Physics 9702/34 — October/November 2024

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

questions

marks

120

minutes

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

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

In this experiment, you will investigate the flow of water through a nozzle.

(a)

(i)

● Remove the plunger from the syringe body.
● Assemble the apparatus as shown in Fig. 1.1 with the bottom of the syringe nozzle approximately $15\text{ cm}$ above the bench.

Fig. 1.1

● Measure and record the height $h_t$ of the $30\text{ cm}^3$ graduation above the bench, as shown in Fig. 1.1.

$h_t$ = ______

● Measure and record the height $h_b$ of the $25\text{ cm}^3$ graduation above the bench, as shown in Fig. 1.1.

$h_b$ = ______

● Calculate the mean $h_m$ of the two values $h_t$ and $h_b$ .

$h_m$ = ______

1M Similar

DifficultyEasy

Worked solution

Working

Measure heights from the bench (example readings to nearest $0.1\ \text{cm}$ ):

$h_t = 32.6\ \text{cm}$

$h_b = 31.0\ \text{cm}$

h_m = \frac{h_t + h_b}{2} = \frac{32.6 + 31.0}{2} = 31.8\ \text{cm}

Answer

$h_t = 32.6\ \text{cm}$

$h_b = 31.0\ \text{cm}$

$h_m = 31.8\ \text{cm}$

Final answer

Example: h_t = 32.6 cm, h_b = 31.0 cm, h_m = 31.8 cm

Detailed explanation

Background Concept

A height measurement in a practical is a measurement of a vertical distance from a clearly defined reference level (here, the bench). To gain credit you must:

measure the correct points (bench to the graduation mark),
use an appropriate instrument (a ruler/metre rule),
record to a sensible precision (typically $\pm 1\ \text{mm}$ , i.e. $0.1\ \text{cm}$ ),
include units.

The mean height $h_m$ is used because the water level changes between the two graduations; averaging gives a representative head during the timing interval.

Understanding the Question

You set up the syringe as shown, then you must record:

$h_t$ : height of the upper graduation (30 cm $^3$ ) above the bench,
$h_b$ : height of the lower graduation (25 cm $^3$ ) above the bench,
$h_m$ : the mean of these two heights.

These values are specific to your apparatus position, so your numerical answers are student-dependent; what matters is correct method, precision and units.

Approach

Place the ruler vertically with its zero at bench level (or measure from bench up to each mark consistently).
Read $h_t$ and $h_b$ at eye level to reduce parallax.
Calculate the mean using $h_m = (h_t + h_b)/2$ .

Step-by-Step Reasoning

Identify the 30 cm $^3$ and 25 cm $^3$ graduation lines on the syringe.
For each graduation, measure the vertical distance from the bench to that graduation.
Record each measurement with a consistent precision (e.g. both to $0.1\ \text{cm}$ ).
Average them:

h_m = \frac{h_t + h_b}{2}

and round $h_m$ consistently with the input measurements.

Key Takeaways

Always quote units and consistent precision for ruler measurements.
The mean is used to represent a quantity that changes during the measurement interval.

Common Mistakes

Measuring from the bench to the nozzle rather than to the graduation line.
Not measuring vertically (measuring along the syringe or at an angle).
Parallax error: reading the ruler from above/below eye level.
Omitting units or recording inconsistent decimal places.
Calculating $h_m$ incorrectly (e.g. subtracting instead of averaging).

Things to Be Careful About

Keep the reference level consistent: both heights must be measured from the same bench level.
Record to the resolution of the instrument (typically $0.1\ \text{cm}$ for a ruler read to $1\ \text{mm}$ ).
When averaging, do not overstate precision: the mean should not have more decimal places than the original measurements.

Techniques used

measure a vertical height from a reference level using a rulerrecord readings to appropriate precision with unitscalculate a mean value from two readings

(ii)

● Place the empty beaker below the syringe nozzle.
● Pour water from the other beaker into the syringe body so that the water level is near the top, then watch the water level fall.
● Start the stop-watch as the water level passes the $30\text{ cm}^3$ graduation, then stop the stop-watch as the water level passes the $25\text{ cm}^3$ graduation.
● Record the stop-watch reading $T$ .

$T$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Start timing as the water level passes the $30\ \text{cm}^3$ mark and stop as it passes the $25\ \text{cm}^3$ mark.

Example reading:

$T = 12.4\ \text{s}$

Final answer

Example: T = 12.4 s

Detailed explanation

Background Concept

A stopwatch measures time intervals, but the uncertainty is often dominated by human reaction time. To reduce this, you must:

use clear start/stop events,
keep your eye level with the graduation marks,
repeat timings (in later parts) or time longer intervals where possible.

Here the event is the water meniscus crossing specific syringe graduations.

Understanding the Question

You fill the syringe, then as water flows out the level falls. You must measure the time $T$ taken for the level to fall from the 30 cm $^3$ mark to the 25 cm $^3$ mark (a fixed volume change of $5\ \text{cm}^3$ ).

Your value of $T$ depends on the nozzle and how high the water is, so it is student-dependent; the mark is for correct timing method and a sensible recorded value with units.

Approach

Watch the meniscus closely as it approaches the 30 cm $^3$ mark.
Start the stopwatch at the instant it passes that mark.
Stop the stopwatch at the instant it passes the 25 cm $^3$ mark.
Record $T$ to the stopwatch resolution (often $0.1\ \text{s}$ or $0.01\ \text{s}$ ).

Step-by-Step Reasoning

Ensure the beaker is in place to catch water so the flow is uninterrupted.
Fill near the top so the water level definitely passes the required graduations.
Use the same viewing position for start and stop to minimise parallax.
Record the stopwatch reading with unit, e.g. $T = 12.4\ \text{s}$ .

Key Takeaways

Timing between two well-defined level crossings gives a reproducible interval.
Good timing technique (clear criteria, correct precision) earns the marks.

Common Mistakes

Starting/stopping when the meniscus is near the mark rather than exactly passing it.
Using different criteria at start and stop (e.g. bottom of meniscus then top of meniscus).
Forgetting the unit or using inconsistent precision.

Things to Be Careful About

Always use the same part of the meniscus (usually the bottom) for both start and stop.
If the flow is very fast, reaction time becomes a larger fraction of $T$ ; later you reduce this by collecting multiple data sets and using a graph.
Make sure the syringe remains vertical; tilting changes the effective reading of the graduation crossing.

Techniques used

time an interval using a stopwatch between two defined eventsuse consistent start and stop criteria for timingrecord time to appropriate precision with units

(b)

Choose two different graduations that are $5\text{ cm}^3$ apart and measure $h_t$ , $h_b$ and $T$ .

Repeat until you have six sets of values of $h_t$ , $h_b$ and $T$ .

Record your results in a table. Include values of $h_m$ and $\frac{1}{T}$ in your table.

9M Similar

DifficultyMedium

Worked solution

Answer

Record six sets of $h_t$ , $h_b$ and $T$ for pairs of graduations $5\ \text{cm}^3$ apart, then calculate

h_m = \frac{h_t + h_b}{2}

and

\frac{1}{T}

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

run	$h_t / \text{cm}$	$h_b / \text{cm}$	$h_m / \text{cm}$	$T / \text{s}$	$(1/T) / \text{s}^{-1}$
1	35.3	33.7	34.5	11.4	0.0877
2	32.6	31.0	31.8	12.4	0.0806
3	29.9	28.3	29.1	13.5	0.0741
4	27.2	25.6	26.4	15.0	0.0667
5	24.4	22.8	23.6	17.2	0.0581
6	21.7	20.1	20.9	20.0	0.0500

Final answer

See working / student-dependent table (must include h_t, h_b, h_m, T and 1/T with units).

Detailed explanation

Background Concept

In Paper 3, a large fraction of marks often come from how you record and present data.
A good results table must:

contain all raw measurements and required calculated quantities,
have clear column headings with quantity and unit (e.g. $T / \text{s}$ ),
use consistent decimal places within each column (raw ruler readings often to $0.1\ \text{cm}$ ; times to $0.1\ \text{s}$ or $0.01\ \text{s}$ depending on the stopwatch),
include an appropriate range and number of readings (here, six sets).

Calculated quantities should be rounded sensibly based on the precision of the raw data.

Understanding the Question

You must choose six different pairs of graduations that are $5\ \text{cm}^3$ apart, and for each pair measure:

$h_t$ (upper graduation height),
$h_b$ (lower graduation height),
$T$ (time for water to fall between those two graduations).

Then you must compute and include in the table:

$h_m$ , the mean of $h_t$ and $h_b$ ,
$1/T$ .

Approach

Pick six different 5 cm $^3$ intervals (e.g. 35–30, 30–25, 25–20, ...).
For each interval, measure $h_t$ and $h_b$ from the bench.
Time the fall of the meniscus between the two marks to obtain $T$ .
Calculate $h_m$ and $1/T$ for each run.
Present all values in one table with correct headings and consistent precision.

Step-by-Step Reasoning

Choosing intervals: Use different parts of the syringe so that $h_m$ changes across a wide range; this helps the graph show a clear trend.
Recording raw data: Put $h_t$ , $h_b$ , and $T$ in separate columns with units.
Mean height: For each run,

h_m = \frac{h_t + h_b}{2}

If $h_t$ and $h_b$ are to $0.1\ \text{cm}$ , then $h_m$ should typically be to $0.1\ \text{cm}$ .

Reciprocal time: For each run,

\frac{1}{T}

Use consistent significant figures (often 3 s.f.) across that column.

Key Takeaways

A single clear table with correct headings/units and consistent precision is essential.
Include both raw and calculated quantities exactly as instructed.

Common Mistakes

Missing units in headings (e.g. writing only $T$ rather than $T / \text{s}$ ).
Mixing decimal places within one column (e.g. some $h$ values to $0.1\ \text{cm}$ and some to $0.01\ \text{cm}$ ).
Forgetting to include $h_m$ or $1/T$ .
Using fewer than six sets of readings.
Calculating $1/T$ but rounding inconsistently (e.g. sometimes 2 s.f., sometimes 4 s.f.).

Things to Be Careful About

The two chosen graduations for each run must be exactly $5\ \text{cm}^3$ apart.
Keep the same measurement technique for all $h$ readings (same reference, same ruler position).
Avoid very small $T$ values where reaction time dominates; choose intervals that give times long enough to measure reliably.

Techniques used

select six suitable intervals for the independent variable and repeat measurementscalculate derived quantities from raw measurementsconstruct a results table with correct headings and unitsuse consistent significant figures and decimal places within columns

(c)

(i)

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

3M Similar

DifficultyMedium

Worked solution

Answer

Plot $h_m$ on the $x$ -axis (unit: $\text{cm}$ ) and $1/T$ on the $y$ -axis (unit: $\text{s}^{-1}$ ).

Use a sensible scale occupying at least half the grid on each axis and plot all six points accurately.

Final answer

Graph of (1/T) against h_m (student-dependent).

Detailed explanation

Background Concept

A graph is used to reveal a relationship between variables and to allow constants to be found from a straight-line fit. For full credit you typically need:

correctly chosen axes (dependent on $y$ , independent on $x$ ),
axes labelled with quantity and unit,
scales that are easy to use and that occupy a large fraction of the grid,
accurate plotting.

Understanding the Question

You are told exactly what to plot: $1/T$ (from your table) against $h_m$ . This sets up a test of a linear relation of the form $1/T = ph_m + q$ .

Approach

Decide axis ranges based on your smallest and largest values of $h_m$ and $1/T$ .
Mark a scale with simple steps (e.g. 1, 2, 5 multiples).
Label axes as $h_m / \text{cm}$ and $(1/T) / \text{s}^{-1}$ .
Plot each point using a sharp pencil and small crosses.

Step-by-Step Reasoning

Find the range of $h_m$ values in your table; choose an $x$ -axis scale that fits them comfortably and uses most of the graph width.
Do the same for $1/T$ on the $y$ -axis.
Plot each pair $(h_m, 1/T)$ carefully (use a ruler to help align if needed).
Ensure every plotted point corresponds to one row of your table.

Key Takeaways

Correct axis choice and labelling are essential and often carry marks.
A good scale makes gradient determination more accurate.

Common Mistakes

Swapping axes (plotting $h_m$ on $y$ instead of $x$ ).
Missing units on axes.
Using awkward scales (e.g. 3 units per big square) or scales that use only a small part of the grid.
Plotting $T$ instead of $1/T$ .

Things to Be Careful About

$1/T$ has unit $\text{s}^{-1}$ .
Plotting accuracy: points should be within about half a small square of their correct position.
Don’t force the graph through the origin unless your plotted trend and theory require it (that is checked later with the best-fit line).

Techniques used

choose suitable axis scales to use most of the graph gridlabel axes with quantity and unitplot points accurately from a results 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

Use a large triangle on the best-fit line:

\text{gradient} = \frac{\Delta (1/T)}{\Delta h_m}

Example (from a suitable large triangle):

\text{gradient} \approx 2.8 \times 10^{-3}\ \text{s}^{-1}\ \text{cm}^{-1}

Read the intercept where the line crosses the $y$ -axis:

\text{$y$-intercept} \approx -8.0 \times 10^{-3}\ \text{s}^{-1}

Answer

gradient $= 2.8 \times 10^{-3}\ \text{s}^{-1}\ \text{cm}^{-1}$

$y$ -intercept $= -8.0 \times 10^{-3}\ \text{s}^{-1}$

Final answer

Example: gradient = 2.8×10^-3 s^-1 cm^-1, y-intercept = −8.0×10^-3 s^-1

Detailed explanation

Background Concept

For a straight-line graph, the gradient (slope) tells you how much $y$ changes per unit change in $x$ :

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

The $y$ -intercept is the value of $y$ when $x = 0$ (where the line crosses the $y$ -axis). Units come from the axis units:

here, $y = 1/T$ has unit $\text{s}^{-1}$ ,
$x = h_m$ has unit $\text{cm}$ ,
so gradient has unit $\text{s}^{-1}\ \text{cm}^{-1}$ .

Understanding the Question

You have drawn a best-fit line on a graph of $1/T$ (vertical) against $h_m$ (horizontal). You must now obtain two numbers from that line:

its gradient,
its $y$ -intercept.

These depend on your plotted line, so your exact values are student-dependent, but the method must be correct.

Approach

Pick two well-separated points on the best-fit line (not necessarily data points).
Read their coordinates accurately.
Compute the gradient using $\Delta y / \Delta x$ .
Read the $y$ -intercept where the best-fit line crosses the $y$ -axis.

Step-by-Step Reasoning

Large triangle: Using two far-apart points reduces percentage reading error.
Suppose the two chosen points on the line are $(h_{m1}, (1/T)_1)$ and $(h_{m2}, (1/T)_2)$ . Then:

\text{gradient} = \frac{(1/T)_2 - (1/T)_1}{h_{m2} - h_{m1}}

Units: Carry units through the calculation to get $\text{s}^{-1}\ \text{cm}^{-1}$ .
Intercept: Extend the best-fit line (if needed) to the $y$ -axis and read the crossing value; unit is $\text{s}^{-1}$ .

Key Takeaways

Always use the best-fit line (not point-to-point) for gradient.
Use a large triangle and quote gradient and intercept with correct units.

Common Mistakes

Calculating $\Delta x / \Delta y$ by accident.
Using two adjacent points giving a small triangle (large uncertainty).
Using raw data points rather than points on the best-fit line.
Missing units for gradient or intercept.

Things to Be Careful About

Read coordinates carefully and consistently with the axis scales.
If the intercept is negative, include the sign.
Do not over-round: typically 2–3 significant figures is appropriate for graph-derived quantities.

Techniques used

determine gradient using a large triangle on a best-fit lineread the y-intercept from the graphcalculate gradient as \Delta y / \Delta x with units

(d)

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

\frac{1}{T} = ph_m + q

where $p$ and $q$ are constants.

Using your answers in (c)(iii), determine the values of $p$ and $q$ .
Give appropriate units.

$p$ = ______
$q$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Given

\frac{1}{T} = ph_m + q

Comparing with $y = mx + c$ for a graph of $y = 1/T$ against $x = h_m$ :

p = \text{gradient}

q = \text{$y$-intercept}

Using (c)(iii) (example values):

p = 2.8 \times 10^{-3}\ \text{s}^{-1}\ \text{cm}^{-1}

q = -8.0 \times 10^{-3}\ \text{s}^{-1}

Answer

$p = 2.8 \times 10^{-3}\ \text{s}^{-1}\ \text{cm}^{-1}$

$q = -8.0 \times 10^{-3}\ \text{s}^{-1}$

Final answer

Example: p = 2.8×10^-3 s^-1 cm^-1, q = −8.0×10^-3 s^-1

Detailed explanation

Background Concept

A linear relationship has the form

y = mx + c

where:

$m$ is the gradient (slope),
$c$ is the $y$ -intercept.

If you plot the correct variables on the axes, you can identify physical constants directly from $m$ and $c$ .

Understanding the Question

You are told that

\frac{1}{T} = ph_m + q

and you have already plotted $1/T$ (as $y$ ) against $h_m$ (as $x$ ). You must use your gradient and intercept from (c)(iii) to state $p$ and $q$ , including units.

Approach

Match $1/T = ph_m + q$ to $y = mx + c$ .
Therefore, $p$ corresponds to the gradient and $q$ corresponds to the intercept.
Units come from the axes: $p$ has units of $y/x$ and $q$ has units of $y$ .

Step-by-Step Reasoning

From the graph:

$y = 1/T$ has unit $\text{s}^{-1}$ ,
$x = h_m$ has unit $\text{cm}$ .
So:

[p] = \frac{\text{s}^{-1}}{\text{cm}} = \text{s}^{-1}\ \text{cm}^{-1}

and

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

Numerically:

p = \text{gradient}

q = \text{$y$-intercept}

using your values obtained in (c)(iii).

Key Takeaways

Correct plotting makes constants easy to obtain: gradient and intercept map directly onto constants in the linear equation.
Units must be consistent with the axis quantities.

Common Mistakes

Swapping $p$ and $q$ .
Giving $p$ the same units as $q$ (forgetting that gradient includes division by $h_m$ ).
Using $h_t$ or $h_b$ instead of $h_m$ .

Things to Be Careful About

If your intercept is negative, include the minus sign.
If your graph uses $h_m$ in different units (e.g. metres), then $p$ units change accordingly; always base units on your actual axis labels.

Techniques used

identify constants in a linear equation from gradient and interceptassign units to constants from plotted axis units

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

In this experiment, you will investigate the conservation of momentum.

(a)

(i)

The apparatus has been partly set up as shown in Fig. 2.1.

Fig. 2.1

● Check that the rod can swing freely on the nail.
● The distance between the nail and the centre of the magnet is $r$ .
Measure and record $r$ .

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

● Record the mass $M$ of the magnet written on the card.

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

2M Similar

DifficultyEasy

Worked solution

Answer

(Example readings)

$\displaystyle r = 25.0\ \text{cm}$

$\displaystyle M = 50.0\ \text{g}$

Final answer

r = 25.0 cm, M = 50.0 g (example)

Detailed explanation

Background Concept

In Paper 3 practical work, marks for measurements are mainly awarded for:

using the correct instrument and reading it correctly,
recording to appropriate precision (matching the instrument resolution),
including units.

Here $r$ is a length (distance from the pivot point at the nail to the centre of the magnet). $M$ is the magnet’s mass as provided on the card.

Understanding the Question

You are told that the rod must swing freely about the nail, and you must:

measure and record the distance $r$ (nail to magnet centre), and
record the mass $M$ written on the card.

The question is checking that you can take and record basic measurements correctly.

Approach

Identify the two points that define $r$ (nail position and centre of magnet).
Use a ruler/metre rule to measure the straight-line distance $r$ .
Record $r$ to the nearest mm (i.e. $0.1\ \text{cm}$ ) if using a standard ruler.
Copy $M$ exactly from the card, including the unit.

Step-by-Step Reasoning

Ensure the rod can swing freely: this reduces systematic effects from friction or obstruction (even though this check is not a numerical mark, it underpins the quality of subsequent readings).
Measure $r$ :

Place the ruler alongside the rod.
Read the position of the nail and the position of the magnet’s centre.
Subtract to obtain $r$ and record it.

Record $M$ :

The value is provided; you simply copy it with unit $\text{g}$ .

(Representative example values were given in the answer; your actual values depend on your apparatus.)

Key Takeaways

Record lengths to the instrument’s resolution (often $0.1\ \text{cm}$ ).
Record given masses exactly as stated and include units.

Common Mistakes

Measuring to the edge of the magnet instead of its centre.
Omitting units ( $\text{cm}$ or $\text{g}$ ).
Giving $r$ with too many decimal places (implies unjustified precision).

Things to Be Careful About

Parallax when reading the ruler: keep your eye directly above the scale mark.
Defining the magnet centre consistently (use its midpoint, not a corner or edge).

Techniques used

measure a length using a ruler with appropriate precisionrecord a stated value with correct unitquote readings to a consistent decimal place

(ii)

● Use small pieces of adhesive putty to fix the $30\text{ cm}$ ruler to the bench with the zero of its scale directly below the centre of the magnet, as shown in Fig. 2.2.
● Attach nut A to the bottom of the magnet. Adjust the position of the boss on the stand until the bottom of the nut is approximately $3\text{ mm}$ above the ruler, as shown in Fig. 2.2.

Fig. 2.2

● Record the mass $m$ of nut A written on the card.

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

● Detach the nut from the magnet.
● Move the rod and hold it so that the bottom of the magnet is directly above the $15.0\text{ cm}$ mark on the ruler scale.
● Place the nut on the ruler so that its centre is above the zero on the ruler scale, as shown in Fig. 2.3.

Fig. 2.3

● Release the rod so that the magnet picks up the nut as it passes and then swings back to a position $x$ on the ruler scale, as shown in Fig. 2.4.

Fig. 2.4

● Read and record $x$ .

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

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

(Example readings)

$\displaystyle m = 5.0\ \text{g}$

$\displaystyle x = 11.0\ \text{cm}$

Final answer

m = 5.0 g, x = 11.0 cm (example)

Detailed explanation

Background Concept

In this experiment the ruler acts as a position scale to measure the horizontal displacement of the swinging magnet+nutt system. The key practical skills being assessed are:

placing the ruler so that $0$ is directly below the magnet’s centre (a defined reference),
releasing the pendulum consistently from a known starting position,
reading the final position $x$ (the turning point after pickup) from the scale.

Understanding the Question

You must:

Fix the ruler on the bench so $0$ is directly below the magnet’s centre.
Use nut A, record its mass $m$ from the card.
Start with the magnet above $15.0\ \text{cm}$ , nut centred above $0$ .
Release the rod, allow pickup, and record the turning point position $x$ .

Approach

Do the alignment first (this affects all $x$ values).
Use the card values for masses.
For $x$ , wait until the system reaches its maximum displacement on the return swing and read the scale at the centre line directly under the magnet/nut.

Step-by-Step Reasoning

Fix the ruler: ensure it cannot move during the run; otherwise you introduce a systematic error in $x$ .
Align zero: the instruction “zero directly below the centre of the magnet” is crucial; it defines the origin for $x$ .
Set the gap (~ $3\ \text{mm}$ ) as instructed so the nut does not scrape the ruler (reduces frictional energy loss and prevents the nut being pushed along).
Record $m$ from the card.
Place magnet above $15.0\ \text{cm}$ and nut at $0.0\ \text{cm}$ as shown.
Release without pushing (avoid adding extra kinetic energy).
Read $x$ at the turning point after pickup; record to the nearest $0.1\ \text{cm}$ if using a mm-scale ruler.

(Representative example values were provided; your measured $x$ depends on your apparatus and release.)

Key Takeaways

Correct alignment of the scale is part of the measurement.
A turning-point reading is often the most uncertain because the system may still be oscillating.

Common Mistakes

Setting the ruler’s zero somewhere other than directly under the magnet centre.
Reading $x$ while the magnet is still moving (not at the maximum displacement).
Measuring $x$ to the wrong reference point (edge of nut/magnet rather than the centre line directly above the scale).

Things to Be Careful About

Ensure the nut is centred above zero before release.
Keep your eye normal to the ruler to reduce parallax.
If the pickup sometimes fails, repeat until you get a clean pickup before recording $x$ .

Techniques used

set up the ruler with a defined zero reference pointcontrol initial conditions before taking a readingread a position from a scale and record with appropriate precision

(iii)

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

percentage uncertainty = ______ $\%$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Take uncertainty in $x$ as $\pm 0.1\ \text{cm}$ .

\%\ \text{uncertainty in }x = \frac{0.1}{11.0} \times 100 = 0.91\%

Answer

$\displaystyle 0.9\%$

Final answer

0.9 %

Detailed explanation

Background Concept

For a direct reading from a ruler with 1 mm divisions:

a typical reading uncertainty is about half the smallest division, i.e. $\pm 0.5\ \text{mm} = \pm 0.05\ \text{cm}$ .
In practice, for a moving turning point, it can be reasonable to take a larger uncertainty such as $\pm 1\ \text{mm} = \pm 0.1\ \text{cm}$ because it is difficult to judge the exact maximum position.

Percentage uncertainty is:

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

Understanding the Question

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

You need an absolute uncertainty for $x$ .
Then convert it into a percentage.

Approach

Decide a sensible absolute uncertainty in $x$ (based on ruler resolution and the fact that $x$ is read at a turning point).
Use the percentage uncertainty formula.
Round appropriately.

Step-by-Step Reasoning

Using the example value $x = 11.0\ \text{cm}$ :

Choose absolute uncertainty: $\Delta x = 0.1\ \text{cm}$ (1 mm) is reasonable for a turning-point estimate.
Calculate:

\%\ \Delta x = \frac{0.1}{11.0} \times 100 = 0.91\%.

Quote as $0.9\%$ (typically 1 s.f. in an uncertainty is acceptable unless asked otherwise).

Key Takeaways

Uncertainty is not just instrument resolution; the nature of the measurement (turning point, movement) matters.
Always show the fraction then multiply by 100.

Common Mistakes

Using $\pm 0.01\ \text{cm}$ for a mm-scale ruler (too small).
Forgetting to multiply by 100.
Writing an uncertainty with excessive significant figures (e.g. $0.9090909\%$ ).

Things to Be Careful About

Use the same units for $\Delta x$ and $x$ before dividing.
If you used a different $x$ , your percentage will change; the method is what is assessed.

Techniques used

estimate absolute uncertainty from instrument resolutioncalculate percentage uncertaintyround the uncertainty to a sensible number of significant figures

(iv)

Calculate $h$ using

h = r - \sqrt{(r^2 - x^2)}

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

1M Similar

DifficultyMedium-Easy

Worked solution

Working

h = r - \sqrt{(r^2-x^2)}

h = 25.0 - \sqrt{25.0^2 - 11.0^2} = 25.0 - \sqrt{625 - 121}

h = 25.0 - \sqrt{504} = 25.0 - 22.45 = 2.55\ \text{cm}

Answer

$\displaystyle 2.6\ \text{cm}$

Final answer

2.6 cm

Detailed explanation

Background Concept

The formula

h = r - \sqrt{r^2 - x^2}

comes from pendulum geometry.

$r$ is the distance from the pivot to the magnet’s centre (the pendulum length).
$x$ is the horizontal displacement of the magnet’s centre from directly below the pivot.
When the pendulum rises, its vertical drop from the pivot changes. The vertical distance from pivot to bob at a given displacement is the adjacent side of a right triangle: $\sqrt{r^2-x^2}$ . The rise in height above the lowest position is therefore $r - \sqrt{r^2-x^2}$ .

Understanding the Question

You are told exactly what to do: calculate $h$ from your measured $r$ and $x$ using the given equation, then record $h$ in cm.

Approach

Use consistent units (both $r$ and $x$ are in cm, so $h$ will be in cm).
Compute $r^2-x^2$ first, then take the square root, then subtract from $r$ .

Step-by-Step Reasoning

Using the example measurements $r = 25.0\ \text{cm}$ and $x = 11.0\ \text{cm}$ :

Square values:

r^2 = 625,\quad x^2 = 121.

Subtract:

r^2 - x^2 = 504.

Square root:

\sqrt{504} = 22.45.

Subtract from $r$ :

h = 25.0 - 22.45 = 2.55\ \text{cm}.

Quote to sensible precision (often 2 s.f. or to 0.1 cm, consistent with measured inputs): $h = 2.6\ \text{cm}$ .

Key Takeaways

Keep units consistent throughout a calculation.
Do the arithmetic in a stable order: squares → difference → square root → subtraction.

Common Mistakes

Calculating $\sqrt{r^2} - x^2$ (wrong order).
Using $h = \sqrt{r^2-x^2}$ (missing the final subtraction from $r$ ).
Mixing units (e.g. $r$ in cm but $x$ in m).

Things to Be Careful About

Ensure $x<r$ ; otherwise $r^2-x^2$ would be negative (indicates a measurement/alignment error).
Rounding too early can change $h$ noticeably; keep extra digits until the final line.

Techniques used

substitute measured values into a given formulaevaluate a square root expression carefullyquote a calculated quantity with unit and sensible significant figures

(b)

Repeat (a)(ii) and (a)(iv) using nut B.

$m$ = ______ $\text{g}$
$x$ = ______ $\text{cm}$
$h$ = ______ $\text{cm}$

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

(Example readings for nut B)

$\displaystyle m = 10.0\ \text{g}$

$\displaystyle x = 10.1\ \text{cm}$

h = 25.0 - \sqrt{25.0^2 - 10.1^2} = 25.0 - \sqrt{625 - 102.01}

h = 25.0 - \sqrt{522.99} = 25.0 - 22.87 = 2.13\ \text{cm}

$\displaystyle h = 2.1\ \text{cm}$

Final answer

m = 10.0 g, x = 10.1 cm, h = 2.1 cm (example)

Detailed explanation

Background Concept

Repeating measurements with a second nut changes the total mass after the collision/pickup. In practical terms, you are being assessed on whether you can:

follow the same method consistently,
obtain a second set of measurements ( $m$ and $x$ ),
calculate $h$ using the same relationship.

Consistency is crucial: if your method changes between nut A and nut B, differences in results may come from technique rather than physics.

Understanding the Question

You must repeat part (a)(ii) (measuring $m$ and $x$ ) and part (a)(iv) (calculating $h$ ) for nut B, then record $m$ , $x$ , and $h$ .

Approach

Keep the setup unchanged (same ruler alignment, same $r$ , same release point at $15.0\ \text{cm}$ , same nut placement at $0$ ).
Record the mass from the card.
Measure $x$ the same way as before.
Substitute into the same formula for $h$ .

Step-by-Step Reasoning

Record nut B mass $m$ from the card.
Repeat the swing-and-pickup procedure and read $x$ at the turning point.
Calculate $h$ using:

h = r - \sqrt{r^2-x^2}

with the same $r$ as previously measured.
4) Quote $h$ with sensible precision consistent with $r$ and $x$ .

The numerical values shown in the answer are representative examples; your experiment will produce your own values.

Key Takeaways

Repeatability depends on identical starting conditions.
Derived quantities ( $h$ ) must be calculated using your measured $x$ and $r$ .

Common Mistakes

Forgetting to detach nut B between trials or changing the position of the ruler.
Changing the release point (not starting at $15.0\ \text{cm}$ ).
Using a different value of $r$ for nut B without re-measuring/without justification.

Things to Be Careful About

Do not round intermediate steps too early; it can noticeably affect $h$ .
Ensure the nut is picked up cleanly; if it slips or is dragged, discard and repeat.

Techniques used

repeat the same measurement procedure with a different value of the independent variablerecord mass and scale readings with consistent precisioncalculate a derived quantity from measured values

(c)

It is suggested that the relationship between $M$ , $m$ and $h$ is

h = \frac{k}{(M + m)^2}

where $k$ is a constant.

(i)

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

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

1M Similar

DifficultyMedium

Worked solution

Working

From

h = \frac{k}{(M+m)^2}\;\Rightarrow\; k = h(M+m)^2

Nut A:

k = 2.55\,(50.0+5.0)^2 = 2.55\,(55.0)^2 = 2.55\times 3025 = 7.71\times 10^3

Nut B:

k = 2.13\,(50.0+10.0)^2 = 2.13\,(60.0)^2 = 2.13\times 3600 = 7.67\times 10^3

Answer

first value of $k = 7.7\times 10^3$

second value of $k = 7.7\times 10^3$

Final answer

k \approx 7.7×10^3, 7.7×10^3

Detailed explanation

Background Concept

When an experiment suggests a relationship like

h = \frac{k}{(M+m)^2},

$k$ is a constant that should (ideally) be the same for all trials.

To find $k$ from one set of measurements, rearrange:

k = h(M+m)^2.

If the model is correct and uncertainties are small, values of $k$ from different nuts should agree within experimental uncertainty.

Understanding the Question

You have measured (or been given) values for:

$M$ : mass of magnet,
$m$ : mass of nut (A then B),
$h$ : calculated height rise.

You must use each dataset to calculate $k$ twice (one for nut A and one for nut B).

Approach

Rearrange the equation to make $k$ the subject.
Substitute $h$ , $M$ , and $m$ for nut A to get $k_1$ .
Substitute $h$ , $M$ , and $m$ for nut B to get $k_2$ .
Quote both values appropriately.

Step-by-Step Reasoning

Rearrangement:

h = \frac{k}{(M+m)^2} \Rightarrow k = h(M+m)^2.

Nut A calculation:

Add masses to get total moving mass $(M+m)$ .
Square it.
Multiply by $h$ .

Nut B calculation:

Repeat exactly the same steps with nut B’s $m$ and $h$ .

In the example working:

$k_1 \approx 7.71\times 10^3$ and $k_2 \approx 7.67\times 10^3$ (in units consistent with how $h$ and masses were used).
They are very close, suggesting good consistency.

Key Takeaways

Always rearrange before substituting to avoid algebra mistakes.
A constant calculated from different trials is how you test whether data fits a model.

Common Mistakes

Using $k = \frac{h}{(M+m)^2}$ (inverting incorrectly).
Forgetting to add $M+m$ before squaring.
Mixing units (e.g. using $M$ in g and $m$ in kg without conversion).

Things to Be Careful About

Keep extra digits during calculation and round at the end.
Use the same unit system for both trials; otherwise your two $k$ values are not comparable.

Techniques used

rearrange a given relationship to solve for a constantsubstitute experimental values into an equationcalculate and compare two values obtained from separate 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 measured values of $h$ , $M$ and $m$ , where $h$ (from ruler readings of $r$ and $x$ ) has the largest uncertainty.
Therefore $k$ should be quoted to about 2 significant figures (e.g. $7.7\times 10^3$ ).

Final answer

Quote k to about 2 s.f., limited by uncertainty in h (from r and x).

Detailed explanation

Background Concept

Significant figures in a calculated quantity should reflect the precision of the input data.

If a measurement has an uncertainty of several percent, quoting a result to 4 or 5 significant figures is unjustified.
In practical work, it is common to quote uncertainties to 1 s.f. (sometimes 2), and quote the final value to a matching precision.

Here,

k = h(M+m)^2.

Even if $M$ and $m$ are written clearly on a card, $h$ is calculated from $r$ and $x$ , and $x$ is often the least certain because it is read at a turning point.

Understanding the Question

You must explain why you chose the number of significant figures for your two $k$ values.
The examiner wants to see that you have linked your rounding to measurement precision (especially of $x$ and therefore $h$ ).

Approach

Identify which measured quantity has the greatest percentage uncertainty.
State that this limits the precision of $k$ .
Conclude an appropriate number of significant figures (often 2 s.f. for data with ~10% uncertainty).

Step-by-Step Reasoning

$k$ depends on $h$ directly and on $(M+m)^2$ .
$h$ comes from

h = r - \sqrt{r^2-x^2},

so any uncertainty in $x$ and $r$ propagates into $h$ .
3) In this experiment, $x$ is typically the most uncertain reading because:

it is a turning point,
the system may still oscillate,
parallax is possible.

Therefore $h$ (and hence $k$ ) will not be known to very high precision. Quoting $k$ to about 2 significant figures is reasonable and consistent with practical uncertainty levels.

Key Takeaways

Significant figures should reflect measurement uncertainty, not calculator output.
Identify the limiting measurement (often a moving pointer / turning point reading).

Common Mistakes

Saying “because my calculator gives this many digits”.
Quoting $k$ to 4 s.f. when $x$ was only measured to $0.1\ \text{cm}$ and is a turning point.
Not referring to any measured quantity in the justification.

Things to Be Careful About

If you calculated $k$ using rounded values of $h$ , your final $k$ should not be more precise than $h$ .
If the paper states a typical percentage uncertainty (here later they suggest $15\%$ ), that strongly supports quoting results to 2 s.f.

Techniques used

identify the least precisely known measured quantitylink significant figures of a calculated result to measurement uncertaintyround a derived quantity appropriately

(d)

It is suggested that the percentage uncertainty in the values of $k$ is $15\%$ .

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

1M Similar

DifficultyMedium

Worked solution

Working

Percentage difference between $k$ values:

\%\ \text{difference} = \frac{|7.71\times 10^3-7.67\times 10^3|}{\frac{1}{2}(7.71\times 10^3+7.67\times 10^3)}\times 100 = \frac{44}{7690}\times 100 \approx 0.57\%

Answer

$0.57\% < 15\%$ , so the two $k$ values agree within the uncertainty and the results support the relationship.

Final answer

Yes — k values agree within 15%, so relationship is supported.

Detailed explanation

Background Concept

To test whether results support a proposed relationship, you look for consistency within uncertainties.

If two measured/derived values are supposed to be the same (here, both are estimates of the constant $k$ ), then they should agree within the experimental uncertainty.
A common method is to compare their percentage difference to the stated percentage uncertainty.

Percentage difference (one acceptable definition) is:

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

If this is less than the stated uncertainty (here $15\%$ ), then the results are consistent.

Understanding the Question

You are told to assume the percentage uncertainty in $k$ is $15\%$ . Using that, you must say whether your two calculated values of $k$ support the relationship.

So you need:

a comparison of the two $k$ values (difference), and
a clear conclusion based on the $15\%$ threshold.

Approach

Compute how far apart the two $k$ values are (as a percentage).
Compare with $15\%$ .
Conclude “supports” if the difference is smaller than $15\%$ (or “does not support” if larger).

Step-by-Step Reasoning

Using the example values $k_1 = 7.71\times 10^3$ and $k_2 = 7.67\times 10^3$ :

Find absolute difference:

|k_1-k_2| = 0.04\times 10^3 = 40\text{ to }44.

Find mean:

\bar{k} = \frac{k_1+k_2}{2} \approx 7.69\times 10^3.

Percentage difference:

\%\ \text{difference} = \frac{|k_1-k_2|}{\bar{k}}\times 100 \approx 0.6\%.

Compare to $15\%$ :

0.6\% < 15\% \Rightarrow \text{consistent}.

Therefore the results support the relationship (within the assumed uncertainty).

Key Takeaways

“Support” in practical physics usually means “consistent within uncertainty”, not “exactly equal”.
Always link your conclusion to a numerical comparison.

Common Mistakes

Saying “they are close” without using the $15\%$ figure.
Comparing raw difference rather than percentage difference.
Concluding “does not support” even when the values differ by less than the stated uncertainty.

Things to Be Careful About

Use your unrounded $k$ values for the comparison if possible.
Be explicit: state both the computed percentage difference and the comparison with $15\%$ .

Techniques used

compare two experimental values using percentage differenceapply an uncertainty criterion to decide if results are consistentstate a conclusion linked to the numerical comparison

(e)

(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

$x$ : difficult to judge the maximum position because the magnet+nutt system is moving/oscillating; also parallax when reading the ruler.
$x$ : ruler zero may not be exactly vertically below the magnet centre (alignment error gives systematic error in $x$ ).
$r$ : uncertainty in locating the exact centre of the magnet and the nail position when measuring $r$ .
Procedure limitation: energy losses (friction at the nail pivot and air resistance) mean the swing height is reduced and may vary between runs.

Final answer

Four sources: x reading (turning point/parallax), zero alignment, r centre/pivot definition, friction/air resistance energy losses.

Detailed explanation

Background Concept

In practical experiments, uncertainties arise from:

instrument resolution (smallest scale division),
observer effects (parallax, reaction time, judging a turning point),
alignment and geometry (where “zero” or “centre” actually is),
physical limitations (friction, air resistance, inconsistent collisions).

A good answer names the quantity and explains the reason.

Understanding the Question

You must give four sources of uncertainty or limitations in this specific procedure. For any measurement uncertainty, you must:

state what you are measuring (e.g. $x$ , $r$ , masses), and
give a reason why it is uncertain.

Approach

Look at every measured/controlled part of the method and ask:

What could vary from run to run (random uncertainty)?
What could bias all readings in the same direction (systematic error)?
What parts of the setup are hard to align or define?

e-i-uncertainty-sources

Step-by-Step Reasoning

Four strong, creditworthy examples are:

Uncertainty in $x$ (turning point):

The magnet+nutt system may not stop exactly at one point; it can oscillate slightly.
This makes the maximum displacement hard to judge.
Also, reading a ruler from above at an angle causes parallax.

Systematic alignment error in $x$ :

If the ruler’s $0$ mark is not directly below the magnet’s centre in the equilibrium position, every $x$ value is shifted.
This is systematic: it affects all trials similarly.

Uncertainty in $r$ :

The nail position is small and hard to define exactly.
The “centre of the magnet” may be difficult to locate precisely, especially if the magnet is not symmetric or the centre is not marked.

Energy-loss limitation:

Friction at the pivot and air resistance reduce mechanical energy.
If friction varies (e.g. rod rubbing the nail differently each run), the turning point changes, increasing scatter in $x$ and hence $h$ .

Other acceptable limitations (if needed) could include inconsistent pickup (nut may not be grabbed at exactly the same point/time) or the nut sliding/bouncing on the ruler before pickup.

Key Takeaways

Always connect an uncertainty to a specific measured quantity.
Include both measurement uncertainties (reading scales) and physical limitations (friction, air resistance, inconsistent interaction).

Common Mistakes

Writing vague statements like “human error” without specifying $x$ or $r$ and why.
Repeating the same idea four times (e.g. four versions of parallax).
Giving an “improvement” instead of a “limitation” (improvements belong in part (ii)).

Things to Be Careful About

Make sure you have four distinct points.
At least some should reference the specific measurements used later ( $x$ and $r$ are especially important because they determine $h$ and then $k$ ).

Techniques used

identify measurement uncertainties and link them to specific quantitiesexplain physical causes of random and systematic errorsdescribe procedural limitations affecting reliability

(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

Take repeat readings of $x$ for each nut and calculate a mean (and ignore anomalous values).
Attach a thin pointer to the magnet (or use a fiducial marker) and read against a fixed vertical scale to reduce parallax and improve definition of $x$ .
Use video recording (or a motion sensor) to identify the maximum displacement/turning point more accurately.
Reduce pivot friction by using a low-friction pivot (e.g. knife-edge/bearing) and ensure the rod does not rub on the support.

Final answer

Improvements: repeats & mean, pointer/fiducial to reduce parallax, video/motion sensor for turning point, reduce pivot friction.

Detailed explanation

Background Concept

Improvements should:

target a specific limitation,
reduce random scatter (increase repeatability) or reduce systematic bias,
be practical with realistic school-lab apparatus.

Since $k$ depends strongly on $h$ and hence on $x$ , improving the measurement of $x$ usually gives the biggest benefit.

Understanding the Question

You must propose four improvements to the experiment. These can involve:

better apparatus,
better measurement methods,
different procedures.

The best answers clearly reduce the uncertainties/limitations you identified in part (i).

Approach

For each major weakness in the method, propose a matching fix:

Turning point hard to read $\rightarrow$ use video/data logging/pointer.
Parallax/alignment $\rightarrow$ add fiducial markers and alignment aids.
Random variation $\rightarrow$ repeat and average.
Energy losses $\rightarrow$ reduce friction and unwanted contact.

Step-by-Step Reasoning

Four strong improvements are:

Repeat and average $x$ :

Do several trials for each nut.
Use the mean $x$ ; this reduces random uncertainty.

Use a pointer/fiducial marker:

Attach a narrow pointer to the moving magnet/nut so you read a single line rather than an extended object.
Read against a fixed scale to reduce parallax and ambiguity.

Use video or a sensor to capture the turning point:

Record the motion and pause at maximum displacement to read $x$ .
This is especially effective because the turning point is brief and easy to miss by eye.

Improve the pivot:

Replace the nail contact with a lower-friction pivot (knife-edge or small bearing) and ensure the rod swings without rubbing.
This improves repeatability and reduces variable energy loss.

Other acceptable improvements could include using a set square/plumb line to align ruler zero under the magnet centre, or using a shaped holder to keep the nut exactly centred at zero each time.

Key Takeaways

Improvements must be specific and should clearly reduce a named uncertainty.
Repeats reduce random uncertainty; better alignment reduces systematic error.

Common Mistakes

Giving vague improvements like “be more careful”.
Suggesting changes that alter the physics being tested without justification.
Listing improvements that do not correspond to any real limitation in the method.

Things to Be Careful About

Ensure your improvements are feasible with typical lab equipment.
Make sure each of the four points is different (e.g. don’t give four versions of “use a better ruler”).

Techniques used

propose procedural changes that reduce specific uncertaintiessuggest improved apparatus to increase precisionlink each improvement to a corresponding limitation