Miscellaneous

Hello World

The Greek Alphabet

${\displaystyle \alpha A}$	\alpha A
${\displaystyle \beta B}$	\beta B
${\displaystyle \gamma \Gamma }$	\gamma \Gamma
${\displaystyle \delta \Delta }$	\delta \Delta
${\displaystyle \epsilon \varepsilon E\;}$	\epsilon \varepsilon E
${\displaystyle \zeta Z}$	\zeta Z
${\displaystyle \eta H}$	\eta H
${\displaystyle \theta \vartheta \Theta }$	\theta \vartheta \Theta

${\displaystyle \iota I}$	\iota I
${\displaystyle \kappa K}$	\kappa K
${\displaystyle \lambda \Lambda \;}$	\lambda \Lambda
${\displaystyle \mu M}$	\mu M
${\displaystyle \nu N}$	\nu N
${\displaystyle \xi \Xi }$	\xi \Xi
${\displaystyle oO\;}$	o O
${\displaystyle \pi \Pi }$	\pi \Pi

${\displaystyle \rho \varrho P\;}$	\rho \varrho P
${\displaystyle \sigma \,\!\Sigma \;}$	\sigma \Sigma
${\displaystyle \tau T}$	\tau T
${\displaystyle \upsilon \Upsilon }$	\upsilon \Upsilon
${\displaystyle \phi \varphi \Phi }$	\phi \varphi \Phi
${\displaystyle \chi X}$	\chi X
${\displaystyle \psi \Psi }$	\psi \Psi
${\displaystyle \omega \Omega }$	\omega \Omega

Chemistry

Conventions

Given some element $\mathrm{X}$

$$\begin{equation} \begin{split} \ce{^{A}_{Z}X} \end{split} \end{equation}$$

$$\begin{equation} \begin{split} \ce{X-A} \end{split} \end{equation}$$

Where

$A = \text{neutrons + protons}$
$Z = \text{protons}$

Cations: Positively (＋) Charged
Anions: Negatively (−) Charged

Mnemonic

Cations are Pawsitive

Conventions on homework

When a question says, determine the energy of 1 ㏖ of photons, the unit will be ᴶ/㏖.

Units

$$\begin{equation} \begin{split} 1 \mol &= \sci{6.022}{23} \end{split} \end{equation}$$

$$\begin{equation} \begin{split} \small 1 \;\mathrm{amu} &\approx \;\text{protons} + \;\text{neutrons} \end{split} \end{equation}$$

SI Prefixes

Value	Prefix	Symbol
$10^{1}$	deca	$\mathrm{da}$
$10^{2}$	hecto	$\mathrm{h}$
$10^{3}$	kilo	$\mathrm{k}$
$10^{6}$	mega	$\mathrm{M}$
$10^{9}$	giga	$\mathrm{G}$
$10^{12}$	tera	$\mathrm{T}$

Value	Prefix	Symbol
$10^{-1}$	deci	$\mathrm{d}$
$10^{-2}$	centi	$\mathrm{c}$
$10^{-3}$	mili	$\mathrm{m}$
$10^{-6}$	micro	$\mathrm{\mu}$
$10^{-9}$	nano	$\mathrm{n}$
$10^{-12}$	pico	$\mathrm{p}$

Classification of Matter

Overview

Mixtures

Heterogeneous mixture

Where the prefix Hetero- means different

Homogeneous mixture

Where the prefix Homo- means same

Quantum Mechanics

The Electromagnetic Spectrum

Overview

Formulas

$$\begin{equation} \begin{split} \mathrm{f} = \nu &= \frac{c}{\lambda} \end{split} \end{equation}$$

$$\begin{equation} \begin{split} \mathrm{E} &= \mathrm{h}\times\mathrm{f}\\ &= \mathrm{h}\frac{\mathrm{c}}{\lambda} \end{split} \end{equation}$$

$$\begin{equation} \begin{split} \mathrm{T} &= \frac{1}{f} \end{split} \end{equation}$$

$$\begin{equation} \begin{split} \lambda &= \frac{c}{f} \end{split} \end{equation}$$

Values

Name	Symbol	Unit	Description	Range
Wavelength	$\mathrm{\lambda}$	Any unit for distance	Distance between two analogous points	Always Positive
Frequency	$f$ or $\nu$ (nu)	$㎐ = \frac{\text{1 cycle}}{\text{second}}$$\mathrm{s}^{-1}$	Number of cycles	Always Positive
Energy	$\mathrm{E}$	$\mathrm{J}$ (joule)	Amount of energy ($\mathrm{E}$) in a light packet

Constants

Name	Symbol	Unit	Value
Speed of Light	$\mathrm{c}$	$\v$	$\mathrm{c} = \sci{3.00}{8}\v$
Planck's constant	$\mathrm{h}$	Energy multiplied by time joule-seconds $\mathrm{J}\cdot\mathrm{s}$	$\mathrm{h} = \sci{6.626}{-34}\;\mathrm{J}\cdot\mathrm{s}$

Other Formulas

de Broglie Relation

$$\begin{equation} \begin{split} \lambda &= \frac{\mathrm{h}}{\mathrm{m}\mathrm{v}}\\ \text{where}&\\ m &= \text{mass}\\ v &= \text{velocity} \neq \nu\;\text{(nu)} \end{split} \end{equation}$$

Heisenberg's Uncertainty Principle

$$\begin{equation} \begin{split} \Delta{x} \times \mathrm{m} \Delta{v} \geq \frac{\mathrm{h}}{4\pi} \end{split} \end{equation}$$

Where

$\Delta{x}$ is the uncertainty in position.
$\Delta{v}$ is the uncertainty in velocity.
$\mathrm{m}$ is the mass of the particle.
$\mathrm{h}$ is the plank's constant.

In general it states that the more you know about an electrons position, the less you know about it's velocity.

Energy of an Electron in an Orbital with Quantum Number $\mathrm{n}$ in a Hydrogen Atom

$$\begin{equation} \begin{split} E_n &= \sci{-2.18}{-18} \mathrm{J} \left(\frac{1}{n^2}\right) \end{split} \end{equation}$$

Change in Energy That Occurs in an Atom When It Undergoes a Transition between Levels

$n_{\small\text{initial}}$ and $n_{\small\text{final}}$

$$\begin{equation} \begin{split} \Delta{E} &= \sci{-2.18}{-18} \mathrm{J} \left(\frac{1}{n^2_f} - \frac{1}{n^2_i}\right) \end{split} \end{equation}$$

If $\Delta{E}$ is negative, energy is being released.
If $\Delta{E}$ is positive, energy is being absorbed.

Electron Configuration

Traditional Chart

Better Method

Examples

Electron configuration for $_{26}\mathrm{Fe}$

$$\begin{equation} \begin{split} \ce{\underbrace{1s^2 2s^2 2p^6 3s^2 3p^6}_{\smallText{Equal to Argon}} 4s^2 3d^6} &= \ce{[Ar] 4s^2 3d^6} \end{split} \end{equation}$$

Since the electron configuration for Argon is

$$\begin{equation} \begin{split} \ce{[Ar] = 1s^2 2s^2 2p^6 3s^2 3p^6} \end{split} \end{equation}$$

Electron configuration for $_{26}\mathrm{Fe}^{+2}$

Beginning with the electron configuration for $_{26}\mathrm{Fe}$

$$\begin{equation} \begin{split} \underbrace{1s^2\; 2s^2\; 2p^6\; 3s^2\; 3p^6}_{\smallText{Equal to Argon}}\; \overbrace{4s^2\; 3d^6}^{\mathclap{ \begin{gathered} \smallText{Which should we remove 2 $\mathrm{e}^{-}$ from?}\\ \smallText{($4s^2$ or $3d^6$)?}\\ \end{gathered} }} &= \ce{[Ar] \underbrace{4s^2\; 3d^6}_{\mathclap{ \begin{gathered} \smallText{Which should we remove 2 $\mathrm{e}^{-}$ from?}\\ \smallText{($4s^2$ or $3d^6$)?}\\ \end{gathered} }} } \end{split} \end{equation}$$

Remove the electrons from the term with the higher electron state. Warning! Do not just remove the electrons from the rightmost term since the rightmost term may be a lower electron state. For instance given $4s^2\; 3d^6$

$4s^2$ is in a higher electron state
$3d^6$ is in a lower electron state

As shown

$$\begin{equation} \begin{split} \underbrace{1s^2\; 2s^2\; 2p^6\; 3s^2\; 3p^6}\; \overbrace{4s^{\left(2 - 2\right)}}^{ \mathclap{ \begin{gathered} \smallText{higher state}\\ \smallText{remove $2\mathrm{e^{-}}$} \end{gathered} }}\; \underbrace{3d^6}_{ \mathclap{\smallText{lower state}}}\; &= [\mathrm{Ar}]\; \overbrace{4s^{\left(2 - 2\right)}}^{ \mathclap{ \begin{gathered} \smallText{higher state}\\ \smallText{remove $2\mathrm{e^{-}}$} \end{gathered} }}\; \underbrace{3d^6}_{ \mathclap{\smallText{lower state}}}\\\\ \underbrace{1s^2\; 2s^2\; 2p^6\; 3s^2\; 3p^6}\; \underbrace{3d^6}_{ \mathclap{\smallText{unchanged}}}\; &= [\mathrm{Ar}]\; \underbrace{3d^6}_{ \mathclap{\smallText{unchanged}}} \end{split} \end{equation}$$

Therefore the electron configuration for $_{26}\mathrm{Fe}^{+2}$ is:

$$\begin{equation} \begin{split} \ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^6} &= \ce{[Ar] 3d^6} \end{split} \end{equation}$$

Electron configuration for $_{24}\mathrm{Cr}$

It would appear that the electron configuration for $_{24}\mathrm{Cr}$ would be

$$\begin{equation} \begin{split} \overbrace{1s^2\; 2s^2\; 2p^6\; 3s^2\; 3p^6}^{\smallText{Equal to Argon}}\; \underbrace{4s^2\; 3d^4}_{ \mathclap{\smallText{this is wrong!}}}\; &= [\mathrm{Ar}]\; \underbrace{4s^2\; 3d^4}_{ \mathclap{\smallText{this is wrong!}}}\; \end{split} \end{equation}$$

But this is wrong! It's actually

$$\begin{equation} \begin{split} \overbrace{1s^2\; 2s^2\; 2p^6\; 3s^2\; 3p^6}^{\smallText{Equal to Argon}}\; \underbrace{4s^1\; 3d^5}_{ \mathclap{\smallText{notice the superscripts}}}\; &= [\mathrm{Ar}]\; \underbrace{4s^1\; 3d^5}_{ \mathclap{\smallText{notice the superscripts}}}\; \end{split} \end{equation}$$

How-tos

What are the valence electrons?

Given

$$\begin{equation} \begin{split} \ce{1s^2 2s^2 2p^6 3s^2 3p^4} \end{split} \end{equation}$$

The valance electrons will be the ones in the highest energy state. Therefore

$$\begin{equation} \begin{split} \overbrace{1s^2\; 2s^2\; 2p^6}^{\mathclap{ \smallText{Core electrons} }}\; \underbrace{3s^2\;3p^4}_{\mathclap{ \begin{gathered} \smallText{Highest state}\\ \smallText{Therefore these are the valence electrons} \end{gathered} }} \end{split} \end{equation}$$

Therefore there are $6$ valence electrons.

Given

$$\begin{equation} \begin{split} \ce{[Ne] 3s^2 3p^2} \end{split} \end{equation}$$

The valance electrons will be the ones in the highest energy state. Therefore

$$\begin{equation} \begin{split} [\mathrm{Ne}] \underbrace{3s^2 3p^2}_{\mathclap{ \begin{gathered} \smallText{Highest state}\\ \smallText{Therefore these are the valence electrons} \end{gathered} }} \end{split} \end{equation}$$

Therefore there are $4$ valence electrons.

Quantum Numbers

Overview

Symbol	Description
$\mathrm{n}$	The principle quantum number
$l$	The angular momentum quantum number
$\mathrm{m}_1$	The magnetic quantum number
$\mathrm{m}_s$	The spin quantum number

The Principle Quantum Number ($\mathrm{n}$)

Value of $\mathrm{n}$	Value of $l$	Orbital Sublevel
$\mathrm{n} = 1$	$l = 0$	$1\mathrm{s}$
$\mathrm{n} = 2$	$l = 0$$l = 1$	$2\mathrm{s}$$2\mathrm{p}$
$\mathrm{n} = 3$	$l = 0$$l = 1$$l = 2$	$3\mathrm{s}$$3\mathrm{p}$$3\mathrm{d}$
$\mathrm{n} = 4$	$l = 0$$l = 1$$l = 2$$l = 3$	$4\mathrm{s}$$4\mathrm{p}$$4\mathrm{d}$$4\mathrm{f}$

Angular Momentum Quantum Number

Value	Result
$l = 0$	$\mathrm{s}$
$l = 1$	$\mathrm{p}$
$l = 2$	$\mathrm{d}$
$l = 3$	$\mathrm{f}$

Value of $l$	Value of $\mathrm{m_l}$
$l = 0$	$\mathrm{m_l} = 0$
$l = 1$	$\mathrm{m_l} = -1$$\mathrm{m_l} = 0$$\mathrm{m_l} = 1$
$l = 2$	$\mathrm{m_l} = -2$$\mathrm{m_l} = -1$$\mathrm{m_l} = 0$$\mathrm{m_l} = 1$$\mathrm{m_l} = 2$
$l = 2$	$\mathrm{m_l} = -3$$\mathrm{m_l} = -2$$\mathrm{m_l} = -1$$\mathrm{m_l} = 0$$\mathrm{m_l} = 1$$\mathrm{m_l} = 2$$\mathrm{m_l} = 3$

Summary

$$\begin{equation} \begin{split} l \leq \mathrm{n} - 1 \end{split} \end{equation}$$

$$\begin{equation} \begin{split} -l \leq \mathrm{m_l} \leq l \end{split} \end{equation}$$

Useful Formulas

The equation for a maximum number of electrons a given energy level can hold given some value for $n$

$$\begin{equation} \begin{split} \text{max}_{\mathrm{e}^{-}} &= 2n^2 \end{split} \end{equation}$$

How many orbitals are possible given some value for $n$

$$\begin{equation} \begin{split} \text{max}_{\smallText{orbitals}} &= n^2 \end{split} \end{equation}$$

Examples

Light

Interference and Diffraction

Constructive Interference

If two waves of equal amplitude are in phase when they interact—that is, they align with overlapping crests—a wave with twice the amplitude results. This is called constructive interference.

Destructive Interference

If two waves are completely out of phase when they interact—that is, they align so that the crest from one overlaps with the trough from the other—the waves cancel by destructive interference.

Mathematics

Algebra

Function Composition And Notation Conveniences

Right to Left Evaluation

$$\begin{equation} \begin{split} f \triangleleft x &= f(x)\\ f \triangleleft g \triangleleft x &= f(g(x))\\ f \circ g \triangleleft x &= f(g(x)) \end{split} \end{equation}$$

(Reading and evaluating expressions from right to left should be the standard in mathematics, as it is in Haskell.)

Left to Right Evaluation

$$\begin{equation} \begin{split} x \triangleright f &= f(x)\\ x \triangleright f \triangleright g &= g(f(x)) \end{split} \end{equation}$$

(Somewhat confusing since it's not right to left, but can be convenient and look nice...)

Parentheses are evil, inelegant, and requires more work to write/type and read. The fewer parentheses the better. When things get complicated, smart use of function composition on the other hand looks beautiful when used well.

Algebra Basics

$$\begin{equation} \begin{split} a^{-x} &= \frac{1}{a^x} \end{split} \end{equation}$$

$$\begin{equation} \begin{split} a^x a^y &= a^{\left(x + y\right)} \end{split} \end{equation}$$

$$\begin{equation} \begin{split} \frac{a^x}{a^y} &= a^{\left(x - y\right)} \end{split} \end{equation}$$

$$\begin{equation} \begin{split} \left(a^x\right)^y &= a^{\left(x y\right)} \end{split} \end{equation}$$

$$\begin{equation} \begin{split} a^0 &= 1 \\ a^1 &= a \\ a^{\frac{1}{n}} &= \sqrt[n]{a} \end{split} \end{equation}$$

$$\begin{equation} \begin{split} \frac{1}{\frac{a}{b}} &= \left(\frac{a}{b}\right)^{-1} &= \frac{b}{a} \end{split} \end{equation}$$

$$\begin{equation} \begin{split} a &= e^x \\ \ln a &= x \end{split} \end{equation}$$

$$\begin{equation} \begin{split} \ln \triangleleft\; e^x &= x \\ e^{\ln{\left(x\right)}} &= x \end{split} \end{equation}$$

$$\begin{equation} \begin{split} \ln \triangleleft\; a \cdot b &= \ln{a} + \ln{b} \end{split} \end{equation}$$

$$\begin{equation} \begin{split} \ln \triangleleft\; \frac{a}{b} &= \ln{a} - \ln{b} \end{split} \end{equation}$$

$$\begin{equation} \begin{split} \ln \triangleleft\; a^n &= n \cdot \ln{a} \end{split} \end{equation}$$

Trigonometry

The Unit Circle & Special Angles In Trig

Warning

Never use Pi ($\pi$)! It makes (thinking in terms of) radians confusing, Tao ($\tau$) is what the enlightened trigonometer uses, and won't screw you over.

To easily memorize the special angles in trig, notice the repeating patterns on the above angles.

For values on the x-axis, anything over $\frac{1}{4}\tau$ and under $\frac{3}{4}\tau$ will be negative
For values on the y-axis, anything over $\frac{1}{2}\tau$ will be negative
Diagonals will be $\pm \frac{\sqrt{2}}{2}$

For ratios of $\frac{1}{12}\tau$ on the sides, i.e. $\frac{1}{12}\tau$, $\frac{2}{12}\tau$, $\frac{4}{12}\tau$, $\frac{5}{12}\tau$, $\frac{7}{12}\tau$, $\frac{8}{12}\tau$, $\frac{10}{12}\tau$, and $\frac{11}{12}\tau$. Draw a circle and dot the point where it occurs (which is pretty easy since the above are simple ratios of a circle when expressed in terms of $\tau$). Then with regards to the $x$ and $y$ axis values:
- The longer size will be $\pm \frac{\sqrt{3}}{2}$
- The shorter side will be $\pm \frac{1}{2}$
See the above examples.

Conventions

$$\begin{equation} \begin{split} \sin\theta &= \frac{b}{c}\\ \cos\theta &= \frac{a}{c}\\ \tan\theta &= \frac{b}{a} \end{split} \end{equation}$$

$$\begin{equation} \begin{split} \csc\theta &= \frac{1}{\sin\theta} = \frac{c}{b}\\ \sec\theta &= \frac{1}{\cos\theta} = \frac{c}{b}\\ \cot\theta &= \frac{1}{\tan\theta} = \frac{a}{b} \end{split} \end{equation}$$

Radians & Radian Conversion

Constants

$$\begin{equation} \begin{split} \tau &= 2\pi = 360^{\circ} \\ \pi &= \frac{1}{2}\tau = 180^{\circ} \end{split} \end{equation}$$

Conversion

Given

$$\begin{equation} \begin{split} {1}^{\circ} &= \frac{{1}}{360} \tau \; {\displaystyle {\mathrm{rad}}} \\ {1} \; \mathrm{rad} &= \frac{{1}}{\tau} \cdot 360^{\circ} = {1} \cdot \frac{360^{\circ}}{\tau} \end{split} \end{equation}$$

Degrees to Radians (in terms of $\tau$)

$$\begin{equation} \begin{split} {x^{\circ}} &= \frac{{x}}{360} \tau \; {\displaystyle {\mathrm{rad}}} \end{split} \end{equation}$$

Radians to (terrible and arbitrary) Degrees (the formula doesn't even look as nice)

$$\begin{equation} \begin{split} {x} \; \mathrm{rad} &= \frac{{x}}{\tau} \cdot 360^{\circ} \\ &= {x} \cdot \frac{360^{\circ}}{\tau} \\ &= {x} \cdot \frac{180^{\circ}}{\pi} \end{split} \end{equation}$$

Geometry

The Circle

$$\begin{equation} \begin{split} \small\text{Area}\;\normalsize &= \pi r^2\\ \small\text{Circumference}\;\normalsize &= 2 \pi r \end{split} \end{equation}$$

Pre-Calculus

Calculus

Function Graphs

Example

Derivative Formulas

$$\begin{equation} \begin{split} \frac{\mathrm{d}}{\mathrm{d}x} x^n &= n \cdot x^{n -1} \end{split} \end{equation}$$

Integral Formulas

$$\begin{equation} \begin{split} \int 1 \;\mathrm{d}x &= x + \mathrm{c} \end{split} \end{equation}$$

$$\begin{equation} \begin{split} \int a \;\mathrm{d}x &= ax + \mathrm{c} \end{split} \end{equation}$$

$$\begin{equation} \begin{split} \int x^n \;\mathrm{d}x &= \frac{x^{x + 1}}{n + 1} + \mathrm{c} \end{split} \end{equation}$$

Physics

Constants

$$\begin{equation} \begin{split} g &= 9.8 \frac{m}{s^2}\;\text{downward} \;\text{where}\; a &= -g \end{split} \end{equation}$$

Tables

Linear Kinematic Equations (1D)

Formula	Missing	Quantities Present
$v_2 = v_1 + a\Delta{t}$	$\Delta{x}$	$v_1$$v_2$$a$$\Delta{t}$
$\Delta{x} = \left(\frac{v_2 + v_1}{2}\right)\Delta{t}$	$a$	$\Delta{x}$$v_1$$\Delta{t}$$v_2$
$\Delta{x} = v_1 \Delta{t} + \frac{1}{2}a\Delta{t^2}$	$v_2$	$\Delta{x}$$v_1$$\Delta{t}$$a$
$\Delta{x} = v_2\Delta{t} - \frac{1}{2}a\Delta{t^2}$	$v_1$	$\Delta{x}$$v_2$$\Delta{t}$$a$
$(v_2)^2 = (v_1)^2 + 2a\Delta{x}$	$\Delta{t}$	$\Delta{x}$$v_1$$v_2$$a$

Rotational Kinematic Equations

Formula	Missing	Quantities Present
$\omega_2 = \omega_1 + \alpha\Delta{t}$	$\Delta\theta$	$\omega_1$$\omega_2$$\alpha$$\Delta{t}$
$\Delta\theta = \left(\frac{\omega_2 + \omega_1}{2}\right)\Delta{t}$	$a$	$\Delta\theta$$\omega_1$$\Delta{t}$$\omega_2$
$\Delta\theta = \omega_1 \Delta{t} + \frac{1}{2}\alpha\Delta{t^2}$	$\omega_2$	$\Delta\theta$$\omega_1$$\Delta{t}$$\alpha$
$\Delta\theta = \omega_2\Delta{t} - \frac{1}{2}\alpha\Delta{t^2}$	$\omega_1$	$\Delta\theta$$\omega_2$$\Delta{t}$$\alpha$
$(\omega_2)^2 = (\omega_1)^2 + 2\alpha\Delta\theta$	$\Delta{t}$	$\Delta\theta$$\omega_1$$\omega_2$$\alpha$

Scalar or Vector

Name	Type
Speed	Scalar
Distance	Scalar
Velocity	Vector
Acceleration	Vector
Average Velocity	Vector
Displacement	Vector
Position	Vector

Kinematic Equations in 1D

Conventions

$$\begin{equation} \begin{split} \bar{v} &= \text{Average velocity}\\ \bar{a} &= \text{Average acceleration}\\ \text{Time} = \Delta{t} &= t_2 - t_1\\ \text{Displacement} = \Delta{x} &= x_2 - x_1\\ \Delta{v} &= v_2 - v_1\\ \end{split} \end{equation}$$

Graphical Representation

Standard Equations

$$\begin{equation} \begin{split} v_2 &= v_1 + a\Delta{t} \end{split} \end{equation}$$

$$\begin{equation} \begin{split} (v_2)^2 &= (v_1)^2 + 2a\Delta{x} \end{split} \end{equation}$$

$$\begin{equation} \begin{split} \Delta{x} &= v_1 \Delta{t} + \frac{1}{2}a\Delta{t^2}\\ &= v_2\Delta{t} - \frac{1}{2}a\Delta{t^2}\\ &= \left(\frac{v_2 + v_1}{2}\right)\Delta{t} \end{split} \end{equation}$$

Summary

Formula	Missing	Quantities Present
$v_2 = v_1 + a\Delta{t}$	$\Delta{x}$	$v_1$$v_2$$a$$\Delta{t}$
$\Delta{x} = \left(\frac{v_2 + v_1}{2}\right)\Delta{t}$	$a$	$\Delta{x}$$v_1$$\Delta{t}$$v_2$
$\Delta{x} = v_1 \Delta{t} + \frac{1}{2}a\Delta{t^2}$	$v_2$	$\Delta{x}$$v_1$$\Delta{t}$$a$
$\Delta{x} = v_2\Delta{t} - \frac{1}{2}a\Delta{t^2}$	$v_1$	$\Delta{x}$$v_2$$\Delta{t}$$a$
$(v_2)^2 = (v_1)^2 + 2a\Delta{x}$	$\Delta{t}$	$\Delta{x}$$v_1$$v_2$$a$

Basics

Constant Velocity

$$\begin{equation} \begin{split} \bar{v} &= \frac{\Delta{x}}{\Delta{t}}\\ \Delta{x} &= \bar{v}\Delta{t} \end{split} \end{equation}$$

Uniform acceleration

$$\begin{equation} \begin{split} \bar{a} &= \frac{\Delta{v}}{\Delta{t}}\\ v_2 &= v_1 + a\Delta{t}\\ \Delta{x} &= v_0\Delta{t}+\frac{1}{2}a\Delta{t^2}\\ (v_2)^2 &= (v_1)^2 - 2\bar{a}\Delta{x} \end{split} \end{equation}$$

Miscellaneous

$$\begin{equation} \begin{split} \bar{v} &= \frac{\Delta{x}}{\Delta{t}} = \frac{v_2 + v_1}{2}\\ \bar{a} &= \frac{\Delta{v}}{\Delta{t}} \end{split} \end{equation}$$

Deriving Displacement Formulas

Displacement when object moves with constant velocity

Deriving $\Delta{x} = \bar{v}\Delta{t}$

$$\begin{equation} \begin{split} \Delta{x} &= \bar{v}\Delta{t} \end{split} \end{equation}$$

Displacement when object accelerates from rest

Deriving $\Delta{x} = \frac{1}{2}\bar{a}\Delta{t^2}$

$$\begin{equation} \begin{split} \Delta{x} &= \frac{1}{2}\Delta{v}\Delta{t}\\ &= \frac{1}{2}\bar{a}\Delta{t^2} \end{split} \end{equation}$$

Displacement when object accelerates with initial velocity

Deriving $\Delta{x} = v_1\Delta{t} + \frac{1}{2}\bar{a}\Delta{t^2}$

$$\begin{equation} \begin{split} \Delta{x} &= v_1\Delta{t} + \frac{1}{2}\Delta{v}\Delta{t} \\ &= v_1\Delta{t} + \frac{1}{2}\bar{a}\Delta{t^2} \end{split} \end{equation}$$

Deriving The Other Kinematic Formulas

Deriving $v_2 = v_1 + \bar{a}\Delta{t}$

Given

$$\begin{equation} \begin{split} \bar{a} &= \frac{\Delta{v}}{\Delta{t}} = \frac{v_2 - v_1}{\Delta{t}}\;(1) \end{split} \end{equation}$$

We can rearrange $v_2$ from equation (1) like so

$$\begin{equation} \begin{split} \bar{a} &= \frac{v_2 - v_1}{\Delta{t}}\\ \bar{a}\Delta{t} &= v_2 - v_1\\ v_1 + \bar{a}\Delta{t} &= v_2\\ \therefore v_2 &= v_1 + \bar{a}\Delta{t} \end{split} \end{equation}$$

Therefore

$$\begin{equation} \begin{split} v_2 &= v_1 + \bar{a}\Delta{t} \end{split} \end{equation}$$

Deriving $v_2^2 = v_1^2 + 2\bar{a}\Delta{x}$

Given

$$\begin{equation} \begin{split} \bar{a} &= \frac{\Delta{v}}{\Delta{t}}\;(1)\\ \bar{v} &= \frac{\Delta{x}}{\Delta{t}}\;(2)\\ &= \frac{v_2 + v_1}{2}\;(3) \end{split} \end{equation}$$

$\Delta{t}$ from equation (1) can be rearranged as

$$\begin{equation} \begin{split} \Delta{t} &= \frac{\Delta{v}}{\bar{a}} = \frac{v_2 - v_1}{\bar{a}}\;(4) \end{split} \end{equation}$$

$\Delta{x}$ from equation (2) can be rearranged like so

$$\begin{equation} \begin{split} \bar{v} &= \frac{\Delta{x}}{\Delta{t}}\\ \bar{v}\Delta{t} &= \Delta{x}\\ \therefore \Delta{x} &= \bar{v}\cdot\Delta{t} \end{split} \end{equation}$$

Using the following equations from above

$\bar{v} = \frac{v_2 + v_1}{2}$ from equation (3)
$\Delta{t} = \frac{v_2 - v_1}{\bar{a}}$ from equation (4)

$$\begin{equation} \begin{split} \Delta{x} &= \bar{v}\cdot\Delta{t}\\ &= \frac{v_2 + v_1}{2}\cdot\frac{v_2 - v_1}{\bar{a}}\\ &= \frac{(v_2)^2 - (v_1)^2}{2\bar{a}}\;(5) \end{split} \end{equation}$$

Rearranging equation (5)

$$\begin{equation} \begin{split} 2\bar{a}\Delta{x} &= (v_2)^2 - (v_1)^2 \end{split} \end{equation}$$

Rearrange again to obtain the more common form

$$\begin{equation} \begin{split} (v_2)^2 &= (v_1)^2 + 2\bar{a}\Delta{x} \end{split} \end{equation}$$

TODO

Two-dimensional Projectile Motion

Conventions

Summary

It's easy to see in the above visualization that $t$ and $x$ increase linearly, while $y$ is non-linear.

Formulas

Displacement & Projectile Position

Generalized

In general (without respect to any $x$ or $y$ axis values)

$$\begin{equation} \begin{split} \text{Displacement} &= \Delta_{\text{general}}\\ &= \bar{v}\cdot\Delta{t} \end{split} \end{equation}$$

Where the distance traveled or displaced is

$$\begin{equation} \begin{split} \text{Displacement} &= \Delta_{\text{general}}\\ &= V_1\cdot\Delta{t} + \frac{1}{2}a\Delta{t^2} \end{split} \end{equation}$$

In terms of $x$ and $y$ axis values

$$\begin{equation} \begin{split} \Delta{x} &= x_2 - x_1\\ \Delta{y} &= y_2 - y_1 \end{split} \end{equation}$$

With respect to the $y$ axis

The displacement of a given projectile in terms of the $y$ axis is

$$\begin{equation} \begin{split} \Delta{y} = V_{y1}\cdot\Delta{t} + \frac{1}{2}a_y\Delta{t^2} \end{split} \end{equation}$$

Since $\Delta{y} = y_2 - y_1$

$$\begin{equation} \begin{split} \Delta{y} &= V_{y1}\cdot\Delta{t} + \frac{1}{2}a_y\Delta{t^2}\\ y_2 - y_1 &= V_{y1}\cdot\Delta{t} + \frac{1}{2}a_y\Delta{t^2}\\ y_2 &= y_1 + V_{y1}\cdot\Delta{t} + \frac{1}{2}a_y\Delta{t^2} \end{split} \end{equation}$$

Which can be read as (in terms of the $y$ axis)

$$\begin{equation} \begin{split} \small{\text{the final position}}\; &= \small{\text{the initial position}}\; + V_{y1}\cdot\Delta{t} + \frac{1}{2}a_y\Delta{t^2} \end{split} \end{equation}$$

With respect to the $x$ axis

The displacement of a given projectile in terms of the $x$ axis is

$$\begin{equation} \begin{split} \Delta{x} = V_{x}\cdot\Delta{t} + \frac{1}{2}a_x\Delta{t^2} (1) \end{split} \end{equation}$$

Note that $a_x = 0$ (because there is no force acting on the projectile in the horizontal direction), and therefore the initial and final velocities are the same. I.e. it's constant throughout. Therefore in summary

$a_x = 0$
$v_{x1} = v_{x2}$ and therefore we will simple refer to the velocity vector as as $v_x$.

Therefore we can simplify equation (1) considerably

$$\begin{equation} \begin{split} \Delta{x} &= V_{x}\cdot\Delta{t} + 0\\ &= V_{x}\cdot\Delta{t} \end{split} \end{equation}$$

Solving Projectile Motion Problems

Projectile Motion

In terms of the $x$ axis

TODO

$$\begin{equation} \begin{split} \Delta{x} &= x_2 - x_1 = V_{0,x} \dot t\\ v_{x} &= v_{0,x} + a_x \end{split} \end{equation}$$

TODO

In terms of the $y$ axis

TODO

$$\begin{equation} \begin{split} \Delta{y} &= y_2 - y_1 = V_{0,x} \dot t\\ v_{y} &= v_{0,y} + a_y \end{split} \end{equation}$$

TODO

In Summary

Initial Quantities

$$\begin{equation} \begin{split} v_x &= v_o\cdot\cos\theta\\ v_y &= v_o\cdot\sin\theta \end{split} \end{equation}$$

$$\begin{equation} \begin{split} a_y &= -g\\ a_x &= 0 \end{split} \end{equation}$$

Derived expressions

$$\begin{equation} \begin{split} \Delta{x} &= v_{0,x} \cdot \Delta{t}\\ v_x &= v_{0,x} \end{split} \end{equation}$$

$$\begin{equation} \begin{split} \Delta{y} &= v_{0,y} \cdot \Delta{t} - \frac{1}{2} g \Delta{t^2}\\ v_y &= v_{0,y} - g \Delta{t} \end{split} \end{equation}$$

Solutions

$$\begin{equation} \begin{split} t_{\text{top}} &= \frac{v_0\cdot\sin\theta}{g}\\ \Delta{y_{\text{max}}} &= \frac{v_0^2 + sin^2\theta}{2g}\\ \text{Range} &= \frac {2 \cdot v_0^2 \cdot \sin\theta \cdot \cos\theta} {g} \end{split} \end{equation}$$

Projectile Motion from an initial height, with given initial velocity and angle

Given

A projectile angle $\theta$
The initial height $y_0$
The initial velocity $v_0$

We can therefore derive the the initial velocities for $x$ and $y$ in terms of the given angle and initial velocity.

$$\begin{equation} \begin{split} v_{0x} &= v_0 \cdot \cos\theta\\ v_{0y} &= v_0 \cdot \sin\theta \end{split} \end{equation}$$

Given the general formulas for displacement and velocity

$$\begin{equation} \begin{split} \small \text{displacement} &= \text{initial displacement} + \text{initial velocity} \cdot \Delta{t} + \frac{1}{2}a\Delta{t^2}\\ \text{velocity} &= \text{initial velocity} + a\cdot\Delta{t} \end{split} \end{equation}$$

Which this information, we will derive specific equations in terms of the $x$ and $y$ axes governing the projectile.

In terms of the $x$ axis

Deriving displacement as a function of time

Using the general formula from above in terms of $x$ as a function of time.

$$\begin{equation} \begin{split} x(t) &= x_0 + v_{0x} t + \frac{1}{2}a_x t^2 \end{split} \end{equation}$$

Which we can simplify using the following facts

From the given depiction of the problem, we know that $x(0) = 0$.
There is no acceleration along the $x$ axis, so $a_x = 0$.
$v_{0x} = v_0\cdot\cos\theta$ as shown above.

Therefore

$$\begin{equation} \begin{split} x(t) &= 0 + v_{0x}\cdot t + \frac{1}{2}0 t^2\\ &= v_{0x} t\\ &= v_0\cos\theta\cdot t \end{split} \end{equation}$$

Deriving velocity

$$\begin{equation} \begin{split} v_x &= v_{0x} + a_x\cdot t\\ &= v_0\cdot\cos\theta + 0\cdot t\\ &= v_0\cdot\cos\theta \end{split} \end{equation}$$

In terms of the $y$ axis

Deriving displacement as a function of time

Using the general formula from above in terms of $y$ as a function of time.

$$\begin{equation} \begin{split} y(t) &= y_0 + v_{0y} t + \frac{1}{2}a_y t^2 \end{split} \end{equation}$$

Which we can simplify using the following facts

Initial height is given to us which we will represent as $y_0$, for the sake of generality.
Acceleration along the $y$ axis is the constant for gravity, so $a_y = -9.8\frac{\mathrm{m}}{\mathrm{s^2}}$.
$v_{0y} = v_0\cdot\sin\theta$ as shown above.

Therefore

$$\begin{equation} \begin{split} y(t) &= y_0 + v_{0y} t + \frac{1}{2}a_y t^2\\ &= y_0 + v_0\cdot\sin\theta\cdot t + \frac{1}{2}\left(-9.8\right) t^2\\ &= y_0 + v_0\cdot\sin\theta\cdot t - 4.9 t^2 \end{split} \end{equation}$$

Deriving velocity

$$\begin{equation} \begin{split} v_y &= v_{0y} + a_y\cdot t &= v_{0y} + g \cdot t\\ &= v_0\cdot\sin\theta - 9.8\frac{\mathrm{m}}{\mathrm{s^2}} \end{split} \end{equation}$$

In summary

$$\begin{equation} \begin{split} x(t) &= v_0\cos\theta\cdot t\;(1)\\ v_x &= v_0\cdot\cos\theta\;(2) \end{split} \end{equation}$$

$$\begin{equation} \begin{split} y(t) &= y_0 + v_0\cdot\sin\theta\cdot t + \frac{1}{2}g t^2\;(3)\\ &= y_0 + v_0\cdot\sin\theta\cdot t - 4.9 t^2\\ v_y &= v_0\cdot\sin\theta + g \cdot t\;(4)\\ &= v_0\cdot\sin\theta - 9.8\frac{\mathrm{m}}{\mathrm{s^2}} \cdot t \end{split} \end{equation}$$

To find the range

We know that at the moment of impact $y = 0$, therefore we can use equation $(3)$

$$\begin{equation} \begin{split} y = y_0 + v_0\cdot\sin\theta\cdot t + \frac{1}{2}g t^2 \end{split} \end{equation}$$

Rearranging a bit and setting $y = 0$, we can see that solving for $t$ will yield the time at which $y = 0$.

$$\begin{equation} \begin{split} 0 = \underbrace{\frac{1}{2}g}_{\text{a}}\;t^2 + \underbrace{v_0\cdot\sin\theta}_{\text{b}}\; t + \underbrace{y_0}_{\text{c}} \end{split} \end{equation}$$

Therefore

$$\begin{equation} \begin{split} t &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\ &= \frac {-v_0\cdot\sin\theta\pm\sqrt{\left(v_0\cdot\sin\theta\right)^2 - 4\left(\frac{1}{2}g\right)y_0}} {2\frac{1}{2}g}\\ &= \frac {-v_0\cdot\sin\theta\pm\sqrt{\left(v_0\cdot\sin\theta\right)^2 - 2 g y_0}} {g} \end{split} \end{equation}$$

Plugging the solution for $t$ (and ignoring the negative or non-real solutions for $t$) into $x(t)$ will yield the horizontal displacement (range) at the time $y = 0$. Therefore:

$$\begin{equation} \begin{split} \text{range}\;=x(t)\;\small\text{where $t$ is the point at which $y=0$} \end{split} \end{equation}$$

To find the maximum vertical displacement (i.e. peak height)

We begin with equation $\text{(4)}$

$$\begin{equation} \begin{split} v_y &= v_0\cdot\sin\theta + g t \end{split} \end{equation}$$

We know that at the moment our projectile crests its trajectory, the vertical component of our projectile will be zero. Therefore $v_y = 0$. To find the time, we simply solve for $t$.

$$\begin{equation} \begin{split} 0 &= v_0\cdot\sin\theta + g \cdot t\\ -v_0\cdot\sin\theta &= g \cdot t\\ \frac{-v_0\cdot\sin\theta}{g} &= t\\ \therefore\;t &= \frac{-v_0\cdot\sin\theta}{g} \end{split} \end{equation}$$

Therefore, knowing the time at which our projectile crests its trajectory, we simply plugin our solution for $t$ into the function given in equation $(4)$. I.e:

$$\begin{equation} \begin{split} y_{\text{max}} &= y_0 + v_0\cdot\sin\theta\cdot t + \frac{1}{2}g t^2\\ &= \small y_0 + v_0\cdot\sin\theta\cdot\left(\frac{-v_0\cdot\sin\theta}{g}\right) + \frac{1}{2}g \left(\frac{-v_0\cdot\sin\theta}{g}\right)^2 \end{split} \end{equation}$$

To find the velocity at a given moment of time

Given some time which we will denote as $t_n$, to find the velocity we simply plug in our given values for $\theta$ and $v_0$ into equations $(2)$ and $(4)$. I.e.

$$\begin{equation} \begin{split} v_x &= v_0\cdot\cos\theta\\ v_y &= v_0\cdot\sin\theta + g \cdot t \end{split} \end{equation}$$

With the given value for $t_n$, yielding the vector at time $t = t_n$, which we will denote as $\vec{v_n}$

$$\begin{equation} \begin{split} A &= v_y(t_n)\\ B &= v_x(t_n)\\ \end{split} \end{equation}$$

To define the vector in terms of engineering notation, (i.e. $v_x\hat{i} + v_y\hat{j}$)

$$\begin{equation} \begin{split} \vec{v_n} &= B\hat{\textbf{i}} + A\hat{\textbf{j}} \end{split} \end{equation}$$

To define the vector in terms of magnitude (which we will denote as $v_n$) and direction (which we will denote as $\theta_n$)

$$\begin{equation} \begin{split} v_n &= \sqrt{A^2 + B^2}\\ \theta_n &= \tan^{-1}\left(\frac{A}{B}\right) \end{split} \end{equation}$$

Range

The distance a projectile travels is called its range.

$$\begin{equation} \begin{split} \underbrace {\text{range}\;=\frac{v^2 \cdot sin\left(2\theta\right)}{g}} _{\small{\text{start/end elevation must be the same}}} \end{split} \end{equation}$$

Only applies in situations where the projectile lands at the same elevation from which it was fired.

Reasoning About Projectile Motion

Notes

An object is in free fall when the only force acting on it is the force of gravity.

Question

Based on the figure, for which trajectory was the object in the air for the greatest amount of time?

Answer

Trajectory A

Explanation

All that matters is the vertical height of the trajectory, which is based on the component of the initial velocity in the vertical direction ($v_0\sin\theta$). The higher the trajectory, the more time the object will be in the air, regardless of the object's range or horizontal velocity.

Problems

The function in this graph represents an object that is speeding up, or accelerating at a constant rate.

When you throw a ball directly upward, what is true about its acceleration after the ball has left your hand?

Answer: The ball’s acceleration is always directed downward.

Wrong: The ball’s acceleration is always directed downward, except at the top of the motion, where the acceleration is zero.

Question

As an object moves in the x-y plane, which statement is true about the object’s instantaneous velocity at a given moment?

Answer

The instantaneous velocity is tangent to the object’s path

Wrong

The instantaneous velocity is perpendicular to the object’s path.
The instantaneous velocity can point in any direction, independent of the object’s path.

Explanation

As an object moves in the x-y plane the instantaneous velocity is tangent to the object‘s path at a given moment. This is because the displacement vector during an infinitesimally small time interval is always directed along the object’s path and the velocity vector always has the same direction as the displacement vector.

Relative Motion

Galilean transformation of velocity

The velocity $\vec{v}$ of some object P as seen from a stationary frame must be the sum of $\vec{w}$ and $\vec{v_F}$

$$\begin{equation} \begin{split} \vec{v} &= \vec{w} + \vec{v_F} \end{split} \end{equation}$$

Where

Symbol	Description
$\vec{v}$	Velocity as measured in a stationary frame
$\vec{w}$	Velocity of an object measured in the moving frame relative to the moving frame
$\vec{v_F}$	velocity of the moving frame - with respect to the stationary frame

Galilean transformation of velocity (alternate notation)

Given two reference frames A and $B$ and some object $O$. The velocity of the object can be defined in terms of $A$ or $B$ as shown

Symbol	Description
$\vec{v_{O,A}}$	The velocity of $O$ relative to $A$
$\vec{v_{O,B}}$	The velocity of $O$ relative to $B$
$\vec{v_{A,B}}$	The velocity of $A$ relative to $B$
$\vec{v_{B,A}}$	The velocity of $B$ relative to $A$. It locates the origin of $A$ relative to the origin of $B$.

Therefore

$$\begin{equation} \begin{split} \vec{v_{O,B}} &= \vec{v_{O,A}} + \vec{v_{A,B}}\\ \vec{v_{O,A}} &= \vec{v_{O,B}} + \vec{v_{B,A}} \end{split} \end{equation}$$

Rotational Motion & Kinematics

Basics

$$\begin{equation} \begin{split} \smallText{Angular velocity} = \omega &= \tau f\\ &= \frac{\tau}{T}\\ \end{split} \end{equation}$$

$$\begin{equation} \begin{split} \smallText{Centripetal acceleration} = a_C &= \frac{v^2}{r}\\ &= \frac{\omega^2 r^2}{r}\\ &= \omega^2 r \end{split} \end{equation}$$

$$\begin{equation} \begin{split} \text{Period} = T &= \frac{1}{f} = \frac{\tau}{\omega} \end{split} \end{equation}$$

$$\begin{equation} \begin{split} \vec{v} &\perp \vec{r}\\ \vec{a} &\perp \vec{v}\\ \end{split} \end{equation}$$

$$\begin{equation} \begin{split} \left. \begin{array}{ll} \vec{a} &\parallel \vec{r}\\ \vec{a} &\propto \vec{r} \end{array} \right\}\;\smallText{They are anti-parallel} \end{split} \end{equation}$$

Auxiliary Formula Reference

$$\begin{equation} \begin{split} \text{Period} = T &= \frac{1}{f} = \frac{\tau}{\omega} \end{split} \end{equation}$$

$$\begin{equation} \begin{split} \theta &= \omega \cdot t\\ &= \omega_1 \cdot t + \frac{1}{2} \alpha t^2\\ \omega_2 &= \omega_1 + \alpha \cdot t\\ \omega_2^2 &= \omega_1^2 + 2\cdot\alpha\cdot\theta\\ v &= \smaller{\frac{1 \text{circumference}}{1 \text{period}}}\\ &= \frac{2 \pi r}{T} \end{split} \end{equation}$$

Formula	Missing	Quantities Present
$\omega_2 = \omega_1 + \alpha\Delta{t}$	$\Delta\theta$	$\omega_1$$\omega_2$$\alpha$$\Delta{t}$
$\Delta\theta = \left(\frac{\omega_2 + \omega_1}{2}\right)\Delta{t}$	$a$	$\Delta\theta$$\omega_1$$\Delta{t}$$\omega_2$
$\Delta\theta = \omega_1 \Delta{t} + \frac{1}{2}\alpha\Delta{t^2}$	$\omega_2$	$\Delta\theta$$\omega_1$$\Delta{t}$$\alpha$
$\Delta\theta = \omega_2\Delta{t} - \frac{1}{2}\alpha\Delta{t^2}$	$\omega_1$	$\Delta\theta$$\omega_2$$\Delta{t}$$\alpha$
$(\omega_2)^2 = (\omega_1)^2 + 2\alpha\Delta\theta$	$\Delta{t}$	$\Delta\theta$$\omega_1$$\omega_2$$\alpha$

$$\begin{equation} \begin{split} \underbrace{\smallText{Arc Length}}_{S} &= \underbrace{\smallText{radius}}_{r} \cdot \underbrace{\smallText{Central angle}}_{\theta}\\ S &= r\cdot\theta\\ \end{split} \end{equation}$$

$$\begin{equation} \begin{split} \underbrace{\smallText{Linear displacement}}_{\Delta{x}} &= \underbrace{\smallText{Angular displacement}}_{\theta} \cdot \underbrace{\smallText{radius}}_{r}\\ \Delta{x} &= \theta \cdot r\\ \underbrace{\smallText{Linear velocity}}_{v} &= \underbrace{\smallText{Angular Velocity}}_{\omega} \cdot \underbrace{\smallText{radius}}_{r}\\ v &= \omega \cdot r\\ \underbrace{\smallText{Linear acceleration}}_{a} &= \underbrace{\smallText{Angular acceleration}}_{\alpha} \cdot \underbrace{\smallText{radius}}_{r}\\ a &= \alpha \cdot r \end{split} \end{equation}$$

$$\begin{equation} \begin{split} \underbrace{\smallText{Angular displacement}}_{\theta} &= \underbrace{\smallText{Angular speed}}_{\omega} \cdot \underbrace{\smallText{time}}_{t}\\ \theta &= \omega \cdot t \end{split} \end{equation}$$

A particle moves with uniform circular motion if and only if its angular velocity V is constant and unchanging.

Uniform Circular Motion

Uniform means content speed

Newton's Laws of Motion

Laws

An object at rest remains at rest, or if in motion, remains in motion at a constant velocity unless acted on by a net external force.

When a body is acted upon by a force, the time rate of change of its momentum equals the force.

$$\begin{equation} \begin{split} \overbrace{\smallText{Force}}^{\mathrm{F}} &= \overbrace{\smallText{Mass}}^{m} \cdot \overbrace{\smallText{Acceleration}}^{a}\\ \therefore\; F &= m \cdot a \end{split} \end{equation}$$

Units

Name	Symbol	Unit
Newtons	$\mathrm{N}$	$\mathrm{kg}\;\cdot\frac{\mathrm{m}}{\mathrm{s^2}}$

$$\begin{equation} \begin{split} a = \frac{\overbrace{\sum \mathrm{F}}^{\mathclap{\smallText{Net force}}}}{m} \end{split} \end{equation}$$

This is the same formula as $\mathrm{F} = m a$, except we've written the force more precisely as the net force $\sum \mathrm{F}$, and we've divided both sides by the mass $m$ to get the acceleration $a$ by itself on one side of the equation.

One advantage of writing Newton's second law in this form is that it makes people less likely to think that $m \cdot a$—mass times acceleration—is a specific force on an object. The expression $m \cdot a$, is not a force, $m \cdot a$, is what the net force equals.

Looking at the form of Newton's second law shown above, we see that the acceleration is proportional to the net force, $\sum \mathrm{F}$, and is inversely proportional to the mass, $m$.

If an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A. I.e. "action equals reaction".

This law represents a certain symmetry in nature: forces always occur in pairs, and one body cannot exert a force on another without experiencing a force itself. We sometimes refer to this law loosely as action-reaction, where the force exerted is the action and the force experienced as a consequence is the reaction.

Terms

Inertia: The property of a body to remain at rest or to remain in motion with constant velocity is called inertia.; Newton's first law is often called the law of inertia; The inertia of an object is measured by its mass.; Mass can be determined by measuring how difficult an object is to accelerate. The more mass an object has, the harder it is to accelerate
Force: A force is a vector.; A force can be either a contact force or a long-range force.
Motion: By "motion", Newton meant the quantity now called momentum, which depends upon the amount of matter contained in a body, the speed at which that body is moving, and the direction in which it is moving.; In modern notation, the momentum of a body is the product of its mass and its velocity: $\vec{p} = m \cdot \vec{v}$

Notes

Question

True of false: A ball is moving upwards and to the left. A net force that points upwards and to the left must be acting on the ball?

Answer: false!

The net force points in the direction of the acceleration, not necessarily in the direction of the velocity.
If the net force points in the direction of velocity, the object will speed up. If the net force points opposite to the velocity, the object will slow down.
Since we don't know the acceleration of this ball which is moving up and left, we can't say anything for sure about the net force.

Question

True of false: A less massive object has more inertia than a more massive object?

Answer: false!

Mass is a measure of an object's inertia (its tendency to resist change in velocity). Less mass means less inertia.

Value	Prefix	Symbol
\(10^{1}\)	deca	\(\mathrm{da}\)
\(10^{2}\)	hecto	\(\mathrm{h}\)
\(10^{3}\)	kilo	\(\mathrm{k}\)
\(10^{6}\)	mega	\(\mathrm{M}\)
\(10^{9}\)	giga	\(\mathrm{G}\)
\(10^{12}\)	tera	\(\mathrm{T}\)

Value	Prefix	Symbol
\(10^{-1}\)	deci	\(\mathrm{d}\)
\(10^{-2}\)	centi	\(\mathrm{c}\)
\(10^{-3}\)	mili	\(\mathrm{m}\)
\(10^{-6}\)	micro	\(\mathrm{\mu}\)
\(10^{-9}\)	nano	\(\mathrm{n}\)
\(10^{-12}\)	pico	\(\mathrm{p}\)

Value of \(\mathrm{n}\)	Value of \(l\)	Orbital Sublevel
\(\mathrm{n} = 1\)	\(l = 0\)	\(1\mathrm{s}\)
\(\mathrm{n} = 2\)	\(l = 0\)\(l = 1\)	\(2\mathrm{s}\)\(2\mathrm{p}\)
\(\mathrm{n} = 3\)	\(l = 0\)\(l = 1\)\(l = 2\)	\(3\mathrm{s}\)\(3\mathrm{p}\)\(3\mathrm{d}\)
\(\mathrm{n} = 4\)	\(l = 0\)\(l = 1\)\(l = 2\)\(l = 3\)	\(4\mathrm{s}\)\(4\mathrm{p}\)\(4\mathrm{d}\)\(4\mathrm{f}\)

Value	Result
\(l = 0\)	\(\mathrm{s}\)
\(l = 1\)	\(\mathrm{p}\)
\(l = 2\)	\(\mathrm{d}\)
\(l = 3\)	\(\mathrm{f}\)

Value of \(l\)	Value of \(\mathrm{m_l}\)
\(l = 0\)	\(\mathrm{m_l} = 0\)
\(l = 1\)	\(\mathrm{m_l} = -1\)\(\mathrm{m_l} = 0\)\(\mathrm{m_l} = 1\)
\(l = 2\)	\(\mathrm{m_l} = -2\)\(\mathrm{m_l} = -1\)\(\mathrm{m_l} = 0\)\(\mathrm{m_l} = 1\)\(\mathrm{m_l} = 2\)
\(l = 2\)	\(\mathrm{m_l} = -3\)\(\mathrm{m_l} = -2\)\(\mathrm{m_l} = -1\)\(\mathrm{m_l} = 0\)\(\mathrm{m_l} = 1\)\(\mathrm{m_l} = 2\)\(\mathrm{m_l} = 3\)

\({\displaystyle \alpha A}\)	\alpha A
\({\displaystyle \beta B}\)	\beta B
\({\displaystyle \gamma \Gamma }\)	\gamma \Gamma
\({\displaystyle \delta \Delta }\)	\delta \Delta
\({\displaystyle \epsilon \varepsilon E\;}\)	\epsilon \varepsilon E
\({\displaystyle \zeta Z}\)	\zeta Z
\({\displaystyle \eta H}\)	\eta H
\({\displaystyle \theta \vartheta \Theta }\)	\theta \vartheta \Theta

\({\displaystyle \iota I}\)	\iota I
\({\displaystyle \kappa K}\)	\kappa K
\({\displaystyle \lambda \Lambda \;}\)	\lambda \Lambda
\({\displaystyle \mu M}\)	\mu M
\({\displaystyle \nu N}\)	\nu N
\({\displaystyle \xi \Xi }\)	\xi \Xi
\({\displaystyle oO\;}\)	o O
\({\displaystyle \pi \Pi }\)	\pi \Pi

\({\displaystyle \rho \varrho P\;}\)	\rho \varrho P
\({\displaystyle \sigma \,\!\Sigma \;}\)	\sigma \Sigma
\({\displaystyle \tau T}\)	\tau T
\({\displaystyle \upsilon \Upsilon }\)	\upsilon \Upsilon
\({\displaystyle \phi \varphi \Phi }\)	\phi \varphi \Phi
\({\displaystyle \chi X}\)	\chi X
\({\displaystyle \psi \Psi }\)	\psi \Psi
\({\displaystyle \omega \Omega }\)	\omega \Omega

Name	Symbol	Unit	Description	Range
Wavelength	\(\mathrm{\lambda}\)	Any unit for distance	Distance between two analogous points	Always Positive
Frequency	\(f\) or \(\nu\) (nu)	\(㎐ = \frac{\text{1 cycle}}{\text{second}}\)\(\mathrm{s}^{-1}\)	Number of cycles	Always Positive
Energy	\(\mathrm{E}\)	\(\mathrm{J}\) (joule)	Amount of energy (\(\mathrm{E}\)) in a light packet

Name	Symbol	Unit	Value
Speed of Light	\(\mathrm{c}\)	\(\v\)	\(\mathrm{c} = \sci{3.00}{8}\v\)
Planck's constant	\(\mathrm{h}\)	Energy multiplied by time joule-seconds \(\mathrm{J}\cdot\mathrm{s}\)	\(\mathrm{h} = \sci{6.626}{-34}\;\mathrm{J}\cdot\mathrm{s}\)

Symbol	Description
\(\mathrm{n}\)	The principle quantum number
\(l\)	The angular momentum quantum number
\(\mathrm{m}_1\)	The magnetic quantum number
\(\mathrm{m}_s\)	The spin quantum number

Formula	Missing	Quantities Present
\(v_2 = v_1 + a\Delta{t}\)	\(\Delta{x}\)	\(v_1\)\(v_2\)\(a\)\(\Delta{t}\)
\(\Delta{x} = \left(\frac{v_2 + v_1}{2}\right)\Delta{t}\)	\(a\)	\(\Delta{x}\)\(v_1\)\(\Delta{t}\)\(v_2\)
\(\Delta{x} = v_1 \Delta{t} + \frac{1}{2}a\Delta{t^2}\)	\(v_2\)	\(\Delta{x}\)\(v_1\)\(\Delta{t}\)\(a\)
\(\Delta{x} = v_2\Delta{t} - \frac{1}{2}a\Delta{t^2}\)	\(v_1\)	\(\Delta{x}\)\(v_2\)\(\Delta{t}\)\(a\)
\((v_2)^2 = (v_1)^2 + 2a\Delta{x}\)	\(\Delta{t}\)	\(\Delta{x}\)\(v_1\)\(v_2\)\(a\)

Formula	Missing	Quantities Present
\(\omega_2 = \omega_1 + \alpha\Delta{t}\)	\(\Delta\theta\)	\(\omega_1\)\(\omega_2\)\(\alpha\)\(\Delta{t}\)
\(\Delta\theta = \left(\frac{\omega_2 + \omega_1}{2}\right)\Delta{t}\)	\(a\)	\(\Delta\theta\)\(\omega_1\)\(\Delta{t}\)\(\omega_2\)
\(\Delta\theta = \omega_1 \Delta{t} + \frac{1}{2}\alpha\Delta{t^2}\)	\(\omega_2\)	\(\Delta\theta\)\(\omega_1\)\(\Delta{t}\)\(\alpha\)
\(\Delta\theta = \omega_2\Delta{t} - \frac{1}{2}\alpha\Delta{t^2}\)	\(\omega_1\)	\(\Delta\theta\)\(\omega_2\)\(\Delta{t}\)\(\alpha\)
\((\omega_2)^2 = (\omega_1)^2 + 2\alpha\Delta\theta\)	\(\Delta{t}\)	\(\Delta\theta\)\(\omega_1\)\(\omega_2\)\(\alpha\)

Symbol	Description
\(\vec{v}\)	Velocity as measured in a stationary frame
\(\vec{w}\)	Velocity of an object measured in the moving frame relative to the moving frame
\(\vec{v_F}\)	velocity of the moving frame - with respect to the stationary frame

Symbol	Description
\(\vec{v_{O,A}}\)	The velocity of \(O\) relative to \(A\)
\(\vec{v_{O,B}}\)	The velocity of \(O\) relative to \(B\)
\(\vec{v_{A,B}}\)	The velocity of \(A\) relative to \(B\)
\(\vec{v_{B,A}}\)	The velocity of \(B\) relative to \(A\). It locates the origin of \(A\) relative to the origin of \(B\).

Colbyn's School Notes Fall 2022

Navigation - Sub Sections

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Miscellaneous

Hello World

Hello World

Hello World

Hello World

Hello World

Hello World

Hello World

Hello World

Hello World

Hello World

Hello World

Hello World

The Greek Alphabet

Chemistry

Conventions

Conventions on homework

Units

SI Prefixes

Classification of Matter

Overview

Mixtures

Heterogeneous mixture

Homogeneous mixture

Quantum Mechanics

The Electromagnetic Spectrum

Overview

Formulas

Values

Constants

Other Formulas

de Broglie Relation

Heisenberg's Uncertainty Principle

Energy of an Electron in an Orbital with Quantum Number \(\mathrm{n}\) in a Hydrogen Atom

Change in Energy That Occurs in an Atom When It Undergoes a Transition between Levels

Electron Configuration

Traditional Chart

Better Method

Examples

Electron configuration for \(_{26}\mathrm{Fe}\)

Electron configuration for \(_{26}\mathrm{Fe}^{+2}\)

Electron configuration for \(_{24}\mathrm{Cr}\)

How-tos

What are the valence electrons?

Quantum Numbers

Overview

The Principle Quantum Number (\(\mathrm{n}\))

Angular Momentum Quantum Number

Summary

Useful Formulas

Examples

Light

Interference and Diffraction

Constructive Interference

Destructive Interference

Mathematics

Algebra

Function Composition And Notation Conveniences

Right to Left Evaluation

Left to Right Evaluation

Algebra Basics

Trigonometry

The Unit Circle & Special Angles In Trig

Warning

Conventions

Radians & Radian Conversion

Constants

Conversion

Geometry

The Circle

Pre-Calculus

Calculus

Function Graphs

Example

Example

Example

Derivative Formulas