Physics 9702/53 — October/November 2024

Cambridge A-Level · Planning, Analysis and Evaluation · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Analysis, Conclusions and Evaluation · Planning

Q115MMedium-HardPlanningAnalysis, Conclusions and Evaluation

A thin cylindrical bar magnet of length $L$ and cross-sectional area $A$ is attached to a block.
An identical magnet is attached to a trolley, as shown in Fig. 1.1.

Fig. 1.1

The trolley is held so that the separation of the N poles of the two magnets is $s$ .

Point P is a distance $D$ from the N pole of the magnet on the stationary trolley.

The trolley is released. The speed $v$ of the trolley at point P is determined using one light gate.

It is suggested that $v$ is related to $s$ by the relationship

\frac{mv^2}{2D} = \frac{KA^2B^2L^2}{s^4} - Q

where $B$ is the magnetic flux density at the N pole of one of the magnets, $m$ is the mass of the trolley, and $K$ and $Q$ are constants.

Plan a laboratory experiment to test the relationship between $v$ and $s$ .

Draw a diagram showing the arrangement of your equipment.

Explain how the results could be used to determine values for $K$ and $Q$ .

In your plan you should include:

● the procedure to be followed
● the measurements to be taken
● the control of variables
● the analysis of the data
● any safety precautions to be taken.

DifficultyMedium-Hard

Worked solution

Answer

Variables

Independent variable: separation of N poles, $s$ .
Dependent variable: trolley speed at point $P$ , $v$ .
Control variables: $D$ (position of $P$ /light gate), same two identical magnets (so $A$ , $L$ , $B$ constant), track angle/level (to keep gravitational component constant), starting from rest with no push, trolley mass $m$ constant.

Apparatus

Low-friction dynamics track and trolley.
Two identical thin cylindrical bar magnets, one fixed to a block/clamp, one fixed to trolley (N poles facing).
Light gate + data logger/timer, interrupt card of known length $\ell$ on trolley.
Metre rule / vernier calipers (for $s$ , $D$ , magnet dimensions), top-pan balance (for $m$ ).
(Optional if required) Hall probe/gaussmeter to measure $B$ at the pole.

diagram-1

Procedure and measurements

Level the track (so trolley does not drift when magnets are far apart).
Fix the block with magnet at one end. Mark point $P$ on the track a distance $D$ from the N pole of the fixed magnet and clamp the light gate at $P$ .
Measure and record $D$ .
Attach an interrupt card of length $\ell$ to the trolley; connect light gate to timer to measure the time $t$ for the card to pass.
Set an initial separation $s$ between the two N poles (measure with a rule/callipers along the track axis). Hold the trolley at this position using a card/spacer and release without a push.
Record $t$ at the light gate and calculate $v = \ell/t$ .
Repeat at least 3 times for the same $s$ and take mean $v$ .
Repeat for at least 6 different values of $s$ over a suitable range (ensuring no collision).
Measure $m$ of the trolley + attached magnet. Measure magnet length $L$ and diameter to obtain cross-sectional area $A$ . Measure $B$ with a Hall probe at a fixed position at the pole (same for both magnets).

Analysis (test the relationship and find $K$ and $Q$ )

Rearrange:

\frac{mv^2}{2D} = \frac{KA^2B^2L^2}{s^4} - Q

Let

y = \frac{mv^2}{2D} \qquad x = \frac{1}{s^4}

Then

y = (KA^2B^2L^2)x - Q

For each run, calculate $x = 1/s^4$ and $y = mv^2/(2D)$ .
Plot a graph of $y$ (vertical) against $x$ (horizontal).
A straight line supports the relationship.
Gradient $G = KA^2B^2L^2$ so

K = \frac{G}{A^2B^2L^2}

Intercept $c = -Q$ so

Q = -c

(Use a best-fit line; estimate uncertainties in $G$ and $c$ using a worst acceptable line if required.)

Safety

Strong magnets: keep fingers clear to avoid trapping; keep away from phones/cards/pacemakers.
Prevent trolley/magnets colliding by using a stop/catcher; keep face/hands clear of moving trolley.

Final answer

See working

Detailed explanation

Background Concept

The suggested relationship is

\frac{mv^2}{2D} = \frac{KA^2B^2L^2}{s^4} - Q

Here $v$ is the speed of the trolley when it reaches a fixed point $P$ a distance $D$ from the fixed magnet’s N pole. The right-hand side depends on the initial separation $s$ between the two N poles when the trolley is released.

To test a relationship experimentally, we typically:

Vary one quantity (independent variable) over a good range.
Measure the response (dependent variable) reliably, with repeats.
Keep other quantities constant (controls), so any change in the dependent variable can be attributed to the independent variable.
Convert the equation into a straight-line form $y = mx + c$ so we can use a graph to check linearity and determine constants from the gradient/intercept.

A light gate gives speed at a point by measuring how long an interrupt card blocks the beam. If the card length is $\ell$ and the blocking time is $t$ , then

v = \frac{\ell}{t}

Understanding the Question

You are asked to plan an experiment where:

you set different initial separations $s$ between the magnets,
release the trolley, and
measure its speed $v$ when it passes a fixed point $P$ .

You must also explain how to process your data to determine numerical values for the constants $K$ and $Q$ .

Crucially, the relationship contains $s^{-4}$ , so plotting $v$ directly against $s$ will not give a straight line. The clue is to use the given equation and rearrange it for a linear graph.

Approach

Choose variables: make $s$ the independent variable; measure $v$ at point $P$ as the dependent variable.
Fix the geometry: keep point $P$ the same for every run, so $D$ is constant. Clamp the light gate at $P$ .
Make other terms constant: use the same trolley and the same pair of identical magnets, with fixed orientation, so $A$ , $L$ and $B$ do not change. Keep the track level so the energy change is due to magnet interaction rather than gravity.
Collect data carefully: measure $v$ using a light gate with an interrupt card; repeat each $s$ value and average.
Linearise: define

y = \frac{mv^2}{2D}, \qquad x = \frac{1}{s^4}

so a plot of $y$ vs $x$ should be a straight line.
6. Extract constants: gradient gives $K$ (after dividing by $A^2B^2L^2$ ), and intercept gives $Q$ .

Step-by-Step Reasoning

diagram-1

Set up a low-friction track and level it. If the track is not level, the trolley gains/loses gravitational potential energy, changing $v$ in a way unrelated to $s$ .
Fix one magnet to a block/clamp so it cannot move. Attach the second identical magnet to the trolley so the N poles face each other (as in the question).
Fix point $P$ by measuring a distance $D$ from the fixed magnet’s N pole along the track and clamping the light gate at this location. Then $D$ is controlled for every run.
Measure speed $v$ at $P$ :
- attach an interrupt card of known length $\ell$ to the trolley,
- the light gate measures the time $t$ the beam is blocked,
- calculate $v = \ell/t$ .
  Repeating each run reduces random timing variation.
Vary $s$ :
- use a ruler/vernier calipers to set and measure the initial separation between the two N poles along the line of motion,
- release the trolley from rest without pushing (a thin card spacer can help you hold and release consistently).
Repeat for a range of $s$ values (e.g. 6–10 values). A wider range helps reveal whether the proposed $s^{-4}$ dependence is correct.
Measure constants needed to calculate $K$ :
- $m$ using a balance (trolley + attached magnet),
- $L$ and magnet diameter with vernier calipers so

A = \pi\left(\frac{d}{2}\right)^2

$B$ with a Hall probe/gaussmeter at a fixed, defined location at the pole (and keep that method identical for all measurements).

Linearise and plot:
For each $s$ , compute

x = \frac{1}{s^4}

and from the mean speed,

y = \frac{mv^2}{2D}

Then plot $y$ against $x$ . If the suggested model is correct, the points should lie close to a straight line.

Determine $K$ and $Q$ :
From

y = (KA^2B^2L^2)x - Q

gradient $G = KA^2B^2L^2$ , hence

K = \frac{G}{A^2B^2L^2}

y-intercept $c = -Q$ , hence

Q = -c

Uncertainty (good practice in Paper 5): estimate uncertainties in $s$ , $D$ , $\ell$ and $t$ ; propagate to $x$ and $y$ if required. For the graph, draw a best-fit line and a worst acceptable line to estimate uncertainties in gradient/intercept (and therefore in $K$ and $Q$ ).

Key Takeaways

Good plans clearly state independent/dependent/control variables and how controls are maintained.
Light gates measure speed at a point via $v = \ell/t$ .
Testing a relationship often means rearranging to $y = mx + c$ and using a straight-line graph.
Constants in a model commonly come from gradient and intercept; here $K$ comes from the gradient after dividing by known constants, and $Q$ comes from the (negative) intercept.

Common Mistakes

Plotting $v$ against $s$ directly (won’t be linear for an $s^{-4}$ law).
Forgetting to keep $D$ fixed (moving the light gate between runs changes $y$ ).
Measuring $s$ from the wrong reference points (must be N-pole to N-pole along the axis).
Releasing the trolley with a push (adds extra kinetic energy, increasing $v$ ).
Not stating how $K$ and $Q$ are obtained from the graph (must link gradient/intercept to the constants).

Things to Be Careful About

Collision risk: for small $s$ the repulsion is strong; ensure the trolley does not crash into the block and that magnets cannot detach.
Consistency of $B$ : $B$ near a magnet pole varies rapidly with position; define a fixed measurement position if measuring $B$ with a Hall probe.
Units: use SI units (e.g. $s$ and $D$ in m, $m$ in kg, $v$ in m s $^{-1}$ ) so the plotted quantities are consistent.
Graph quality: use a wide scale, plot enough points, and use large triangles for gradients; if asked for uncertainty, use worst acceptable line methods.

Techniques used

identify independent, dependent and control variablesmeasure speed using a light gate and an interrupt cardlinearise the relationship to the form y = mx + cplot a graph and determine gradient and interceptdetermine constants from gradient and intercept with consistent units

Answer

gradient $= \dfrac{3000}{3E - E_s}$

$y$ -intercept $= \dfrac{2Z}{3E - E_s}$

Final answer

gradient = 3000/(3E - Es), y-intercept = 2Z/(3E - Es)

Detailed explanation

Background Concept

To test a suggested relationship experimentally, we often rearrange it into the straight-line form

y = mx + c

where $m$ is the gradient and $c$ is the $y$ -intercept. If a graph of $y$ against $x$ is a straight line, the relationship is supported, and values of physical constants can be found from $m$ and $c$ .

Here, the suggested equation is

3E - E_s = I(3R + 2Z)

where $E$ and $E_s$ are e.m.f.s, $I$ is current, and $R$ and $Z$ are resistances.

Understanding the Question

You are told the student plots $\frac{1}{I}$ (vertical axis) against $R$ (horizontal axis). You must rearrange the given equation into an expression for $\frac{1}{I}$ in terms of $R$ , then identify:

the coefficient of $R$ (this becomes the gradient),
the constant term (this becomes the $y$ -intercept).

Also note the graph uses $R/\text{k}\Omega$ on the $x$ -axis, so the numerical gradient is different by a factor of $1000$ compared with using $R$ in $\Omega$ .

Approach

Rearrange to make $\frac{1}{I}$ the subject.
Compare with $\frac{1}{I} = mR + c$ .
Adjust $m$ if the plotted $x$ variable is $R$ in $\text{k}\Omega$ rather than $\Omega$ .

Step-by-Step Reasoning

Starting with

3E - E_s = I(3R + 2Z)

divide both sides by $I(3E - E_s)$ :

\frac{1}{I} = \frac{3R + 2Z}{3E - E_s}

Split the fraction:

\frac{1}{I} = \frac{3}{3E - E_s}R + \frac{2Z}{3E - E_s}

So if the $x$ -axis were $R$ in $\Omega$ , then

\text{gradient} = \frac{3}{3E - E_s}, \quad \text{intercept} = \frac{2Z}{3E - E_s}.

But the graph uses $x = R/\text{k}\Omega$ , so $R = 1000x$ :

\frac{1}{I} = \frac{3}{3E - E_s}(1000x) + \frac{2Z}{3E - E_s}

Hence

\text{gradient (for }x=R/\text{k}\Omega\text{)} = \frac{3000}{3E - E_s},

and the intercept is unchanged:

\text{y-intercept} = \frac{2Z}{3E - E_s}.

Key Takeaways

Rearrange into $y = mx + c$ to read off gradient and intercept.
Always check what units/scaling are used on graph axes (here $\text{k}\Omega$ introduces a factor of $1000$ ).

Common Mistakes

Giving gradient as $\frac{3}{3E - E_s}$ even though the $x$ -axis is in $\text{k}\Omega$ .
Mixing up intercept and gradient terms.

Things to Be Careful About

The intercept does not change when changing the unit of the $x$ -axis; only the gradient changes.
Keep $E_s$ and $E$ in volts and resistances consistently when later using the gradient/intercept to find constants.

Techniques used

rearrange an equation into straight-line formmatch coefficients with y = mx + caccount for axis scaling in gradient units

(b)

Values of $R$ and $I$ are given in Table 2.1.

Table 2.1

$R / \text{k}\Omega$	$I / \mu\text{A}$	$\frac{1}{I} / \text{A}^{-1}$
1.50	$194 \pm 2$
1.75	$180 \pm 2$
1.92	$172 \pm 2$
2.22	$160 \pm 2$
2.48	$150 \pm 2$
2.72	$144 \pm 2$

Calculate and record values of $\frac{1}{I} / \text{A}^{-1}$ in Table 2.1.

Include the absolute uncertainties in $\frac{1}{I}$ .

2M Similar

DifficultyMedium

Worked solution

For $y = \dfrac{1}{I}$ ,

\Delta y = \frac{\Delta I}{I^2}

(using $I$ in A and $\Delta I = 2\,\mu\text{A} = 2.0 \times 10^{-6}\,\text{A}$ )

Values:

$I = 194\,\mu\text{A}$ : $\dfrac{1}{I} = 5.15 \times 10^{3}\,\text{A}^{-1}$ , $\Delta\!\left(\dfrac{1}{I}\right)= 5.3 \times 10^{1}\,\text{A}^{-1}$
$I = 180\,\mu\text{A}$ : $\dfrac{1}{I} = 5.56 \times 10^{3}\,\text{A}^{-1}$ , $\Delta\!\left(\dfrac{1}{I}\right)= 6.2 \times 10^{1}\,\text{A}^{-1}$
$I = 172\,\mu\text{A}$ : $\dfrac{1}{I} = 5.81 \times 10^{3}\,\text{A}^{-1}$ , $\Delta\!\left(\dfrac{1}{I}\right)= 6.8 \times 10^{1}\,\text{A}^{-1}$
$I = 160\,\mu\text{A}$ : $\dfrac{1}{I} = 6.25 \times 10^{3}\,\text{A}^{-1}$ , $\Delta\!\left(\dfrac{1}{I}\right)= 7.8 \times 10^{1}\,\text{A}^{-1}$
$I = 150\,\mu\text{A}$ : $\dfrac{1}{I} = 6.67 \times 10^{3}\,\text{A}^{-1}$ , $\Delta\!\left(\dfrac{1}{I}\right)= 8.9 \times 10^{1}\,\text{A}^{-1}$
$I = 144\,\mu\text{A}$ : $\dfrac{1}{I} = 6.94 \times 10^{3}\,\text{A}^{-1}$ , $\Delta\!\left(\dfrac{1}{I}\right)= 9.6 \times 10^{1}\,\text{A}^{-1}$

Answer

$\dfrac{1}{I}$ values and absolute uncertainties as above (to be entered in Table 2.1).

Final answer

1/I = 5.15e3, 5.56e3, 5.81e3, 6.25e3, 6.67e3, 6.94e3 A^-1 with uncertainties ±5.3e1, ±6.2e1, ±6.8e1, ±7.8e1, ±8.9e1, ±9.6e1 A^-1

Detailed explanation

Background Concept

If a measured quantity $I$ has an absolute uncertainty $\Delta I$ , then for a derived quantity

y = \frac{1}{I}

the uncertainty can be found using differentiation (or the standard rule for powers):

\Delta y \approx \left|\frac{\mathrm{d}}{\mathrm{d}I}\left(I^{-1}\right)\right|\Delta I = \frac{\Delta I}{I^2}.

This gives an absolute uncertainty in the same units as $y$ (here, $\text{A}^{-1}$ ).

Understanding the Question

The table gives $I$ in $\mu\text{A}$ with $\pm 2\,\mu\text{A}$ . You must:

convert each $I$ into amperes,
calculate $1/I$ in $\text{A}^{-1}$ ,
calculate the absolute uncertainty in $1/I$ using $\Delta(1/I)=\Delta I/I^2$ .

Approach

For each row:

Use $I = (\text{value in }\mu\text{A}) \times 10^{-6}\,\text{A}$ .
Compute $1/I$ .
Use $\Delta I = 2.0 \times 10^{-6}\,\text{A}$ and calculate $\Delta(1/I)$ .

Step-by-Step Reasoning

Example for $I = 194\,\mu\text{A}$ :

Convert:

I = 194 \times 10^{-6}\,\text{A} = 1.94 \times 10^{-4}\,\text{A}

Reciprocal:

\frac{1}{I} = \frac{1}{1.94 \times 10^{-4}} = 5.1546 \times 10^3\,\text{A}^{-1}

Uncertainty (with $\Delta I = 2.0 \times 10^{-6}\,\text{A}$ ):

\Delta\!\left(\frac{1}{I}\right) = \frac{2.0 \times 10^{-6}}{(1.94 \times 10^{-4})^2} = 5.315 \times 10^{1}\,\text{A}^{-1}

Repeat the same steps for each current value. Notice that as $I$ gets smaller, $1/I$ gets larger and its uncertainty also gets larger (because $\Delta(1/I) \propto 1/I^2$ ).

Key Takeaways

Always convert to SI units before calculating.
For reciprocals, absolute uncertainty is $\Delta(1/I) = \Delta I / I^2$ .
Smaller currents lead to larger uncertainty in $1/I$ .

Common Mistakes

Using $\Delta(1/I) = \Delta I / I$ (that is a fractional uncertainty idea, not correct for the absolute uncertainty here).
Forgetting to convert $\mu\text{A}$ to A, giving answers off by a factor of $10^6$ .
Writing uncertainties as percentages when the question asks for absolute uncertainties.

Things to Be Careful About

Quote $1/I$ in $\text{A}^{-1}$ (not $\mu\text{A}^{-1}$ ).
Keep rounding sensible: uncertainties to 1 or 2 significant figures, and the value of $1/I$ to a matching decimal place/significant figure.

Techniques used

convert microampere readings into amperescalculate reciprocals of measured quantitiespropagate uncertainty through a reciprocal using \Delta(1/I) = \Delta I / I^2quote absolute uncertainties to appropriate significant figures

(c)

(i)

Plot a graph of $\frac{1}{I} / \text{A}^{-1}$ against $R / \text{k}\Omega$ . Include error bars for $\frac{1}{I}$ .

Graph 2.1

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Plot points of $\frac{1}{I}\, /\, \text{A}^{-1}$ against $R/\text{k}\Omega$ from Table 2.1.

Draw vertical error bars of magnitude $\pm\Delta\!\left(\frac{1}{I}\right)$ for each point.

Final answer

See graph (points plotted with vertical error bars).

Detailed explanation

Background Concept

A graph is used to test whether two quantities are linearly related. Experimental uncertainty should be shown using error bars so that later judgments (such as a worst acceptable line) are justified.

In this question:

$x$ is $R/\text{k}\Omega$ .
$y$ is $1/I$ in $\text{A}^{-1}$ .
Only $y$ has uncertainties provided (from $I \pm 2\,\mu\text{A}$ ), so only vertical error bars are required.

Understanding the Question

You must transfer your calculated $1/I$ values (and their absolute uncertainties) onto Graph 2.1 by plotting:

each $(R, 1/I)$ point,
an error bar for each point extending up and down by the absolute uncertainty in $1/I$ .

Approach

Choose a sensible scale (Graph 2.1 is already scaled).
Plot each point carefully.
For each point, draw a vertical line centered on the point with total height $2\Delta(1/I)$ .

Step-by-Step Reasoning

Read $R$ directly in $\text{k}\Omega$ from the table (e.g. $1.50, 1.75, \dots$ ).
Use your computed $1/I$ in $\text{A}^{-1}$ .
For each, draw the error bar from

\left(\frac{1}{I} - \Delta\!\left(\frac{1}{I}\right)\right) \text{ to } \left(\frac{1}{I} + \Delta\!\left(\frac{1}{I}\right)\right).

Key Takeaways

Correct units and labels are essential: $R/\text{k}\Omega$ and $1/I\,/\,\text{A}^{-1}$ .
Error bars represent uncertainty and will be used to judge the range of acceptable straight lines.

Common Mistakes

Plotting $I$ instead of $1/I$ .
Plotting $R$ in $\Omega$ when the axis is $\text{k}\Omega$ .
Drawing error bars of incorrect size (e.g. using $\pm 2\,\mu\text{A}$ directly on the $1/I$ axis).

Things to Be Careful About

Use a sharp pencil and plot points to within half a small square.
Error bars must be centered on the plotted point and must be vertical (since $x$ uncertainties are not given).

Techniques used

plot calculated values with correct axis labels and unitsadd vertical error bars from absolute uncertaintiesuse an appropriate scale that uses most of the graph grid

(ii)

Draw the straight line of best fit and a worst acceptable straight line on your graph. Label both lines.

2M Similar

DifficultyMedium

Worked solution

Answer

Draw a straight line of best fit through the plotted points.

Draw a worst acceptable straight line (steepest or shallowest) that still passes through all the error bars.

Label the lines "best fit" and "worst".

Final answer

See graph (best-fit line and worst acceptable line drawn and labelled).

Detailed explanation

Background Concept

A best-fit line represents the most likely linear relationship between $x$ and $y$ . Because data have uncertainty, many straight lines might be consistent with the measurements. A worst acceptable line is chosen to estimate uncertainty in gradient and intercept:

It must still be consistent with the data within error bars.
It should be the steepest or shallowest line that is still acceptable.

Understanding the Question

After plotting points with error bars, you must:

draw the best-fit straight line,
draw one worst acceptable straight line,
label both, because later parts use them to calculate uncertainties in gradient and intercept.

Approach

Best-fit line: balance the points (roughly equal scatter above and below, not forced through the origin unless justified).
Worst acceptable line: rotate the line to make it as steep (or as shallow) as possible while still intersecting every vertical error bar.

Step-by-Step Reasoning

Use a ruler to draw the best-fit line through the trend of the points.
For the worst acceptable line:
- pick two points near the ends of the graph range,
- ensure the line goes through the top of one end error bar and the bottom of the other end error bar (for maximum change in slope),
- check it still intersects all intermediate error bars.
Label each line clearly on the graph.

Key Takeaways

Worst acceptable line is defined by error bars, not by how far it is from the best-fit line.
The uncertainty in gradient/intercept comes from comparing best-fit with worst-fit.

Common Mistakes

Drawing a second line that does not pass through all error bars.
Choosing a line that is only slightly different from the best-fit (not truly “worst”).
Forgetting to label the two lines.

Things to Be Careful About

Use end points far apart to make gradient calculations reliable.
Ensure the worst acceptable line is still straight and consistent across the whole data range.

Techniques used

draw a straight line of best fit through scattered datachoose a worst acceptable line consistent with all error barslabel best-fit and worst-fit lines clearly

(iii)

Determine the gradient of the line of best fit. Include the absolute uncertainty in your answer.

gradient = ______

2M Similar

DifficultyMedium

Worked solution

Working

Using two points on the best-fit line (e.g. near the ends):

m = \frac{\Delta(1/I)}{\Delta R}

Example:

m = \frac{6940 - 5150}{2.72 - 1.50} \approx 1.47 \times 10^{3}\ \text{A}^{-1}\ \text{k}\Omega^{-1}

Worst acceptable gradient (from worst line) gives about $m_w \approx 1.59 \times 10^{3}\ \text{A}^{-1}\ \text{k}\Omega^{-1}$ (or shallowest $\approx 1.34 \times 10^{3}$ ).

\Delta m = |m_w - m| \approx 0.12 \times 10^{3}\ \text{A}^{-1}\ \text{k}\Omega^{-1}

Answer

m = (1.47 \pm 0.12) \times 10^{3}\ \text{A}^{-1}\ \text{k}\Omega^{-1}

Final answer

(1.47 ± 0.12) × 10^3 A^-1 kΩ^-1

Detailed explanation

Background Concept

For a straight-line graph, the gradient is

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

To reduce the effect of reading uncertainty, you should use two points far apart on the line (a large triangle). In Paper 5 analysis, uncertainty in gradient is usually estimated using a worst acceptable line:

\Delta m = |m_{\text{worst}} - m_{\text{best}}|.

Understanding the Question

You have drawn:

a best-fit line,
a worst acceptable line.

You must find the best-fit gradient and include an absolute uncertainty based on the difference between the gradients of these two lines.

Approach

Choose two well-separated points on the best-fit line (not necessarily data points).
Compute $m_{\text{best}}$ .
Choose two well-separated points on the worst line and compute $m_{\text{worst}}$ .
Uncertainty is the absolute difference.

Step-by-Step Reasoning

Read two points on the best-fit line, for example near $R=1.50\,\text{k}\Omega$ and $R=2.72\,\text{k}\Omega$ .
Calculate:

m_{\text{best}} = \frac{y_2 - y_1}{x_2 - x_1}.

Repeat for worst line (using points on that line).
Then:

\Delta m = |m_{\text{worst}} - m_{\text{best}}|.

This is why part (c)(ii) is important: your chosen worst line controls the uncertainty you obtain.

Key Takeaways

Use a large triangle for gradients.
The uncertainty in gradient is obtained by comparison with the worst acceptable line.

Common Mistakes

Using two nearby points, giving a large percentage error.
Calculating $\Delta x/\Delta y$ instead of $\Delta y/\Delta x$ .
Taking half the difference between gradients (Cambridge typically expects the full difference here).

Things to Be Careful About

Keep units as $\text{A}^{-1}\,\text{k}\Omega^{-1}$ because the $x$ -axis is in $\text{k}\Omega$ .
Ensure you read from the line, not from the plotted points unless they lie exactly on the line.

Techniques used

calculate gradient from two well-separated points on a straight lineuse a large triangle to reduce reading uncertaintydetermine gradient uncertainty from best-fit vs worst-fit gradients

(iv)

Determine the $y$ -intercept of the line of best fit. Include the absolute uncertainty in your answer.

$y$ -intercept = ______

2M Similar

DifficultyMedium

Worked solution

Working

From the best-fit line, $y$ -intercept $c \approx 2.96 \times 10^{3}\ \text{A}^{-1}$ .

From the worst acceptable line, $c_w \approx 2.72 \times 10^{3}\ \text{A}^{-1}$ (or $3.19 \times 10^{3}\ \text{A}^{-1}$ ).

\Delta c = |c_w - c| \approx 0.24 \times 10^{3}\ \text{A}^{-1}

Answer

c = (2.96 \pm 0.24) \times 10^{3}\ \text{A}^{-1}

Final answer

(2.96 ± 0.24) × 10^3 A^-1

Detailed explanation

Background Concept

For a straight line

y = mx + c,

the $y$ -intercept $c$ is the value of $y$ when $x=0$ . On a graph, it is where the line crosses the $y$ -axis.

Because the line is determined from uncertain data, the intercept also has uncertainty. As with the gradient, Paper 5 commonly estimates this by comparing the best-fit and worst acceptable lines:

\Delta c = |c_{\text{worst}} - c_{\text{best}}|.

Understanding the Question

You must read the intercept of the best-fit line and include an absolute uncertainty. The uncertainty comes from how much the intercept changes when using the worst acceptable line.

Approach

Extend the best-fit line to the $y$ -axis (at $R=0$ ).
Read the intercept $c_{\text{best}}$ .
Extend the worst acceptable line to the $y$ -axis and read $c_{\text{worst}}$ .
Compute $\Delta c = |c_{\text{worst}} - c_{\text{best}}|$ .

Step-by-Step Reasoning

Even if $R=0$ is not on the grid, you can extrapolate the straight line back to the $y$ -axis.
Read carefully using the grid scale.
The worst line generally gives either a larger or smaller intercept than the best line.
Take the difference as the absolute uncertainty.

Key Takeaways

Intercept is found at $x=0$ .
Uncertainty is determined by comparing best-fit and worst-fit intercepts.

Common Mistakes

Reading the intercept at the smallest $R$ value instead of at $R=0$ .
Forgetting to include an uncertainty.

Things to Be Careful About

Extrapolation should be along the straight line you have drawn (use a ruler).
Keep units: intercept is in $\text{A}^{-1}$ .

Techniques used

read the y-intercept from a straight-line graphestimate intercept uncertainty using best-fit vs worst-fit linesquote intercept with an absolute uncertainty

(d)

(i)

Using your answers to (a), (c)(iii) and (c)(iv), determine the values of $E$ and $Z$ . Include appropriate units.

Data: $E_s = (2.20 \pm 0.05)\ \text{V}$

$E$ = ______
$Z$ = ______

2M Similar

DifficultyMedium-Hard

Worked solution

Using (a) with $R/\text{k}\Omega$ on $x$ -axis:

m = \frac{3000}{3E - E_s}

3E - E_s = \frac{3000}{m}

With $m = 1.47 \times 10^{3}\ \text{A}^{-1}\ \text{k}\Omega^{-1}$ :

3E - E_s = \frac{3000}{1.47 \times 10^{3}} = 2.04\ \text{V}

E = \frac{E_s + 2.04}{3} = \frac{2.20 + 2.04}{3} = 1.41\ \text{V}

Intercept $c$ :

c = \frac{2Z}{3E - E_s}

With $c = 2.96 \times 10^{3}\ \text{A}^{-1}$ and $3E-E_s = 2.04\ \text{V}$ :

Z = \frac{c(3E - E_s)}{2} = \frac{(2.96 \times 10^{3})(2.04)}{2} = 3.02 \times 10^{3}\ \Omega

Answer

$E = 1.41\ \text{V}$

$Z = 3.02 \times 10^{3}\ \Omega$

Final answer

E = 1.41 V, Z = 3.02 × 10^3 Ω

Detailed explanation

Background Concept

Once a straight-line graph has been made, the measured gradient $m$ and intercept $c$ can be matched to theoretical expressions to determine unknown constants.

Here, after linearising, we have

\frac{1}{I} = mx + c

with $x = R/\text{k}\Omega$ . From part (a):

m = \frac{3000}{3E - E_s}, \quad c = \frac{2Z}{3E - E_s}.

These allow you to find $3E - E_s$ first from $m$ , then $E$ , and finally $Z$ from $c$ .

Understanding the Question

You are given:

$E_s = 2.20\ \text{V}$ ,
your measured gradient $m$ and intercept $c$ from the graph.

You must determine:

the cell e.m.f. $E$ (in V),
the resistor value $Z$ (in $\Omega$ ).

Approach

Use $m = 3000/(3E - E_s)$ to calculate $(3E - E_s)$ .
Rearrange to get $E$ .
Use $c = 2Z/(3E - E_s)$ to get $Z$ .

Step-by-Step Reasoning

From

m = \frac{3000}{3E - E_s}

rearrange:

3E - E_s = \frac{3000}{m}.

Substitute your gradient to obtain a numerical value for $(3E - E_s)$ in volts. Then

3E = E_s + (3E - E_s)

E = \frac{E_s + (3E - E_s)}{3}.

Finally, from

c = \frac{2Z}{3E - E_s}

rearrange:

Z = \frac{c(3E - E_s)}{2}.

Note: $c$ has units $\text{A}^{-1}$ and $(3E - E_s)$ has units V, so $c(3E - E_s)$ has units $\text{V A}^{-1} = \Omega$ , as required.

Key Takeaways

Gradient gives $(3E - E_s)$ directly.
Intercept then gives $Z$ once $(3E - E_s)$ is known.
Dimensional checking confirms the algebra ( $\text{V}/\text{A} = \Omega$ ).

Common Mistakes

Using $m = 3/(3E - E_s)$ (forgetting the factor $1000$ from $\text{k}\Omega$ ).
Substituting $E_s$ with the wrong sign.
Not giving units for $E$ and $Z$ .

Things to Be Careful About

Ensure your gradient unit matches your formula (because $x$ was $R/\text{k}\Omega$ ).
Keep consistent significant figures (usually 3 s.f. is acceptable for derived constants).

Techniques used

equate experimental gradient and intercept to theoretical expressionssolve simultaneous expressions for unknown constantsconvert between k\Omega and \Omega where needed

(ii)

Determine the absolute uncertainty in $E$ .

absolute uncertainty in $E$ = ______

1M Similar

DifficultyMedium

Worked solution

From (d)(i):

A = (3E - E_s) = \frac{3000}{m}

Fractional uncertainty:

\frac{\Delta A}{A} = \frac{\Delta m}{m}

With $m = 1.47\times 10^{3}$ and $\Delta m = 0.12\times 10^{3}$ :

\Delta A = 2.04\times\frac{0.12}{1.47} = 0.17\ \text{V}

Since

E = \frac{E_s + A}{3}

\Delta E = \frac{\Delta E_s + \Delta A}{3} = \frac{0.05 + 0.17}{3} = 0.07\ \text{V}

Answer

absolute uncertainty in $E = 0.07\ \text{V}$

Final answer

0.07 V

Detailed explanation

Background Concept

Uncertainty propagation rules commonly used at A level:

For $y = 1/x$ , the fractional uncertainties are equal:

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

For $y = x + a$ , absolute uncertainties add:

\Delta y = \Delta x + \Delta a.

For $y = x/n$ (division by an exact constant $n$ ), the absolute uncertainty scales:

\Delta y = \frac{\Delta x}{n}.

Understanding the Question

You have found $E$ using

E = \frac{E_s + A}{3}, \quad A = \frac{3000}{m}.

You are asked for the absolute uncertainty in $E$ . The uncertainty in $E$ comes from:

uncertainty in the gradient $m$ (from best vs worst line),
uncertainty in $E_s$ (given as $\pm 0.05\,\text{V}$ ).

Approach

Find uncertainty in $A = 3000/m$ using fractional uncertainty in $m$ .
Combine $\Delta A$ with $\Delta E_s$ as a sum.
Divide by 3 to get $\Delta E$ .

Step-by-Step Reasoning

Because

A = \frac{3000}{m},

and $3000$ is exact (from unit conversion), the fractional uncertainty is

\frac{\Delta A}{A} = \frac{\Delta m}{m}.

\Delta A = A\frac{\Delta m}{m}.

Next,

E = \frac{E_s + A}{3}.

The uncertainty in the numerator is

\Delta(E_s + A) = \Delta E_s + \Delta A,

\Delta E = \frac{\Delta E_s + \Delta A}{3}.

Key Takeaways

Use fractional uncertainties for inverse relationships.
Use absolute uncertainty addition for sums.
Division by an exact number divides the absolute uncertainty by that number.

Common Mistakes

Adding fractional uncertainties when the relationship is a sum.
Forgetting to include the uncertainty in $E_s$ .
Using $\Delta A = \Delta m/m^2$ (that would be a calculus absolute method but must still be applied correctly; the fractional method is simpler here).

Things to Be Careful About

Make sure $m$ and $\Delta m$ are in the same units.
Quote the final uncertainty to 1 significant figure (or 2 if it begins with 1 or 2); $0.07\,\text{V}$ is appropriate.

Techniques used

propagate uncertainty through an inverse relationship using fractional uncertaintiescombine absolute uncertainties in a sumscale uncertainty through division by a constant

(e)

The experiment is repeated. Determine the resistance $R$ that gives a value of $I$ of $250\ \mu\text{A}$ .

$R$ = ______ $\Omega$

1M Similar

DifficultyMedium-Easy

Worked solution

Using $\dfrac{1}{I} = m\left(\dfrac{R}{\text{k}\Omega}\right) + c$ .

For $I = 250\,\mu\text{A}$ :

\frac{1}{I} = \frac{1}{250\times 10^{-6}} = 4.00\times 10^{3}\ \text{A}^{-1}

\frac{R}{\text{k}\Omega} = \frac{4.00\times 10^{3} - 2.96\times 10^{3}}{1.47\times 10^{3}} = 0.711

R = 0.711\ \text{k}\Omega = 7.11\times 10^{2}\ \Omega

Answer

R = 7.1 \times 10^{2}\ \Omega

Final answer

7.1 × 10^2 Ω

Detailed explanation

Background Concept

Once you have a linear relationship

y = mx + c,

you can predict the value of $x$ needed to produce a desired $y$ by rearranging:

x = \frac{y - c}{m}.

In this question, the graph relates $y = 1/I$ to $x = R/\text{k}\Omega$ .

Understanding the Question

You are asked for the resistance $R$ that would give a current of $250\,\mu\text{A}$ . Using the straight-line relationship from the graph (your best-fit line), you:

convert $I$ to $1/I$ ,
use the line equation to find $R/\text{k}\Omega$ ,
convert to $\Omega$ .

Approach

Calculate $y = 1/I$ for $I = 250\,\mu\text{A}$ .
Substitute into $1/I = m(R/\text{k}\Omega) + c$ .
Solve for $R$ .

Step-by-Step Reasoning

Compute the reciprocal current:

I = 250\times 10^{-6}\,\text{A} \Rightarrow \frac{1}{I} = 4.00\times 10^{3}\,\text{A}^{-1}.

Now rearrange the line equation:

\frac{R}{\text{k}\Omega} = \frac{\frac{1}{I} - c}{m}.

Substitute the numerical values of $m$ and $c$ from your best-fit line to obtain $R/\text{k}\Omega$ , then multiply by $1000$ to convert $\text{k}\Omega$ to $\Omega$ .

Key Takeaways

A straight-line graph lets you interpolate/extrapolate to find values not directly measured.
Here, you must convert from current to $1/I$ before using the graph equation.

Common Mistakes

Using $I$ instead of $1/I$ .
Forgetting that the $x$ -axis is in $\text{k}\Omega$ and giving $R$ too large by a factor of $1000$ .

Things to Be Careful About

Keep consistent significant figures (typically 2 s.f. is fine for a value derived from a graph).
If your own $m$ and $c$ differ slightly from the representative values here, your $R$ will differ correspondingly; method marks come from correct substitution and rearrangement.

Techniques used

use the straight-line equation to predict an unmeasured valuesubstitute a target y-value and solve for xconvert between k\Omega and \Omega

Physics 9702/53 — October/November 2024

Answer

Variables

Apparatus

Procedure and measurements

Analysis (test the relationship and find KKK and QQQ)

Safety

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Analysis (test the relationship and find $K$ and $Q$ )