PHYSICS
CHAPTER 1
CHAPTER 1
PHYSICAL
QUANTITIES
AND
CHAPTER 1:
Physical quantities and
MEASUREMENTS
measurements
(3 Hours)
UNIT FIZIK
KOLEJ MATRIKULASI MELAKA
1
PHYSICS
CHAPTER 1
Learning Outcome:
1.1 Physical Quantities and Units (1 hours)
At the end of this chapter, students should be able to:
 State basic quantities and their respective SI units: length
(m), time (s), mass (kg), electrical current (A), temperature
(K), amount of substance (mol) and luminosity (cd).
( Emphasis on units in calculation)
2
State derived quantities and their respective units and
symbols: velocity (m s-1), acceleration (m s-2), work (J),
force (N), pressure (Pa), energy (J), power (W) and
frequency (Hz).

State and convert units with common SI prefixes.
2
PHYSICS
CHAPTER 1
1.1 Physical Quantities and Units





Physical quantity is defined as a ……………………………….
It can be categorized into 2 types
 Basic (base) quantity
 Derived quantity
Basic quantity is defined as …………………………………………….
………………………………………………………………………………..
Table 1.1 shows all the basic (base) quantities.
Quantity
Symbol
SI Unit
Symbol
Length
l
metre
m
Mass
m
……………….
kg
Time
t
second
s
T/
kelvin
K
I
ampere
…………..
……….
mole
mol
Temperature
Electric current
Amount of substance
Table 1.1
3
PHYSICS


CHAPTER 1
Derived quantity is defined as a quantity which can be expressed
in term of base quantity.
Table 1.2 shows some examples of derived quantity.
Derived quantity
Symbol
Formulae
Unit
Velocity
v
s/t
m s-1
Volume
lwt
v/t
m3
Acceleration
……..
a
m s-2
Density

m/V
…………….
p
………
W
P
f
…………
kg m s-1
ma
Fs
F/A
kg m s-2 @ N
N m-2 @ ……
1/T
s-1 @ ……..
Momentum
Force
Table 1.2
Work
Pressure
Frequency
……….. @ J
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PHYSICS
CHAPTER 1
1.1.1 Unit Prefixes


It is used for presenting larger and smaller values.
Table 1.3 shows all the unit prefixes.
Prefix
Table 1.3

Multiple
Symbol
tera
 1012
T
giga
 …….
G
mega
 106
M
kilo
 103
………..
deci
 101
d
centi
 102
c
milli
 103
m
micro
 106
………
nano
 ,,,,,,,
n
pico
 1012
p
Examples:
 5740000 m = 5740 km = 5.74 Mm
 0.00000233 s = 2.33  106 s = 2.33 s
5
PHYSICS
CHAPTER 1
Example 1.1 :
Solve the following problems of unit conversion.
a. 15 mm2 = ? m2
b. 65 km h1 = ? m s1
c. 450 g cm3 = ? kg m3
Solution :
a. 15 mm2 = ? m2
1 mm2  ......m2
1 mm2  106 m 2
b. 65 km h-1 = ? m s-1
1st method :
3

65

10
m
1

65 km h  

1
h


3

65

10
m
1

65 km h  
..........s 

1
1
65 km h  ........ m s
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PHYSICS
CHAPTER 1
2nd method :
65 km h
1
 65 km 


 1h 
 65 km  .......m .... h 
65 km h  



 1 h  1 ......  3600s 
1
65 km h 1  18 m s 1
c. 450 g cm-3 = ? kg m-3
3
450g cm 
450 g cm3  4.5 105 kg m 3
7
PHYSICS
CHAPTER 1
Follow Up Exercise
1. A hall bulletin board has an area of 250 cm2. What is this area in
square meters ( m2 ) ?
2.
The density of metal mercury is 13.6 g/cm3. What is this density
as expressed in kg/m3
3.
A sheet of paper has length 27.95 cm, width 8.5 cm and
thickness of 0.10 mm. What is the volume of a sheet of paper in
m3 ?
4.
Convert the following into its SI unit:
(a) 80 km h–1 = ? m s–1
(b) 450 g cm–3 = ? kg m–3
(c) 15 dm3 = ? m3
(d) 450 K = ? ° C
8
PHYSICS
CHAPTER 1
Learning Outcome:
1.2 Scalars and Vectors (2 hours)
At the end of this chapter, students should be able to:
a) Define scalar and vector quantities,
b) Perform vector addition and subtraction operations
graphically.
c) Resolve vector into two perpendicular components (2-D)
 Components in the x and y axes.

Components in the unit vectors in Cartesian
coordinate.
9
PHYSICS
CHAPTER 1
Learning Outcome:
1.2 Scalars and Vectors
At the end of this topic, students should be able to:
d) Define and use dot (scalar) product;
 
A  B  AB cos θ   B A cos θ 
e) Define and use cross (vector) product;
 
A  B  AB sin θ   B A sin θ 
Direction of cross product is determined by corkscrew
method or right hand rule.
10
PHYSICS
CHAPTER 1
1.2 Scalars and Vectors

Scalar quantity is defined as a quantity with magnitude only.
 e.g. mass, time, temperature, pressure, electric current,
work, energy and etc.
 Mathematics operational : ordinary algebra

Vector quantity is defined as a quantity with both magnitude
& direction.
 e.g. displacement, velocity, acceleration, force, momentum,
electric field, magnetic field and etc.
 Mathematics operational : vector algebra
11
PHYSICS
CHAPTER 1
1.2.1 Vectors
Vector A
Length of an arrow– magnitude of vector A
Direction of arrow – direction of vector A

Table 1.4 shows written form (notation) of vectors.
displacement
velocity

s

v
v
s
Table 1.4

v (bold)
s (bold)
acceleration

a
a
a (bold)
Notation of magnitude of vectors.

v v

a a
12
PHYSICS

CHAPTER 1
Two vectors equal if both magnitude and direction are the same.
(shown in figure 1.1)
Figure 1.1


Q

P
 
PQ
If vector A is multiplied by
 a scalar quantity k

Then, vector A is
kA

kA

A

A

if k = +ve, the vector is in the same direction as vector A.

if k = -ve, the vector is in the opposite direction of vector A.
13
PHYSICS
CHAPTER 1
1.2.2 Direction of Vectors

Can be represented by using:
a) Direction of compass, i.e east, west, north, south, north-east,
north-west, south-east and south-west
b) Angle with a reference line
e.g. A boy throws a stone at a velocity of 20 m s-1, 50 above
horizontal.
y

v
50
0
x
14
PHYSICS
CHAPTER 1
c) Cartesian coordinates
 2-Dimension (2-D)

s  ( x, y )  (1 m, 5 m)
y/m
5
0

s
1
x/m
15
PHYSICS

CHAPTER 1
3-Dimension (3-D)

s  ( x, y, z )  (4, 3, 2) m
s  ...i +...j + ..k
y/m
3

s
2
z/m
0
4
x/m
16
PHYSICS
CHAPTER 1
Unit vectors
A unit vector is a vector that has a magnitude of 1 with no units.
Are use to specify a given direction in space.
i , j & k is used to represent unit vectors
pointing in the positive x, y & z directions.
| iˆ | = | ˆj | = | kˆ | = 1
17
PHYSICS
CHAPTER 1
d) Polar coordinates


F  30 N,150


F
150
e) Denotes with + or – signs.
+
+
-
-
18
PHYSICS
CHAPTER 1
1.2.3 Addition of Vectors


There are two methods involved in addition of vectors graphically i.e.
 Parallelogram
 Triangle 

For example : A  B

A

B
Parallelogram
 
A B
 
A B

B
O
Triangle

A
O

A

B
19
PHYSICS

CHAPTER 1
Triangle of vectors method:
a) Use a suitable scale to draw vector A.
b) From the head of vector A draw a line to represent the vector B.
c) Complete the triangle. Draw a line from the tail of vector A to the
head of vector B to represent the vector A + B.
   
A B  B  A

B
Commutative Rule

A
 
B A
O
20
PHYSICS

CHAPTER 1
If there are more than 2 vectors therefore
 Use vector polygon and associative rule. E.g.

Q

P

  
PQ R

R

  
PQ  R

 
PQ

P

Q
     
PQ  R  P Q R






R
Associative Rule
21
PHYSICS

CHAPTER 1
Distributive Rule :


 


a.  A  B  A  B



b.    A  A  A

For example :
Proof of case a: let

 
=2
 
 
 A B  2 A B
 
A B
O
 ,  are real number

A

B


 
2 A B

22
PHYSICS
CHAPTER 1




A  B  2 A  2B


2 A  2B
O

2A


 


2 A  B  2 A  2B

2B

23
PHYSICS
CHAPTER 1
Proof of case b: let
 = 2 and  = 1



   A  2 1A  3A

A




  3A
A  A  2 A 1A

2A

3A

 
2 1A  2 A  1A



A
24
PHYSICS
CHAPTER 1
1.2.4 Subtraction
 of
 Vectors

For example :
CD
......

C

D
 
  

C DC  D
Parallelogram

C
O

D
 
CD

C
O
Triangle
 
CD

D
25
PHYSICS

CHAPTER 1
Vectors subtraction can be used
 to determine the velocity of one object relative to another object
i.e. to determine the relative velocity.
 to determine the change in velocity of a moving object.
Exercise 1 :
1. Vector A has a magnitude of 8.00 units and 45 above the positive x
axis. Vector B also has a magnitude of 8.00 units and is directed along
the negative x axis. Using graphical methods and suitable scale to
determine
 
 
a) A  B
b) A  B


A  2B
c)
(Hint : use 1 cm = 2.00 units)
d)
 
2A  B
26
PHYSICS
CHAPTER 1
1.2.5 Resolving a Vector
1st method :

2nd method :

y

Ry
0
y

R


Rx

Ry
x
0


R


Rx
x
Rx
Rx
 cos θ  Rx  ..........
 sin  Rx  R sin 
R
R
Ry
Ry
 cos   ....................
 sin θ  ....  R sin θ
R
R
27
PHYSICS

CHAPTER 1
The magnitude of vector
R:

R or R  ................

Direction of vector
tan θ 

R:
Ry
Rx
or
1 
Ry 
θ  tan  
 Rx 
Vector R in terms of unit vectors written as

R  ...............
28
PHYSICS
CHAPTER 1
Example 1.2 :
A car moves at a velocity of 50 m s-1 in a direction north 30 east.
Calculate the component of the velocity
a) due north.
b) due east.
Solution :

N
a) vN  v sin 60 or

vN
30

v
60
W

vE
E
b)
S
29
PHYSICS
CHAPTER 1
Example 1.3 :

F
150
x
S
A particle S experienced a force of 100 N as shown in figure above.
Determine the x-component and the y-component of the force.
Solution :
Vector
x-component
y-component
y

…………………………
x
F  F cos30

F
30

Fx

Fy
150
S
x

F
Fx  86.6 N
or
…………………………
Fy  50 N
or
Fx  F cos150 Fy  F sin150
Fx  100 cos150 Fy  100sin150
Fx  .............

Fy  ........
30
PHYSICS
CHAPTER 1
Example 1.4 :
y

F1 (10N)
30o
x
O
30o

F3 (40N)

F2 (30N)
The figure above shows three forces F1, F2 and F3 acted on a particle
O. Calculate the magnitude and direction of the resultant force on
particle O.
31
PHYSICS
CHAPTER 1
y
Solution :

F2

F3 x

F2 x
30o
60o
30o

F3

F2 y

F1
x
O

F3 y
32
PHYSICS
CHAPTER 1
Solution :
Vector

F1
x-component
y-component
F1x  0 N
F1 y  F1
F1 y  10 N
F2 y  30 sin 60 
F2 y  26 N

F2

F3
Vector
sum
F3 x  40 cos 30 
F3 x  34.6 N
F
x
 .................
F
y
 ..................
33
PHYSICS
CHAPTER 1
Solution :
The magnitude of the resultant force is
Fr 
 F    F 
2
2
x
y
Fr  ................
Fr  52.1 ....
and

1 
θ  tan


1 


Fy 

Fx 
16 

θ  tan 
  18
  49.6 
y

Fr

162 
18


Fx

Fy
x
O
Its direction is 162 from positive x-axis OR 18 above negative x-axis.
34
PHYSICS
CHAPTER 1
Exercise 2 :

B
1. Vector A has components Ax = 1.30 cm, Ay = 2.25 cm; vector
has components Bx = 4.10 cm, By = -3.75
 cm.Determine
a) the components of the vector sum
 A B,
b) the magnitude and direction ofA B ,
c) the components of the vector B A,
d) the magnitude and direction of B  A. (Young & freedman,pg.35,no.1.42)
ANS. : 5.40 cm, -1.50 cm; 5.60 cm, 345; 2.80 cm, -6.00 cm;
6.62 cm, 295


2. For the vectors A and B in Figure 1.2, use the method of vector
resolution to determine
 themagnitude and directionyof
a) the vector sum A

  B ,
b) the vector sum B  A
B 18.0 m s -1
 , 
c) the vector difference A
  B ,
d) the vector difference B  A. 

(Young & freedman,pg.35,no.1.39)
s-1,

A 12.0 m s -1
ANS. : 11.1 m
77.6; U think;
28.5 m s-1, 202; 28.5 m s-1, 22.2

37.0
0
Figure 1.2
x
35

PHYSICS
CHAPTER 1
Exercise 2 :
3.


Vector A points in the negative x direction. Vector
 B points at an
angle of 30 above the positive x axis. Vector C has a magnitude of
15 m 
and points

 in a direction 40 below the positive x axis. Given
that A  B  C  0, determine the magnitudes of A and B .
(Walker,pg.78,no. 65)
ANS. : 28 m; 19 m
4. Given three vectors P, Q and R as shown in Figure 1.3.


Q 24 m s 2


R 10 m s 2


P 35 m s 2
y



50
0
x
Figure 1.3
Calculate the resultant vector of P, Q and R.
ANS. : 49.4 m s2; 70.1 above + x-axis
36
PHYSICS
CHAPTER 1
1.2.6 Unit Vectors
aˆ, bˆ, cˆ

notations –

E.g. unit vector a – a vector with a magnitude of 1 unit in the direction
of vector A.

A
aˆ    1
A

Unit vectors are dimensionless.

Unit vector for 3 dimension axes :

A
aˆ
aˆ   1
x - axis ⇒iˆ @ i(bold)
y - axis ⇒ˆj @ j (bold)
z - axis ⇒kˆ @ k (bold)
iˆ  ˆj  kˆ  1
37
PHYSICS
CHAPTER 1
y
kˆ
ˆj
iˆ
x
z

Vector can be written in term of unit vectors as :

r  rx iˆ  ry ˆj  rz kˆ

Magnitude of vector,
r
rx 2  ry 2  rz 2
38
PHYSICS

E.g. :

CHAPTER 1


s  4iˆ  3 ˆj  2kˆ m
s
4
2
 3  2  5.39 m
2
2
y/m
3 ˆj

s
2kˆ
0
4iˆ
x/m
z/m
39
PHYSICS
CHAPTER 1
Example 1.5 :
Two vectors are given as:



 ˆ
a  i  2 ˆj  6kˆ m

b  4iˆ  3 ˆj  kˆ m

Calculate
 
a) the vector a
  b and its magnitude,
b) the vector b  aand its magnitude,

c) the vector 2a  b and its magnitude.
Solution :
a)
a  b  ........................


 a b   ........................

a  b   a  b  6 1  7kˆ
x
y
z
z
z
a  b  .........................
The magnitude, a  b  .....................  9.95 m
40
PHYSICS
b)
CHAPTER 1
b  a   b  a  ............
b  a   b  a  ................
x
x
b  a 
y
y
z
x
y
 bz  az  ..................
b  a  .............. m
The magnitude,
c)
b  a  .....................
 2a  b   .....................
 2a b   .........................

2a  b   2a  b  26 1  13kˆ
x
y
z
z
z
2 a  b  ....................... m
The magnitude, 2a  b 
62   72  132
 15.9 m
41
PHYSICS
CHAPTER 1
1.2.7 Multiplication of Vectors
Scalar (dot) product
 The physical meaning ofthe scalar
 product can be explained by
considering two vectors A and B as shown in Figure 1.4a.

A

Figure 1.4a


B

Figure 1.4b
 shows
 the projection of vector B onto the direction
 of
vector A .
A  B  A component of B parallel to A


A
B cos θ
Figure 1.4b


A

B


 Acos θ B
Figure 1.4c
A onto the direction of

A  B  B component of A parallel to B 42
Figure 1.4c
the projection of vector
 shows


vector B .


PHYSICS

CHAPTER 1
From the Figure 1.4b, the scalar product can be defined as
 
A  B  AB cos θ 
meanwhile from the Figure 1.4c,
 
B  A  B Acos θ 


where θ : angle between two vectors
The scalar product is a scalar quantity.
The angle  ranges from 0 to 180 .


 When 0  θ  90
scalar product is positive
90   θ  180
θ  90 

scalar product is negative
scalar product is zero
The scalar product obeys the commutative law of multiplication i.e.
   
A B  B  A
43
PHYSICS
CHAPTER 1
Example of scalar product is work done by a constant force where the
expression is givenby


W  F  s  F s cosθ   sF cosθ 
The scalar product of the unit vectors are shown below :

y
kˆ
2
iˆ  iˆ  i 2 cos 0 o  1 1  1
ˆj  ˆj  j 2 cos 0o  12 1  1
2
kˆ  kˆ  k 2 cos 0o  1 1  1
ˆj
iˆ
x
iˆ  iˆ  ˆj  ˆj  kˆ  kˆ  1
z
iˆ  ˆj  11cos 90 o  0
ˆj  kˆ  11cos 90o  0
iˆ  kˆ  11cos 90o  0
iˆ  ˆj  ˆj  kˆ  iˆ  kˆ  0
44
PHYSICS
CHAPTER 1
Example 1.6
 :
Calculate the A  B and the angle 
following
 problems.
a) A  iˆ  ˆj  kˆ

B  4iˆ  2 ˆj  3kˆ


between vectors A and B for the
b)

A
  4iˆ  3 ˆj  kˆ
B  2 ˆj  3kˆ
Solution :
a)
A  B  .........iˆ  iˆ  ........ ˆj 
A  B  .............
A B  3
The magnitude of the vectors: A 
The angle  ,
B
ANS.:3; 99.4
ˆj  .......kˆ  kˆ
12   12  12 
42   22   32

A  B  AB cos θ 


3 
1 A  B
1 



θ  cos 
 cos 

AB 
3 29 


θ  71.2
3
 29
45
PHYSICS
CHAPTER 1
Example 1.7 : 
C 1 m 
y
25
0
19 x
D2 m 
Figure 1.5
Referring to the vectors in Figure 1.5,
a) determine the scalar product between them.
b) express the resultant vector of C and D in unit vector.
Solution :
a) The angle between vectors C and D is
θ  180  25  19  174

Therefore 
C  D  CD cos θ
 ...................
C  D  1.99 .......
46
PHYSICS
CHAPTER 1
b) Vectors C and
 D in unit vector are
C  Cxiˆ  Cy ˆj

 .........iˆ  .......... ˆj


C   0.91iˆ  0.42 ˆj m
and
Hence

 


D  2 cos19 iˆ   2 sin19 ˆj
D  .....................m
 
C  D   0.91  1.89iˆ  0.42  0.65 ˆj
 0.98iˆ  0.23 ˆj m


47
PHYSICS
CHAPTER 1
Vector (cross) product
 Consider two vectors :


A  xiˆ  yˆj  zkˆ

B  piˆ  qˆj  rkˆ
In general, the vector product
as
  is defined

A B  C
and its magnitude is given by
 

 
A  B  C  A B sin θ  AB sin θ
θ : angle between two vectors
The angle  ranges from 0 to 180  so the vector product always
where



positive value.
Vector product is a vector
 quantity.
The direction of vector C is determined by
RIGHT-HAND RULE
48
PHYSICS

CHAPTER 1
For example:
 How to use right hand rule :
 Point the 4 fingers to the direction of the 1st vector.
 Swept the 4 fingers from the 1st vector towards the 2nd vector.
 The thumb shows the direction of the vector product.

C

A

  
A B  C

B
   
A B  B  A
but

B

A

C  

B A  C


 
 
A B   B  A

(C ) always perpendicular

Direction of the vector product
to the plane containing the vectors
A
and
B.
49
PHYSICS
CHAPTER 1
The vector product of the unit vectors are shown below :

y
ˆj
kˆ
iˆ
z
x
iˆ  ˆj   ˆj  iˆ  kˆ
ˆj  kˆ  kˆ  ˆj  iˆ
kˆ  iˆ  iˆ  kˆ  ˆj
iˆ  iˆ  i 2 sin 0 o  0
iˆ  iˆ  ˆj  ˆj  kˆ  kˆ  0
ˆj  ˆj  j 2 sin 0 o  0
kˆ  kˆ  k 2 sin 0 o  0

Example of vector product is a magnetic force on the straight
conductor carrying current places in magnetic field where the
expression is given by 
 


F  I l B
F  IlB sin θ
50
PHYSICS

CHAPTER 1
The vector product can also be expressed in determinant form as
iˆ
ˆj kˆ
 
A B  x y
p q
z
r
1st method :

 
A B   yr  zq iˆ  xr  zp  ˆj  xq  ypkˆ

2nd method :
 
A B   yr  zq iˆ  zp  xr  ˆj  xq  ypkˆ

Note :
 The angle between two vectors can only be determined by
using the scalar (dot) product.
51
PHYSICS
CHAPTER 1
Example 1.8 : 
ˆ  2 ˆj  kˆ
A


3
i

B  iˆ  5kˆ
Determine
 
 
a) A  B and its magnitude 
b) A  B
Given two vectors :
c) the angle between vectors A and
Solution :
iˆ ˆj kˆ
a)
B
.
A B 
A B 
 
A  B  10iˆ  16 ˆj  2kˆ
The magnitude,
A B 
 
A  B  19
52
PHYSICS
b)


CHAPTER 1
 
A  B   3iˆ  2 ˆj  kˆ  iˆ  0 ˆj  5kˆ

 
A B  2
c) The magnitude of vectors,
…………………………………………..
………………………………………….
Using the scalar (dot) product formula,
A  14
B  26
 
A  B  AB cos θ
θ  84 
53
PHYSICS
CHAPTER 1
Exercise 3 :

1. If vector a = 3iˆ + 5 ˆj

a) a  b ,
ANS. : 2kˆ; 26; 46
and vector
 
b) a  b ,

b = 2iˆ + 4 ˆj , determine


c) a  b  b .


2. Three vectors are given as follow :

 ˆ ˆ

ˆ
ˆ
ˆ
ˆ
a  3i  3 j  2k ; b  i  4 j  2k and c  2iˆ  2 ˆj  kˆ
 
Calculate
 


a) a  b  c ,
b) a 
ANS. :  21;  9; 5iˆ  11 ˆj  9kˆ
3.

 
b c

,

 

c) a  b  c

.


ˆ
ˆ
ˆ
If vector P  3i  2 j  k and vector Q  2iˆ  4 ˆj  3kˆ,
determine
 
a) the direction of P  Q


b) the angle between P and Q .
ANS. : U think, 92.8
54
PHYSICS
CHAPTER 1
THE END…
Next Chapter…
CHAPTER 2 :
Kinematics of Linear Motion
55