📐 < and >
< means "is less than" and > means "is greater than". The open end points at the bigger number. 0.5 A < 2 A and 90 J > 40 J.
Maths Skills » Maths Skills: Algebra and Units
What you'll learn this session
Study time: 30 minutes
Edexcel spec: Mathematical skills (Appendix 4), second part
- Use the symbols <, >, ∝ and ~
- Change the subject of an equation, including squares and fractions, with a method that always works
- Convert units before substituting, and give a unit with every answer
- Set out calculations the way examiners reward
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Unlock This CourseFour symbols you must know
There is no new physics here. These are the algebra skills that let you use every equation you have already met. Equations in this lesson come from other lessons, so the physics is not explained again.
Key terms:
- Proportional symbol (∝): means 'is directly proportional to'. Remember from Force, Mass and Acceleration: if one quantity doubles, the other doubles too, so the graph is a straight line through the origin.
- Approximately (~): roughly equal to, not exact.
🔄 ∝ and ~
F ∝ a means force is proportional to acceleration when the mass stays the same. If F doubles, a doubles. ~ means about: g ~ 10 N/kg.
Changing the subject of an equation
The subject is the quantity on its own on one side. In V = I × R, the subject is V. To find R you need R on its own.
A method that works every time
An equation is a balance. Whatever you do to one side, do the same to the other side.
- Write the equation.
- Undo what is happening to the quantity you want, one step at a time, by doing the opposite to both sides. The opposite of × is ÷, of ÷ is ×, and of squaring is a square root.
- Write the new equation with your quantity on its own.
Formula triangles can help with the simplest equations, but they fail on equations with squares or fractions. Learn the method.
Worked example: one step
A resistor has a voltage of 6.0 V across it and a current of 0.25 A. Calculate its resistance.
V = I × R. Divide both sides by I: R = V ÷ I
R = 6.0 ÷ 0.25 = 24 Ω
Worked example: two steps with a square
A ball has kinetic energy 90 J and mass 0.20 kg. Calculate its speed. KE = ½ × m × v2
Multiply both sides by 2: 2 × KE = m × v2
Divide both sides by m: v2 = (2 × KE) ÷ m
v2 = (2 × 90) ÷ 0.20 = 900
Take the square root: v = √900 = 30 m/s
Worked example: an equation with fractions
A fixed volume of gas has a pressure of 150 kPa at 300 K. It is heated to 360 K. Use p1 ÷ T1 = p2 ÷ T2 to find the new pressure.
Multiply both sides by T2: p2 = (p1 × T2) ÷ T1
p2 = (150 × 360) ÷ 300 = 180 kPa
Solving simple equations
Sometimes you substitute first and then solve, because the unknown is not on its own. The same balance method applies.
Worked example: solving
An 800 kg car accelerates at 0.50 m/s2. The engine's driving force is 1200 N. Find the resistive force R.
Resultant force = mass × acceleration = 800 × 0.50 = 400 N
Resultant force = driving force − resistive force, so 1200 − R = 400
Add R to both sides and subtract 400 from both sides: R = 1200 − 400 = 800 N
Units: convert first
Equations only work when the units match. Convert every quantity to the standard unit (metres, kilograms, seconds, newtons) before you substitute.
| Change | How | Example |
|---|---|---|
| milli (m) to the base unit | ÷ 1000 | 45 mA = 0.045 A |
| kilo (k) to the base unit | × 1000 | 2.5 kW = 2500 W |
| mega (M) to the base unit | × 1 000 000 | 3 MJ = 3 000 000 J |
| g to kg | ÷ 1000 | 250 g = 0.25 kg |
| cm to m | ÷ 100 | 30 cm = 0.30 m |
| cm2 to m2 | ÷ 10 000 | 200 cm2 = 0.0200 m2 |
| minutes to s | × 60 | 15 min = 900 s |
| hours to s | × 3600 | 2 h = 7200 s |
Areas are the trap. One metre is 100 cm, so one square metre is 100 × 100 = 10 000 cm2. Dividing by 100 gives the wrong answer.
The set-out examiners reward
Use four lines every time: equation, substitution, answer, unit. Even if the final answer is wrong, you can still earn marks for a correct equation and substitution.
Worked example: convert, then calculate
A 250 g ball moves at 12 m/s. Calculate its kinetic energy.
Convert: 250 g = 0.25 kg
Equation: KE = ½ × m × v2
Substitution: KE = ½ × 0.25 × 122 = ½ × 0.25 × 144
Answer: KE = 18 J
Worked example: a force on an area
A force of 600 N acts on an area of 200 cm2. Calculate the pressure. Pressure = force ÷ area
Convert: 200 cm2 = 0.0200 m2
Pressure = 600 ÷ 0.0200 = 30 000 Pa
Worked example: power and time
A 2.5 kW heater is on for 15 minutes. Calculate the energy transferred. Energy = power × time
Convert: 2.5 kW = 2500 W and 15 minutes = 900 s
Energy = 2500 × 900 = 2 250 000 J (or 2.25 MJ)
Which equations you must memorise, and which are printed on the formulae page, is covered in Equations to Learn and Equations Given.
Common mistakes
Common mistakes
- Forgetting to convert g to kg, or cm to m, before substituting.
- Dividing by 100 instead of 10 000 to change cm2 to m2.
- Squaring only the number part, or forgetting to square at all. In ½ × m × v2, square v first, then multiply.
- Finishing at v2 = 900 and forgetting the square root.
- Leaving off the unit, or using a unit that does not match the quantity.
Exam-style question
(a) Complete the statement using < or >: 0.8 kA ____ 700 A. [1 mark]
(b) A heater has a voltage of 230 V across it and a resistance of 46 Ω. Calculate the current in the heater. [3 marks]
(c) A block of mass 450 g moves at 4.0 m/s. Calculate its kinetic energy. [3 marks]
(d) A fixed volume of gas has a pressure of 150 kPa at 250 K. It is heated to 300 K. Calculate the new pressure. [3 marks]
Model answer
(a) 0.8 kA = 800 A, so the answer is > (1)
(b) V = I × R, so I = V ÷ R (1). I = 230 ÷ 46 (1) = 5.0 A (1)
(c) 450 g = 0.45 kg (1). KE = ½ × 0.45 × 4.02 (1) = 3.6 J (1)
(d) p2 = (p1 × T2) ÷ T1 (1). p2 = (150 × 300) ÷ 250 (1) = 180 kPa (1)
Exam tip
In part (c), the first mark is for converting 450 g to 0.45 kg. If you skip that step, you lose it and the answer too. Write the converted value on its own line.