Physics 9702/38 — October/November 2025

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

questions

marks

120

minutes

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

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

In this experiment, you will investigate the 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

Answer

Recorded added mass:

z = 40.0\ \text{g}

Calculated mass:

M = 200\ \text{g} + z = 200\ \text{g} + 40.0\ \text{g} = 240.0\ \text{g}

Measured time between successive occasions when the two masses are in phase:

P = \text{(stopwatch reading)}\ \text{s} \quad \text{e.g. } P = 12.3\ \text{s}

Final answer

z = 40.0 g, M = 240.0 g, P = student-dependent (e.g. 12.3 s)

Detailed explanation

Background Concept

Two oscillators with slightly different frequencies will not stay in phase. The phase difference increases with time and they periodically return to being in phase. The time between successive occasions when they are in phase is the beat period (here labelled $P$ ).

The mass for system B is increased by adding $z$ , so the total oscillating mass is

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

Understanding the Question

You are told that the added mass is $z = 40\ \text{g}$ and asked to (1) record it, (2) calculate $M$ , and (3) time $P$ , where $P$ is the time from one “in phase” moment to the next “in phase” moment.

In practice, “in phase” means both masses are at the same part of the motion together (e.g. both at the lowest point at the same time, or both moving upward through equilibrium together).

Approach

Record $z$ to the appropriate precision (typically the balance resolution or the stated value).
Add $200\ \text{g}$ and $z$ to get $M$ .
Start both masses together, then use a stopwatch to measure the time between two successive clear “in phase” events.

To reduce reaction-time uncertainty, it is good practice to time several beat periods (e.g. 3 to 5 times the in-phase-to-in-phase interval) and divide by the number of periods.

Step-by-Step Reasoning

The question states $z = 40\ \text{g}$ , so you record this (often as $40.0\ \text{g}$ if masses are quoted to 0.1 g).
Calculate

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

For $P$ , you observe the motion: initially in phase, then out of phase, then in phase again. Choose a repeatable in-phase indicator (e.g. both masses simultaneously at the lowest point). Start timing at one in-phase moment and stop timing at the next matching in-phase moment.

Key Takeaways

Use the definition of a derived quantity (here $M = 200\ \text{g} + z$ ).
Time an interval defined by a repeated event; using multiple cycles reduces percentage uncertainty.

Common Mistakes

Using the wrong event for $P$ (e.g. timing from “in phase” to “out of phase”).
Calculating $M$ incorrectly (subtracting $z$ instead of adding).
Quoting $P$ with no unit or with inappropriate precision.

Things to Be Careful About

Keep the definition consistent: $P$ is from “in phase” to the next time in phase.
Ensure both masses are released together with small amplitude so that the motion is clear and similar for each run.
Record units explicitly: $z$ and $M$ in $\text{g}$ (as used throughout the question) and $P$ in $\text{s}$ .

Techniques used

read a mass value from the stated added massescalculate a derived mass by additionmeasure a time interval corresponding to a repeated observable event

(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

Take six different values of $z$ and for each value measure $P$ .

Calculate for each set:

M = (200 + z)\ \text{g},\qquad \frac{1}{\sqrt{M}}\ (\text{g}^{-1/2}),\qquad \frac{1}{P}\ (\text{s}^{-1}).

Record all results in one table, e.g.

$z\ /\ \text{g}$	$M\ /\ \text{g}$	$P\ /\ \text{s}$	$1/\sqrt{M}\ /\ \text{g}^{-1/2}$	$1/P\ /\ \text{s}^{-1}$
$40.0$	$240.0$	$\cdots$	$\cdots$	$\cdots$
$\cdots$	$\cdots$	$\cdots$	$\cdots$	$\cdots$
$\cdots$	$\cdots$	$\cdots$	$\cdots$	$\cdots$
$\cdots$	$\cdots$	$\cdots$	$\cdots$	$\cdots$
$\cdots$	$\cdots$	$\cdots$	$\cdots$	$\cdots$
$\cdots$	$\cdots$	$\cdots$	$\cdots$	$\cdots$

(Consistent decimal places within each column; derived quantities to an appropriate number of significant figures.)

Final answer

Single results table with 6 sets of z, M, P, 1/sqrt(M), 1/P (student-dependent values)

Detailed explanation

Background Concept

A practical investigation needs:

a suitable independent variable (here $z$ , hence $M$ ),
a measured dependent variable (here $P$ ),
enough data points (here six sets) over a sensible range,
derived quantities calculated consistently,
data presented in a clear table with headings and units.

When you later plot a graph, the quality of the straight line depends strongly on the range and spread of the independent variable and on reducing random uncertainty (repeat readings / longer timing intervals).

Understanding the Question

You must change $z$ and measure $P$ each time until you have six pairs of values. You must also include in your table:

$M = 200\ \text{g} + z$ ,
$1/\sqrt{M}$ ,
$1/P$ .

So each row of the table corresponds to one chosen $z$ (and hence one $M$ ) and the resulting measured $P$ .

Approach

Choose at least six values of $z$ spanning a reasonable range (for example, start at $40\ \text{g}$ and increase in equal steps).
For each $z$ $z$ :
- calculate $M$ ,
- start both oscillations together and measure $P$ ,
- repeat the timing (or time several beat periods) and use a mean $P$ .
Compute $1/\sqrt{M}$ and $1/P$ for each row.
Present everything in a single table with correct headings, units, and consistent precision.

Step-by-Step Reasoning

Choosing $z$ values: Six distinct values are required; using equal steps helps produce evenly spaced $x$ -values on the graph.
Timing $P$ well: Reaction time can be a large fraction of a single timing. A better method is:
- time $N$ beat periods (e.g. from in-phase to in-phase $N$ times),
- record total time $t$ ,
- then $P = t/N$ .
  This reduces the percentage uncertainty in $P$ .
Calculating derived columns:
- Use $M = 200 + z$ (in $\text{g}$ as defined).
- Then compute $1/\sqrt{M}$ . Units follow algebraically: if $M$ is in $\text{g}$ then $1/\sqrt{M}$ has unit $\text{g}^{-1/2}$ .
- Compute $1/P$ with unit $\text{s}^{-1}$ .
Tabulation conventions:
- Put quantity and unit in the column heading (e.g. $P/\text{s}$ ).
- Keep raw readings to the instrument precision.
- Keep consistent decimal places within each column (especially $P$ and $M$ ).

Key Takeaways

Collect multiple data points with a good range.
Reduce random error by repeating readings or timing longer intervals.
Present raw and derived quantities clearly in one table with correct units.

Common Mistakes

Fewer than six sets of values.
Missing derived columns, or missing units in headings.
Inconsistent precision (e.g. some $P$ values to 0.1 s and others to 0.01 s without reason).
Calculating $1/\sqrt{M}$ incorrectly (e.g. using $1/M$ ).

Things to Be Careful About

Use the same mass unit consistently throughout the table (here the question uses $\text{g}$ ).
Do not round intermediate results too aggressively; calculate, then round the final table entry.
Ensure the “in phase” event you time is the same each run (same visual cue).

Techniques used

vary the independent variable in suitable steps to obtain a range of valuesrepeat timing measurements and average to reduce random errorcalculate derived quantities from raw measurementsrecord data in a single 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

Worked solution

Answer

Plot a graph with

$y$ -axis: $\dfrac{1}{P}\ /\ \text{s}^{-1}$
$x$ -axis: $\dfrac{1}{\sqrt{M}}\ /\ \text{g}^{-1/2}$

Use a suitable scale (at least half the graph paper in each direction) and plot all six points accurately.

Final answer

Graph of 1/P (s^-1) against 1/sqrt(M) (g^-1/2) plotted

Detailed explanation

Background Concept

A graph is used to test whether two variables have a linear relationship. If plotting the specified quantities produces points close to a straight line, the suggested equation is supported.

Graphing marks in Paper 3 typically depend on:

correct axes and labels (quantity and unit),
sensible scales,
accurate plotting.

Understanding the Question

You are instructed to plot $1/P$ (vertical axis) against $1/\sqrt{M}$ (horizontal axis). The values come from your table in part (b).

So each data pair is:

\left(x, y\right) = \left(\frac{1}{\sqrt{M}},\ \frac{1}{P}\right).

Approach

Label axes with the correct quantities and units.
Choose scales so the data fill most of the available grid.
Plot each of the six points as small, neat crosses (or dots in circles), using a sharp pencil.

Step-by-Step Reasoning

From the table, take each row’s calculated values of $1/\sqrt{M}$ and $1/P$ .
Decide axis ranges: pick minima slightly below the smallest value and maxima slightly above the largest value, avoiding awkward scales (e.g. 3 squares = 1 unit).
Plot each point by locating its $x$ -coordinate, then $y$ -coordinate.

graph-axes

Key Takeaways

Always plot exactly what the question specifies.
Correct labels and units are essential for marks.

Common Mistakes

Swapping axes (plotting $1/\sqrt{M}$ on $y$ ).
Missing units, or writing units incorrectly.
Using a cramped scale so the points occupy only a small corner of the grid.

Things to Be Careful About

Use the same units as in the table headings (if $M$ is in $\text{g}$ , then $1/\sqrt{M}$ is in $\text{g}^{-1/2}$ ).
Plot all points; do not omit a point because it looks “off”.

Techniques used

choose appropriate axes consistent with the variables specifiedselect a sensible scale that uses a large proportion of the gridplot experimental points accurately using calculated coordinates

(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})}

with units $\text{s}^{-1} / \text{g}^{-1/2} = \text{g}^{1/2}\ \text{s}^{-1}$ .

Read the $y$ -intercept at $x = 0$ from the best-fit line; unit $\text{s}^{-1}$ .

Answer

gradient $= \text{(from graph)}$ (e.g. $2.1\ \text{g}^{1/2}\ \text{s}^{-1}$ )

$y$ -intercept $= \text{(from graph)}$ (e.g. $0.030\ \text{s}^{-1}$ )

Final answer

Gradient and y-intercept: student-dependent (with units g^{1/2} s^{-1} and s^{-1})

Detailed explanation

Background Concept

For a straight-line graph, the gradient $m$ and intercept $c$ in

y = mx + c

are found from the best-fit line (not from individual data points).

Gradient:

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

$y$ -intercept: value of $y$ where the line crosses the $y$ -axis (i.e. at $x=0$ ).

Units come from the plotted variables.

Understanding the Question

Your graph has:

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

You must determine the gradient and the $y$ -intercept of the best-fit line you drew.

Approach

Choose two points on the best-fit line that are far apart (to reduce percentage reading uncertainty).
Read their coordinates accurately.
Compute $\Delta y$ and $\Delta x$ and then $m = \Delta y / \Delta x$ .
Find the intercept by reading where the line crosses the $y$ -axis (extend the line back to $x=0$ if necessary).

Step-by-Step Reasoning

Choosing points: Pick points on the drawn line, ideally where it crosses grid intersections, not necessarily your measured data points.
Gradient calculation: $m = \frac{(1/P)_2 - (1/P)_1}{(1/\sqrt{M})_2 - (1/\sqrt{M})_1}.$
Units: $[m] = \frac{\text{s}^{-1}}{\text{g}^{-1/2}} = \text{g}^{1/2}\ \text{s}^{-1}.$
Intercept: At $x=0$ , $y=c$ . Read the crossing point and quote it with unit $\text{s}^{-1}$ .

Key Takeaways

Use the best-fit line for gradient/intercept, and use a large triangle.
Always attach units derived from the axis quantities.

Common Mistakes

Using a data point that is not on the best-fit line.
Using a small triangle, giving a large percentage uncertainty.
Calculating gradient as $\Delta x / \Delta y$ (inverted).
Forgetting units or giving incorrect units.

Things to Be Careful About

Read values to about half a small square if possible.
Make sure you use the same scale units as your axes.
If the line does not cross the axis within the plotted region, extend it neatly with a ruler before reading the intercept.

Techniques used

construct a large gradient triangle on the best-fit linecalculate gradient using \Delta y/\Delta x with unitsread y-intercept by extrapolating the best-fit line to 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 $y = mx + c$ using $y = 1/P$ and $x = 1/\sqrt{M}$ .

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

Units:

[a] = \frac{\text{s}^{-1}}{\text{g}^{-1/2}} = \text{g}^{1/2}\ \text{s}^{-1},\qquad [b] = \text{s}^{-1}.

Answer

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

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

Final answer

a = gradient (g^{1/2} s^{-1}), b = y-intercept (s^{-1})

Detailed explanation

Background Concept

If a relationship can be written in the form

y = mx + c,

then plotting $y$ against $x$ gives a straight line with:

gradient $m$ ,
$y$ -intercept $c$ .

The constants in the original physical equation can be identified by matching the equation to this straight-line form.

Understanding the Question

You are given

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

and in part (c) you plotted $1/P$ (as $y$ ) against $1/\sqrt{M}$ (as $x$ ). You have already found the gradient and intercept from your graph. Now you must use those to state $a$ and $b$ , including units.

Approach

Rewrite the given equation to show it is already linear in $1/\sqrt{M}$ .
Identify which term multiplies $x$ (that is the gradient) and which term is added (that is the intercept).
Determine units from the axes units.

Step-by-Step Reasoning

Match term-by-term:

y = \frac{1}{P}

x = \frac{1}{\sqrt{M}}

Then the equation becomes

y = a x + b.

So directly:

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

For units:

$y = 1/P$ has unit $\text{s}^{-1}$ .
$x = 1/\sqrt{M}$ has unit $\text{g}^{-1/2}$ (since $M$ was in $\text{g}$ in your table).
Therefore

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

and

[b] = \text{s}^{-1}.

Key Takeaways

When you choose $x$ and $y$ well, the graph gives constants immediately.
Units of constants should be consistent with the units used on the graph axes.

Common Mistakes

Swapping $a$ and $b$ .
Giving $a$ the wrong units (often forgetting the $\text{g}^{1/2}$ factor).
Using $M$ in $\text{kg}$ in calculations but still writing units based on $\text{g}$ (or vice versa) without consistency.

Things to Be Careful About

Your units for $a$ depend on how you expressed $M$ when calculating $1/\sqrt{M}$ . If you used $M$ in $\text{kg}$ , then $[a]$ would be $\text{kg}^{1/2}\ \text{s}^{-1}$ instead.
Quote $a$ and $b$ to a sensible number of significant figures consistent with the precision of your graph reading.

Techniques used

match a linear equation to y = mx + cidentify constants from gradient and interceptdeduce units of constants from plotted variables

Q2MediumManipulation, 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

Answer

Measure the separation of the two lower nails with a ruler (to the nearest $1\ \text{mm}$ ).

Example recorded value:

d = 37.0\ \text{cm}

Final answer

d = 37.0 cm (example)

Detailed explanation

Background Concept

In Paper 3, many marks are for how you measure and how you record. A length $d$ should be measured between two clearly defined reference points (here, the centres of the two lower nails). A typical metre rule has $1\ \text{mm}$ smallest divisions, so readings are recorded to $\pm 1\ \text{mm}$ (i.e. to $0.1\ \text{cm}$ ).

Understanding the Question

You are told the distance between the two lower nails is $d$ . You must measure this distance on the apparatus shown and record it in $\text{cm}$ .

Approach

Decide exactly which two points define $d$ (the centres of the two lower nails).
Place a ruler alongside the line joining them.
Read at eye level to avoid parallax.
Record in $\text{cm}$ to $0.1\ \text{cm}$ .

Step-by-Step Reasoning

Align the ruler so that its scale is parallel to the distance between the nails.
If the ruler cannot start at zero exactly at the first nail, take two readings (at each nail) and subtract.
Record the result with one decimal place in $\text{cm}$ , e.g. $d = 37.0\ \text{cm}$ .

Key Takeaways

Use the correct reference points (centres of nails, not edges of bosses/clamps).
Record to a precision matching the instrument (typically $0.1\ \text{cm}$ ).

Common Mistakes

Measuring between the outside edges of the nails rather than the centres.
Reading the scale at an angle (parallax error).
Recording too many/few decimal places (e.g. $37\ \text{cm}$ or $37.00\ \text{cm}$ without justification).

Things to Be Careful About

Ensure the ruler is not tilted relative to the measured line.
If you subtract two readings, the uncertainty is larger than for a single reading (relevant later when discussing uncertainties).

Techniques used

identify the two reference points to be measured from the apparatus diagramuse a ruler to take a length reading with appropriate precisionrecord a measurement with unit and consistent decimal places

(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

Answer

Pull horizontally with the newton meter until $F = 5.0\ \text{N}$ and measure the horizontal deflection.

Example recorded value (to $0.1\ \text{cm}$ ):

x = 2.5\ \text{cm}

Final answer

x = 2.5 cm (example)

Detailed explanation

Background Concept

A displacement like $x$ is the change in position from an initial reference line to a new position. In this set-up, you are pulling the string sideways at the midpoint between the two lower nails, so $x$ is the horizontal distance between the original (undeflected) line of the string and the new position of the string at the point of pull.

Understanding the Question

You must:

apply a horizontal force $F = 5.0\ \text{N}$ using the newton meter, and
measure and record the resulting deflection $x$ in $\text{cm}$ .

Approach

Hook the newton meter at the midpoint between the nails.
Pull until the reading is exactly $5.0\ \text{N}$ .
Keep the pull horizontal and wait for the string to be steady.
Measure $x$ relative to the original string line.

Step-by-Step Reasoning

Set up the string and masses so the string is initially straight between the two lower nails.
Attach the newton meter halfway between them.
Pull sideways so the meter reads $5.0\ \text{N}$ (not $5\ \text{N}$ ; one decimal place matters).
Measure the horizontal separation $x$ with a ruler (typically to $0.1\ \text{cm}$ ).
Record a value such as $x = 2.5\ \text{cm}$ (your value depends on your apparatus).

Key Takeaways

$x$ is a horizontal displacement, so the ruler should be aligned horizontally.
Allow the system to settle before reading.

Common Mistakes

Measuring along the string instead of horizontally.
Not pulling exactly at the midpoint between nails.
Reading the newton meter while it is changing (oscillations).
Recording $x$ with no unit or unsuitable precision.

Things to Be Careful About

Ensure the newton meter is level (use a set square or visual alignment).
The string has thickness; measure consistently from the same side/centre each time.
The stand or nails may move slightly if not clamped well, affecting $x$ .

Techniques used

apply a specified horizontal force using a newton meteridentify the reference line for a displacement measurementmeasure a small horizontal displacement with a ruler and record it with unit

(ii)

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

percentage uncertainty = ______ $\%$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Ruler resolution $= 1\ \text{mm} = 0.1\ \text{cm}$ , so each reading $\pm 0.1\ \text{cm}$ .

If $x$ is found from two readings,

\Delta x \approx 0.1 + 0.1 = 0.2\ \text{cm}

\%\ \text{uncertainty in } x = \frac{\Delta x}{x} \times 100 = \frac{0.2}{2.5} \times 100

\%\ \text{uncertainty} = 8.0\%

Answer

8.0\%

Final answer

8.0%

Detailed explanation

Background Concept

An absolute uncertainty is the estimated uncertainty in the measurement itself (e.g. $\pm 0.1\ \text{cm}$ ). A percentage uncertainty compares this to the size of the measurement:

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

If a length is obtained by subtracting two ruler readings, the absolute uncertainties add.

Understanding the Question

You must estimate the percentage uncertainty in your measured $x$ and show working. The mark is mainly for a correct method.

Approach

Identify the resolution of the instrument used to measure $x$ .
Decide whether $x$ was measured directly or by subtraction of two readings.
Convert to an absolute uncertainty $\Delta x$ .
Calculate $\Delta x / x \times 100$ .

Step-by-Step Reasoning

If using a ruler with $1\ \text{mm}$ divisions, take each reading uncertainty as $\pm 1\ \text{mm} = \pm 0.1\ \text{cm}$ .
If $x$ is the difference between two positions (original and deflected), then:

\Delta x \approx 0.1 + 0.1 = 0.2\ \text{cm}

Then for a value such as $x = 2.5\ \text{cm}$ :

\%\ \text{uncertainty} = \frac{0.2}{2.5} \times 100 = 8.0\%

(Your numerical percentage will depend on your own $x$ .)

Key Takeaways

Use the instrument resolution to estimate uncertainty.
If $x$ comes from two scale readings, add the uncertainties.

Common Mistakes

Using $\pm 0.05\ \text{cm}$ for a mm ruler (too optimistic unless explicitly justified).
Forgetting to multiply by $100$ .
Using the uncertainty of one reading when $x$ is obtained from two readings.

Things to Be Careful About

If you measured $x$ directly from a fixed zero reference (rather than subtracting), then $\Delta x$ could be $\pm 0.1\ \text{cm}$ instead of $\pm 0.2\ \text{cm}$ .
Quote the percentage to a sensible precision (usually 1–2 s.f.).

Techniques used

estimate absolute uncertainty from instrument resolutionpropagate uncertainty for a difference of two ruler readingscalculate percentage uncertainty using a ratio

(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 $x = 2.5\ \text{cm}$ and $d = 37.0\ \text{cm}$ ,

y = \sqrt{\left(2.5^2 + \frac{37.0^2}{4}\right)} = \sqrt{\left(6.25 + 342.25\right)}

y = \sqrt{348.5} = 18.67\ \text{cm}

Answer

y = 18.7\ \text{cm}

Final answer

18.7 cm

Detailed explanation

Background Concept

When a string is pulled sideways at its midpoint between two fixed points, each half of the string forms a straight segment. If the two fixed points are separated by distance $d$ , then from the midpoint to either nail the horizontal displacement is $x$ and the vertical separation is $d/2$ . This gives a right-angled triangle, so the sloping length $y$ (midpoint to nail) is found using Pythagoras:

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

diagram-ciii-1

Understanding the Question

You are given the formula for $y$ in terms of your measured $x$ and $d$ . You must calculate $y$ and record it in $\text{cm}$ .

Approach

Use the provided equation directly.
Keep $x$ and $d$ in the same units (here, both in $\text{cm}$ ).
Calculate $x^2$ , calculate $d^2/4$ , add, then take the square root.

Step-by-Step Reasoning

With example values $x = 2.5\ \text{cm}$ and $d = 37.0\ \text{cm}$ :

Square $x$ :

x^2 = (2.5)^2 = 6.25\ \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 and square root:

y = \sqrt{6.25 + 342.25} = \sqrt{348.5} = 18.67\ \text{cm}

Round sensibly (often to 3 s.f. or to match input precision):

y = 18.7\ \text{cm}

Key Takeaways

The expression is just Pythagoras on a right triangle with legs $x$ and $d/2$ .
Use consistent units throughout.

Common Mistakes

Using $d^2/2$ instead of $d^2/4$ (forgetting the $d/2$ in Pythagoras).
Mixing units (e.g. $x$ in $\text{cm}$ and $d$ in $\text{m}$ ).
Rounding too early, giving noticeable numerical error.

Things to Be Careful About

$y$ comes out in $\text{cm}$ because you used $x$ and $d$ in $\text{cm}$ .
Keep enough significant figures in intermediate steps before final rounding.

Techniques used

substitute measured values into a given formulaevaluate a square root expression carefullypresent a calculated quantity with correct unit and 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

Worked solution

Working

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

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

Example measured value:

x = 1.5\ \text{cm}

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

y = \sqrt{\left(1.5^2 + \frac{37.0^2}{4}\right)} = \sqrt{\left(2.25 + 342.25\right)}

y = \sqrt{344.5} = 18.56\ \text{cm}

Answer

T = 6.87\ \text{N}

x = 1.5\ \text{cm}

y = 18.6\ \text{cm}

Final answer

T = 6.87 N, x = 1.5 cm (example), y = 18.6 cm

Detailed explanation

Background Concept

Increasing the hanging mass increases the tension in the string. The same relationships apply:

Tension from the hanging mass (equilibrium):

T = Mg

Geometry of the deflected string (right triangle):

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

As tension increases, for the same horizontal pull $F$ , the deflection $x$ is expected to decrease.

Understanding the Question

You must change $M$ to $0.700\ \text{kg}$ and repeat:

(b): calculate $T$
(c)(i): measure $x$ for $F = 5.0\ \text{N}$
(c)(iii): calculate $y$ using your $x$ and the same $d$ .

Approach

Compute $T$ from the new $M$ .
Repeat the pulling method exactly and measure $x$ with the same technique.
Substitute into the given expression for $y$ .

Step-by-Step Reasoning

Calculate the new tension:

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

Apply $F = 5.0\ \text{N}$ horizontally at the midpoint and measure $x$ (example: $1.5\ \text{cm}$ ).
With the same measured $d$ (example: $37.0\ \text{cm}$ ), calculate:

y = \sqrt{1.5^2 + \frac{37.0^2}{4}} = \sqrt{2.25 + 342.25} = \sqrt{344.5} = 18.56\ \text{cm}

so $y \approx 18.6\ \text{cm}$ .

Key Takeaways

Repeating measurements under changed conditions is essential for testing a relationship.
Keep the method consistent so differences in $x$ are due to tension changes, not technique changes.

Common Mistakes

Forgetting to include the mass hanger in the total mass $M$ .
Not keeping $F$ at $5.0\ \text{N}$ when re-measuring $x$ .
Using a different value of $d$ (it should not change unless you physically move nails).

Things to Be Careful About

Let oscillations die out before reading $x$ .
Record $x$ and $d$ to the same stated precision each time (e.g. to $0.1\ \text{cm}$ ).

Techniques used

calculate tension using T = Mg for a new value of massrepeat a measurement consistently for a changed conditionsubstitute repeated measurements into a given formula to compute a derived quantity

(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

Worked solution

Working

From

ky = Tx

k = \frac{Tx}{y}

Using $T = 3.92\ \text{N}$ , $x = 2.5\ \text{cm}$ , $y = 18.7\ \text{cm}$ :

k_1 = \frac{3.92 \times 2.5}{18.7} = 0.524\ \text{N}

Using $T = 6.87\ \text{N}$ , $x = 1.5\ \text{cm}$ , $y = 18.6\ \text{cm}$ :

k_2 = \frac{6.87 \times 1.5}{18.6} = 0.554\ \text{N}

Answer

First value of $k$ :

0.52

Second value of $k$ :

0.55

Final answer

k1 = 0.52, k2 = 0.55

Detailed explanation

Background Concept

If a relationship contains an unknown constant $k$ , you find $k$ by rearranging to make $k$ the subject.

Given

ky = Tx

we can solve for $k$ :

k = \frac{Tx}{y}

If the suggested relationship is correct and the experiment is good, the two values of $k$ found from two different runs should agree within experimental uncertainty.

Understanding the Question

You have two sets of data (one for $M = 0.400\ \text{kg}$ and one for $M = 0.700\ \text{kg}$ ). Using each set, calculate a value of $k$ .

Approach

Rearrange $ky = Tx$ to $k = Tx/y$ .
Substitute each set of $T$ , $x$ , $y$ values.
Quote each $k$ to a sensible number of significant figures (addressed in part (ii)).

Step-by-Step Reasoning

Rearrangement:

ky = Tx \Rightarrow k = \frac{Tx}{y}

First run (example numbers):

k_1 = \frac{(3.92)(2.5)}{18.7} = \frac{9.80}{18.7} = 0.524

Second run:

k_2 = \frac{(6.87)(1.5)}{18.6} = \frac{10.305}{18.6} = 0.554

These are reasonably close, suggesting $k$ is approximately constant.

Key Takeaways

Always rearrange so the constant is the subject before substituting.
Use consistent units (here $x$ and $y$ are both in $\text{cm}$ so they cancel).

Common Mistakes

Rearranging incorrectly (e.g. $k = Ty/x$ ).
Mixing units (e.g. converting only one of $x$ or $y$ to metres).
Writing $k$ with inappropriate precision.

Things to Be Careful About

The unit of $k$ depends on the units used for $x$ and $y$ . If both are in the same length units, $x/y$ is dimensionless and $k$ has unit of newtons (often omitted in final box if the question format expects just a numerical constant).

Techniques used

rearrange an equation to make the constant the subjectsubstitute experimental values into the rearranged expressioncompare calculated constants from two different trials

(ii)

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

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

$k$ is calculated from $k = Tx/y$ . The least precise values used are $x$ (and hence $y$ ) measured with a ruler (typically $2\ \text{s.f.}$ such as $2.5\ \text{cm}$ or $1.5\ \text{cm}$ ), so $k$ should be given to $2\ \text{s.f.}$ (e.g. $0.52$ and $0.55$ ).

Final answer

k given to 2 s.f. to match least precise measurements (e.g. x).

Detailed explanation

Background Concept

Significant figures communicate the precision of a value. For calculated quantities, you should not quote more significant figures than justified by the least precise measurement used in the calculation.

In many practicals, ruler measurements (like $x$ and $d$ ) limit the precision, and there can also be substantial experimental scatter/uncertainty (here later suggested as about $20\%$ for $k$ ).

Understanding the Question

You must explain why you chose the number of significant figures for your calculated $k$ values.

Approach

Identify the input quantity with the lowest precision/significant figures.
State that the calculated result should be rounded to that precision.
Optionally mention that large experimental uncertainty also makes extra digits meaningless.

Step-by-Step Reasoning

$k$ is calculated using:

k = \frac{Tx}{y}

$T$ comes from $Mg$ and can be quoted to 3 s.f. ( $6.87\ \text{N}$ etc.).
The limiting quantities are typically the ruler measurements, especially $x$ (e.g. $2.5\ \text{cm}$ or $1.5\ \text{cm}$ , usually 2 s.f.) and $y$ derived from them.
Therefore $k$ should be quoted to 2 significant figures, e.g. $0.52$ and $0.55$ . Giving $0.524$ would imply unjustified precision.

Key Takeaways

Final precision is set by the least precise measurement.
Extra digits do not add accuracy; they can lose marks in practical papers.

Common Mistakes

Quoting $k$ to 3 or 4 s.f. just because a calculator shows them.
Rounding each intermediate step too early instead of rounding only the final answer.

Things to Be Careful About

If your measured $x$ happened to be recorded to 3 s.f. (e.g. $2.50\ \text{cm}$ ) with justification, then 3 s.f. for $k$ might be acceptable; otherwise 2 s.f. is safer, especially with large overall uncertainty.

Techniques used

identify the least precise measured quantity used in a calculationmatch the significant figures of a derived value to the raw data precisionavoid over-precise reporting given experimental uncertainty

(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

Working

Using $k_1 = 0.52$ and $k_2 = 0.55$ ,

\%\ \text{difference} = \frac{|0.55 - 0.52|}{(0.55 + 0.52)/2} \times 100 = 5.6\%

Since $5.6\% < 20\%$ , the values agree within the stated uncertainty.

Answer

Yes. The two values of $k$ are consistent within $20\%$ , so the results support $ky = Tx$ .

Final answer

Yes, consistent within 20%.

Detailed explanation

Background Concept

To test whether results support a relationship that predicts a constant $k$ , you check whether repeated values of $k$ agree within the experimental uncertainty.

If the percentage uncertainty in $k$ is $20\%$ , then two measured values can be considered consistent if their difference is not larger than about $20\%$ of their size.

A common way to compare two values is the percentage difference:

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

Understanding the Question

You are told to assume a $20\%$ uncertainty in $k$ and to decide whether your two calculated $k$ values support the suggested relationship $ky = Tx$ .

Approach

Calculate how far apart the two $k$ values are in percentage terms.
Compare this percentage difference to $20\%$ .
Conclude whether they agree within uncertainty.

Step-by-Step Reasoning

Using example values $k_1 = 0.52$ and $k_2 = 0.55$ :

Mean value:

\bar{k} = \frac{0.52 + 0.55}{2} = 0.535

Difference:

|\Delta k| = |0.55 - 0.52| = 0.03

Percentage difference:

\frac{0.03}{0.535} \times 100 \approx 5.6\%

Since $5.6\%$ is less than $20\%$ , the results are consistent and therefore support the suggested relationship.

Key Takeaways

You must use the uncertainty to justify your conclusion, not just say “they look close”.
Agreement within uncertainty supports (but does not prove) the model.

Common Mistakes

Comparing absolute difference instead of percentage difference.
Forgetting to use the given $20\%$ uncertainty in the argument.
Saying “it proves the relationship” (experiments support or do not support; they do not prove).

Things to Be Careful About

Use your own $k$ values when doing this in the exam.
Quote a clear, logical conclusion linked to the uncertainty.

Techniques used

compare two experimental values using percentage differencejudge agreement using a stated percentage uncertaintydraw a conclusion about whether data supports a proposed relationship

(g)

(i)

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

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

4M Similar

DifficultyMedium

Worked solution

Answer

Uncertainty in $x$ : difficult to judge the exact original and deflected position of the string (string has thickness and may not be steady), giving parallax when reading the ruler.
Uncertainty in $d$ : difficult to measure between the centres of two nails accurately; ruler may not align exactly with the nail centres.
Force $F$ not exactly horizontal / applied not exactly at the midpoint between the two lower nails, so the geometry assumed for calculating $y$ is not exact.
Friction at the nails (string rubbing on nails) means tension may not be the same on both sides / tension not equal to $Mg$ , affecting the assumed value of $T$ .

Final answer

See working (four limitations listed).

Detailed explanation

Background Concept

In evaluating an experiment, you gain credit for identifying specific uncertainties/limitations and explaining why they occur. Good answers:

name the quantity affected (e.g. $x$ , $d$ , $F$ , $T$ ), and
give a reason (parallax, movement, friction, alignment, equilibrium not reached, etc.).

Understanding the Question

You must describe four sources of uncertainty or limitations. If you mention measurement uncertainty, you must state what is being measured and why it is uncertain.

Approach

Pick four different issues from different parts of the procedure:

measuring $x$
measuring $d$
controlling/applying $F$
assumptions about tension $T$ (e.g. friction, motion)
For each, write one clear sentence naming the quantity and the reason.

Step-by-Step Reasoning

Possible creditworthy limitations include:

$x$ measurement: The string can oscillate after pulling; the position is not perfectly steady. The string has thickness, so “the position of the string” is ambiguous. Reading a ruler at an angle causes parallax.
$d$ measurement: The nails are round; you want centre-to-centre distance, which is hard to judge. The ruler may not sit exactly alongside the line joining nail centres.
Applying $F$ : Keeping the newton meter exactly horizontal is difficult by eye. Also, pulling exactly at the midpoint is difficult, changing the geometry and making the $y$ formula less exact.
Tension not equal to $Mg$ everywhere: The string contacts nails, so friction can make the tension differ along the string. Also, if the mass is moving/oscillating, instantaneous tension differs from $Mg$ .

Any four such distinct, clearly explained points earn the marks.

Key Takeaways

Always name the quantity and the reason.
Limitations can be measurement issues or validity issues (assumptions not met).

Common Mistakes

Vague statements like “human error” or “reaction time” (reaction time is not relevant here unless timing is involved).
Listing the same idea four times (e.g. parallax for everything) without variety.
Not stating which quantity the uncertainty refers to.

Things to Be Careful About

Do not claim “ $g$ is uncertain” (it is given).
Ensure each point is distinct (e.g. $x$ parallax is different from $F$ not horizontal).

Techniques used

identify specific measured quantities affected by uncertaintylink each limitation to a physical reason for measurement difficultydescribe practical issues that affect repeatability and validity

(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

Use low-friction pulleys at the nail positions (or smooth rollers) to reduce friction so tension is closer to uniform and equal to $Mg$ .
Use a set square / horizontal guide to ensure the newton meter pulls exactly horizontally.
Mark the midpoint of the string and attach the newton meter at this mark each time to ensure the force is applied at the same point.
Repeat measurements of $x$ for the same $F$ and average, taking readings only when the string is steady (reduces random scatter).

Final answer

See working (four improvements listed).

Detailed explanation

Background Concept

Improvements should either reduce uncertainty (better measurement) or improve validity (better control of variables / meeting assumptions). The best responses explicitly address the limitations you identified.

Understanding the Question

You need four improvements. You may suggest different apparatus or procedures.

Approach

For each improvement, think: “What limitation does this fix?” and state the improvement clearly and practically.

Step-by-Step Reasoning

Examples of strong improvements:

Reduce friction / make tension uniform: Replace nails with small pulleys or rollers so the string slides with minimal friction. This makes the tension more uniform and closer to $Mg$ throughout.
Make pull truly horizontal: Use a set square against the string or a fixed horizontal track/guide for the newton meter so the applied force is horizontal.
Ensure midpoint pulling: Measure and mark the midpoint between the lower nails (or mark the midpoint of the string segment) so the hook position is consistent each time.
Improve repeatability: Take several readings of $x$ for the same conditions and average; wait for oscillations to stop before reading.

Other valid improvements could include: using a pointer attached to the string and a fixed scale to reduce parallax, using a clamp to hold the newton meter position once $5.0\ \text{N}$ is reached, or ensuring the stand is firmly clamped so nails do not move.

Key Takeaways

Improvements must be specific and feasible.
Best answers directly target named uncertainties/limitations.

Common Mistakes

Stating “use better equipment” without naming what and how it helps.
Repeating the same improvement in different words.
Suggesting irrelevant changes (e.g. data logger for timing when no timing is done).

Things to Be Careful About

Improvements should not change the intended physics model (still measuring the same $x$ for the same $F$ and $T$ ).
If you propose extra measurements (more values of $M$ ), that improves the strength of the conclusion, but you still need measurement-quality improvements too.

Techniques used

propose apparatus changes that reduce a named uncertaintymodify procedure to improve repeatabilitylink improvements directly to stated limitations