Physics 9702/51 — 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

diagram-1

Procedure and measurements

Set up a low-friction, level track with the fixed magnet rigidly clamped at one end and the second identical magnet on a trolley, with like poles facing.
Place a light gate at point $P$ and connect it to a timer/data logger.
Attach an interrupt card (flag) of measured length $l$ to the trolley.
Measure and keep constant the distance $D$ from the fixed magnet’s N pole to the light gate at $P$ .
Set the initial separation $s$ (distance between the two N poles) using a ruler/vernier and spacers; hold the trolley at this position and release it without a push.
Record the time $t$ for the card to pass the light gate and calculate

v = \frac{l}{t}

Repeat at least 3 times for each $s$ and take the mean $v$ .
Change $s$ over a wide range (at least 6 values) and repeat.
Measure $m$ of the trolley (balance). Measure $L$ and diameter to find $A$ (e.g. with calipers, $A=\pi r^2$ ). Measure $B$ at the pole with a Hall probe/teslameter (same position/orientation each time).

Control of variables

Use the same two magnets throughout (so $A$ , $L$ , and $B$ are constant) and keep temperature similar.
Keep $D$ fixed; keep magnets coaxial and at the same height.
Ensure track is horizontal; keep friction as constant as possible.
Use the same release method each time (no initial push).

Analysis (to find $K$ and $Q$ )

From

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

calculate for each run:

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

Plot $y$ against $x$ . A straight line is expected:

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

Gradient $G = KA^2B^2L^2$ and y-intercept $c=-Q$ .

K = \frac{G}{A^2B^2L^2} \qquad Q = -c

Safety

Keep fingers clear when bringing magnets close (pinch hazard); do not allow magnets to snap together.
Keep magnets away from sensitive electronics/magnetic media and from people with pacemakers.
Use an end-stop so the trolley does not run off the bench.

Final answer

See working

Detailed explanation

Background Concept

The suggested relationship links the trolley’s speed $v$ at a fixed point $P$ to the initial separation $s$ between the like poles of two identical magnets. The left-hand side

\frac{mv^2}{2D}

uses the trolley’s kinetic energy term $\tfrac{1}{2}mv^2$ scaled by the fixed distance $D$ to point $P$ . The right-hand side contains a term proportional to $\tfrac{1}{s^4}$ and a constant offset $Q$ .

A key Paper 5 skill is to test an equation by rearranging it into a straight-line form $y = mx + c$ , so that a graph can be plotted and constants can be obtained from the gradient and intercept.

Understanding the Question

You have:

two identical bar magnets (length $L$ , cross-sectional area $A$ ), one fixed, one on a trolley,
initial separation of the N poles $s$ that you can vary,
a fixed point $P$ at distance $D$ from the fixed magnet’s N pole,
a method to measure the trolley’s speed $v$ at $P$ using one light gate.

You are asked to plan an experiment to test how $v$ depends on $s$ and to show how to obtain numerical values of the constants $K$ and $Q$ .

Approach

Choose variables: vary $s$ (independent), measure $v$ at $P$ (dependent), keep $D$ and all magnet properties constant.
Collect data: for each chosen $s$ , release the trolley from rest and measure $v$ at the light gate.
Linearise the provided equation by defining

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

so that plotting $y$ vs $x$ should give a straight line.
4. Extract constants: use gradient and intercept to determine $K$ and $Q$ .

Step-by-Step Reasoning

diagram-1

1) Setting up and measuring speed

A light gate measures the time $t$ for an interrupt card of length $l$ to pass through. The instantaneous speed at the gate is then

v = \frac{l}{t}

So you must:

fix the light gate at point $P$ ,
measure $l$ with a ruler/calipers,
ensure the flag passes cleanly through the gate each time.

2) Choosing values of $s$

You need enough different $s$ values to see a trend and produce a meaningful line (typically $\ge 6$ values). Because the equation involves $1/s^4$ , small changes in $s$ at small separations produce large changes in the force/energy, so:

use spacers or marked positions to set $s$ repeatably,
measure $s$ between the N poles as stated, not between magnet ends or trolley edges.

3) Controlling variables

The equation assumes $A$ , $B$ , $L$ , $m$ and $D$ are constants.

$m$ : keep the trolley unchanged; measure $m$ once with a balance.
$D$ : keep the light gate fixed relative to the fixed magnet’s N pole.
$A$ and $L$ : use the same magnets throughout and measure their dimensions.
$B$ : use the same magnet(s) and avoid heating; if measuring $B$ with a Hall probe, always measure at the same location/orientation.
reduce unwanted effects: keep the track level, keep magnets aligned coaxially, and release without pushing.

4) Linearising and plotting

Start from

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

Define

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

Then

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

So a plot of $y$ (vertical axis) against $x$ (horizontal axis) should give a straight line.

If the points form a straight line, the relationship is supported.
If it is not straight, the suggested model is not consistent with the measurements (or controls are poor).

5) Finding $K$ and $Q$

From the straight-line graph:

gradient $G = KA^2B^2L^2$
intercept $c = -Q$

Therefore

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

and

Q = -c

If uncertainties are required/desired, you would add error bars (notably from $s$ and timing) and draw best-fit and worst-acceptable lines to estimate uncertainties in $G$ and $c$ .

Key Takeaways

A good plan clearly identifies independent, dependent and controlled variables.
Use a light gate + interrupt card to measure speed reliably: $v=l/t$ .
Turn the given model into a straight-line form by choosing
$y=\dfrac{mv^2}{2D}$ and $x=\dfrac{1}{s^4}$ .
Determine constants from a graph: gradient gives $K$ (after dividing by known factors) and the intercept gives $Q$ .

Common Mistakes

Plotting $v$ against $s$ directly (this will not generally be a straight line).
Forgetting that the graph intercept is $-Q$ (so $Q$ is the negative of the intercept).
Not keeping $D$ fixed, or measuring $D$ from the wrong point (must be from the fixed magnet’s N pole to the light gate).
Measuring $s$ between the wrong reference points (must be N pole to N pole as defined).
Releasing the trolley with an extra push, changing the initial kinetic energy.

Things to Be Careful About

Because $x=1/s^4$ , small percentage errors in $s$ produce large percentage errors in $x$ ; measure $s$ as accurately as possible and use a wide range of $s$ .
Ensure the trolley passes the light gate at the same height and without rubbing (otherwise $t$ is unreliable).
Use a large triangle when calculating gradient; do not use two adjacent points.
Keep magnets aligned; misalignment changes the effective magnetic interaction and makes results scatter.
Safety: magnets can snap together suddenly and cause injury or damage; also control the trolley motion with an end stop.

Techniques used

identify independent, dependent and controlled variablesmeasure speed using a light gate and interrupt cardlinearise the given relationship by selecting suitable plotted variablesdetermine constants from the gradient and y-intercept of a best-fit linereduce random error by repeating readings and taking a mean

Working

From

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

\frac{1}{I} = \frac{3R + 2Z}{3E - E_s} = \left(\frac{3}{3E - E_s}\right)R + \left(\frac{2Z}{3E - E_s}\right)

Answer

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

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

Final answer

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

Detailed explanation

Background Concept

To analyse experimental data, we often try to convert a relationship into the straight-line form

y = mx + c

where $m$ is the gradient and $c$ is the $y$ -intercept. If we can rewrite the given physics equation so that the measured quantities appear as $y$ and $x$ , then we can read off expressions for $m$ and $c$ .

Understanding the Question

You are told the suggested relationship between current $I$ and resistance $R$ is

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

A graph is plotted of $\frac{1}{I}$ (vertical axis) against $R$ (horizontal axis). So we want an equation of the form

\frac{1}{I} = (\text{something})\,R + (\text{something}).

The “something” multiplying $R$ is the gradient, and the constant term is the $y$ -intercept.

Approach

Start with the given equation.
Divide both sides by $I(3E - E_s)$ to make $1/I$ the subject.
Expand into a term proportional to $R$ plus a constant.
Compare directly with $y = mx + c$ .

Step-by-Step Reasoning

Starting equation:

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

Divide both sides by $I$ :

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

Now divide by $(3E - E_s)$ :

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

Split the fraction into two terms:

\frac{1}{I} = \left(\frac{3}{3E - E_s}\right)R + \left(\frac{2Z}{3E - E_s}\right)

Comparing with $y = mx + c$ , with $y = 1/I$ and $x = R$ :

gradient $m = \dfrac{3}{3E - E_s}$
intercept $c = \dfrac{2Z}{3E - E_s}$

Key Takeaways

Linearising means rewriting the physics in $y = mx + c$ form.
The coefficient of the plotted $x$ -quantity is the gradient.
The constant term is the $y$ -intercept.

Common Mistakes

Leaving the equation as $I$ against $R$ instead of making $1/I$ the subject.
Forgetting to split $\dfrac{3R + 2Z}{3E - E_s}$ into a term in $R$ plus a constant.
Mixing up the intercept with the gradient.

Things to Be Careful About

The gradient/intercept expressions depend on what exactly is plotted on the axes. If the horizontal axis is labelled in $\text{k}\Omega$ , a factor of $1000$ affects the numerical gradient; the algebraic comparison should always reflect the axis units used.

Techniques used

rearrange the given equation into the form y = mx + cidentify the gradient and y-intercept by comparison with a straight-line equation

(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

Working

I = (I_{\mu\text{A}})\times 10^{-6}\ \text{A}, \quad \frac{1}{I} = \frac{1}{I_{\mu\text{A}}\times 10^{-6}} = \frac{10^{6}}{I_{\mu\text{A}}}\ \text{A}^{-1}

For $y = 1/I$ , fractional uncertainty:

\frac{\Delta y}{y} = \frac{\Delta I}{I} \;\Rightarrow\; \Delta\left(\frac{1}{I}\right)=\left(\frac{\Delta I}{I}\right)\left(\frac{1}{I}\right)

Example (for $I=194\ \mu\text{A}$ , $\Delta I=2\ \mu\text{A}$ ):

\frac{1}{I} = \frac{10^6}{194}=5.15\times 10^3\ \text{A}^{-1}

\Delta\left(\frac{1}{I}\right)=\frac{2}{194}(5.15\times 10^3)=5.3\times 10^1\ \text{A}^{-1}

Answer

\begin{array}{c|c} I / \mu\text{A} & \frac{1}{I} / \text{A}^{-1} \\\hline 194\pm 2 & (5150 \pm 53) \\ 180\pm 2 & (5560 \pm 62) \\ 172\pm 2 & (5810 \pm 68) \\ 160\pm 2 & (6250 \pm 78) \\ 150\pm 2 & (6670 \pm 89) \\ 144\pm 2 & (6940 \pm 96) \end{array}

Final answer

1/I values (A^-1): 5150±53, 5560±62, 5810±68, 6250±78, 6670±89, 6940±96

Detailed explanation

Background Concept

When you transform data (for example, taking a reciprocal), the uncertainty also changes.

y = \frac{1}{x},

then a small change $\Delta x$ produces a change in $y$ of approximately

\Delta y \approx \left|\frac{\mathrm{d}y}{\mathrm{d}x}\right|\Delta x = \frac{\Delta x}{x^2}.

Equivalently, for reciprocals the fractional (percentage) uncertainty stays the same:

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

Understanding the Question

You are given current readings $I$ in $\mu\text{A}$ with an absolute uncertainty $\pm 2\ \mu\text{A}$ . You must fill the third column with:

$1/I$ in $\text{A}^{-1}$ (so you must convert $\mu\text{A}$ to $\text{A}$ ), and
the absolute uncertainty in $1/I$ .

Approach

Convert each current from $\mu\text{A}$ to $\text{A}$ , or use the shortcut $1/I = 10^6/I_{\mu\text{A}}$ .
Compute $1/I$ .
Find $\Delta(1/I)$ $Δ (1/ I)$ using either:
- $\Delta(1/I)=\Delta I/I^2$ , or
- $\Delta(1/I) = (\Delta I/I)(1/I)$ .
Round the uncertainty sensibly (usually 1 s.f. or 2 s.f.) and round $1/I$ to match.

Step-by-Step Reasoning

Take the first row:

$I = 194\ \mu\text{A} = 194\times 10^{-6}\ \text{A}$ .
Reciprocal:

\frac{1}{I} = \frac{1}{194\times 10^{-6}} = \frac{10^6}{194} = 5154.6\ \text{A}^{-1}.

Uncertainty in the reciprocal:
Using fractional uncertainty,

\frac{\Delta I}{I} = \frac{2}{194} = 0.0103

\Delta\left(\frac{1}{I}\right)=0.0103\times 5154.6 = 53.1\ \text{A}^{-1}.

You repeat the same process for each row.

Key Takeaways

Always ensure units are correct before transforming data.
For $y=1/x$ , the fractional uncertainty in $y$ equals the fractional uncertainty in $x$ .
Quote $1/I$ in $\text{A}^{-1}$ , not $\mu\text{A}^{-1}$ .

Common Mistakes

Calculating $1/I$ using $I$ still in $\mu\text{A}$ without the $10^{-6}$ conversion.
Writing the uncertainty in $1/I$ as $\pm 2$ (copying the current uncertainty directly).
Rounding $1/I$ too aggressively (losing useful precision for plotting).

Things to Be Careful About

Keep rounding consistent: if the uncertainty is, for example, $\pm 96\ \text{A}^{-1}$ , then quoting $1/I$ to the nearest unit is unnecessary; nearest $10$ is usually fine.
Uncertainty bars on the graph should use the absolute uncertainty in $1/I$ , not percentage uncertainty.

Techniques used

convert microampere readings into ampere valuescalculate a reciprocal quantitypropagate absolute uncertainty through a reciprocal using \Delta y = (\Delta x/x) y

(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 1

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Axes: $y$ -axis $\frac{1}{I} / \text{A}^{-1}$ , $x$ -axis $R / \text{k}\Omega$ .

Plot points:

(1.50,5150),\ (1.75,5560),\ (1.92,5810),\ (2.22,6250),\ (2.48,6670),\ (2.72,6940)

Draw vertical error bars (in $\text{A}^{-1}$ ):

\pm 53,\ \pm 62,\ \pm 68,\ \pm 78,\ \pm 89,\ \pm 96

Final answer

Graph plotted with points and vertical y-error bars for 1/I as calculated.

Detailed explanation

Background Concept

A good graph in Paper 5 must communicate:

correct variables on axes with units,
accurate plotting,
uncertainties shown as error bars.

Error bars show the range of plausible values of a point. If the uncertainty is only in $y$ , the error bars are vertical with half-length equal to the absolute uncertainty in $y$ .

Understanding the Question

You have already calculated $1/I$ and its absolute uncertainty for each value of $R$ . You must now:

plot $1/I$ (vertical) against $R$ (horizontal), and
include error bars for $1/I$ .
The provided grid suggests appropriate scales.

Approach

Check axis labels and units match the instruction: $R/\text{k}\Omega$ on $x$ , $1/I\ (\text{A}^{-1})$ on $y$ .
Plot each point carefully.
For each point, draw a vertical error bar from $(1/I - \Delta(1/I))$ to $(1/I + \Delta(1/I))$ .

Step-by-Step Reasoning

Put $R$ values directly on the $x$ -axis because the axis is in $\text{k}\Omega$ .
Use your computed $1/I$ values for the $y$ -axis.
For the first point at $R=1.50\ \text{k}\Omega$ , $1/I = 5150\ \text{A}^{-1}$ with uncertainty $\pm 53\ \text{A}^{-1}$ , so the error bar extends from $5097$ to $5203\ \text{A}^{-1}$ .
Repeat similarly for each point.

Key Takeaways

Error bars must correspond to the calculated absolute uncertainties.
Axis labels must include both quantity and unit.
Accurate plotting is essential because gradient and intercept come from the graph.

Common Mistakes

Drawing error bars with the wrong size (e.g. using $\pm 2\ \mu\text{A}$ instead of converting to $\pm$ in $1/I$ ).
Missing units on axes.
Plotting $R$ in $\Omega$ when the axis is labelled in $\text{k}\Omega$ .

Things to Be Careful About

Use a sharp pencil and plot to better than half a small square.
Error bars should be centred on the plotted point.
Do not force the graph to pass through the origin; you will test that with best-fit/worst-fit lines later.

Techniques used

label axes with quantities and unitsplot points accurately from a results tabledraw vertical error bars using absolute uncertainties

(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 trend of points.

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

Label the lines clearly as “best fit” and “worst acceptable”.

Final answer

Best-fit and worst acceptable straight lines drawn and labelled.

Detailed explanation

Background Concept

A line of best fit represents the overall linear trend of the data. Because each point has uncertainty, many straight lines could plausibly represent the relationship.

A “worst acceptable line” is a line chosen to be as different as possible from the best-fit line (usually maximum or minimum gradient) while still being consistent with the data, meaning it passes through (or at least intersects) every error bar.

Understanding the Question

You have plotted $1/I$ vs $R$ with vertical error bars. Now you must:

draw the single straight line that best represents the data trend, and
draw one additional straight line that is still plausible given the error bars but gives the most extreme gradient.
These two lines will be used in later parts to estimate uncertainties in gradient and intercept.

Approach

Best-fit line: draw by eye so that points are roughly balanced (similar scatter above and below).
Worst acceptable line: choose either the steepest or the shallowest line that still intersects every error bar.
Label both lines on the graph.

Step-by-Step Reasoning

Start with the best-fit line: it should follow the middle of the vertical spread of the points.
To get a worst acceptable line, use the end points of the data range:
- for a steep line, aim to go through a low value within the leftmost error bar and a high value within the rightmost error bar;
- for a shallow line, aim to go through a high value within the leftmost error bar and a low value within the rightmost error bar.
After drawing, check that your worst acceptable line still intersects every error bar. If it misses any, adjust.

diagram-c-ii-1

Key Takeaways

Best-fit: balanced through the scatter.
Worst acceptable: extreme gradient but still consistent with all error bars.
These two lines are the standard Paper 5 method for gradient/intercept uncertainty.

Common Mistakes

Drawing the worst line through the outermost plotted points but missing some error bars.
Drawing two best-fit lines (not making the second one “worst”).
Forgetting to label which line is which.

Things to Be Careful About

The worst acceptable line should be a straight line, not a curve.
Use the full length of the graph when drawing lines to reduce reading error.
The “worst” line is not chosen randomly; it must be justified by the error bars.

Techniques used

draw a straight line of best fit through plotted pointsconstruct a worst acceptable line consistent with all error barsuse best and worst lines to estimate uncertainty

(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 (well separated), e.g.

(x_1,y_1)=(1.50,5178),\quad (x_2,y_2)=(2.70,6955)

\text{gradient } m = \frac{\Delta y}{\Delta x} = \frac{6955-5178}{2.70-1.50} = 1.48\times 10^3\ \text{A}^{-1}\ \text{k}\Omega^{-1}

Worst acceptable gradients (from worst line):

m_{\max}\approx 1.59\times 10^3,\quad m_{\min}\approx 1.34\times 10^3

\Delta m \approx \max\left(|1.48-1.59|,|1.48-1.34|\right)\times 10^3 \approx 0.14\times 10^3\ \text{A}^{-1}\ \text{k}\Omega^{-1}

Answer

m = (1.48 \pm 0.14)\times 10^3\ \text{A}^{-1}\ \text{k}\Omega^{-1}

Final answer

(1.48 ± 0.14) × 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 percentage reading error, you choose two points that are far apart on the line (not necessarily your raw data points, but points on the drawn line).

In Paper 5, the uncertainty in gradient is commonly found by comparing the best-fit gradient with the gradient of a worst acceptable line:

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

using the worst line that gives the largest difference.

Understanding the Question

You must find:

the gradient of the best-fit line on your $1/I$ vs $R$ graph, and
an absolute uncertainty in this gradient, using your worst acceptable line.

Approach

Read two well-separated points on the best-fit line.
Compute $m_{\text{best}} = \Delta y/\Delta x$ .
Repeat for the worst acceptable line to find an extreme gradient ( $m_{\text{max}}$ or $m_{\text{min}}$ ).
Take the absolute difference from the best-fit gradient as the uncertainty.

Step-by-Step Reasoning

Pick two points on the best-fit line that are near the ends of the plotted range (this makes $\Delta x$ large).
Subtract to find $\Delta y$ and $\Delta x$ and divide.
Check units: $y$ is in $\text{A}^{-1}$ and $x$ is in $\text{k}\Omega$ , so gradient unit is $\text{A}^{-1}\ \text{k}\Omega^{-1}$ .
For the uncertainty, do the same calculation using the worst acceptable line. If you drew the steepest worst line, that gives $m_{\text{max}}$ ; if you drew the shallowest, that gives $m_{\text{min}}$ .
The uncertainty is the larger difference between $m_{\text{best}}$ and whichever extreme you found.

Key Takeaways

Use a large triangle on the line to reduce reading error.
Always calculate gradient as $\Delta y/\Delta x$ .
Worst acceptable line provides an experimental uncertainty estimate.

Common Mistakes

Using two nearby points so $\Delta x$ is small (gives a very uncertain gradient).
Calculating $\Delta x/\Delta y$ by accident.
Using raw data points rather than points on the drawn line (the mark scheme expects a line-based gradient).

Things to Be Careful About

Make sure $\Delta x$ uses the same axis units you plotted (here $\text{k}\Omega$ ).
Quote an absolute uncertainty (same units as the gradient), not a percentage, unless asked.
Keep the number of significant figures sensible: uncertainty typically to 1 s.f. and value to match.

Techniques used

determine gradient from two well-separated points on a straight linecalculate gradient as \Delta y/\Delta xestimate gradient uncertainty from best-fit and worst acceptable lines

(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 best-fit line, using a point on the line (e.g. $x=1.50$ , $y=5178$ ) and $m=1.48\times 10^3$ :

c = y - mx = 5178 - (1.48\times 10^3)(1.50) = 2.96\times 10^3\ \text{A}^{-1}

Worst acceptable intercepts (from worst line):

c_{\min}\approx 2.72\times 10^3,\quad c_{\max}\approx 3.19\times 10^3

\Delta c \approx \max\left(|2.96-2.72|,|2.96-3.19|\right)\times 10^3 \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 plotted graph you can find it by extending the line to cross the $y$ -axis.

You can also calculate it from any point on the line:

c = y - mx.

The uncertainty in the intercept is estimated by comparing the best-fit intercept with the intercept from the worst acceptable line.

Understanding the Question

You must determine the $y$ -intercept of your best-fit line on the graph of $1/I$ against $R$ , and include an absolute uncertainty based on your worst acceptable line.

Approach

Find $c$ for the best-fit line either by reading at $x=0$ or calculating $c=y-mx$ .
Find $c$ for the worst acceptable line in the same way.
Uncertainty is the absolute difference between best and worst values (take the larger difference if you have both extremes).

Step-by-Step Reasoning

Because extrapolating to $x=0$ can be hard to read accurately, using $c=y-mx$ from a clear point on the line is often more accurate.
Choose a point on the best-fit line that lies exactly on grid intersections if possible.
Substitute into $c=y-mx$ .
Repeat using a point on the worst acceptable line to get $c_{\text{worst}}$ .
Compute

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

Key Takeaways

Intercept is the value at $x=0$ .
$c=y-mx$ is a reliable way to compute intercept.
Worst acceptable line gives uncertainty in $c$ .

Common Mistakes

Using the intercept from the worst acceptable line as the final answer instead of the best-fit intercept.
Reading $c$ from the graph without extending the line properly.
Mixing up intercept units; it must be in $\text{A}^{-1}$ .

Things to Be Careful About

Keep consistent units: since $x$ is in $\text{k}\Omega$ , do not convert $x$ to $\Omega$ in $c=y-mx$ unless you also convert $m$ .
Quote uncertainty to 1 s.f. (or 2 s.f.) and the intercept to match that precision.

Techniques used

extrapolate a straight line to determine the y-interceptcompute the intercept using c = y - mxestimate intercept uncertainty using best-fit and worst acceptable lines

(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

Worked solution

Working

From (a):

\frac{1}{I} = \left(\frac{3}{3E-E_s}\right)R + \left(\frac{2Z}{3E-E_s}\right)

Since the graph uses $R/\text{k}\Omega$ on the $x$ -axis, the gradient is

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

3E - E_s = \frac{3000}{m} = \frac{3000}{1.48\times 10^3} = 2.03\ \text{V}

E = \frac{E_s + (3E-E_s)}{3} = \frac{2.20 + 2.03}{3} = 1.41\ \text{V}

Using $y$ -intercept $c$ :

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

Z = \frac{(2.96\times 10^3)(2.03)}{2} = 3.00\times 10^3\ \Omega

Answer

E = 1.41\ \text{V}

Z = 3.00\times 10^3\ \Omega

Final answer

E = 1.41 V, Z = 3.00 × 10^3 Ω

Detailed explanation

Background Concept

Once you have a straight-line graph, you can extract constants by comparing your experimental straight-line equation with the theoretical one.

y = mx + c,

then $m$ and $c$ are measured from the graph. If physics predicts

y = (\text{expression in constants})x + (\text{expression in constants}),

you can equate coefficients to solve for the constants.

Understanding the Question

You have:

a best-fit gradient $m$ from the graph of $1/I$ vs $R$ (with $R$ plotted in $\text{k}\Omega$ ),
a best-fit intercept $c$ from the same graph,
a known supply e.m.f. $E_s = (2.20 \pm 0.05)\ \text{V}$ ,
and you must determine the cell e.m.f. $E$ and the resistor value $Z$ .

Approach

Start from the linear form from (a).
Adjust the gradient expression to match the fact that $R$ is plotted in $\text{k}\Omega$ (this introduces a factor of $1000$ ).
Use the measured gradient to find $(3E - E_s)$ .
Use $E_s$ to find $E$ .
Use the intercept to find $Z$ .

Step-by-Step Reasoning

From (a):

\frac{1}{I} = \left(\frac{3}{3E-E_s}\right)R + \left(\frac{2Z}{3E-E_s}\right)

If the horizontal axis used $R$ in $\Omega$ , then gradient would be $\frac{3}{3E-E_s}$ . But the actual plotted $x$ is $R/\text{k}\Omega$ , i.e. $R = 1000x$ .
Substitute $R = 1000x$ :

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

So the gradient read from your graph is

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

Rearrange:

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

Then use $3E = E_s + (3E-E_s)$ to get $E$ :

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

For $Z$ , the intercept $c$ is still

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

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

Because $c$ is in $\text{A}^{-1}$ and $(3E-E_s)$ in V, their product has units $\text{V}\,\text{A}^{-1} = \Omega$ , as expected.

Key Takeaways

Always match the algebra to the axis units (here $\text{k}\Omega$ introduces a factor $1000$ ).
Gradient gives $(3E-E_s)$ , then $E$ follows using the known $E_s$ .
Intercept gives $Z$ once $(3E-E_s)$ is known.

Common Mistakes

Forgetting the $1000$ factor from plotting $R$ in $\text{k}\Omega$ , which makes $E$ wrong.
Using $c = Z/(3E-E_s)$ instead of $c = 2Z/(3E-E_s)$ .
Giving $Z$ in $\text{k}\Omega$ without stating the unit, or mixing $\Omega$ and $\text{k}\Omega$ inconsistently.

Things to Be Careful About

Keep enough significant figures in intermediate steps (especially $(3E-E_s)$ ) before rounding.
Check unit consistency: if you use $m$ in $\text{A}^{-1}\ \text{k}\Omega^{-1}$ , then you must use $3000$ (not $3$ ) in the formula.

Techniques used

relate graph gradient and intercept to physical constants by algebraic comparisonsubstitute measured gradient/intercept into derived expressionsconvert between kΩ and Ω consistently

(ii)

Determine the absolute uncertainty in $E$ .

absolute uncertainty in $E$ = ______

1M Similar

DifficultyMedium

Worked solution

Working

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

\Delta(3E-E_s)=\Delta\left(\frac{3000}{m}\right)=\frac{3000\,\Delta m}{m^2}

With $m=(1.48\pm 0.14)\times 10^3$ :

\Delta\left(\frac{3000}{m}\right)=\frac{3000\times 0.14\times 10^3}{(1.48\times 10^3)^2}=0.19\ \text{V}

For $E = \dfrac{E_s + (3E-E_s)}{3}$ , absolute uncertainties add in the numerator:

\Delta E = \frac{\Delta E_s + \Delta(3E-E_s)}{3} = \frac{0.05 + 0.19}{3} = 0.08\ \text{V}

Answer

absolute uncertainty in $E$ $= 0.08\ \text{V}$

Final answer

0.08 V

Detailed explanation

Background Concept

Uncertainty rules used here:

For a reciprocal (or more generally $y = k/x$ ), fractional uncertainties add:

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

which can be rearranged to

\Delta\left(\frac{k}{x}\right) = \frac{k\,\Delta x}{x^2}.

For addition/subtraction ( $q = a \pm b$ ), absolute uncertainties add:

\Delta q = \Delta a + \Delta b.

For division by a constant ( $q = a/3$ ), absolute uncertainty scales the same way:

\Delta q = \Delta a/3.

Understanding the Question

You already found $E$ in (d)(i). Now you must find the absolute uncertainty in $E$ .

$E$ depends on:

the measured gradient $m$ (with its uncertainty from worst line), and
the given supply e.m.f. $E_s$ (with uncertainty $\pm 0.05\ \text{V}$ ).

Approach

Express $(3E-E_s)$ in terms of $m$ .
Find the uncertainty in $(3E-E_s)$ due to uncertainty in $m$ .
Use the relationship between $E$ , $E_s$ and $(3E-E_s)$ , and add absolute uncertainties in the numerator.

Step-by-Step Reasoning

From the graph with $R$ in $\text{k}\Omega$ :

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

Differentiate/uncertainty-propagate for $3000/m$ :

\Delta\left(\frac{3000}{m}\right)=\frac{3000\,\Delta m}{m^2}.

Then

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

Since the numerator is a sum, the absolute uncertainty in the numerator is

\Delta E_s + \Delta(3E-E_s).

Finally divide by 3 to get $\Delta E$ .

Key Takeaways

Reciprocal relationships amplify uncertainty when the denominator is uncertain.
For sums, add absolute uncertainties.
Dividing by 3 reduces the absolute uncertainty by a factor of 3.

Common Mistakes

Adding percentage uncertainties when the quantities are being added (should add absolute uncertainties for addition).
Forgetting to divide the final uncertainty by 3 when calculating $\Delta E$ .
Using the intercept uncertainty instead of the gradient uncertainty (here $E$ depends on $m$ and $E_s$ ).

Things to Be Careful About

Use $m$ in the correct units corresponding to the $\text{k}\Omega$ axis, otherwise $(3E-E_s)=3000/m$ is not valid.
Round uncertainties sensibly; typically 1 s.f. is acceptable for the final absolute uncertainty.

Techniques used

propagate uncertainty through a reciprocalcombine absolute uncertainties for sums and differencesscale uncertainties 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

Working

Using

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

R = \frac{1}{3}\left(\frac{3E-E_s}{I} - 2Z\right)

With $3E-E_s = 2.03\ \text{V}$ , $Z = 3.00\times 10^3\ \Omega$ , $I = 250\ \mu\text{A} = 2.50\times 10^{-4}\ \text{A}$ :

R = \frac{1}{3}\left(\frac{2.03}{2.50\times 10^{-4}} - 2(3.00\times 10^3)\right)

R = \frac{1}{3}\left(8120 - 6000\right)=7.05\times 10^2\ \Omega

Answer

R = 705\ \Omega

Final answer

705 Ω

Detailed explanation

Background Concept

Once you have determined constants from a calibration graph (here $E$ and $Z$ ), you can use the original relationship to predict what value of an input (here $R$ ) will produce a desired output (here $I$ ).

The key skill is rearranging the equation correctly and keeping units consistent.

Understanding the Question

You repeat the experiment and want $I = 250\ \mu\text{A}$ . Using the same circuit constants ( $E$ , $Z$ , and known $E_s$ ), you must find the required resistance $R$ .

Approach

Start from

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

Make $R$ the subject.
Substitute $I = 250\ \mu\text{A}$ converted into amperes.

Step-by-Step Reasoning

Rearrange:

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

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

and

R = \frac{1}{3}\left(\frac{3E-E_s}{I} - 2Z\right).

Convert the target current:

250\ \mu\text{A} = 250\times 10^{-6}\ \text{A} = 2.50\times 10^{-4}\ \text{A}.

Substitute values (using your earlier best estimates of $3E-E_s$ and $Z$ ) to obtain $R$ .

Key Takeaways

Always convert $\mu\text{A}$ to $\text{A}$ before substituting into equations with volts and ohms.
Rearranging carefully avoids sign errors.
A larger current here requires a smaller $R$ (a useful reasonableness check).

Common Mistakes

Substituting $I = 250$ instead of $250\times 10^{-6}$ .
Forgetting the factor 3 in $3R$ .
Using $Z$ in $\text{k}\Omega$ while other quantities are in SI units, without conversion.

Things to Be Careful About

Keep enough significant figures in $(3E-E_s)$ and $Z$ before the final rounding.
Check the answer magnitude: it should be below the earlier $R$ range (1.5 to 2.7 k $\Omega$ ) because $250\ \mu\text{A}$ is larger than the earlier currents.

Techniques used

rearrange the governing equation to make the required variable the subjectsubstitute a target value into a calibrated linear relationshipconvert microampere to ampere consistently

Physics 9702/51 — October/November 2024

Procedure and measurements

Control of variables

Analysis (to find KKK and QQQ)

Safety

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

1) Setting up and measuring speed

2) Choosing values of sss

3) Controlling variables

4) Linearising and plotting

5) Finding KKK and QQQ

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

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

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Analysis (to find $K$ and $Q$ )

2) Choosing values of $s$

5) Finding $K$ and $Q$