $M$ prematurely before calculating $1/\sqrt{M}$ .
Ensure the same in-phase reference is used every time when timing $P$ .

Techniques used

vary the independent variable over a suitable rangerepeat measurements and take a mean to improve reliabilitycalculate derived quantities for each data setrecord results in a table with correct headings and units

(c)

(i)

Plot a graph of $\frac{1}{P}$ on the $y$ -axis against $\frac{1}{\sqrt{M}}$ on the $x$ -axis.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Plot a graph with:

$y$ -axis: $\dfrac{1}{P}$ in $\text{s}^{-1}$
$x$ -axis: $\dfrac{1}{\sqrt{M}}$ in $\text{g}^{-1/2}$ (or $\text{kg}^{-1/2}$ if $M$ is in kg)

Use a suitable scale (at least half the grid) and plot all six points.

Final answer

Graph of 1/P (y) against 1/√M (x), correctly labelled with units and sensible scales.

Detailed explanation

Background Concept

To test a proposed linear relationship, you plot the variables in a form that should give a straight line. Good graphs in Cambridge practical papers require:

correct axis choice,
correct labels including units,
scales that make good use of the page,
points plotted accurately and clearly.

Understanding the Question

You have a results table containing $M$ , $P$ , and the derived quantities $1/\sqrt{M}$ and $1/P$ . You are asked specifically to plot:

vertical axis: $1/P$ ,
horizontal axis: $1/\sqrt{M}$ .

Approach

Decide the units you are using for $M$ (commonly g in this experiment). Keep them consistent.
Choose axis ranges that cover all your values with some margin.
Pick simple scales (e.g. 1 big square = 0.01) rather than awkward ones.
Plot each point as a small cross.

Step-by-Step Reasoning

From the table, identify the minimum and maximum values of $1/\sqrt{M}$ and $1/P$ .
Draw axes and label:
- $x$ as $1/\sqrt{M}$ with unit $\text{g}^{-1/2}$ (if $M$ in g).
- $y$ as $1/P$ with unit $\text{s}^{-1}$ .
Choose scales so the plotted data fills at least half of each axis.
Plot all six points carefully (use a ruler to read grid lines; avoid large blobs).

Key Takeaways

Correct labels include both the quantity and its unit.
Using most of the grid improves the accuracy of gradient and intercept.

Common Mistakes

Swapping axes (plotting $1/\sqrt{M}$ on $y$ ).
Missing units or writing units wrongly (e.g. $\text{s}$ instead of $\text{s}^{-1}$ ).
Using a tiny section of the graph paper, making the best-fit line uncertain.

Things to Be Careful About

If $M$ is in g, then $\sqrt{M}$ has unit $\text{g}^{1/2}$ and $1/\sqrt{M}$ has unit $\text{g}^{-1/2}$ .
Keep consistent significant figures in the plotted values to match your table.

Techniques used

choose appropriate axes and scales that use most of the graph paperlabel axes with quantity and unitplot experimental 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 well-separated points on the best-fit line:

\text{gradient} = \frac{\Delta (1/P)}{\Delta (1/\sqrt{M})}

Read the $y$ -intercept from where the best-fit line crosses the $y$ -axis ( $x=0$ ).

Answer

gradient = (from graph)

$y$ -intercept = (from graph)

Final answer

Gradient and y-intercept read from the best-fit line (student-dependent).

Detailed explanation

Background Concept

For a straight-line graph:

y = mx + c,

$m$ is the gradient (slope): $m = \Delta y / \Delta x$ ,
$c$ is the $y$ -intercept: the value of $y$ when $x=0$ .

In practical graphs, you must calculate the gradient using two points on the best-fit line, not necessarily two data points, and they should be far apart to reduce percentage reading error.

Understanding the Question

Your graph has:

$y = 1/P$ ,
$x = 1/\sqrt{M}$ .

You must determine:

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

Approach

Draw a large triangle on the best-fit line (use points that are widely separated).
Read the coordinates of the two triangle points.
Compute gradient using $\Delta y/\Delta x$ .
Read the intercept by extending the line to the $y$ -axis and reading the value.

Step-by-Step Reasoning

diagram-ciii

Pick two points A and B on the best-fit line, ideally near the ends of the drawn line.
Read their coordinates: $(x_1, y_1)$ and $(x_2, y_2)$ where $x = 1/\sqrt{M}$ and $y = 1/P$ .
Calculate:

\Delta y = y_2 - y_1, \quad \Delta x = x_2 - x_1,

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

For the intercept, extend the best-fit line until it crosses the $y$ -axis (where $x=0$ ). Read this value as $c$ .

Units:

$y$ has unit $\text{s}^{-1}$ .
If $M$ is in g, $x$ has unit $\text{g}^{-1/2}$ .
So gradient unit is:

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

The intercept has the same unit as $y$ , i.e. $\text{s}^{-1}$ .

Key Takeaways

Use the best-fit line and a large triangle for a more accurate gradient.
Always quote gradient and intercept with correct units.

Common Mistakes

Using two adjacent points (small triangle) leading to large uncertainty.
Calculating $\Delta x/\Delta y$ by accident (inverting the gradient).
Reading the intercept from a data point rather than from the best-fit line.

Things to Be Careful About

Ensure you use the same scale divisions when reading coordinates.
Don’t round intermediate readings too aggressively; keep enough precision from the graph readings to give a sensible final gradient.
If the best-fit line does not reach the $y$ -axis on your plotted range, extend it carefully with a ruler to read the intercept.

Techniques used

construct a large gradient triangle on a best-fit linecalculate gradient using \u0394y/\u0394x with unitsread y-intercept from the graph at x = 0

(d)

It is suggested that the quantities $P$ and $M$ are related by the equation

\frac{1}{P} = \frac{a}{\sqrt{M}} + b

where $a$ and $b$ are constants.

Use your answers in (c)(iii) to determine the values of $a$ and $b$ .
Give appropriate units.

$a$ = ______
$b$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Given

\frac{1}{P} = \frac{a}{\sqrt{M}} + b

Compare with $y = mx + c$ where $y = \dfrac{1}{P}$ and $x = \dfrac{1}{\sqrt{M}}$ .

a = \text{gradient}, \qquad b = y\text{-intercept}.

Units:

$b$ has same unit as $1/P$ : $\text{s}^{-1}$ .
$a$ has units of (gradient) = $\text{s}^{-1} \div \text{g}^{-1/2} = \text{g}^{1/2}\ \text{s}^{-1}$ (or $\text{kg}^{1/2}\ \text{s}^{-1}$ if $M$ in kg).

Answer

a = \text{gradient}\ \left(\text{g}^{1/2}\ \text{s}^{-1}\right)

b = y\text{-intercept}\ \left(\text{s}^{-1}\right)

Final answer

a = gradient (units g^{1/2} s^{-1} or kg^{1/2} s^{-1}); b = y-intercept (units s^{-1}).

Detailed explanation

Background Concept

If a relationship can be written in the form

y = mx + c,

then a plot of $y$ against $x$ gives a straight line with:

gradient $m$ ,
intercept $c$ .

Matching an experimental straight line to this form is how you extract constants from data.

Understanding the Question

You are told the suggested model:

\frac{1}{P} = \frac{a}{\sqrt{M}} + b.

From part (c)(iii) you already have the gradient and $y$ -intercept of the graph of $1/P$ (y-axis) against $1/\sqrt{M}$ (x-axis). You must use these to find $a$ and $b$ , and include appropriate units.

Approach

Identify what you plotted: $y = 1/P$ and $x = 1/\sqrt{M}$ .
Rewrite the model as $y = ax + b$ .
Therefore, $a$ is the gradient and $b$ is the y-intercept.
Work out units from the axis units.

Step-by-Step Reasoning

Start with the given equation:

\frac{1}{P} = \frac{a}{\sqrt{M}} + b.

Define:

y = \frac{1}{P}, \quad x = \frac{1}{\sqrt{M}}.

Then the equation becomes:

y = ax + b.

So by comparison with $y = mx + c$ :

gradient $m = a$ ,
intercept $c = b$ .

Units:

$P$ is in seconds, so $1/P$ is in $\text{s}^{-1}$ . Hence $b$ is in $\text{s}^{-1}$ .
If $M$ is in grams, then $x = 1/\sqrt{M}$ has units $\text{g}^{-1/2}$ .
So

[a] = \frac{[y]}{[x]} = \frac{\text{s}^{-1}}{\text{g}^{-1/2}} = \text{g}^{1/2}\ \text{s}^{-1}.

If you used kilograms for $M$ , replace $\text{g}^{1/2}$ by $\text{kg}^{1/2}$ .

Key Takeaways

A graph is most useful when it linearises the relationship into $y = mx + c$ .
Constants are read directly as gradient and intercept when variables match this form.
Units of constants come from the plotted quantities’ units.

Common Mistakes

Swapping $a$ and $b$ (remember: $a$ multiplies $x$ , $b$ is the intercept).
Giving $a$ the same units as $b$ (they are generally different).
Mixing g and kg between table, graph, and final units.

Things to Be Careful About

Use the gradient and intercept values from your best-fit line (not from two data points).
State units clearly using index notation (e.g. $\text{s}^{-1}$ , $\text{g}^{1/2}\ \text{s}^{-1}$ ).
If you converted $M$ to kg at any stage, keep that consistent through to the units for $a$ .

Techniques used

match a straight-line graph to the form y = mx + cidentify constants from gradient and interceptdetermine units of constants from axis units

Q220MMediumManipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and Evaluation

In this experiment, you will investigate the tension in a string.

(a)

• Set up the apparatus as shown in Fig. 2.1.

Fig. 2.1

• The mass hanger and masses should have a total mass $M$ of $0.400\text{ kg}$ .
• The distance between the two lower nails is $d$ , as shown in Fig. 2.1.
Measure and record $d$ .

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

1M Similar

DifficultyEasy

Worked solution

Measure the vertical distance between the two lower nails with a ruler, to the nearest $0.1\ \text{cm}$ .

Example recorded value:

d = 37.0\ \text{cm}

Final answer

d = 37.0 cm (example)

Detailed explanation

Background Concept

In Paper 3, marks for a measurement are awarded for (i) measuring the correct quantity shown on the diagram and (ii) recording it with appropriate precision for the instrument used.

A ruler typically has $1\ \text{mm}$ smallest divisions, so a sensible recorded precision is to the nearest $1\ \text{mm}$ , i.e. $0.1\ \text{cm}$ .

Understanding the Question

You are told that $d$ is the distance between the two lower nails in Fig. 2.1. So you must locate the two lower nails and measure the separation between them (along the stand, i.e. the vertical separation shown).

Approach

Place the ruler alongside the stand.
Read the positions of the two lower nails (or their boss positions) against the ruler.
Subtract to get $d$ .
Record $d$ in $\text{cm}$ to $0.1\ \text{cm}$ .

Step-by-Step Reasoning

Align the ruler parallel to the stand so that it measures the separation vertically.
Avoid parallax: your eye should be level with the mark you are reading.
Read the position of the middle nail and the bottom nail, then compute the difference.
Record a value such as $d = 37.0\ \text{cm}$ (your value depends on your set-up).

Key Takeaways

Measure the correct distance indicated by the diagram.
Record to a precision appropriate to the instrument (typically $0.1\ \text{cm}$ for a ruler).

Common Mistakes

Measuring the wrong distance (e.g. from the top nail to a lower nail).
Recording with inappropriate precision (e.g. $37\ \text{cm}$ with no decimal, or $37.00\ \text{cm}$ implying unrealistic precision).

Things to Be Careful About

Ensure you measure between the two lower nails only.
Keep the ruler aligned with the direction of the distance being measured and avoid parallax.

Techniques used

identify the correct distance to be measured from the apparatus diagrammeasure a length with a ruler and record to an appropriate precision

(b)

The tension in the string is $T$ .
Calculate $T$ using $T = Mg$ , where $g = 9.81\text{ N kg}^{-1}$ .

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

1M Similar

(c)

(i)

• Hook the newton meter on the string half-way between the two lower nails and pull it horizontally with a force $F$ of $5.0\text{ N}$ , as shown in Fig. 2.2.

Fig. 2.2

• The force $F$ causes the string to deflect a distance $x$ , as shown in Fig. 2.2.
Measure and record $x$ .

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

2M Similar

DifficultyMedium-Easy

Worked solution

Apply $F = 5.0\ \text{N}$ with the newton meter and measure the horizontal deflection $x$ with a ruler, to the nearest $0.1\ \text{cm}$ .

Example recorded value:

x = 6.2\ \text{cm}

Final answer

x = 6.2 cm (example)

Detailed explanation

Background Concept

A displacement measurement in practical work needs:

a clear reference position (here, the undeflected string position),
a clear measured point (here, the midpoint of the string segment between the two lower nails),
an instrument of suitable resolution (ruler/metre rule),
a steady reading (no oscillations).

Understanding the Question

You must pull the string horizontally at the midpoint between the two lower nails with a force $F = 5.0\ \text{N}$ using a newton meter. This produces a sideways deflection $x$ . You then measure $x$ and record it in $\text{cm}$ .

Approach

Locate the midpoint between the two lower nails.
Hook the newton meter at this midpoint.
Pull until the newton meter reads $5.0\ \text{N}$ .
Measure the horizontal shift $x$ from the original (undeflected) string line to the pulled position.

Step-by-Step Reasoning

With no pull, note the vertical line of the string between the lower nails (this is your zero position).
Pull sideways and keep the newton meter reading at $5.0\ \text{N}$ .
Wait for the string to stop oscillating.
Use a ruler (or set square) to measure the perpendicular horizontal distance from the original vertical line to the new position at the midpoint.
Record to $0.1\ \text{cm}$ , e.g. $x = 6.2\ \text{cm}$ .

Key Takeaways

Always measure a displacement relative to a defined reference.
For ruler measurements, record to $0.1\ \text{cm}$ (typical).

Common Mistakes

Measuring along the string (gives a longer distance than the horizontal deflection).
Measuring from the nail rather than from the original straight line of the string.
Not keeping $F$ at $5.0\ \text{N}$ while measuring.

Things to Be Careful About

Parallax: read the ruler at eye level.
Ensure you are measuring the deflection at the midpoint between nails (as instructed).

Techniques used

apply a specified horizontal force using a newton metermeasure a displacement from an initial reference positionrecord a reading with appropriate precision

(ii)

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

percentage uncertainty = ______ $\%$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Assume ruler uncertainty in $x$ is $\pm 0.1\ \text{cm}$ .

\%\ \text{uncertainty} = \frac{0.1}{6.2} \times 100

\%\ \text{uncertainty} = 1.6\%

Answer

$1.6\%$

Final answer

1.6%

Detailed explanation

Background Concept

For a directly measured quantity, an absolute uncertainty can be estimated from the instrument resolution.

If a ruler is read to the nearest $1\ \text{mm}$ , a common estimate is an absolute uncertainty of about $\pm 1\ \text{mm} = \pm 0.1\ \text{cm}$ for a single reading.

Percentage uncertainty is

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

where $\Delta x$ is the absolute uncertainty.

Understanding the Question

You must estimate the percentage uncertainty in your measured value of $x$ and show working. So you need an assumed absolute uncertainty and then compute the percentage.

Approach

Choose a sensible absolute uncertainty for $x$ based on how you measured it.
Use $\Delta x / x \times 100$ .

Step-by-Step Reasoning

If $x$ is measured with a ruler to $0.1\ \text{cm}$ , take $\Delta x \approx 0.1\ \text{cm}$ .
With example $x = 6.2\ \text{cm}$ :

\%\ \text{uncertainty} = \frac{0.1}{6.2} \times 100 = 1.61\%

Quote sensibly (typically 2 s.f.): $1.6\%$ .

(If you judged the reading to be harder, e.g. because the string is thick or moving, you might justify a larger $\Delta x$ ; the key is to show a reasonable method.)

Key Takeaways

Percentage uncertainty comes from $\Delta x / x$ .
Instrument resolution is the starting point for $\Delta x$ .

Common Mistakes

Using $\Delta x = 0.05\ \text{cm}$ without justification when readings are only to $0.1\ \text{cm}$ .
Forgetting to multiply by 100.

Things to Be Careful About

If $x$ is found from two readings (initial and final position), you should consider adding the absolute uncertainties of both readings. Only do this if it matches how you actually measured $x$ and you state it clearly.

Techniques used

estimate absolute uncertainty from instrument resolutioncalculate percentage uncertainty using a ratiopresent uncertainty to an appropriate number of significant figures

(iii)

Calculate $y$ , where

y = \sqrt{\left( x^2 + \frac{d^2}{4} \right)}

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

1M Similar

DifficultyMedium-Easy

Worked solution

Working

y = \sqrt{\left(x^2 + \frac{d^2}{4}\right)}

Using $x = 6.2\ \text{cm}$ and $d = 37.0\ \text{cm}$ :

y = \sqrt{\left(6.2^2 + \frac{37.0^2}{4}\right)}

y = \sqrt{\left(38.44 + 342.25\right)} = \sqrt{380.69} = 19.51\ \text{cm}

Answer

y = 19.5\ \text{cm}

Final answer

19.5 cm

Detailed explanation

Background Concept

This question gives you a defined mathematical relationship:

y = \sqrt{\left(x^2 + \frac{d^2}{4}\right)}

This has the form of Pythagoras’ theorem, where $y$ is the hypotenuse of a right triangle with perpendicular sides $x$ and $d/2$ . As long as $x$ and $d$ are in the same units, $y$ comes out in those same units.

Understanding the Question

You have measured $x$ (the horizontal deflection) and $d$ (the separation of the two lower nails). You are asked to calculate $y$ using the provided formula and to record it in $\text{cm}$ .

Approach

Ensure $x$ and $d$ are both in $\text{cm}$ .
Compute $x^2$ .
Compute $d^2/4$ .
Add them.
Take the square root to get $y$ .

Step-by-Step Reasoning

diagram-ciii-1

Using the example values $x = 6.2\ \text{cm}$ and $d = 37.0\ \text{cm}$ :

Square $x$ :

x^2 = 6.2^2 = 38.44\ \text{cm}^2

Compute $d^2/4$ :

\frac{d^2}{4} = \frac{37.0^2}{4} = \frac{1369}{4} = 342.25\ \text{cm}^2

Add:

x^2 + \frac{d^2}{4} = 38.44 + 342.25 = 380.69\ \text{cm}^2

Square root:

y = \sqrt{380.69} = 19.51\ \text{cm}

Quote appropriately (matching measurement precision): $y = 19.5\ \text{cm}$ .

Key Takeaways

Keep units consistent before substitution.
When squaring lengths you get $\text{cm}^2$ , and the square root returns to $\text{cm}$ .

Common Mistakes

Using $d^2/2$ instead of $d^2/4$ .
Mixing units (e.g. $x$ in cm and $d$ in m).
Forgetting the square root at the end.

Things to Be Careful About

Significant figures: $x$ might be 2 s.f., so quoting $y$ to 3–4 s.f. is usually unnecessary.
Calculator brackets: ensure you calculate $d^2/4$ correctly, not $d^{2/4}$ .

Techniques used

substitute measured values into a given expressionsquare, add and take a square root with consistent unitspresent a calculated length to appropriate significant figures

(d)

• Add slotted masses to the mass hanger so that the total mass $M$ is $0.700\text{ kg}$ .
• Repeat (b), (c)(i) and (c)(iii).

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

3M Similar

DifficultyMedium-Easy

Worked solution

For $M = 0.700\ \text{kg}$ :

T = Mg = 0.700 \times 9.81 = 6.867\ \text{N} \approx 6.87\ \text{N}

Apply $F = 5.0\ \text{N}$ and measure $x$ (example):

x = 3.5\ \text{cm}

Using $d = 37.0\ \text{cm}$ :

y = \sqrt{\left(3.5^2 + \frac{37.0^2}{4}\right)} = \sqrt{\left(12.25 + 342.25\right)} = \sqrt{354.50} = 18.83\ \text{cm}

y = 18.8\ \text{cm}

Final answer

T = 6.87 N, x = 3.5 cm, y = 18.8 cm (example)

Detailed explanation

Background Concept

Changing the hanging mass changes the tension via $T = Mg$ . A larger $M$ gives a larger tension, so for the same sideways force $F$ the deflection $x$ is typically smaller.

You then recompute $y$ from the same geometrical relationship:

y = \sqrt{\left(x^2 + \frac{d^2}{4}\right)}

Understanding the Question

You add masses so the total is $0.700\ \text{kg}$ , then you must repeat:

(b): calculate $T$ using $T=Mg$ ,
(c)(i): measure the new $x$ when $F=5.0\ \text{N}$ ,
(c)(iii): calculate the corresponding $y$ .

Approach

Compute $T$ from $M$ .
Perform the pull at $5.0\ \text{N}$ and record the new $x$ .
Substitute $x$ and the same measured $d$ into the formula for $y$ .

Step-by-Step Reasoning

Tension:

T = 0.700 \times 9.81 = 6.867\ \text{N} \approx 6.87\ \text{N}

Deflection: with greater tension the string resists sideways displacement, so $x$ is usually less than before (example $x = 3.5\ \text{cm}$ ).
Calculate $y$ :

y = \sqrt{\left(3.5^2 + \frac{37.0^2}{4}\right)}

y = \sqrt{\left(12.25 + 342.25\right)} = \sqrt{354.50} = 18.83\ \text{cm} \approx 18.8\ \text{cm}

Key Takeaways

Repeating measurements for a new value of $M$ gives another data set to test the proposed relationship.
Consistent methods (same $d$ , same $F$ ) are essential.

Common Mistakes

Forgetting to include the mass hanger in the total mass.
Allowing $F$ to differ from $5.0\ \text{N}$ when measuring $x$ .
Using a new value of $d$ without re-measuring or stating it.

Things to Be Careful About

Keep the same midpoint position for pulling.
Record $x$ to consistent precision (e.g. $0.1\ \text{cm}$ ) so later significant figures in $k$ are justified.

Techniques used

calculate tension from a changed loadmeasure a new displacement while keeping the applied force constantsubstitute new measurements into a given formula

(e)

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

ky = Tx

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-Easy

Worked solution

Working

From $ky = Tx$ :

k = \frac{Tx}{y}

For $M = 0.400\ \text{kg}$ :

k_1 = \frac{(3.92)(6.2)}{19.5} = 1.25\ \text{N} \approx 1.2\ \text{N}

For $M = 0.700\ \text{kg}$ :

k_2 = \frac{(6.87)(3.5)}{18.8} = 1.28\ \text{N} \approx 1.3\ \text{N}

Answer

first value of $k = 1.2\ \text{N}$

second value of $k = 1.3\ \text{N}$

Final answer

k ≈ 1.2 N and 1.3 N (example)

Detailed explanation

Background Concept

When a relationship includes an unknown constant, you can find the constant by rearranging and substituting measured values.

Given

ky = Tx

$k$ is constant if the relationship is correct. Rearranging gives

k = \frac{Tx}{y}

Because $x$ and $y$ are both lengths measured in the same units, the ratio $x/y$ is dimensionless, so $k$ has the same unit as $T$ , i.e. $\text{N}$ .

Understanding the Question

You have two sets of data (one for $M=0.400\ \text{kg}$ and one for $M=0.700\ \text{kg}$ ). You must calculate $k$ for each set to see whether it stays about the same.

Approach

Rearrange to $k = Tx/y$ .
Use your calculated $T$ and calculated $y$ together with your measured $x$ .
Compute two values, $k_1$ and $k_2$ .

Step-by-Step Reasoning

Using the example data:

First run ( $M=0.400\ \text{kg}$ $M = 0.400 kg$ ):
- $T = 3.92\ \text{N}$ , $x = 6.2\ \text{cm}$ , $y = 19.5\ \text{cm}$

k_1 = \frac{Tx}{y} = \frac{(3.92)(6.2)}{19.5} = 1.25\ \text{N}

Second run ( $M=0.700\ \text{kg}$ $M = 0.700 kg$ ):
- $T = 6.87\ \text{N}$ , $x = 3.5\ \text{cm}$ , $y = 18.8\ \text{cm}$

k_2 = \frac{(6.87)(3.5)}{18.8} = 1.28\ \text{N}

You then round both values to the appropriate significant figures (see next part).

Key Takeaways

Always rearrange first so you substitute correctly.
Check units: here $k$ comes out in $\text{N}$ .

Common Mistakes

Using $k = Ty/x$ (wrong rearrangement).
Mixing units (e.g. converting only $x$ to m but leaving $y$ in cm).

Things to Be Careful About

Use the corresponding $T$ , $x$ , and $y$ from the same trial.
If your calculated $k$ values differ slightly, that is expected; later you compare within uncertainty.

Techniques used

rearrange an algebraic relationship to make the constant the subjectsubstitute experimental values into an equationcompare values from two trials

(ii)

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

1M Similar

DifficultyMedium-Easy

Worked solution

The limiting data are $F = 5.0\ \text{N}$ (2 s.f.) and the measured $x$ values (typically 2 s.f.).
So $k$ is given to 2 significant figures.

Final answer

k to 2 significant figures

Detailed explanation

Background Concept

A calculated quantity should not be quoted to more significant figures than the least precise quantity used in the calculation.

If you multiply/divide quantities, the result should usually be given to the same number of significant figures as the input with the fewest significant figures.

Understanding the Question

You calculated $k$ from

k = \frac{Tx}{y}

You must justify how many significant figures you used for $k$ .

Approach

Identify the least precisely known quantities that affect $k$ .
Choose significant figures for $k$ to match those.

Step-by-Step Reasoning

$T$ is calculated from $M$ and $g$ . Here $g = 9.81$ (3 s.f.) and $M$ is given (often 3 s.f.), so $T$ can reasonably be 3 s.f.
$x$ is a measured deflection and is commonly recorded to $0.1\ \text{cm}$ ; values like $6.2\ \text{cm}$ or $3.5\ \text{cm}$ are typically 2 s.f.
$y$ is calculated from $x$ and $d$ , but its meaningful precision is still limited by the precision of $x$ and $d$ .
Also, the applied force is stated as $F = 5.0\ \text{N}$ (2 s.f.), and in practice you cannot hold it exactly at more than that.

Therefore, quoting $k$ to 2 significant figures is justified.

Key Takeaways

Significant figures in derived constants should reflect measurement precision.

Common Mistakes

Quoting $k$ to 4 s.f. from a calculator display.
Rounding inconsistently (e.g. one $k$ to 2 s.f. and the other to 3 s.f. without reason).

Things to Be Careful About

If your measured values are recorded to 3 s.f. (e.g. $x = 6.20\ \text{cm}$ ), then you may justify 3 s.f.; the justification must match what you actually recorded.

Techniques used

identify the limiting measurement in a calculationmatch significant figures of a derived constant to raw data precision

(f)

It is suggested that the percentage uncertainty in the values of $k$ is $20\%$ .
Using this uncertainty, explain whether your results support the relationship in (e).

1M Similar

DifficultyMedium

Worked solution

Using $k_1 = 1.2\ \text{N}$ and $k_2 = 1.3\ \text{N}$ :

\text{percentage difference} = \frac{0.1}{1.25} \times 100 = 8\%

$8\% < 20\%$ , so the two values agree within the stated uncertainty and support $ky = Tx$ .

Final answer

Yes, supports within 20% uncertainty

Detailed explanation

Background Concept

Experimental results rarely match exactly. We decide whether they support a suggested relationship by checking whether differences are within the stated uncertainty.

If the percentage uncertainty in $k$ is $20\%$ , then two values of $k$ are consistent if their difference is not more than about $20\%$ (a common simple test is to compare percentage difference with the uncertainty).

Understanding the Question

You have two values of $k$ from part (e). You are told to assume a $20\%$ uncertainty in $k$ and to decide whether your results support the relationship $ky = Tx$ (i.e. whether $k$ is approximately constant).

Approach

Find how different the two $k$ values are (percentage difference).
Compare this with $20\%$ .
Conclude whether they agree within uncertainty.

Step-by-Step Reasoning

Using example rounded values $k_1 = 1.2\ \text{N}$ and $k_2 = 1.3\ \text{N}$ :

Difference:

\Delta k = 1.3 - 1.2 = 0.1\ \text{N}

Mean value:

\bar{k} = \frac{1.2 + 1.3}{2} = 1.25\ \text{N}

Percentage difference:

\frac{\Delta k}{\bar{k}} \times 100 = \frac{0.1}{1.25} \times 100 = 8\%

Since $8\%$ is less than the stated $20\%$ , the two values of $k$ are consistent within uncertainty. That means the data support the proposed relationship.

Key Takeaways

“Supports the relationship” means “consistent within uncertainty,” not “exactly equal.”
Use a clear quantitative comparison when an uncertainty is given.

Common Mistakes

Concluding “does not support” just because $k_1 \neq k_2$ exactly.
Comparing the absolute difference ( $0.1$ ) to $20\%$ without converting to a percentage.

Things to Be Careful About

Use your own calculated $k$ values (not the example ones).
State the conclusion explicitly and link it to the inequality (e.g. “difference < 20% therefore supports”).

Techniques used

compare two calculated values using an allowed percentage uncertaintyuse percentage difference to judge consistencystate a conclusion linked to uncertainty

(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

Measurement of $x$ : difficult to judge the undeflected reference line and the position of the string (string has thickness / may be moving), giving parallax and reading uncertainty.
Maintaining $F = 5.0\ \text{N}$ : newton meter reading fluctuates and is hard to hold steady while measuring $x$ .
Tension not uniform: friction where the string contacts the nails means the tension between the lower nails may not equal $Mg$ .
String stretches / oscillates: applying the pull can cause the string to stretch or vibrate so the measured $x$ is not a steady equilibrium value.

Final answer

Four valid limitations/uncertainties (see working)

Detailed explanation

Background Concept

Uncertainties and limitations come from:

measurement resolution and judgement (random uncertainty),
difficulty maintaining constant conditions,
systematic effects (e.g. friction),
assumptions in the physics model (e.g. uniform tension).

Good practical answers:

name the quantity affected, and
give a physical reason why it is uncertain or biased.

Understanding the Question

You must give four sources of uncertainty/limitations in this specific string-tension experiment. For measurement uncertainties, you must state what is being measured and why it is uncertain.

Approach

Think through each stage:

measuring $d$ ,
applying and reading $F$ ,
measuring $x$ ,
assuming $T = Mg$ everywhere,
stability of the string.
Choose four distinct issues and describe them clearly.

Step-by-Step Reasoning

Examples of creditworthy points:

Uncertainty in $x$ : the string is not a sharp line; it has thickness, may not remain still, and the “original” vertical reference position is hard to mark, so $x$ has parallax/judgement uncertainty.
Uncertainty in $F$ : the newton meter pointer may fluctuate and it is hard to keep exactly $5.0\ \text{N}$ while also reading $x$ . This changes $x$ .
Friction at nails (systematic): the string rubbing on nails means the tension may differ on each side of a contact point; therefore the tension in the lower section may not equal the simple value $Mg$ .
Oscillation/stretching: pulling can make the string vibrate; if you measure before it settles you do not get the true equilibrium $x$ . Also, elastic stretching changes the geometry during the measurement.

Other acceptable limitations could include difficulty identifying the midpoint between nails, nails not perfectly aligned, or the hanging mass swinging.

Key Takeaways

Always link an uncertainty to a specific measured quantity.
Include both random (reading/judgement) and systematic (friction, non-uniform tension) limitations.

Common Mistakes

Writing vague statements like “human error” or “parallax” with no quantity stated.
Repeating the same idea four times (e.g. four versions of “hard to measure”).

Things to Be Careful About

Make sure each point is distinct.
Include the reason (e.g. “string oscillates so reading changes”), not just the effect.

Techniques used

identify measurement uncertainties and link them to specific quantitiesexplain physical or procedural causes of scatter or systematic errorstate limitations clearly and distinctly

(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

Measure $x$ with a fixed ruler scale and a perpendicular set square/pointer attached at the midpoint of the string to reduce parallax and improve repeatability.
Use a clamp/retort stand to hold the newton meter position so that $F$ is steady at $5.0\ \text{N}$ while $x$ is read.
Reduce friction at contact points by using smooth pulleys/low-friction rings instead of nails (or lubricate/ensure smooth nail surfaces).
Take repeated readings of $x$ for each mass and average (and wait for oscillations to die away before recording).

Final answer

Four valid improvements (see working)

Detailed explanation

Background Concept

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

are specific and practical,
explain how the change reduces uncertainty or systematic error,
ideally link back to a limitation from part (g)(i).

Understanding the Question

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

Approach

Match improvements to common issues:

improve how $x$ is measured,
keep $F$ constant,
reduce friction so tension is closer to $Mg$ everywhere,
reduce random scatter with repeats and averaging.

Step-by-Step Reasoning

Examples:

Better measurement of $x$ : attach a small pointer at the midpoint of the string and place a ruler behind it; use a set square to ensure the horizontal distance is measured perpendicularly. This reduces judgement and parallax.
Keep $F$ steady: clamp the newton meter or use a fixed horizontal pulling arrangement so the force remains at $5.0\ \text{N}$ while you read $x$ .
Reduce friction: replace nails with smooth pulleys or low-friction rings so the tension is more uniform along the string, making $T = Mg$ a better assumption.
Repeats and averaging: repeat $x$ measurements several times for each $M$ and average; also wait until oscillations die away before taking readings. This reduces random uncertainty.

Other improvements could include using a digital force sensor, increasing the number of different masses to collect more data, or securing the mass to prevent swinging.

Key Takeaways

Improvements must be realistic and should directly address the causes of uncertainty.

Common Mistakes

Giving an “improvement” that is just a restatement of a limitation without a fix.
Suggesting irrelevant changes (e.g. “use a micrometer” for a large displacement measurement).

Things to Be Careful About

Do not repeat essentially the same improvement in different words.
If you suggest additional readings, state explicitly that you would average them and keep conditions the same.

Techniques used

propose procedural changes to reduce random uncertaintypropose apparatus changes to reduce systematic errorlink each improvement to a corresponding limitation