physics chapter 1 2014

8/12/2019 PHYSICS Chapter 1 2014

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1. 1.

1.1 1.1


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.

)

: (), (), (), (), (),

(), ().

)

: ( 1),

( 2), (), (), (),

(), () ().

) .


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2


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1.1 ()

Quantity Symbol SI Unit Symbol

Length l metre m

Time t second s

Temperature T/ kelvin K

Electric current I ampere A

Amount of substance N mole mol

Luminous Intensity candela cd


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Derived quantity Symbol Formulae Unit

Velocity v s/t m s-1

Volume V l w t M3

1.2 .

Acceleration a v/t m s-2

Density m/V kg m-3

Momentum p m v kg m s-1

Force F m a kg m s-2 @ N

Work W F s kg m2 s-2 @ J


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Unit Prefixes

It is used for presenting larger and smaller values.for presenting larger and smaller values. Table 1.3 shows all the unit prefixes.

Prefix Value Symbol

tera 1012 T

giga 109 G

mega 106 M

kilo 103

kdeci 10-1 d

centi 10-2 c

milli 10-3 m

micro 10-6

nano 10-9 n

pico 10-12 p


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:

Solve the following problems of unit conversion

a. 30 mm2 = ? m2 b. 865 m h-1 = ? m s-1

c. 300 g cm-3 = ? kg m-3 d. 2.4 x 10-5 cm3 = ? m3


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Solution :Solution :

(a) 30 mm2 = ? m2

( ) ( )232 m10mm1 =262 m10mm1 =

25262 m103.0orm1030mm30 =


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()865 m h

-1

= ? m s-1

11stst methodmethod

=

h1

m10865hkm865

31

=

s3600m10865hkm865

3

1

11 sm240hkm865 =

22ndnd methodmethod

=

s3600

h1

km1

m1000

h1

km865hkm865 1

=

s3600

h1

km1

m1000

h1

km865

hkm865

1

11 sm240hkm865 =


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(c) 300 g cm-3

= ? kg m-3

() 2.4 x 10-5 cm3 = ? m3

( )

=332-

33-

3

3-

m10

cm1

g1

kg10

cm1

g300cmg300

-353 mkg103.0cmg300 =

( )

311

65

325

m104.2

10104.2

10104.2

=

==


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1.21.2 Scalars and Vectors


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a) Define scalar and vector quantities.

b) Perform vector addition and subtraction

operations graphically.

c eso ve vec or n o wo perpen cu arcomponents (x and y axes).

d) Illustrate unit vectors ( ) in Cartesian

coordinate.

, ,i j k


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) ():

) ()

)cos()cos( ABBABA ==

rr

:

.

)sin()sin( ABBABA == rr


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ScalarScalarquantity is defined as a quantity withquantity with

magnitudemagnitude only.

e.g. mass, time, temperature, pressure, electriccurrent, work, energy and etc.

Mathematics operational : ordinary algebra

VectorVector quantity is defined as a quantity with bothquantity with both

magnitude & direction.magnitude & direction.

e.g. displacement, velocity, acceleration, force,

momentum, electric field, magnetic field and etc.

Mathematics operational : vector algebra


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1.6 () .

Vector A LengthLength of an arrow magnitudemagnitude of vector A

displacement velocity acceleration

DirectionDirection of arrow directiondirection of vector A

15

.

sr vr ar

s av

vv =r

aa =r

s (bold)v (bold)

a (bold)

Table 1.6Table 1.6


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r

Qr

QP

rr

=

16

k ,

kk , .

kk , .

Akr

Akr

r

r


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..

ParallelogramParallelogram TriangleTriangle

:

Addition of Vectors

Brr

+

17


rr

r

r

Brr +

O

r

r

Brr +

O


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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

18

vec or .

BBrrrr

+=+ Commutative RuleCommutative Rule

Br

r

B

rr

+O


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If there are more than 2 vectors therefore

Use vector polygon and associative rule. E.g. RQPrrr

++

rQr

r

19

r

Qr

r QPrr

+

RQPRQP

rrrrrr

++=++ Associative RuleAssociative Rule

RQPrrr

++


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:.

.

:

BABArrrr

+=+

( ) AAA

rrr

+=+

numberrealare,

20

BABArrrr

+=+ 2

r

r

Brr

+

O BArr

+2


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Br

2

Brr

22 +

BABA rrrr 22 +=+

21

r

2O

BABA rrrr 222 +=+


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r

( ) ( ) AAArrr

312 =+=+

r

22

r

3

AAAA rrrr 12 +=+r

2r+

=( ) AAA

rrr

1212 +=+


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:1.2.4

rCr

DCrr

r

DCDCrrrr

+=

23


O OCr

Dr

Crr

Cr

r

DCrr


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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

24

.


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:Resolving a Vector

Dr

yDr

y :

Dr

yDr

y

25

Dr

0x

Dx cos= DD cos=

Dy

sin= DDy sin=

Dr

0x

= sinDx = sinDDx

= cosDy = cosDDy


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D:

D:

( ) ( )2y2

x DDDD +=orr

yD= = y

D1

26

D

xD xD

jDiDD yx +=

r


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Example 1 : y

45o

)( N30F2r

)( N10F1r

x20

27

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.

)( N40F3r


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O

y

x45o30o

20

1F

r

y1F

r

x1Fr

x3Fr

x2Fr

28

3Fr

2Fr

y2Fr y3

Fr

++== 321r FFFFF

rrrrr

+= yxr FFFrrr

x3x2x1x FFFFrrrr

++=y3y2y1y FFFF

rrrr

++=


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Vector x-component y-component

o20FF 1x1 cos=1F

r

r

o2010Fx1 cos=N9.40=x1F

o20FF 1y1 sin=o

2010Fy1 sin=N3.42=y1F

o= o4530F si=

29

3Fr

2F N21.2=x2F N21.2=y2Fo3040Fx3 cos=

N34.6=3F

o3040Fy3 sin=N20.0=y3F

VectorVector

sumsum

( )34.621.29.40 ++= xFN4.00= xF

( ) ( )20.021.23.42 ++= yFN37.8= yF


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y

Solution :Solution :The magnitude of the resultant force is

( ) ( )22 += yxr FFF

N38.0=rF

( ) ( )22 37.84.00 +=rF

o264

30

xO

and its direction is

=

x

y

F

F 1tan

( )iseanticlockwaxis-xpositivefrom264or84.0 oo

=

= 4.00

37.8tan 1

rFr

y

Fr

xF

r

84.0


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(3)

3

m2)3,4,(),,(),,( === kjizyxsr

y/m

31

sr

2

4x/m

z/m

0


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Scalar (dot) productScalar (dot) product The physical meaning of the scalar productphysical meaning of the scalar product can be explained by

considering two vectors and as shown in figure 1.3a.

r r

r

r

r

Figure 1.3aFigure 1.3a

32

Figure 1.3b shows the projection of vector onto the direction of

vector .

Figure 1.3c shows the projection of vector onto the direction of

vector .

r

r

r

r

r

cosFigure 1.3bFigure 1.3b

r

r

cos

Figure 1.3cFigure 1.3c

ABABArrrr

toparallelofcomponent=

BABBA

rrrr

toparallelofcomponent=


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From the figure 1.3b, the scalar product can be defined as

meanwhile from the figure 1.3c,

where

( )BABA cos= rr

vectorsobetween twangle:

( )ABAB cos= rr

33

.

The angle ranges from 0 to 180 . When

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

oo 900


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Calculate the and the angle between vectors and for thefollowing problems.

a)

Example 1 : rrr

r

kjiA 32 +=r

kjiB 52 +=

r

35


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SolutionSolution ::a)

The magnitude of the vectors:

( )( ) ( )( ) ( )( ) kkjjiiBA 531221 ++= rr

1522 =BA

rr

19= rr

36

The angle ,

( ) ( ) ( ) 14321222

=++=A( ) ( ) ( ) 30512 222 =++=B

ABBA cos=

rr

=

=

3014

19coscos 11

AB

BA

rr

o

158=


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Vector (cross) productVector (cross) product

Consider two vectors :

In general, the vector product is defined as

and its magnitudemagnitude is given by

krjqipB ++=r

kzjyixA ++=

r

CBArrr

=

ABBACBA sinsin === rrrrr

37

where

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

positivepositive value.

Vector product is a vector quantityvector quantity.

The direction of vector is determined by

vectorsobetween twangle:

RIGHTRIGHT--HAND RULEHAND RULE

Cr


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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.

Cr

rCBArrr

=

38

Direction of the vector product always perpendicular to the

plane containing the vectors and .

r

Br

r

Cr

CABrrr

=

rrrr but ( )ABBArrrr =

Br

)(Cr

r


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The vector product of the unit vectors are shown below :

x

y

kj

i

ijkkj

==

kijji ==

jkiik ==

39

Example of vector product is a magnetic force on the straighta magnetic force on the straight

conductor carrying current places in magnetic fieldconductor carrying current places in magnetic field where the

expression is given by

z0 === kkjjii

0in == o2

0siii

0in == o2 0sjjj

0in == o2 0skkk

BlIFrrr

=

IlBF sin=


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The vector product can also be expressed in determinant form as

11stst method :method :

rqp

zyx

kji

BA

= rr

rr

40

22ndnd method :method :

Note :Note :

The angle between two vectorsThe angle between two vectors can only be determined by

using the scalar (dot) product.scalar (dot) product.

zz =

( ) ( ) ( )kypxqjxrzpizqyrBA

++=

rr


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Given two vectors :

Determine

a) and its magnitude

b)

Example 2:

B

rr

Brr

kjiA 425 +=r

kjiB 5 ++=r

r r

41

c) the angle between vectors and .

B


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SolutionSolution ::a)

511

425

=kji

BArr

42

The magnitude,

( )( ) ( )( )( ) ( )( ) ( )( )( ) ( )( ) ( )( )( )kjiBA 121514551452 += rr

kjiBA 72141 +=

rr

( ) ( ) ( )kjiBA 25425410 ++= rr

2.26=BArr

( ) ( ) ( )222 72114 ++=BArr


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b)

c) The magnitude of vectors,

kjikjiBA 5425 +++= rr

23=Brr

( )( ) ( )( ) ( )( ) kkjjiiBA 541215 ++= rr

2025 +=BArr

222

43

Using the scalar (dot) productscalar (dot) product formula,

ABBA cos=

rr

=

=

2745

23coscos 11

AB

BA

rr

o

7.48=

==

( ) ( ) ( ) 27511 222 =++=B

physics chapter 1 2014

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