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Physics 51: Introduction to Classical Mechanics, Fluids and Thermodynamics 1st Exam Reviewer FS AY 2015-2016

Course Outline (1st Exam Scope): 1. Introduction

1.1. Science and Creativity 1.2. Physics and Its Relation to Other Fields 1.3. Models, Theories and Laws 1.4. Measurement and Uncertainty 1.5. Unit, Standards and the SI System 1.6. Converting Units 1.7. Order of Magnitude: Rapid Estimation 1.8. Mathematics in Physics

2. Describing Motion: Kinematics in One Dimension 2.1. Reference Frames and Displacement 2.2. Average Velocity and Scalars 2.3. Instantaneous Velocity 2.4. Acceleration 2.5. Motion at a Constant Acceleration 2.6. Free Fall

3. Kinematic in Two Dimension: Vectors 3.1. Vectors and Scalars 3.2. Graphical Addition and Refraction of Vectors 3.3. Multiplication of a Vector with A Scalar 3.4. Adding Vectors by Components 3.5. Projectile Motion 3.6. Relative Velocities

4. Motion and Force: Dynamics 4.1. Force 4.2. Newtons First Law of Motion 4.3. Mass 4.4. Newtons Second Law of Motion 4.5. Newtons Third Law of Motion 4.6. Weight and Normal Force 4.7. Friction 4.8. Free Body Diagrams

5. Circular Motion: Gravitation 5.1. Kinematics of Uniform Circular Motion 5.2. Dynamics of Uniform Circular Motion 5.3. Newtons Law of Universal Gravitation 5.4. Satellites and Weightlessness

5.5. Keplers Law and Newtons Synthesis

Reminders: Count proper significant figures Always indicate units whenever necessary Scientific Calculator (mode in DEG not RAD)

Class Standing:

75% Exams (we only have 3 DepExs ) 25% Others (Problem Sets, HW,

Recitation, Reports, Attendance)

Finals Exemption: 1. No missed exam 2. No exam < 50% 3. Class Standing of 60% (~2.50)

or better

References:

Giancoli, D. C. (2014). Physics Principles With Applications (7th ed.). San Francisco, CA: Pearson Prentice Hall.

Giancoli, D. C., Davis, B., & Hendrickson, J. E. (2014). Physics Principles With Applications Instructor Solutions Manual (7th ed.). San Francisco, CA: Pearson Prentice Hall.

Young, H. D., & Freedman, R. (2012). University

Physics with Modern Physics (13th ed.). San

Francisco, CA: Addison-Wesley.

Personal Note: For questions/ clarifications/ concerns,

magsabi lang po sa Acads Comm Members.

1st time ko pong gumawa ng reviewer so comments and suggestions are very much appreciated. G lang po

And check niyo po yung Giancoli and Young kasi the

same po yung order ng contents sa course outline natin.

GLHF and God Bless sa aral time! (FYI ^ Good Luck, Have Fun hahaha)

\(^_^)/

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Introduction

Physics is the most basic of the sciences. It deals with the behavior and structure of matter (Giancoli, 2014). Branches:

1. Classical Physics include:

motion

fluids

heat

sound

light

electricity

magnetism 2. Modern Physics include:

relativity

atomic structure

quantum theory

condensed matter

nuclear physics

elementary particles

cosmology and astrophysics Accuracy VS Precision

Accuracy - how close a measurement is to the true value (~ bullseye)

Precision - repeatability of the measurement using a given instrument

Scalar VS Vector

Scalar quantity w/ magnitude only

Vector qty w/ magnitude and direction

Significant Figures (Applies only to final answer)

Multiplication or Division: check given with the fewest number of significant figures. Example: 1.60 x 2.296 = 3.67 (1.32578 x107) x (4.11 x 10-3) = 5.45 x 104

Addition of Subtraction: check given with fewest number of decimal places. Example: 27.153 + 138.2 - 11.74 = 153.6 12 + 9.8 + 76.00 = 98

Further reading: Giancoli (2014) pages 6-8 or Young and Freedman (2012) pages 8-9

Measurement

Quantity Unit Unit

Abbreviation

Length meter m Time second s Mass kilogram kg

Electric Current

ampere A

Temperature kelvin K Amount of substance

mole mol

Luminous intensity

candela cd

Metric Prefixes

Prefix Abbreviation Value

yotta Y 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 10-6

nano n 10-9

pico p 10-12

femto f 10-15

atto a 10-18

zepto z 10-21

yocto Y 10-24

Steps in Units Conversion: 1. Identify conversion factor (i.e. 1in = 2.54 cm) 2. Setup equation and multiply to conversion

factor Example: 27.0 cm __ ft 1 ft = 12 in

1 in = 2.54 cm

27.0 cm x1 in

2.54 cmx

1 ft

12 in = 0.886 ft

Note: Conversion factors are constants. When checking for significant figures,

always refer to the given.

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1-D Kinematics

Any measurement of position, distance, or speed must be made with respect to a reference frame, or frame of reference [Giancoli (2014) page 22]. Example:

A person walks towards the front of a train at 5km/h. The train is moving 80km/h with respect to the ground, so the walking persons speed, relative to the ground, is 85 km/h.

In Physics, a frame of reference is a set of coordinate axes (x and y).

Or can be expressed by specifying directions (N, S, E, W, NS, SW etc.) Origin = (0,0) For one-dimensional motion, the x-axis (horizontal axis) serves as the frame of reference.

Displacement change in position of the object; how far the object is from its starting point.

Example : A person walks 70 m east, then 30 m west.

The total distance traveled is 100 m (path is shown dashed in black); but the displacement is 40 m to the east (as shown by the blue arrow).

Average speed VS Average velocity

Average speed is the total distance traveled along its path divided by the time it takes to travel this distance.

Average velocity is the displacement divided by the elapsed time

Average speed = distance travelled

time elapsed

Average velocity ( )= x

t =

displacement

time elapsed

Applying Example (t = 80s):

Average speed = 100 m

80 s = 1.25

Average velocity ( ) = 70m30m

80s =

40m

80s

= 0.5 east

Note: Speed is scalar but velocity is a vector. Never forget to indicate the direction.

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Instantaneous velocity the average velocity over an infinitesimally short time interval.

Further reading: Giancoli (2014) page 25

or Young and Freedman (2012) pages 38-41 Acceleration and Deceleration

Acceleration - how rapidly the velocity of an object is changing.

Average acceleration = v

t =

change in velocity

time elapsed

Deceleration velocity and acceleration

point in opposite directions; does not mean that the acceleration is necessarily negative.

Graphical Analysis of Linear Motion

Given the graph of distance as a function of time, we can derive the graph for velocity and acceleration since they are just the slope of the former. x vs t graphs slope is average velocity since

Average velocity ( )= x

t

v vs t graph = slope is average acceleration

Average acceleration = v

t

Simple Example: Graph A:

distance time Interpretation: Object in Constant forward motion

Graph B: velocity time Interpretation: Since the object is constantly moving forward, it has a constant positive velocity (derived from slope of Graph A). Graph C: acceleration

time Interpretation: Because of the constant velocity, acceleration is zero (derived from slope of Graph B).

Further reading: Detailed discussion in Giancoli (2014) pages 39-40

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Motion at Constant Acceleration Derivations can be found in: Giancoli (2014) pages 28-29 or Young and Freedman (2012) pages 46-49

Final equations at constant a:

(1) v = v0 + at

(2) x = x0 + v0t + at2

(3) v2 = v02 + 2a(x-x0)

(4) x - x0 =(v0x + vx

2)t

Eqtn # Quantities present

1 t v a

2 t x a

3 x v a

4 t x v

Free Fall At a given location on the Earth and in the absence of air resistance, all objects fall with the same constant acceleration. Acceleration due to gravity (g) = 9.8 m s 2

Example: A coin is dropped from the Leaning Tower of Pisa and falls freely from rest. What are its position and velocity after 5.00 s?

Solution:

First, we need to identify and set-up the given. falls freely means falls with constant acceleration due to gravity allowing us to use the constant-acceleration equations.

(Setting the frame of reference) We take the origin O at the starting point and

the upward direction as positive. The initial coordinate and initial y-velocity are both zero. The y-acceleration is downward

We are asked for the position (y) at 5s, From the given we have:

a = -9.8m/s2 v0 = 0 m/s y0 = from origin = 0 t = 5.00 s

We use x = x0 + v0t + at2 but replace x with y (bc vertical axis).

y = y0 + v0t + at2

y = 0 + (0) (5s) + (-9. 8m/s2) (5s)2

y = -122.5 = The coin is 123 m below the origin

after 5.00 seconds.

Note: Check the units because te given g may be

expressed in 9.8 or in 32

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The Clich Example: A person throws a ball upward into the air with an initial velocity 15.0 m/s. Calculate how high it goes. Ignore air resistance.

Up = Positive Y Down = Negative Y

g = a = -9.8 m/s 2

a = -9.8 m/s 2

@ Point A:

t0 = 0

y0 = 0 m V0 = 15.0 m/s

@ Point B:

t = ?

yB = ? = max height VB = 0

@Point C:

tC = ?

yC = 0 m VC = -15.0 m/s

Recall the contant-acceleration equations.

There are 3 approaches to solve this problem. 1. Use Eqtn (3) [replace x to y]

(3) v2 = v02 + 2a(y-y0)

(y-y0) = v^2 v0^2

2a

(y - 0) = (0)2 - (15.0 m/s)2 2(-9.8m/s2) y = 11.5m

2. Use Eqtns (2) (to find t) [replace x to y] then (2) (to find y)

(2) y = y0 + v0t + at2

0 = 0 + (15.0 m/s)t + (9.80 m/s2)t2

Factor out t and solve. t = 0 and t = 3.06 s t = 0 is Point A t = 3.06 is Point C Point B = 3.06s / 2 = 1.53s

(2) y = y0 + v0t + at2

y = 0 m + (15.0m/s)(1.53s)

+ ()(-9.8m/s2) (1.53s)2

y = 11.5m

3. Use Eqtn (1) (to find t) then (2) (to find y) [replace x to y]:

(1) v = v0 + at

t = v v0

a

t = 0 15.0 m/s

9.8 m/s^2

t = 1.53s

(2) y = y0 + v0t + at2

y = 0 m + (15.0 m/s) (1.53 s)

+ ()(-9.8 m/s2) (1.53 s)2

y = 11.5m

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2-D Kinematics

Vector Quantities

A quantity that has direction as well as magnitude (Vector A denoted as: ) o Displacement o Velocity o Force o Momentum o Etc.

Scalar Quantities

A quantity that has only magnitude (Magnitude of A denoted as: | |) o Mass o Time o Temperature o Etc.

Vector Addition

Vectors in one-dimension o Simple arithmetic

Example: A person walks 8 km east one day, and 6 km east the next day, the persons net or resultant displacement is 14 km East of the origin. But if he walks 8 km east on the first day, and 6 km west (in the reverse direction) on the

second day, then the person will end up 2 km East of the origin.

Vectors in two-dimensions Example: A person walks 10.0 km east and then 5.0 km north.

o Graphical Method: - Accurate drawing using ruler

and protractor to measure length and angle but is not always sufficient.

1. Tail-to-tip method The resultant is drawn from the tail of the first vector to the tip of the last one added.

Example:

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2. Parallelogram method The two vectors are drawn starting from a common origin, and a parallelogram is constructed using these two vectors as adjacent sides

Example:

o Analytical Method - Identify and use components

x = | |cos

y = | |sin y = x + y

x

Pythagorean: c2 = a2 + b2

| | = | x|2 + | y|2 | y| = | |sin | x| = | |cos

= tan-1 Ay Ax

Further reading: Giancoli (2014) pp. 49-57 or Young and Freedman (2012) pp. 10-18

Multiplication by a Scalar Multiplication of a vector by a positive scalar c changes the magnitude of the vector by a factor c but doesnt alter the direction. If c is a negative scalar, the magnitude of the product is changed by the factor. Example:

Subtraction of Vectors

Vector addition with opposite direction

2 - 1 = 2 + (- 1)

Example:

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Vectors in three-dimension

= | |, , (if 2D: = 90 = 0) Final Equations:

| | = Ax2 + Ay2 + Az2

= tan-1 Ay Ax

= cos-1 Az2

Ax2 + Ay2 + Az2

| | = | |cos sin

| | = | |sin sin | | = | |cos

Unit Vectors Unit vectors describe directions in space. A unit vector has a magnitude of 1, with no units. The unit vectors are aligned with the x-, y-, and z-axes of a rectangular coordinate system.

Further reading: Young and Freedman (2012) pp. 19-24

Projectile Motion

A projectile is any body that is given an initial velocity and then follows a path determined entirely by the effects of gravitational acceleration (air resistance oftentimes neglected). The path followed by a projectile is called its trajectory. The x-distance travelled is called the horizontal range (R).

R = X = V02 sin20 g

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Since acceleration is due to gravity which is constant (ignoring effects of wind, air resistance, etc.) we can use the constant-acceleration equations and separate each coordinate plane.

Kinematic Equations for Projectile Motion:

Max height (y) of projectile motion:

V02sin2 2g

Max range (x) of projectile motion:

0 = 45 Rmax = V02/g

Derivations and further readings can be found in: Giancoli (2014) pp. 60-64

or Young and Freedman (2012) pp. 77-85

Example: A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0m from the base of the cliff where the cameras are? Ignore air resistance.

Problem Solving Steps:

1. and 2. Read, choose the object, and draw a diagram.

Object: motorcycle and driver, taken as a single unit.

3. Choose a coordinate system.

Origin: edge of cliff

y is positive: upwards

x is positive: to the right

4. Choose a time interval.

t = 0: when the motorcycle leaves the cliff top

t = end: before the motorcycle touches the ground

5. Examine x and y motions.

Horizontal (x):

ax = 0 vx is constant

Xat the ground = +90.0m Vx = unknown

Vertical (y):

ay = -g = -9.80 m/s2

Yat the ground = +50.0m

Vy = 0 6. List knowns and unknowns.

Aside from Vx, we also do not know the time (t) when the motorcycle reaches the ground.

7. Apply relevant equations.

The motorcycle maintains constant Vx as long as it is in

the air. The time it stays in the air is determined by the y

motionwhen it reaches the ground.

Horizontal Motion

ax = 0, Vx = constant

Vertical Motion

ay = -g = constant

V0x = |Vo| cos V0y = |V0| sin

Vx = V0x Vy = V0y - gt

X = X0 + V0xt Y = Y0 + V0yt gt2

V2y = V20y 2g (y - y0)

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So we first find the time using the y motion, and then use

this time value in the x equations. To find out how long it

takes the motorcycle to reach the ground below, we use

the modified Equation #2 with Y = 0 and Vy0 = 0.

Note: In the time interval of the projectile motion, the

only acceleration is gin the negative y direction. The

acceleration in the x direction is zero.

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Motion & Force: Dynamics

Force as any kind of a push or a pull on an object Newtons First Law of Motion (Law of Inertia)

Every object continues in its state of rest, or of uniform velocity in a straight line, as long as no net force acts on

it. Mass (kg) measure of the inertia of an object; The more mass an object has, the greater the force needed to give it a particular acceleration Newtons Second Law of Motion (Law of Acceleration) The acceleration of an object is directly proportional to the net force acting on it, and is inversely proportional

to the objects mass. The direction of the acceleration is in the direction of the net force acting on the object.

Mathematically:

F = ma Units of Force = Newton (N) = kg m / s2

Newtons Third Law of Motion (Law of Interaction)

Whenever one object exerts a force on a second object, the second object exerts an equal force in the

opposite direction on the first. Weight and Normal Force

Mass VS Weight Mass is a property of an object itself Weight is a force, the pull of gravity acting on an object.

Mathematically:

Mass = m Weight = mg (g = acceleration due to gravity)

Contact Force - force when two objects are in contact Normal Force (FN) - Contact force that is perpendicular

to the common surface of contact

Example: A 65-kg woman descends in an elevator that briefly accelerates at 0.20g* downward. She stands on a scale that reads in kg. (a) During this acceleration, what is her weight and what does the scale read? (b) What does the scale read when the elevator descends at a constant speed of 2.0 m/s ? *acceleration due to gravity; not grams (g) Solution: (a) Use Newtons 2nd Law

F = ma

mg - FN = m (0.20 g) FN = mg - m (0.20 g) = 0.80mg Actual weight = mg

= (65 kg)(9.8 m/s2) = 640 N

Force exerted by scale = 0.80m

= (0.8)(65 kg) = 52 kg

(b) constant speed of 2.0 m/s means no acceleration. Using Newtons 2nd Law mg - FN = 0 and mg = FN scale reading is 65 kg.

Further reading: Giancoli (2014) pp. 84-86 or Young and Freedman (2012) pp. 117-120

Free Body Diagrams

a diagram showing all the forces acting on each object involved; Indicates the direction of each force and its relationship to the other forces.

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Friction Another type of contact force. It is always

perpendicular to the normal force.

1. Kinetic Friction: Acts when a body slides over a surface. The magnitude of the kinetic friction force usually increases when the normal force increases.

fk = kn

fk = magnitude of kinetic friction force (N) k = coefficient of kinetic friction (No unit) n = normal force (Unit: N)

2. Static Friction: Acts when there is no relative motion. Friction force exerted with an equal magnitude and opposite direction

fs sn

fs = magnitude of static friction force (N) s = coefficient of static friction (No unit) n = normal force (Unit: N)

Example: You want to move a crate by pulling upward on the rope

at an angle of 30 above the horizontal. How hard must you pull to keep it moving with constant velocity?

Assume that k = 0.40.

Illustration:

Free-body diagram of the crate:

We need the magnitude of the tension force T. Using Newtons 2nd Law, we get the Force per component.

Substitute the value of n to solve for T

Further reading: Giancoli (2014) pp.93-98 or Young and Freedman (2012) pp. 146-151

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Circular Motion: Gravitation

Kinematics of Uniform Circular Motion 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. It has centripetal acceleration (center-pointing acceleration) or radial acceleration (since it is directed along the radius, toward the center of the

circle) (ar)

Frequency (f) = number of revolutions per second Period (T) = time required to complete one revolution. Equations:

v = distance

time=

2r

T

ar = v2

r =

42r

T2

T = 1

f

Derivations & further readings can be found in: Giancoli (2014) pages 108-112 or Young and Freedman (2012) pages 154-157

Dynamics of Uniform Circular Motion

Applying Newtons 2nd Law, uniform circular motions net force must be directed toward the center of the circle

F = ma = mv2

r

Derivations & further readings can be found in: Giancoli (2014) pages 112-115

Newtons Law of Universal Gravitation Every particle in the universe attracts every other

particle with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. This force acts along the

line joining the two particles.

FG = G m1m2 r2

G = 6.67 x 10-11 N m2 / kg

Keplers Law and Newtons Synthesis Keplers Laws of Planetary Motion

Keplers first law: The path of each planet around the Sun is an ellipse with the Sun at one focus

Keplers second law:Each planet moves so that an imaginary line drawn from the Sun to the planet sweeps out equal areas in equal periods of time

Keplers third law:The ratio of the squares of the periods Tof any two planets revolving around the Sun is equal to the ratio of the cubes of their mean distances from the Sun

Derivations & further readings can be found in: Giancoli (2014) pages 122-129