physics reviewer
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Physics ReviewerTRANSCRIPT
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LCB 2015 Page 1 of 14
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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LCB 2015 Page 2 of 14
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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LCB 2015 Page 3 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]. 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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LCB 2015 Page 4 of 14
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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LCB 2015 Page 5 of 14
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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LCB 2015 Page 6 of 14
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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LCB 2015 Page 7 of 14
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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LCB 2015 Page 8 of 14
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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LCB 2015 Page 9 of 14
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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LCB 2015 Page 10 of 14
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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LCB 2015 Page 11 of 14
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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LCB 2015 Page 12 of 14
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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LCB 2015 Page 13 of 14
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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LCB 2015 Page 14 of 14
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