A-Level · 25 September 2026 · 6 min read
AS Physics maths skills: a practical readiness checklist
Being able to follow a calculation in class is different from setting one up independently. AS Physics exposes that difference because quantities arrive in unfamiliar forms: a wire diameter in millimetres, a gradient multiplied by a power of ten, or a force acting at an angle.
Use this checklist as a short diagnostic, not a list to memorise. Try each example on paper before reading the calculation. The skills support Edexcel International AS Physics and provide a foundation for the more demanding graph and exponential work that follows at A2.
Rearrange before inserting the numbers
Start from E = ½mv² and make v the subject. Multiplying by two and dividing by m gives v² = 2E/m; taking the positive root gives the speed v = √(2E/m). For E = 18 J and m = 0.40 kg, v = √90 = 9.49 m s⁻¹. Writing the rearrangement first makes a missing factor of two much easier to catch.
Now ask what the equation assumes and what the answer means. Kinetic energy gives a speed magnitude, not a direction. If a question requires velocity, direction must come from the motion description or your sign convention. Mathematical manipulation should preserve the physics rather than quietly adding information that the equation cannot supply.
Convert powers of units, not just prefixes
A diameter of 0.50 mm is 5.0 × 10⁻⁴ m. The radius is half that, 2.5 × 10⁻⁴ m, so the circular area is πr² = 1.96 × 10⁻⁷ m². Two separate operations are involved: converting millimetres to metres and converting diameter to radius. Skipping either one can produce an answer that looks plausible but is many times too large.
Remember that 1 mm² equals 10⁻⁶ m² because both dimensions change by 10⁻³. Similarly, 1 cm³ equals 10⁻⁶ m³. Before calculating, estimate the size: the cross-sectional area of a thin wire should be a small fraction of a square metre. This simple judgement can detect an exponent error that a calculator happily accepts.
Use ratios to expose proportional relationships
For a wire at constant temperature, R = ρL/A. If the length doubles and the diameter also doubles, the area becomes four times larger. The new resistance is therefore 2/4 = ½ of its original value. No original length, diameter or resistivity is needed. Ratios often show the dependence more clearly than substituting invented numbers.
State the quantities kept constant. This comparison assumes the same material and temperature; changing either may change resistivity. Also distinguish inverse proportionality from a general decrease: saying resistance falls as diameter rises does not yet establish that it follows 1/d². A useful explanation includes the intermediate link from diameter to area and then from area to resistance.
Resolve vectors using the marked angle
A 12 N force acts at 30° above the horizontal. Its horizontal component is 12 cos 30° = 10.4 N and its vertical component is 12 sin 30° = 6.0 N. If the angle were measured from the vertical, the role of sine and cosine would swap. Draw the right triangle before choosing the function; the diagram is more reliable than memorising a phrase about horizontal forces.
Check that each component is smaller than the original magnitude and that the squares sum to the square of the original force, allowing for rounding. Keep signs consistent with your chosen axes. A component pointing left is negative when right is positive, even if the force magnitude supplied in the question is a positive number.
Read the axes before finding a gradient or area
A velocity–time graph rises from 2.0 to 8.0 m s⁻¹ between 1.0 and 4.0 s. Its gradient is (8.0 − 2.0)/(4.0 − 1.0) = 2.0 m s⁻², the acceleration. Over that interval the area is the trapezium ½(2.0 + 8.0) × 3.0 = 15 m, the displacement. Dividing the final velocity by the final time would use the wrong changes.
For experimental graphs, choose well-separated points on the best-fit line rather than two nearby measured points. Include any axis scale factors and derive the gradient unit from vertical units divided by horizontal units. An area below the time axis represents negative displacement; total distance requires adding the magnitudes of the separate areas.
Finish with a combined readiness problem
A 0.20 kg trolley starts from rest and reaches 3.0 m s⁻¹ in 1.5 s with constant acceleration. Find the acceleration, resultant force and distance travelled. The answers are a = 3.0/1.5 = 2.0 m s⁻², F = ma = 0.40 N, and s = ½ × 3.0 × 1.5 = 2.25 m, usually reported as 2.3 m to match two-significant-figure data.
If you struggled, identify which step failed: interpreting 'from rest', dividing changes, applying Newton's law, or recognising the triangular graph area. Practise that operation with a new context the next day. Keep extra digits during working and round at the end. Pearson's mathematical-skills appendix is the official scope reference; this checklist is a starting diagnostic rather than every skill the qualification can assess.
Questions, explained
Choose a question for a direct answer, then explore the explanation and supporting resources. Each answer has its own link to save or share.
How can I improve the maths in AS Physics without rereading everything?
Keep a record of the exact operations that fail, such as rearranging a squared quantity or converting a wire diameter. Practise a small set of those operations without a worked solution beside you, then apply them in unfamiliar physics questions. Check the model, units and answer size as well as the number. This makes practice respond to your errors rather than repeating calculations you already understand.
Should my calculator be in degrees or radians for Physics?
Use the mode appropriate to the angle or formula. A force resolved using an angle labelled 30° requires degree mode. In expressions such as x = A cos(ωt), angular frequency in rad s⁻¹ makes ωt a radian angle, so use radian mode. Check the display before a calculation rather than leaving one mode selected for the entire course.
How do I work out the units of a graph gradient?
Divide the unit on the vertical axis by the unit on the horizontal axis, including any scale factors when calculating the numerical value. A force–extension gradient has units N m⁻¹, while a velocity–time gradient has units m s⁻². Compare the result with the equation you are using: this can reveal whether the gradient is the quantity you want or its reciprocal.