Physics 9702/33 — October/November 2019

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)

You have been provided with a metre rule with a $100\text{ g}$ mass attached to it.

Set up the apparatus as shown in Fig. 1.1.

Fig. 1.1

The distance between the end of the rule and the string loop from which mass P is suspended is $x$ , as shown in Fig. 1.1.

The distance between the same end of the rule and the string loop suspended from the rod of the clamp is $y$ .

Position mass P so that $x$ is approximately $30\text{ cm}$ .
Without changing $x$ , adjust the position of the rule until it balances.
Measure and record $x$ and $y$ .

$x$ = ______
$y$ = ______

DifficultyMedium-Easy

Worked solution

Answer

(Example readings, to nearest $1\ \text{mm}$ )

$x = 30.0\ \text{cm}$

$y = 30.6\ \text{cm}$

Final answer

x = 30.0 cm, y = 30.6 cm (example)

Detailed explanation

Background Concept

A metre rule is in equilibrium when the net force is zero and the net turning effect (moment) about any point is zero. In practice, you find equilibrium by adjusting the support position until the rule is horizontal and does not rotate.

Distances such as $x$ and $y$ are measured from the same reference end of the rule, so you must keep that end fixed as the zero point for all readings.

Understanding the Question

You are told to:

set up the apparatus as in the diagram,
place mass $P$ so that $x \approx 30\ \text{cm}$ ,
adjust the rule until it balances (horizontal, not turning),
then measure and record $x$ and $y$ .

So the only “answers” here are your measured values of $x$ and $y$ (with sensible precision).

diagram-1

Approach

Set up the rule, clamp, and loops as shown.
Set $x$ by sliding the loop for mass $P$ to about $30\ \text{cm}$ from the chosen end.
Without changing $x$ , slide/adjust the suspension point (or the rule position relative to the suspension loop) until the rule is balanced.
Read $x$ and $y$ from the same end of the rule, at eye level to reduce parallax, and record to the nearest millimetre (i.e. $0.1\ \text{cm}$ ).

Step-by-Step Reasoning

Choose the left end of the metre rule as the reference (the end indicated in the figure).
Move mass $P$ until the distance from that end to its loop is about $30\ \text{cm}$ ; this is $x$ .
Keep $x$ fixed.
Adjust the suspension position until the rule is horizontal and remains at rest when released gently (no rotation).
Measure:
- $x$ : from the reference end to the loop holding mass $P$ .
- $y$ : from the same reference end to the suspension loop attached to the clamp rod.
Record values with consistent precision, typically to $0.1\ \text{cm}$ .

(Example only: $x = 30.0\ \text{cm}$ and $y = 30.6\ \text{cm}$ .)

Key Takeaways

Equilibrium requires careful balancing before taking readings.
Always measure all distances from the same reference end.
Record lengths to appropriate precision and avoid parallax.

Common Mistakes

Measuring $x$ and $y$ from different ends of the rule.
Reading the scale at an angle (parallax error).
Recording excessive precision (e.g. $30.03\ \text{cm}$ ) or too little (e.g. $30\ \text{cm}$ ) without instrument justification.
Letting $x$ change while trying to balance the rule.

Things to Be Careful About

Ensure the rule is truly balanced (not slowly rotating).
Check the loops are vertical and not rubbing against the rule.
Record units (cm) and keep consistent decimal places (usually 1 d.p. for cm).

Techniques used

set up the apparatus to match the given diagramadjust the rule to achieve balance (equilibrium)read distances on a metre rule with appropriate precisionavoid parallax when taking scale readings

(b)

Change $x$ . Adjust the position of the rule until it balances. Measure and record $x$ and $y$ .

Repeat until you have six sets of values.

Record your results in a table.

DifficultyMedium-Easy

Worked solution

Answer

Record six sets of $(x,y)$ (both in cm) with $x$ covering a suitable range.

(Example of an acceptable results table; values are illustrative.)

set	$x$ / cm	$y$ / cm
1	20.0	25.4
2	30.0	30.6
3	40.0	35.8
4	50.0	41.0
5	60.0	46.2
6	70.0	51.4

Final answer

See working (table of six (x, y) readings).

Detailed explanation

Background Concept

In practical physics, you must collect enough data points (here, six) over a range of the independent variable to reveal a trend and support later graph work. A good table makes the data easy to read and reduces the chance of losing marks for presentation.

Key table features:

One table for all readings.
Clear column headings with quantity and unit.
Consistent decimal places based on instrument precision.

Understanding the Question

You are asked to change $x$ , rebalance the rule each time, and measure $x$ and $y$ until you have six pairs of values. Then you must present them in a results table.

So the marks are mainly for:

having six sets,
a sensible range of $x$ values,
correct and consistent recording (headings/units/precision).

Approach

Choose a range of $x$ values (e.g. from about $20\ \text{cm}$ to $70\ \text{cm}$ ).
For each chosen $x$ $x$ :
- position mass $P$ at that $x$ ,
- adjust the rule/support position until balanced,
- read $y$ .
Record $x$ and $y$ immediately in a single table with units.

Step-by-Step Reasoning

Select $x$ values that are well-spaced (not clustered), because this improves the reliability of the best-fit line later.
Each time you change $x$ , you must re-establish equilibrium before reading $y$ . If you read $y$ while the rule is not balanced, you are not measuring the correct equilibrium position.
Use consistent precision:
- If reading to the nearest $1\ \text{mm}$ , record as $0.1\ \text{cm}$ .
- Keep the same number of decimal places for all $x$ readings, and similarly for all $y$ readings.
A suitable table layout includes either a “set number” column or just the two columns $x$ and $y$ .

Key Takeaways

Six well-spaced readings are needed for a meaningful graph.
Good tabulation (headings, units, consistent precision) is assessed directly.

Common Mistakes

Fewer than six sets of readings.
No units in the headings.
Inconsistent precision (e.g. mixing $30.0\ \text{cm}$ and $31\ \text{cm}$ ).
Choosing $x$ values too close together, making the graph unreliable.

Things to Be Careful About

Always rebalance before measuring $y$ .
Ensure $x$ really changed between readings (don’t accidentally repeat the same $x$ ).
Write the table neatly; do not scatter readings around the page.

Techniques used

vary the independent variable over a suitable rangebalance the system before each readingrecord results in a table with headings and unitsuse consistent precision within each column

(c)

(i)

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

DifficultyMedium-Easy

Worked solution

Answer

Horizontal axis: $x / \text{cm}$ .
Vertical axis: $y / \text{cm}$ .
Use a sensible linear scale (at least half the grid on each axis).
Plot all six points accurately.

Final answer

Graph of y (cm) against x (cm) plotted.

Detailed explanation

Background Concept

A graph is used to reveal the relationship between two measured quantities. For a likely linear relationship, plotting $y$ against $x$ should give points close to a straight line.

Good graphing practice is assessed in Paper 3:

correct axes and labels (with units),
sensible scales (not cramped, not awkward like 3 cm = 7 units),
accurate plotting.

Understanding the Question

You must plot $y$ on the vertical axis against $x$ on the horizontal axis using your six measured pairs from part (b).

The examiner is looking for correct graph conventions and accurate plotting.

Approach

Decide the range of $x$ and $y$ from your table.
Choose linear scales so the plotted points fill most of the graph paper.
Label each axis with the symbol and unit.
Plot each point as a small cross; if using crosses, ensure they are small enough for accuracy.

Step-by-Step Reasoning

Suppose your $x$ values run from about $20\ \text{cm}$ to $70\ \text{cm}$ . Choose an axis from, say, $0$ to $80\ \text{cm}$ or from $10$ to $80\ \text{cm}$ depending on your data.
Suppose your $y$ values run from about $25\ \text{cm}$ to $52\ \text{cm}$ . Choose a y-axis range that comfortably includes these values and uses most of the grid.
Label axes clearly:
- x-axis: $x / \text{cm}$
- y-axis: $y / \text{cm}$
Plot all six points carefully. Each plotted point should correspond to a single row in your table.

Key Takeaways

Axes must be labelled with quantities and units.
Scales must be simple and should use much of the grid.
Accurate plotting is essential for a reliable gradient later.

Common Mistakes

Swapping axes (plotting $x$ on y-axis).
Missing units on axes.
Awkward scales (e.g. 1 big square = 3 cm) or scales that use only a small portion of the grid.
Plotting blobs that are too large to judge accuracy.

Things to Be Careful About

Start axes at a convenient value (not necessarily zero) but show a clear scale.
Keep the same linear scale throughout each axis.
Ensure every table pair is plotted exactly once.

Techniques used

choose sensible axis scales that use most of the gridlabel axes with quantity and unitplot experimental points accuratelyuse a sharp pencil and small plotting marks

(ii)

Draw the straight line of best fit.

DifficultyEasy

Worked solution

Answer

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

Final answer

Straight line of best fit drawn.

Detailed explanation

Background Concept

A best-fit line represents the overall trend of the data and is used to determine parameters like gradient and intercept. It should not be forced through every point because experimental data contains random error.

Understanding the Question

After plotting the six points, you must draw the straight line that best represents the linear trend.

Approach

Use a ruler to draw a single straight line.
Place it so that points are reasonably balanced: similar numbers above and below, with similar sized deviations.

Step-by-Step Reasoning

Do not join points dot-to-dot.
Do not force the line through an outlier if most points follow a different trend.
Extend the line across the full useful width of the graph to allow accurate reading of gradient and intercept.

Key Takeaways

Best-fit means “overall trend”, not “through every point”.
A long, well-placed line improves the accuracy of later calculations.

Common Mistakes

Dot-to-dot joining.
Drawing a line that goes through the first and last points only, ignoring the rest.
Drawing a very short line segment instead of a full best-fit line.

Things to Be Careful About

Use a sharp pencil and ruler.
Ensure the line is straight and not kinked.

Techniques used

draw a single straight best-fit line through the trendbalance points above and below the lineavoid joining dot-to-dot

(iii)

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

gradient = ______
y-intercept = ______

DifficultyMedium-Easy

Worked solution

Working

Using two points on the best-fit line, e.g.

$(x_1,y_1) = (20.0,25.4)\ \text{and}\ (x_2,y_2) = (70.0,51.4)$

\text{gradient} = \frac{\Delta y}{\Delta x} = \frac{51.4 - 25.4}{70.0 - 20.0} = \frac{26.0}{50.0} = 0.520

$y$ -intercept from graph at $x = 0$ :

$y = 15.0\ \text{cm}$

Answer

$\text{gradient} = 0.520$

$y$ -intercept $= 15.0\ \text{cm}$

Final answer

gradient = 0.520, y-intercept = 15.0 cm (example from best-fit line)

Detailed explanation

Background Concept

For a straight-line graph of $y$ against $x$ , the line can be written as

y = mx + c

where:

$m$ is the gradient (slope):

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

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

In practical work, you should calculate the gradient using two points on the best-fit line, not necessarily two measured points.

Understanding the Question

You must find:

the gradient of your best-fit line on the $y$ vs $x$ graph,
the y-intercept of that line.

These will be used in later parts to identify constants $A$ and $B$ .

Approach

Choose two points far apart on the best-fit line (to reduce percentage uncertainty).
Read their coordinates carefully.
Compute $m = \Delta y / \Delta x$ .
Find the intercept by reading where the line crosses the y-axis (or by substituting a point into $y = mx + c$ ).

Step-by-Step Reasoning

Pick widely separated points: using a large triangle reduces the effect of small reading errors.
Example (illustrative): points on the line at $x=20.0\ \text{cm}$ and $x=70.0\ \text{cm}$ .
Compute differences:
- $\Delta y = 51.4 - 25.4 = 26.0\ \text{cm}$
- $\Delta x = 70.0 - 20.0 = 50.0\ \text{cm}$
Gradient:

m = \frac{26.0}{50.0} = 0.520

Because both axes are in cm, the unit cancels, so the gradient is dimensionless.

y-intercept: extend the best-fit line to $x=0$ and read $y$ where it crosses; here $15.0\ \text{cm}$ .

Key Takeaways

Use two points on the best-fit line, far apart.
Gradient is $\Delta y / \Delta x$ , not $y/x$ .
Intercept is the value at $x=0$ .

Common Mistakes

Using two neighbouring points, giving a large uncertainty.
Using two raw data points that are not on the best-fit line.
Inverting the gradient (using $\Delta x/\Delta y$ ).
Forgetting that the gradient here has no unit (cm cancels).

Things to Be Careful About

Read coordinates accurately from the graph scale.
Keep consistent decimal places when reading values.
If the graph does not include $x=0$ , you must carefully extend the best-fit line to the y-axis to estimate the intercept.

Techniques used

select two widely separated points on the best-fit linecalculate gradient using \Delta y/\Delta xread the y-intercept from the graph at x = 0use consistent units when calculating slope

(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.

$A$ = ______
$B$ = ______

DifficultyEasy

Worked solution

Answer

From $y = Ax + B$ and the graph of $y$ against $x$ :

$A = \text{gradient} = 0.520$

$B = y\text{-intercept} = 15.0\ \text{cm}$

Units: $A$ has no unit; $B$ is in $\text{cm}$ .

Final answer

A = 0.520 (no unit), B = 15.0 cm

Detailed explanation

Background Concept

If a graph of $y$ (vertical axis) against $x$ (horizontal axis) is a straight line, it can be written in the form

y = mx + c

Comparing with

y = Ax + B

gives a direct correspondence:

$A$ is the gradient,
$B$ is the y-intercept.

Units:

$A$ has units of $y/x$ . If $y$ and $x$ are both in cm, $A$ is dimensionless.
$B$ has the same unit as $y$ .

Understanding the Question

You are told that $y$ and $x$ satisfy $y = Ax + B$ . Using your values from (c)(iii), you must state numerical values for $A$ and $B$ and include appropriate units.

Approach

Take $A$ directly as the gradient you found.
Take $B$ directly as the y-intercept you found.
Decide units by comparing with $y = Ax + B$ .

Step-by-Step Reasoning

From (c)(iii), suppose the gradient is $0.520$ $0.520$ .
- Then $A = 0.520$ .
From (c)(iii), suppose the y-intercept is $15.0\ \text{cm}$ $15.0 cm$ .
- Then $B = 15.0\ \text{cm}$ .
Check units:
- $Ax$ must have the same unit as $y$ .
- If $x$ and $y$ are both in cm, then $A$ must be unitless so that $A \times x$ is in cm.

Key Takeaways

Straight-line constants come from gradient and intercept.
Always consider units: gradient units are (vertical units)/(horizontal units).

Common Mistakes

Giving $A$ the unit cm.
Swapping $A$ and $B$ .
Using a gradient calculated from two data points rather than from the best-fit line.

Things to Be Careful About

If you used metres instead of cm on one axis, the numerical value and unit of $A$ would change. Always keep consistent units with your graph.

Techniques used

match the graph equation to y = Ax + Bidentify constants from gradient and interceptstate appropriate units for constants

(e)

Theory suggests that

A = \frac{2M}{3M + Q}

where $M$ is the mass of the metre rule and $Q = 0.100\text{ kg}$ .

Determine a value for $M$ .
Give your answer to three significant figures. Include an appropriate unit.

$M$ = ______

DifficultyMedium

Worked solution

Working

Given

A = \frac{2M}{3M + Q}

A(3M + Q) = 2M

3AM + AQ = 2M

(3A - 2)M = -AQ

M = \frac{AQ}{2 - 3A}

With $Q = 0.100\ \text{kg}$ and $A = 0.520$ :

M = \frac{0.520 \times 0.100}{2 - 3(0.520)} = \frac{0.0520}{0.440} = 0.118\ \text{kg}

Answer

$M = 0.118\ \text{kg}$

Final answer

0.118 kg

Detailed explanation

Background Concept

This part connects experimental results to a theoretical model. Once you have measured $A$ from the graph, you can use the theoretical expression

A = \frac{2M}{3M + Q}

to estimate the metre rule mass $M$ .

Here:

$A$ is a dimensionless constant from your graph.
$Q$ is a known mass ( $0.100\ \text{kg}$ ).
$M$ is the unknown mass of the metre rule.

Understanding the Question

You must calculate $M$ using your measured value of $A$ and the given constant $Q = 0.100\ \text{kg}$ . The answer must be:

to three significant figures,
with an appropriate unit.

Approach

Rearrange the given formula to make $M$ the subject.
Substitute numerical values for $A$ and $Q$ .
Quote $M$ to 3 s.f. with unit kg.

Step-by-Step Reasoning

Start with

A = \frac{2M}{3M + Q}

Multiply both sides by $(3M+Q)$ :

A(3M + Q) = 2M

Expand the bracket:

3AM + AQ = 2M

Collect terms in $M$ on one side:

3AM - 2M = -AQ

Factor out $M$ :

(3A - 2)M = -AQ

Divide by $(3A-2)$ (or equivalently by $(2-3A)$ ):

M = \frac{AQ}{2 - 3A}

Now substitute $Q = 0.100\ \text{kg}$ and your experimental $A$ (illustrative value $A=0.520$ ):

M = \frac{0.520 \times 0.100}{2 - 3(0.520)} = \frac{0.0520}{0.440} = 0.118\ \text{kg}

Finally, quote to three significant figures:

M = 0.118\ \text{kg}

Key Takeaways

Practical graphs often give constants that can be substituted into theory.
Rearranging equations cleanly is a key analysis skill.
Significant figures and units are part of the marks.

Common Mistakes

Algebra error when rearranging (especially sign mistakes).
Using $M = \frac{AQ}{3A-2}$ without handling the negative sign correctly.
Forgetting that $Q$ is already in kg and converting it incorrectly.
Rounding too early, leading to a final value not consistent with 3 s.f.

Things to Be Careful About

Ensure $2 - 3A$ is positive (it will be if $A < 2/3$ ); otherwise check your gradient.
Use your own value of $A$ from (c)(iii), not the illustrative value shown here.
Quote the final answer with unit kg and 3 significant figures as instructed.

Techniques used

rearrange the given equation to make M the subjectsubstitute the experimental value of Aquote the final answer to the required significant figures

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Q2Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation · Presentation of Data and Observations20M

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