Measurements and errors: A-level Physics AQA revision notes

3.1.1

SI units and prefixes

SI units 3.1.1

Definitions

Base SI unit: one of the seven units from which every other unit in this course is built — metre (m), kilogram (kg), second (s), ampere (A), kelvin (K), mole (mol), candela (cd).

Key results

Newton (force), from $F = ma$ : $N = kg \cdot m \cdot s^{- 2}$ .
Joule (energy), from $W = F d$ : $J = kg \cdot m^{2} \cdot s^{- 2}$ .
Watt (power), from $P = \frac{W}{t}$ : $W = kg \cdot m^{2} \cdot s^{- 3}$ .

Notes

Every derived unit's combination of base units follows directly from the equation that defines the quantity — it is never arbitrary.
Checking that both sides of an equation reduce to the same combination of base units — homogeneity of units — is one of the fastest ways to catch an algebra error before reaching for a calculator.
If a derived expression for a speed comes out with units of $m^{2}$ , something has gone wrong upstream, regardless of how confident the algebra looked.
This dimensional check works both ways: it can confirm a suspicious-looking formula is at least dimensionally plausible, though it can never confirm the formula's numerical constant (e.g. a factor of $\frac{1}{2}$ ) is correct.

Worked examples

Worked example 3.1.1 · 4 marks

A student derives an expression for the terminal velocity of a small sphere falling through a viscous fluid and obtains:

v = \frac{2 r ^{2} ( ρ _{s} - ρ _{f} ) g}{9 η}

where $r$ is the sphere's radius (m), $ρ_{s}$ and $ρ_{f}$ are densities (kg m $^{- 3}$ ), $g$ is the acceleration due to gravity (m s $^{- 2}$ ) and $η$ is the fluid's viscosity (Pa s, equivalent to kg m $^{- 1}$ s $^{- 1}$ ).

Show by dimensional analysis that the right-hand side of this expression has the units of velocity, m s $^{- 1}$ .

Show worked solution

Numerator units (the constants 2 and 9 carry no units): $r^{2} \to m^{2}$ ;

(ρ_{s} - ρ_{f}) \to kg m^{- 3}

$g \to m s^{- 2}$ .

Combining:

m^{2} \times kg m^{- 3} \times m s^{- 2} = kg m^{(2 - 3 + 1)} s^{- 2} = kg s^{- 2}

Dividing by the units of $η$ , $kg m^{- 1} s^{- 1}$ , is equivalent to multiplying by $kg^{- 1} m s$ :

kg s^{- 2} \times kg^{- 1} m s = m s^{- 1}

This is the unit of velocity, so the expression is dimensionally homogeneous.

Mark scheme · 4 marks

Writes the units of $r^{2}$ as $m^{2}$ 1 mark
Combines the units of $r^{2}$ , $(ρ_{s} - ρ_{f})$ and $g$ to give $kg s^{- 2}$ for the numerator 1 mark
Writes the units of $η$ as $kg m^{- 1} s^{- 1}$ and divides the numerator by it correctly 1 mark
Shows the result simplifies to $m s^{- 1}$ , confirming the expression is homogeneous 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.1.1 · 3 marks

One derivation of the pressure of an ideal gas gives:

P = \frac{1}{3} \frac{N m ⟨ c ^{2} ⟩}{V}

where $N$ is the number of molecules (no units), $m$ is the mass of one molecule (kg), $⟨ c^{2} ⟩$ is the mean square speed of the molecules (m $^{2}$ s $^{- 2}$ ) and $V$ is the volume occupied (m $^{3}$ ).

Determine the base SI units this expression predicts for $P$ , and state whether this is consistent with the pascal.

Show worked solution

Combining $m$ and $⟨ c^{2} ⟩$ :

kg \times m^{2} s^{- 2} = kg m^{2} s^{- 2}

Dividing by the units of $V$ :

\frac{kg m ^{2} s ^{- 2}}{m ^{3}} = kg m^{- 1} s^{- 2}

( $N$ and $\frac{1}{3}$ are dimensionless).

The known unit of pressure is:

Pa = N m^{- 2} = (kg m s^{- 2}) m^{- 2} = kg m^{- 1} s^{- 2}

which matches exactly, so the expression is dimensionally consistent with the pascal.

Mark scheme · 3 marks

Combines the units of $m$ and $⟨ c^{2} ⟩$ to give $kg m^{2} s^{- 2}$ 1 mark
Divides by the units of $V$ to give $kg m^{- 1} s^{- 2}$ for $P$ 1 mark
Derives $kg m^{- 1} s^{- 2}$ for the pascal from $N m^{- 2}$ and confirms the two match 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.1.1 · 3 marks

A student is checking two proposed equations for the range $R$ of a projectile launched at speed $u$ and angle $θ$ : Equation A:

R = \frac{u ^{2} sin ( 2 θ )}{g}

Equation B:

R = \frac{u ^{2} sin ( 2 θ )}{g ^{2}}

where $u$ is speed (m s $^{- 1}$ ) and $g$ is the acceleration due to gravity (m s $^{- 2}$ ).

Using dimensional analysis, determine which equation (if either) is dimensionally consistent with $R$ being a distance (m), and explain your reasoning.

Show worked solution

$sin (2 θ)$ is dimensionless, so it can be ignored in a unit check.

Equation A:

\frac{u ^{2}}{g} \to \frac{m ^{2} s ^{- 2}}{m s ^{- 2}} = m

which is a unit of distance, so Equation A is dimensionally consistent.

Equation B:

\frac{u ^{2}}{g ^{2}} \to \frac{m ^{2} s ^{- 2}}{m ^{2} s ^{- 4}} = s^{2}

a unit of time squared, not distance — so Equation B cannot be correct, regardless of how the algebra that produced it looked.

Mark scheme · 3 marks

States that $sin (2 θ)$ is dimensionless and can be ignored in the unit check 1 mark
Derives the units of $u^{2} / g$ as $m$ , confirming Equation A is dimensionally consistent 1 mark
Derives the units of $u^{2} / g^{2}$ as $s^{2}$ , showing Equation B is not dimensionally consistent with a distance 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

3.1.2

Limitation of physical measurements

Uncertainty 3.1.2

Definitions

Absolute uncertainty, $Δ x$ : the range within which a measurement's true value is believed to lie, expressed in the same units as the quantity itself.
Percentage uncertainty: $\frac{Δ x}{x} \times 100%$ — lets the precision of very different measurements be compared on equal terms.

Key results

A single reading's uncertainty is usually half the smallest division on the instrument (e.g. $\pm 0.5 mm$ on a millimetre scale).
A repeated measurement's uncertainty is usually half the range of the repeats: $Δ x = \frac{x _{m a x} - x _{m i n}}{2}$ .
Multiplying/dividing ( $Q = \frac{A B}{C}$ ): percentage uncertainties add, $\frac{Δ Q}{Q} = \frac{Δ A}{A} + \frac{Δ B}{B} + \frac{Δ C}{C}$ .
Raising to a power ( $Q = A^{n}$ ): percentage uncertainty is multiplied by that power, $\frac{Δ Q}{Q} = n \frac{Δ A}{A}$ .
Adding/subtracting ( $Q = A + B$ ): absolute uncertainties add, $Δ Q = Δ A + Δ B$ .

Notes

An uncertainty is not a mistake — it is an honest statement of how precisely a quantity is known, and percentage uncertainty is what you carry through a calculation once quantities combine.
Common pitfall: reporting a final answer to more significant figures than its uncertainty justifies. If the percentage uncertainty is 8%, quoting a result to five significant figures claims a precision the data does not support.
Round the value so its last significant figure is roughly where the uncertainty starts to bite.
When a quantity is raised to a power inside a larger expression (not on its own), still apply the power rule to that quantity's own percentage uncertainty before combining it with the others by addition — don't try to apply the addition and power rules simultaneously in one step.

Data analysis 3.1.2

Method

Identify the relationship theory predicts (e.g. $y = m x + c$ ) and choose axes so the two plotted quantities are linearly related — not necessarily the raw measured quantities.
Plot the linearised quantities; a straight line through the data confirms the relationship and lets gradient and intercept be read directly.
Estimate the gradient's uncertainty from the steepest and shallowest lines that still fit within the data's error bars (the 'worst-fit' lines), taking half the difference between their gradients.

In practiceFor a pendulum, the gradient of $T^{2}$ against $l$ is 4.02 s $^{2}$ m $^{- 1}$ ; the worst-fit lines give 3.90 and 4.16. Find $g$ with its uncertainty.

$T^{2} = \frac{4 π ^{2}}{g} l$ plot $T^{2}$ against $l$ : a line through the origin
$g = \frac{4 π ^{2}}{gradient} = \frac{39.48}{4.02} = 9.82 m s^{- 2}$
$Δ (gradient) = \frac{1}{2} (4.16 - 3.90) = 0.13 (3.2%)$ half the spread of the worst-fit lines
$g = 9.8 \pm 0.3 m s^{- 2}$ the same percentage uncertainty

Notes

A straight-line graph is the most information-dense way to test a proposed relationship — a constant gradient and intercept are easy to read and easy to check against theory.
Example: if $s = u t + \frac{1}{2} a t^{2}$ and $u = 0$ , plotting $s$ against $t^{2}$ gives a straight line through the origin with gradient $\frac{a}{2}$ , far easier to extract a reliable value of $a$ from than fitting a parabola to $s$ against $t$ directly.
A percentage uncertainty quoted without any indication of how it was obtained is a claim, not evidence.
The 'worst-fit line' method for gradient uncertainty is preferred over a purely statistical calculation at this level because it visibly respects each point's own error bar, rather than treating every point as equally reliable regardless of its stated uncertainty.

Worked examples

Worked example 3.1.2 · 3 marks

A student times 20 oscillations of a pendulum with a stopwatch, repeating the measurement five times and averaging the results to reduce random error.

She later discovers the stopwatch was started 0.3 s late on every single trial.

(a) State and explain whether repeating and averaging the readings removes the effect of this late start.

(b) State the type of error the late start represents, and explain how it differs from random error.

Show worked solution

(a) No.

Repeating and averaging only reduces random error, since random fluctuations scatter above and below the true value and partly cancel when averaged.

The 0.3 s late start is applied identically to every single trial, so it shifts the average by the same fixed amount and is not reduced by averaging.

(b) This is a systematic error: it displaces every reading in the same direction by the same amount, so the measured value is consistently offset from the true value.

This differs from random error, which causes readings to scatter unpredictably both above and below the true value, and whose effect can therefore be reduced by taking more readings and averaging.

Mark scheme · 3 marks

States that averaging repeated readings does not remove the effect of the late start 1 mark
Explains that the 0.3 s offset is applied identically to every trial, so averaging cannot cancel it 1 mark
Identifies the error as systematic and explains it shifts every reading the same way, unlike random error which scatters and partially cancels on averaging 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.1.2 · 6 marks

Trial	1	2	3	4	5
Time for 20 oscillations / s	36.4	36.0	35.8	36.2	36.1

A student investigates a simple pendulum of length:

l = (0.800 \pm 0.002) m

She times 20 complete oscillations in five separate trials, recording the results shown in the table below.

Using:

g = \frac{4 π ^{2} l}{T ^{2}}

(a) calculate the mean period $T$ and its absolute uncertainty;

(b) calculate $g$ and the percentage uncertainty in $g$ .

Show worked solution

Mean time for 20 oscillations:

= \frac{36.4 + 36.0 + 35.8 + 36.2 + 36.1}{5} = 36.1 s

so:

T = \frac{36.1}{20} = 1.805 s

The uncertainty in the 20-oscillation time is half the range,

\frac{36.4 - 35.8}{2} = 0.3 s

dividing by 20 gives:

Δ T = \frac{0.3}{20} = 0.015 s

(b):

g = \frac{4 π ^{2} \times 0.800}{1.80 5 ^{2}} = \frac{31.58}{3.258} = 9.69 m s^{- 2}

Percentage uncertainty in $l$ :

\frac{0.002}{0.800} \times 100% = 0.25%

Percentage uncertainty in $T$ :

\frac{0.015}{1.805} \times 100% = 0.83%

since $g \propto T^{- 2}$ , the power rule doubles this to:

2 \times 0.83% = 1.66%

Adding (product/quotient rule): percentage uncertainty in:

g = 0.25% + 1.66% \approx 1.9%

Mark scheme · 6 marks

Calculates the mean time for 20 oscillations = 36.1 s 1 mark
Calculates the period $T = 36.1/20 = 1.805 s$ 1 mark
Calculates $Δ T$ as half the range of the 20-oscillation times divided by 20, giving 0.015 s 1 mark
Calculates $g = 4 π^{2} l / T^{2} \approx 9.69 m s^{- 2}$ 1 mark
Calculates percentage uncertainty in $l$ (0.25%) and doubles the percentage uncertainty in $T$ using the power rule (2×0.83% = 1.66%) 1 mark
Adds the two percentage uncertainties to give a total percentage uncertainty in $g$ of approximately 1.9% 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.1.2 · 4 marks

In an experiment to find the acceleration $a$ of a trolley released from rest, a student measures the distance travelled $s$ for various times $t$ and plots $s$ against $t^{2}$ , since:

s = \frac{1}{2} a t^{2}

predicts a straight line through the origin of gradient $\frac{a}{2}$ .

The best-fit line through the data has gradient $2.40 m s^{- 2}$ .

The steepest line that still passes through every point's error bars ('worst fit') has gradient $2.58 m s^{- 2}$ , and the shallowest such line has gradient $2.22 m s^{- 2}$ .

(a) Calculate $a$ from the best-fit gradient.

(b) Calculate the absolute uncertainty in $a$ .

Show worked solution

(a) Gradient:

= \frac{a}{2} = 2.40 m s^{- 2}

so:

a = 2 \times 2.40 = 4.80 m s^{- 2}

(b) The uncertainty in the gradient is half the difference between the two worst-fit gradients:

\frac{2.58 - 2.22}{2} = 0.18 m s^{- 2}

Since $a$ is twice the gradient, its uncertainty scales by the same factor:

Δ a = 2 \times 0.18 = 0.36 m s^{- 2}

So:

a = (4.80 \pm 0.36) m s^{- 2}

Mark scheme · 4 marks

Calculates $a = 2 \times 2.40 = 4.80 m s^{- 2}$ from the best-fit gradient 1 mark
Calculates the gradient uncertainty as half the difference between the worst-fit gradients: $(2.58 - 2.22) /2 = 0.18 m s^{- 2}$ 1 mark
Scales the gradient uncertainty by the same factor (×2) used to convert gradient to $a$ , giving $Δ a = 0.36 m s^{- 2}$ 1 mark
States the final result $a = (4.80 \pm 0.36) m s^{- 2}$ 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

3.1.3

Estimation of physical quantities

Estimation of physical quantities 3.1.3

Definitions

Order-of-magnitude estimate: a value quoted to roughly one significant figure or the nearest power of ten, used when only a rough approximation is needed or possible.

Notes

Estimating a physical quantity means combining a small number of known or reasonably assumed facts to reach a sensible approximate value — not carrying out a precise calculation from exact data.
A good estimate breaks an unfamiliar quantity into simpler quantities that can each be judged with reasonable confidence, then combines them (sometimes called Fermi estimation) — e.g. estimating the number of air molecules in a room from its volume, atmospheric pressure, temperature and the ideal gas equation, rather than looking the number up directly.
Sensible reference values for everyday quantities are worth keeping in mind as anchors: an adult's mass ( $\sim 70 kg$ ), walking speed ( $\sim 1.5 m s^{- 1}$ ), the diameter of a human hair ( $\sim 0.1 mm$ ), atmospheric pressure ( $\sim 1 0^{5} Pa$ ), and the speed of sound in air ( $\sim 340 m s^{- 1}$ ) — comparing an unfamiliar estimate against one of these is often the fastest way to sanity-check it.
Examiners mark estimation questions on the validity of the method and the reasonableness of any assumed values, not on matching one single 'correct' numerical answer — showing the reasoning and assumptions explicitly is essential, since two candidates with different (but both reasonable) assumed values can both earn full credit.
A result that comes out many orders of magnitude away from a sensible reference value is a strong signal of an arithmetic or unit error, even before checking the working in detail.

Worked examples

Worked example 3.1.3 · 5 marks

Estimate the number of air molecules in a school classroom of dimensions approximately $8 m \times 6 m \times 3 m$ , at atmospheric pressure and room temperature.

State any assumed values, show your method clearly, and comment on which assumption most affects your answer.

Show worked solution

Assume atmospheric pressure $p \approx 1 \times 1 0^{5} Pa$ and room temperature $T \approx 293 K$ ( $20\,^\circ\text{C}$ ).

Volume of the room:

V = 8 \times 6 \times 3 = 144 m^{3}

Treating air as an ideal gas, $p V = N k T$ with the Boltzmann constant:

k = 1.38 \times 1 0^{- 23} J K^{- 1}

so:

N = \frac{p V}{k T} = \frac{1 \times 1 0 ^{5} \times 144}{1.38 \times 1 0 ^{- 23} \times 293} = \frac{1.44 \times 1 0 ^{7}}{4.04 \times 1 0 ^{- 21}} \approx 3.6 \times 1 0^{27}

molecules.

The method's validity matters more than matching this exact figure: the assumed temperature affects the answer only slightly, but if the room were not sealed at full atmospheric pressure (e.g. a door left open changing local pressure) the estimate would be affected more strongly, since $N$ is directly proportional to $p$ .

Mark scheme · 5 marks

States reasonable assumed values for pressure (~ $1 \times 1 0^{5} Pa$ ) and temperature (~290–293 K) 1 mark
Calculates the room's volume $V = 8 \times 6 \times 3 = 144 m^{3}$ 1 mark
Quotes and uses $N = p V / (k T)$ with a correct value for the Boltzmann constant 1 mark
Calculates $N$ to the correct order of magnitude, $\sim 1 0^{27}$ molecules, consistent with the candidate's own assumed values 1 mark
Comments that the method and reasonableness of assumptions are what is credited, and identifies which assumption most affects the result 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.1.3 · 4 marks

Estimate the average power output of a person walking up a flight of stairs.

State any assumed values, show your method clearly, and note that a numerical answer alone will not gain full credit.

Show worked solution

Assume a typical adult mass $m \approx 70 kg$ , a flight height $h \approx 3 m$ , and a time to climb it $t \approx 8 s$ .

Work done against gravity:

W = m g h = 70 \times 9.81 \times 3 \approx 2.1 \times 1 0^{3} J

Average power:

P = \frac{W}{t} = \frac{2.1 \times 1 0 ^{3}}{8} \approx 260 W

i.e. of order $1 0^{2} W$ .

This is credited for the method and the reasonableness of the assumed values (a candidate assuming a faster or slower climb, or a taller flight, would reach a different but equally valid figure), not for matching one fixed numerical answer.

Mark scheme · 4 marks

States reasonable assumed values for mass (~70 kg), height climbed (~3 m) and time taken (~8–10 s) 1 mark
Uses $W = m g h$ to calculate work done, of order $1 0^{3} J$ 1 mark
Divides by the assumed time to calculate power, of order $1 0^{2} W$ 1 mark
States that the method and reasonableness of the assumptions are what is credited, not one fixed numerical answer 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.1.3 · 4 marks

Estimate the pressure exerted on the ground by a person of mass 70 kg standing still on one foot.

State any assumed values, show your method clearly, and use a sensible reference value to check your answer is reasonable.

Show worked solution

Assume mass $m \approx 70 kg$ and a foot's contact area with the ground $A \approx 0.02 m^{2}$ (roughly $20 cm \times 10 cm$ ).

Weight:

W = m g = 70 \times 9.81 \approx 690 N

Pressure:

P = \frac{W}{A} = \frac{690}{0.02} \approx 3.4 \times 1 0^{4} Pa

of order $1 0^{4} Pa$ .

This is roughly a third of atmospheric pressure ( $\sim 1 0^{5} Pa$ ), which is a sensible size for a person standing on one foot rather than being many orders of magnitude away — a result differing from this reference by several orders of magnitude would suggest an arithmetic or unit error rather than a genuinely different physical situation.

Mark scheme · 4 marks

States reasonable assumed values for mass (~70 kg) and the contact area of one foot (~0.02 m², order $1 0^{- 2} m^{2}$ ) 1 mark
Uses $W = m g$ to calculate the person's weight, $\approx 690 N$ 1 mark
Uses $P = W / A$ to calculate pressure, of order $1 0^{4} Pa$ 1 mark
Compares the result against a sensible reference value (e.g. atmospheric pressure $\sim 1 0^{5} Pa$ ) to check the order of magnitude is reasonable 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Scalars and vectors

Vectors and scalars 3.4.1.1

Definitions

Scalar: fully described by magnitude alone — e.g. mass, energy, temperature, speed.
Vector: needs magnitude and direction — e.g. displacement, velocity, acceleration, force, momentum.

Key results

Resolving into components: $F_{x} = F cos θ$ and $F_{y} = F sin θ$ .
Recombining components: $F = F_{x}^{2} + F_{y}^{2}$ , with direction $θ = tan^{- 1} (\frac{F _{y}}{F _{x}})$ .

Notes

Confusing a scalar with a vector is one of the most common sources of sign errors in mechanics — a car that changes direction at constant speed has zero change in speed but a large change in velocity, because velocity is a vector.
Adding vectors graphically (tip-to-tail) and adding them by resolving into components are the same operation viewed two ways.
Component addition is almost always faster once more than two vectors are involved — sum all the $x$ -components and all the $y$ -components separately before recombining.
A common exam trap: forgetting the sign convention when resolving. Decide once (e.g. right and up are positive) and keep every component consistent with it throughout a single calculation.