Physics Reviewer

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. Newton’s First Law of Motion 4.3. Mass 4.4. Newton’s Second Law of Motion 4.5. Newton’s 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. Newton’s Law of Universal Gravitation 5.4. Satellites and Weightlessness 5.5. Kepler’s Law and Newton’s 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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Measurement

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 LCB 2015

Quantity

Unit

Length Time Mass Electric Current Temperature Amount of substance Luminous intensity

meter second kilogram

Unit Abbreviation m s kg

ampere

A

kelvin

K

mole

mol

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 μ 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 x

1 in 2.54 cm

x

1 ft 12 in

= 0.886 ft

Note: Conversion factors are constants. When checking for significant figures, always refer to the given. Page 2 of 14

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].

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.

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 person’s speed, relative to the ground, is 85 km/h.

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).

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

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 ( )=

Δx Δt

=

displacement time elapsed

Applying Example ☼ (t = 80s):

Or can be expressed by specifying directions (N, S, E, W, NS, SW etc.)

Average speed =

100 m 80 s

Average velocity ( ) = Origin = (0,0) For one-dimensional motion, the x-axis (horizontal axis) serves as the frame of reference.

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= 1.25 𝑚⁄𝑠

70m−30m 80s

=

40m 80s

= 0.5 𝑚⁄𝑠 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.

Graph B:

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 = 

Δv Δt

velocity

change in velocity

=

time

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.

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

Graph C:

acceleration x vs t graph’s slope is average velocity since Average velocity ( )=

Δx Δt

v vs t graph = slope is average acceleration Average acceleration =

Δv Δt

time

Simple Example:

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

Graph A:

distance

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

time Interpretation: Object in Constant forward motion

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Motion at Constant Acceleration

(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

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 + ½at2

(3) v2 = v02 + 2a(x-x0) v0x + vx

(4) x - x0 =(

2

Eqtn # 1 2 3 4

)t

Quantities present t t t

v x x x

v v

a a a

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.

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 + ½at2 but replace x with y (bc vertical axis).

y = y0 + v0t + ½at2 y = 0 + (0) (5s) + ½ (-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 𝒎⁄𝒔 𝟐 or in 32 𝒇𝒕⁄𝒔 𝟐

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The Cliché 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.

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 + ½at2 0 = 0 + (15.0 m/s)t + ½(–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

Up = Positive Y Down = Negative Y g = a = -9.8 m/s 2 a = -9.8 m/s 2 @ Point A: t0 = 0

(2) y = y0 + v0t + ½at2 y = 0 m + (15.0m/s)(1.53s) + (½)(-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=

y0 = 0 m V0 = 15.0 m/s @ Point B: t=?

yB = ? = max height VB = 0

@Point C:

t=

v− v0 a

0 − 15.0 m/s

−9.8 m/s^2 t = 1.53s (2) y = y0 + v0t + ½at2 y = 0 m + (15.0 m/s) (1.53 s) + (½)(-9.8 m/s2) (1.53 s)2 y = 11.5m

tC = ?

yC = 0 m VC = -15.0 m/s LCB 2015

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2-D Kinematics

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

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.

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

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 person’s net or resultant displacement is 14 km East of the origin.

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:

But if he walks 8 km east on the first day, and 6 km west (in the reverse direction) on the

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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

Multiplication by a Scalar Multiplication of a vector by a positive scalar c changes the magnitude of the vector by a factor c but doesn’t alter the direction. If c is a negative scalar, the magnitude of the product is changed by the factor. Example:

Analytical Method - Identify and use components

x

= | |cosϴ

y

= | |sinϴ

Subtraction of Vectors Vector addition with opposite direction 2 1= 2 + (1) Example:

y

=

x

+

ϴ

y

x Pythagorean: √c 2 = √a2 + b 2 | | = √| x|2 + | y|2

| |

y|

= | |sinϴ | = | |cosϴ x ϴ = tan-1 Ay

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

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

Vectors in three-dimension

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

√Ax2 + Ay2 + Az2

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

ϴ = tan-1 Ay

Ax Az2

Φ = cos-1

√Ax2 + Ay2 + Az2 | | = | |cosϴ sinΦ | | = | |sinϴ sinΦ | | = | |cosΦ

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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 sin2ϴ0 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:

Horizontal Motion ax = 0, Vx = constant

Vertical Motion ay = -g = constant

V0x = |Vo| cos ϴ Vx = V0x X = X0 + V0xt

V0y = |V0| sin ϴ Vy = V0y - gt Y = Y0 + V0yt – ½gt2 V2y = V20y – 2g (y - y0)

Max height (y) of projectile motion:

V02sin2 ϴ 2g Max range (x) of projectile motion:

ϴ0 = 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 motion—when it reaches the ground.

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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

Newton’s 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.”

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)

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

Solution: (a) Use Newton’s 2nd Law

Newton’s 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 object’s mass. The direction of the acceleration is in the direction of the net force acting on the object.”

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

∑ F = ma

Actual weight = mg = (65 kg)(9.8 m/s2) = 640 N

Mathematically:

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

Newton’s 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.

Force exerted by scale = 0.80m = (0.8)(65 kg) = 52 kg (b) constant speed of 2.0 m/s means no acceleration. Using Newton’s 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.

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

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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.

Free-body diagram of the crate:

fk = 𝜇 kn fk = magnitude of kinetic friction force (N) 𝜇 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

We need the magnitude of the tension force T. Using Newton’s 2nd Law, we get the Force per component.

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

Substitute the value of n to solve for T

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 𝜇 k = 0.40.

Illustration:

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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.

Newton’s 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 Kepler’s Law and Newton’s Synthesis Kepler’s Laws of Planetary Motion  Kepler’s first law: The path of each planet around the Sun is an ellipse with the Sun at one focus

Equations:

v=

distance time

ar =

v2 r

=

T=

=

2πr T

4π2 r T2

1 f

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



Kepler’s 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



Kepler’s 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

Dynamics of Uniform Circular Motion Applying Newton’s 2nd Law, uniform circular motion’s net force must be directed toward the center of the circle

∑ F = ma = m

Derivations & further readings can be found in: Giancoli (2014) pages 112-115 LCB 2015

v2 r

Derivations & further readings can be found in: Giancoli (2014) pages 122-129 Page 14 of 14