PHYSICS INTRODUCTION
Connecting Physics to Our World.
How do you see Physics?
-I see Physics through Quantitative means.
-I see Physics as the science that explains the movement of matter in time and space.
Where do you see Physics?
-I see physics everywhere.
EVERYTHING = MATTER
What is Physics?
-The branch of science which deals with the interaction of matter and energy.
In relation to Chemistry, Chemistry is the study of matter.
Branches of Physics
1. Classical Physics
 Mechanics
 Heat and Thermodynamics
 Optics
 Acoustics
 Electricity and Magnetism
2. Modern Physics – Post Newtonian
 Relativity
 Quantum Mechanics
 Nuclear Physics
Classical Physics: Past -1900 AD
Modern Physics: 1900 AD – Present
Why is Physics important?



In our daily life we hardly find a device in which laws of physics are not involved.
All the Technologies you might take for granted exploit the rules of physics.
A possible career.
PHYSICS IS THE FOUNDATION OF ALL SCIENCES.
PHYSICS AND CHEMISTRY ALWAYS WORK TOGETHER.
Electrostatic Force = Force responsible for Chemical Bonding.

+ - Ionic charges get attracted to one another then they form bonds.
MEASUREMENT
Gospel Reading: Rev 21: 15-17. ”The measuring of the New Jerusalem”
Measurement
Early system of measurement.



Hands
Feet
Arm Span
Physics Is an experimental science. Experiments require measurement, and we generally use
numbers to describe the results of measurement.
*Physical Quantity

Reference Standard = Refers to the Unit
o 15 Meters
o 25 mL
o 4.5 in
Question: How do we measure things?
2 SYSTEMS OF MEASUREMENT
Physics is the only quantitative if the sciences. It deals with the concepts that can be measured. Two
measurement system are commonly used Systeme International D’ Unites (SI) or International
System of Units and the English System (ES).
Quantity
Length
Mass
Weight
Time
Luminous
Intensity
Temperature
Electric
Current
Amount of
Substance
Area
Volume
Speed &
Velocity
Description
the interval or distance between
two points
The amount of matter in an
object.
Symbol
d
Unit of Measure
Meter (m)
m
Kilogram (kg)
The measure of the force exerted
by Earth on an object
the interval between two events
at the same place in space
The quantity of visible light that is
emitted in unit time per unit solid
angle. The unit for the quantity of
light flowing from a source in any
one second
The degree or intensity of heat
present in a substance or object,
especially as expressed according
to a comparative scale and shown
by a thermometer or perceived by
touch.
A stream of charged particles,
such as electrons or ions, moving
through an electrical conductor or
space.
The amount of substance in a
given sample of matter is defined
as the number of discrete atomicscale particles in it divided by the
Avogadro constant NA.
The quantity that expresses the
extent of a two-dimensional
region, shape, or planar lamina, in
the plane.
the amount of space that a
substance or object occupies, or
that is enclosed within a
container
A scalar quantity that refers to
"how fast an object is moving."
Speed can be thought of as the
rate at which an object covers
distance.
m
Kilogram (kg)
t
Seconds (s)
lv
Candela (cd)
T
Kelvin (K)
I, i
Ampere (A)
n
Mole (mol)
A
Square meter (m²)
V
Cubic meter (m³)
V, v
Meter per second
(m/s)
Density
The degree of compactness of a
substance.
Concentration The abundance of a constituent
divided by the total volume of a
mixture.
Mass and
he mass concentration ρᵢ is
Concentration defined as the mass of a
constituent mᵢ divided by the
volume of the mixture V.
Acceleration
The rate of change of the velocity
of an object with respect to time.
Force /
Any influence that, when
Weight
unopposed, will change the
motion of an object.
Pressure
The force applied perpendicular
to the surface of an object per
unit area over which that force is
distributed.
Work
The energy transferred to or from
an object via the application of
force along a displacement.
Power
The amount of energy transferred
or converted per unit time.
p
Kilogram per cubic
meter (kg /m³)
Mole per cubic
meter (mol /m³)
c
p, v
Kilogram per cubic
meter (kg /m³)
a
Meter per second
square (m/s²)
Newton (N) (kg •
m/s²)
F, W
P
Pa, Pascal (N/m²)
W
J, Joule (N • m)
P
W, Watt (N • m / s)
METRIC (SI) PREFIXES TABLE
Prefix
yotta
Abbreviation
Y
Value
1024
zetta
Z
1021
exa
E
1018
peta
P
1015
tera
T
1012
giga
G
109
mega
M
106
kilo
k
103
hecto
h
102
Deka
da
101
deci
d
10−1
centi
c
10−2
milli
m
10−3
Micro
μ
10−6
nano
n
10−9
Pico
p
10−12
femto
f
10−15
Atto
a
1018
Zetto
z
1021
yocto
y
1024
ENGLISH SYSTEM
MEASURES (ENGLISH, METRIC, AND EQUIVALENTS)
UNITS OF LENGTH IN THE U.S. SYSTEM
1 foot (ft) = 12 inches (in)
1 yard (yd) = 3 feet (ft)
1 yard (yd) = 36 inches (in)
1 mile (mi) = 5,280 feet (ft)
UNITS OF LENGTH IN THE METRIC SYSTEM
SYSTEM TO SYSTEM CONVERSIONS FOR LENGTH
1,000 millimeters (mm) = 1 meter
1 inch = 2.54 centimeters
100 centimeters (cm) = 1 meter
1 meter ≈ 3.28 feet
10 decimeters (dm) = 1 meter
1 mile ≈1.61 km
1 dekameter (dam) = 10 meters
1 foot ≈ 0.30 meters (m)
1 hectometer (hm) = 100 meters
1 yard≈0.91 meters
1 kilometer (km) = 1000 meters
1 km ≈ 0.62 mi
UNITS OF WEIGHT IN THE METRIC SYSTEM
UNITS OF WEIGHT IN THE U.S. SYSTEM
1 gram ≈ the weight of a paper clip
1 gram≈the weight of a raisin
1 pound (lb) = 16 ounces (oz)
1 gram = 1,000 milligrams (mg)
1 ton (T) = 2,000 pounds (lb)
SYSTEM TO SYSTEM CONVERSIONS FOR WEIGHT
1 ounce ≈ 28.3 grams
1 pound ≈ 0.45 kg
1 gram = 100 centigrams (cg)
1 kilogram (kg) = 1,000 grams
1 metric ton (t) = 1,000 kilograms
UNITS OF AREA IN THE U.S. SYSTEM
2
2
1 square foot (𝑓𝑡 ) = 144 square inches (𝑖𝑛 )
UNITS OF AREA IN THE METRIC SYSTEM
1 𝑐𝑚2 = 100 𝑚𝑚2
1 square yard (𝑦𝑑 ) = 9 square feet (𝑓𝑡 )
1 𝑑𝑚2 = 100 𝑐𝑚2
1 acre = 43,560 square feet (𝑓𝑡 2 )
1 𝑚2 = 100 𝑑𝑚2
2
2
SYSTEM TO SYSTEM CONVERSIONS FOR AREA
1 𝑖𝑛2 ≈ 6.45 𝑐𝑚2
1 𝑚2 ≈ 1.196 𝑦𝑑2
1 𝑎𝑟𝑒 (𝑎) = 100 𝑚2
1 square mile (𝑚𝑖 2 ) = 640 acres
1 ℎ𝑎 ≈ 2.47 𝑎𝑐𝑟𝑒𝑠
100 ℎ𝑒𝑐𝑡𝑎𝑟𝑒 (ℎ𝑎) = 100 𝑎
UNITS OF VOLUME IN THE METRIC SYSTEM
UNITS OF VOLUME IN THE U.S. SYSTEM
1 cubic foot (𝑓𝑡 3 ) = 1,728 cubic inches (𝑖𝑛3 )
1 cc = 1 𝑐𝑚
3
1 milliliter (mL) = 1 𝑐𝑚
SYSTEM TO SYSTEM CONVERSIONS FOR VOLUME
1 𝑖𝑛3 ≈ 16.39 𝑚𝐿
3
1 𝑙𝑖𝑡𝑒𝑟 ≈ 1.06 𝑞𝑡
1 cubic yard (𝑦𝑑 ) = 27 cubic feet (𝑓𝑡 )
1 liter (L) = 1,000 millimeters (mL)
1 𝑔𝑎𝑙𝑙𝑜𝑛 ≈ 3.79 𝑙𝑖𝑡𝑒𝑟𝑠
1 cord = 128 cubic feet (𝑓𝑡 3 )
1 hectoliter (hL) = 100 liters (mL)
1 𝑚3 ≈ 35.31 𝑓𝑡 3
1 kiloliter (kL) = 1,000 liters (L)
1 𝑞𝑢𝑎𝑟𝑡 ≈ 0.95 𝐿
UNITS OF FLUID VOLUME IN THE U.S. SYSTEM
UNITS OF TIME IN BOTH SYSTEMS
SYSTEM TO SYSTEM CONVERSIONS FOR TEMPERATURE
1 tablespoon = 3 teaspoons
1 millisecond = 1,000 microseconds
3
3
1 fl oz = 2 tablespoons
1 second =1,000 milliseconds
1 cup (c) = 8 fluid ounces (fl oz)
1 minute = 60 seconds
℉→℃
1 pint (pt) = 2 cups (c)
1 hour = 60 minutes
1 quart (qt) = 2 pints (pt)
1 day ≈ 24 hours
1 gallon (gal) = 4 quarts (qt)
1 month ≈ 30 days
1 gallon (gal) = 128 fluid ounces (fl oz)
1 year ≈ 365 days
1 banking year = 360 days
℃→℉
℃=
5
(𝐹 − 32)
9
℉=
9
𝐶 + 32
5
CONVERSION OF UNIT
Conversion Factor
Smaller- Large prefix = Larger value is on the denominator
Larger- smaller prefix = Larger value is on the numerator
ACCURACY AND PRECISION
Accuracy –Describes the closeness of measurement to the true value of the quantity that was
measured.
Precision- Refers to how closely two or more measurements of the same quantity agree with one
another.
Errors in Measurement


An error or fault can be describes as the disparity between the calculated worth and exact
worth
Different types of Measurement errors are significant in decreasing the statement effort and
offer more independence to the result
Systematic Error
-consistent and repeatable error associated with faulty equipment or flawed experiment design
-Divided into three:



Observational Error – faulty observation of instrument.
Environmental Error – Environment can always skew the results.
Instrumental Error – Instrument might be lacking of something, or the physical limitation of
the equipment.
Gross Error


Physical errors in analysis apparatus calculating the measurement of the experiment
Big Error
Random Error

Statistical fluctuation (in either direction) in the measured data due to the precision
limitations of the measurement device.
SCALAR AND VECTORS
Scalar:





Mass
Volume
Temperature
Speed
Distance
Vector:
 Acceleration
 Momentum
 Displacement
 Velocity
 Force
REPRESENTING VECTORS.
Vector Diagram




Cartesian Plane
Vector
Vector
Scale
Direction of a Vector


The direction of a vector is often expressed as an angle of rotation of the vector about its
“tail” from east, west, north, or south.
The direction of a vector is often expressed as a counter clockwise angle of rotation of the
vector about its tail. From due East.
Magnitude of a Vector.


The magnitude of a vector is a number for comparing one vector to another. In the
geometric interpretation of a vector the vector is represented by an arrow.
The arrow has two parts that define it. The two parts are its length which represents the
magnitude and its direction with respect to some set of coordinate axes.
Vector Addition
C> = A> + B>


Add the sum of two parallel vectors
The sum of two antiparallel vectors are the same
Graphical Method


Use measurement tools to find vectors
Involves the use of rulers and protractors without computation
Head to Tail Method Steps




Draw an arrow to represent the first vector
Draw an arrow to represent the second vector. Place the tail of the second vector at the
head of the first vector
Draw and arrow from the tail of the first vector to the head of the last vector
Measure the magnitude using a ruler.
Analytical Method

The analytical method of vector addition and subtraction involves using the Pythagorean
Theorem and trigonometric identities to determine the magnitude and direction of a resultant
vector.
ADDING VECTORS USING ANALYTICAL METHODS
Add the vector A to the vector B shown in Figure 8, using perpendicular components along the x–
and y-axes. The x– and y-axes are along the east–west and north–south directions, respectively.
Vector A represents the first leg of a walk in which a person walks 53.0 m in a direction 20.0º north
of east. Vector B represents the second leg, a displacement of 34.0 m in a direction 63.0º north of
east.
Vector A has magnitude 53.0 m and direction 20.0º north of the x-axis. Vector B has magnitude 34.0
m and direction 63.0º north of the x-axis. You can use analytical methods to determine the
magnitude and direction of R.
Strategy
The components of A and B along the x– and y-axes represent walking due east and due north to
get to the same ending point. Once found, they are combined to produce the resultant.
Following the method outlined above, we first find the components of A and B along the x– and yaxes. Note that 𝑨 = 𝟓𝟑. 𝟎 𝒎, 𝜽𝑨 = 𝟐𝟎. 𝟎°, 𝑩 = 𝟑𝟒. 𝟎 𝒎, 𝒂𝒏𝒅 𝜽𝑩 = 𝟔𝟑. 𝟎°. We find the xcomponents by using 𝑨𝒙 = 𝑨 𝐜𝐨𝐬 𝜽
, which gives
𝑨𝒙 = 𝑨𝐜𝐨𝐬𝜽𝑨 = (𝟓𝟑. 𝟎𝒎)(𝐜𝐨𝐬𝟐𝟎. 𝟎 ∘ ) = (𝟓𝟑. 𝟎𝒎)(𝟎. 𝟗𝟒𝟎) = 𝟒𝟗. 𝟖 𝐦
And
𝑩𝒙 = 𝑩 𝒄𝒐𝒔𝜽 𝑩 = (𝟑𝟒. 𝟎𝒎)(𝒄𝒐𝒔 𝟔𝟑. 𝟎 ∘ ) = (𝟑𝟒. 𝟎𝒎)(𝟎. 𝟒𝟓𝟒) = 𝟏𝟓. 𝟒 𝒎
Similarly, the y-components are found using
𝑨𝒚 = 𝑨 𝒔𝒊𝒏𝜽 𝑨
:
𝑨𝒚 = 𝑨𝐬𝐢𝐧𝜽𝑨 = (𝟓𝟑. 𝟎𝒎)(𝐜𝐨𝐬𝟐𝟎. 𝟎 ∘ ) = (𝟓𝟑. 𝟎𝒎)(𝟎. 𝟑𝟒𝟐) = 𝟏𝟖. 𝟏 𝐦
And
𝑩𝒚 = 𝑩𝐬𝐢𝐧𝜽𝑩 = (𝟑𝟒. 𝟎𝒎)(𝐜𝐨𝐬𝟔𝟑. 𝟎 ∘ ) = (𝟑𝟒. 𝟎𝒎)(𝟎. 𝟖𝟗𝟏) = 𝟑𝟎. 𝟑 𝐦
The x– and y-components of the resultant are thus
𝐑𝐱 = 𝐀𝐱 + 𝐁𝐱 = 𝟒𝟗. 𝟖 𝐦 + 𝟏𝟓. 𝟒 𝐦 = 𝟔𝟓. 𝟐 𝐦
And
𝐑𝐲 = 𝐀𝐲 + 𝐁𝐲 = 𝟏𝟖. 𝟏 𝐦 + 𝟑𝟎. 𝟑 𝐦 = 𝟒𝟖. 𝟒 𝐦
Now we can find the magnitude of the resultant by using the Pythagorean theorem:
𝑹 = √𝑹𝒙𝟐 + 𝑹𝒚𝟐 = √(𝟔𝟓. 𝟐)𝟐 + (𝟒𝟖. 𝟒)𝟐 𝒎
So that
𝐑 = 𝟖𝟏. 𝟐 𝐦.
Finally, we find the direction of the resultant:
𝛉 = 𝐭𝐚𝐧−𝟏 (𝐑𝐲/𝐑𝐱) = + 𝐭𝐚𝐧−𝟏 (𝟒𝟖. 𝟒/𝟔𝟓. 𝟐)
Thus,
𝛉 = 𝐭𝐚𝐧−𝟏 (𝟎. 𝟕𝟒𝟐) = 𝟑𝟔. 𝟔º
Discussion
This example illustrates the addition of vectors using perpendicular components. Vector subtraction
using perpendicular components is very similar—it is just the addition of a negative vector.
Subtraction of vectors is accomplished by the addition of a negative vector. A − B ≡ A + (–B). Thus,
the method for the subtraction of vectors using perpendicular components is identical to that for
addition. The components of –B are the negatives of the components of B. The x– and ycomponents of the resultant A−B = A are thus
𝑹𝒙 = 𝑨𝒙 + (−𝑩𝒙)
And
𝑹𝒚 = 𝑨𝒚 + (−𝑩𝒚)
And the rest of the method outlined above is identical to that for addition.
1-D KINEMATICS
Keyword in 1-D Kinematics


Linear Motion
Uniformly Accelerated Motion
We discussed in detail the graphical side of kinematics, but now let’s focus on the equations. The goal
of kinematics is to mathematically describe the trajectory of an object over time. To do that, we use
three main equations. However, I will include two more for the sake of convenience. Remember that
the acceleration is assumed to be constant!
Instantaneous Speed.




The speed at any instant When the speed of an object is constantly changing, the
instantaneous speed is the speed of an object at a particular moment (instant) in time.
Travels at constant speed (speedometer)
1𝑠 = 2 𝑚/𝑠, 2𝑠 = 4𝑚/𝑠 , 3𝑠 = 8𝑚/𝑠
𝑇𝑟𝑎𝑓𝑓𝑖𝑐 𝐿𝑖𝑔ℎ𝑡𝑠.
Average Speed.


The distance travelled divided by the total time elapsed in travelling that distance.
When an object gains speed at a constant rate, the average speed can be calculated on the
basis of two speeds. (vi and vf)
Deriving the Kinematics Equations:
We use 4 quantities to describe kinematics:
1.
2.
3.
4.
Position (x or y)
Velocity (m/s)
Acceleration (m/s2)
Time (s)
Note: Constant acceleration ⇒ 𝑎̅ = 𝑎 at all times.
a̅ = a =
Δv
v − v0
=
Δt
t − t0
let to = 0s ⇒ a =
v − v0
⇒ v = at + v0
t
By mathematical definition:
v̅ =
v + v0
← (mid point of a line)
2
⇒
and by a physics definition: 𝑣̅ =
𝛥𝑥
𝛥𝑡
v + v0
1
x − x0
1
= (v + v0 ) =
⇒ x = (v + v0 ) t + x0
2
2
t
2
Substitute 𝑣 = 𝑎𝑡 + 𝑣0 into the above result to yield:
x =
1
1
1
(at + v0 + v0 )t + x0 = (at + 2v0 )t + x0 ⇒ x = at 2 + v0 t + x0
2
2
2
Note that:
𝑣 = 𝑎𝑡 + v0 ⇒ 𝑡 =
𝑣 − v0
𝑎
Substitute the above equation into
x =
1
(v + v0 )t + x0
2
To yield
Δx =
1
v − v0
(v + v0 ) (
)
2
a
⇒ 2aΔx = (v + v0 )(v − v0 ) = v 2 + v02
⇒ 𝑣 2 = 2aΔx + v02
The use of the x direction in these derivations was completely arbitrary. Therefore, these equations
apply for the y-direction as well.
KEYNOTES:
Kinematic Equations
𝟏
𝒙 = 𝒙𝟎 + 𝐯𝟎 𝒕 + 𝒂𝒕𝟐
𝟐
𝒗 = 𝐯𝟎 + 𝒂𝒕
𝟐
𝒗 =
𝐯𝟎𝟐
+ 𝟐𝒂(𝒙 − 𝒙𝟎 ) )
𝒙 = 𝒙𝟎 + 𝒗𝒕 −
Meaning of Variables
𝒙𝟎
𝒙
∆𝒙
𝐯𝟎
𝒗
̅
𝒗
𝒂
𝒕
Initial position
Final position
Displacement
Initial Velocity
Final Velocity
Average Velocity
Acceleration
Time
𝟏 𝟐
𝒂𝒕
𝟐
𝜟𝒙 = 𝒗̅𝒕
𝜟𝒙 = (𝒙 − 𝒙𝟎 )
̅=
𝒗
∆𝒙
∆𝒕
̅=
𝒂
∆𝒗
∆𝒕
Tips On How to Use The Equations:
 First make a quick sketch of the situation
(optional but strongly recommended).
o Include your coordinate system!
 Label the quantities you know and
identify the quantity (or quantities) you
want to find.
 Select the equation(s) that contains the
variables you know and the variable you
want to find.
o You may need to use more than
one equation to find your
desired quantity.
 Solve for the desired quantity and make
sure the value seems physically
reasonable!
Sample Problem:
You are driving home from school steadily at 105 kph for 180. km. It then begins to rain, and you slow
to 75 kph. You arrive home after driving 5.0 h.
a) How far is your hometown from school?
b) What was your average speed?
Given:
𝑣1 = 105 𝑘𝑝ℎ
𝑑1 = 180 𝑘𝑚
𝑣2 = 75 𝑘𝑝ℎ
𝑡 = 5.0 ℎ
Solution:
𝑡1 =
180 𝑘𝑚
= 1.7 ℎ
𝑘𝑚
105
ℎ
𝑡2 = 5.0 ℎ − 1.7 ℎ = 3.3 ℎ
Unknown:
𝑑 =?
𝑣𝑎𝑣 = ?
𝑑 = 180 𝑘𝑚 + 75
Formula:
𝑣𝑎𝑣 =
Driving time at 105 kph:
𝑡1 =
𝑑1
𝑣1
𝑘𝑚
∙ 3.3 ℎ = 427.5 𝑘𝑚
ℎ
427.5 𝑘𝑚
= 85.5 𝑘𝑝ℎ
5.0 ℎ
Answers:
Total time is 5.0 h. therefore driving time at 75
kph is:
Distance of hometown from school:
𝑑 = 428 𝑘𝑚
𝑡2 = 𝑡 − 𝑡1
Therefore the travelled distance from school is:
𝑑 = 𝑑1 + 𝑑2 = 𝑑1 + 𝑣2 𝑡2
Average speed
𝑣𝑎𝑣 = 85.5 𝑘𝑝ℎ
Average speed
𝑣𝑎𝑣 =
𝑑
𝑡
2-D KINEMATICS: PROJECTILE MOTION
Motion in Two Dimensions
Projectile


An object upon which the only force acting is gravity.
Any object which once projected or dropped continues in motion by its own inertia and is
influenced only by the downward force of gravity.
Projectile Motion


The motion of an object thrown or projected into the air, subject to only the acceleration of
gravity.
Parabolic
o Symmetrical
o Zero Velocity at Peak
Elements of Projectile
Projectile Element
Time of Flight
Symbol
𝑡𝑓
Range
𝑥
Maximum Height
𝑦
Description
Duration while the
projectile is in its
trajectory
Maximum horizontal
distance travelled by a
projectile
Maximum vertical
displacement travelled
by a projectile in its
trajectory
Projectile Motion and Inertia

Since a projectile is in free fall after it is launched, the only force acting on it is gravity, which
influence the vertical motion of the projectile, causing a vertical acceleration.

The horizontal motion of the projectile is the result of inertia. There is no horizontal force.

The combination of inertia and gravity causes the parabolic trajectory that is characteristic of
projectiles
Projectile Motion: Horizontally Launched Projectile
Figure Analysis:

𝑣𝑦 = 0 since the initial velocity of the projectile BEFORE launch is zero

𝑣𝑥 = 20

𝑣𝑖𝑥 = 𝑣𝑓𝑥 = 𝑣𝑥 Once again, because horizontal velocity doesn’t change.


𝑎𝑥 = 0 Since the velocity didn’t change therefore no acceleration happened.
𝑎𝑦 = 𝑔 the downward acceleration is the same as gravitational acceleration.
𝑚
𝑠
𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 since the horizontal velocity of the projectile is CONSTANT
Acceleration
Velocity
Displacement
x
Uniform motion
𝑎𝑥 = 0
Y
Accelerated motion
𝑎𝑦 = 𝑔
𝑚
𝑎𝑦 = 9.81 2
𝑠
𝑣𝑦 = 𝑔𝑡
𝑣𝑖𝑥 = 𝑣0 = 𝑣𝑓𝑥 = 𝑣𝑥
𝑥 = 𝑣0 𝑡
1
𝑑 = v0 𝑡 + 𝑎𝑡 2
2
*Since 𝑎𝑥 = 0
1
𝑦 = 𝑣𝑖𝑦 + 𝑔𝑡 2
2
horizontal velocity
component:
sin 𝜃 =
𝑉𝑦
𝑉
cos 𝜃 =
𝑉𝑥
𝑉
vertical velocity component:
tan 𝜃 =
𝑉𝑦
𝑉𝑥
magnitude of resultant vector
𝑣𝑥 = 𝑣 cos 𝜃
𝑣𝑦 = 𝑣 sin 𝜃
|𝑣| = √𝑣2𝑥 + 𝑣2𝑦
direction of resultant vector
2
𝑉 =
Sample Problem:
𝑉𝑋2
+
𝑉𝑌2
𝜃 = tan−1
𝑣𝑦
𝑣𝑥
A plane traveling with a horizontal velocity of 100 m/s is 500 m above the ground. At some point the
pilot decided to drop some supplies to designated target below.
 How long is the drop in the air?
 How far away from point where it was launched will it land?
Given:
2𝑦
√𝑡 2 =
𝑦 = 500 𝑚
𝑔
𝑣 =0
𝑖𝑦
𝑣𝑥 = 100
𝑚
𝑠
𝟐𝒚
𝒕=√
𝒈
𝑚
𝑔 = 9.8 2
𝑠
Solution:
Unknown:
𝑡 =?
𝑥 =?
2𝑦
𝑡=√
𝑔
Formula:
Find t using:
1
𝑦 = 𝑣𝑖𝑦 + 𝑔𝑡 2
2
2(−500𝑚)
𝑡=√
9.8 𝑚/𝑠 2
Since 𝑣𝑖𝑦 is zero
𝑡 = 10.1 𝑠
1
𝑦 = 𝑣𝑖𝑦 + 𝑔𝑡 2
2
𝑥 = 𝑣0 𝑡
1
𝑦 = 𝑔𝑡 2
2
Derive formula and isolate t
1
2 (𝑦 = 𝑔𝑡 2 ) 2
2
2𝑦 𝑔𝑡 2
=
𝑔
𝑔
𝑚
𝑥 = (100 ) 10.1 𝑠
𝑠
𝑥 = 1010 𝑚
Answers:
𝑡 = 10.1 𝑠
𝑥 = 1010 𝑚
Projectile Motion: Angular
1
𝑦 = 𝑣𝑜𝑦 𝑡 + 𝑔𝑡 2
2
𝑥 = 𝑣𝑎𝑥 𝑡
𝑣𝑎𝑥 = 𝑣𝑜 cos 𝜃
𝑣𝑜𝑦 = 𝑣𝑜 sin 𝜃
Sample Problem:
Given:
Solution:
𝑚
𝑣 = 20.0
𝑠
𝜃 = 37.0°
𝑡=
𝑚
𝑠
𝑚
= 𝑣 sin 37.0 = 12.0
𝑠
𝑣𝑖𝑥 = 𝑣 cos 37.0 = 12.0
𝑣𝑖𝑦
𝑡=
𝑚
𝑔 = 9.8 2
𝑠
𝑣𝑓𝑦 = −12.0

𝑚
𝑠
Since it is parabolic inverse of initial
velocity in y axis
Unknown:
𝑡 =?
𝑥 =?
𝑦 =?

−𝑣𝑖𝑦
1
2𝑔
−12.0
𝑚
𝑠
−4.9 𝑚/𝑠 2
Negative since downward
𝑡 = 2.45 𝑠
𝑥 = 𝑣𝑖 𝑡
𝑥 = (16.0
𝑚
) (2.45𝑠)
𝑠
𝑥 = 39.2 𝑚
Formula:
Find t using:
1
𝑦 = 𝑣𝑖𝑦 𝑡 + 𝑔𝑡 2
2
𝑡𝑢𝑝 =
2.45 𝑠
2
𝑡𝑢𝑝 = 1.23 𝑠
Let y be zero

1
−𝑣𝑖𝑦 𝑡 = + 𝑔𝑡 2
2
Derive formula and isolate t
1 2
𝑔𝑡
−𝑣𝑖𝑦 𝑡
= +2
1
1
2 𝑔𝑡
2 𝑔𝑡
𝒕=
Time when object reached the peak
1
𝑦 = 𝑣𝑜𝑦 𝑡 + 𝑔𝑡 2
2
𝑚
1
𝑚
𝑦 = (12.0 ) (1.23 𝑠) + (−9.8 2 ) (1.23 𝑠)2
𝑠
2
𝑠
Answers:
−𝒗𝒊𝒚
𝟏
𝟐𝒈
Uniform Circular Motion
 An object that moves in a circle AT A CONSTANT SPEED.
Equations for Uniform Circular Motions
𝑣2 + 𝑣1
𝑡
∆𝑣 change in velocity
𝑡 short time interval
𝑎=


𝑥 = 39.2 𝑚
𝑦 = 7.35 𝑚
𝑡 = 2.45 𝑠
𝑎𝑅 =
𝑣
𝑣
𝑟
or 𝑎𝑅 =
𝑣
∆𝑙
𝑟
∆𝑣 𝑣 ∆𝑙
𝑎𝑅 =
=
∆𝑡 𝑟 ∆𝑡
𝑣2
𝑟
∆𝑣 =

Linear speed
𝐶𝐵 ⊥ 𝑣2
Since perpendicular:
𝜃 𝑓𝑜𝑟 𝐶𝐴 𝑎𝑛𝑑 𝐶𝐵
= 𝜃 𝑓𝑜𝑟 𝑣1 𝑎𝑛𝑑 𝑣2
𝑎 = 𝑎𝑅 ← 𝑅𝑎𝑑𝑖𝑎𝑙 𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛



Similar to ∠𝐴𝐵𝐶
𝜃 = 𝑣𝑒𝑟𝑦 𝑠𝑚𝑎𝑙𝑙 𝑣𝑎𝑙𝑢𝑒
∆𝑣
∆𝑙
𝑣 = 𝑣1 = 𝑣2 = 𝑣 ≈ 𝑟

An object moving in a circle of radius r at constant speed v has an acceleration whose direction
is toward the center of the circle and whose magnitude is;
𝑎𝑟 =



Direction
Radius of Circle
It is often described in terms of the frequency f, the number of revolutions per second. The
period T of an object revolving in a circle is the time required for one complete revolution.
𝑣=
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 2𝜋𝑟
=
→ 𝑚/𝑠 2
𝑡𝑖𝑚𝑒
𝑇
𝑇=


𝑣2
𝑟
One complete Revolution
Revolution per Second
1
→ 𝑚/𝑠
𝑓
Sample Problem
A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m. The
ball makes 2.00 revolutions in a second. What is its centripetal acceleration?
Given:
Solution:
𝑟 = 0.600𝑚
𝑓 = 2.00 𝑟𝑒𝑣𝑠/𝑠
𝑇=
1
= 0.500 𝑠
2.00
Unknown:
𝑇 =?
𝑣 =?
𝑎 =?
𝑣=
𝑎𝑟 =
Formula:
1
𝑓
2𝜋𝑟
𝑣=
𝑇
𝑇=
𝑎𝑟 =
𝑣2
𝑟
2𝜋 (0.600 𝑚)
= 7.54 𝑚/𝑠
0.500 𝑠
(7.54 𝑚/𝑠 )2
0.600𝑚
= 9.54 𝑚/𝑠 2
Answers:
𝑇 = 0.500 𝑠
𝑣 = 7.54 𝑚/𝑠
𝑎 = 94.8 𝑚/𝑠2
Relative Velocity





When two observers measure the velocity of a moving body, they get different results if one
observer is moving relative to the other.
When two observers measure the velocity of a moving body, they get different results if one
observer is moving relative to the other.
Relative Velocity
The velocity seen by a particular observer.
When two observers measure the velocity of a moving body, they get different results if one
observer is moving relative to the other.
KEYNOTE:

Relative velocity ALWAYS DEPEND ON THE OBSERVER
Let's use the symbol A for the cyclist's frame of
reference (at rest with respect to the ground) and
the symbol B for the frame of reference of the
moving train.
Equation for Relative Velocity
𝑋𝑃/𝐴 = 𝑋𝑃/𝐵 + 𝑋𝐵/𝐴
𝑣𝑃/𝐴−𝑋 = 𝑣𝑃/𝐵−𝑋 + 𝑣𝐵/𝐴−𝑋


X is Velocity
Relative Velocity along a line
Given:
P (Passenger) = 1.00 m/s
B (Train) = 3.00 m/s
Unknown:
𝑣𝑃/𝐴−𝑋 = ?
Formula:
𝑣𝑃/𝐴−𝑋 = 𝑣𝑃/𝐵−𝑋 + 𝑣𝐵/𝐴−𝑋
Solution:
𝑣𝑃/𝐵−𝑋 = 1.00 𝑚/𝑠
𝑣𝐵/𝐴−𝑋 = 3.00 𝑚/𝑠
𝑣𝑃/𝐴−𝑋 = 1.00 𝑚/𝑠 + 3.00 𝑚/𝑠
𝑣𝑃/𝐴−𝑋 = 4.00 𝑚/𝑠
Answer:
𝑣𝑃/𝐴−𝑋 = 4.00 𝑚/𝑠
Note: Different directions implies inverse signs
Equation for Relative Velocity (Field)
𝑟⃗𝑃/𝐴 = 𝑟⃗𝑃/𝐵 + 𝑟⃗𝐵/𝐴
𝑣⃗𝑃/𝐴−𝑋 = 𝑣⃗𝑃/𝐵−𝑋 + 𝑣⃗𝐵/𝐴−𝑋

Velocity relative to space
Since a RIGHT triangle has been made within the
space / field we must use the Pythagorean
Theorem to determine the relative velocity
Sample Problem:
Given:
𝑣⃗𝐵/𝐴−𝑋 = 1.00 𝑚/𝑠
𝑣⃗𝑃/𝐵−𝑋 = 3.00 𝑚/𝑠
Unknown:
𝑣⃗𝑃/𝐴−𝑋 = ?
Formula:
𝑣𝑃/𝐴−𝑋 = √𝑣𝑃/𝐵−𝑋 2 + 𝑣𝐵/𝐴−𝑋 2
 Squared because it is in a field or space
Solution:
𝑣𝑃/𝐴−𝑋 = √𝑣𝑃/𝐵−𝑋 2 + 𝑣𝐵/𝐴−𝑋 2
𝑣𝑃/𝐴−𝑋 = √(3.00 𝑚/𝑠)2 + (1.00 𝑚/𝑠)2
𝑣𝑃/𝐴−𝑋 = 3.2 𝑚/𝑠
Answer:
𝑣𝑃/𝐴−𝑋 = 3.2 𝑚/𝑠
DYNAMICS
Forces and Motion
Forces
Basic Concepts and Free-Body Diagrams
Force
- Vector quantity
- Push or pull of an object
- interaction between two objects or between an object and its environment
- measured with a spring balance



Force is measured in terms of Newton with the symbol N
In the smaller scale. it is expressed in dynes
100,000 dynes = 1 N
Types of Forces:

Contact Force
- Ex: An Ice Skater on a Turn
- The ice below the skater supplies both an upward component and an inward
component of force on the skater.
 Force that arises because of the physical contact between the object from
which the force originates and the object to which the force acts.
a) Normal force 𝑛
⃗⃗⃗⃗ : When an object rests or pushes on a surface, the surface exerts a push on
it that is directed perpendicular to the surface.
𝐹𝑛𝑜𝑟𝑚
b) Friction force ⃗⃗⃗⃗
𝑓 : In addition to the normal force, a surface may exert a frictional force on an
object, directed parallel to the surface.
𝐹𝑓𝑟𝑖𝑐𝑡
c) Tension force ⃗⃗⃗⃗⃗
𝑇 : A pulling force exerted on an object by a rope, cord, etc.
𝐹𝑡𝑒𝑛𝑠
d) Weight 𝑤
⃗⃗⃗ : The pull of gravity on an object is a long-range force (a force that acts over a
distance).
𝐹𝑔𝑟𝑎𝑣
Forces involved in an object:
Free-body Diagram
A diagram used to show relative magnitude and direction of all forces acting upon an object
in a given situation.
𝐹𝑛𝑜𝑟𝑚
𝐹𝑓𝑟𝑖𝑐𝑡
𝐹𝑎𝑝𝑝
applied force
𝐹𝑔𝑟𝑎𝑣
Drawing a Free-Body Diagram:
Examples:
A college student rests a backpack upon his shoulder. The pack is suspended motionless by one strap
from one shoulder.
𝐹𝑡𝑒𝑛𝑠
𝐹𝑔𝑟𝑎𝑣
Examples:
A rightward force is applied to a book in order to move it across a desk at constant velocity. Consider
frictional forces. Neglect air resistance.
𝐹𝑛𝑜𝑟𝑚
𝐹𝑓𝑟𝑖𝑐𝑡
𝐹𝑎𝑝𝑝
𝐹𝑔𝑟𝑎𝑣
The concept of force and net force:
a push or a pull exerted on some objects (force do not always cause motion)
Operationally defined based on observed effects:
1. Net force causes an object at rest to start moving.
2. Net force causes a moving object to stop.
3. Force causes a moving object to change its direction.
ALL of them CHANGE the STATE of MOTION of an OBJECT
Net Force


a net force is a physical quantity that can change an object's state of motion.
Net force is the sum or the resultant of all forces acting on an object or a system.
𝐹𝑛𝑒𝑡 = F₁ + F2 ₁ + F3 .... F7
Laws of Motion
Basic Concepts and Free-Body Diagrams
Sir Isaac Newton



English Mathematician and Physicist
formulated the 3 laws of motion and universal laws of gravitation
explain the motion of falling objects on Earth’s surface, the concept of weight, and the
motions of the Moon, the planets and the comets.
Newton's 1st Law of Motion
Law of Inertia
"Every object continues to remain at rest or uniform motion in a straight line unless a force acts on it
to change its state."
The Law of Inertia
Forces are Balanced
Objects at Rest
Objects at Motion
(v = 0m/s)
(v≠0m/s)
a = 0 m/s
a = 0 m/s
Stay at Rest
Stay in Motion
Inertial Reference Frame
"a frame of reference that is neither moving nor moving at a constant pace."
Newton's 2nd Law of Motion
Law of Acceleration
"The acceleration of a body is directly proportional and is in the same direction as the net force
applied to it and is inversely proportional to the mass of the body."
𝐹 = 𝑚𝑎
(𝑘𝑔)(𝑚/𝑠^2)
𝐹 = 𝑘𝑔 • 𝑚/𝑠^2 𝑜𝑟 𝑁
𝑎 = 𝐹/𝑚
Forces are Unbalanced
There is an acceleration
The acceleration
The acceleration
depends directly
depends inversely
upon the "net force"
upon the object's mass.
Sample Problems:
Estimate the net force needed to accelerate:
A 1000-kg car at ½ g;
A 200-g apple at the same rate.
a.)
b.)
Given:
𝑚 = 1000 𝑘𝑔
Given:
1
𝑎 = 𝑔 = 4.9𝑚/𝑠 2
2
𝑎 = 4.9𝑚/𝑠 2
Unknown:
𝐹 =?
Unknown:
𝐹 =?
Formula:
𝐹 = 𝑚𝑎
Formula:
𝐹 = 𝑚𝑎
Solution:
Solution:
𝐹 = (1000 𝑘𝑔)(4.9𝑚/𝑠 2 )
𝐹 = (0.2 𝑘𝑔)(4.9𝑚/𝑠 2 )
Answer:
𝐹 = 4900 𝑁
Answer:
𝐹 = 0.98 𝑁
𝑚 = 200𝑔 = 0.2 𝑘𝑔
What average net force is required to bring a 1500-kg car to rest from a speed of 100 kph within a
distance of 55 m?
Given:
𝑣𝑓 = 0 𝑚/𝑠
𝑣𝑖 = 100 𝑘𝑝ℎ
𝑑 = 55𝑚
𝑚 = 1500 𝑘𝑔
Unknown:
𝑎 =?
𝐹⃗ =?
Formula:
Find a using Eq. 6
𝑑=
𝑣𝑓 2 − 𝑣𝑖 2
2𝑎
𝑎=
𝑣𝑓 2 − 𝑣𝑖 2
2𝑑
𝐹 = 𝑚𝑎
27.8 𝑚/𝑠
Solutions:
0 − (27.8𝑚/𝑠)2
𝑎=
2(55𝑚)
𝐹 = (1500𝑘𝑔)(−7.0 𝑚/𝑠 2 )
Answers:
𝑎 = −7.0 𝑚/𝑠 2
𝐹 = −10,500 𝑁 𝑜𝑟 − 1.1 × 104
Newton's 3rd Law of Motion
Law of Interaction/ Action and Reaction
"For every action, there is an equal and opposite reaction."



Action Force
Reaction Force
Action-Reaction pair
Example:
Horizontal force exerted on the ground by person's foot. & Horizontal force exerted on the person's
foot by the ground.
𝐹⃗𝐺𝑃 = −𝐹⃗𝑃𝐺
Weight

Not to be confused with mass
Mass

Property of an object itself: quantity of matter
Weight


A force, the pull of gravity acting on an object
Gravitational Force
Applications of Laws of Motion:
FIRST LAW OF MOTION
Designing safety devices:



headrests
seatbelts
automatic inflating bags
SECOND LAW OF MOTION

structural design of race cars mass is reduced
THIRD LAW OF MOTION

operation of rocket engines; action force is provided by the burned fuel, downward force or
thrust produces an equal but opposite upward force
Notes and Additional Info by: James Luke A. Añonuevo
Special Thanks to: Willian Joan Garcia