1A Physical quantities & units
Learning outcomes
▢ understand physical quantities and units
▢ recall SI base quantities and units, and express derived units in terms of
base
▢ understand homogeneity of equations
▢ understand scientific notation
▢ recall prefixes and be able to manipulate prefixes
▢ define scalar and vector quantities
▢ know how to add and subtract vectors
▢ know how to resolve a vector to two perpendicular components
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1A.1 Physical quantities & units
• Physical quantities are quantities that have units, such as distance,
time and speed.
• A physical quantity consists of a numerical magnitude and a unit.
E.g. length of table ℓ = 1.5 m,
speed of car 𝑣 = 50 km h-1.
• Some quantities have no units so they are not physical, e.g. number
of electrons, angle (degrees & radians are not real units), ratios (e.g.
𝑡2 /𝑡1 ), efficiency and strain.
Physical
quantities/units
BASE
DERIVED
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SI base quantities
• In the SI system, 7 quantities and 7 units are chosen as base.
• All other quantities/units are called derived quantities/units:
• derived quantities: energy, speed, frequency etc.
• derived units: m s −1 , Hz (s −1 ), newton etc.
Base quantities
Base units
1. time
second, s
2. length
metre, m
3. mass
kilogram, kg
4. (thermodynamic) temperature
kelvin, K
5. (electric) current
ampere, A
6. amount of substance
mole, mol
7. luminous intensity
candela, cd
Required in
A Level.
(mole in A2)
3
Derived quantities
• Quantities which are not base are called derived quantities.
• Any derived units can be expressed in terms of base units.
• Some examples:
Derived quantities
SI units
Derived units
frequency
hertz (Hz)
s−1
speed
m s−1
m s−1
acceleration
m s−2
m s−2
force
newton (N)
kg m s−2
energy
joule (J)
kg m2 s−2
power
watt (W)
kg m2 s−3
electric charge
coulomb (C)
As
potential difference
volt (V)
kg m2 s−3 A−1
specific heat capacity
J kg−1 K−1
m−2 s−2 K−2
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Expressing derived units in base
Some important derived quantities:
• force (newton, N)
• pressure (pascal, Pa)
• energy or work (joule, J)
• power (watt, W)
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Example 1
(a) Determine base units of pressure.
Ans
𝐹
𝑝=
𝐴
SI unit of pressure = pascal,
1 Pa = 1 N m−2
Since 𝐹 = 𝑚𝑎, 1 N = 1 kg m s-2,
1 Pa = 1 kg m s −2 ⋅ m−2
= 1 kg m−1 s −2
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Example 1
(b) Derive base units for energy.
Ans
work done = force  displacement
𝑊 = 𝐹𝑠
1J=1Nm
= 1 kg m s−2 ⋅ m
= 1 kg m2 s −2
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Example 1
(c) Derive base units for power.
Ans
𝑊
𝑃=
𝑡
work
power =
time
1 W = 1 J s−1
= 1 N m s −1
= 1 kg m s −2 ⋅ m s −1
= 1 kg m2 s −3
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Example 2
Which of the following definitions is correct and uses only
quantities rather than units?
A
B
C
D
Density is mass per cubic metre.
Potential difference is energy per unit current.
Pressure is force per unit area.
Speed is distance travelled per unit second.
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Homogeneity of equations
• A correct equation must be homogeneous, i.e. all the terms must
have the same base units.
• E.g. kinetic energy of a mass 𝑚 moving with speed 𝑣 is given by:
1
𝐾 = 𝑚𝑣 2
2
J=Nm
= kg m s −2 ⋅ m
= kg m2 s −2
1
𝑚𝑣 2
2
= kg m s −1 2
= kg m2 s −2
units of
LHS and RHS have same base units, so equation is homogeneous.
• Notice that all pure numbers are ignored e.g. ½, 𝜋, e.
• A homogeneous equation is not necessarily correct, e.g. 𝐾 = 2𝑚𝑣 2
is wrong for KE even though it is homogenous.
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Use of homogeneity
Homogeneity can be used to determine unknown units of a physical
quantity or unknown exponent.
Example 1
The viscous force 𝐹 on a sphere falling through a fluid is given by the
formula
𝐹 = 6𝜋𝑎𝜂𝑣
where 𝑎 is the radius of the sphere,  is the viscosity of the fluid, and 𝑣
is the velocity of the sphere. What are the units of ?
𝐹
𝜂=
6𝜋𝑎𝑣
N
kg m s−2
units of 𝜂 =
= 2 −1 = kg m−1 s −1
−1
m⋅m s
m s
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Example 2
An equation relating period 𝑇 of oscillation, angle 𝜃, and radius 𝑟 is
given by
𝐴
2
𝑇 2 + 2 = 𝐵 cos 𝜃
𝑟
where 𝐴 and 𝐵 are constants.
Determine the base units of 𝐴 and 𝐵.
A:
B:
m2
s2
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Example 3
The speed 𝑣 of a liquid leaving a tube depends on the change in pressure
Δ𝑃 and the density 𝜌 of the liquid. The speed is given by the equation
𝑣=𝑘
Δ𝑃 𝑛
𝜌
where 𝑘 is a constant that has no units. What is the value of 𝑛?
−2
N
m
m s −1 =
kg m−3
𝑛
kg m s −2 m−2
=
kg m−3
𝑛
m s −1 = m2 s −2 𝑛 = m2𝑛 s −2𝑛
1 = 2𝑛
1
𝑛=
2
Alternative: 𝑝 = 𝜌𝑔ℎ ⇒
𝑝
= 𝑔ℎ.
𝜌
So, m s −1 = m s −2 ⋅ m 𝑛 = m2 s −2 2 = m2𝑛 s −2𝑛
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Scientific notation
In physics the quantities we encounter range from very small to very large.
So scientific notation (or standard form) is used, by separating the significant
figures and the order of magnitude (i.e. power of ten).
E.g.
charge of electron ≈ −0.000 000 000 000 000 000 16 C = −1.6 × 10−19 C
speed of light 𝑐 = 299 792 458 ≈ 300 000 000 m s −1 = 3.0 × 108 m s−1
Rule for scientific notation:
𝐴 × 10𝑛
where
1 ≤ |𝐴| < 10,
𝑛 ∈ ℤ (integers)
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Prefixes
Prefixes offer another way to express very large and very small numbers.
Multiple
Sub-multiple
Prefix
Symbol
Multiplying
factor
Prefix
Symbol
Multiplying
factor
tera
T
1012
deci
d
10−1
giga
G
109
centi
c
10−2
mega
M
106
milli
m
10−3
kilo
k
103
micro
𝜇
10−6
nano
n
10−9
pico
p
10−12
Some common units with a prefix:
mm, cm, km, kg, Mbps, TB
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Working with prefixes
Often we need to convert units with different prefixes, e.g. cm2 to m2, mm3 to
m3, kg m−3 to g cm−3 etc.
Examples
a) Write 5.0 cm2 in m2.
b) How many micrograms are there in 1.0 milligram?
c) A car travels with speed of 90 km h–1. What is the speed in m s–1?
d) Density of water is 1.0 g cm–3. What is it in kg m–3?
e) Calculate the time of 1 day in ks.
a) 5.0 cm2 = 5.0 × 10−2 m 2 = 5.0 × 10−4 m2
b) 1.0 mg = 1.0 × 10−3 g = 1.0 × 103 × 10−6 g = 1000 μg
90 km
90×103 m
90
–1
c) 90 km h =
=
= m s −1 = 25 m s −1
1h
3600 s
90 km
1000 m
Or, 90 km h–1 =
×
1h
1 km
3.6
×
1h
3600 s
=
90
m s −1 = 25 m s −1
3.6
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d) 1.0 g cm–3 = 1.0 × 10−3 kg 10−2 m −3
= 1.0 × 10−3 kg 106 m−3
= 1.0 × 103 kg m−3 (1000 kg m−3 )
e) 1 day = 24 × 60 × 60 = 86,400 s ×
= 86.4 ks
1 ks
1000 s
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Conventions on typeface
To distinguish physical quantities and units, by convention,
• symbols for physical quantities are italicized
• symbols for units are upright
For example, speed 𝑣 of a car may be written as:
𝑣 = 60 km h−1
italic
upright
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Conventions for tables and graphs
For A Level, the following heading style is preferred:
ℓ/m
𝑡/s
ℓ / cm
𝑰 / 10−𝟑 A
0.010
0.21
20.0
12.3
0.020
0.35
25.0
20.5
0.030
0.45
30.0
33.4
The same convention is used for labelling axes of graphs.
𝑡/s
20
30
40
ℓ/𝑚
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Order of magnitudes
• In physics, we routinely deal with wide-ranging quantities.
– one light year ∼ 9.5 × 1015 m
– diameter of nucleus ∼ 6 × 10−15 m
• For many practical reasons, ability to estimate the rough size of a quantity
is important, e.g. to enable some quick back-of-envelope calculation, or to
plan for experiment.
• Usually it suffices to know the order of magnitude of the quantity, which is
defined as the power of ten of a value after rounded to the closest 10.
• E.g. diameter of a hydrogen atom is approx. 3.2 × 10−11 m, we’d say the
diameter is of the order of 10−11 m, and write as 𝑑 ∼ 10−11 m.
• Knowledge of order of magnitude is also useful to check mistakes in
calculations.
– e.g. mass of Earth is 6.0 × 1024 kg (∼ 1025 kg), if we do calculation and got
answer of 6.0 × 1020 kg (∼ 1021 kg), which is not the expected order, then we
know there must have been mistakes.
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From miniscule to astronomical
distance / m
distance from Earth to edge of observable Universe
4.3 × 1026
diameter of a galaxy
1.2 × 1021
distance from Earth to the Sun
1.5 × 1011
distance from KL to London
1.1 × 107
distance from KL to Singapore (or London to Paris)
3.5 × 105
length of a car
4
diameter of hair
5 × 10−4
diameter of water molecule
2.8 × 10−10
diameter of hydrogen atom
3.2 × 10−11
diameter of proton
1.7 × 10−15
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Homework
Estimate the following quantities, then look up the actual values online.
My estimate
Actual
Speed of Olympic sprinter
m s −1
m s −1
Speed of commercial plane
km h−1
km h−1
Mass of a sedan car
kg
kg
Mass of a 40-seater bus
kg
kg
Temperature of blue candle flame
C
C
Temperature of Earth’s core
C
C
Temperature of Sun’s core
C/K
C/K
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1A.2 Scalar and vector quantities
A scalar is a quantity which has magnitude only.
A vector is a quantity which has magnitude and direction.
SCALAR
distance
speed
angle
time
pressure
work
energy
power
mass
intensity
frequency
density
VECTOR
length
area
volume
amount of
substance
temperature
emf
displacement
current
voltage
resistance
velocity
acceleration
momentum
force
moment
angular
displacement
angular velocity
electric field strength
magnetic flux density
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Vector representation
Vectors are represented using arrows, where
• direction of arrow represents the direction of vector,
• length of arrow represents the magnitude.
A scale is needed.
E.g. to represent a force of 50 N due east, we can use a scale of
1 cm: 10 N.
N
50 N
scale:
1 cm: 10 N
5 cm
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Direction of vectors
v1: 40 north of east, or 50 east of north.
v2: __________________, or ___________________
v3: __________________, or ___________________
v4: due west.
north (N)
v1
west (W)
v4
40
20
v3
east (E)
60
v2
south (S)
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Adding coplanar vectors
Coplanar vectors are vectors which lie on the same plane.
Coplanar vectors can be added using 2 methods:
1. Triangle law (head-to-tail)
2. Parallelogram law
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1. Triangle law (head-to-tail)
Given two vectors 𝐴 and 𝐵, and
𝐶 =𝐴+𝐵
How to find 𝐶?
𝐴
Draw 𝐴 first, then draw 𝐵 such that
the head of 𝐴 joins with tail of B
𝐵
𝐴
𝐴+𝐵
𝐵
OR
Draw 𝐵 first, then draw 𝐴 such that
the head of 𝐴 joins with tail of 𝐵
𝐵+𝐴
Note: 𝐴Ԧ + 𝐵 = 𝐵 + 𝐴Ԧ
𝐴
𝐵
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2. Parallelogram law
Combining the two possible triangles we get a parallelogram.
𝐵
𝐴
𝐶
𝐴
𝐵
Resultant vector is given by the vector pointing from initial
point to opposite vertex.
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Subtracting vectors
Suppose 𝐶 = 𝐴 – 𝐵. How to find 𝐶 ?
To subtract vectors, we actually perform addition of 𝐴 and −𝐵, as
follows:
𝐶 = 𝐴 − 𝐵 = 𝐴 + (−𝐵)
−𝐵 means a vector with same magnitude as 𝐵 but pointing in
opposite direction.
𝐴
−𝐵
𝐵
𝐴−𝐵
𝐴
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Vector subtraction is needed, e.g., when velocity of a body changes
and we want to determine the change in velocity:
Δ𝑣 = 𝑣 − 𝑢
Suppose a ping pong ball hits a pole with velocity 𝑢 and deflects to the
right with velocity 𝑣. The change in velocity is:
Δ𝑣 = 𝑣 + (−𝑢)
𝑢
Δ𝑣
𝑣
𝑣
−𝑢
Top view
(not to scale)
Velocity triangle
(to scale)
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Trigonometric rules
Two useful trigonometric rules in solving vector problems.
1. Sine Rule
sin 𝐴 sin 𝐵 sin 𝐶
=
=
𝑎
𝑏
𝑐
2. Cosine Rule
𝑎2 = 𝑏 2 + 𝑐 2 − 2𝑏𝑐 cos 𝐴
𝐴
𝑏
𝑐
𝐵
𝐶
𝑎
𝑏 2 = 𝑎2 + 𝑐 2 − 2𝑎𝑐 cos 𝐵
𝑐 2 = 𝑎2 + 𝑏 2 − 2𝑎𝑏 cos 𝐶
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Example
An eagle is flying due east in a speed of 10 m s–1 when suddenly
a strong wind of speed 3.0 m s–1 blows from north-east. The
velocity vector diagram is shown to scale.
(a) Draw an arrow to indicate the resultant velocity of the eagle.
(b) What is the resultant velocity?
wind
eagle
𝜃
v
resultant velocity
45
45
magnitude of v:
𝑣 2 = 102 + 32 − 2 10 3 cos 45°
𝑣 = 8.2 m s−1
direction:
sin 𝜃 sin 45°
=
3
8.2
𝜃 = 15° (S of E)
∴ resultant velocity is 8.2 m s–1 at a
direction of 15° S of E.
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Resolving vectors
The opposite of addition of vectors is resolution.
A vector can be resolved (i.e. split) into 2 or more vectors called components.
𝑣 = 𝑣1 + 𝑣2 + 𝑣3 + ⋯
Usually we resolve to 2 perpendicular components to solve problem.
𝑣 = 𝑣𝑥 + 𝑣𝑦
𝑦
𝑣
𝑣𝑦
𝑣𝑥 = 𝑣 cos 𝜃 = 𝑣 sin 𝛼
𝑣𝑦 = 𝑣 sin 𝜃 = 𝑣 cos 𝛼
𝛼
𝜃
𝑣𝑥
𝑥
33
Trigonometric functions
Sine, cosine and tangent are the 3 main functions in
trigonometry and are based on right-angled triangle.
SOH, CAH, TOA
“Oscar Had A Heap Of Apples”
34
Example 1
What is the component of this displacement vector in the
direction XY?
𝑠XY = 𝑠 cos 𝜃
= 5.0 cos 53°
= 3.0 km
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Resolving vectors into components can be useful when adding
vectors.
Example 2
A ship is pulled by two small boats. The tension in each cable is
4000 N. Find the resultant force of the pulls of the boats on the
ship.
Ans
𝐹 = 4000 cos 40° + 4000 cos 40° = 6128 ≈ 6100 N
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Exercise
A boat can be rowed at a speed of 7.0 km h–1 in still water. A river
flows at a constant speed of 1.5 km h–1. Determine the angle to
the bank at which the boat must be rowed so that the boat
travels directly across the river. What is the resultant velocity?





river flow
resultant
v

boat,
7.0 km h–1
𝜃
river,
1.5 km h–1
1.5
cos 𝜃 =
7.0
𝜃 = 77.6° ≈ 78°
𝑣=
72 − 1.52 = 6.8 km h−1
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