PHYSICS
CHAPTER 1
PHYSICS
1.1.1 Unit Prefixes
Example 1.1 :
It is used for presenting larger and smaller values. Table 1.3 shows all the unit prefixes.
tera
×
1012
T
Solve the following problems of unit conversion. a. 15 mm2 = ? m2 b. 65 km h−1 = ? m s −1 − − 3 3 c. 450 g cm = ? kg m Solution :
giga
× 109
G
a. 15 mm2 = ? m2
mega
× 106
M
kilo
× 10
k
deci
× 10−1
d
centi
× 10−2
c
milli
×
10−3
m
micro
× 10−6
µ
nano
× 10−9
n
10−12
p
Prefix
Table 1.3
Examples:
Multiple
pico
×
Symbol
5740000 m = 5740 km = 5.74 Mm 0.00000233 s = 2.33 × 10−6 s = 2.33 µs
PHYSICS
2
−1
2
65 × 103 m 65 km h = 1 h 65 × 103 m −1 65 km h = 3600 s −1 −1 −1
65 km h
5
= 18 m s
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6
CHAPTER 1
Follow Up Exercise
=
1. A hall bulletin board has an area of 250 cm 2. What is this area in square meters ( m 2 ) ? 2.
65 km h −1 = 18 m s −1 c. 450 g cm-3 = ? kg m -3
−3
b. 65 km h-1 = ? m s-1 1st method :
65 km 1 h 65 km 1000 m 1 h 1 65 km h − = 1 h 1 km 3600 s
65 km h
=
1 mm 2 = 10 −6 m 2
CHAPTER 1
2nd method :
CHAPTER 1
3. − 3 10 3 kg 450 g 1 cm 3 3 1 cm 1 g (10 − 2 ) m 3
450 g cm −3 =
450 g cm −3 = 4.5 × 105 kg m −3
The density of metal mercury is 13.6 g/cm 3. What is this density as expressed in kg/m3
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 °
7
8
PHYSICS
CHAPTER 1
PHYSICS
1.2.2 Direction of Vectors
CHAPTER c) Cartesian coordinates 2-Dimension (2-D)
Can be represented by using: a) Direction of compass, compass i.e east, west, north, south, north-east, north-west, south-east and south-west b) Angle with a reference line
s = ( x, y ) = (1 m, 5 m) y /m
e.g. A boy throws a stone at a velocity of 20 m s -1, 50° above horizontal.
5
v
50°
0
x
0
1
13
PHYSICS
CHAPTER 1 3-Dimension (3-D)
PHYSICS
CHAPTER
Unit vectors
s = ( x , y , z ) = ( 4, 3, 2) m
A unit vector is a vector that has a magnitude of 1 with n
s = 4 i +3j + 2k y /m
Are use to specify a given direction in space.
3 i , j & k is used to represent unit vectors pointing in the positive x, y & z directions.
s
4
PHYSICS
CHAPTER 1 d) Polar coordinates
(
F = 30 N,150
PHYSICS 1.2.3 Addition of Vectors
)
F
CHAPTER
150°
There are two methods involved in addition of vecto Parallelogram Triangle For example : A + B
A e) Denotes with + or – – signs signs.
B
+ Parallelogram
A + B
A
B
+
-
O 17
-
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
A
PHYSICS
CHAPTER If there are more than 2 vectors therefore Use vector polygon and associative rule.PE.g. +
Q
P
R
Commutative Rule
A
O
A + B = B + A
B
Trian Triangle
(P + Q ) + R
PHYSICS
CHAPTER 1
PHYSICS
Distributive Rule :
(
a. α A + B
α , β are real number
(α + β ) A = α A + β A
2 A + 2 B
For example : Proof of case a: let
α A + α B = 2 A + 2 B
) = α A + α B
b.
CHAPTER
α α = 2
α A + B = 2 A + B O
2 A
A + B
(
B O
(
2 A + B
A
)
2 A + B = 2 A + 2 B
)
21
PHYSICS
CHAPTER 1
Proof of case b: let
PHYSICS
1.2.4 Subtraction of Vectors For example : C − D
β β = 1 α = 2 and β
CHAPTER
(α + β ) A = (2 + 1) A = 3 A
C
A
α A + β A = 2 A + 1 A
2 A
( )
C − D = C + − D
D
Parallelogram
+
Trian Triangle
A
O
C
C O
PHYSICS
CHAPTER 1
PHYSICS
CHAPTER
1.2.5 Resolving a Vector
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.
1st method :
2nd method
y
y
Exercise 1 :
c) A + 2 B (Hint : use 1 cm = 2.00 units)
d)
R
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 e nega ve x ax s . s n g grap ca me o s an su a e sca e o determine a) A + B b) A − B
R y
R y
θ
x
0
φ R
0
R x
R x
2 A − B
R x R R y R
= cos θ ⇒ R x = R cos θ = sin θ
⇒ R y = R sin θ
R x R R y R
= sin φ ⇒ = cosφ ⇒
25
PHYSICS
CHAPTER 1
R or R =
CHAPTER
Example 1.2 :
The magnitude of vector R :
PHYSICS
A car moves at a velocity of 50 m s-1 in a direction north 3 Calculate the component of the velocity a) due north. b) due east. Solution :
( R x )2 + (R y )2
N
Direction of vector R :
tan θ =
y
R x
or
θ =
R y
tan −1
v N
R x
W
a) v N
= v sin 60 v N = 50 sin 60
30°
60°
v N = 43.3 m s
v
E
−
or
PHYSICS
CHAPTER 1
PHYSICS
Example 1.3 :
CHAPTER
Example 1.4 :
y
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 : ec or x-componen y-componen y
F
F y
150°
30°
S
F x
F
x
F y = F sin 30 F x = − F cos 30 F x = −100 cos 30 F y = 100 sin 30 F y = 50 N F x = −86.6 N
F 1 (10N)
or
30o
F 3 (40N)
or
F 2 (30 N
F x = F cos150 F y = F sin 150 F F y = 100 sin 150 x = 100 cos150 F x = −86.6 N F y = 50 N
The figure above shows three forces F1, F2 and F3 acted O. Calculate the magnitude and direction of the resultant particle O.
29
PHYSICS
CHAPTER 1 y
Solution :
F 3 x
Vector
30o
F 1
60o
30o
x O
x-component
y-com
F 1
F 1 x = 0 N
F 1 y = F 1 y =
F 2 x = −30 cos 60 2 x = −
F 2 x
F
F 3
F 3 y
CHAPTER
Solution :
F 2 y
F 2
PHYSICS
F 3
F 2 y = 30 2 y =
F 3 x = −40 cos 30 F 34.6 N
F 3 y = − F =
PHYSICS
CHAPTER 1
PHYSICS
Exercise 2 : 1. Vector A has components A x = 1.30 cm, A y = 2.25 cm
Solution : The magnitude of the resultant force is
(∑ F ) + (∑ F ) 2
F r =
2
x
(− 49.6)
F r =
has components B x = 4.10 cm, B y = -3.75 cm. Determ
y
2
2
+ (16)
y
F r = 52.1 N
∑
and
F r
∑ F ∑ F
θ = tan −1
y
F y
162
18°
x
∑
−1 16 θ = tan = −18 − 49.6
CHAPTER
x
O
F x
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 & f ANS. : 5.40 cm -1.50 cm 5.60 cm 345° 2.80 2.80 cm 6.62 cm, 295° 2. For the vectors A and B in Figure 1.2, use the resolution to determine themagnitude and direction o a) the vector sum A , + B b) the vector sum B + A , c) the vector difference A − B , d) the vector difference B − A. (Young & freedman,pg.35,no.1.39)
x-axis OR 18 ° above negative xx-axis. Its direction is 162° from positive x33
PHYSICS
CHAPTER 1
Exercise 2 :
s-1,
ANS. : 11.1 m 77.6°; U think; 28.5 m s -1, 202°; 28.5 m s-1, 22.2°
PHYSICS
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
E.g. unit vector a – a vector with a magnitude of 1 un of vector A.
aˆ =
(
Q 24 m s −2
y
)
(
A
=1
A
. .
P 35 m s −2
)
ANS. : 28 m; 19 m
Figur
1.2.6 Unit Vectors ˆ , bˆ, cˆ notations – a
(Walker,pg.78,no. 65)
,
-1
CHAPTER
15 m and points axis. Given in a direction 40° below the positive x that A + B + C = 0 , determine the magnitudes of A and B .
.
(
A 12.0 m s
Unit vectors are dimensionless.
[aˆ ] = 1
A
aˆ
PHYSICS
CHAPTER 1
PHYSICS
y
CHAPTER 1
E.g. :
)
2
2
(4 )
s=
jˆ
ˆ k
(
ˆ m s = 4iˆ + 3 jˆ + 2k
2
+ (3) + (2 ) = 5.39 m
y/ m
x
iˆ 3 jˆ
z
s
Vector can be written in term of unit vectors as : ˆ ˆ ˆ r = r x i + r y j + r z k
ˆ 2k
x/ m
4iˆ
0
Magnitude of vector,
r =
z/ m
(r x )2 + (r y )2 + (r z )2 37
PHYSICS
CHAPTER 1
PHYSICS
Two vectors are given as:
(
)
b)
ˆ m a = iˆ − 2 jˆ + 6k ˆ m b = 4 iˆ − 3 jˆ + k
(
x
x
z
)
(b − a )
)
b−a =
2
2
3 + −1 + − 5
2
= 5.92 m
c)
2a + b x = 2a x + b x = 2(1) + 4 = 6iˆ
2a + b y = 2a y + b y = 2(− 2) + (− 3) = −7 ˆj
(2a + b )
z
)
(5)2 + (− 5)2 + (7 )2
(
The ma nitude,
ˆ = 2a z + b z = 2(6) + 1 = 13k
(
)
ˆ m 2 a + b = 6iˆ − 7 jˆ + 13k
ˆ m a + b = 5iˆ − 5 jˆ + 7k
=
ˆ = b z − a z = 1 − 6 = −5k
ˆ m b − a = 3iˆ − jˆ − 5k
ˆ = a z + b z = 6 + 1 = 7 k
The magnitude, a + b
z
x
(
b − a x = b x − a x = 4 − 1 = 3iˆ
b − a y = b y − a y = −3 − (− 2) = − ˆj
a + b y = a y + b y = −2 − 3 = −5 jˆ
(a + b )
Calculate a) the vector a and its magnitude, +b b) the vector b − a and its magnitude, c the vector 2a + b and its ma nitude. Solution : a) a + b = a + b = 1 + 4 = 5iˆ
)
CHAPTER 1
Example 1.5 :
(
38
= 9.95 m
The magnitude, 2a + b = 39
(6)2 + (− 7 )2 + (13)2
= 15.9 m 40
PHYSICS
CHAPTER 1
PHYSICS
1.2.7 Multiplication of Vectors Scalar (dot) product The physical meaning ofthe scalar product can be explained by
where θ : angle between
θ
.
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
θ ranges from 0° to 180 °. scalar product is positive 0 < θ < 90 scalar product is negative 9 0 < θ < 180 scalar product is zero θ = 90
The angle
When
θ
θ
Figure 1.4c
B
A
B cos θ
The scalar product obeys the commutative law of multiplication i.e.
A • B = B • A
A cos θ B
Figure 1.4c shows the projection of vector A onto the direction of vector B . 41
(
A • B = B component of A parallel to B
PHYSICS
)
CHAPTER 1
42
PHYSICS
CHAPTER 1
Example of scalar product is work done by a constant force where the expression isgiven by
Example 1.6 :
θ between vectors A and B for the
The scalar product of the unit vectors are shown below :
Calculate the A • B and the angle following problems. a)
b)
W = F • s = F (s cos θ ) = s (F cos θ )
two vectors
B • A = B( A cos θ )
B
Figure 1.4b
meanwhile from the Figure 1.4c,
A
From the Figure 1.4b, the scalar product can be defined as
A • B = A( B cos θ )
considering two vectors A and B as shown in Figure 1.4a.
Figure 1.4a
CHAPTER 1
y
2 iˆ • iˆ = i 2 cos 0 o = (1) (1) = 1 2 2 o jˆ • jˆ = j cos 0 = (1) (1) = 1
ˆ ˆ k
iˆ
x
ANS.:−3; 99 99..4°
(1)(4)iˆ • iˆ + (− 1)(− 2) jˆ • jˆ + (1)(− 3)k ˆ • k ˆ
=
−
The magnitude of the vectors: A = The angle θ ,
o iˆ • jˆ = (1)(1)cos 90 = 0
ˆ = iˆ • k ˆ=0 iˆ • jˆ = jˆ • k 43
ˆ ˆ ˆ A = 4i − 3 j + k ˆ B = 2 jˆ + 3k
A • B = 3 ˆ • k ˆ =1 iˆ • iˆ = jˆ • jˆ = k
z
ˆ = (1)(1)cos 90 o = 0 jˆ • k ˆ = (1)(1)cos 90 o = 0 iˆ • k
ˆ ˆ ˆ A = i − j + k ˆ B = 4iˆ − 2 jˆ − 3k
Solution : a) A • B =
ˆ • k ˆ = k 2 cos 0 o = 1 2 1 = 1 k
B =
(1)2 + (− 1)2 + (1)2
=
(4)2 + (− 2)2 + (− 3)2
3 =
A • B = AB cos θ 3 −1 A • B −1 θ = cos AB = cos 3 29 θ = 71 2
44
29
PHYSICS
CHAPTER 1
PHYSICS
y
Example 1.7 :
CHAPTER 1
D in unit vector are b) Vectors C and
C (1 m )
C = C x iˆ + C y jˆ
25° 19°
0 Figure 1.5
and
Referring to the vectors in Figure 1.5, a) determine the scalar product between them. b) express the resultant vector of
(
) (
)
= − 1cos 25 iˆ + 1sin 25 jˆ C = − 0.91iˆ + 0.42 ˆj m
x D(2 m )
(
(
Hence
)
) (
)
D = 2 cos19 iˆ + − 2 sin 19 jˆ = . ˆ− . ˆ m
C and D in unit vector.
C + D = (− 0.91 + 1.89)iˆ + (0.42 − 0.65) jˆ
(
Solution :
)
= 0.98iˆ − 0.23 ˆj m
C and D is θ = (180 − 25) + 19 = 174
a) The angle between vectors
Therefore
C • D = CD cos θ = (1)(2) cos174
C • D = −1.99 m 2
45
PHYSICS
CHAPTER 1
PHYSICS
Vector (cross) product Consider two vectors : A = xiˆ + y jˆ
ˆ + zk ˆ B = piˆ + q jˆ + r k
46
CHAPTER 1 For example: How to use right hand rule :
In general, the vector product is defined as
A × B = C
×
=
A × B = C
C
=
and its magnitude is given by
Point the 4 fingers to the direction of the 1 st vector. Swept the 4 fingers from the 1 st vector towards the 2 nd vector. The thumb shows the direction of the vector product.
B
sn =
sn
where θ : angle between
A
two vectors
The angle θ ranges from 0° to 180 ° so the vector product always positive value. quantity Vector product is a vector quantity. The direction of vector C is determined by
B
A
A × B ≠ B × A
but
(
Direction of the vector product
(C )
to the plane containing the vectors 47
)
A × B = − B × A
RIGHT--HAND RULE RIGHT
C B × A = C
always perpendicular
A
and
B. 48
PHYSICS
y
CHAPTER 1 The vector product of the unit v ectors are shown below :
b)
(
)(
)
c) The magnitude of vectors,
=
B =
ˆ × k ˆ=0 iˆ × iˆ = jˆ × jˆ = k
jˆ × jˆ = j 2 sin 0 o = 0
−
2
+ −
2
2
+ −
(1)2 + (0)2 + (− 5)2
=
=
26
Using the scalar (dot) product formula,
ˆ × k ˆ = k 2 sin 0 o = 0 k
A • B = 2
iˆ × iˆ = i 2 sin 0 o = 0
z
ˆ • iˆ + 0 jˆ − 5k ˆ A • B = − 3iˆ − 2 jˆ − k
A • B = −3 + 0 + 5
ˆ × iˆ = −iˆ × k ˆ = jˆ k
x
iˆ
CHAPTER 1 ˆ • k ˆ A • B = (− 3)(1)iˆ • iˆ + (− 2)(0) jˆ • jˆ + (− 1)(− 5) k
ˆ iˆ × jˆ = − jˆ × iˆ = k ˆ = −k ˆ × jˆ = iˆ jˆ × k
jˆ
ˆ k
PHYSICS
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
PHYSICS
49
CHAPTER 1
Exercise 3 :
1. If vector a = 3iˆ + 5 jˆ and vector a) a × b , b) a • b , ˆ; 26; 46 ANS. : 2k
A • B = AB cos θ 2 −1 A • B −1 θ = cos AB = cos 14 26 θ = 84
PHYSICS
50
CHAPTER 1
b = 2iˆ + 4 jˆ , determine c) a + b • b .
THE END…
2. Three vectors are given as follow :
ˆ ; b = −iˆ − 4 jˆ + 2k ˆ and c = 2iˆ + 2 jˆ + k ˆ a = 3iˆ + 3 jˆ − 2k
Calculate a) a • b × c , b) a • ˆ ANS. : − 21; − 9; 5iˆ − 11 jˆ − 9k
(
)
3. If vector P determine
b +c
, c)
a× b +c
…
.
CHAPTER 2 : Kinematics of Linear Motion
ˆ and vector Q = −2iˆ + 4 jˆ + 3k ˆ, = 3iˆ + 2 jˆ − k
a) the direction of P × Q b) the angle between P and ANS. : U think, 92.8°
Q
.
51
52
