Physics Equations & Answers QuickStudy .pdf



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WORLD’S #1 ACADEMIC OUTLINE

BarCharts,Inc.®


Essential Tool for Physics Laws, Concepts, Variables & Equations, Including ✎ Sample Problems,

BASICS

A. Units for Physical Quantities
Base Units

Length

Symbol
l, x

Meter - m

T

Kelvin - K

Mass

m, M

Time

t

Temperature
Electric Current

Derived Units

Acceleration

I

Symbol
a

ω

radian/sec

C

Farad F (C/V)

θ, ϕ

Charge

Q, q, e

Capacitance
Density

Displacement

Electric Field
Electric Flux

Electromotive
Force (EMF)
Energy

Entropy
Force

Frequency
Heat

Magnetic Field
Magnetic Flux
Momentum
Potential
Power

Pressure

Resistance

Torque

Velocity

Volume

Wavelength
Work

m2/s

V/m

Φe

S

J/K

Gravitation Constant

G

f, ν
Q
B

Φm
p

V

P, P
P
R

τ

v

V

λ

W

Newton - N (kg

m/s2

Hertz - Hz (cycle/s)

= J/m)

Joule - J

Tesla (Wb/m2)

Weber Wb (kg m2/A s2)
kg m/s

Voltage V (J/C)
Watt - W (J/s)

Joule - J (N m)

Energy

Electron
Volt

Force

Dyne

Volume

Liter

8.314 J

k

mol–1

K–1

1.38×10–23 J K–1
6.67×10–11

µ0

m3

4¹×10–7 N/A2

kg–1 s–2

A=3i+4j–3k

|A| =

B= –2i+6j+5k

|B| =

=
=

Ay

2

Ax

65 = 8.06

1 +100 +4

1

Angstrom

1 Bar = 105 Pa

1 Å = 1×10–10m

sinθ = 5/( 5 × 10 ) = 5/ 50 = 1/ 2
< AB: θ = 45º = ¹/2 radians
c. The Right-Hand Rule gives the orientation of
vector e fig 4
B. Trigonometry
1. Basic relations for a triangle fig 5
y
sin θ = r
Values Of sin, cos and tan
θ rad [º] sin θ cos θ tan θ
x
cos θ = r
0 [0º]
0.00 1.00 0.00
sin
tan θ = x = cos
y

sin2 + cos2 = 1

¹/6 [30º] 0.50

0.866 0.577

¹/3 [60º] 0.866

0.50

¹/4 [45º] 0.707 0.707

1.00

1.732


2. Sinand Cos waves ¹/2 [90º] 1.00 0.00
¹ [180º] 0.00 –1.00 0.00
fig 6

3

e

4

5

sin and cos waves

A ×B

37°

sin

B

A

90° = ¹/2

1 L = 1 dm3

• If A and B are in x-y plane, A × B is along
the + z direction
C
• θ is the angle formed by AB: sinθ =
A B
Given: |A| = 5
|B| = 10 |C| = 5

4 +36 +25

B

A

CGS unit (g cm/s2 = erg/cm)
1 dyne = 10–5 N

A = 2i + j
B = i + 3j
i j k
A × B = 2 1 0 = (6 – 1) k = 5k
1 3 0

34 = 5.83

y

CGS unit (g cm2/s2)
1 erg = 10–7 J

✎ Sample: Vector Product:

|B|= 9 +25 = 34 = 5.831

1

180º = ¹ rad

b. Cross or Vector Product:
C = A × B = |A| |B|sinθ e
θ – Angle between A and B, vector e is
perpendicular to A and B
i j k
A × B = Ax Ay Az
Bx By Bz

9 +16 +9

25
cosθ = A :B =
= 0.796
A B 5.385 #5.83
-1
θ = cos (0.796 ) = 37º = 0.2¹ rad fig 3

Description

1 eV = 1.602×10–19 J

Pressure Bar
Length

8.85×10–12 F/m

ε0

|A|= 25 +4 = 29 = 5.385

Nm

meter - m

96,485 C/mol

= 105 = 10.25
Note: |A| + |B| ≥ |A + B|
3. Multiply A & B:
a. Dot or scalar product: A • B = | A| |B|cosθ =
( Ax Bx) + ( Ay By ) + (Az Bz )
Note: θ is the angle between A and B;
A • B = 0, if θ = ¹ /2 fig 2
✎ Sample: Scalar product:
A = 5i + 2j
B = 3i + 5j
A • B = 3 × 5 + 2 × 5 = 15 + 10 = 25

Ohm Ω (V/A)

m3

Erg

3×108 m s –1

A+B=i+10 j+2k |A+B| =

Pascal - Pa (N/m2)

m/s

Energy

c. Length of A = |A| = Ax2 +Ay2 +Az2
2. Addition of vectors A & B, add components:
A + B = (Ax+Bx) i + (Ay+By) j + (Az+Bz)k
✎ Sample Addition and Length Calculations:

E, U, K Joule J (kg m2 s–2)
F

6.022×1023 mol–1

A. Vector Algebra
1. Vector: Denotes directional character using (x, y, z)
components fig 1
a. Unit vectors: i along x, j along y, k along z
b. Vector A = Ax i + Ay j + Az k

Vm

Volt - V

R

Unit

º (degree)

kg

MATHEMATICAL CONCEPTS

meter - m

E

Molar Gas Constant

kg

1.602×10–19 C

c

Permittivity of Space

Coulomb C (A s)

Ε

s, d, h

F

Permeability of Space

radian

kg/m3

Faraday Constant

Helpful Hints

Angle

1.67×10–27

e

Boltzmann Constant

radian/s2

ρ

NA

Speed of light

Unit

Ang. Velocity
Angle

Avogadro Constant

mp

Elementary charge

Ampere - A (C/s)

kg

me

Common Pitfalls &

C. Conversion factors and alternative units

Unit

9.11×10–31

Mass of proton

Second - s

m/s2

Symbol

Mass of electron

Kilogram - kg

α
L

Base Units

Unit

Ang. Accel.

Ang. Momentum

B. Fundamental Physical Constants

!

x

θ

A

Right-Hand
Rule

cos

r

y
x

θ
sinθ

cosθ

6

MATHEMATICAL CONCEPTS (cont.)

C. Geometry
Circle: Area = ¹r 2; Circumference = 2¹r
Sphere: Volume = 4/3¹r 3; Area = 4¹r 2

PHYSICS & MEASUREMENT
7

r

Cylinder: Volume =
Triangle: Sum of angles = 180º fig 7
r
D. Coordinate Systems
1. One dimension (1-D): position = x fig 8
• The x position is described relative to an origin
h
2. Two dimensions (2-D) fig 9
r
x = r cosθ, y = r sinθ, r2 = x2 + y2
a. Calculate (r, θ) from (x, y):
y
r= x2 +y2 ; θ = sin-1 c r m
b. Calculate (x, y) from (r, θ), or x and y
8
components of a vector “r” with angle θ;
x = rcosθ; y = r sinθ
–x 0 +x
✎ Sample: Generate x and y vector v = dx/dt;
a = dv/dt
components, given: r = 5.0, θ = r (30º)
6
y Polar 9
x = r cos b r l = 5.0 × 0.866 = 4.33
6
r
h¹r 2

y = r sin b r l = 5 × 0.50 = 2.50
6
θ
Check your work: x 2 + y 2 = r 2
x
2.52 + 4.332 = 6.25 + 18.75 = 25.00
Cartesian: (x, y)
2
It checks, r = 25.00
Polar: (r, θ)
3. Three Dimensions (3-D)
10
a. Cartesian (x, y, z): The basic coordinate
system
Cylindrical
z
b. Cylindrical: (r, θ, z) fig 10
z
• Polar coordinates, with a z axis

•Calculate (r, θ) from (x, y);
y
calculate (x, y) from (r, θ)
•Same process as for 2-d polar;
θ
z: same as Cartesian
r
x
c. Spherical: (r, θ, ϕ)
x = rcosθ,
x = r sinϕ cosθ, y = r sinϕ sinθ,
y = rsinθ,
z = r cosϕ, r2 = x2 + y2 + z2 fig 11
r2 = x2 + y2
•Calculate (r, θ, ϕ) from (x, y, z)
11
•Calculate (x, y, z) from (r, θ, ϕ)
Spherical
Hint: Follow the strategy for 2-d
z
polar coordinates
r
ϕ
E. Use of Calculus in Physics
1. Methods from calculus are used in
y
physics definitions, and the derivations
of equations and laws
θ
Physical meanings of calculus expressions: x
dF] xg x = r sinϕ cosθ,
a. Derivative - slope of the curve:
dx
y = r sinϕ sinθ,
b. Integral - area under the curve: ∫F(x)dx z = r cosϕ,
r2 = x2 + y2 + z2
✎ Samples:
• Position: x or F(x)
Velocity: v(x) = dF] xg
dt
Common derivatives
and integrals
Acceleration: a = dv] xg
• Power and work: dt
F(x) dF] xg ∫F(x) dx
dx
P = dW
dt
constant 0
constant x
• Energy and force:
E = ∫ Fdx
1 x2
x
1
2. Other useful expressions:
2
d ] F :G g
=F dG +G dF
a.
1 x3
dx
dx
dx
2x
x2
3
d (F 'G) 1 dF F dG
=
b.
1
G dx G2 dx
dx
xn
n x n –1 n +1 x n +1
c. Partial derivative:
-1
1
2F (x, y, z) dF
ln x
= ,
x
x2
2x
dx
hold y & z constant
1
x ln x – x
x
d. Gradient Operator ∇ (Del) = ln x
∂/∂x + ∂/∂y + ∂/∂z
ex
ex
ex
e. Integration by parts:
∫udv = uv – ∫vdu
sin(x) cos(x) –cos(x)
f. Symbol for integration of
cos(x) –sin(x) sin(x)
closed surface or volume: #

A. Understand Your Data
1. Vector vs. scalar
a. Vector: Has magnitude and direction
b. Scalar: Magnitude only, no direction
2. Number and unit
a. Physical data, constants and equations have
numerical values and units
b. A correct answer must include the correct numerical
value PLUS the correct unit
3. Significant figures (sigfig)
a. The # of sigfigs reflects the accuracy of experimental
data; calculations must accommodate this
uncertainty
b. For multiplication: The # of sigfigs in the final
answer is limited by the entry with the fewest sigfigs
c. For addition: The # of decimal places in the final answer
is given by the entry with the fewest decimal places
d. Rules for “rounding sigfigs”
• If the last digit is >5, round up
• If the last digit is <5, round down
• If digit = 5, round up if preceding digit is odd
✎ Samples:
1.245 + 0.4 = 1.6 (1 decimal place)
1.345 × 2.4 = 3.2 (2 sigfigs)

• Electrostatic potential energy:
q1 q2
r
Uc = 1/(4¹ε0 ) r
q1 ... ... q2 14
constant: 1/4¹ε0; units are J m/C 2
r in m, q1 and q2 in Coulomb

Units of Uc = (J m/C2)C2/m = J fig 14
5. Using Conversion Factors
a. Purpose: Modify experimental data to
match the units of constants and equations
b. SI units: MKS (m-kg-s) and CGS (cm-g-s)
c. Common English units: Foot, pounds,
BTU, calories
d. Conversion factors are obtained from an
equality of two units
✎ Sample: 100 cm = 1 m
• This equality gives two conversion factors:
1m & 100cm
100cm
1m
•Use the 1st factor to convert “cm” to “m”
1m
= 0.54 m
✎ Sample: 54 cm × 100
cm

• Use the 2nd to convert “m” to “cm”
cm = 230 cm
✎ Sample: 2.3 m × 100
Units of basic variables
1m
B. Solve the Problem Strategically
time: second s
position: meter m
a. Two key issues:
1. Understand the physics principles
mass: kilogram kg
volume: m3
2. Have a correct mathematical strategy
b. Useful steps in problem solving:
Temperature: K
density: kg/m3
1. Prepare a rough sketch of the problem
velocity: m/s
2. Identify relevant physical variables,
acceleration: m/s2
physical concepts and constants
energy: Joule J = kg m2/s2 force: Newton N = kg m/s2
! Pitfall - do not simply search for the
“right” equation in your notes or text
! Pitfall: If the units are wrong, the answer is
3. Describe the physics using a
wrong!
mathematical
diagram,
with
appropriate symbols and a coordinate
Hint: Before doing the calculation:
system
* Check all constants and variable units
4. Obtain the relevant physical constants
* Take special care if you derive the equation
Do you have all the essential data?
4. Dimensional analysis
Hint: You may have extra
Verify that constants and variables in an equation result
information
in the correct overall unit
5. The hard part: Derive or obtain a
✎ Samples: The energy unit is Joules for kinetic,
mathematical expression for the
gravitational and Coulombic energy
problem; use dimensional analysis to
check the equation, constants and data
• Kinetic: K = 1 mv2
2
m
6. The easy part: Plug numbers into the
v 12
m in kg, v in m/s
Units of K = kg
equation and use the calculator to obtain
m2/s2 = J fig 12
the numerical answer
• Gravitational potential energy:
7. Check the final answer, using the
13
m
Ug = m g h
Constant: g = 9.8 m/s2
original statement of the problem, your
sketch and common sense; are the
m in kg, h in m
h
units & sign correct?
Units of U = kg m2/s2 = J fig 13
g

MECHANICS

A. Motion along a Straight Line
B. Motion in Two and Three Dimensions
1. Goal: Determine position, velocity, acceleration
1. Goal: Similar to “A,” with 2 or 3 dimensions
2. Key terms: Acceleration: a = dv/dt; velocity: v = dx/dt
2. Key concept: Select Cartesian, polar or
spherical coordinates, depending on the type of
3. Key Equations: x = vi t + 1 at 2
vf = vi + at
2
motion
x(t), v(t) for variable a fig 15
Sample: A projectile is launched at angle θ
with vri ; how do we set up the problem?
x
x
x
15
Step 1. Define x as horizontal and y vertical
Step 2. Determine initial vxi and vyi fig 16
a>0
a<0
a=0
vxi = vri cosθ
vyi = vri sinθ
t
t
t
16
vy
v
v
v
r
vi
a=0

v0

a>0
t

v0
2

t

θ
a<0
t

vx

Step 3. Identify ax - Gravitational force => ay = –g

MECHANICS (continued)

Step 4. Identify ay - No horizontal force => ax = 0
Step 5. Develop x- and y-equations of motion
x = vixt + 1 a xt 2 = vit
2
y = viy t + 1 ayt 2 = viy t – 1 gt 2
2
2
C. Newton’s Laws of Motion
1. Goal: Examine force and acceleration
2. Key concepts: Newton’s Laws:
Law #1. A body remains at rest or in motion unless
influenced by a force
Law #2. Forces acting on a body equal the mass
multiplied by the acceleration; force and
acceleration determine motion
Law #3. Every action is countered by an opposing
action
3. Key equations:
a. Law #2: F = m a or ΣF = m a

m

F

f

Hint: Forces are vectors!
b. Types of forces: Body - gravity: Fg = m g
•Surface - friction: = Ff = µ Fn
✎ Sample: Ft exerted on
Fn
object on a horizontal plane
Ff
Ff = µ Fn = µ Fg =µ m g
m
Net force = Ft – Ff fig 17
✎ Sample: Object on plane
inclined at angle θ;
examine Fg & Ff
18
Fn = Fg cosθ = m g cosθ
Ff = µ Fn = µ m g cosθ
Fn
Ft = m g sinθ fig 18
θ
e. Law #3:
Fg
F12 = –F21 or m1 a1 = – m2 a2 Ft
Ft
✎ Sample: Examine recoil of
θ
bullet fired from a rifle
Rifle recoil = a(bullet) × m (bullet)
D. Circular Motion fig 19
1. Goal: Examine body moving in a circular path;
use 2-d polar coordinates: (r, θ)

c. Power = Work/time: W = Power∆t or ∫P(t) dt
d. Wnet = K final – K initial ; K is converted to work
✎ Sample: Determine the work expended in lifting
a 50 kg box 10 m ; given: a = g = 9.8 m/s2
Equations: F = m g => W = m g d
Calculation: W = 50 kg × 9.8 m/s2 × 10 m = 4,900 J
F. Potential Energy & Energy Conservation
1. Goal: Use energy conservation to study the interplay
of potential and kinetic energy
2. Key Equations
a. Potential energy: Energy of position: U (r);
gravitation (U = mgh),
electrostatic (U a qq/r)
b. E = K + U Conservative system: No external force
✎ Sample: Examine K & U for a launched rocket
Initial: h = 0, therefore, U = m g h = 0 ❶
E = Ki = 1 m vi 2
2
Next, resolve into x and y components: Kxi & Kyi
Note: Kx is constant during
y
20
the flight
2
17
At max height: Ky = 0; U =
m g h = Kxi ❷
Ft
Final state: Rocket hits the
x
1
3
ground: U = 0, K = Ki ❸
fig 20
G. Collisions and Linear Momentum fig 21
1. Goal: Examine momentum of colliding bodies

r

Key variables:

θ
ω

m

rad

angle with
reference (x) axis

rad/s angular velocity

α rad/s2
s

distance from
rotation center

m

angular
acceleration

at

19

ac

s

θ
r

motion arc;
s = r θ (θ in rad)

Hint: For a full rotation, s = 2¹r = circumference
of a circle of radius r
2. Tangential acceleration & velocity: vt = r ω; a t = r α;
along path of motion arc
2
3. Centripetal acceleration: ac = vr ; directed towards
the center fig 19
✎ Sample: Determine vt at the Earth’s equator
Equation: vt = rω
Data: r = 6.378 × 106 m
ω = 2¹ rad/day;
1 day = 24 × 60 × 60 sec = 86,400 s
Convert ω to SI: ω = 2¹ rad/day × 1 day/86,400 s =
7.3 × 10-5 rad/s
Calculate vt:

vt = rω = 6.378 × 106 m × 7.3 × 10-5 rad/s vt = 465 m/s
E. Energy and Work
1. Goal: Examine the energy and work associated
with forces acting on an object
2. Key equations:
a. Kinetic energy: 1 mv 2; energy of motion
2
b. Work: Force acting over a distance
•For F(x): Work = ∫F(x) dx
•For a constant force: W = F d cosθ = F × r
•θ is the angle between the F and r
•W maximum for θ = 0 (note: sin(θ = 0) = 1)

Hint: For 2-D or 3-D, use
21
Cartesian components
m
1
2. Key Variables and Equations
a. Types of collisions:
• Elastic: Conserve energy
• Inelastic: Energy lost as heat or m2
deformation
b. Relative motion and frames of reference: A body
moves with velocity v in frame S; in frame S’, the
velocity is v’; if Vs’ is the velocity of frame S’
relative to S, then v = Vs’ + v’
c. Linear Momentum: p = m v
d. Conserve K & p for conservative system (no
external forces):
Σ 1 mvi 2 = Σ 1 mvf 2
Σmvi = Σmvf
2
2
Sample
1-d
problem:
Two
bodies collide, stick

together and move away from the collision site fig 22
Conservation of momentum:
m1 v1i + m2 v2i = (m1 + m2)vf
v2
v
22 m2 1
m1 before collision
m1 m2

vf

after collision

f. Impulse: I = F∆t or ∫F(t)dt
g. Momentum change: pfin = pinit + I
H. Rotation of a Rigid Object
1. Goal: Examine the rotation of a rigid body of a
defined shape and mass
2. Key variables and equations :
a. Center of mass: xcm , ycm , zcm
Rmi yi
Rmi xi
Rmi zi
xcm =
ycm =
zcm =
Rmi
Rmi
Rmi
✎ Sample: Calculate the center of mass for a 1 kg
& a 2 kg ball connected by a 1.00 m bar
ball 1: x1 = 0.00, m1 = 1 kg; m1 x1 = 0.00 kg m
ball 2: x2 = 1.00 m 2 = 2 kg; m2 x 2 = 2.00 kg m
Σm i = 1 kg + 2 kg = 3 kg
Σm i xi = m1 x1 + m2 x2 = 0.00 + 2.00 = 2.00 kg m
2.00kg m
Rmi xi
xcm =
=
= 0.66 m
3.00kg
Rmi
Hint: The center of mass is nearer the heavier
ball fig 23
0.66m

1 kg

23

0.33m

center
of mass
3

2 kg

b. Moment of inertia:
I = Σmi ri 2 , with ri about the center of mass along a
specific axis

Hint: I functions as the effective mass for
rotational energy and momentum
✎ Sample: I for bodies of mass m: fig 24

Twirling thin rod of length, L
I = 1 m L2
12
Rotating cylinder of radius, R
I = 1 m R2
2
Rotating sphere of radius, R
I = 2 m R2
5

L

24

R

R

✎ Sample: Determine the I for a spherical Earth,

assume uniform M ;
Data: M = 6 ×1024 kg, r = 6.4×106 m
I = 2 M r 2 = 2 × 6×1024 kg × (6.4×106 m)2
5
5
= 9.8 × 1037 kg m2
e. Rotational Energy = 1 I ω2
2
f. Torque: τ = Iω = r × F (ang. acceleration force)
I. Angular Momentum
1. Goal: Quantify the force, energy
25
and momentum of rotating objects τ
2. Key variables and equations
a. Angular momentum:
L = Iω = r × p = ∫r × vdm
r
b. Torque: τ = r × F = dL/dt;
note: vector cross product fig 25 F
J. Static Equilibrium and Elasticity
1. Case 1: Examine several forces acting on a body
• Guiding principles: Equilibrium is defined as:
Σforce = 0 & Σtorque = 0
The point of balance is the center of mass
Hint: Evaluate each
component; any net force
m1
m
moves the object, any net
x1
x2 2
torque rotates the object
26
✎ Sample: Beam balance
For equilibrium:
fig 26
F
A
m1 x1 1= m2 x2
2. Case 2. Examine deformation
27a
of a solid body
Key Equation: Stress l0= elastic ∆lmodulus × strain;
F1
A
modulus: stress/strain
(Hooke’s
Law)
Note: Force=Fforce/change
l is longitudinal
27a
• Linear (Tensile) Stress: Young’s Modulus Y
l0
∆l
F /A
Y= 1
Dl/l0
Note: Force Fl isAlongitudinal
F1
fig 27a
27a
l∆x
∆l
0
Note: Force Fl is longitudinal
Ft
• Shape Stress: Shear ModulusA S
27b
F /A
∆x
S= t
Dx/h
h
Ft
fig 27b
A
27b
Fixed
face
∆x
h Force Ft is tangential to face A
Note:
Ft
A
27b
Fixed face
Note: Force Ft is tangential to face A
h
• Volume Stress: Bulk Modulus B
V
F /A
B= n
Fixed face
D V/ V
Note: Force Ft is tangential to face A
fig 27c
27c
V
Fn
27c
VV – ∆V F
n
Note: Force Fn is normal to face A
27c
V – ∆V

MECHANICS (continued)

K. Universal Gravitation
1. Goal: Examine gravitational energy and force fig 28
2. Case 1: Bodies of mass M1 & M2 separated by r
3. Key equations:
GM M
a. Gravitational Energy: Ug = r1 2
b. Gravitational force: Fg =

WAVE MOTION

r
M1 ... ... M2 28
m

29
h

GM1 M2
r2

c. Acceleration due to gravity: g = G M(earth)/r 2
For objects on the Earth’s surface, g = 9.8 m/s2

✎ Sample: Verify “g” at the Earth’s surface

Equation: g = G M(earth)/r2
Given: M = 6 ×1024 k, r = 6.4×106 m
6.67 #10-11 m3 kg-1 #6 #1024 kg
Calculation: =
= 9.8 ms 2
]6.4 #106 mg2
4. Case 2: A body interacts with the Earth fig 29
5. Key Equation:
a. Gravitational potential energy: Ug = m g h; object on
the Earth’s surface, h = 0; Ug = 0

b. Weight = gravitational force; Fg = m g

30

k

Harmonic Wave Properties

λ (m)
T (sec)

Wavelength
Period
Frequency

m
HOOKE’S
LAW

l

A. Descriptive Variables
1. Types: Transverse, longitudinal, traveling, standing, harmonic
a. General form for transverse traveling wave: y = f (x – vt) (to the
right) or y = f (x + vt) (to the left)
b. General form of harmonic wave: y = Asin(kx – ωt) or y = Acos(kx – ωt)
c. Standing wave: Integral multiples of m fit the length of the
2
oscillating material
d2 y 1 d2 y
=
d. General wave equation:
dx2 v2 dt2
e. Superposition Principle: Overlapping waves interact => constructive
and destructive interference

31

✎ Sample: Calculate escape velocity, vesc , for an orbiting
m
rocket of mass m at altitude h
F1
32
Hint: K = Ug at point of escape; r = h + r (earth)
1 m v 2 = GmM ; therefore, v = 2GM
esc
r
r
A2
2 esc
A1
Note: vesc varies with altitude, but not rocket mass
L. Oscillatory Motion
F2
P
1. Goal: Study motion & energy of oscillating body
2. Simple harmonic motion (1-d)
P1
33
Air
a. Force: F = –k∆ x (Hooke’s Law)
Surface
b. Potential Energy: Uk = 1 k∆ x2
2
ρ
h
Liquid
k fig 30
c. Frequency = 1
P2
2r m
3. Simple Pendulum
Air 34
a. Period: T = 2¹ gl
Surface
b. Potential energy: Ug = m g h
V ρ Liquid
g
1
c. Frequency =
fig 31
2r l
Fb
4. For both cases:
a. Kinetic energy: K = 1 m v2
2
35
P1
b. Conservation of Energy: E = U + K
P2
M. Forces in Solids and Liquids
V1
V2
1. Goal 1: Examine properties of solids & liquids
mass
A
a. Density of a solid or liquid: ρ =
2
A1
volume
Flow
Through
a Hose
3
3
•Common unit: g/cm ; g/L; kg/m

• . Sample: A piece of metal, 1.5 cm × 2.5 cm × 4.0 cm, has a mass of 105.0 g;
determine ρ
Equation: ρ = m
V
Data: m = 105.0 g, V = 1.5 × 2.5 × 4.0 cm3 = 15 cm3
Calculate: ρ = 105.0/15.0 g/cm3 = 7.0 g/cm3
b. Pressure exerted by a fluid: P = force
area
c. Pascals’s Law: For an enclosed fluid, pressure is equal at all points in the vessel
. Sample: Hydraulic press: F = P/A for enclosed liquid; A is the surface area of
the piston inserted into the fluid
Equation: A1F1 = A2 F2; cylinder area determines force fig 32
d. A column of water generates pressure, P increases with depth;
Equation: P2 = P1 + ρgh fig 33
e. Archimedes’ Principle: Buoyant force, Fb , on a object of volume V submerged

in liquid of density ρ: Fb = ρVg fig 34
2. Goal 2: Examine fluid motion & fluid dynamics
a. Properties of an Ideal fluid: Non-viscous, incompressible, steady flow, no turbulence
At any point in the flow, the product of area and velocity is constant: A1v1 = A2v2

b. Variable density: ρ1A1v1 = ρ2 A2v2; illustrations: gas flow through a smokestack,
water flow through a hose fig 35
c. Bernoulli’s Equation: For any point y in the fluid flow, P + 1 ρv 2 + ρ g y = constant
2
•Special case: Fluid at rest P1 – P2 = ρ g h
4

f (Hz)

Angular Frequency

ω (rad/s)

Wave number

k (m–1)

Wave Amplitude
Speed

A
v (m/s)

Distance between peaks
Time to travel one λ
f= 1
T
ω = 2r = 2¹ f
T
Height of wave
v = λf
k = 2r
m

2. ✎ Sample: Determine the velocity and period of a wave with
λ = 5.2 m and f = 50.0 Hz
T= 1
Equations: v = λ f
f
Data: λ = 5.20 m; f = 50.0 Hz
Calculations: v = λ f = 5.20 m × 50.0 = 260 m/s
T = 1 = 1 Hz = 0.02 s
50
f
B. Sound Waves
F1
32
1. Wave nature
of sound: Compression wave displaces the medium
carrying the wave
B;
2. General speed of sound: vA=2 t
A
note: B = Bulk
Modulus (measure of volume compressibility)
1
cRT F2
Cp
For a gas: v = P
; note: γ =
(ratio of gas heat capacities)
M
Cv
✎ Sample: Calculate speed of sound in Helium at 273 K
Helium: Ideal gas, γ = 1.66; M = 0.004 kg/mole
cRT
v=
M
1.66 # 8.314 kg m2 /s2 # 273K
=
0.004kg
= 941, 900 m 2 /s 2 = 971 m/s note:
applies to the units
3. Loudness as intensity and relative intensity
a. Absolute Intensity (I = Power/Area) is an inconvenient measure of
loudness
b. Relative loudness: Decibel scale (dB): β = 10 log I ; I0 is the
I0
threshold of hearing; β(I0) = 0
c. Samples: Jet plane: 150 dB; Conversation: 50 dB; a change in 10 dB
represents a 10−fold increase in I
f'
due to relative motion
4. Doppler effect: The sound frequency shifts
f
of source and listener;
v0 - listener speed; vs - source speed; v - speed of sound
f ' v +v0
f ' v -v0
= v -v
= v +v
f
f
s
s
Key: Identify relative speed of source and listener

THERMODYNAMICS

A. Goal: Study of work, heat and energy of a system fig 36

Key Variables
Heat: Q

+Q added to the system

Energy: E

System internal E

Work: W

Enthalpy: H
Entropy: S

Temperature: T
Pressure: P
Volume: V

+W done by the system
H = E + PV

Thermal disorder

Measure of thermal E

Force exerted by a gas
Space occupied

36
Q

W
system
∆E

THERMODYNAMICS (continued)
Types of Processes

Isothermal

∆T = 0

Adiabatic

Q=0

Isobaric:

∆P = 0

Isochoric

∆V =0

∆E = 0, Q =W
PV = constant

∆E = –W
PV γ = constant
W = P∆V,
∆H = Q
∆E = Q;
W=0

E. The Kinetic Theory of Gases
1. Goal: Examine kinetic energy of gas molecules
2. Key Equations: E = 1 Mv 2 and E = 3 RT
2
2
a. Speed, vrms = 3RT
M
✎ Sample: Calculate the speed of
Helium at 273 K
Helium: M = 0.004 kg/mole
vrms = 3RT =
M

Volume (L)

Volume (L)

3 # 8.314 kg m2 /s2 # 273 K
B. Temperature & Thermal Energy
0.004 kg
1. Goal: Temperature is in Kelvin, absolute
temperature: T(K) = T( ºC) + 273.15
vrms = 1, 702, 292 m/s = 1305 m/s
Note: T(K) is always positive; lab temperature
must be converted from ºC to Kelvin (K)
b. Kinetic energy for Ideal Gas: K = 3 RT
2
✎ Sample: Convert 35º C to Kelvin:
c. For real gas: Add terms for vibrations
T(K) = T(ºC) + 273.15 = 35 + 273.15 =
and rotations
308.15 K
F.
Entropy
& 2nd Law of Thermodynamics
2. Thermal Expansion of Solid, Liquid or Gas
1. Goal: Examine the driving force for a process
a. Goal: Determine the change in the length
2. Key Variables:
(L) or volume (V ) as a function of
a. Entropy: S, thermal disorder; dS = dQ
temperature
T
b. S(univ) = S(system) + S(thermal bath)
b. Solid: DL = α∆T
L
3. Guiding Principle: 2nd Law of
c. Liquid: DV = β∆T
Thermodynamics:
V
For any process, ∆Suniv > 0; one exception:
^T2 -T1h nR
d. Gas: ∆V =
P
∆Suniv = 0 for a reversible process
Examples:
4.
Q
3. Heat capacity: C =
or Q = C∆T
a. Natural heat flow: Q flows
DT
from Thot to Tcold fig 38 38
a. Special cases: Cp–constant P; Cv–constant V
Thot
∆Suniv = ∆Shot + ∆Scold =
• Ideal Gas:
Cp
Q
Q
T -T
Cp = 5 R; Cv = 3 R; γ =
= 5 = 1.667
Q

+
= Q hot cold
2
2
3
Cv
Thot
Tcold
Thot Tcold
b. Carnot’s Law: For ideal gas: Cp – Cv = R
• ∆E = Cv ∆T; ∆H = Cp∆T
Tcold
hint: ∆Suniv > 0 for a
• Exact for monatomic gas, modify for
natural process
molecular gases
b. Phase change: ∆S =
Q (phase change )
37
T (phase change )
Boyle's Law
Charles' Law
T
c. Ideal Gas S(T ): ∆S = nCp ln 2
T1
d. Ideal Gas: S(V ): ∆S = n R ln

V2
V1

G. Heat Engines
1. Goal: Examine Q
Q
and W of an 39 hot
engine
2. Thermal Engine:
gas (mol); gas constant R = 8.314 J mol–1 K–1
The engine
! Pitfall: Common errors in T, P or V units
transfers Q from a
3. Key Applications:
hot to a cold
reservoir, and
Qcold
a. P Ä 1 , T fixed: Boyle’s Law
V
produces W fig 39
b. P Ä T, V fixed
3. Efficiency
of
c. V Ä T, P fixed: Charles’ Law
Qcold
W
=1–
engine: p =
Qhot
Qhot
d. Derive thermodynamic relationships
D. Enthalpy and 1st Law of Thermodynamics
1. Goal: Determine Q, ∆E and W; W and Q
Carnot Cycle
P
depend on path; ∆ E is a state variable,
A
Qhot
independent of path
2. Guiding Principle:
B ∆T = 0
a. 1st Law of Thermodynamics: ∆E = Q – W
C. Ideal Gas Law; PV = nRT fig 37
1. Goal: Simple equation of state for a gas
2. Key Variables: P(Pa), V(m3), T (K ), n moles of
Pressure (Pa)

Temperature (K)

• Key idea: Conservation of Energy
b. Examine the T, P, W & Q for the problem
3. Enthalpy: H = E + PV; ∆H = ∆E + P∆V
a. ∆H = Q for ∆P = 0 (constant pressure)
b. Variable temperature: ∆H = ∫Cp d T
c. For constant Cp: ∆H = Cp ∆T
4. Work: W = ∫PdV
a. W depends on the path or process
b. Ideal Gas, Reversible, Isothermal:
W = n RTln V2
V1
c. Ideal Gas, Isobaric: W = P∆V

W

Thot
W

Tcold

40

Thot

∆T = 0

D

Tcold
V
4. Idealized heat engine: Carnot Cycle fig 40
a. Four steps in the cycle: two isothermal,
two adiabatic; for overall cycle: ∆E = 0
and ∆S = 0
T
b. Efficiency = 1 – cold
Thot
Qcold

C

5

ELECTRICITY & MAGNETISM

A. Electric Fields and Electric Charge
1. Goal: Examine the nature of the field generated by an
electric charge, and forces between charges
2. Key Variables and Equations
a. Coulomb C : “ampere sec” of charge
b. e - charge on an electron; 1.6022 × 10–19 C
q1 q2
c. Coulomb’ Law - electrostatic force: F = 1
e
4rf0 r2
• Vector direction defined by e
d. Electric Field: E = F
q
Hint: Calculation shortcut:
q1 ]C gq2 ]C g
r]m g2
Note: q in Coulombs and r in meters
3. Superposition Principle: Forces and fields are
composites of contributions from each charge
F = 9 × 109N

F = R Fi , E = R Ei ;

Hint: Forces and electric fields

are vectors
B. Gauss’s Law
1. Goal: Define electric flux, Φe
2. Key Variables and Equations:
a. Gauss’s Law: Φe =

# E × dA =

Q
f0
b. The electric flux, Φe , depends on the total charge in
the closed region of interest
C. Electric Potential & Coulombic Energy
1. Goal: Determine Coulombic potential energy
2. Key Variables and Equations:
a. Potential energy: U = 1 q1 q2
4rf0 r
U
b. Potential: V(q1) = q = 1 q1
4rf0 r
2
Note: The potential is scalar, depending on |r|
c. For an array of charges, qi, V total = ΣVi
q1 (C) q2 (C)
d. Shortcut to U(r ): U = 9 × 109J
r (m )
3. Continuous charge distributions: V = 1 ∫ dq fig 41
4rf0 r
Sample:
Conducting
sphere,

41
R
Radius R, Charge Q
Q
V= 1
for r ≤ R
4rf0 R
V
kQ
V = 1 Q for r > R
4rf0 r
r
kQ R
4. Dielectric effect: V & F depend on R
the dielectric constant, κ; replace
r
ε0 with κε0 for the material;
1
V(κ) = V (vacuum)
Conducting Sphere
l
1
F(κ) = F (vacuum)
42
d
l
D. Capacitance and Dielectrics
Q
V
1. Goal: Study capacitors, plates with
charge Q separated by a vacuum or
A
dielectric material fig 42
2. Key Equations:
a. Capacitance, C = Q , V is the
V
measured voltage
b. Parallel plate capacitor, vacuum, with area A,
Q
spacing d : C = ε0 A ; E =
A
f0
d
c. Parallel plate capacitor, dielectric κ, with area A,
spacing d: C = κε0 A
d
3. Capacitors in series: 1 = / 1
Ci
Ctot
4. Capacitors in parallel: Ctot = ΣCi fig 43
Two Capacitors in Series
C1
C2

1 = 1 + 1
Ctot C1 C 2

or C tot

Two Capacitors in Parallel
C1 C2 C = C +C
tot
1
2

43
C1 C2
=
C1 +C2

# BdA

c. Force on charge, q and v, moving in B:
F = qv × B = qvBsinθ; v parallel to B => F = 0 ; v
perpendicular to B => F = qvB
d. Magnetic Moment of a Loop: M = I A
e. Torque on a loop: τ = M × B
f. Magnetic Potential Energy: U = – M • B
g. Lorentz Force: Charge interacts with E and B;
F = qE + qv × B
H. Faraday’s Law and Electromagnetic Induction Key Equations:
1. Faraday’s Law: Induced EMF:

E

=

# E ds

=

Plane mirror: Law of Reflection

50a
s
O

–d Φm /dt
2. Biot-Savart Law: Conductor induces B; current I,
n0
Plane mirror: Law of Reflection
length dL: dB = 4r IdL
× rr3
50a

s'

s
I

virtual object

erect

inverted

real image
erect

virtual image
inverted

50b
h'

θ
θ

Image

Customer Hotline #
1.800.230.9522
hundreds of titles at

free downloads &

Spherical Concav

Converging Lens
s'

h
Object

real object

Note: Due to the condensed nature of this guide, use as a quick reference guide, not as a replacement for assigned course work.
All rights reserved. No part of this publication may be reproduced or transmitted in any form, or by any means, electronic or mechanical, including photocopy, recording, or any information
storage and retrieval system, without written permission from the publisher. © 2005 BarCharts, Inc. Boca Raton, FL 1007

b. Magnetic Flux: Φm =

d. Laws of Geometric Optics:
s object dist.
1 + 1 = 1 ; s =- h
s’
image dist.
s s' f s'
h'
e. Combination of 2 thin lenses:
h object size
1 = 1 + 1 or f= f1 f2
h’ image size
f f1 f2
f1 +f2
3. ✎ Sample Guidelines for ray tracing:
a. Rays that parallel optic axis pass through “f ”
b. Rays pass through center of the lens unchanged
c. Image forms at convergence of ray tracings
✎ Sample ray tracings: fig 50, a,b,c

quickstudy.com

a. EMF: Circuit voltage; E = Vb + IR; battery voltage
Vb = I r, r internal battery resistance
b. Junction: Connection of 3 or more conductors
c. Loop: A closed conductor path
d. Resistors in series or parallel => replace with Rtot
e. Capacitors in series or parallel => replace with Ctot
3. Kirchoff’s Circuit Rules
I2 45
a. For any Loop: ΣV = Σ I R;
I1
Hint: Conserve energy
I3
b. For any Junction: Σ I = 0;
I1 = I2 + I3
Hint: Conserve charge;
define “+” flow fig 45
G. Magnetic Field: B
1. Key concepts:
a. Moving charge => Magnetic Field B

A. Basic Properties of Light
Reflection and Refraction
Incident
1. Goal: Examine light and its interaction with matter
47
Ray
2. Key variables:
θ1
a. c: speed of light in a vacuum
θ2
θr
b. Index of refraction: n; nc = speed of light in medium Reflected
Refracted
Ray
Ray
n1 n2
c. Light as electromagnetic wave: λf = c
Normal
Light characterized by “color” or wavelength
d. Light as particle: e = h f ; energy of photon
3. Reflection and Refraction of Light fig 47
θ
a. Law of Reflection: θ1 = θr fig 48
θ1 r
b. Refraction: Bending of light ray as it passes from n1 to n2
48
•Snell’s Law: n1sinθ1 = n2sinθ2 , n1, n2 are the indices of refraction of
two materials fig 49
Normal
n
c. Internal Reflectance: sinθc = n2
1
Light passing from material of higher n to a lower n may be
trapped in the material
θ1
θ1 > θ2
4. Polarized light: E field is not spherically symmetric
n1
Air
a. Examples: Plane/linear polarized, circularly polarized
b. Polarization by reflection from a dielectric surface at angle θc ; Glass
n2
n
Brewster’s Law: tanθc = n2
θ2
1
B. Lenses and Optical Instruments
49
1. Goal: Lenses and mirrors generate images of objects
2. Key concepts and variables
Lens and Mirror Properties
a. Radius of curvature: R = 2f
b. Optic axis: Line from base of object Parameters
+ sign
- sign
through center of lens or mirror
converging lens diverging lens
f focal length
c. Magnification: M = ss'
concave mirror convex mirror

PRICE
U.S.$5.95 CAN.$8.95

E. Current and Resistance
1. Goal: Examine the current, I, quantity of charge, Q,
resistance, R; determine the voltage and power dissipated
2. Key Equations:
44
a. Total charge, Q = It
Two Resistors in Series
b. V = I R, or R = V
I
R1
c. Resistors in Series:
Rtot = R1+R2
R2
Rtot = Σ Ri fig 44
d. Resistors in Parallel:
Two Resistors in Parallel
1 = Σ 1 fig 44
R1 R2
Rtot
Ri
1 = 1 + 1
2
Rtot R1 R2
e. Power = IV = I R
F. Direct Current Circuit
R1 R2
1. Goal: Examine a circuit
or R tot =
R
1 +R2
containing
battery,
resistors and capacitors;
determine voltage and current properties
2. Key Equations and Concepts:

CREDITS
Author: Mark Jackson, PhD. Chemical Physics
M.A. Physics, Harvard University

BEHAVIOR OF LIGHT

f
f
I

Spherical Concave Mirror
Converging Lens
n0 I
3. Sample: Long conducting wire: B(r) = 4r r
s
s'
C.
of Light Waves
s Interference
s'
50c
50b
I. Electromagnetic Waves- Key Equations and Concepts:
1. Goal: Examine constructive and destructive interference of
O
I
1. Transverse B and E fields; E = c
h
h'
light waves
B
f
θ2. Key Variables and Concepts:
y
f
θ
Object
Image
E/B Wave
46
interference: fig 51
a. Constructive
O
R
f
I
b. Destructive interference: fig 52
c. Huygens’ Principle: Each portion of wave front acts as a
I
source of new waves
x
3. Diffraction of light, from a grating with spacing d, produces
B
an interference pattern; dsinθ = mλ; (m = 0, 1, 2, 3,...)
z
4. Single-slit experiment, slit width a ; destructive interference
1
2. c =
n0 f0
for sinθ = mm
a ; (m = 0, ±1, ±2...)
3. Electromagnetic Wave: c = f λ fig 46
5.
X-ray
diffraction
from a crystal with
dy ; 2dsinθ
=ymλ;
(m = 0, 1, 2, 3,...)
y
y atomic spacing
y1
J. Maxwell’s Equations:
2
51
52
Q
y
1. Gauss’s Law for Electrostatics: # E • d A = f ;
0
key: Charge gives rise to E
y1
2. Gauss’s Law for Magnetism: # B • d A = 0 ;
key: Absence of magnetic charge
x
x
3. Ampere-Maxwell Law:
dU e
# B • ds = µ0 I + µ0ε0 dt ;
1
y2
key: Current + change in electric flux => B
3
dU m
4. Faraday’s Law: # E • dS = –
;
6
2
dt

O

R

f
I

ISBN-13: 978-142320695-8
ISBN-10: 142320695-9

ELECTRICITY &
MAGNETISM (continued)


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