CHAPTER 1 INTRODUCTION TO PHYSICS

PHYSICS CHAPTER 1

CHAPTER 1:
Introduction to Physics

1

PHYSICS CHAPTER 1

Learning Outcome:

1.1 Physics Understanding

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).

 State derived quantities (in terms of basic 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).

 Perform conversion between SI units.

2

PHYSICS CHAPTER 1

1.1 Physical Quantities and Units

 Physical quantity is defined as a quantity which can be measured and
observed.

 It can be categorised into 2 types

 Basic (base) quantity

 Derived quantity

 Basic quantity is defined as a quantity which cannot be derived from
any physical quantities.

 Table 1.1 shows all the basic (base) quantities.

Quantity Symbol SI Unit Symbol

Length l metre m

Mass m kilogram kg
Time t second s Table 1.1
Temperature K
T/ kelvin

Electric current I ampere A

Amount of substance n mole mol 3
Luminous Intensity candela cd
Iv

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.

Table 1.2 Derived quantity Symbol Formulae Unit
Velocity m s-1
Volume v s/t m3
V lwt m s-2
Acceleration a kg m-3
Density v/t kg m s-1
Momentum  kg m s-2 @ N
Force m/V kg m2 s-2 @ J
Work p mv
F ma
W Fs

4

PHYSICS CHAPTER 1

 Unit is defined as a standard size of measurement of
physical quantities.

 Examples :
 1 second is defined as the amount of time it takes light
in a vacuum to travel 299,792,458 metres.
 1 kilogram is defined as the mass of a platinum-iridium
cylinder kept at International Bureau of Weights and
Measures Paris.
 1 meter is defined as the length of the path travelled by
light in vacuum during a time interval of

1s
299,792,458

5

PHYSICS CHAPTER 1

 The unit of basic quantity is called base unit
 addition unit for base unit:
 unit of plane angle - radian (rd)

 rad 180o

1 rad  180o  57.296o



 unit of solid angle- steradian (sr)

 The common system of units used today are S.I unit (System 6
International/metric system) and cgs unit - UK.

 The unit of derived quantity – called derived unit

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 Multiple Symbol

tera 1012 T

giga 109 G

mega 106 M

kilo 103 k

deci  101 d

centi  102 c

milli  103 m

micro  106 

Table 1.3 nano  109 n
 Examples: pico  1012 p

 5740000 m = 5740 km = 5.74 Mm

 0.00000233 s = 2.33  106 s = 2.33 s 7

PHYSICS CHAPTER 1

1.1.2 Conversion of Unit

 Table 1.4 shows the conversion factors between SI and British units for
length and mass only.

Length Mass

1 m = 39.37 in = 3.281 ft 1 kg = 103 g
1 in = 2.54 cm 1 slug = 14.59 kg
1 km = 0.621 mi 1 lb = 0.453 592 kg
1 mi = 5280 ft = 1.609 km 1 kg = 0.0685 slug
1 angstrom (Å) = 1010 m

Table 1.4

8

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 d. 29 cm = ? in

e. 12 mi h1 = ? m s1

Solution :

   a. 15 mm2 = ? m2
1 mm 2  103 m 2

1 mm 2  10 6 m2

15 mm2  15 106 m2 or 1.5 105 m2

b. 65 km h-1 = ? m s-1  65 103 m 
1st method : 1h
65 km h 1 

65 km h 1   65 10 3m 
3600 s

65 km h 1  18 m s1 9

PHYSICS CHAPTER 1

2nd method : 65 km h1   65 km 
 1h 

65 km h1   65 km 1000 m  1h 
 1h  1km  3600 s 

65 km h 1  18 m s1

c. 450 g cm-3 = ? kg m-3

cm3   450 g  10 3 kg  1 cm3 
 1 cm3 1g 10 2 3 m3 
 450g

450 g cm3  4.5105 kg m3

10

PHYSICS CHAPTER 1

d. 29 cm = ? in 1
2.54
29 cm  29 cm in 

1 cm

29 cm  11.4 in

e. 12 mi h-1 = ? m s-1

12 mi h1  12 mi 1.609 km 1000 m  1h 
 1h  1mi  1km  3600 s 

12 mi h 1  5.36 m s1

11

PHYSICS CHAPTER 1

Learning Outcome:

1.2 Scalars and Vectors

At the end of this chapter, students should be able to:

 Define scalar and vector quantities.

 Compare scalar and vector quantities.

 Resolve vector into two perpendicular components (x and
y axes).

 Determine resultant vector of two vector component

 Write a laboratory report (Experiment 1: Physical
Measurement) (Experiment 2: Plotting and interpreting
linear graph)

12

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

13

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 1.6 shows written form (notation) of vectors.

displacement velocity acceleration
s a
 a
s v
v a (bold)

Table 1.6 s (bold) v (bold) 14

 Notation of magnitude of vectors.
v v


a a

PHYSICS CHAPTER 1

 Two vectors equal if both magnitude and direction are the same.

(shown in figure 1.1) 

 Q 
P PQ

Figure 1.1

 If vector A is multiplied by 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. 15

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 x
v

50

0

16

PHYSICS CHAPTER 1

c) Cartesian coordinates
 2-Dimension (2-D)


s  (x, y)  (1m, 5 m)
y/m

5
s

01 x/m

17

PHYSICS CHAPTER 1

 3-Dimension (3-D)
s  (x, y, z)  (4, 3, 2) m

y/m

3 4 x/m


s

0
2

z/m

18

PHYSICS CHAPTER 1

 d) 
Polar coordinates F  30 N,150 


F

150

e) Denotes with + or – signs. + +
-
- 19

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 Triangle

   
B A B A B

O O B
A A
20

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 Commutative Rule


A


B 

B A

O

21

PHYSICS CHAPTER 1

 If there are more than 2 vectors therefore 

 Use vector polygon and associative rule. E.g. P  Q  R

 Q
P
R

   

PQ R


PQ
  
R
P



Q 22

        

P  Q  R  P  Q  R Associative Rule

PHYSICS CHAPTER 1

 Distributive Rule :  ,  are real number

    

a.  A  B AB

b.    A  A  A

 For example :

   Proof of case a: let  = 2

 AB 2 AB


A B

 B  
O 2 AB
A

23

PHYSICS CHAPTER 1
 

AB  2A 2B

 
2A 2B 2B

O
2A

    

2 A B  2A 2B

24

PHYSICS CHAPTER 1

Proof of case b: let  = 2 and = 1

   A  2 1A  3A



A


    3A
A  A  2A 1A
 
 A
2A


3A
2 1A  2  
A 1A

25

PHYSICS CHAPTER 1

1.2.4 Subtraction of Vectors
 For example : C  D

 
C D

    D

CDC D

Parallelogram O  Triangle 
C C D

O  
 CD CD

D

26

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 1.2 :

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
determine
a) A  B 
b) A  B
 
c) A  2B d) 2 A  B

(Hint : use 1 cm = 2.00 units)

27

PHYSICS CHAPTER 1

1.2.5 Resolving a Vector

 1st method :  2nd method :

yy

     
Ry R Ry R

 x  x
0 Rx Rx
0

Rx  cosθ  Rx  R cosθ Rx  sin  Rx  Rsin 
R R

Ry  sin θ  Ry  Rsin θ Ry  cos  Ry  R cos

R R

28

PHYSICS CHAPTER 1

 The magnitude of vector R :

 

R or R 
Rx 2  Ry 2

 Direction of vector R :

tan θ  Ry or θ  tan1 Ry 
Rx Rx

 Vector R in terms of unit vectors written as

R  Rxiˆ  Ry ˆj

29

PHYSICS CHAPTER 1

Example 1.6 :

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 : a) vN  vsin60 or vN  vcos30
vN  50sin60 vN  50cos30
N

 30  vN  43.3 m s1
vN v

W 60 E
vE
b) vE  v cos60 or vE  vsin 30
vE  50cos60 vE  50sin 30

S vE  25 m s1

30

PHYSICS CHAPTER 1

Example 1.7:
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 : x-component y-component

y Vector  F cos30  F sin 30
 100 cos30  100 sin 30
 Fx Fy
F Fy Fx Fy

30 150  Fx  86.6 N Fy  50 N
F or
x Fy or s in 150 
Fx S Fx  F cos150 
Fx 100 cos150 F

Fy  100 sin150 

Fx  86.6 N Fy  50 N
31

PHYSICS CHAPTER 1

Example 1.8 : y


F1(10N)

30o O x


30o F2 (30N)

F3 (40 N)

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.

32

PHYSICS CHAPTER 1

Solution : y

  F2 y
F2 30o F1


F F3 x 2x 60o
x
30o O
33
 
 F3    F3 y

Fr  F  F1 F2  F3
 Fr   Fx  Fy
 Fx  F1x  F2x  F3x
 Fy  F1y  F2 y  F3 y

PHYSICS CHAPTER 1

Solution : y-component

Vector x-component F1y  F1
F1y  10 N
 F1x  0 N F2 y  30 sin 60
F1 F2 y  26 N
F3y  40sin 30
 F2x  30 cos60 F3y  20 N
F2 F2x  15 N
  Fy 10  26   20.0
F3 F3x  40 cos30  Fy 16 N
F3x  34.6 N
Vector
sum  Fx  0  15  34.6

 Fx  49.6 N

34

PHYSICS CHAPTER 1

Solution :
The magnitude of the resultant force is

Fr   Fx 2   Fy 2

Fr   49.62  162

Fr  52.1 N y


Fr  Fy
and
θ  Fy  18 162
 tan 1  Fx 
 O
θ  tan1 16   18 x

Fx

  49.6 

Its direction is 162 from positive x-axis OR 18 above negative x-axis.

35

PHYSICS CHAPTER 1

Exercise1.3 : 
1. Vector Ahas components Ax = 1.30 cm, Ay = 2.25 cm; vector B

has components Bx = 4.10 cm, By = -3.75 cm. Determine
a) the components of the vector sum A B,
b) the magnitude and direction ofA B ,
c) the components of the vector B A,
B  Ad) the magnitude and direction of
. (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 ,   
c) the vector difference A  B ,
d) the vector difference B  A. B 18.0 m s-1

(Young & freedman,pg.35,no.1.39)   37.0

ANS. : 11.1 m s-1, 77.6; U think; A 12.0 m s-1 x

28.5 m s-1, 202; 28.5 m s-1, 22.2 0 36

Figure 1.2

PHYSICS CHAPTER 1

Exercise 1.3 : 
3. Vector A points in the negative x direction. Vector Bpoints at an
angle of 30 above the positive x axis. Vector C has a magnitude of

15 m and pointsin a direction 40 below the positive x axis. G iven
that A  B  C  0, determine the magnitudes of A and B .

(Walker,pg.78,no. 65)

ANS. : 28 m; 19 m

vectors P, Q Ryas in


 4. Given three and shown
Figure 1.3.
 Q 24 m s2 P 35 m s 2

  50

R 10 m s2 x

Figure 1.3 0

Calculate the resultant vector of P, Q and R.

ANS. : 49.4 m s2; 70.1 above + x-axis

37

PHYSICS CHAPTER 1

1.2.6 Unit Vectors

 notations – aˆ, bˆ, cˆ

 E.g. unit vector a – a vector with a magnitude of 1 unit in the direction

of vector A. 
aˆ  A  1 A

A aˆ

 Unit vectors are dimensionless.

aˆ  1

 Unit vector for 3 dimension axes : iˆ  ˆj  kˆ  1

x - axis⇒iˆ@i(bold)
y - axis⇒ ˆj @ j(bold)
z - axis⇒kˆ@k(bold)

38

PHYSICS CHAPTER 1

y

ˆj x
kˆ iˆ

z

 Vector can be wrirtten inrxteiˆrmorf yunˆjit vectors as :

 rzkˆ

 Magnitude of vector,

 r  rx 2  ry 2  rz 2

39

PHYSICS CHAPTER 1


E.g. : s 
 
4iˆ  3 ˆj  2kˆ m

s  42  32  22  5.39m

y/m

3ˆj 4iˆ x/m
s

2kˆ 0

z/m

40

PHYSICS CHAPTER 1

1.2.7 Multiplication of Vectors

Scalar (dot) product

 The physical meaning ofthe scalar 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 showsthe projection of vector B onto the direction of

 vector A. A B  A component of B parallelto A
AA

B cosθ 

Figure 1.4b   
B B
Figure 1.4c A cosθ
 Figure 1.4cshow s the projection of vector A onto the direction of

 vector B. A B  B component of A parallel to B 41

PHYSICS CHAPTER 1

 From the Figure 1.4b,the scalar product can be defined as

 A B  A Bcosθ

meanwhile from the Figure1.4c,

B  A  BAcosθ

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 scalar product is negative

θ  90 scalar product is zero

 The scalar product obeys the commutative law of multiplication i.e.

 
AB  B A

42

PHYSICS CHAPTER 1

Vector (cross) product  xiˆ  yˆj  zkˆ
 Consider two vectors : A 
piˆ  qˆj  rkˆ
B

 In general, the vector product is defined as
AB C

and its magnitude is given by

   
A B  C  A B sinθ  ABsinθ

where θ : angle between two vectors

 The angle  ranges from 0 to 180  so the vector product always

positive value.

 Vector product is a vector quantity.
 The direction of vector C is determined by

RIGHT-HAND RULE

43

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B  C 

 B
A
 
B C  

B A  C

A    
 A B  B  A but A B   B  A

  Direction of the vector product (C ) always perpendicular
A Bto the plane containing the vectors
and .

44

PHYSICS CHAPTER 1

 The vector product of the unit vectors are shown below :

y

iˆ  ˆj   ˆj  iˆ  kˆ

ˆj ˆj  kˆ  kˆ  ˆj  iˆ
kˆ iˆ x kˆ  iˆ  iˆ  kˆ  ˆj

iˆ  iˆ  i2sin 0o  0

z ˆj  ˆj  j2sin 0o  0 iˆ  iˆ  ˆj  ˆj  kˆ  kˆ  0

kˆ  kˆ  k 2sin 0o  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  IlBsinθ 45

PHYSICS CHAPTER 1

THE END…

Next Chapter…

CHAPTER 2 :
Linear Kinematics

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