IB Physics SL Tool 3 — Mathematics Paper 1 & 2 Select · manipulate · derive ~8 min read

Applying General Mathematics in Physics

Physics is maths with a story attached. Behind every calculation question is a short chain of moves: spot the quantities you’re given, find the equation that links them, rearrange it, and put the numbers in. The good news — you don’t have to memorise the equations, they’re in the data booklet. What you’re really being tested on is your ability to choose, rearrange, and combine them. This page gets those moves feeling automatic.

📘 What you need to know

Algebra and rearranging equations

Solving physics problems with algebra comes down to two skills: applying the ordinary rules of arithmetic to expressions, and manipulating those expressions to get the variable you want on its own. Whatever you do to one side of an equation, you do to the other — that’s the rule that keeps everything balanced.

Say you’re given the equation of motion v = u + at and asked to make time the subject. Peel the equation apart one operation at a time:

Rearranging for t vu = at  →  t = (vu) ÷ a
Think of the variable you want as a parcel wrapped in layers. To unwrap t, you reverse whatever was done to it, working from the outside in: first move the u across, then undo the multiplication by a. Do the operations in reverse order and you can’t go wrong.

Areas, volumes and other shape formulas

Geometry sneaks into physics all the time — the cross-section of a wire, the volume of a sphere of gas, the area a pressure acts on. Most of these are handed to you in the data booklet, so the skill is recognising which one you need and substituting correctly.

These combine with definitions to solve real problems. Density is mass per unit volume, so for a cylindrical wire of mass m, radius r and length L, substitute the cylinder volume straight into the density formula:

Density of a wire ρ = m ÷ V = m ÷ (πr2L)

Fractions, percentages, reciprocals and exponents

A few everyday number skills carry most of the load in physics. Fractions turn up in algebra and uncertainty work — and remember the S⇔D button on your calculator flips a fraction to a decimal. Percentages appear as percentage change, difference, error and uncertainty. Ratios let you compare quantities cleanly.

Two more worth naming outright. A reciprocal is just one divided by the number, which is why period and frequency are reciprocals of each other: T = 1 ÷ f. An exponent is a power a number is raised to — as in the SHM acceleration relationship a = −ω2x.

Trigonometric ratios

Right-angled triangles are everywhere in physics, especially when you resolve vectors into components. The three ratios link an angle to the sides of the triangle:

SOH CAH TOA sin θ = O ÷ H  ·  cos θ = A ÷ H  ·  tan θ = O ÷ A
Quick recap: rearrange by reversing operations, pull the right shape or equation from the data booklet, and keep SOH CAH TOA ready for anything with an angle.

The calculation recipe

Almost every “calculate” question yields to the same four-step routine. Learn it once and it works whatever the topic.

🧭 Solving a calculation question

  1. List the known quantities — write each one with its symbol and unit so nothing hides
  2. Identify the equation in the data booklet that connects them
  3. Rearrange it for the quantity you want, if it isn’t already the subject
  4. Substitute the values and compute the final answer with its unit
List knowns
Pick equation
Rearrange
Substitute
WE 1

A go-kart travelling at constant velocity covers 45 m in 3.0 s. The driving force from the engine is 1.2 kN. Calculate the power output of the go-kart.

Two power equations exist: P = ΔW ÷ Δt and P = Fv. We’re given force, distance and time, so combine the first with work done W = Fs.

List & select s = 45 m, t = 3.0 s, F = 1.2 kN = 1200 N P = Fs ÷ t Substitute P = (1200 × 45) ÷ 3.0 P = 18 000 W = 18 kW Same answer via P = Fv, since v = s/t = 15 m s⁻¹ and 1200 × 15 = 18 000 W.

Deriving relationships

Deriving means building a new equation by combining ones you already trust. Start by naming the fundamental principle at play — a force balance, energy conservation, momentum — then list the relevant equations and manipulate them together. If it’s a “show that” question, the target equation is printed for you, so you just steer the algebra towards it.

WE 2

A copper cylinder of cross-sectional area A and length L has resistance 12 Ω. A second copper cylinder has twice the area (2A) and three times the length (3L). Find its resistance.

Resistivity gives ρ = RA ÷ L, so R = ρL ÷ A. Since ρ is the same for both, write it as a ratio to cancel the constant.

As a proportion R ∝ L ÷ A R₂ ÷ R₁ = (L₂ ÷ L₁) × (A₁ ÷ A₂) Substitute the changes R₂ ÷ R₁ = 3 × (1 ÷ 2) = 1.5 R₂ = 1.5 × 12 R₂ = 18 Ω Tripling length raises R; doubling area lowers it — the net effect is a 1.5× increase.

💡 Top tips

⚠ Common mistakes

Up next: Scalar & Vector Quantities — where these trig ratios and rearranging skills start doing real work, splitting forces and displacements into components.

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