What the current equation list supports

Imagine an exam question gives the mass and volume of a metal block, then asks for its density. Finding $\rho = m/V$ on a sheet is useful, but the sheet does not identify which values are mass and volume, convert their units, operate the calculator or judge whether the answer is plausible. Those are the physics and mathematical decisions this lesson reviews.

Appendix 2 of the specification divides equations into two original groups:

• list (a): recall and apply - students may be asked to recall an equation, recall and apply it, or apply it when it is supplied in the question;
• list (b): select and apply - the equation is selected from a formula list and then applied.

Pearson’s May–June 2026 W84389RA list supplies these equations with the question papers. Pearson also confirms equation-sheet support for 2027. Ofqual has confirmed continued sheets from 2028 onwards for the lifetime of the current qualifications, so support does not end in 2027. Use the clean sheet issued for your own examination series; do not assume a downloaded practice edition is the document to take into an examination.

These arrangements change the practical recall demand, not the meaning of an equation or which specification content belongs to a tier. The original Appendix 2 labels remain useful provenance; supplied-equation practice must assess selection and application. Sources: Pearson support confirmation and Ofqual continuation decision. Even with the list, you must know what each quantity means and choose a relationship that contains the target quantity and the known data.

Foundation and Higher

On W84389RA, an unlabelled row is available to both tiers. The following five rows are labelled HT, so they are Higher tier only and must not be used for Foundation questions:

This is a separate GCSE Physics 1PH0 lesson, so it also includes the final Physics only section of the sheet. Three of those extra equations are unlabelled and therefore apply to both Foundation and Higher Physics: moment, pressure $P = F/A$, and $P_{1} \times V_{1} = P_{2} \times V_{2}$ for a fixed mass of gas at constant temperature. The other two Physics-only equations, transformer turns ratio and liquid-column pressure, are HT as shown above.

A complete equation map

An equation is a compact statement about physical quantities. Read its words before its symbols, because the same letter can mean different quantities in different contexts. The map below groups every relationship on the inspected W84389RA list by the kind of situation it describes.

Motion, forces, energy and waves

Work, springs and moments

Electricity, magnetism and transformers

Matter and pressure

The equation’s conditions matter: $F=ma$ uses resultant force; $v^2-u^2=2ax$ assumes constant acceleration; $E=Fd$ uses a constant force component along displacement; spring energy uses the linear elastic region from zero extension; the transformer power equation assumes 100% efficiency; and the gas law uses fixed mass, constant temperature and absolute pressures. For efficiency as a percentage, multiply the energy ratio by 100. An HT label does not turn a valid physical relationship into a false one; it identifies the assessed tier.

The list gives $P = E/t$ twice: once in the language of work done and once in the language of energy transferred. It is the same rate relationship. The context decides whether E describes work done or another energy transfer.

Selecting the right equation

Do not begin with a keyword such as "energy" and choose the first equation containing E. Several equations can involve energy. Begin with the target quantity and the known quantities instead.

Use this selection route:

1. Write the target quantity and its requested unit.
2. Translate every given value into a quantity symbol, with its unit.
3. First look for a direct equation using known quantities. If one intermediate quantity is missing, consider whether another equation can determine it.
4. Choose an equation containing the target and all the known quantities, ideally with only the target unknown.
5. Check that the physical situation fits the relationship's condition, such as a conductor being at right angles to a magnetic field or a gas having fixed mass and constant temperature.

For example, suppose a lamp operates with a potential difference of 12 V, a current of 0.50 A and a time of 180 s, and the target is energy in joules. The knowns are V, I and t, so $E = I \times V \times t$ contains exactly those three knowns and the target E. An equally valid linked route is $Q = It$ followed by $E = QV$; it first calculates the charge that was not supplied directly.

Symbols that need context

Upper- and lower-case letters are not interchangeable. p for momentum is not P for power, and a numerical answer followed by W uses watt as a unit rather than weight as a quantity.

Aligning units before substitution

Numbers can be substituted safely only when their units match the equation. In most GCSE calculations, use the unprefixed SI or standard derived units shown below unless the question deliberately supplies a consistent alternative.

Convert before substitution and show the factor:

• $3.2 \mathrm{kJ} = 3.2 \times 10^{3} \mathrm{J} = 3200 \mathrm{J}$;
• 2.5 min = 2.5 × 60 s = 150 s;
• $25 \mathrm{cm} = 25 \times 10^{-2} \mathrm{m} = 0.25 \mathrm{m}$;
• $6.0 \mathrm{mm} = 6.0 \times 10^{-3} \mathrm{m} = 0.0060 \mathrm{m}$.

Areas and volumes need the conversion factor squared or cubed:

• $1 \mathrm{cm}^{2} = (10^{-2} \mathrm{m})^{2} = 10^{-4} \mathrm{m}^{2}$;
• $1 \mathrm{cm}^{3} = (10^{-2} \mathrm{m})^{3} = 10^{-6} \mathrm{m}^{3}$.

This is why converting $20 \mathrm{cm}^{2}$ by multiplying only by $10^{-2}$ is wrong. The correct value is $20 \times 10^{-4} \mathrm{m}^{2} = 2.0 \times 10^{-3} \mathrm{m}^{2}$.

Worked example: a wavelength conversion and rearrangement

A sound wave travels at 340 m/s and has a wavelength of 25 cm. Calculate its frequency.

The target is frequency f in hertz. The knowns are wave speed v and wavelength λ, so select:

$v = f \times \lambda$

Convert the wavelength first:

$25 \mathrm{cm} = 25 \times 10^{-2} \mathrm{m} = 0.25 \mathrm{m}$

Rearrange by dividing both sides by λ:

$f = v/\lambda$

Substitute and calculate:

$f = 340 \mathrm{m}/s / 0.25 \mathrm{m} = 1360 \mathrm{Hz}$

To two significant figures, $f = 1.4 \times 10^{3} \mathrm{Hz}$. The unit check works because $(m/s)/m = 1/s = \mathrm{Hz}$. Substituting back gives approximately $1.4 \times 10^{3} \mathrm{Hz} \times 0.25 \mathrm{m} = 350 \mathrm{m}/s$; the small difference from 340 m/s comes from rounding the final frequency. Using the unrounded 1360 Hz recovers 340 m/s exactly.

Rearranging equations transparently

An equals sign says that the expression on its left has the same value as the expression on its right. Rearranging preserves that equality. A symbol does not jump across the equals sign and change operation; you apply the same operation to both sides.

For density,

$\rho = m/V$

suppose volume V is the target. Multiply both sides by V:

$\rho \times V = m$

Then divide both sides by ρ:

$V = m/\rho$

This form makes physical sense: for a fixed mass, a greater density means a smaller volume.

For power,

$P = E/t$

suppose energy E is the target. Multiply both sides by t:

$P \times t = E$

Swap the equal sides to write the target first:

$E = P \times t$

Squared quantities

When the target is squared, isolate the square before taking a square root. From

$\mathrm{KE} = 1/2 \times m \times v^{2}$,

multiply both sides by 2, divide both sides by m, then take the square root of both sides:

$2 \times \mathrm{KE} = m \times v^{2}$

$2 \times \mathrm{KE}/m = v^{2}$

$v = \sqrt{2 \times \mathrm{KE}/\mathrm{m}}$

Kinetic energy determines the speed, so use the non-negative square root. It does not reveal the direction of a velocity.

Worked example: finding speed from kinetic energy

A cyclist and bicycle have a total mass of 80 kg and kinetic energy of 1.6 kJ. Calculate their speed.

Convert the energy:

$1.6 \mathrm{kJ} = 1600 \mathrm{J}$

Use the rearranged relationship and substitute:

$v = \sqrt{2 \times \mathrm{KE}/\mathrm{m}}$

$v = \sqrt{(2 \times 1600 \mathrm{J})/80 \mathrm{kg}}$

$v = \sqrt{40 \mathrm{J}/\mathrm{kg}} = 6.324... m/s$

To two significant figures, the speed is 6.3 m/s. The unit is sensible because J/kg is equivalent to $\mathrm{m}^{2}/\mathrm{s}^{2}$, and its square root is m/s. Substitution back gives $1/2 \times 80 \times 6.3^{2}$ approximately $1.6 \times 10^{3} \mathrm{J}$, so the rounded answer returns the stated kinetic energy.

For equations containing signed velocities, choose a positive direction before substitution. In $v^{2} - u^{2} = 2ax$, an object slowing while moving in the positive direction has a negative acceleration. Keeping the signs consistent prevents an impossible negative distance caused only by a sign error.

Linking equations and checking answers

Some questions need more than one relationship. Work backwards from the final target and identify an intermediate quantity that connects the equations. Keep full calculator precision for the intermediate result, then round only the final answer.

Worked example: current and energy for a heater

An electric heater transfers energy at a power of 2.0 kW when connected to a potential difference of 230 V. It operates for 3.0 min. Calculate the current and the energy transferred.

The current can be found from electrical power:

$P = I \times V$

Convert power, rearrange and substitute:

$2.0 \mathrm{kW} = 2000 \mathrm{W}$

$I = P/V = 2000 \mathrm{W}/230 \mathrm{V} = 8.695... A$

To two significant figures, $I = 8.7 \mathrm{A}$.

For the energy, convert the time and rearrange $P = E/t$:

3.0 min = 180 s

$E = P \times t$

$E = 2000 \mathrm{W} \times 180 \mathrm{s} = 360 000 \mathrm{J} = 3.6 \times 10^{5} \mathrm{J}$

The equation $E = I \times V \times t$ gives the same energy if the unrounded current is used. This agreement is a useful cross-check. The scale is also sensible: 2.0 kW means 2.0 kJ each second, so in 180 s the heater should transfer 360 kJ, which is $3.6 \times 10^{5} \mathrm{J}$.

A reliable final routine

Before leaving any calculation, check the whole chain:

1. Target: what quantity and unit are requested?
2. Knowns: which physical quantities do the data represent?
3. Equation: can this equation or linked set determine the target from the known quantities?
4. Conditions: does the relationship apply to this situation and tier?
5. Units: have prefixes, minutes, lengths, areas or volumes been converted consistently?
6. Algebra: has the same operation been applied to both sides, including a square or square root where needed?
7. Substitution: are the equation and numerical substitution visible?
8. Presentation: is the final unit present, with rounding done only at the end and to the requested or justified precision?
9. Sense-check: do the unit, sign, direction and order of magnitude fit? Does reverse substitution approximately recover the data?

A correct unit cannot prove that every number is right, but a wrong unit proves that something in the route needs attention. Likewise, a calculator can evaluate the numbers entered, but it cannot tell whether cm should have been converted to m, whether a negative acceleration is expected, or whether a Foundation question has accidentally used an HT-only relationship.