What are the key steps in processing results in Practical Skills in Physics I for Edexcel International A Levels?

Processing results involves several key steps: collecting data accurately, organizing data in tables, performing calculations, analyzing trends, and drawing conclusions. It's crucial to ensure that measurements are precise and that calculations are checked for errors. This process helps in understanding the underlying physics concepts and validating experimental hypotheses.

How can I effectively organize data collected during a physics experiment?

To effectively organize data, use tables to systematically record measurements. Label each column with the variable name and units, and ensure that data is recorded consistently. This organization aids in identifying patterns and simplifies the process of graphing and further analysis, which are essential components of Practical Skills in Physics I for Edexcel International A Levels.

Why is graphing important in processing results for physics experiments?

Graphing is crucial because it visually represents data, making it easier to identify trends and relationships between variables. In the context of Edexcel International A Levels, graphing helps in interpreting results and supports the analysis of experimental data. It also aids in determining the accuracy and reliability of the results by allowing for the identification of anomalies.

What common errors should I avoid when processing results in physics experiments?

Common errors include incorrect unit conversions, misreading instruments, and calculation mistakes. It's important to double-check all measurements and calculations. Additionally, ensure that graphs are plotted accurately with appropriate scales and labels. Avoiding these errors is essential for achieving reliable results in Practical Skills in Physics I for Edexcel International A Levels.

How do I draw conclusions from experimental data in Practical Skills in Physics I?

Drawing conclusions involves analyzing the processed data to determine if it supports the initial hypothesis. Look for patterns or correlations in the data, and consider any anomalies or unexpected results. Discuss the implications of the findings in the context of the physics principles being studied. This step is crucial for demonstrating a comprehensive understanding of the experiment and its outcomes in the Edexcel International A Levels curriculum.

Why is "Using a small gradient triangle (a fraction of the line) to read the gradient" a common mistake in Practical Skills - Processing Results?

Why it happens: Students pick two convenient nearby grid points instead of two well-separated points on the line. How to avoid it: Draw a **large triangle spanning at least half the plotted line** and read the coordinates to the scale's precision. A big triangle reduces the fractional error in x and y and protects the accuracy mark. Source: Edexcel IAL Physics Unit 3 (WPH13) examiner reports

Why is "Joining the plotted points dot-to-dot instead of drawing a single line of best fit" a common mistake in Practical Skills - Processing Results?

Why it happens: Students think every point must be touched, or don't trust a line that misses points. How to avoid it: Draw **one thin, straight (or smoothly curved) best-fit line** with points balanced above and below it. The whole purpose of a best-fit line is to **average out random scatter** — a zig-zag reproduces the noise. Source: Edexcel IAL Physics Unit 3 (WPH13) examiner reports

Why is "Quoting a gradient (or a quantity derived from it) with no units" a common mistake in Practical Skills - Processing Results?

Why it happens: The gradient is treated as a pure number rather than a ratio of two measured quantities. How to avoid it: Always give the gradient's units as **(y-axis units) ÷ (x-axis units)** — e.g. a v–t gradient is in m s^-2. Stating the units also acts as a check that the gradient is the physical quantity you expect. Source: Edexcel IAL Physics Unit 3 (WPH13) examiner reports

Why is "Not multiplying the percentage uncertainty by the power for a squared (or cubed) term" a common mistake in Practical Skills - Processing Results?

Why it happens: Students add each measurement's percentage uncertainty once, ignoring that some are raised to a power. How to avoid it: For a term x^n the percentage uncertainty is |n|×\%x. A **squared** quantity (e.g. A d^2, or T^2) contributes **twice** its percentage uncertainty; a cube contributes three times. Source: Edexcel IAL Physics Unit 3 (WPH13) examiner reports

Why is "Confusing ln and log (base 10) when linearising, e.g. plotting lg for an exponential" a common mistake in Practical Skills - Processing Results?

Why it happens: Both ‘log’ buttons exist on the calculator and the difference in method is not memorised. How to avoid it: Use **ln (base e) for exponentials** y = Ae^kx (plot y vs x, gradient = k), and **lg (base 10) for power laws** y = Ax^n (plot y vs x, gradient = n). Mixing them scales the gradient by 10 2.303. Source: Edexcel IAL Physics Unit 3 (WPH13) examiner reports

Why is "Mixing up validity and reliability — e.g. saying ‘repeating makes the result more valid’" a common mistake in Practical Skills - Processing Results?

Why it happens: The two evaluation terms sound similar and are used loosely in everyday speech. How to avoid it: **Reliability** = consistent, repeatable results (improved by repeating/averaging, which cuts random error). **Validity** = measuring the right thing in a fair test (improved by controlling variables/removing systematic error). Repeating improves reliability, NOT validity. Source: Edexcel IAL Physics Unit 3 (WPH13) examiner reports

Edexcel International A Levels Physics (XPH11-YPH11)

Practical Skills - Processing Results

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Detailed Study Notes

Detailed notes on Practical Skills in Physics I for Edexcel International A Levels Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Practical Skills — Processing Results — Pearson Edexcel International A Level Physics (XPH11/YPH11) Study Notes (Unit 3, WPH13)

How to plot a graph that scores full marks (sensible scales, correct plotting, a single line of best fit), find a gradient from a large triangle and give its units, relate the gradient and intercept to physics, linearise relationships including ln and log graphs, calculate and combine percentage uncertainties (doubling for squared terms), estimate the uncertainty in a gradient, and evaluate an experiment (systematic vs random error, validity vs reliability, specific improvements).

At a glance

Plotting: independent variable on the x-axis, dependent variable on the y-axis; use easy scales (multiples of 1/2/5) so the points fill at least half the grid each way.
Line of best fit: one thin line with points balanced above and below — never dot-to-dot. It averages out random scatter.
Gradient: use a large triangle (≥ half the line), read coordinates carefully, and always state the units (= y-units ÷ x-units). The gradient usually IS a physical quantity.
Linearise into $y = mx + c$ : for $s = \tfrac12 gt^2$ plot $s$ vs $t^2$ (gradient $= \tfrac12 g$ ); for $y = Ae^{kx}$ plot $\ln y$ vs $x$ (gradient $= k$ ); for $y = Ax^n$ plot $\lg y$ vs $\lg x$ (gradient $= n$ ).
Absolute uncertainty $= \tfrac12\times$ resolution (single reading) or $\dfrac{x_{\max}-x_{\min}}{2}$ (repeats). Percentage uncertainty $= \dfrac{\Delta x}{x}\times100\%$ .
Combining: ADD the percentage uncertainties for products/quotients; multiply by the power for a raised term (a squared term contributes $2\times$ its %).
Gradient uncertainty: draw max and min lines through the error bars; $\Delta m = \dfrac{m_{\max}-m_{\min}}{2}$ .
Evaluation: systematic = same-direction offset (zero error, parallax); random = scatter (reaction time). Repeating improves reliability, not validity. Suggest specific improvements that target the named error.

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XPH11-YPH11 syllabus (2018-onwards).

Plot a graph with the independent variable on the x-axis, sensible linear scales that fill at least half the grid, correct plotting and a single line/curve of best fit.
Determine a gradient from a large triangle, quote its units, and relate the gradient and y-intercept to the physical quantities being investigated.
Rearrange a relationship into the straight-line form $y = mx + c$ , including using $\ln$ for exponentials and $\lg$ for power laws, and interpret the resulting gradient and intercept.
Calculate absolute and percentage uncertainties, and combine them correctly for products, quotients and powers.
Estimate the uncertainty in a gradient using the maximum/minimum line method.
Evaluate an experiment: distinguish systematic from random errors and validity from reliability, identify anomalies, and suggest specific, feasible improvements.

Plotting a graph that scores full marks

Right axes, easy scales that fill the grid, correct points, one best-fit line.

Graph questions on Unit 3 (WPH13) reward a few disciplined habits. Get these right and the plotting marks are almost automatic.

Which variable goes where. The independent variable — the one you deliberately change — goes on the x-axis (horizontal). The dependent variable — the one you measure in response — goes on the y-axis (vertical). (One exception: sometimes you deliberately plot a linearised quantity like $t^2$ or $\ln A$ on an axis — see the linearisation section.)

Choosing the scale. Two rules:

The plotted points must fill at least half of the grid in both directions. A scale that squashes the data into one corner throws away precision and loses a mark.
Use an easy scale — 1, 2, 5 (or 10, 20, 50…) units per large square. Awkward scales like 3 or 7 per square make every point hard to plot and read.

Find your largest value, divide the number of squares available, and round up to the next easy value. For extensions up to 84 mm on a 10-square axis, $84/10 = 8.4$ , so choose 10 mm per square — the points then fill about 8.4 of the 10 squares.

Plotting the points to better than half a small square, using a sharp pencil and small neat crosses or dots.

The line of best fit. Draw a single thin line — straight if the trend is linear — with roughly as many points above it as below. It need not touch any point or pass through the origin unless the physics demands it. What you must never do is join the points dot-to-dot: the whole reason for drawing a best-fit line is that it averages out the random scatter in the individual readings.

Independent variable on x, dependent variable on y.
Scale: easy multiples (1/2/5) and points fill ≥ half the grid each way.
Plot to better than half a small square.
One best-fit line, points balanced above/below — never dot-to-dot.

See the full worked example for practical skills - processing results →

Gradients — big triangle, units, and the physics

Read the slope from a large triangle, give its units, and link it to a quantity.

The gradient of a straight-line graph carries the physics. Reading it well is worth several marks.

Use a large triangle. Choose two points on your best-fit line (not raw data points) that are far apart — the triangle should span at least half the plotted line. Read their coordinates to the precision of the scale, then

$\text{gradient} = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}.$

A large triangle makes the fractional error in $\Delta x$ and $\Delta y$ small; a tiny triangle magnifies reading errors and can cost the accuracy mark.

Always state the units. The gradient's units are (y-axis units) ÷ (x-axis units). On a velocity–time graph the gradient is in $\text{m s}^{-1} \div \text{s} = \text{m s}^{-2}$ — i.e. an acceleration. Quoting a bare number is a guaranteed lost mark.

Relate the gradient to the physics. After linearising, compare your equation with $y = mx + c$ and read off what the gradient equals:

Graph plotted	Underlying relation	Gradient equals
$s$ vs $t^2$ (free fall from rest)	$s = \tfrac12 g t^2$	$\tfrac12 g$ , so $g = 2\times$ gradient
stress vs strain (elastic region)	stress $= E\times$ strain	Young modulus $E$
$R$ vs $L$ (fixed area $A$ )	$R = \dfrac{\rho}{A}L$	$\dfrac{\rho}{A}$ , so $\rho = \text{gradient}\times A$
$T^2$ vs $m$ (mass–spring)	$T^2 = \dfrac{4\pi^2}{k}m$	$\dfrac{4\pi^2}{k}$

The y-intercept (where the line crosses $x = 0$ ) is the constant $c$ in $y = mx + c$ and often has a physical meaning — for example the $\ln A_0$ term on a log graph, or a fixed offset. If the theory says the line should pass through the origin and yours does not, that hints at a systematic error.

Pick two well-separated points on the best-fit line, mark their coordinates, and use the large triangle to compute the gradient Δy/Δx. State the units (y-units ÷ x-units) and relate the value to the physics.

Take two points on the best-fit line, far apart (triangle ≥ half the line).
gradient = Δy/Δx, quoted with units = y-units ÷ x-units.
Compare with $y = mx + c$ to identify the physical quantity.
y-intercept = c; a non-zero intercept where the origin is expected hints at systematic error.

See the full worked example for practical skills - processing results →

Linearisation — turning any relationship into a straight line

Rearrange to y = mx + c; use ln for exponentials and lg for power laws.

A straight line is the easiest graph to analyse, so a common skill is to rearrange a relationship into the form $y = mx + c$ and plot the combination that makes it linear.

Simple algebraic linearisation. If $s = \tfrac12 g t^2$ , then plotting $s$ (y) against $t^2$ (x) gives $y = mx$ with gradient $\tfrac12 g$ . You manufacture a straight line by plotting a power or a reciprocal of one variable.

Exponentials — use natural logs (ln). For $y = A e^{kx}$ (e.g. radioactive decay, capacitor discharge, cooling), take natural logs of both sides:

$\ln y = kx + \ln A.$

Plot $\ln y$ (y-axis) against $x$ (x-axis): the gradient is $k$ (negative for decay) and the intercept is $\ln A$ , so $A = e^{\text{intercept}}$ .

Power laws — use logs base 10 (lg). For $y = A x^{n}$ (e.g. period vs length), take logs of both sides:

$\lg y = n\lg x + \lg A.$

Plot $\lg y$ against $\lg x$ : the gradient is the power $n$ and the intercept is $\lg A$ , so $A = 10^{\text{intercept}}$ . A straight line confirms the power law.

Which log to use? The variable that appears in the exponent ⇒ take a log of the whole side (ln for base $e$ ). The variable that is raised to an unknown power ⇒ take logs of both $y$ and $x$ (log–log). Calculate ln/log values to 3 significant figures before plotting.

Plotting ln A against t straightens an exponential decay A = A₀e^(−λt). The gradient is −λ (negative because A falls) and the y-intercept is ln A₀. For a power law you would instead plot lg y against lg x, where the gradient is the power n.

Rearrange to $y = mx + c$ and plot the combination that is linear.
Exponential $y = Ae^{kx}$ : plot $\ln y$ vs $x$ , gradient $= k$ , intercept $= \ln A$ .
Power law $y = Ax^n$ : plot $\lg y$ vs $\lg x$ , gradient $= n$ , intercept $= \lg A$ .
Compute ln/log values to 3 s.f. before plotting.

See the full worked example for practical skills - processing results →

Uncertainties — finding, converting and combining them

Half the resolution, turn it into a %, then add %s (and double for squares).

Every measurement carries an uncertainty, and you must be able to find it, express it as a percentage, and combine several into the uncertainty of a final result.

Absolute uncertainty of a reading.

Single reading: $\Delta x = \tfrac12\times\text{resolution}$ . A metre rule (1 mm divisions) gives $\pm0.5\ \text{mm}$ ; a micrometer (0.01 mm) gives $\pm0.005\ \text{mm}$ .
Repeated readings: $\Delta x = \dfrac{x_{\max}-x_{\min}}{2}$ (half the spread), quoted with the mean.
A difference of two readings (e.g. a titre, or a length from two rule readings) carries twice the single-reading uncertainty, because each end contributes.

Percentage uncertainty.

$\%\,x = \frac{\Delta x}{x}\times100\%.$

Converting to a percentage lets you compare and combine uncertainties in different quantities and units.

Combining uncertainties.

Products and quotients: ADD the percentage uncertainties. If $\rho = \dfrac{RA}{L}$ then $\%\rho = \%R + \%A + \%L$ .
Powers: multiply the percentage uncertainty by the power. A squared term contributes twice its % (area $A\propto d^{2}$ means $\%A = 2\,\%d$ ; and $g\propto 1/T^{2}$ means $T$ contributes $2\,\%T$ ). A cube contributes three times.
Constants like $4\pi^2$ or $\tfrac43\pi$ have no uncertainty.

Judging the result. A total percentage uncertainty above about 5% signals significant error and a design worth improving. Look at which term dominates — reducing the biggest contributor (often a squared or small-value quantity) improves the result most. For example, in a pendulum $g$ , the period term (doubled) usually dominates, so timing more oscillations is the best improvement.

Uncertainty in a gradient — the max/min line method. With error bars on your points, draw the steepest and the shallowest straight lines that still pass through (most of) the error bars. Then

$\Delta m = \frac{m_{\max} - m_{\min}}{2},$

and the percentage uncertainty is $\dfrac{\Delta m}{m_{\text{best}}}\times100\%$ (divide by the best-fit gradient, not the maximum).

The steepest (max) and shallowest (min) lines that still fit the error bars bracket the best-fit gradient. The gradient uncertainty is half the difference of their gradients; dividing by the best-fit gradient gives the percentage uncertainty.

Single reading: $\Delta x = \tfrac12\times$ resolution; repeats: $(x_{\max}-x_{\min})/2$ ; a difference: double it.
$\%\,x = (\Delta x / x)\times100\%$ .
Products/quotients: ADD the %s. Powers: multiply the % by the power (square ⇒ ×2).
Gradient uncertainty: $\Delta m = (m_{\max}-m_{\min})/2$ , then ÷ best-fit gradient for the %.

See the full worked example for practical skills - processing results →

Evaluating an experiment — errors, validity and improvements

Systematic vs random, validity vs reliability, and specific fixes that target the error.

The final part of a practical question asks you to evaluate — to judge the quality of the data and suggest improvements. Vague answers score nothing; the marks are for precise reasoning.

Systematic vs random error.

A systematic error shifts every reading by the same amount in the same direction — the results are consistently too high or too low. Causes: a zero error (instrument doesn't read zero when it should), consistent parallax, a mis-calibrated scale. It affects accuracy and is not reduced by repeating; you must find and correct its cause (e.g. subtract the zero error). On a graph, it can show up as an unexpected non-zero intercept.
A random error makes readings scatter unpredictably above and below the true value. Causes: reaction time when timing by hand, fluctuating air currents, small variations between repeats. It affects precision/reliability and is reduced by repeating and averaging. On a graph, it shows up as scatter of points about the best-fit line.

Anomalies. An anomalous result lies clearly off the trend. Check/repeat it if possible, and ignore it when drawing the best-fit line — do not let it drag the line towards it.

Validity vs reliability (a favourite trap):

Validity = does the experiment measure what it is meant to (a fair test, other variables controlled, systematic errors removed)? Repeating readings does not make an experiment more valid.
Reliability = are the results consistent and repeatable (small random scatter)? Repeating and averaging improves reliability.

Suggesting improvements. Each improvement must be specific and feasible, and must target a named error:

To remove reaction-time (random) error: use light gates / a data-logger for electronic timing, or time more oscillations and divide.
To reduce random scatter: repeat and take a mean; take readings over a range of the independent variable and draw a best-fit line.
To remove a systematic zero error: check and subtract the zero reading, or calibrate the instrument.
To improve resolution: use a more precise instrument (micrometer instead of ruler) for the quantity with the largest percentage uncertainty.

Answers like "measure more carefully" or "use better equipment" earn nothing — you must say what to change and how it reduces which error.

Systematic = same-direction offset (zero error/parallax); affects accuracy; not fixed by repeats.
Random = scatter (reaction time/air currents); affects reliability; reduced by averaging.
Validity = measuring the right thing (fair test); reliability = consistent/repeatable — repeating improves reliability, NOT validity.
Improvements must be specific, feasible, and target the named error.

See the full worked example for practical skills - processing results →

Quick recap

Independent variable on x, dependent on y; easy scales (1/2/5) that fill ≥ half the grid; one best-fit line, never dot-to-dot.
Gradient from a large triangle (≥ half the line), quoted with units (y-units ÷ x-units) and linked to the physics via $y = mx + c$ .
Linearise: $s$ vs $t^2$ for $s=\tfrac12 gt^2$ ; $\ln y$ vs $x$ for exponentials (gradient $= k$ ); $\lg y$ vs $\lg x$ for power laws (gradient $= n$ ).
Absolute uncertainty $= \tfrac12\times$ resolution (single) or $(x_{\max}-x_{\min})/2$ (repeats); percentage $= (\Delta x/x)\times100\%$ .
Combine: ADD %s for products/quotients; multiply % by the power (squared term ⇒ ×2). Total > ~5% signals significant error.
Gradient uncertainty $= (m_{\max}-m_{\min})/2$ from max/min lines through the error bars, divided by the best-fit gradient for the %.
Evaluate: systematic vs random error, validity vs reliability, ignore anomalies for the best-fit line, and give specific improvements that target the named error.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Independent variable on the x-axis; scales in multiples of 1/2/5 filling ≥ half the grid; single line of best fit (not dot-to-dot).
Gradient from a LARGE triangle (≥ half the line); units = y-units ÷ x-units; ALWAYS state them.
Exponential $y = Ae^{kx}$ : plot $\ln y$ vs $x$ , gradient $= k$ , intercept $= \ln A$ . Power law $y = Ax^n$ : plot $\lg y$ vs $\lg x$ , gradient $= n$ , intercept $= \lg A$ .
Absolute uncertainty $= \tfrac12\times$ resolution (single reading) or $(x_{\max}-x_{\min})/2$ (repeats); a difference doubles it.
Percentage uncertainty $= \dfrac{\Delta x}{x}\times100\%$ .
Products/quotients: ADD the %s. Powers: multiply the % by the power (square ⇒ 2×, cube ⇒ 3×). Constants have no uncertainty.
Gradient uncertainty $\Delta m = \dfrac{m_{\max}-m_{\min}}{2}$ (max/min lines through the error bars).
Systematic = same-direction offset (accuracy, not fixed by repeats); random = scatter (reliability, reduced by averaging). Repeating improves reliability, NOT validity.

How it’s examined

Processing Results is examined on Unit 3 (WPH13), Practical Skills in Physics I — a written paper based on the Core Practicals and on unfamiliar experiments. Typical items: complete or critique a graph (choose axes/scale, plot points, draw a line of best fit) for 3–4 marks; ‘determine the gradient’ and state its units and physical meaning for 2–3 marks; ‘state what should be plotted to obtain a straight line’ for a linearisation (log/ln) question for 3–4 marks; calculate and combine percentage uncertainties (with a squared term) for 3–4 marks; find the uncertainty in a gradient via max/min lines; and an evaluation asking you to identify errors (systematic vs random) and suggest specific improvements for up to 6 marks. Discipline that earns the marks: sensible scales filling the grid, a single best-fit line, a large gradient triangle with units, doubling the % for squared terms, using ln for exponentials and lg for power laws, and improvements that name the error they remove. Vague evaluation (‘be more careful’) and dropped gradient units are the commonest lost marks.

Sources: Pearson Edexcel International Advanced Level Physics (YPH11/XPH11) specification (Issue 3, July 2021); Edexcel IAL Physics Unit 3 (WPH13) examiner reports; Edexcel IAL Physics Unit 3 (WPH13) sample assessment materials and mark schemes. Last reviewed 2026-07-05.

Take this whole topic with you

Step-by-step worked examples — Practical Skills - Processing Results

Step-by-step solutions to past-paper-style questions on practical skills - processing results, written exactly the way a tutor would explain them at the board.

Question type:

1Choosing sensible graph scales (3 marks)
Getting startedGraph or diagram• Edexcel IAL Physics Unit 3 (WPH13) style• graph plotting, scales, Unit 3
Question
A student measures the extension $x$ of a spring for loads giving forces from 0 N to 5.0 N, and records extensions from 0 mm up to 84 mm. The grid provided is 12 large squares wide and 10 large squares tall. State two rules the student must follow when choosing the axes, and suggest a suitable scale for the extension axis. (3 marks)
Step-by-step solution
1. Step 1
  The independent variable (the force the student sets) goes on the x-axis; the dependent variable (the measured extension) goes on the y-axis.
2. Step 2
  The plotted points must fill at least half of the grid in each direction, and the scale must be easy to read — use multiples of 1, 2 or 5 per square, never awkward values like 3 or 7.
3. Step 3
  Extension runs 0–84 mm over 10 squares. A scale of 10 mm per large square reaches 100 mm and uses about 8.4 of the 10 squares — more than half the grid, with an easy value per square.
  $\text{extension axis: } 10\ \text{mm per square} \;\Rightarrow\; 0\text{–}100\ \text{mm over } 10 \text{ squares}$
Answer
Independent variable (force) on the x-axis, dependent variable (extension) on the y-axis; points must occupy at least half the grid each way with an easy-to-read scale. A scale of 10 mm per large square is suitable (points fill ~8.4 of the 10 squares).
Examiner tip
Marks: (1) IV on x, DV on y; (2) scale fills ≥ half the grid AND is a sensible multiple (1/2/5); (3) a workable value such as 10 mm/square. The commonest lost mark is a scale that crams the points into a corner (e.g. 20 mm/square, using < half the grid) or an awkward scale (e.g. 8.4 mm/square) that makes plotting and reading unreliable.
2Line of best fit vs dot-to-dot (2 marks)
Getting startedGraph or diagram• Edexcel IAL Physics Unit 3 (WPH13) style• line of best fit, graph plotting, Unit 3
Question
A set of eight plotted points shows a clear straight-line trend but with some scatter. Describe how the student should draw the trend line, and state one thing they must not do. (2 marks)
Step-by-step solution
1. Step 1
  Draw a single straight line of best fit — a thin line with roughly as many points above it as below it, balancing the scatter. It need not pass through any particular point or through the origin unless the physics requires it.
2. Step 2
  The student must not join the points dot-to-dot (a zig-zag), and must not force the line through the first and last points only. A single smooth best-fit line represents the underlying trend and averages out random scatter.
Answer
Draw one thin, straight best-fit line with a balanced number of points on each side; do NOT connect the points dot-to-dot or force it through the end points only.
Examiner tip
Marks: (1) single best-fit line balancing points above/below; (2) explicitly not dot-to-dot. Examiner reports repeatedly flag ‘dot-to-dot’ lines and lines drawn thick or kinked. A best-fit line is the whole point of plotting — it lets random errors average out.
3Gradient with a large triangle and its units (3 marks)
Getting startedGraph or diagram• Edexcel IAL Physics Unit 3 (WPH13) style• gradient, units, Unit 3
Question
On a velocity–time graph (velocity in m s⁻¹, time in s), a straight best-fit line passes through the points $(1.0\ \text{s},\ 4.0\ \text{m s}^{-1})$ and $(9.0\ \text{s},\ 28.0\ \text{m s}^{-1})$ , which lie near the two ends of the line. Determine the gradient, including its units, and state what the gradient represents. (3 marks)
Step-by-step solution
1. Step 1
  Use a large triangle spanning most of the line — here the two given points are near the ends, so the triangle is large and the reading is precise.
  $\text{gradient} = \frac{\Delta v}{\Delta t} = \frac{28.0 - 4.0}{9.0 - 1.0}$
2. Step 2
  Evaluate the gradient.
  $\text{gradient} = \frac{24.0}{8.0} = 3.0$
3. Step 3
  The units of the gradient are (y-units) ÷ (x-units) $= \text{m s}^{-1} \div \text{s} = \text{m s}^{-2}$ . On a v–t graph the gradient is the acceleration.
  $\text{gradient} = 3.0\ \text{m s}^{-2} = \text{acceleration}$
Answer
Gradient $= 3.0\ \text{m s}^{-2}$ ; it represents the acceleration of the object.
Examiner tip
Marks: (1) large triangle / correct coordinates read; (2) gradient $= 3.0$ ; (3) correct units (m s⁻²) AND identifying it as acceleration. Dropping the units is the single most common lost mark on gradient questions; a small triangle (a fraction of the line) loses precision and can lose the accuracy mark.
4Gradient of a stress–strain graph gives the Young modulus (3 marks)
Building confidenceGraph or diagram• Edexcel IAL Physics Unit 3 (WPH13), Core Practical link• gradient, Young modulus, Unit 3
Question
A stress–strain graph for a metal wire is a straight line through the origin in its elastic region. A large gradient triangle gives a change in stress of $1.20\times10^{8}\ \text{Pa}$ for a change in strain of $6.0\times10^{-4}$ . Determine the Young modulus of the metal, with units. (3 marks)
Step-by-step solution
1. Step 1
  The Young modulus is defined as stress ÷ strain, which for a linear stress–strain graph is exactly the gradient.
  $E = \frac{\text{stress}}{\text{strain}} = \text{gradient}$
2. Step 2
  Substitute the values read from the large triangle.
  $E = \frac{1.20\times10^{8}}{6.0\times10^{-4}}$
3. Step 3
  Evaluate. Strain is a ratio (no units), so the gradient has the units of stress, pascals (Pa).
  $E = 2.0\times10^{11}\ \text{Pa}$
Answer
Young modulus $E = 2.0\times10^{11}\ \text{Pa}$ (equivalently $200\ \text{GPa}$ ).
Examiner tip
Marks: (1) $E =$ gradient of the stress–strain line; (2) correct substitution; (3) $2.0\times10^{11}\ \text{Pa}$ with units. Because strain is dimensionless, the gradient carries the units of the y-axis (Pa). A common error is to invert (strain/stress) or to forget the power of ten.
5Linearising a power law — what to plot (4 marks)
Building confidenceShow that / prove• Edexcel IAL Physics Unit 3 (WPH13) style• linearisation, log graph, Unit 3
Question
The period $T$ of a pendulum is thought to follow a power law with its length $L$ : $T = kL^{n}$ , where $k$ and $n$ are constants. State what quantities the student should plot to obtain a straight line, and explain how the gradient and the intercept give $n$ and $k$ . (4 marks)
Step-by-step solution
1. Step 1
  Take logarithms of both sides (base 10) to turn the product and power into a sum.
  $\lg T = \lg(kL^{n}) = \lg k + n\lg L$
2. Step 2
  Compare with the straight-line form $y = mx + c$ : identify $y = \lg T$ , $x = \lg L$ , gradient $m = n$ and intercept $c = \lg k$ .
  $\underbrace{\lg T}_{y} = \underbrace{n}_{m}\underbrace{\lg L}_{x} + \underbrace{\lg k}_{c}$
3. Step 3
  So plot $\lg T$ on the y-axis against $\lg L$ on the x-axis. A straight line confirms the power law.
4. Step 4
  The gradient equals $n$ (the power). The y-intercept equals $\lg k$ , so $k = 10^{\text{intercept}}$ . (For a pendulum the theory predicts $n = 0.5$ .)
  $n = \text{gradient}, \qquad k = 10^{\,\text{intercept}}$
Answer
Plot $\lg T$ (y-axis) against $\lg L$ (x-axis). The gradient equals the power $n$ ; the y-intercept equals $\lg k$ , so $k = 10^{\text{intercept}}$ .
Examiner tip
Marks: (1) take logs of both sides; (2) rearrange to $\lg T = n\lg L + \lg k$ ; (3) state axes (lg T vs lg L); (4) gradient $= n$ and intercept $= \lg k$ . For a power law you use log–log (any base). Students who plot $T$ vs $L$ (a curve) or confuse this with the ln method for exponentials lose marks.
6Linearising an exponential decay — using ln (4 marks)
Building confidenceShow that / prove• Edexcel IAL Physics Unit 3 (WPH13) style• linearisation, ln graph, Unit 3
Question
The count rate $A$ of a radioactive source is expected to fall exponentially with time: $A = A_{0}e^{-\lambda t}$ . Readings of $A$ at $t = 0,\ 10,\ 20,\ 30$ s are $200,\ 110,\ 60,\ 33$ counts s⁻¹. State what to plot to get a straight line, and calculate the decay constant $\lambda$ from the graph. (4 marks)
Step-by-step solution
1. Step 1
  For an exponential, take natural logs of both sides (not log base 10, because the base is $e$ ).
  $\ln A = \ln A_{0} - \lambda t$
2. Step 2
  Compare with $y = mx + c$ : plot $\ln A$ (y-axis) against $t$ (x-axis). The gradient is $-\lambda$ and the intercept is $\ln A_{0}$ .
  $\underbrace{\ln A}_{y} = \underbrace{-\lambda}_{m}\,\underbrace{t}_{x} + \underbrace{\ln A_{0}}_{c}$
3. Step 3
  Compute $\ln A$ (to 3 s.f.): $\ln 200 = 5.30$ , $\ln 110 = 4.70$ , $\ln 60 = 4.09$ , $\ln 33 = 3.50$ . These fall on a straight line.
4. Step 4
  Gradient from the first and last points: $\dfrac{3.50 - 5.30}{30 - 0} = -0.0600\ \text{s}^{-1}$ . The gradient is $-\lambda$ , so $\lambda = 0.0600\ \text{s}^{-1}$ .
  $\lambda = -\text{gradient} = -(-0.0600) = 0.0600\ \text{s}^{-1}$
Answer
Plot $\ln A$ (y-axis) against $t$ (x-axis); gradient $= -\lambda$ , so $\lambda = 0.0600\ \text{s}^{-1}$ (intercept $= \ln A_{0} = 5.30$ ).
Examiner tip
Marks: (1) take ln of both sides; (2) axes ln A vs t and gradient $= -\lambda$ ; (3) ln values to 3 s.f.; (4) $\lambda = 0.0600\ \text{s}^{-1}$ . Use ln (not lg) for a base- $e$ exponential — mixing them up scales the gradient by $\ln 10 \approx 2.303$ and is a classic error.
7Combining uncertainties with a squared term (4 marks)
StretchMulti-step problem• Edexcel IAL Physics Unit 3 (WPH13) style• percentage uncertainty, combining, Unit 3
Question
A pendulum experiment finds $g$ from $g = \dfrac{4\pi^{2}L}{T^{2}}$ . The length is $L = 0.850 \pm 0.001\ \text{m}$ and the period is $T = 1.85 \pm 0.02\ \text{s}$ . Calculate the value of $g$ and its percentage uncertainty. (4 marks)
Step-by-step solution
1. Step 1
  Calculate $g$ from the mean values.
  $g = \frac{4\pi^{2}(0.850)}{(1.85)^{2}} = \frac{33.56}{3.4225} = 9.80\ \text{m s}^{-2}$
2. Step 2
  Find the percentage uncertainty in each measured quantity (absolute ÷ value × 100).
  $\frac{0.001}{0.850}\times100 = 0.118\%, \qquad \frac{0.02}{1.85}\times100 = 1.08\%$
3. Step 3
  For a product/quotient add the percentage uncertainties, but $T$ is squared, so its percentage uncertainty is multiplied by the power 2. (The constant $4\pi^{2}$ has no uncertainty.)
  $\%\,g = \%L + 2\times\%T = 0.118 + 2(1.08)$
4. Step 4
  Evaluate the total percentage uncertainty.
  $\%\,g = 0.118 + 2.16 = 2.28\% \approx 2.3\%$
Answer
$g = 9.80\ \text{m s}^{-2}$ with a percentage uncertainty of about $2.3\%$ (i.e. $9.80 \pm 0.23\ \text{m s}^{-2}$ ).
Examiner tip
Marks: (1) $g = 9.80\ \text{m s}^{-2}$ ; (2) both %uncertainties; (3) recognising the factor of 2 for the squared $T$ ; (4) total $\approx 2.3\%$ . The dominant error is forgetting to double the period's contribution — a squared term contributes twice its percentage uncertainty. Note $T$ dominates: measuring more oscillations to reduce $\%T$ is the best improvement.
8Percentage uncertainty of a resistivity result (4 marks)
StretchMulti-step problem• Edexcel IAL Physics Unit 3 (WPH13) style• percentage uncertainty, resistivity, Unit 3
Question
Resistivity is found from $\rho = \dfrac{RA}{L}$ with a wire of cross-sectional area $A = \tfrac{\pi d^{2}}{4}$ . The measurements are: $R = 3.4 \pm 0.1\ \Omega$ ; diameter $d = 0.38\ \text{mm}$ read on a micrometer with resolution $0.01\ \text{mm}$ ; length $L = 1.000\ \text{m}$ read on a metre rule to the nearest mm. Estimate the percentage uncertainty in $\rho$ and comment on whether the experiment is acceptable. (4 marks)
Step-by-step solution
1. Step 1
  Percentage uncertainty in $R$ is given directly.
  $\%R = \frac{0.1}{3.4}\times100 = 2.9\%$
2. Step 2
  For the micrometer, a single reading has absolute uncertainty $= \tfrac{1}{2}\times$ resolution $= 0.005\ \text{mm}$ . Since $A \propto d^{2}$ , the diameter's percentage uncertainty is doubled.
  $\%d = \frac{0.005}{0.38}\times100 = 1.3\%, \qquad \%A = 2\times1.3 = 2.6\%$
3. Step 3
  For the metre rule, absolute uncertainty $= \tfrac{1}{2}\times1\ \text{mm} = 0.5\ \text{mm} = 0.0005\ \text{m}$ .
  $\%L = \frac{0.0005}{1.000}\times100 = 0.05\%$
4. Step 4
  $\rho = RA/L$ is a product/quotient, so add the percentage uncertainties (using $\%A$ , already doubled).
  $\%\rho = \%R + \%A + \%L = 2.9 + 2.6 + 0.05 = 5.6\%$
Answer
Percentage uncertainty in $\rho \approx 5.6\%$ . This is just above the ~5% rule-of-thumb, so the result is borderline; the diameter (via the $d^{2}$ term) and $R$ dominate, so improving those — e.g. measuring $d$ at several points and using a larger current range for $R$ — would most reduce it.
Examiner tip
Marks: (1) $\%R = 2.9\%$ ; (2) $\%d = 1.3\%$ with $\%A = 2\%d = 2.6\%$ ; (3) $\%L = 0.05\%$ ; (4) add to $\approx 5.6\%$ with a valid comment. The two traps: forgetting to double $\%d$ because $A\propto d^{2}$ , and using the full resolution (0.01 mm) instead of half of it for a single reading. A total above ~5% signals the experiment needs improvement.
9Uncertainty in a gradient by the max/min line method (5 marks)
StretchGraph or diagram• Edexcel IAL Physics Unit 3 (WPH13) style• gradient uncertainty, max/min line, Unit 3
Question
From plotted data with error bars, a student draws a best-fit line of gradient $0.420\ \text{V A}^{-1}$ , a steepest reasonable (maximum) line of gradient $0.450\ \text{V A}^{-1}$ , and a shallowest (minimum) line of gradient $0.390\ \text{V A}^{-1}$ , both still passing through the error bars. Determine the absolute and percentage uncertainty in the gradient. (5 marks)
Step-by-step solution
1. Step 1
  The max and min lines are the steepest and shallowest straight lines that still pass through (most of) the error bars. Their gradients bracket the best-fit gradient.
  $m_{\text{best}} = 0.420, \quad m_{\text{max}} = 0.450, \quad m_{\text{min}} = 0.390\ \text{V A}^{-1}$
2. Step 2
  The absolute uncertainty in the gradient is half the range of the max and min gradients.
  $\Delta m = \frac{m_{\text{max}} - m_{\text{min}}}{2} = \frac{0.450 - 0.390}{2}$
3. Step 3
  Evaluate the absolute uncertainty.
  $\Delta m = \frac{0.060}{2} = 0.030\ \text{V A}^{-1}$
4. Step 4
  The percentage uncertainty is the absolute uncertainty divided by the best-fit gradient.
  $\%\,m = \frac{\Delta m}{m_{\text{best}}}\times100 = \frac{0.030}{0.420}\times100$
5. Step 5
  Evaluate the percentage uncertainty.
  $\%\,m = 7.1\%$
Answer
Gradient $= 0.420 \pm 0.030\ \text{V A}^{-1}$ ; percentage uncertainty $\approx 7.1\%$ .
Examiner tip
Marks: (1) max/min lines defined via error bars; (2) $\Delta m = (m_{\max}-m_{\min})/2$ ; (3) $\Delta m = 0.030$ ; (4) %method; (5) $7.1\%$ . Common slips: taking the whole range $(m_{\max}-m_{\min})$ instead of half, and dividing by the max gradient rather than the best-fit gradient. A gradient uncertainty above ~5% suggests significant scatter worth reducing.

Model Answers — Practical Skills - Processing Results

High-scoring sample answers for practical skills - processing results on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
2 marks
Q (2 marks). State what is meant by a ‘line of best fit’ and why drawing one is preferable to joining the plotted points.
Model answer
A line of best fit is a single smooth line (straight or curved) drawn through a set of plotted points so that the points are balanced about it — roughly as many above as below — representing the underlying trend.

It is preferable to joining the points because it averages out the random scatter in individual readings, giving a more reliable estimate of the relationship (and of any gradient or intercept read from it) than a dot-to-dot line, which would just reproduce the noise.
Why this scores
Why this scores 2/2. (1) single line balancing points above and below; (2) it averages out random error / represents the trend. Describing it as ‘connecting the dots’ contradicts the definition and scores nothing.
Question 2
2 marks
Q (2 marks). A length is measured once with a metre rule graduated in millimetres, giving 42.6 cm. State the absolute uncertainty in this reading and hence write the result in the form (value ± uncertainty).
Model answer
For a single reading on an analogue scale the absolute uncertainty is half the smallest division. The smallest division is 1 mm, so the uncertainty is $\tfrac{1}{2}\times1\ \text{mm} = 0.5\ \text{mm} = 0.05\ \text{cm}$ .

The result is therefore $\mathbf{42.60 \pm 0.05\ \text{cm}}$ .
Why this scores
Why this scores 2/2. (1) uncertainty $= \tfrac12\times$ resolution $= 0.5\ \text{mm}$ ; (2) quoted as $42.60 \pm 0.05\ \text{cm}$ with consistent units. If the length is a difference of two rule readings the uncertainty would be twice this ( $\pm 0.1\ \text{cm}$ ) — a common follow-up.
Question 3
3 marks
Q (3 marks). Distinguish between a systematic error and a random error, giving one example of each, and state how each type affects a set of results.
Model answer
A systematic error shifts every reading by the same amount in the same direction, so all results are too high or too low by a fixed offset. An example is a zero error on a micrometer (or parallax when a scale is read consistently off-line). It reduces accuracy and cannot be reduced by repeating.

A random error causes readings to scatter unpredictably above and below the true value. An example is human reaction time when timing with a stopwatch (or fluctuating air currents). It reduces precision/reliability but can be reduced by repeating and averaging.
Why this scores
Why this scores 3/3. (1) systematic = same-direction offset with a valid example (zero error/parallax); (2) random = scatter with a valid example (reaction time); (3) effect on results — systematic affects accuracy and is not reduced by repeats, random is reduced by averaging. Naming the error type without the mechanism/effect loses the third mark.
Question 4
Edexcel IAL Physics Unit 3 (WPH13) style4 marks
Q (4 marks). In an experiment to find the resistivity of a wire, resistance $R$ is plotted against length $L$ for a wire of constant cross-sectional area $A = 2.0\times10^{-7}\ \text{m}^{2}$ . The best-fit straight line through the origin has a gradient of $0.85\ \Omega\,\text{m}^{-1}$ , found from a large triangle. (a) State why plotting $R$ against $L$ gives a straight line. (b) Determine the resistivity of the wire.
Model answer
(a) From $R = \dfrac{\rho L}{A}$ , with $\rho$ and $A$ constant, $R$ is directly proportional to $L$ . Comparing with $y = mx$ , a graph of $R$ (y) against $L$ (x) is therefore a straight line through the origin with gradient $= \dfrac{\rho}{A}$ .

(b) Rearranging, $\rho = \text{gradient}\times A$ :

$\rho = 0.85 \times (2.0\times10^{-7}) = 1.7\times10^{-7}\ \Omega\,\text{m}$

The resistivity of the wire is $\mathbf{1.7\times10^{-7}\ \Omega\,\text{m}}$ .
Why this scores
Why this scores 4/4. (a) 2 marks: $R = \rho L/A$ so $R \propto L$ , gradient $= \rho/A$ . (b) 2 marks: $\rho = \text{gradient}\times A = 1.7\times10^{-7}\ \Omega\,\text{m}$ with units. The gradient's units ( $\Omega\,\text{m}^{-1}$ ) times $A$ ( $\text{m}^{2}$ ) correctly give $\Omega\,\text{m}$ — a useful units check.
Question 5
Edexcel IAL Physics Unit 3 (WPH13) style4 marks
Q (4 marks). The volume of a sphere is found from $V = \tfrac{4}{3}\pi r^{3}$ . The radius is measured as $r = 12.0 \pm 0.2\ \text{mm}$ . Calculate the volume and its percentage uncertainty.
Model answer
The volume is:

$V = \tfrac{4}{3}\pi (12.0)^{3} = \tfrac{4}{3}\pi (1728) = 7.24\times10^{3}\ \text{mm}^{3}$

The percentage uncertainty in $r$ is:

$\%r = \frac{0.2}{12.0}\times100 = 1.67\%$

Because $V \propto r^{3}$ , the radius's percentage uncertainty is multiplied by the power 3 (the constant $\tfrac{4}{3}\pi$ has no uncertainty):

$\%V = 3\times\%r = 3\times1.67 = 5.0\%$

So $V = 7.24\times10^{3}\ \text{mm}^{3}$ with a percentage uncertainty of about 5.0%.
Why this scores
Why this scores 4/4. (1) $V = 7.24\times10^{3}\ \text{mm}^{3}$ ; (2) $\%r = 1.67\%$ ; (3) recognising the factor of 3 for the cube; (4) $\%V = 5.0\%$ . Forgetting to multiply by the power 3 (quoting 1.7%) is the standard error — a cubed term contributes THREE times its percentage uncertainty.
Question 6
4 marks
Q (4 marks). The intensity $I$ of light from a lamp is thought to vary with distance $d$ as $I = \dfrac{k}{d^{2}}$ , where $k$ is a constant. A student wants to test this using a log–log graph. State what should be plotted, what a straight line would confirm, and what the gradient should equal if the relationship holds.
Model answer
Rewrite the relationship as a power law, $I = k\,d^{-2}$ , and take logs (base 10) of both sides:

$\lg I = \lg k - 2\lg d$

Comparing with $y = mx + c$ , the student should plot $\lg I$ (y-axis) against $\lg d$ (x-axis).

A straight line would confirm that $I$ is a power of $d$ (i.e. that the log–log relationship holds). If the inverse-square law is correct, the gradient should equal $-2$ , and the intercept would be $\lg k$ .
Why this scores
Why this scores 4/4. (1) take logs / rearrange to $\lg I = -2\lg d + \lg k$ ; (2) plot $\lg I$ vs $\lg d$ ; (3) straight line confirms the power law; (4) gradient $= -2$ . Log–log (any base) is for power laws; using ln vs $d$ would be wrong here because the relationship is a power, not an exponential.
Question 7
Edexcel IAL Physics Unit 3 (WPH13) style4 marks
Q (4 marks). In an evaluation, a student writes that repeating readings makes an experiment ‘more valid’. Explain the difference between the validity and the reliability of an experiment, and state correctly what repeating readings improves.
Model answer
Validity concerns whether the experiment actually measures what it is intended to measure — i.e. whether it tests the stated relationship with all other variables controlled (a fair test). An invalid experiment gives the ‘right’ answer to the wrong question, no matter how many times it is repeated.

Reliability concerns whether the results are consistent and repeatable — whether repeating the measurement gives closely agreeing values (small random scatter).

The student is incorrect: repeating readings and averaging reduces the effect of random error, so it improves the reliability/precision, not the validity. Validity is improved by controlling variables and removing systematic errors, not by repeating.
Why this scores
Why this scores 4/4. (1) validity = measures what is intended / fair test; (2) reliability = consistent, repeatable results; (3) repeating reduces random error → improves reliability; (4) correcting the student: it does NOT improve validity. This validity/reliability mix-up is one of the most common evaluation errors flagged by examiners.
Question 8
Edexcel IAL Physics Unit 3 (WPH13) style6 marks
Q (6 marks). A student times 10 complete oscillations of a pendulum once with a hand-held stopwatch and divides by 10 to find the period $T = 1.85\ \text{s}$ . The percentage uncertainty in $T$ is estimated at about 5%. (a) Identify the main type of error in the timing and explain why it is significant here. (b) Describe two specific improvements to the method that would reduce the uncertainty in $T$ , explaining how each helps.
Model answer
(a) The main error is a random error from human reaction time at starting and stopping the stopwatch (about 0.2 s at each press). Because the timed interval for one oscillation is only ~1.85 s, a reaction-time uncertainty of a few tenths of a second is a large fraction of the reading, giving a percentage uncertainty of several percent. (There may also be a systematic component if the student consistently anticipates the stop.)

(b) Two specific improvements:

Time more oscillations (e.g. 20 or more) and divide by that number. The reaction-time uncertainty (a fixed ~0.2 s) is then spread over a much longer total time, so its percentage contribution to $T$ falls (e.g. over 20 oscillations it is roughly halved). Repeating the whole timing several times and taking a mean further reduces random scatter.

Use light gates / a photogate connected to a data-logger triggered as the pendulum passes a fixed point. Electronic timing removes the human reaction-time error entirely, cutting the dominant source of uncertainty in $T$ .
Why this scores
Why this scores 6/6. (a) 2 marks: random error from reaction time AND why it matters (large fraction of a short time). (b) 4 marks: two SPECIFIC, feasible improvements EACH with the mechanism — time more oscillations (spreads the fixed error, lowering %), and electronic timing (removes reaction time). Vague answers (‘be more careful’, ‘use a better stopwatch’) without the mechanism score nothing. Improvements must target the NAMED error.
Question 9
Edexcel IAL Physics Unit 3 (WPH13) style6 marks
Q (6 marks). A student plots current $I$ (y-axis) against voltage $V$ (x-axis) for a metal wire at constant temperature and obtains a straight line through the origin, except for one point that lies clearly off the line. (a) State what this off-line point is called and how the student should treat it when drawing the best-fit line. (b) The best-fit line passes through $(0,0)$ and $(6.0\ \text{V},\ 0.30\ \text{A})$ . Determine the gradient with units, and state what physical quantity its reciprocal represents.
Model answer
(a) The off-line point is an anomaly (anomalous result). The student should ignore it when drawing the best-fit line (draw the line through the trend of the remaining points), and ideally check it by repeating that reading; it should not be allowed to pull the line towards it. If time allows, the cause (e.g. a misread scale or a loose contact) should be identified.

(b) Using a large triangle from the origin to $(6.0\ \text{V},\ 0.30\ \text{A})$ :

$\text{gradient} = \frac{\Delta I}{\Delta V} = \frac{0.30 - 0}{6.0 - 0} = 0.050\ \text{A V}^{-1}$

For an $I$ – $V$ graph the gradient is $\dfrac{I}{V} = \dfrac{1}{R}$ , so its reciprocal is the resistance $R$ of the wire:

$R = \frac{1}{\text{gradient}} = \frac{1}{0.050} = 20\ \Omega$
Why this scores
Why this scores 6/6. (a) 3 marks: named as an anomaly; ignore it for the best-fit line / do not let it pull the line; check by repeating. (b) 3 marks: gradient $= 0.050\ \text{A V}^{-1}$ with units from a large triangle; reciprocal $= R$ ; $R = 20\ \Omega$ . Including an anomaly in the best-fit line, or omitting the gradient's units, are the usual lost marks.
Question 10
Edexcel IAL Physics Unit 3 (WPH13) style6 marks
Q (6 marks). A student investigates how the time period $T$ of a mass–spring oscillator depends on the mass $m$ , expecting $T = 2\pi\sqrt{\dfrac{m}{k}}$ . (a) Show that plotting $T^{2}$ against $m$ should give a straight line through the origin, and state what the gradient represents. (b) The best-fit line has gradient $2.5\ \text{s}^{2}\,\text{kg}^{-1}$ , with a maximum line of $2.7$ and a minimum line of $2.3\ \text{s}^{2}\,\text{kg}^{-1}$ . Determine the spring constant $k$ and the percentage uncertainty in the gradient.
Model answer
(a) Squaring the relationship:

$T^{2} = \frac{4\pi^{2}}{k}\,m$

Comparing with $y = mx$ (with $y = T^{2}$ , $x = m$ ), a graph of $T^{2}$ against $m$ is a straight line through the origin with gradient $= \dfrac{4\pi^{2}}{k}$ .

(b) Rearranging for $k$ using the best-fit gradient:

$k = \frac{4\pi^{2}}{\text{gradient}} = \frac{4\pi^{2}}{2.5} = 15.8\ \text{N m}^{-1}$

The uncertainty in the gradient (max/min method):

$\Delta m = \frac{2.7 - 2.3}{2} = 0.20\ \text{s}^{2}\,\text{kg}^{-1}$

$\%\,\text{gradient} = \frac{0.20}{2.5}\times100 = 8.0\%$

So $k = 15.8\ \text{N m}^{-1}$ and the gradient (and hence $k$ ) has a percentage uncertainty of about 8.0%.
Why this scores
Why this scores 6/6. (a) 3 marks: square to $T^{2} = (4\pi^{2}/k)m$ , so $T^{2}\propto m$ , gradient $= 4\pi^{2}/k$ . (b) 3 marks: $k = 4\pi^{2}/\text{gradient} = 15.8\ \text{N m}^{-1}$ ; $\Delta m = (2.7-2.3)/2 = 0.20$ ; $\%\text{gradient} = 8.0\%$ . Because $k = 4\pi^{2}/\text{gradient}$ (a single power of the gradient), the percentage uncertainty in $k$ equals that in the gradient (8.0%). Taking the full max−min range, not half, is the classic slip.

Key Formulae — Practical Skills - Processing Results

The formulae you need to memorise for practical skills - processing results on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Absolute uncertainty of a single reading
$\Delta x = \tfrac{1}{2}\times\text{resolution}$
$\Delta x$
absolute uncertainty in the reading
$resolution$
smallest division / step of the instrument
When to use
For a single reading on an analogue or digital scale (e.g. a metre rule to 1 mm gives $\Delta x = 0.5\ \text{mm}$ ). Must be recalled for Unit 3.
Example
Micrometer resolution $0.01\ \text{mm}$ : single reading uncertainty $= 0.005\ \text{mm}$ .
Absolute uncertainty from repeated readings
$\Delta x = \frac{x_{\text{max}} - x_{\text{min}}}{2}$
$x_{\text{max}}$
largest repeated value
$x_{\text{min}}$
smallest repeated value
$\Delta x$
absolute uncertainty (half the spread)
When to use
When several repeats of the same measurement are taken; the mean is quoted with this ± uncertainty (half the range).
Example
Times 20.1, 20.3, 20.5 s: mean 20.3 s, $\Delta t = (20.5-20.1)/2 = 0.2\ \text{s}$ .
Percentage uncertainty
$\%\,x = \frac{\Delta x}{x}\times100\%$
$\Delta x$
absolute uncertainty
$x$
measured value
When to use
Converting an absolute uncertainty into a fraction of the reading, so uncertainties in different quantities can be combined.
Example
$R = 3.4 \pm 0.1\ \Omega$ : $\%R = (0.1/3.4)\times100 = 2.9\%$ .
Combining uncertainties (products and quotients)
$\text{if } Q = \frac{a\,b}{c} \;\Rightarrow\; \%Q = \%a + \%b + \%c$
$Q$
derived quantity
$\%a,\ \%b,\ \%c$
percentage uncertainties of the measured quantities
When to use
For any quantity formed by multiplying and/or dividing measurements: ADD the percentage uncertainties (never the absolute ones).
Example
$\rho = RA/L$ : $\%\rho = \%R + \%A + \%L$ .
Combining uncertainties (powers)
$\text{if } Q = x^{n} \;\Rightarrow\; \%Q = |n|\times\%x$
$n$
power to which the quantity is raised
$\%x$
percentage uncertainty in $x$
When to use
When a measured quantity is raised to a power. A squared term ( $n=2$ ) contributes TWICE its percentage uncertainty; area $A\propto d^{2}$ means $\%A = 2\,\%d$ .
Example
$T$ has $\%T = 1.08\%$ ; in $g\propto 1/T^{2}$ it contributes $2\times1.08 = 2.16\%$ .
Gradient of a straight-line graph
$m = \frac{\Delta y}{\Delta x} = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$m$
gradient (units = y-units ÷ x-units)
$(x_{1},y_{1}),(x_{2},y_{2})$
two points on the best-fit line, far apart
When to use
Reading a rate/constant from a linear graph. Use a LARGE triangle (≥ half the line) and always STATE the units of the gradient.
Example
v–t line through $(1,4)$ and $(9,28)$ : $m = 24/8 = 3.0\ \text{m s}^{-2}$ .
Uncertainty in a gradient (max/min lines)
$\Delta m = \frac{m_{\text{max}} - m_{\text{min}}}{2}$
$m_{\text{max}}$
gradient of the steepest reasonable line
$m_{\text{min}}$
gradient of the shallowest reasonable line
$\Delta m$
absolute uncertainty in the gradient
When to use
Estimating the uncertainty in a best-fit gradient from data with error bars; the max/min lines still pass through the error bars.
Example
$m_{\max}=0.450$ , $m_{\min}=0.390$ : $\Delta m = 0.060/2 = 0.030$ .
Linearising a power law (log–log)
$y = Ax^{n} \;\Rightarrow\; \lg y = n\lg x + \lg A$
$n$
power = gradient of the $\lg y$ vs $\lg x$ line
$A$
constant; $\lg A$ = y-intercept, so $A = 10^{\text{intercept}}$
When to use
Testing/using a power-law relationship. Plot $\lg y$ against $\lg x$ ; gradient gives the power $n$ , intercept gives $\lg A$ .
Example
$T = kL^{0.5}$ : plot $\lg T$ vs $\lg L$ , gradient $= 0.50$ .
Linearising an exponential (ln)
$y = A e^{kx} \;\Rightarrow\; \ln y = kx + \ln A$
$k$
constant = gradient of the $\ln y$ vs $x$ line
$A$
constant; $\ln A$ = y-intercept, so $A = e^{\text{intercept}}$
When to use
Testing/using an exponential relationship (decay or growth). Plot $\ln y$ against $x$ ; gradient $= k$ (negative for decay), intercept $= \ln A$ . Use ln (base $e$ ), NOT lg.
Example
$A = A_{0}e^{-\lambda t}$ : plot $\ln A$ vs $t$ , gradient $= -\lambda$ .

Key Definitions and Keywords — Practical Skills - Processing Results

Definitions to memorise and the exact keywords mark schemes credit for practical skills - processing results answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Independent variable
Examiner keyword
The quantity the experimenter deliberately changes/sets. It is plotted on the x-axis (horizontal) of a results graph.
Dependent variable
Examiner keyword
The quantity measured in response to the independent variable. It is plotted on the y-axis (vertical) of a results graph.
Line of best fit
Examiner keyword
A single smooth line (straight or curved) drawn through plotted points so they are balanced about it, representing the trend and averaging out random scatter. It is NOT a dot-to-dot line.
Gradient
Examiner keyword
The slope of a graph, $m = \dfrac{\Delta y}{\Delta x}$ , read from a large triangle spanning at least half the line. Its units are the y-axis units divided by the x-axis units, and it often equals a physical quantity.
y-intercept
Examiner keyword
The value of $y$ where the best-fit line crosses the y-axis ( $x = 0$ ). In $y = mx + c$ it is the constant $c$ , and it often has a physical meaning (e.g. $\ln A_{0}$ on a log graph).
Linearisation
Examiner keyword
Rearranging a non-linear relationship into the straight-line form $y = mx + c$ so a graph gives a constant from its gradient/intercept (e.g. plotting $s$ vs $t^{2}$ , or $\ln y$ vs $x$ ).
Absolute uncertainty
Examiner keyword
The size of the possible error in a reading, in the same units as the measurement: $\tfrac{1}{2}\times$ resolution for a single reading, or $\dfrac{x_{\text{max}}-x_{\text{min}}}{2}$ for repeats. Quoted as value ± uncertainty.
Percentage uncertainty
Examiner keyword
The absolute uncertainty expressed as a percentage of the reading, $\dfrac{\Delta x}{x}\times100\%$ . Percentage uncertainties are added for products/quotients and multiplied by the power for a raised term.
Systematic error
Examiner keyword
An error that shifts all readings by the same amount in the same direction (e.g. a zero error or consistent parallax). It affects accuracy and is NOT reduced by repeating.
Random error
Examiner keyword
An error causing readings to scatter unpredictably about the true value (e.g. reaction time, air currents). It affects precision/reliability and is reduced by repeating and averaging.
Zero error
A type of systematic error where an instrument does not read zero when the true value is zero (e.g. a micrometer offset). It offsets every reading and should be subtracted.
Validity
Examiner keyword
Whether an experiment actually measures what it is intended to measure — a fair test with the other variables controlled and systematic errors removed. Repeating readings does NOT improve validity.
Reliability
Examiner keyword
Whether results are consistent and repeatable — small random scatter when the measurement is repeated. Improved by repeating readings and taking a mean.
Anomaly (anomalous result)
Examiner keyword
A result that lies clearly off the trend of the other data. It should be checked/repeated and ignored when drawing the line of best fit, not allowed to pull the line towards it.
Resolution
The smallest change in a quantity an instrument can detect — the smallest scale division (analogue) or the last digit (digital). It sets the absolute uncertainty of a single reading (half the resolution).
Maximum/minimum line
The steepest and shallowest straight lines that still pass through (most of) the error bars. Their gradients bracket the best-fit gradient and give the gradient uncertainty, $\Delta m = \dfrac{m_{\text{max}}-m_{\text{min}}}{2}$ .

Common Mistakes and Misconceptions — Practical Skills - Processing Results

The traps other students keep falling into on practical skills - processing results questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Using a small gradient triangle (a fraction of the line) to read the gradient
Edexcel IAL Physics Unit 3 (WPH13) examiner reports
Why it happens
Students pick two convenient nearby grid points instead of two well-separated points on the line.
How to avoid it
Draw a large triangle spanning at least half the plotted line and read the coordinates to the scale's precision. A big triangle reduces the fractional error in $\Delta x$ and $\Delta y$ and protects the accuracy mark.
✕Joining the plotted points dot-to-dot instead of drawing a single line of best fit
Edexcel IAL Physics Unit 3 (WPH13) examiner reports
Why it happens
Students think every point must be touched, or don't trust a line that misses points.
How to avoid it
Draw one thin, straight (or smoothly curved) best-fit line with points balanced above and below it. The whole purpose of a best-fit line is to average out random scatter — a zig-zag reproduces the noise.
✕Quoting a gradient (or a quantity derived from it) with no units
Edexcel IAL Physics Unit 3 (WPH13) examiner reports
Why it happens
The gradient is treated as a pure number rather than a ratio of two measured quantities.
How to avoid it
Always give the gradient's units as (y-axis units) ÷ (x-axis units) — e.g. a v–t gradient is in $\text{m s}^{-2}$ . Stating the units also acts as a check that the gradient is the physical quantity you expect.
✕Not multiplying the percentage uncertainty by the power for a squared (or cubed) term
Edexcel IAL Physics Unit 3 (WPH13) examiner reports
Why it happens
Students add each measurement's percentage uncertainty once, ignoring that some are raised to a power.
How to avoid it
For a term $x^{n}$ the percentage uncertainty is $|n|\times\%x$ . A squared quantity (e.g. $A\propto d^{2}$ , or $T^{2}$ ) contributes twice its percentage uncertainty; a cube contributes three times.
✕Confusing ln and log (base 10) when linearising, e.g. plotting lg for an exponential
Edexcel IAL Physics Unit 3 (WPH13) examiner reports
Why it happens
Both ‘log’ buttons exist on the calculator and the difference in method is not memorised.
How to avoid it
Use ln (base $e$ ) for exponentials $y = Ae^{kx}$ (plot $\ln y$ vs $x$ , gradient $= k$ ), and lg (base 10) for power laws $y = Ax^{n}$ (plot $\lg y$ vs $\lg x$ , gradient $= n$ ). Mixing them scales the gradient by $\ln 10 \approx 2.303$ .
✕Mixing up validity and reliability — e.g. saying ‘repeating makes the result more valid’
Edexcel IAL Physics Unit 3 (WPH13) examiner reports
Why it happens
The two evaluation terms sound similar and are used loosely in everyday speech.
How to avoid it
Reliability = consistent, repeatable results (improved by repeating/averaging, which cuts random error). Validity = measuring the right thing in a fair test (improved by controlling variables/removing systematic error). Repeating improves reliability, NOT validity.

Practical Skills - Processing Results — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Practical Skills - Processing Results

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Choosing sensible graph scales (3 marks)

2Line of best fit vs dot-to-dot (2 marks)

3Gradient with a large triangle and its units (3 marks)

4Gradient of a stress–strain graph gives the Young modulus (3 marks)

5Linearising a power law — what to plot (4 marks)

6Linearising an exponential decay — using ln (4 marks)

7Combining uncertainties with a squared term (4 marks)

8Percentage uncertainty of a resistivity result (4 marks)

9Uncertainty in a gradient by the max/min line method (5 marks)

Absolute uncertainty of a single reading

Absolute uncertainty from repeated readings

Percentage uncertainty

Combining uncertainties (products and quotients)

Combining uncertainties (powers)

Gradient of a straight-line graph

Uncertainty in a gradient (max/min lines)

Linearising a power law (log–log)

Linearising an exponential (ln)

Independent variable

Dependent variable

Line of best fit

Gradient

y-intercept

Linearisation

Absolute uncertainty

Percentage uncertainty

Systematic error

Random error

Zero error

Validity

Reliability

Anomaly (anomalous result)

Resolution

Maximum/minimum line

✕Using a small gradient triangle (a fraction of the line) to read the gradient

✕Joining the plotted points dot-to-dot instead of drawing a single line of best fit

✕Quoting a gradient (or a quantity derived from it) with no units

✕Not multiplying the percentage uncertainty by the power for a squared (or cubed) term

✕Confusing ln and log (base 10) when linearising, e.g. plotting lg for an exponential

✕Mixing up validity and reliability — e.g. saying ‘repeating makes the result more valid’

Practical Skills - Processing Results — frequently asked questions