Physics 9702/33 — May/June 2015

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 how the position of a suspended card varies with the distribution of masses attached to it.

(a)

(i)

Use the nail to make two holes in the card as shown in Fig. 1.1 and Fig. 1.2.

Fig. 1.1

Fig. 1.2

The holes should be approximately 1 cm from the edges of the card as shown in Fig. 1.2.

Each hole should be big enough for the card to swing freely when the nail is inserted in the hole.

DifficultyEasy

Worked solution

Answer

Make two holes at the positions shown (about $1\ \text{cm}$ from each adjacent edge). Ensure each hole is large enough that the card swings freely on the nail.

Final answer

Two holes made ~1 cm from edges; holes allow free swinging.

Detailed explanation

Background Concept

When an object is suspended from a point, it can rotate until its centre of mass lies vertically below the point of support. For this to happen reliably, the support point must act like a free pivot (minimal friction and no snagging), otherwise the card may stick and settle in an incorrect position.

Understanding the Question

You are being asked to prepare the card by making two suspension holes at specified positions. The exact positions matter because later you will draw plumb-line lines from each suspension point and find their intersection.

Approach

Use the diagram as a template to place the holes consistently: each is approximately $1\ \text{cm}$ from the top edge and $1\ \text{cm}$ from the side edge. Make the hole just big enough to rotate around the nail without rubbing.

Step-by-Step Reasoning

Measure/estimate $1\ \text{cm}$ from the top edge and $1\ \text{cm}$ from the side edge for each of the two top corners, as in Fig. 1.2.
Use the nail to pierce the card at each position.
Enlarge each hole slightly (if needed) so that, when hung, the card can swing smoothly and does not jam on the nail.

Key Takeaways

A good pivot (low friction) is essential for the card to align correctly under gravity.
The position of the holes must match the diagram so the later geometry is consistent.

Common Mistakes

Making the holes too small so the card catches on the nail and does not settle properly.
Placing holes at noticeably different distances from the edges, leading to inconsistent results.

Things to Be Careful About

Do not make the holes so large that the suspension point is ill-defined (this increases scatter in $y$ ).
Avoid tearing the card around the hole; damage can change how the card hangs.

Techniques used

follow the diagram to locate positions on the cardcreate clearance holes so the card can rotate freely about the nailcontrol the hole size to reduce friction and sticking

(ii)

Record the mass $C$ of the card shown on the base of the stand.

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

DifficultyEasy

Worked solution

Answer

$C = 12.4\ \text{g}$ (example, recorded to the balance resolution).

Final answer

C = 12.4 g (example)

Detailed explanation

Background Concept

A balance measures mass. The key skill is recording the measurement with appropriate precision: you should not claim more precision than the instrument can provide.

Understanding the Question

You must record the mass $C$ of the card as shown on the base of the stand (i.e. the balance reading). This value will later be used in calculations such as $y(C+m)$ .

Approach

Read the display carefully and copy it exactly, including the unit $\text{g}$ . Record the value to the number of decimal places shown by the balance.

Step-by-Step Reasoning

Ensure the balance is stable and reading a steady value.
Read $C$ directly from the display.
Write it down with the unit $\text{g}$ .

Key Takeaways

Precision comes from the instrument resolution.
Always include units.

Common Mistakes

Rounding incorrectly (e.g. writing $12\ \text{g}$ when the balance reads $12.4\ \text{g}$ ).
Omitting the unit.

Things to Be Careful About

If the display fluctuates, wait for it to stabilise or take a sensible average.
Do not record extra decimal places that are not shown on the balance.

Techniques used

read the mass from a balance displayrecord a measurement to the correct resolutioninclude the correct unit with the recorded value

(b)

(i)

Set up the apparatus as shown in Fig. 1.3. Suspend the card from the nail through one of the holes. Hang the plumb-line from the nail. Mark the card at a point along the plumb-line as shown in Fig. 1.3.

Fig. 1.3

DifficultyEasy

Worked solution

Answer

Suspend the card from one hole on the nail, hang the plumb-line from the same nail, allow the card to come to rest, then mark the card at a point along the plumb-line.

Final answer

Card suspended; plumb-line hung; mark made on card.

Detailed explanation

Background Concept

A plumb-line aligns itself with the gravitational field, so it shows the true vertical direction through the suspension point. When the card is at rest, the vertical line through the nail passes through the centre of mass.

Understanding the Question

You must set up the apparatus exactly as shown and create one mark on the card that lies on the vertical line (the plumb-line) passing through the suspension point.

Approach

Suspend the card freely so it can rotate, wait until it stops moving, then mark where the plumb-line lies against the card.

Step-by-Step Reasoning

Put the nail through one hole and support it on the stand.
Hang the plumb-line from the nail so it hangs beside/in front of the card.
Wait until the card stops oscillating.
Make a small clear mark on the card at a convenient point on the plumb-line.

Key Takeaways

The plumb-line provides a reference vertical.
Waiting for oscillations to die away improves accuracy.

Common Mistakes

Marking while the card is still moving.
Letting the plumb-line touch the card and be deflected sideways.

Things to Be Careful About

Ensure the card can swing freely (no rubbing on the stand/boss).
Make the mark fine (not thick), to reduce reading uncertainty later.

Techniques used

suspend an object from a pivot pointuse a plumb-line to establish the vertical directionmark a point where the plumb-line crosses the card

(ii)

Remove the card. Draw a line on the card through the hole and the mark. This line should go just over half the length of the card as shown in Fig. 1.4.

DifficultyEasy

Worked solution

Answer

Remove the card and draw a straight line through the hole and the mark. Extend the line to just over half the card length.

Final answer

Line drawn through hole and mark; extended over half the card.

Detailed explanation

Background Concept

The line you draw represents the vertical line through the suspension point. Doing this for two different suspension points gives two verticals that should intersect at the centre of mass.

Understanding the Question

You must join the suspension hole and your plumb-line mark with a straight line, and extend it far enough that it will later cross the line from the second hole.

Approach

Use a ruler to draw a thin, accurate line through the two points, and extend it beyond the midpoint of the card so the intersection is clear.

Step-by-Step Reasoning

Place the card on a flat surface.
Align a ruler through the centre of the hole and the marked point.
Draw a thin straight line and extend it to slightly more than half the card length.

Key Takeaways

Two suspension positions give two vertical lines; their intersection locates the centre of mass.

Common Mistakes

Drawing a short line so the intersection is hard to see.
Drawing a thick line that makes the intersection position uncertain.

Things to Be Careful About

Ensure the line passes through the centre of the hole (not the edge).
Keep the card flat to avoid ruler slipping.

Techniques used

draw a straight line through two marked pointsextend a construction line sufficiently for intersectionuse a ruler to improve line accuracy

(iii)

Repeat (b)(i) and (b)(ii) using the other hole in the card.

DifficultyEasy

Worked solution

Answer

Repeat the suspension and plumb-line marking using the other hole, then draw the second straight line through that hole and its corresponding mark.

Final answer

Procedure repeated with second hole; second line drawn.

Detailed explanation

Background Concept

One vertical line alone does not locate the centre of mass uniquely (it only tells you the centre of mass lies somewhere on that line). A second suspension point gives a second vertical; the centre of mass must lie on both lines, so it is at their intersection.

Understanding the Question

You must repeat parts (b)(i) and (b)(ii) using the other hole so you obtain a second line on the same card.

Approach

Perform the same careful suspension-and-marking procedure, then draw the second line accurately and ensure it crosses the first line.

Step-by-Step Reasoning

Suspend the card from the second hole and hang the plumb-line from the nail.
Wait for the card to come to rest, then make the mark.
Remove the card and draw the second straight line through the second hole and its mark.

Key Takeaways

Two lines are needed to determine a single point (intersection).

Common Mistakes

Forgetting to wait for the card to settle, giving inconsistent lines.
Mixing up marks (using the wrong mark with the wrong hole).

Things to Be Careful About

Keep the marking method consistent (same type of mark and thickness).
Ensure both lines are long enough to cross clearly.

Techniques used

repeat a measurement procedure for a second configurationuse two independent suspension points to find an intersectionreduce random error by careful repeated setup

(iv)

Measure and record the distance $y$ as shown in Fig. 1.4.

Fig. 1.4

$y$ = ______

DifficultyMedium-Easy

Worked solution

Answer

Measure the vertical distance from the top edge of the card to the intersection of the two lines.

$y = 9.7\ \text{cm}$ (example, to the nearest $1\ \text{mm}$ ).

Final answer

y = 9.7 cm (example)

Detailed explanation

Background Concept

A measured length should be taken between clearly defined reference points. Here the reference points are: (1) the top edge of the card, and (2) the intersection of the two plumb-line lines.

Understanding the Question

You have drawn two lines on the card (one from each suspension hole). Where they cross is the determined position (centre of mass). You must measure $y$ , the vertical distance from the top edge down to that intersection point, as shown in Fig. 1.4.

Approach

Locate the intersection point accurately, then use a ruler to measure vertically from the top edge to that point. Record to the ruler resolution.

Step-by-Step Reasoning

diagram-biv

Identify the point where the two drawn lines cross (use the centre of the crossing, not the edge of a thick line).
Place a ruler with its zero aligned with the top edge of the card.
Measure straight down (perpendicular to the top edge) to the intersection point.
Record $y$ with unit (commonly $\text{cm}$ ) and appropriate precision (e.g. nearest $0.1\ \text{cm}$ ).

Key Takeaways

Use the correct reference: top edge to intersection point.
Thin lines and careful reading reduce uncertainty.

Common Mistakes

Measuring from the hole instead of from the top edge.
Measuring along a slanted line rather than vertically.
Recording without a unit.

Things to Be Careful About

Avoid parallax: keep your eye directly above the ruler mark.
If the intersection is a small region (due to thick pencil lines), estimate the centre.
Record $y$ to a consistent precision to match later calculations.

Techniques used

identify the intersection point of two linesmeasure a length using a ruler with appropriate precisionrecord the measurement with a unit

(c)

(i)

Using some Blu-Tack, attach one of the 10 g slotted masses to the card. The position of the slotted mass should be half-way along the edge of the card and touching the edge as shown in Fig. 1.5.

Fig. 1.5

DifficultyEasy

Worked solution

Answer

Attach one $10\ \text{g}$ mass with Blu-Tack half-way along the edge of the card, touching the edge, as shown in Fig. 1.5.

Final answer

10 g mass attached at midpoint of edge, touching the edge.

Detailed explanation

Background Concept

Changing where mass is located changes the centre of mass of the card–mass system. To study the relationship reliably, the mass must be placed in a consistent, repeatable position each time.

Understanding the Question

You must attach a single $10\ \text{g}$ slotted mass at a specified location (half-way along the edge, touching the edge). This standardises the geometry so that only the total mass $m$ changes later.

Approach

Use Blu-Tack to hold the mass firmly, and align it carefully to the midpoint so its position does not vary between trials.

Step-by-Step Reasoning

Find the midpoint of the specified edge of the card.
Press a small amount of Blu-Tack onto the mass.
Fix the mass so it is centred on the midpoint and touches the edge, as shown.
Check it does not slip when the card is suspended.

Key Takeaways

Keep the mass position constant so $m$ is the only variable changed in part (d).

Common Mistakes

Placing the mass off-centre or not touching the edge.
Using too little Blu-Tack so the mass moves during suspension.

Things to Be Careful About

Ensure the mass is fixed at the same position when adding additional masses (stack directly above/behind).

Techniques used

attach a known mass securely using adhesiveposition the mass at a specified location on the cardensure the mass touches the edge to standardise the geometry

(ii)

Repeat (b) using the card with the mass attached.

$y$ = ______

DifficultyMedium-Easy

Worked solution

Answer

Repeat (b) with the $10\ \text{g}$ mass attached and measure $y$ .

$y = 12.1\ \text{cm}$ (example).

Final answer

y = 12.1 cm (example)

Detailed explanation

Background Concept

Adding a mass at the edge shifts the combined centre of mass, so the intersection point of the two vertical lines (and hence $y$ ) changes.

Understanding the Question

With one $10\ \text{g}$ mass attached (so $m=10\ \text{g}$ ), you repeat the same suspension-and-line procedure to find the new $y$ value.

Approach

Do exactly what you did in part (b), but ensure the added mass does not move and the card still swings freely. Then measure $y$ from the top edge to the new intersection.

Step-by-Step Reasoning

Suspend the card from the first hole, hang the plumb-line, and mark.
Draw the line through the first hole and mark.
Repeat for the second hole to get the second line.
Measure $y$ from the top edge to the intersection.

Key Takeaways

Consistency of method between trials is crucial for a meaningful graph later.

Common Mistakes

Forgetting to redraw/identify the correct intersection for the new setup.
Allowing the mass to shift position between suspensions.

Things to Be Careful About

Use the same units and precision for all $y$ values.
Ensure the plumb-line is not obstructed by the attached mass.

Techniques used

repeat the centre-of-mass line method with modified mass distributionmeasure the new intersection distance y consistentlyrecord a value with correct precision and unit

(d)

The mass attached to the card is $m$ . Increase $m$ by fixing another 10 g slotted mass on top of, or behind, the first mass.

Record $m$ and repeat (b) until you have six sets of readings of $m$ and $y$ . Include your results from (b) and (c).

Include values of $y(C + m)$ in your table.

10M

DifficultyMedium

Worked solution

Answer

Obtain 6 readings including $m=0\ \text{g}$ and $m=10\ \text{g}$ , then increase $m$ in $10\ \text{g}$ steps up to $50\ \text{g}$ ; measure $y$ each time and calculate $y(C+m)$ .

Example table (using $C = 12.4\ \text{g}$ ):

$m / \text{g}$	$y / \text{cm}$	$y(C+m) / \text{g cm}$
$0$	$9.7$	$120$
$10$	$12.1$	$271$
$20$	$13.0$	$421$
$30$	$13.4$	$568$
$40$	$13.7$	$718$
$50$	$13.9$	$867$

(Values shown are examples; candidates use their own readings.)

Final answer

Six sets of m, y with calculated y(C+m) in one table (student-dependent).

Detailed explanation

Background Concept

In practical work you must present results clearly so patterns can be analysed. A good table:

has clear column headings with quantity and unit,
has a sensible range and number of readings,
shows consistent precision,
includes any derived quantity required (here $y(C+m)$ ).

Understanding the Question

You must increase the attached mass $m$ by adding $10\ \text{g}$ masses until you have six pairs of $(m,y)$ , including the earlier cases with no added mass (from part (b), so $m=0$ ) and one mass (from part (c), so $m=10\ \text{g}$ ). You also must calculate $y(C+m)$ for each row and include it in the same table.

Approach

Use $m$ as the independent variable. Choose six values: $0,10,20,30,40,50\ \text{g}$ (or similar). For each $m$ :

measure $y$ using the same method,
compute $C+m$ ,
compute $y(C+m)$ , and record with appropriate units and s.f.

Step-by-Step Reasoning

Start with $m=0$ (no slotted mass attached) and record your measured $y$ .
Attach one $10\ \text{g}$ mass ( $m=10\ \text{g}$ ) and record $y$ .
Add additional $10\ \text{g}$ masses one at a time, keeping them stacked in the same position so only $m$ changes.
For each row calculate

y(C+m)

using your measured $y$ and your measured $C$ .
5. Record all values in one table with headings such as $m/\text{g}$ , $y/\text{cm}$ , and $y(C+m)/\text{g cm}$ .

Key Takeaways

Six readings across a range improve the reliability of the graph.
Derived quantities must be calculated correctly and consistently.

Common Mistakes

Forgetting to include the $m=0$ data from (b) or the $m=10\ \text{g}$ data from (c).
Splitting results into multiple tables.
Missing units in headings, or writing headings like “m (g)” incorrectly without the quantity symbol.
Calculating $y(C+m)$ incorrectly (e.g. $yC+m$ ).

Things to Be Careful About

Use consistent precision for $y$ across all rows (e.g. all to $0.1\ \text{cm}$ ).
Use the same unit for all masses (g) and do not mix g and kg.
When multiplying, the unit becomes $\text{g cm}$ (or $\text{g mm}$ if you used mm consistently).

Techniques used

vary the independent variable in equal steps over a suitable rangerecord repeated sets of readings in a structured tablecalculate a derived quantity for each rowmaintain consistent significant figures within each column

(e)

(i)

Plot a graph of $y(C + m)$ on the $y$ -axis against $m$ on the $x$ -axis.

DifficultyMedium-Easy

Worked solution

Answer

Plot $y(C+m)$ on the $y$ -axis against $m$ on the $x$ -axis.

Label axes: $m/\text{g}$ and $y(C+m)/\text{g cm}$ .
Use a suitable scale (at least half the grid in each direction).
Plot all six points with small, neat crosses.

Final answer

Graph of y(C+m) vs m plotted (student-dependent).

Detailed explanation

Background Concept

A graph reveals the relationship between variables. Good graph technique includes correct axis labels with units, an appropriate scale (not cramped), accurate plotting, and clear points.

Understanding the Question

You must plot a graph with:

horizontal axis: $m$
vertical axis: $y(C+m)$
using your tabulated values from (d).

Approach

Decide axis ranges that cover all your data with some margin, choose scales that use most of the grid, label with quantity/unit, then plot each data point.

Step-by-Step Reasoning

diagram-ei

On the $x$ -axis write $m/\text{g}$ and mark a scale covering your minimum to maximum $m$ .
On the $y$ -axis write $y(C+m)/\text{g cm}$ and choose a scale covering your computed values.
Plot each pair $(m,\ y(C+m))$ as a small cross.
Check for any obvious mis-plotted point by comparing with the expected trend.

Key Takeaways

Axes must be labelled with quantity and unit.
A good scale improves the precision of gradients and intercepts.

Common Mistakes

Swapping axes (plotting $m$ on the $y$ -axis).
Writing units incorrectly or missing them.
Using awkward scales (e.g. 3 squares = 7 units) that reduce accuracy.

Things to Be Careful About

Keep the same derived-unit convention throughout (e.g. if $y$ in cm, then $y(C+m)$ in $\text{g cm}$ ).
Plot with sharp pencil; large blobs reduce accuracy for best-fit lines.

Techniques used

choose appropriate axis variables and labels with unitsselect a sensible scale that uses most of the graph gridplot points accurately from a results table

(ii)

Draw the straight line of best fit.

DifficultyEasy

Worked solution

Answer

Draw a single straight line of best fit (not dot-to-dot), with roughly equal scatter of points above and below the line, extended across the full range of the plotted data.

Final answer

Straight line of best fit drawn.

Detailed explanation

Background Concept

When theory suggests a linear relationship, experimental points may scatter due to random uncertainties. A best-fit line represents the underlying trend rather than connecting every point.

Understanding the Question

After plotting your points, you must draw the best straight line representing the trend of $y(C+m)$ against $m$ .

Approach

Use a ruler to draw one straight line that is a good compromise: the points should be distributed fairly evenly above and below.

Step-by-Step Reasoning

Visually judge the overall trend of the points.
Place a ruler so that the line passes through the central tendency of the points.
Draw the straight line and extend it across the main cluster (and usually slightly beyond the extreme points).

Key Takeaways

Best-fit is about balance, not passing through every point.

Common Mistakes

Joining points dot-to-dot.
Forcing the line through the origin when it does not match the data.

Things to Be Careful About

Do not let one outlier point dominate the line position.
Use a thin pencil line to make later gradient/intercept readings more accurate.

Techniques used

judge a balanced straight line through scattered pointsignore minor scatter rather than joining dot-to-dotextend the best-fit line across the full data range

(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.

\text{gradient} = \frac{\Delta\big(y(C+m)\big)}{\Delta m}

From the best-fit line (example):

\text{gradient} \approx 15.0\ \text{cm}

$y$ -intercept (example):

\text{intercept} \approx 120\ \text{g cm}

Answer

gradient $= 15.0\ \text{cm}$

$y$ -intercept $= 120\ \text{g cm}$

Final answer

gradient = 15.0 cm, y-intercept = 120 g cm (example)

Detailed explanation

Background Concept

For a straight-line graph, the gradient is the rate of change of $y$ with $x$ :

\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 axes: here $y(C+m)$ has units $\text{g cm}$ and $m$ has units $\text{g}$ , so the gradient has units $\text{g cm} / \text{g} = \text{cm}$ .

Understanding the Question

You must find numerical values for:

the gradient of your best-fit line on the graph of $y(C+m)$ against $m$ ,
the $y$ -intercept of that best-fit line.

Approach

Pick two points that lie on the drawn best-fit line (not necessarily measured points), far apart to reduce percentage reading error. Compute gradient using changes in coordinates. Then read the intercept where the line crosses the $y$ -axis.

Step-by-Step Reasoning

Choose two well-separated points on the line (e.g. near the left and right ends of the plotted range).
Read their coordinates: $(m_1,\ Y_1)$ and $(m_2,\ Y_2)$ where $Y=y(C+m)$ .
Calculate

\text{gradient} = \frac{Y_2 - Y_1}{m_2 - m_1}

Determine the intercept by reading the value of $Y$ at $m=0$ (the crossing with the $y$ -axis).
Quote units: gradient in $\text{cm}$ and intercept in $\text{g cm}$ .

Key Takeaways

Use two points on the best-fit line, far apart.
Always compute gradient as $\Delta y / \Delta x$ .
Units of gradient come from axis units.

Common Mistakes

Using two experimental points that are close together, leading to a large gradient uncertainty.
Doing $\Delta x / \Delta y$ by mistake.
Giving gradient with the wrong unit (e.g. $\text{g cm}$ ).

Things to Be Careful About

Do not use the small triangle between adjacent grid lines; use a large triangle.
Read values carefully; a small intercept reading error can propagate into part (f).

Techniques used

calculate the gradient from two well-separated points on the best-fit lineread the y-intercept from where the line crosses the y-axisuse correct units derived from axis labels

(f)

It is suggested that the quantities $y$ , $C$ and $m$ are related by the equation

y(C + m) = Am + \frac{AB}{2}

where $A$ and $B$ are constants.

Use your answers in (e)(iii) to determine the values of $A$ and $B$ . Give appropriate units.

$A$ = ______
$B$ = ______

DifficultyMedium

Worked solution

Working

Given

y(C+m) = Am + \frac{AB}{2}

Comparing with $Y = mx + c$ for a graph of $Y=y(C+m)$ against $x=m$ :

A = \text{gradient}

\frac{AB}{2} = \text{y-intercept}

B = \frac{2(\text{y-intercept})}{A}

Using gradient $= 15.0\ \text{cm}$ and intercept $= 120\ \text{g cm}$ (example):

A = 15.0\ \text{cm}

B = \frac{2\times 120}{15.0} = 16.0\ \text{g}

Answer

$A = 15.0\ \text{cm}$

$B = 16.0\ \text{g}$

Final answer

A = 15.0 cm, B = 16.0 g (example)

Detailed explanation

Background Concept

If experimental data follow a linear form

Y = mx + c

then the gradient of a graph of $Y$ against $x$ is $m$ , and the $y$ -intercept is $c$ .

Here the suggested relationship is

y(C+m) = Am + \frac{AB}{2}

which is already in linear form if we define $Y = y(C+m)$ and $x=m$ .

Understanding the Question

You have already plotted $y(C+m)$ against $m$ and found the gradient and $y$ -intercept. You must now use those two graph features to determine the constants $A$ and $B$ , including their units.

Approach

Identify what corresponds to the gradient and intercept by comparing directly with $Y = mx + c$ .
Use the intercept expression to solve for $B$ .
Determine units from the equation: $y(C+m)$ has units $\text{cm}\cdot\text{g} = \text{g cm}$ .

Step-by-Step Reasoning

Let

Y = y(C+m)

and $x=m$ . Then

Y = A x + \frac{AB}{2}

So:

gradient $=A$
intercept $=AB/2$

Rearrange the intercept to find $B$ :

\text{intercept} = \frac{AB}{2} \Rightarrow B = \frac{2(\text{intercept})}{A}

Units:

$Y$ is in $\text{g cm}$ and $x$ is in $\text{g}$ , so gradient has units $\text{cm}$ , hence $A$ is in $\text{cm}$ .
Intercept is $\text{g cm}$ , so $AB$ has units $\text{g cm}$ ; since $A$ is $\text{cm}$ , $B$ must be $\text{g}$ .

Key Takeaways

Matching to $Y=mx+c$ is a powerful way to extract constants.
Always check units as a consistency test.

Common Mistakes

Swapping $A$ and $B$ (thinking $B$ is the gradient).
Using $AB$ instead of $AB/2$ for the intercept.
Giving incorrect units (e.g. $B$ in cm).

Things to Be Careful About

Use your own measured gradient and intercept, not the example values.
Keep consistent units for $y$ (cm or mm) throughout; units of $A$ and the intercept follow from that choice.

Techniques used

match a linear equation to y = mx + cidentify constants from gradient and interceptrearrange an equation to solve for an unknown constantdeduce units from the plotted quantities

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

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