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LESSON 1: MEASUREMENTS
Introduction
Physics is a body of facts giving an exact description of man’s experiences regarding the
physical world pertaining to matter, energy and their interactions not involving chemical
and biological changes
...
Hence, it is a numerical description
...

Learning Outcomes
After successful completion of this lesson, you should be able to:
Understand the concept of significant figures and how to handle them when
carrying out simple arithmetic operations
...

Describe the three widely used systems of units
...

Discussion
1
...
Exact numbers are numbers with an infinite number of significant figures while
approximate numbers are numbers with limited number of significant figures
...

1
...
1

Rules for Counting Significant Figures
✓ Nonzero digits are always significant
...

✓
Zeros to the left of nonzero digits (called leading zeros) are never
significant
...
If a number contains a decimal point, the zeros to the right of it
are always significant
...
00 has four significant figures
...
The number 93,000 has at least two and at
most, five significant figures
...


1
...
2

1
...
3

2005
10
...
4
0
...
0

Rules for Rounding Off
✓

If the digit you remove from the right-hand end of the number is 0, 1, 2, 3
or 4, then the last digit you keep is unchanged
...


Significant Figures in Calculations
There are a few simple rules to follow in deciding how many significant figures to

keep
...
That is, x = 3m has only one significant figure,
and expressing this value as x = 0
...
If we instead wrote x = 3
...
0030 km), we would imply that we know the value of x to two
significant figures
...

Rule 2 — When multiplying or dividing, keep a number of significant figures in the
product or quotient no greater than the number of significant figures in
the least precise of the factors
...
3

3
...
2

A bit of good judgment is occasionally necessary when applying this
rule:
9
...
03 = 10
...
8 technically has only two significant figures, it
is very close to being a number with three significant figures
...

Rule 3 — In adding or subtracting, the least significant digit of the sum or
difference occupies the same relative position as the least significant
digit of the quantities being added or subtracted
...
For example, suppose we wish to find the sum of three
numbers as follows:
103
...
10
+ 0
...
319 or 106
...
By
rule 1, we should include only one doubtful digit; thus the result should
be expressed as 106
...

*Note:

1
...


Scientific Notation

Scientific Notation is a method of writing numbers in the following form:
n 10m, where n is any number between 1 and 10 (1 n 10) while m is any positive or
negative integer
...
234 105
0
...
010 10− 4
This method is used in expressing very large or very small numbers and in
indicating whether the zeros at the end of a number are significant or not
...
0 103
5,000 = 5
...
3

only the first zero is significant
all zeros are significant

Measurement of Physical Quantities

Physical quantity is a quantity, which actually exist and is therefore directly or
indirectly measurable in the laboratory
...
A unit is a measure of a quantity that is equal to 1
...

*Note: Units and standards can be defined freely
...

Two kinds of physical quantities according to units:
✓

Fundamental quantities — quantities that are relatively easy to define and
choose a suitable standard
...


In practice, the smallest possible number of fundamental quantities are chosen
and the rest of the other quantities are merely defined in terms of them
...
4

Three Widely Used Systems of Units

1
...
1

The International System of Units (SI)
SI System — abbreviation of the French equivalent of the English term
International System of Units
— metric system or the MKS system
— internationally accepted system of units
SI Fundamental Units
Quantity
time
length
mass
amount of substance
temperature
electric current
luminous intensity

Name

Symbol

second
meter
kilogram
mole
kelvin
ampere
candela

s
m
kg
mol
k
A
cd

7

✓

mass — resistance to force, which causes objects to move (the more
massive the object, the more force is needed to move the object
from rest)

✓

amount of substance — amount of material contained in an object in terms
of the number of atoms or molecules

✓

temperature — hotness or coldness of a substance

✓

electric current — amount of negative charges flowing

✓

luminous intensity — brightness of light

✓

MKS — stands for the first letters of the three base units of length (meter),
mass (kilogram), and time (seconds)

✓

standard of time in the SI system — second (s)
— the time required for a cesium – 133
atom to undergo 9,192,631,770
vibrations

Two advantages of the SI system:
1
...
conveniently relates different units of measure using the SI prefixes added to
the unit
...
4
...
It is also known
as the CGS system, a name derived the first letter of the three units of length, mass and
time
...
4
...
It is also known as the FPS system, a name derived from the
first letter of the fundamental unit of length (feet), the derived unit of force (pound), and
the fundamental unit of time (second)
...
5

Dimensional Consistency and Conversion of Units

The dimensional consistency is the consistency of units used in manipulating
different physical quantities
...

Example 1:
Given the following physical quantities:
A = length in meters, B = time in seconds and C = length in meters
A + B → wrong!
A + C → correct!

Since A is in unit of length while B is in unit of time
Since both are in units of length

Example 2:
Given the following measurements of the sides of a rectangle:
A = length in meters
C = width in mm
B = length in feet
D = width in meters
perimeter = 2 (A + B) → wrong! Since A is in SI while B is in the FPS
system
perimeter = 2 (A + C) → wrong! Since A is in meters while C is in mm
perimeter = 2 (A + D) → correct! Since bot are in meters
Similarly,
9

area = A • B → wrong! Since A is in SI while B is in the FPS system
area = A • C → wrong! Although both are in SI system, since A is in
meters while C is in mm
area = A • D → correct!
✓ Lack of dimensional consistency — requires conversion of units from one
system to another to achieve consistency
...

✓ Conversion factor — a ratio of units that is equal to one
...
5 min = 2
...
5 min • 60 s min = 15 s

multiplication

Equivalently,
2
...
5 min 2
...
5 • 60 s = 15 s
=
1 min
CV2
60 s

division

Example 4:
Express 36
...


1 fathom (fath) = 6 ft → conversion factor of 6 ft fath
60 s = 1 min → conversion factor of 60 s min
3
...
28 ft m

36
...
5 fath min )(6 ft fath ) = 1
...
28 ft m )

10

Note that the conversion factor can be multiplied

(like the 60 s min or 3
...


11

LESSON 2: VECTORS AND SCALARS
Introduction
This lesson dealt with scalar and vector quantities
...

Learning Outcomes:
After successful completion of this lesson, you should be able to:
Know the difference between vectors and scalar quantities
...

2
...
Some examples of scalar quantities are:
✓
✓
✓
✓
✓

the number of chairs in a room
the distance between two points
the temperature of an object
the age of a person
the speed of a moving object

Vectors are physical quantities with both magnitude and direction
...

✓ the velocity of a moving object, specifying the speed of the object and the direction
the motion is taking place
...

The normal rules of algebraic addition, subtraction, multiplication and division are used
with scalars, while special mathematical rules are needed to manipulate vectors
...
2

Addition of Vectors

Vector addition is the process of finding a single vector which will produce the same
effect produced by the given vectors
...
The vector,
which is opposite to the resultant but of the same magnitude is called the equilibriant
...

2
...
1

The Graphical Method

In this method, a directed line segment is used to represent vector quantities
...
An appropriate scale is used in drawing the length of the arrow
...
2
...
Vector A is 2
...
5 cm long represents a 15
km displacement 300 north of east
...
2
...
Parallelogram Method — this method is applicable to two vectors at a time
...
With these two
vectors as adjacent sides of a parallelogram, construct the
parallelogram
...
For the
direction of R , measure the angle

with side of the parallelogram
...
Using
the second resultant and the fourth vector, apply the parallelogram
method again to get the third resultant vector
...
The last resultant vector
obtained is the total resultant vector of all the given vectors
...

14

2
...
This is done by
drawing the vectors head–to–tail one at a time (as illustrated in Fig
...
3a)
...
The resultant vector R goes from the tail of
the first vector to the head of the last vector
...
2
...
2
...
2
...
The vectors need
not be drawn to scale, but a working diagram must be drawn
...
The Pythagorean Theorem — this method is limited to two vectors that are
perpendicular with each other as shown in Fig
...
4
...
2
...
Cosine Law and Sine Law — this is used when the two vectors are not
perpendicular with each other as illustrated in Fig
...
5
...
Cosine law is used for
solving the magnitude of the resultant R
15

R2 = A2 + B2 – 2ABcos

R2 = A2 + B2 + 2ABcosφ

or

and sine law for the direction
...
2
...
Component Method — a more accurate, faster and more convenient method for
solving variety of interesting problems involving vectors is the
component method because it is applicable to all types of vectors
...
2
...
The
sign of Fx and Fy depends on the direction of the projection of F on the
x and y axes respectively
...

Fiy = F1y + F2y + F3y +
...
The quadrant in which R points is determined from
a rough drawing of R based on its components values
...


Calculate the magnitude and direction of the resultant of the following two
displacement vectors:
a
...
50 km E
40 km W
70 km 60° N of E
Solutions:
N
a)
Fig
...
7

R 2 = 302 + 402
R = 50 km
30
= tan -1
40
= 36
...
2
...
4km
70km sin1200
sin =
104
...
50

R

70 km
0

60

R = 104
...
5° N of E

E
50 km

2
...
2
...

Solutions:

Fx = 200 cos30 0 − 100 cos450 − 150 cos0 0 + 200 cos37 0
= 173
...
7 − 150 + 159
...
2 N
17

Fy = 200 sin 30 0 + 100 sin 450 + 150 sin 0 0 − 200 sin 37 0
= 100 + 70
...
36
= 50
...
2) + (50
...
9N
50
...
2
θ R = 24
...
There are several types of motion like motion along a straight line, circular
motion and rotational motion
...
To understand motion fully, it is a must to know distance and displacement, speed
and velocity, and acceleration
...
1

Distance and Displacement

Distance is a scalar quantity that measures the position of a moving object from a
reference point, while displacement refers to the change in position
...

3
...
The
significant difference is that velocity is a vector quantity while speed is a scalar quantity, which
means that velocity specifies both how fast and in what direction an object moves, whereas
speed specifies only how fast an object moves
...
Velocity is the time rate of change of displacement
...


v=

s
t

(3
...
2
...
2
...
2)

Instantaneous Speed and Instantaneous Velocity

Instantaneous speed, or velocity, is the speed or velocity at any given instant
...
3)
v=
t
Sample Problems with Solutions:
1
...
4 s
...
24m
=
32
...
0 m s

v1 =

s2
t
224 m
=
33
...
7 m s

v2 =

At t = 32 s, the distance traveled by the slower runner is

d 2 = (v 2 )(32s )

= (6
...
4 m

21

so that

d = s1 − d 2
= 224 m - 214
...
6 m
2
...
0
average speed of 8
...
Calculate the total time required and the average speed
...
0 m s
= 133
...
33 s + 100 s
= 233
...


t2 =

s2
v2

800m
8
...
33 s
= 6
...
Your average velocity is 2
...
Calculate
a) your walking time
b) your average speed
Solution:
a)

s
t
s
t=
v
360 m
=
2 ms
= 180s

v=

22

b)

total distance s t
=
total time
t
s t = s1 + s 2
v=

(

)

(

= (360 m ) cos 530 + (360 m ) sin 530
= 216
...
5m
= 504
...
15 m
v=
180 s
= 2
...
3

)

Acceleration

Acceleration is the time rate of change of velocity, where the change in velocity may
be a change in magnitude, direction or both
...
When an object is increasing its velocity, it has a positive acceleration
...
Negative acceleration
is also called deceleration
...
3
...
The acceleration in this case is taken as the average acceleration, which is equal to the
change in velocity divided by the time required for that change
...
The change in velocity during that time interval is vf – vo
...
4)
a= f
t
from which we get

vf = vo + at
The average velocity of the body during the interval is

v=

1
2

(vo + vf )

(3
...
6)
23

Substituting the expression for t from equation (3
...
6),

s=

1
2

(vo + vf )
2

2 a s = vf − vo

vf + vo
a

2

vf2 = vo2 + 2as

(4
...
4) and (3
...
8)

Sample Problems with Solutions:
1
...
Calculate

s2

Solution:

(15 mi h )(5280 ft mi )

1h
= 22 ft s
3600 s

(25 mi h )(5280 ft mi )

1h
= 36
...
67 ft s )2 − (22 ft s )2
=
(2)(50 ft )

a=

= 0
...


The driver of a car traveling at 30

m

s

needs to quickly reduce the car’s speed to 20

to round a curve in the highway
...
0

m

s

m 2
s

, how far does it travel while reducing its speed?

24

Solution:
2

2

vf = vo + 2 a s
2

2

vf − vo
2a
(20 m s )2 − (30 ft s )2
=
(2) - 6 m s 2
= 41
...


)

An airplane taking off from a landing field has a run of 500 m
...


v f = v o + at

(

)

= 0 + 1
...
3 m s

= 1
...
It then

m 2
s

until it stops at

the next station
...

Solution:
a1 = 2

a3 = 2

t1 = 15s
s1

vo = 0

s t = s1 + s 2 + s3

v1

s3

v3 = 0

1
2

m 2
s

2

m 2
s

v3 − v 2
0 − v1
- (30 m s )
=
=
= 150m
2a 3
2a 3
2 - 3ms 2
2

s3 =

v2

(2 )(15s) = 225m
)(15s)(25s) = 750m
= (2

s1 = v o t + 12 a1t 2 = 0 +
s 2 = v1t 2 = (a1t1 )t 2

t2 = 25s
s2

2

2

2

(

)

s t = 225m + 750m + 150m = 1125m

25

5
...
Just as he passes the stop, he begins to
decelerate
...

a)
b)

Calculate the speed of the bus as it passed the bus stop
...
4
5
= -2
...
4)
= 10 + 12
= 22
v 0 = 22 m s
6
...
At the same instant,

overtakes and passes the automobile
...
3
...
The acceleration is caused by the
gravitational force of the earth pulling down all objects
...
The magnitude
of this acceleration g has the following values on the earth’s surface:
g = 9
...
That is,

v f = v o + gt
v=

1
2

(v o + v f )

H = vt
2

2

v f = v o + 2gH
H = v o t + 12 gt 2
Acceleration of gravity g is positive for bodies that are falling and negative for bodies
that are thrown upward provided the +y axis is chosen downward and the –y axis upward
...


A ball is dropped from a height of 200 m
...
0 + gt
1
2

= 12 gt

v f = v o + gt

= 0 + (9
...
4)
= 62
...
8
= 6
...


At what velocity must a stone be thrown vertically upward to reach a height of 20 m?
Solution:
At H = 20 m, vf = 0
2

2

2

2

v f = vo + 2 gH
vo = v f − 2 gH

= 0 − 2(- 9
...
8 m s
3
...
How far below the top of the cliff
will the second stone overtake the first?

28

Solution:
For the first stone,

H1 = vot + 12 gt 2
= 0+

1
2

(9
...
9t 2
For the second stone,

H 2 = vo (t − 1) + 12 g (t − 1)

2

= 20(t − 1) +

1
2

(9
...
9 t 2 − 2t + 1
= 4
...
2t − 15
...
9t 2 = 4
...
2t − 15
...
2t = 15
...
1
t =
10
...
48
time the second stone overtakes the first
so that

H = vot + 12 gt 2
= 0+

1
2

(9
...
48)2

= 10
...
73

4
...
The window is 30 ft above the
ground
...
6 ft s
2

2

v f = vo + 2 gH
2

2

2

2

2 gH = v f − vo
H =

v f − vo
2g

0 − (46
...
9

2

=

H = 33
...
5

t = 0
...
3
...
This type of motion follows a parabolic path like a
baseball having been hit into the air by a bat, or by an ice skater jumping over some barrels
...


30

3
...
3
...
At
any point in its path, this velocity has a
horizontal component, vo, the initial
velocity, which is always constant
because
there
is
no
horizontal
acceleration, and a vertical component,
vy, which always increases as the body
increases its distance of fall H
...


vo
x

0
v

o

vy

vo

vy

v y = 2 gH
y

The resultant velocity at any point is given by
2

v = vo + v y

2

The time of fall of this projectile on the ground is the same time of fall it has when freely
dropped:
t =

2H
g

The range R or the horizontal distance traveled is,

R = vo
3
...
3
...


A stone is thrown horizontally from the top of a building 20 m high with a velocity of 8
m
...


2H
=
g

2 • 20m
= 4
...
8 m s 2

a)

t=

b)

R = vo

c)

v y = 2 gH = 2(9
...
08s ) = 32
...
If

the airplane is 10,000 ft above the ground, how far from the target must the bomb be
released in order to make a direct hit? Neglect the resistance
...


A ball was thrown with a velocity of 150
a)
b)

ft

s

at an angle of 45° above the horizontal
...

What is the maximum height reached?

Solution:
2

vo sin 2
g

R=

(150 ft s )2 (sin 450 )

=

32 ft s 2

= 703
...
63s
2

H =
=

vo sin 2
2g

(150 ft s )2 (sin 2 450 )
(2)(32 ft s )
2

= 175
...


A batted baseball, which leaves the bat at an angle of 30° above the horizontal, is
caught by an outfielder 120 m from the plate
...
8)(t )

(

= vo sin 30

0

)

120
120
− (4
...
9)(120)
120 sin 300
= 2
0
cos 30
vo cos 2 300
2

2

vo =

(4
...
93
sin 300 cos 300
= 36
...
3m
c)

t=

2vo sin
g

t = 3
...
85)2 sin 2 300

→

=

2 • 9
...
3

maximum height reached by the ball

(2)(36
...
76
9

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