Maths Skills for Physics

Physics is the most mathematical of the three GCSE sciences: at least 20% of the marks reward mathematical skills, and in practice calculations and data-handling appear throughout both papers. The good news is that the maths tested is a defined, learnable set — units and prefixes, standard form, significant figures, rearranging equations, graphs, and ratios and percentages. Master these and you unlock a fifth or more of the paper that pure content-learners routinely leave on the table.

By the end of this lesson you should be confident with SI units and prefixes, standard form and significant figures, rearranging and substituting into equations, reading and drawing graphs (gradient, area, line of best fit), and using ratios and percentages — with a worked example of each.

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Units and SI Prefixes

Physics runs on SI units, and getting units right is often worth a mark in itself. The core ones:

Quantity	SI unit	Symbol
Length	metre	m
Mass	kilogram	kg
Time	second	s
Current	ampere	A
Energy	joule	J
Force	newton	N
Power	watt	W

Quantities also come with prefixes that scale them. You must be able to convert between them:

Prefix	Symbol	Multiplier	Example
kilo	k	$\times 10^{3}$×103	1 km $= 1000$=1000 m
—	—	$\times 1$×1	1 m
centi	c	$\times 10^{-2}$×10−2	1 cm $= 0.01$=0.01 m
milli	m	$\times 10^{-3}$×10−3	1 mm $= 0.001$=0.001 m
micro	μ	$\times 10^{-6}$×10−6	1 μs $= 0.000001$=0.000001 s
mega	M	$\times 10^{6}$×106	1 MW $= 1\,000\,000$=1000000 W
giga	G	$\times 10^{9}$×109	1 GW $= 10^{9}$=109 W

Worked example: convert 250 mm to metres. $250\text{ mm} \times 10^{-3} = 0.25\text{ m}$250 mm×10−3=0.25 m. Always convert to base SI units before substituting into an equation.

Exam Tip: The commonest maths error in physics is a unit slip — using grams instead of kilograms, or centimetres instead of metres. Convert every value to base SI units before you calculate, and write the conversion in your working so a slip is easy to catch.

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Standard Form

Physics quantities range from the tiny (the wavelength of light, $\sim 10^{-7}$∼10−7 m) to the astronomical (the distance to the Sun, $\sim 10^{11}$∼1011 m), so standard form — writing a number as $A \times 10^{n}$A×10n with $1 \le A < 10$1≤A<10 — is essential.

• $3\,000\,000\text{ m/s} = 3 \times 10^{6}\text{ m/s}$3000000 m/s=3×106 m/s
• $0.000\,05\text{ m} = 5 \times 10^{-5}\text{ m}$0.00005 m=5×10−5 m

To multiply in standard form, multiply the numbers and add the powers; to divide, divide the numbers and subtract the powers.

Worked example: the speed of light is $c = 3 \times 10^{8}$c=3×108 m/s. How far does light travel in $2 \times 10^{-3}$2×10−3 s?

$s = v \times t = (3 \times 10^{8}) \times (2 \times 10^{-3}) = 6 \times 10^{5}\text{ m}$s=v×t=(3×108)×(2×10−3)=6×105 m

(Multiply $3 \times 2 = 6$3×2=6; add the powers $8 + (-3) = 5$8+(−3)=5.)

Exam Tip: Enter standard form on your calculator with the EXP / ×10ˣ button, not by typing "× 10 ^" by hand. Mis-keying is a silent source of factor-of-ten errors. And always check your answer is written with $1 \le A < 10$1≤A<10 — $60 \times 10^{4}$60×104 should be tidied to $6 \times 10^{5}$6×105.

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Significant Figures

Answers should be given to a sensible number of significant figures — usually the same as the least precise value in the question, or as the question specifies. Rounding rules:

• Count from the first non-zero digit: $0.00340$0.00340 has 3 sig figs (3, 4, 0).
• $45\,600$45600 has 3 sig figs; $45\,678$45678 rounded to 3 s.f. is $45\,700$45700.
• If the next digit is 5 or more, round up.

Worked example: a calculation gives $4.637\text{ m/s}$4.637 m/s from data quoted to 2 s.f. Give the answer to 2 s.f.: $4.6$4.6 m/s.

Exam Tip: Do not round during a multi-step calculation — carry the full value on your calculator and round only the final answer. Rounding partway through introduces errors that can cost the accuracy mark. And never give an answer to more sig figs than the data justifies.

A subtlety worth mastering is the difference between significant figures and decimal places, since questions specify one or the other and they are not interchangeable. To 2 significant figures, $0.00456$0.00456 becomes $0.0046$0.0046 (the leading zeros are not significant — you count from the first non-zero digit), whereas to 2 decimal places the same number becomes $0.00$0.00, which is useless. Conversely, a large number like $45\,678$45678 to 3 significant figures is $45\,700$45700, but "to 3 decimal places" makes no sense for a whole number. Read the instruction carefully: "give your answer to 2 significant figures" and "to 2 decimal places" ask for genuinely different things, and answering with the wrong one loses the accuracy mark even when your physics and arithmetic are perfect. When no instruction is given, matching the significant figures of the least precise data in the question is the safe default.

Exam Tip: Watch whether the question asks for significant figures or decimal places — they are not the same. For small numbers like 0.00456 the two give wildly different answers. When unsure and no instruction is given, match the significant figures of the least precise value you were given.

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Rearranging and Substituting

The mechanics of rearranging were covered in the equation-sheet lesson; here is the maths habit that makes it reliable: do the same operation to both sides of the equation until the unknown is alone.

Worked example: rearrange $v = f\lambda$v=fλ to make $\lambda$λ the subject, then find $\lambda$λ when $v = 340$v=340 m/s and $f = 170$f=170 Hz.

$v = f\lambda \quad\Rightarrow\quad \lambda = \dfrac{v}{f} = \dfrac{340}{170} = 2\text{ m}$v=fλ⇒λ=fv​=170340​=2 m

For an equation with a square, take the root last:

$E_k = \tfrac{1}{2}mv^2 \quad\Rightarrow\quad v = \sqrt{\dfrac{2E_k}{m}}$Ek​=21​mv2⇒v=m2Ek​​​

Exam Tip: Rearrange the letters first, substitute the numbers last. Manipulating symbols is far less error-prone than juggling digits, and it keeps your working clean for the examiner to award method marks even if the final arithmetic slips.

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Graphs — Gradient, Area and Line of Best Fit

Graph skills are heavily tested. Three techniques recur:

1. Line of best fit. Draw a single straight (or smooth curved) line with roughly equal points either side. Ignore anomalies — do not force the line through every point, and do not "dot-to-dot".

2. Gradient. For a straight-line graph, the gradient often is a physical quantity (e.g. the gradient of a distance–time graph is speed). Calculate it as:

$\text{gradient} = \dfrac{\text{change in } y}{\text{change in } x} = \dfrac{\Delta y}{\Delta x}$gradient=change in xchange in y​=ΔxΔy​

Use a large triangle spanning most of the line for accuracy, and read the values off the line, not off a data point.

3. Area under the line. For some graphs the area is a physical quantity — the area under a velocity–time graph is the distance travelled. Split the area into rectangles and triangles and add them.

Worked example (gradient): on a distance–time graph, the line rises from $(0\text{ s}, 0\text{ m})$(0 s,0 m) to $(4\text{ s}, 20\text{ m})$(4 s,20 m).

$\text{speed} = \text{gradient} = \dfrac{20 - 0}{4 - 0} = 5\text{ m/s}$speed=gradient=4−020−0​=5 m/s

Here is a simple velocity–time graph whose area gives the distance travelled: