distance travelled = average speed × time

acceleration = change in velocity ÷ time
a  = (v – u) ÷ t

weight= mass × gravitational field strength
W = m × g

momentum = mass × velocity
ρ = m × v

change in gravitational potential energy = mass × gravitational field strength × change in vertical height
ΔGPE = m × g × Δh

kinetic energy = ½ × mass × velocity²
KE = ½ × m × v²

efficiency = (useful energy transferred by the device) ÷ (total energy supplied to the device)

wave speed = frequency × wavelength
v = f × λ

wave speed = distance ÷ time
v = x ÷ t

work done = force × distance moved in the direction of the force
E = F × d

power = work done ÷ time taken
P = E ÷ t

energy transferred = change moved × potential difference
E = Q × V

charge = current × time
Q = I × t

potential difference = current × resistance
V = I × R

power = energy transferred ÷ time taken
P = E ÷ t

electrical power = current × potential difference
P = I × V

electrical power = (current)² × resistance
P = I² × R

density = mass ÷ volume
ρ = m ÷ v

force exerted on a spring = spring constant × extension
F = k × x 

(final velocity)² – (initial velocity)² = 2 × acceleration × distance
v² – u² = 2 × a × x

force = change in momentum ÷ time
F = (mv – mu) ÷ t

energy transferred = current × potential difference × time
E = I ×V × t

force on a conductor at right angles to a magnetic field carrying a current = magnetic flux density × current × length
F = B × I × L

for transformers with 100% efficiency,
potential difference across primary coil × current in primary coil = potential difference across secondary coil × current in secondary coil
V_p × I_p = V_s × I_s

change in thermal energy = mass × specific heat capacity × change in temperature
ΔQ = m × c × Δθ

thermal energy for a change of state = mass × specific latent heat
Q = m × L

energy transferred in stretching = 0.5 × spring constant × (extension)²
E = ½ × k × x²

unit	what is measured with it
m (metres)	distance or length equation symbol d or displacement, х wavelength, λ change in height, Δh
s (seconds)	time equation symbol t
m/s (metres per second)	velocity or speed equation symbol v, u or s or wave speed
m/s² (metres per second squared)	acceleration equation symbol a or acceleration due to gravity, g
kg (kilograms)	mass equation symbol m
N (Newtons)	force equation symbol F or weight, W
N/kg (Newtons per kilogram)	gravitational field strength equation symbol g (g on earth = 10 N/kg or 10 m/s²)
kgm/s (kilogram metres per second)	momentum equation symbol p
J (Joules)	energy equation symbol E or work done, E
Hz (Hertz)	frequency equation symbol f
A (Amps or Amperes)	current equation symbol I
C (Coulombs)	charge equation symbol Q
V (Volts)	potential difference equation symbol V or supply voltage, V
Ω (Ohms)	resistance equation symbol R
°C (degrees Celsius)	temperature equation symbol θ
K (Kelvin)	temperature equation symbol θ
J/kg°C (Joules per kilogram degrees Celsius)	specific heat capacity equation symbol c
J/kg (Joules per kilogram)	latent heat (of fusion or of vaporisation) equation symbol L
kg/m³ (kilograms per metres cubed)	density equation symbol ρ
g/cm³ (grams per centimetres cubed)	density equation symbol ρ
W (Watts)	power equation symbol P
T (Teslas)	magnetic flux density equation symbol B
N/m (Newtons per metre)	spring constant equation symbol k

Metric prefixes sit in front of units to indicate how many multiples of ten larger or smaller than the standard unit they are.
The table below shows metric prefixes from largest to smallest, decreasing by a factor of ×1000 each row

prefix	name	example in metres (m)
T	Tera-	1 Tm = 1 000 000 000 000 m
G	Giga-	1 Gm = 1 000 000 000 m
M	Mega-	1 Mm = 1 000 000 m
k	kilo-	1 km = 1000 m
(blank)	(no prefix)	1 m
m	milli-	1 mm = 0.001 m
μ	micro-	1 μm = 0.000 001 m
n	nano-	1 nm = 0.000 000 001 m
p	pico-	1 pm = 0.000 000 000 001 m

Memorise the order of the prefixes above. Then to convert units from one metric prefix to another ×1000 for each step down, or ÷1000 for each step up.

Worked example 1:
Q: Convert 0.5 MJ to J?
A: M (Mega) to no prefix (blank) is two steps down. So the answer is 0.5 × 1000 × 1000 = 500 000 J

Worked example 2:
Q: What is 300 μg in kg?
A: μ (micro) to k (kilo) is three steps up. So the answer is 300 ÷ 1000 ÷ 1000 ÷ 1000 = 0.000 000 3 kg

You also need to know that the prefix c (centi-) which doesn’t fit this pattern and is instead ×100 smaller than no prefix. So 1 cm = 0.01 m

The table below shows units of time from largest to smallest. The conversion factor indicates how to convert between neighbouring units.

prefix	name	conversion factor
wk	week	↓ ×7
d	day	↓ ×24
h or hr	hour	↓ ×60
min	minute	↓ ×60
s	seconds	↓ ×1000
ms	millisecond	↓ ×1000
μs	microsecond	↓ ×1000
ns	nanosecond	↓ ×1000
ps	picosecond

Then to convert from smaller to larger units divide by the conversion factor instead of multiplying.

Worked example 1:
Q: Convert 5 hours to seconds?
A: Convert hours to minutes first (×60), then minutes to seconds (×60). So 5 × 60 × 60 = 18 000 seconds

Worked example 2:
Q: What is 500 ms in minutes?
A: Convert microseconds to seconds first (÷1000), then seconds to minutes (÷60). So 500 ÷ 1000 ÷ 60 = 0.00833 minutes

-273 °C (degrees Celsius) is the coldest temperature possible.
This temperature is equal to 0 K (Kelvin).
A change of 1 °C is equal to a change of 1 K.
So to convert between °C and K add 273.
To convert between K and °C subtract 273.

Worked example 1:
Q: What is 20 °C in Kelvin?
A: 20 + 273 = 293 K

Worked example 2:
Q: Convert 50 K to °C?
A: 50 – 273 = -223 °C

The HEIST approach is a method of guaranteeing maximum marks on any question involving calculations

H	Highlight values and units in the question and answer line. Convert values so that you have matching units.
E	Write out the equation from the equation list that contains what you are calculating and what the question has given you (words or symbols).
I	Insert values into the equation by replacing the words or symbols of the equation with numbers.
S	Solve the equation. Using algebra if necessary.
T	Top off your answer by rewriting it to the correct number of significant figures and adding a unit.

Worked example 1:
Q: A cyclist travelling at 6 m/s applies their brakes. After 4 seconds the cyclist is at rest. What was the acceleration of the cyclist?

Highlight

6 m/s is speed or velocity. 4 seconds is time. ‘at rest’ means a speed or velocity of 0 m/s. We want acceleration.

Equation

acceleration = change in velocity ÷ time

Insert

acceleration = (0 – 6) ÷ 4

Solve

-6 ÷ 4 = -1.5

Top off

acceleration = -1.5 m/s²

Worked example 2:
Q: When a rock falls 8 m from a cliff to the beach below it loses 249 J of potential energy. What must the mass of the rock have been? Give your answer to three significant figures.

Highlight

8 m distance or a change in height. 249 J is change in gravitational potential energy. We want mass. We also know that gravitational field strength on Earth = 10 N/kg.

Equation

ΔGPE = m × g × Δh

Insert

Energy is lost so the change is negative. Height decreases so the change is also negative. g = 10 N/kg.
-248 = mass × 10 × -8

Solve

-249 = mass × 80
(solve by making one side of the equation be only the thing we want and the equals sign. This can be done by clearing ‘× -80’ by dividing by -80. Remember to do the same to the other side of the equals sign to keep everything equal)
-249 ÷ -80 = mass × -80 ÷ -80
-249÷ -80 = mass
3.1125 = mass

Top off

mass = 3.11 kg

Worked example 3:
Q: How much time would it take a sound wave travelling at 330 m/s to cover a distance of 5.1 km?

Highlight

330 m/s is speed or velocity. 5.1 km is distance but does not match the metres part of ‘m/s’ that the speed is given in so we need to convert it.
5.1 × 1000 = 5100 m
We want time.

Equation

wave speed = distance ÷ time

Insert

330 = 5100 ÷ time

Solve

(solve by making one side of the equation be only the thing we want and the equals sign. When the thing we want comes after a ÷ we should first move it to the other side of the equation by multiplying both sides by it)
330 × time = 5100 ÷ time × time
330 × time = 5100
(when there is a × we are allowed to divide by the number either side of it. So now we can ÷330 to leave time on its own. Remember to do the same to the other side of the equals sign to keep everything equal)
330 × time ÷ 330 = 5100 ÷ 330
time = 5100 ÷ 300
time = 15.45454545

Top off

Unless told otherwise, round long answers to three significant figures.
time = 15.5 s