Physics 9702/35 — May/June 2018

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

Q1Manipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and EvaluationFree sample

In this experiment, you will investigate the equilibrium of a metre rule.

(a)

(i)

You have been provided with some masses.

Set up the apparatus as shown in Fig. 1.1.

Fig. 1.1

• Mass Q should be $200\text{ g}$ .
• The distance between the $50\text{ cm}$ mark on the rule and the string loop supporting the rule is $x$ . Adjust the position of the metre rule so that $x$ is approximately $15\text{ cm}$ .
• The distance between the string loop supporting mass P and the string loop supporting the rule is $z$ . Adjust the position of mass P so that $z$ is approximately $30\text{ cm}$ .
• The distance between the string loop supporting the rule and the string loop supporting mass Q is $y$ . Adjust the position of mass Q until the rule is balanced.
• Measure and record $z$ .

$z$ = ______

DifficultyEasy

Worked solution

Answer

Measure $z$ (distance between the two string loops for the rule and mass $P$ ) on the metre rule.

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

$z = 30.0\ \text{cm}$

Final answer

z = 30.0 cm (example)

Detailed explanation

Background Concept

In equilibrium, the metre rule is balanced (no turning effect overall). In this practical, you adjust positions until the rule is horizontal and steady, then you measure distances along the rule.

A distance such as $z$ is read directly from the scale on the rule by taking the difference between the two scale readings at the relevant points (here, the positions of the two string loops).

Understanding the Question

You are given the apparatus shown: a metre rule supported by a string loop (near the 50 cm region) with two hanging masses $P$ and $Q$ . You are instructed to:

choose $Q = 200\ \text{g}$ ,
set $x \approx 15\ \text{cm}$ ,
set $z \approx 30\ \text{cm}$ ,
then adjust $y$ until the rule is balanced,
and finally measure and record $z$ .

So the mark is for a clear, sensible measurement and recording of $z$ .

diagram-a1

Approach

Identify the two points that define $z$ (the loop supporting the rule and the loop supporting mass $P$ ).
Read their positions on the metre rule scale.
Calculate $z$ as the difference (or read directly if you can align one at a known mark).
Record $z$ with a sensible precision (typically to the nearest mm or $0.1\ \text{cm}$ ).

Step-by-Step Reasoning

Ensure the rule is steady and horizontal.
Look straight down at the scale to avoid parallax.
Note the scale position of the support loop for the rule (call it $s_\text{rule}$ ).
Note the scale position of the support loop for mass $P$ (call it $s_P$ ).
Compute

z = |s_\text{rule} - s_P|

Record $z$ with unit.

A typical value consistent with the instruction “approximately $30\ \text{cm}$ ” is $z = 30.0\ \text{cm}$ .

Key Takeaways

Identify precisely which two points define the distance.
Read a metre rule correctly (eye normal to scale).
Record to appropriate precision with units.

Common Mistakes

Measuring from the wrong reference point (e.g. from the 50 cm mark instead of between the two loops).
Parallax error from viewing the scale at an angle.
Writing no unit, or recording an over-precise value (e.g. many decimal places) inconsistent with a metre rule.

Things to Be Careful About

Make sure $z$ is the separation between the two loops, not the position of one loop.
If the loops are thick, decide a consistent reference (e.g. centre of the loop) each time.
Keep the rule steady before reading.

Techniques used

set up the apparatus following a diagrambalance the metre rule by adjusting a hanging massmeasure a distance on a metre rule using the correct scale and avoiding parallaxrecord a length to an appropriate resolution with unit

(ii)

• Measure and record $x$ .

$x$ = ______

• Measure and record $y$ .

$y$ = ______

DifficultyEasy

Worked solution

Answer

Measure $x$ (from the $50\ \text{cm}$ mark to the rule support loop) and $y$ (from the rule support loop to the mass $Q$ loop).

Example values (to nearest $0.1\ \text{cm}$ ):

$x = 15.0\ \text{cm}$

$y = 33.8\ \text{cm}$

Final answer

x = 15.0 cm, y = 33.8 cm (example)

Detailed explanation

Background Concept

$x$ and $y$ are distances along the metre rule defined relative to specific points:

$x$ is measured from the 50 cm mark to the rule’s support loop.
$y$ is measured from the rule’s support loop to the loop holding mass $Q$ .

In Paper 3, marks are typically for correct identification of what to measure, sensible precision, and clear recording with units.

Understanding the Question

You are asked to:

measure and record $x$ after setting up the apparatus,
then measure and record $y$ when the rule is balanced.

The instruction earlier says make $x \approx 15\ \text{cm}$ , so your measured value should be around this.

Approach

Locate the 50 cm mark and the support loop position on the rule.
Read their scale positions to get $x$ .
Locate the loop holding $Q$ and the support loop position.
Read their positions to get $y$ .
Record each value with consistent precision and units.

Step-by-Step Reasoning

For $x$ $x$ :
- Read the scale at the support loop position: $s_\text{rule}$ .
- Use the 50 cm mark as $s_{50} = 50.0\ \text{cm}$ .

x = s_\text{rule} - 50.0\ \text{cm}

If the support is to the right of 50 cm, $x$ is positive.

For $y$ $y$ :
- Read the scale at the mass $Q$ loop position: $s_Q$ .

y = s_Q - s_\text{rule}

Record with unit. A typical consistent set might be $x = 15.0\ \text{cm}$ and (after balancing) $y = 33.8\ \text{cm}$ .

Key Takeaways

Use the correct reference points (50 cm mark and loop positions).
Use a consistent sign convention (important later when $x$ can be negative).
Record with appropriate precision and units.

Common Mistakes

Measuring $x$ from the end of the rule instead of from the 50 cm mark.
Forgetting that $y$ is measured from the support loop (not from the 50 cm mark).
Omitting units.

Things to Be Careful About

If the support is left of 50 cm in later readings, $x$ must be recorded as negative.
Avoid parallax: eye directly above the scale mark being read.
Use the same reference point on each loop (e.g. centre of the loop) each time.

Techniques used

measure a distance from a reference mark on a scaleuse sign convention to record a negative displacement when measured to the leftrecord lengths to consistent resolution and include units

(b)

• Write down your value of $z$ from (a)(i).

$z$ = ______

• Keeping $z$ constant, change $x$ and adjust $y$ until the rule is balanced. Repeat until you have six sets of values of $x$ and $y$ . Record your results in a table.

You may include readings where $x$ is measured to the left of the $50\text{ cm}$ mark. In such cases $x$ has a negative value.

DifficultyMedium

Worked solution

Answer

Write the previously measured constant value of $z$ .

Example: $z = 30.0\ \text{cm}$ (kept constant).

Record six sets of corresponding $x$ and balanced $y$ values in one table (with units in headings and consistent dp).

Example table:

$x$ / $\text{cm}$	$y$ / $\text{cm}$
$-5.0$	$18.8$
$0.0$	$22.5$
$5.0$	$26.3$
$10.0$	$30.0$
$15.0$	$33.8$
$20.0$	$37.5$

Final answer

z constant; 6 (x, y) readings recorded in a table (student-dependent)

Detailed explanation

Background Concept

To test a relationship between two quantities experimentally, you must:

vary the independent variable over a useful range,
keep other relevant variables constant,
measure the dependent variable for each setting,
present results clearly.

Here, you keep $z$ constant, change $x$ , and each time adjust $y$ until the rule is in equilibrium (balanced). Each balanced position gives a paired data point $(x, y)$ .

Understanding the Question

You must:

Copy your value of $z$ from (a)(i).
Keep $z$ constant throughout.
Change $x$ and, for each $x$ , adjust $y$ until the rule balances.
Repeat until you have six pairs of $(x, y)$ .
Record in a table.

You are allowed to take readings with the support point to the left of the 50 cm mark; then $x$ is negative.

Approach

Choose values of $x$ that span a reasonable range (including possibly negative $x$ ).
For each chosen $x$ , adjust mass $Q$ position until the metre rule is level and stationary.
Measure $y$ once balanced.
Record all results in one clearly laid-out table with headings of the form “quantity / unit”.

Step-by-Step Reasoning

Set $z$ to your chosen fixed value (around $30\ \text{cm}$ ) and do not move mass $P$ relative to the rule support loop.
Move the support point (or rule position) so that $x$ changes; measure $x$ from the 50 cm mark:

x = s_\text{rule} - 50.0\ \text{cm}

Slide mass $Q$ until the rule balances (horizontal, not rotating).
Measure $y$ between the rule support loop and the $Q$ loop:

y = s_Q - s_\text{rule}

Repeat until six sets are collected.

A good table has:

two columns ( $x$ and $y$ ),
units in the headings,
consistent decimal places down each column,
values that vary across a range.

Key Takeaways

Control variables: keep $z$ constant.
Collect enough data (six points) with a good spread.
Present results in a properly formatted table.

Common Mistakes

Fewer than six sets of readings.
Changing $z$ accidentally while changing $x$ .
Missing units in headings, or putting units in every cell instead of the heading.
Inconsistent precision (e.g. mixing $15.0$ and $15.23$ when using a metre rule).

Things to Be Careful About

Sign convention: if the support is left of 50 cm, $x$ must be negative (do not write a positive number).
Ensure the rule is truly balanced before reading $y$ (wait for oscillations to die out).
Read positions at eye level to reduce parallax.

Techniques used

keep one variable constant while varying anotherobtain repeated sets of readings covering a suitable rangerecord data in a single table with correct headings and unitsuse a sign convention for negative values where appropriate

(c)

(i)

Plot a graph of $y$ on the $y$ -axis against $x$ on the $x$ -axis.

DifficultyMedium-Easy

Worked solution

Answer

Plot $y$ on the $y$ -axis against $x$ on the $x$ -axis.

Axes labelled $x / \text{cm}$ and $y / \text{cm}$ .
Use a sensible scale (at least half the graph paper).
Plot all six points accurately.

Final answer

See graph

Detailed explanation

Background Concept

A graph is used to reveal whether two measured quantities are related linearly. If $y$ depends on $x$ , plotting $y$ (dependent variable) against $x$ (independent variable) is the standard way to test whether a straight-line relationship is plausible.

Understanding the Question

You have collected six pairs of $(x, y)$ values. You are asked to plot $y$ on the vertical axis and $x$ on the horizontal axis. This is the standard “ $y$ against $x$ ” convention.

Approach

Put $x$ on the horizontal axis and $y$ on the vertical axis.
Choose scales that fit all your data and spread points out (avoid cramped plots).
Label axes correctly, including units.
Plot points with small, neat crosses.

Step-by-Step Reasoning

Determine the minimum and maximum values of $x$ and $y$ from your table.
Choose axis limits slightly beyond these extremes.
Choose a scale such as $1\ \text{cm}$ on paper representing $2\ \text{cm}$ or $5\ \text{cm}$ of measured value, so the plotted range uses most of the available grid.
Label axes as:

horizontal: $x / \text{cm}$
vertical: $y / \text{cm}$

Plot each point carefully using the scale.

diagram-c1

Key Takeaways

Independent variable on $x$ -axis; dependent variable on $y$ -axis.
Axes must have quantity and unit.
Good scales and accurate plotting earn marks.

Common Mistakes

Swapping axes (plotting $x$ against $y$ ).
Missing units on axes.
Using awkward scales (e.g. 3 squares = 2 cm) that make plotting inaccurate.

Things to Be Careful About

Include negative $x$ values if you took them (do not force the axis to start at zero if it wastes space).
Plot crosses, not blobs; ensure each cross is centred at the correct coordinate.

Techniques used

choose appropriate axis scales that use most of the gridlabel axes with quantity and unitplot experimental points accurately

(ii)

Draw the straight line of best fit.

DifficultyEasy

Worked solution

Answer

Draw one straight line of best fit (not dot-to-dot) through the plotted points, with roughly equal scatter about the line.

Final answer

See graph (best-fit line drawn)

Detailed explanation

Background Concept

A best-fit line represents the overall trend of the data when random uncertainties cause scatter. For a proposed linear relationship, the best-fit line should be straight and placed so that points are distributed approximately evenly above and below it.

Understanding the Question

You have already plotted $y$ against $x$ . Now you must draw the straight line that best represents the relationship suggested by the data.

Approach

Use a ruler.
Draw one straight line.
Do not force the line through every point.
Aim for balanced residuals (similar scatter on both sides).

Step-by-Step Reasoning

Visually judge the trend of the points.
Place a ruler so that the line passes through the “middle” of the scatter.
Check that no single outlier dominates the line placement.
Draw the line clearly across the full range of the data.

Key Takeaways

Best-fit line is about trend, not connecting points.
A good best-fit line is crucial because gradient and intercept come from it.

Common Mistakes

Joining points dot-to-dot.
Drawing a line that goes through the origin without justification.
Drawing the line only between two middle points instead of extending across the range.

Things to Be Careful About

If you have an obvious outlier, you still draw the best-fit line for the main trend (unless instructed otherwise).
Use a sharp pencil so the line is thin and readings from it are accurate.

Techniques used

draw a single straight line of best fit through scattered pointsbalance points above and below the best-fit lineavoid joining point-to-point

(iii)

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

gradient = ______

$y$ -intercept = ______

DifficultyMedium-Easy

Worked solution

Working

Use two well-separated points on the best-fit line, e.g.
$(x_1, y_1) = (-5.0\ \text{cm},\ 18.8\ \text{cm})$ and $(x_2, y_2) = (20.0\ \text{cm},\ 37.5\ \text{cm})$ .

\text{gradient} = \frac{\Delta y}{\Delta x} = \frac{37.5 - 18.8}{20.0 - (-5.0)} = \frac{18.7}{25.0} = 0.748 \approx 0.75

At $x = 0$ ,

y\text{-intercept} \approx 22.5\ \text{cm}

Answer

gradient $= 0.75$

y-intercept $= 22.5\ \text{cm}$

Final answer

gradient = 0.75; y-intercept = 22.5 cm (example)

Detailed explanation

Background Concept

For a straight-line graph,

y = mx + c

$m$ is the gradient (slope):

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

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

If both axes are lengths in cm, the gradient has units $\text{cm}/\text{cm}$ , so it is dimensionless.

Understanding the Question

After drawing a straight line of best fit on your $y$ vs $x$ graph, you must:

determine the gradient of the line,
determine the $y$ -intercept.

These must be taken from the best-fit line, not by joining two data points at random.

Approach

Choose two points on the best-fit line that are far apart to reduce percentage reading uncertainty.
Compute gradient using $\Delta y/\Delta x$ .
Find the intercept by reading where the line crosses the $y$ -axis (at $x = 0$ ).

Step-by-Step Reasoning

Pick two widely separated points on the drawn line (they do not need to be actual plotted data points).
Read their coordinates carefully from the axes.
Calculate differences:

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

Calculate gradient:

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

For the intercept, set $x=0$ and read $y$ where the line crosses the vertical axis.

Using the example values shown:

\text{gradient} = \frac{37.5 - 18.8}{20.0 - (-5.0)} \approx 0.75

and $y$ -intercept $\approx 22.5\ \text{cm}$ .

diagram-c3

Key Takeaways

Always use a large triangle for gradient.
Gradient is $\Delta y/\Delta x$ , not $x/y$ .
Intercept is read at $x=0$ .

Common Mistakes

Using two nearby points, giving a very uncertain gradient.
Calculating $\Delta x/\Delta y$ (inverting the gradient).
Using two raw data points rather than points on the best-fit line.
Forgetting units for the intercept.

Things to Be Careful About

Use consistent units: if axes are in cm, keep readings in cm.
If your best-fit line does not cross the $y$ -axis within the plotted area, extend it carefully with a ruler.
Quote gradient to sensible significant figures (typically 2–3 sf).

Techniques used

use a large triangle on the best-fit line to calculate the gradientcalculate gradient as $\Delta y / \Delta x$read the y-intercept from the best-fit line at $x = 0$

(d)

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

y = Ax + B

where $A$ and $B$ are constants.

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

$A$ = ______

$B$ = ______

DifficultyMedium-Easy

Worked solution

Working

Given

y = Ax + B

Comparing with $y = mx + c$ :

A = \text{gradient},\qquad B = y\text{-intercept}

Using (c)(iii):

A = 0.75\ (\text{no unit}),\qquad B = 22.5\ \text{cm}

Answer

$A = 0.75$

$B = 22.5\ \text{cm}$

Final answer

A = gradient; B = y-intercept (with units)

Detailed explanation

Background Concept

A straight line has equation

y = mx + c

$m$ is the gradient.
$c$ is the $y$ -intercept.

If an experiment suggests

y = Ax + B,

then $A$ plays the role of $m$ and $B$ plays the role of $c$ .

Units come from the axes:

If $y$ is in cm and $x$ is in cm, then $A$ has units $\text{cm}/\text{cm}$ (dimensionless), and $B$ has units of $y$ (cm).

Understanding the Question

You have already found the gradient and the $y$ -intercept from your graph. This part simply asks you to use these to state $A$ and $B$ (with appropriate units).

Approach

Recognise the direct correspondence: $A = \text{gradient}$ and $B = \text{intercept}$ .
Transfer the numerical values.
Attach correct units: none for $A$ (here), cm for $B$ .

Step-by-Step Reasoning

From (c)(iii), suppose your results are:

gradient $= 0.75$
$y$ -intercept $= 22.5\ \text{cm}$

Then

A = 0.75

and

B = 22.5\ \text{cm}

Key Takeaways

Constants in a linear relationship come directly from the graph.
Intercept has the same unit as $y$ .

Common Mistakes

Swapping $A$ and $B$ .
Giving a unit to $A$ when $x$ and $y$ have the same unit.
Forgetting units for $B$ .

Things to Be Careful About

If you plotted in metres instead of cm, your numerical values for $A$ and $B$ would change accordingly; always use consistent units with your graph axes.

Techniques used

match a straight-line equation to the form y = mx + cidentify constants as gradient and intercept from a graphstate appropriate units from axis units

(e)

The mass of P is $p$ . The mass of Q is $q$ , where $q = 0.200\text{ kg}$ .

The constants $A$ and $B$ are related to $p$ , $q$ and $z$ by

A = \frac{p}{q} \text{ and } B = \frac{pz}{q}

Calculate $p$ .

$p$ = ______ $\text{kg}$

DifficultyEasy

Worked solution

Working

Given

A = \frac{p}{q}

p = Aq

With $A = 0.75$ and $q = 0.200\ \text{kg}$ :

p = 0.75 \times 0.200 = 0.150\ \text{kg}

Answer

$p = 0.150\ \text{kg}$

Final answer

0.150 kg

Detailed explanation

Background Concept

You are given relationships linking the straight-line constant $A$ to the masses:

A = \frac{p}{q}

This is a direct proportionality: for fixed $q$ , the gradient-related constant $A$ tells you the unknown mass $p$ .

Understanding the Question

Mass of $Q$ is known: $q = 0.200\ \text{kg}$ .
You have already determined $A$ from the graph.
You must calculate the mass of $P$ , $p$ .

Approach

Rearrange the given equation for $A$ to make $p$ the subject, then substitute your values for $A$ and $q$ .

Step-by-Step Reasoning

Starting with

A = \frac{p}{q}

Multiply both sides by $q$ :

p = Aq

Substitute $A = 0.75$ (example from the graph) and $q = 0.200\ \text{kg}$ :

p = 0.75 \times 0.200 = 0.150\ \text{kg}

Key Takeaways

Gradient/intercept constants can be used to determine physical quantities.
Rearranging simple algebraic relationships is a common Paper 3 skill.

Common Mistakes

Using $q = 200\ \text{g}$ without converting to kg when the answer is required in kg.
Dividing instead of multiplying (writing $p = q/A$ ).

Things to Be Careful About

Keep mass units consistent: since $q$ is given in kg and the answer requires kg, use kg throughout.

Techniques used

substitute a measured gradient into a given formularearrange to solve for an unknown massuse consistent units for masses

(f)

The experiment is repeated using the same equipment but a smaller value of $z$ . For this experiment, draw a second line on the graph to show the expected results. Label this line W.

DifficultyMedium-Easy

Worked solution

Answer

For smaller $z$ , $A$ is unchanged and $B$ decreases, so draw line $W$ parallel to the original best-fit line but with a smaller $y$ -intercept (shifted downward).

Final answer

Line W: same gradient, smaller y-intercept

Detailed explanation

Background Concept

The straight-line form is

y = Ax + B

and you are given

A = \frac{p}{q}, \qquad B = \frac{pz}{q}

If you repeat the experiment with the same equipment and masses $p$ and $q$ unchanged, then:

$A = p/q$ stays the same (so the gradient is unchanged),
$B = pz/q$ depends on $z$ , so reducing $z$ reduces $B$ (so the intercept decreases).

Understanding the Question

You must add a second predicted line on the same $y$ vs $x$ graph for an experiment where $z$ is smaller than before. This is not new data; it is a prediction based on how the linear relationship changes.

Approach

Keep the same gradient as the original line.
Move the line down so it crosses the $y$ -axis at a smaller value.
Label this new line clearly as $W$ .

Step-by-Step Reasoning

Recognise that smaller $z$ means smaller $B$ :

B \propto z

Recognise that $A$ does not contain $z$ , so $A$ is unchanged.
On the graph, draw a new straight line with the same slope as the original (parallel).
Ensure it crosses the $y$ -axis below the original intercept.
Label it $W$ .

diagram-f1

Key Takeaways

Changing a parameter that appears only in the intercept shifts the line up/down without changing slope.
Parallel lines correspond to the same gradient.

Common Mistakes

Drawing a line with a different gradient.
Shifting the line upward (would correspond to larger $z$ , not smaller).
Forgetting to label the line $W$ .

Things to Be Careful About

The line must remain straight and parallel; do not pivot it around a point.
The predicted line should be drawn over the same $x$ -range as the data for clear comparison.

Techniques used

predict how changing one parameter affects gradient and interceptsketch an additional straight line with altered interceptlabel a predicted line on an existing graph

The rest of this paper

1 more questions

Q2Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation20M

Loading the full paper…