July 2022 – Physicspupil

July 11, 2022October 10, 2025

202207110941 Pupil Physics

Table A: Definitions & Terms

(A001) Scalar ( $v$ ). A quantity with magnitude only. Read More

$l$

$\textrm{m}$

$m$

$\textrm{kg}$

$t$

$\textrm{s}$

$T$

$\mathrm{K}$

$I$

$\textrm{A}$

$n$

$\textrm{mol}$

$I_\textrm{V}$

$\textrm{cd}$

$[L]$

$[M]$

$[T]$

$[\Theta]$

$[I]$

$[N]$

$[J]$

(A002) Vector ( $\mathbf{v}$ , $\vec{v}$ ). A quantity with magnitude (size) and direction. Read More

vector

$\mathbf{r}=x\,\hat{\mathbf{i}}+y\,\hat{\mathbf{j}}+z\,\hat{\mathbf{k}}$

$r=|\mathbf{r}|=\sqrt{x^2+y^2+z^2}$

ii.

$\mathbf{s}$

$\mathbf{s}=\Delta \mathbf{r}=\mathbf{r_2}-\mathbf{r_1}=\mathbf{r}(t_2)-\mathbf{r}(t_1)$

iii

$\displaystyle{\mathbf{v}_{\textrm{avg}}=\frac{\Delta\mathbf{r}}{\Delta t}=\frac{\mathbf{r}(t_1+\Delta t)-\mathbf{r}(t_1)}{\Delta t}}$

$\Delta t=t_2-t_1$

iv.

$\displaystyle{\mathbf{v}\equiv \lim_{\Delta t\to 0}\frac{\Delta r}{\Delta t}=\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}=v_x\,\hat{\mathbf{i}}+v_y\,\hat{\mathbf{j}}+v_z\,\hat{\mathbf{k}}}$

$v=|\mathbf{v}|$

$\displaystyle{\mathbf{a}_{\textrm{avg}}=\frac{\Delta \mathbf{v}}{\Delta t}=\frac{\mathbf{v}(t_1+\Delta t)-\mathbf{v}(t_1)}{\Delta t}}$

vi.

$\displaystyle{\mathbf{a}=\lim_{\Delta t\to 0}\frac{\Delta \mathbf{v}}{\Delta t}=\frac{\mathrm{d}\mathbf{v}}{\mathrm{d} t}=\frac{\mathrm{d}^2\mathbf{r}}{\mathrm{d}t^2}}$

(A003) Distance. How far something travels, a scalar. Read More

$\mathrm{m}$

$[L]$

(A004) Displacement ( $\mathbf{s}$ , $\vec{s}$ ). How far something travels in a given direction, a vector. Read More

vector

$\mathbf{s}$

$s=|\mathbf{s}|$

scalar

$\mathrm{m}$

$[L]$

(A005) Speed. How fast something is moving, a scalar. Read More

$\displaystyle{\textrm{speed}=\frac{\textrm{distance}}{\textrm{time}}}$

$\mathrm{m\, s^{-1}}$

$\displaystyle{\frac{[L]}{[T]}}$

(A006) Velocity ( $\mathbf{v}$ , $\vec{v}$ ). How fast something is moving in a given direction, a vector. Read More

Velocity

$v$

displacement

$\mathbf{s}$

$t$

$\leftrightarrow$

displacement

$\leftrightarrow$

distance

vector

$\mathbf{v}$

$\mathrm{m\,s^{-1}}$

$\displaystyle{\frac{[L]}{[T]}}$

$v=|\mathbf{v}|$

scalar

$\displaystyle{\textrm{velocity }v=\frac{\textrm{displacement } s}{\textrm{time }t}}\quad \textrm{\scriptsize{OR}}\quad \boxed{\displaystyle{\mathbf{v}=\frac{\mathrm{d}\mathbf{s}}{\mathrm{d}t}}}$

(A007) Acceleration ( $\mathbf{a}$ , $\vec{a}$ ). The rate at which the velocity changes during a given amount of time, a vector denoted by $\mathbf{a}$ . Read More

Its unit being $\mathrm{m\,s^{-2}}$ in dimensions $\displaystyle{\frac{[L]}{[T]^2}}$ and its magnitude $a=|\mathbf{a}|$ a scalar, $\displaystyle{a=\frac{\mathrm{d}v}{\mathrm{d}t}=\frac{\mathrm{d}^2s}{\mathrm{d}t^2}}\qquad \textrm{\scriptsize{OR}}\qquad \boxed{\displaystyle{\mathbf{a}=\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}=\frac{\mathrm{d}^2\mathbf{s}}{\mathrm{d}t^2}}}$

(A008) Free Fall. The movement of an object in response to a gravitational attraction.

(A009) Projectile. An object that moves through space acted upon only by the Earth’s gravity.

(A010) Rectilinear Motion. ( $\ni$ straight-line, one-dimensional, axial) Read More

For uniformly accelerated motion, i.e., $\mathbf{a}=\textrm{Const.}$ , the equations of rectilinear motion are

$\boxed{\begin{aligned}\mathbf{v} & =\mathbf{u}+\mathbf{a}t \\\mathbf{s} & =\mathbf{u}t+\frac{1}{2}\mathbf{a}t^2\\\mathbf{v}^2 & =\mathbf{u}^2+2\mathbf{a}\mathbf{s} \\\end{aligned}}$

(A011) Curvilinear Motion. ( $\ni$ parabolic, projectile, planar) Read More

For uniformly accelerated motion, i.e., $\mathbf{a}=\textrm{Const.}$ , the equations of curvilinear motion are

$\boxed{\begin{aligned}&\quad\, \forall\, i\in\{ x,y\} \\\mathbf{v}_i & =\mathbf{u}_i+\mathbf{a}_it \\\mathbf{s}_i & =\mathbf{u}_it+\frac{1}{2}\mathbf{a}_it^2\\\mathbf{v}_i^2 & =\mathbf{u}_i^2+2\mathbf{a}_i\mathbf{s}_i \\\end{aligned}}$

where

$\boxed{\begin{aligned}\mathbf{s}(t) & = s_x(t)\,\hat{\mathbf{i}} + s_y(t)\,\hat{\mathbf{j}} \\\mathbf{u}(t) & = u_x(t)\,\hat{\mathbf{i}} + u_y(t)\,\hat{\mathbf{j}} \\\mathbf{v}(t) & = v_x(t)\,\hat{\mathbf{i}} + v_y(t)\,\hat{\mathbf{j}} \\\mathbf{a}(t) & = a_x(t)\,\hat{\mathbf{i}} + a_y(t)\,\hat{\mathbf{j}} \\\end{aligned}}$

such that

$\boxed{\begin{aligned}s & = |\mathbf{s}| \\& = \sqrt{(s_x)^2+(s_y)^2} \\& = \sqrt{s_x^2+s_y^2} \\u & = |\mathbf{u}| \\& = \sqrt{(u_x)^2+(u_y)^2} \\& = \sqrt{u_x^2+u_y^2} \\v & = |\mathbf{v}| \\& = \sqrt{(v_x)^2+(v_y)^2} \\& = \sqrt{v_x^2+v_y^2} \\a & = |\mathbf{a}| \\& = \sqrt{(a_x)^2+(a_y)^2} \\& = \sqrt{a_x^2+a_y^2} \\\end{aligned}}$

(A012) Rotational Motion. ( $\ni$ uniform circular, orbital, planetary) Read More

$\alpha$

$\omega = \omega_0+\alpha t$

$\theta =\omega_0t+\frac{1}{2}\alpha t^2$

$\omega^2 = \omega_0^2+2\alpha\theta$

$\theta$

$\omega$

$t$

$\textrm{mass }m\leftrightarrow\textrm{rotational inertia }I=k\cdot mR^2$
$\textrm{acceleration }a=\alpha R\leftrightarrow\textrm{angular acceleration }\alpha =a/R$
$\textrm{velocity }v=\omega R\leftrightarrow\textrm{angular velocity }\omega =v/R$
$\textrm{displacement }s=\theta R\leftrightarrow\textrm{angular displacement }\theta =s/R$
$\textrm{force }F=ma\leftrightarrow\textrm{torque }\tau =I\alpha$
$\textrm{work }W=Fs\leftrightarrow\textrm{work }W=\tau\theta$
$\textrm{KE}=\frac{1}{2}mv^2\leftrightarrow\textrm{KE}=\frac{1}{2}I\omega^2$
$\textrm{momentum }p=mv\leftrightarrow\textrm{angular momentum }L=I\omega$

Circular motion examples. Aircraft turning a flight, vehicles rounding bends with or without banking, loop motion, satellite orbits, centrifuge, etc.

(A013) Periodic Motion. ( $\ni$ simple harmonic, pendulum, oscillating) Read More

Periodic motion is motion repeated in equal time interval, period. Oscillation is the back-and-forth motion about a fixed point called the equilibrium position, e.g., simple, compound, or torsion pendulum, spring-mass, and alternating current. If there is no damping force disturbing the body, it oscillates forever at the natural frequency $f$ that is characteristic of the system. If a system whose acceleration $\mathbf{a}$ is in magnitude proportional to displacement $\mathbf{s}$ , and yet in opposite direction to it, i.e., $a=|\mathbf{a}|\propto |\mathbf{s}|=s$ and $\mathbf{\hat{a}}=-\mathbf{\hat{s}}$ , it is said to be in a simple harmonic motion.

Hooke’s law states that the restoring force $F$ is proportional to the displacement from its equilibrium position:

$F=-kx$

(A014) Force ( $F$ ). A push or a pull, vector denoted by $\mathbf{F}$ . Read More

Its magnitude being a scalar $F=|\mathbf{F}|$ , the unit being Newton $N$ , or $\mathrm{kg\, m\, s^{-2}}$ , in dimensions $\displaystyle{\frac{[M][L]}{[T]^2}}$ . E.g., gravitational force, electromagnetic force, frictional force, viscous force, upthrust, etc. The properties of force are implicitly stated in Newton’s Laws of Motion:

- Newton’s 1^st Law. A body remains at rest, or in motion at a constant speed in a straight line, unless acted upon by a force.
  - Also called “Law of Inertia” as discovered by Galileo.
  - Inertia, aka vis insita or innate force of matter, is the reluctance of a body to change its state of rest or motion.
  - Inertia is quantified by mass.
- Newton’s 2^nd Law. When a body is acted upon by a force, the time rate of change of its momentum equals the force.
  - $\displaystyle{\textrm{Net }F = \frac{\mathrm{d}p}{\mathrm{d}t} = \frac{\mathrm{d}(mv)}{\mathrm{d}t} = m\cdot \frac{\mathrm{d}v}{\mathrm{d}t} + \frac{\mathrm{d}m}{\mathrm{d}t}\cdot v= ma}$
- Newton’s 3^rd Law. If two bodies exert forces on each other, these forces have the same magnitude but opposite directions.
  - For every action, there is an equal but opposite reaction.
  - $\text{}_AF_B = \text{}_BF_A$

(A015) Friction ( $f$ ). The force that acts to oppose the motion between two materials moving past each other. Read More

Static (resp. kinetic) friction $f_{\textrm{s}(\textrm{k})}$ is given by $f_{\textrm{s, max}}=\mu_\textrm{s}N$ (resp. $f_{\textrm{k}}=\mu_\textrm{k}N$ ). By definition, friction $f=f(\mu ,N)$ is $\textrm{\scriptsize{ONLY}}$ dependent on the material properties ( $\mu$ ) of contact surfaces and the normal reaction ( $N$ ), $\textrm{\scriptsize{BUT}}$ independent of the speed ( $v$ ) of motion and the area ( $A$ ) of contact.

(A016) Static Friction ( $f_\textrm{s}$ ). The resistance force that must be overcome to start an object in motion, a vector. Read More

$0\le f_\textrm{s}\le f_{\textrm{s, max}}$

$f_{\textrm{s, max}}=\mu_\textrm{s}N$

$\mu_\textrm{s}$

$N$

When net force $F (=f_\textrm{s})\le f_{\textrm{s, max}}$ the object will not move. But if net force $F>f_{\textrm{s, max}}$ it will be moved.

(A017) Kinetic Friction ( $f_\textrm{k}$ ). The resistance force between two surfaces already in motion, a vector. Read More

$f_\textrm{k}=\mu_\textrm{k}N$

$\mu_\textrm{k}$

$N$

(A018) Statics.The study of forces in equilibrium, i.e., $\textrm{\scriptsize{NO}}$ rotation $\textrm{\scriptsize{NO}}$ acceleration. Read More

For static equilibrium, i. the net force acting on the object must be zero, i.e., $\sum F_x=0$ , $\sum F_y=0$ , and $\sum F_z=0$ ; and ii. the net torque acting on the object must be zero $\sum \tau =0$ .

A couple, torque, or moment of force satisfies $\textrm{\scriptsize{NOT}}$ ii. $\textrm{\scriptsize{BUT}}$ i.

(A019) Dynamics. The study of cause and effect of motions, namely, of the agent force (contact) and the field force (non-contact).

(A020) Kinematics. The study of motions proper to the movements, being concerned with displacement $\mathbf{s}(t)$ , velocity $\mathbf{v}(t)$ , acceleration $\mathbf{a}(t)$ and time $t$ only, without reference to force $F$ or mass $m$ .

(A021) Pressure. The force per unit area. Read More

Cf. Archimedes’s principle, Pascal’s principle, Bernoulli’s principle, and Stokes’s law.

(A022) Momentum ( $\mathbf{p}$ , $p$ ). A measure of how difficult it is to stop a moving object, aka “quantity of motion” by Newton, a vector. Read More

The product of mass (aka quantity of matter) and velocity.

(A023) Impulse ( $\mathbf{Imp}$ , $\boldsymbol{J}$ , $J$ , $\Delta \mathbf{p}$ , $\Delta p$ ). The product of the force exerted on an object and the time interval during which it acts, or, in essence, the change in momentum. Impulse is a vector. Read More

I.e.

$\mathbf{Imp}=\Delta\mathbf{p}=m\mathbf{v}-m\mathbf{u}=\mathbf{F}_{\textrm{avg}}\Delta t$

$\mathrm{N\, s}$

$\mathrm{kg\, m\, s^{-1}}$

(A024) Elastic Collision. A collision in which objects collide and bounce apart with no energy loss. Read More

$\textrm{\scriptsize{AND}}$

$\textrm{CoM: }m_1\mathbf{u}_1+m_2\mathbf{u}_2=m_1\mathbf{v}_1+m_2\mathbf{v}_2$ ; $\textrm{CoE: }\displaystyle{\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2}$

You might wish to further know of Newton’s law of restitution:

$\mathbf{v}_2-\mathbf{v}_1 = -e(\mathbf{u}_2-\mathbf{u}_1)$

For two bodies impinging directly or obliquely, their relative velocity (wrt to one) after impact is equal to $e$ times that before impact in the opposite direction, the constant $e\in [0,1]$ being the coefficient of restitution ranging from $0$ (perfectly inelastic) to $1$ (perfectly elastic).

(A025) Inelastic Collision. A collision in which objects collide and some mechanical energy is transformed into thermal energy. Read More

$\textrm{\scriptsize{YES}}$

$\textrm{\scriptsize{NO}}$

$\checkmark\enspace\textrm{CoM: }m_1\mathbf{u}_1+m_2\mathbf{u}_2=m_1\mathbf{v}_1+m_2\mathbf{v}_2$ $\times\enspace\textrm{CoE: }\displaystyle{\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2\neq \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2}$

Perfectly inelastic collision in addition is such that the colliding objects stick together after they hit each other.

$\checkmark\enspace\textrm{CoM: }m_1\mathbf{u}_1+m_2\mathbf{u}_2=(m_1+m_2)\mathbf{v}_3=m_3\mathbf{v}_3$ $\times\enspace\textrm{CoE: }\displaystyle{\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2 > \frac{1}{2}(m_1+m_2)v_3^2=\frac{1}{2}m_3v_3^2}$

Super inelastic collision as an aside is such that one object explodes into granules gaining kinetic energy by losing potential energy.

$\checkmark\enspace\textrm{CoM: }m_1\mathbf{u}_1+m_2\mathbf{u}_2=(m_1+m_2)\mathbf{v_3}=m_3\mathbf{v_3}$ $\times\enspace\textrm{CoE: }\displaystyle{\frac{1}{2}Mu^2 = \frac{1}{2}\Bigg(\sum_{i=1}^{N}m_i\Bigg) u^2 < \frac{1}{2}\sum_{i=1}^{N}m_iv_i^2}$

Nevertheless, in the absence of external force, $\textrm{\scriptsize{TOTAL}}$ energy of a closed system is conserved, as stated by the law of conservation of energy.

(A026) Work ( $W$ ). The product of the component of the force exerted on an object in the direction of displacement and the magnitude of the displacement, a scalar. Read More

$\boxed{W=\mathbf{F}\cdot\mathbf{s}=Fs\cos\theta}$

(A027) Power ( $P$ ).The rate at which work is done. Read More

Generally, $\boxed{\displaystyle{P=\frac{E}{t}}}$ ; for mechanical power, $\boxed{P=Fv}$ ; and for power in a circuit, $\boxed{\displaystyle{P=IV=I^2R=\frac{V^2}{R}}}$

(A028) Energy ( $E$ ). The ability to do work.

(A029) Potential Energy ( $\textrm{PE}$ , $E_{\textrm{p}}$ , $U$ , $V$ ). Energy of position, or stored energy. Read More

$\boxed{\displaystyle{E_\textrm{p}=mgh}}$

(A030) Kinetic Energy ( $\textrm{KE}$ , $E_{\textrm{k}}$ , $T$ ). Energy of motion. Read More

$\boxed{\displaystyle{E_\textrm{k}=\frac{1}{2}mv^2}}$

(A031) Machine. A device that helps to do work by changing the magnitude or direction of the applied force. E.g., lever, pulley, and incline.

(A032) Efficiency. The ratio of the work output to the work input.

(A033) Period ( $T$ ). The time it takes for one full rotation or revolution of an object, and also the time it takes for a vibrating object to repeat its motion. Read More

$\displaystyle{\textrm{period }(T)=\frac{1}{\textrm{frequency }(f)}}$

(A034) Frequency ( $f$ ). The number of rotations or revolutions per unit time, and also the number of vibrations made per unit time. Read More

$\displaystyle{\textrm{frequency }(f)=\frac{1}{\textrm{period }(T)}}$

(A035) Torque ( $\boldsymbol{\tau}$ , $\tau$ ). A measurement of the tendency of a force to produce a rotation about an axis. Read More

Torque is the moment of force, e.g. $\tau =rF$ , where $r$ is the perpendicular distance of a point, say $O$ , from $F$ .

(A036) Center of Gravity ( $\textrm{CG}$ ). The point on any object that acts like the place at which all the weight is concentrated.

(A037) Moment of Inertia ( $I$ ). The resistance of an object to changes in its rotational motion.

(A038) Angular Momentum ( $L$ ). The measure of how difficult it is to stop a rotating object, a (pseudo-)vector. Read More

Angular momentum is defined the moment of momentum, e.g., $L=rp=r(mv)$ , where $r$ is the perpendicular distance of a point, say $O$ , from $v$ .

(A039) Law of Universal Gravitation. Every particle attracts every other particle with a force that is proportional to the mass of the particles and inversely proportional to the square of the distance between them. Read More

I.e., $\boxed{\displaystyle{F_G=G\frac{m_1m_2}{r^2}}}$ where $G$ is the universal gravitational constant.

(A040) Escape Speed. The minimum speed an object must possess in order to escape from the gravitational pull of a body. Read More

$\textrm{KE}=\textrm{PE}$

$\textrm{\scriptsize{OR}}$

$\displaystyle{\frac{1}{2}mv_e^2=G\frac{Mm}{r}}$

$\textrm{\scriptsize{OR}}$

$v_e=\displaystyle{\sqrt{\frac{2GM}{r}}}$

(A041) Density ( $\rho$ ). A measure of how much mass occupies a given space. Read More

$\displaystyle{\textrm{Density }(\rho )= \frac{\textrm{mass }(m)}{\textrm{volume }(V)}}$

(A042) Stress. The force exerted on an area divided by the area. Read More

$\displaystyle{\textrm{Stress }(\sigma )=\frac{\textrm{Force }(F)}{\textrm{Cross-sectional area }(A)}}$

Its unit is either $\mathrm{N\, m^{-2}}$ or $\mathrm{Pa}$ .

(A043) Strain. The ratio of change in dimension to original dimension. Read More

$\displaystyle{\textrm{Strain }(\varepsilon)=\frac{\textrm{extension }(e)}{\textrm{original length }(l)}}$

It has no unit.

(A044) Temperature. A quantity that you can measure with a thermometer. Read More

Cf. average kinetic energy, $\textrm{KE}_\textrm{avg}$ .

(A045) Heat ( $Q$ ). The transfer of energy between two objects that differ in temperature. Read More

For energy transfer by conduction, the rate is given by

$\displaystyle{\frac{Q}{t}=\kappa\frac{A(T_H-T_C)}{d}}$

(A046) Specific Heat. A measure of the amount of heat needed to raise the temperature of $1\,\mathrm{kg}$ of a substance by $1\,^\circ\mathrm{C}$ . Read More

$\boxed{\displaystyle{c=\frac{Q}{m\Delta T}}}$

(A047) Latent Heat of Fusion. The quantity of heat needed per kilogram to melt a solid (or solidify a liquid) at a constant temperature and atmospheric pressure. Read More

$\displaystyle{L_f=\frac{Q}{m}}$

(A048) Latent Heat of Vaporization. The quantity of heat needed per kilogram to vaporize a liquid (or liquidize a gas) at a constant temperature and atmospheric pressure. Read More

$\displaystyle{L_v=\frac{Q}{m}}$

(A049) Doppler Effect. A change in the apparent frequency of sound due to the motion of the source ( $\textrm{s}$ ) or the observer ( $\textrm{o}$ ). Read More

$\displaystyle{f'=\bigg( \frac{1\pm u_o/v}{1\mp u_s/v}\bigg) f}$

(A050) Reflection. The bouncing of light. Read More

$\theta_i =\theta_r$

(A051) Mirror/Thin Lens Equation. Read More

$\boxed{\displaystyle{\frac{1}{\textrm{focal length }(f)}=\frac{1}{\textrm{object distance }(u)}+\frac{1}{\textrm{image distance }(v)}}}$

(A052) Wave Equation. Read More

$\boxed{y(x,t)=y_m\sin (kx-\omega t+\phi) }$

where $k=\frac{2\pi}{\lambda}$ is the angular wave number, $\omega =2\pi f$ the frequency, and $\phi$ the phase angle.

(A053) Refraction. The change in direction of light due to a change in speed as it passes from one medium to another. Read More

$\boxed{\displaystyle{\frac{\sin\theta_1}{\sin\theta_2}=\frac{v_1}{v_2}=\frac{n_2}{n_1}}}$

(A054) Diffraction. The spreading of a wave as it passes around an obstacle or through an opening. Read More

The diffraction grating equation is

$\boxed{d\sin\theta =n\lambda}$ .

(A055) Interference. When two waves overlap to produce one new wave. Read More

Cf. constructive (/superposition) as light fringes and increased intensity of sound; destructive (/neutralization) as dark fringes and silence.

Fringe separation in double-slit interference is given by

$\boxed{\displaystyle{\Delta y=\frac{\lambda D}{a}}}$ .

(A056) Electrostatics. The study of electric charges at rest, electric forces in equilibrium, and electric field under invariance.

(A057) Coulomb’s Law. Two charged objects attract each other with a force that is proportional to the charge on the objects and inversely proportional to the square of the distance between them. Read More

$\boxed{\displaystyle{F=\frac{Qq}{4\pi\varepsilon_0r^2}}}$

(A058) Electric Field. An area of influence around a charged object. The magnitude of the field is proportional to the amount of electrical force exerted on a positive test charge placed at a given point in the field. Read More

Electric field strength due to a point charge:

$\boxed{\displaystyle{E=\frac{Q}{4\pi\varepsilon_0r^2}}}$

Electric field between parallel plates:

$\boxed{\displaystyle{E=\frac{V}{d}}}$

(A059) Potential Difference. The work done to move a test charge (resp. mass) from one location to another, denoted by $\Delta V_e$ (resp. $\Delta V_g$ ).

(A060) Current ( $I$ ). The amount of charge that passes through an area in a given amount of time. Read More

$\boxed{\displaystyle{I=\frac{Q}{t}}}$

Current flows from point $A$ to point $B$ iff $\exists\,\Delta V=V_A-V_B>0$ .

(A061) Resistance ( $R$ ). An opposition to the flow of charge. Read More

$\displaystyle{R=\frac{V}{I}}$

(A062) Capacitor. A device that stores charge on conductors that are separated by an insulator. Read More

The capacitance $C$ of a capacitor is defined the amount of electric charges $Q$ stored per unit volt $V$ .

$\displaystyle{C=\frac{Q}{V}}$

Its unit being $\mathrm{C\, V^{-1}}$ or Farad ( $\mathrm{F}$ ).

(A063) Inductor. A device that stores energy to oppose the current flowing through it. Read More

$\displaystyle{L:=\frac{\Phi_\mathbf{B}}{I}}$

(A064) Kirchhoff’s Laws. Read More

Voltage (/loop) rule: The algebraic sum of the changes in potential encountered in a complete traversal of any loop of a circuit must be zero. I.e., $\Sigma V=0$
Current (/junction) rule: The sum of the currents entering any junction must be equal to the sum of the currents leaving that junction. I.e., $\Sigma I=0$
(Note: sign convention applies.)
In sum, $\Sigma [\textrm{emf}]=\Sigma [IR]$ .

(A065) Magnetic Field. An area of influence around a moving charge. The size of the field is related to the amount of magnetic force experienced by the moving charge when it is at a given location in the field. Read More

Magnetic field due to a long straight wire:

$\boxed{\displaystyle{B=\frac{\mu_0I}{2\pi r}}}$

Magnetic field inside a long solenoid:

$\boxed{\displaystyle{B=\frac{\mu_0NI}{l}}}$

Magnetic field at the centre of a toroid:

$\boxed{\displaystyle{B=\frac{\mu_0NI}{2\pi r}}}$

(A066) Flux. The number of field lines passing through a given area.

(A067) Faraday’s Law. If the flux through a given area changes over time, a voltage will be induced in the wire and a current will momentarily flow. If the number of turns is increased, the voltage will increase proportionally. Read More

$\displaystyle{E=-N\frac{\mathrm{d}\Phi}{\mathrm{d}t}}$

(A068) Lenz’s Law. An induced voltage always produces a magnetic field that opposes the field that originally produced it. Read More

By $\displaystyle{\Psi = \oint_{\Sigma} \mathbf{B}\cdot\mathrm{d}\mathbf{A}}$ , write $\displaystyle{\varepsilon =\frac{\mathrm{d}\Psi}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}(\mathbf{B}\cdot\mathbf{A})=\frac{\mathrm{d}}{\mathrm{d}t}(BA\cos\theta )}$ where $B(t)$ , $A(t)$ , and $\theta (t)$ might some or other be constant(s).

E.g. As translational motion in a $\mathbf{B}$ -field might be horizontal, i.e., $\theta =0$ or at an angle $\theta$ , in either case constant, so a current-carrying rod of length $l$ sweeping at a speed $v$ an area $A(t)=l\cdot vt$ will induce an electromotive force (emf) $\varepsilon =Blv$ . Similarly, a current-carrying rod rotating horizontally ( $\theta =0$ ) at angular speed $\omega$ about its centre will sweep an area $A(t)=\displaystyle{\frac{(l/2)^2\theta}{2}=\frac{l^2\omega t}{8}}$ $\textrm{\scriptsize{AS}}$ $\displaystyle{\omega =\frac{2\pi}{T}}$ $\textrm{\scriptsize{SO}}$ $\displaystyle{A(t)=\frac{l^2\pi t}{4T}}$ , and thus $\displaystyle{\varepsilon =\frac{B\pi l^2}{4T}}$ .

(A069) Transformer. A device that produces a change in voltage in an alternating current circuit. Read More

The ratio of secondary voltage to primary voltage in a transformer is given by

$\boxed{\displaystyle{\frac{V_{\mathrm{s}}}{V_{\mathrm{p}}}\approx\frac{N_\mathrm{s}}{N_\mathrm{p}}}}$

(A070) Quantum. A packet of energy that exhibits both particle and wave properties. Read More

The macrostate of a particle (say, electron, photon, neutron) is a physically measurable phenomenon theorized by a wavefunction $\Psi =\Psi (\mathbf{r},t)$ , the probability density $|\Psi |^2$ of which satisfies i. microstate superposition in that $\Psi =c_1\Psi_1+c_2\Psi_2+\cdots +c_n\Psi_n$ ; ii. probablistic normalization such that $\displaystyle{\int_{-\infty}^{\infty}|\Psi |^2\,\mathrm{d}^3x=1}$ ; and iii. boundary conditions by that $\displaystyle{\int_{-x}^{x}|\Psi |^2\,\mathrm{d}^3x=}\textrm{ finite}$ and $\Psi (x=\infty ,t)=0$ . To find some particle somewhere in the universe, we do space-and-time integration over the probability density, i.e., $\displaystyle{\int_{-\infty}^{\infty}}\int_{-\infty}^{\infty}|\Psi |^2\,\mathrm{d}^3x\,\mathrm{d}t$ , however easier said than done. We hope it thus much time-independent and evolves in the simplest one dimension, say $\hat{\imath}$ . The probability of finding a particle in some observable state $|A\rangle$ (e.g. position, momentum, energy) is so called the probability amplitude $\langle A|\Psi\rangle$ . The measurement order does matter, viz., $[\hat{x},\hat{p}]\neq [\hat{p},\hat{x}]$ being non-commutative. As for the dynamics of quantum system, one should read Schrödinger equation.

(A071) De Broglie Wavelength ( $\lambda$ ). The effective wavelength of a moving particle. Read More

$\boxed{\displaystyle{\lambda=\frac{h}{p}=\frac{h}{mv}}}$

(A072) Radioactivity. E.g., alpha ( $\alpha$ -) decay, beta ( $\beta$ -) decay, and gamma ( $\gamma$ -) decay. Read More

Alpha decay. ${}^{A}_{Z}\textrm{N}\to {}^{A-4}_{Z-2}\textrm{N}+{}^{4}_{2}\textrm{He}$ ,

Beta decay. $\textrm{n} =\textrm{p}+\textrm{e}^{-}$ ,

Gamma decay. ${}^{A}_{Z}\textrm{N'}\to {}^{A}_{Z}\textrm{N}+\gamma$ ,

(A073) Activity. The rate at which a radioactive sample decays. Read More

Activity and the number of undecayed nuclei are related by

$\boxed{A=\lambda N}$

(A074) Decay Constant, Lambda ( $λ$ ). The probability of disintegration per unit time. Read More

The law of radioactive decay is given by

$\boxed{N=N_0e^{-\lambda t}}$

(A075) Half-life ( $t_{1/2}$ ). The time it takes for half of a radioactive sample to decay. Read More

Half-life and decay constant are related by

$\boxed{\displaystyle{t_{1/2}=\frac{\ln 2}{\lambda}}}$

TABLE B: Theorems & Practicums

(B001) Logic Gates. Read More

(by analogy with switches)

$\textrm{AND}$ -gate: The bulb emits light if and only if switches $A$ and $B$ are both closed.

$\textrm{OR}$ -gate: The bulb emits light if either switch $A$ or switch $B$ is closed.

$\textrm{NOR}$ -gate: The bulb emits light if neither switch $A$ nor switch $B$ is closed.

$\textrm{NOT}$ -gate: The bulb emits light if switch $A$ is not closed.

$\textrm{NAND}$ -gate: The bulb does not emit light if both switch $A$ and switch $B$ are closed.

(B002) Work-Energy Theorem. Read More

$W=\Delta E_{\textrm{k}}=-\Delta E_{\textrm{p}}$

(B003) Minimum Total Potential Energy Principle. Read More

There is by nature a tendency to stability than ability.

(B004) Gradient, Divergence, and Curl. Read More

$\boxed{\begin{aligned}\textrm{grad}\, f & = \nabla f \\\textrm{div}\, \mathbf{F} & = \nabla \cdot \mathbf{F}\\\textrm{curl}\, \mathbf{F} & = \nabla \times \mathbf{F} \\\boldsymbol{\Delta} [\cdots ] & = \nabla^2 [\cdots ] \\\end{aligned}}$

E.g., conservative force as the negative gradient of potential: $\mathbf{F}=-\nabla U$ ; outward flux as the positive divergence of field; the curls of $\mathbf{E}$ -field and $\mathbf{B}$ -field as in Faraday’s law and Ampère’s law; and the Laplacian as in Schrödinger equation.

(B005) Principle of Least Action. Read More

There is by nature a tendency to the optimum than the extremum.

(B006) Heisenberg Uncertainty Principle. Read More

$\boxed{ \Delta x\Delta p \ge \displaystyle{\frac{\hbar}{2}} }$ $\textrm{\scriptsize{OR}}$ $\Delta x(F\Delta t) \ge \displaystyle{\frac{\hbar}{2}}$ $\textrm{\scriptsize{OR}}$ $(F\Delta x)\Delta t \ge \displaystyle{\frac{\hbar}{2}}$ $\textrm{\scriptsize{OR}}$ $\boxed{\Delta E\Delta t\ge \displaystyle{\frac{\hbar}{2}}}$

(B007) Noether’s Theorem. Read More

$\begin{array}{|c|c|}\hline\textrm{invariance} & \textrm{conservation} \\\hline\textrm{spatial translation} & \textrm{linear momentum} \\\textrm{spatial rotation} & \textrm{angular momentum} \\\textrm{time} & \textrm{energy} \\\hline\end{array}$

(B008) Wave-Particle Duality. Read More

Wave manifests itself in reflection, refraction, interference, diffraction, and polarisation whereas particle exhibits reflection, refraction, and photoelectric effect. Both wave-like and particle-like properties of a quantum, e.g., photon, were demonstrated by experiments with photoelectric effect and electron diffraction.

Einstein’s photoelectric equation:

$\boxed{\displaystyle{\frac{1}{2}m_ev_{\textrm{max}}^2=hf-\phi}}$

(B009) Observational Error. Read More

For a reading $\boxed{\underbrace{R}_{\textrm{best estimate}}\pm \underbrace{\Delta R}_{\textrm{uncertainty}}}$ , its absolute error is $\boxed{\Delta R}$ , its fractional error $\boxed{\displaystyle{\frac{\Delta R}{R}}}$ , and its percentage error $\boxed{\displaystyle{\frac{\Delta R}{R}\times 100\%}}$ .

$\begin{array}{|l|l|l|}\hline\textrm{addition} & R=mA+nB & \Delta R=|m|\Delta A+|n|\Delta B \\\textrm{subtraction} & R = mA - nB & \Delta R=|m|\Delta A+|n|\Delta B \\\textrm{scaling} & R=mA & \Delta R=|m|\Delta A \\\textrm{power} & R=kA^m & \frac{\Delta R}{R}=|m|\frac{\Delta A}{A} \\\textrm{product} & R=kA^m\times B^n & \frac{\Delta R}{R}=|m|\frac{\Delta A}{A}+|n|\frac{\Delta B}{B} \\\textrm{quotient} & R=kA^m\div B^n & \frac{\Delta R}{R}=|m|\frac{\Delta A}{A}+|n|\frac{\Delta B}{B} \\\textrm{others} & R=\sin x,\ln x,\textrm{ etc.} & \Delta R=\frac{R_{\textrm{max}}-R_{\textrm{min}}}{2} \\\textrm{specials} & R=\sqrt{A} &\Delta R=\sqrt{A\pm\Delta A}=\sqrt{A}\pm\frac{\Delta A}{2\sqrt{A}} \\\hline\end{array}$

(B010) Fundamental Theorem of Calculus. Read More

Let $f$ be a function continuous on $[a,x]$ if $x\ge a$ (or $[x,a]$ if $x\le a$ ). The function $\displaystyle{F(x)=\int_a^xf(t)\,\mathrm{d}t}$ is a primitive function of $f(x)$ , i.e.,

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}x}\int_a^xf(t)\,\mathrm{d}t=f(x)}$ .

For any primitive function $F(t)$ of $f(t)$ ,

$\displaystyle{\int_a^xf(t)\,\mathrm{d}t=[F(t)]_{a}^{x}=F(x)-F(a)}$

TABLE C: Rules &TOOLS

(C001) Exemplification.

Ex. 1 of 3, Mechanics Read More

On a slope of inclination angle $30^\circ$ , a block of mass $m$ is kept at rest by static friction $f_\textrm{s}\, (=?)$ . At time $t=0$ applied to the block is a force of magnitude $F\, (=?)$ starting it in motion with the minimally possible velocity of magnitude $u\, (=?)$ . The block meets with kinetic friction $f_\textrm{k}\, (=?)$ along the displacement. After some time $\Delta t_1\, (=?)$ , it has travelled a distance of $s_1\, (=?)$ long to the slope bottom, its velocity being then of magnitude $v\, (=?)$ . The block moves further $s_2\, (=?)$ for a duration of $\Delta t_2\, (=?)$ on the ground level until it stops.

Ex. 2 of 3, Mechanics Read More

Assume that the collision between two billiard balls of identical mass $m_1=m_2$ conserve momentum and energy, and that the connecting strings be inextensible and massless. At $t=t_0$ , ball $m_1$ hanging at an angle $\alpha$ is released from rest to the lowest level. It then bombards with ball $m_2$ , such that at time $t_1\, (=?)$ , they bounce off each other with linear speed $v_1\, (=?)$ and $v_2\, (=?)$ and angular speed $\dot{\beta}\, (=?)$ . At time $t_2\, (=?)$ , they make with the vertical angles $\gamma\, (=?)$ and $\theta\, (=?)$ and stop motion there temporarily.

Ex. 3 of 3, Mechanics Read More

One artificial satellite-to-be of mass $m$ is launched from the ground $r=R$ into an ideally empty space at a height of some $h\,\textrm{(=?)}\,\mathrm{km}$ in a geostationary low Earth orbit (LEO) where $\exists\, !\, h\in [180, 2000]$ . In order to stay in orbit perpetually, the orbital period of this satellite needs to be $T\,\textrm{(=?)}\,\mathrm{h}$ , the orbital speed $v\,\textrm{(=?)}\,\mathrm{km/h}$ , and the launch velocity $u\,\textrm{(=?)}\,\mathrm{km/h}$ in magnitude and perpendicular $\textrm{(why?)}$ to the Earth’s surface. If the satellite can reach an altitude of $h'$ , i.e., $h'\le r\le \max (h)$ , the launch succeeds and the satellite revolves around the Earth in uniform circular motion; else, i.e., $r<\min (h)\le h'$ , the launch fails and the satellite falls down on the Earth in projectile motion, its range being $\Delta s_x=R\omega\,\textrm{(=?)}$ .

(C002) Correction.

Dimensional analysis. Dimensional analysis is a procedure to check the validity of any equation by dimensional consistency. Read More

i.e.

$\textrm{SI unit of LHS}=\textrm{SI unit of RHS}$

E.g. The SUVAT equations comprise the sum, the difference, the product, and the power law in functions of variables $s$ displacement, $u$ initial velocity, $v$ final velocity, $a$ acceleration, and $t$ time, namely, $v = u + at$ , $s = ut+\frac{1}{2}at^2$ , $s = vt-\frac{1}{2}at^2$ , $s = \frac{1}{2}(u+v)t$ , and $v^2 = u^2+2as$ where the corresponding dimensions for both sides, $(\mathsf{L}\mathsf{T}^{-1})=(\mathsf{LT^{-1}})+(\mathsf{LT^{-2}})(\mathsf{T})$ , $(\mathsf{L})=(\mathsf{LT^{-1}})(\mathsf{T})+\frac{1}{2}(\mathsf{LT^{-2}})(\mathsf{T^2})$ , $(\mathsf{L})=(\mathsf{LT^{-1}})(\mathsf{T})-\frac{1}{2}(\mathsf{LT^{-2}})(\mathsf{T^2})$ , $(\mathsf{L})=\frac{1}{2}((\mathsf{LT^{-1}})+(\mathsf{LT^{-1}}))(\mathsf{T})$ , and $(\mathsf{LT^{-1}})^2=(\mathsf{LT^{-1}})^2+2(\mathsf{LT^{-2}})(\mathsf{L})$ are equivalent.

(C003) Deduction.

Deduce the law of conservation of linear momentum subsequent to all 1^st, 2^nd, and 3^rd of Newton’s laws of motion. Read More

By Newton’s 1^st law, if there is no $\textrm{\scriptsize{NET}}$ force (from without) acting externally on the system, the objects within will remain at rest or in uniform motion, i.e., $\mathbf{u}_1, \mathbf{u}_2, \mathbf{v}_1, \mathbf{v}_2=\textrm{Const.}$ when $\mathbf{a},\boldsymbol{\alpha} =0$ .

By Newton’s 3^rd law, an action-reaction pair of forces is opposite in direction $\mathbf{F}_1=-\mathbf{F}_2$ and equal in magnitude $F_1=|\mathbf{F}_1|=|-\mathbf{F}_2|=F_2$ .

By Newton’s 2^nd law, from $\mathbf{F}_1=-\mathbf{F}_2$ $\textrm{\scriptsize{OR}}$ $\displaystyle{\frac{\mathrm{d}\mathbf{p}_1}{\mathrm{d}t}=-\frac{\mathrm{d}\mathbf{p}_2}{\mathrm{d}t}}$ $\textrm{\scriptsize{OR}}$ $\displaystyle{\int \frac{\mathrm{d}\mathbf{p}_1}{\mathrm{d}t}\,\mathrm{d}t = -\int \frac{\mathrm{d}\mathbf{p}_2}{\mathrm{d}t}\,\mathrm{d}t}$ $\textrm{\scriptsize{OR}}$ $\Delta \mathbf{p}_1=-\Delta\mathbf{p}_2$ $\textrm{\scriptsize{OR}}$ $m_1\mathbf{v}_1-m_1\mathbf{u}_1=-(m_2\mathbf{v}_2-m_2\mathbf{u}_2)$ , we have $m_1u_1+m_2u_2=m_1v_1+m_2v_2$ .

Derive Snell’s Law from Fermat’s principle of least time. Read More

Setting zero ( $=0$ ) the derivative $\frac{\mathrm{d}t}{\mathrm{d}x}$ of time $t$ wrt to path $x$ , you shall have for reflection $\theta_i=\theta_r$ and for refraction $n_1\sin\theta_1=n_2\sin\theta_2$ .

E.g. Let the origin $O$ be $O(0,0)$ , the start point $A$ be $A(x_1,y_1)$ , and the end point $B$ be $B(x_2,y_2)$ , where $x_1,y_1<0$ and $x_2,y_2>0$ such that light travels from $A$ in quadrant III via the origin $O$ to $B$ in quadrant I. Paths $AO$ and $OB$ are thus $\sqrt{x_1^2+y_1^2}$ and $\sqrt{x_2^2+y_2^2}$ long. Let the speed of light in the initial and the final medium be $u$ and $v$ . The time $t$ needed is thus $\displaystyle{t=\frac{\sqrt{x_1^2+y_1^2}}{u}+\frac{\sqrt{x_2^2+y_2^2}}{v}}$ . Since $\displaystyle{0\equiv\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{x_1}{u\sqrt{x_1^2+y_1^2}}+\frac{x_2}{v\sqrt{x_2^2+y_2^2}}}$ , we have $\displaystyle{\frac{-\sin\theta_1}{u}+\frac{\sin\theta_2}{v}=0}$ $\textrm{\scriptsize{OR}}$ $\displaystyle{\frac{n_1}{n_2}=\frac{\sin\theta_2}{\sin\theta_1}}$ .

Derive the laws of reflection and refraction by Huygen’s principle. Read More

_{This file is licensed under the Creative Commons Attribution-Share Alike 2.5 Generic license.}
_{Author: Arne Nordmann from Renningen, Germany}

Deduce that electromagnetic waves are sinusoidal by observing induction of the magnetomotive and the electromotive in time-varying $\mathbf{E}$ -field and $\mathbf{B}$ -field. Read More

Let electromagnetic (EM) waves in a time-varying field be described by $f_{\EM}(\mathbf{r},t)$ a linear superposition of electric field component $f_{\textrm{E}}$ and magnetic field component $f_{\textrm{B}}$ , i.e.,

$\boxed{f(\mathbf{r},t)=f_{\textrm{E}}(\mathbf{r},t)+f_{\textrm{B}}(\mathbf{r},t)}$ .

By experimentation on electromagnetic induction, we see that

$\begin{aligned}\frac{\partial f_\textrm{E}(\mathbf{r},t)}{\partial t} & = f_\textrm{B}(\mathbf{r},t) \\\frac{\partial f_\textrm{B}(\mathbf{r},t)}{\partial t} & = f_\textrm{E}(\mathbf{r},t) \\\end{aligned}$

Hence

$\begin{aligned}\frac{\partial^2}{\partial t^2}(f_\textrm{E}) =\frac{\partial}{\partial t}\bigg( \frac{\partial f_\textrm{E}}{\partial t} \bigg) & = \frac{\partial}{\partial t}(f_\textrm{B}) = f_\textrm{E} \\\frac{\partial^2}{\partial t^2}(f_\textrm{B}) =\frac{\partial}{\partial t}\bigg( \frac{\partial f_\textrm{B}}{\partial t} \bigg) & = \frac{\partial}{\partial t}(f_\textrm{E}) = f_\textrm{B} \\\end{aligned}$

$\boxed{\begin{aligned}\because \enspace & f'' = f \\\therefore \enspace & f\textrm{ is sinusoidal}\end{aligned}}$

Derive Kepler’s Third Law by Newton’s law of universal gravitation. Read More

Let the mass of the Sun be $M$ and the mass of a planet $m$ .

$\begin{aligned}{}_mF_M & = {}_MF_m \\G\frac{Mm}{r^2} & = mr\omega^2 = mr\bigg( \frac{2\pi}{T}\bigg)^2 \\T^2 & = \frac{4\pi^2}{GM}r^3 \\T^2 & \propto r^3 \\\end{aligned}$

Deduce the existence of a net force (unnamed centripetal still) in uniform circular motion by Newton’s first law. Read More

Uniform circular motion is at uniform speed $v$ , its direction changing in a circle. By Newton’s first law,

$\boxed{\displaystyle{\nexists\,\mathbf{F} \Leftrightarrow \mathbf{v}=0\textrm{ \scriptsize{OR} }\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}=0}}$

$\begin{aligned}\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} & = \bigg( \frac{\mathrm{d}}{\mathrm{d}t}v_x \bigg)\,\hat{\mathbf{i}} + \bigg( \frac{\mathrm{d}}{\mathrm{d}t}v_y \bigg)\,\hat{\mathbf{j}} \\& = \bigg( \frac{\mathrm{d}}{\mathrm{d}t}(r\cos\theta )\bigg)\,\hat{\mathbf{i}} + \bigg( \frac{\mathrm{d}}{\mathrm{d}t}(r\sin\theta )\bigg)\,\hat{\mathbf{j}} \\& = -r\dot{\theta}\sin\theta\,\hat{\mathbf{i}} + r\dot{\theta}\cos\theta\,\hat{\mathbf{j}} \\& \neq 0 \\\end{aligned}$

$\boxed{\displaystyle{\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}\neq 0\Leftrightarrow \exists\, \mathbf{F}}}$

Deduce the nature of acceleration in uniform circular motion by vector analysis. Read More

$\begin{aligned}\mathbf{r} & = (R\cos\theta , R\sin\theta ) \\r & = |\mathbf{r}| = \sqrt{(R\cos\theta)^2+(R\sin\theta)^2} = R \\\hat{\mathbf{r}} & = \frac{\mathbf{r}}{r} = (\cos\theta , \sin\theta ) \\\mathbf{v} & = (-R\dot{\theta}\sin\theta ,R\dot{\theta}\cos\theta ) \\v & = |\mathbf{v}| = \sqrt{(-R\dot{\theta}\sin\theta )^2+(R\dot{\theta}\cos\theta )^2} = R\dot{\theta} \\\hat{\mathbf{v}} & = \frac{\mathbf{v}}{v} = (-\sin\theta , \cos\theta ) \\\mathbf{a} & = -R(\dot{\theta}^2\cos\theta +\ddot{\theta}\sin\theta , \dot{\theta}^2\sin\theta -\ddot{\theta}\cos\theta ) \\\textrm{By }& \textrm{definition, }\ddot{\theta}=0 \\& = -R(\dot{\theta}^2\cos\theta , \dot{\theta}^2\sin\theta ) \\a & = |\mathbf{a}| = R\sqrt{(\dot{\theta}^2\cos\theta )^2+(\dot{\theta}^2\sin\theta )^2} \\& = R\dot{\theta}^2 \\\hat{\mathbf{a}} & = \frac{\mathbf{a}}{a} = (-\cos\theta , -\sin\theta ) \\\hat{\mathbf{a}} & =- \hat{\mathbf{r}} \textrm{ \scriptsize{AND} }\mathbf{a}\parallel\mathbf{r}\perp\mathbf{v}\\\end{aligned}$

$\therefore$ In uniform circular motion there is $\textrm{\scriptsize{ONLY}}$ centripetal acceleration; if there be tangential acceleration, it is non-uniform.

Deduce the use of a rheostat by definition of path of least resistance.

Derive Bernoulli’s equation from Work-Energy Theorem for fluids. Read More

$\Delta W=\Delta\mathrm{KE}+\Delta\mathrm{PE}$
$\displaystyle{\Delta (PV)=\Delta \bigg(\frac{1}{2}mv^2\bigg) +\Delta (mgh)}$
$\displaystyle{\Delta P=\Delta \bigg(\frac{mv^2}{2V}\bigg) +\Delta \bigg(\frac{mgh}{V}}\bigg)$
$\displaystyle{\rho=\frac{m}{V}}$
$\Delta P=\Delta (\frac{1}{2}\rho v^2) +\Delta (\rho gh)$
$P_1+\frac{1}{2}\rho v_1^2+\rho gh_1=P_2+\frac{1}{2}\rho v_2^2+\rho gh_2 = \textrm{Const.}$

Deduce Newton’s second law from Euler-Lagrange equation. Read More

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}\bigg( \frac{\partial \mathcal{L}}{\partial \dot{x}}\bigg) - \frac{\partial \mathcal{L}}{\partial x}=0}$ $\textrm{\scriptsize{OR}}$ $\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}\bigg( \frac{\partial (T-V)}{\partial \dot{x}}\bigg) - \frac{\partial (T-V)}{\partial x}=0}$ $\textrm{\scriptsize{OR}}$ $\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}\bigg( \frac{\partial T}{\partial \dot{x}} - \frac{\partial V}{\partial \dot{x}} \bigg)-\bigg( \frac{\partial T}{\partial x}-\frac{\partial V}{\partial x} \bigg) =0}$ $\textrm{\scriptsize{AS}}$ $V=V(x)$ is $\dot{x}$ -independent s.t. $\displaystyle{\frac{\partial V}{\partial \dot{x}}=0}$ $\textrm{\scriptsize{SO}}$ $\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial T}{\partial \dot{x}}-\frac{\partial T}{\partial x}=-\frac{\partial V}{\partial x}=F}$ $\textrm{\scriptsize{AS}}$ $\displaystyle{T=T(\dot{x})=\frac{1}{2}m\dot{x}^2}$ is $x$ -independent s.t. $\displaystyle{\frac{\partial T}{\partial x}=0}$ $\textrm{\scriptsize{SO}}$ $\displaystyle{F=\frac{\mathrm{d}}{\mathrm{d}t}\bigg[ \frac{\partial}{\partial \dot{x}} \bigg( \frac{1}{2}m\dot{x}^2\bigg) \bigg] =\frac{\mathrm{d}}{\mathrm{d}t}(m\dot{x})=m\ddot{x}=ma}$

Deduce the speed of light from Maxwell’s equations.

Deduce the ideal gas law, and hence the equation of state, from Boyle’s law, Charles’s law, and Gay-Lussac’s law. Read More

By assumption

$\begin{aligned}\frac{\partial}{\partial t}(PV)\bigg|_{T=\textrm{ Const.}} = 0 \Leftrightarrow PV= c_1\textrm{ (Const.)} \\\frac{\partial}{\partial t}\bigg(\frac{V}{T}\bigg)\bigg|_{P=\textrm{ Const.}} = 0\Leftrightarrow \frac{V}{T}= c_2\textrm{ (Const.)} \\ \\\frac{\partial}{\partial t}\bigg(\frac{P}{T}\bigg)\bigg|_{V=\textrm{ Const.}} = 0\Leftrightarrow \frac{P}{T}= c_3\textrm{ (Const.)} \\ \\\end{aligned}$

multiplying together,

$\displaystyle{\frac{P^2V^2}{T^2}}=c_1c_2c_3$ .

hence $\displaystyle{\frac{PV}{T}=\sqrt{c_1c_2c_3}}=C\,\textrm{ (Const.)}$ .

One can countercheck as follows

$f=f(P,V,T)=PV-CT=0$ s.t.

$\displaystyle{\frac{\partial f}{\partial t}=\frac{\partial f}{\partial P}\frac{\partial P}{\partial t}+\frac{\partial f}{\partial V}\frac{\partial V}{\partial t}+\frac{\partial f}{\partial T}\frac{\partial T}{\partial t}=0}$

agrees with

$\begin{aligned}P\bigg( \frac{\partial V}{\partial t} \bigg) + \bigg( \frac{\partial P}{\partial t}\bigg) V & = 0\\V\frac{\partial}{\partial t}\bigg( \frac{1}{T}\bigg) + \frac{\partial V}{\partial t}\bigg( \frac{1}{T}\bigg) & = 0 \\P\frac{\partial}{\partial t}\bigg( \frac{1}{T}\bigg) + \frac{\partial P}{\partial t}\bigg( \frac{1}{T}\bigg) & = 0 \\\end{aligned}$

Deduce the kinetic gas equation from conservation of momentum by assumption of elastic collision. Read More

For a cubic container of volume $V=L^3$ , the average squared speed $\overline{v^2}$ of one particle of mass $m$ in random motion along any $x$ -, $y$ -, $z$ -directions is identical $\overline{v_x^2}=\overline{v_y^2}=\overline{v_z^2}$ $\textrm{\scriptsize{OR}}$ $\overline{v^2}=\overline{v_x^2}+\overline{v_y^2}+\overline{v_z^2}=3\overline{v_x^2}$ $\textrm{\scriptsize{OR}}$ $\displaystyle{\overline{v_x^2}=\frac{1}{3}\overline{v^2}}$ . For impulse $\Delta p=p_{i,x}-p_{f,x}=p_{i,x}-(-p_{i,x})=2p_{i,x}=2mv_x$ during $\displaystyle{\Delta t=\frac{2L}{v_x}}$ , write $\displaystyle{F=\frac{\Delta p}{\Delta t}=(2mv_x)\bigg(\frac{v_x}{2L}\bigg) =\frac{mv_x^2}{L}}$ . With $N$ particles of rms speed $\overline{v_x^2}$ , rewrite $\displaystyle{F=\frac{Nm\overline{v_x^2}}{L}=\frac{Nm\overline{v^2}}{3L}}$ , the pressure $P$ being $\displaystyle{P=\frac{F}{A}=\frac{F}{L^2}=\frac{Nm\overline{v^2}}{3L}\bigg(\frac{1}{L^2}\bigg)=\frac{Nm\overline{v^2}}{3V}}$ , thus the kinetic gas equation, $\boxed{\displaystyle{PV=\frac{1}{3}Nm\overline{v^2}}}$ .

Deduce the magnetic field strength $B$ , due to a long straight wire and to a long tightly-packed solenoid both with current carrying, from Biot–Savart law and also from Ampere’s law. Read More

Ampere’s law states that the magnetic flux summed over a closed surface is proportional to the current enclosed by the closed path, i.e.,

$\boxed{\displaystyle{\oint\mathbf{B}\cdot\mathrm{d}\mathbf{l}=\mu_0I_{\textrm{enclosed}}}}$

By the use of Gaussian pillbox, the magnetic field at a distance $r$ from a long straight wire is given by

$\begin{aligned}\oint\mathbf{B}\cdot\mathrm{d}\mathbf{l} & = B(2\pi r) = \mu_0I \\B & = \frac{\mu_0I}{2\pi r} \\\end{aligned}$

By the use of amperian loop of length $L$ with $N$ turns of coils, the coil density being $n=\frac{N}{L}$ ,

$\begin{aligned}\oint\mathbf{B}\cdot\mathrm{d}\mathbf{l} & = \sum_{\textrm{side }1}B_{\parallel}\Delta L + \sum_{\textrm{side }2}B_{\parallel}\Delta L + \sum_{\textrm{side }3}B_{\parallel}\Delta L + \sum_{\textrm{side }4}B_{\parallel}\Delta L \\& = BL + 0 + 0 + 0 \\& = \mu_0I_{\textrm{enclosed}} \\BL & = \mu_0IN=\mu_0I(nL) \\\end{aligned}$

$\boxed{\displaystyle{B=\mu_0nI}}$

Biot–Savart law of magnetic field states that

$\boxed{\displaystyle{\mathbf{B}=\oint_{\Sigma}\frac{\mu_0I}{4\pi}\frac{\mathrm{d}\mathbf{s}\times\hat{\mathbf{r}}}{r^2}}}$

At a distance $r\textrm{ (Const.)}$ away from a point $m=m(0,0,0)$ , let there be a (infinitely) long vertical line $l=l(r,0,s)$ for $s\in (-\infty ,\infty)$ . The locus, joining point $m$ and any a point on line $l$ , is given by $\mathbf{r}(\theta )=(r,0,r\tan\theta )$ s.t. $|\mathbf{r}|=\sqrt{(r)^2+(0)^2+(r\tan\theta )^2}=r\sec\theta$ and $\hat{\mathbf{r}}=(\cos\theta ,0, \sin\theta )$ where $\theta$ is the angle of elevation or depression. So, $s=r\tan\theta$ , $\mathrm{d}s=r\sec^2\theta\,\mathrm{d}\theta$ , and $\mathrm{d}\mathbf{s}=(r,0,\mathrm{d}s) = (r,0,r\sec^2\theta\,\mathrm{d}\theta )$ .

$\begin{aligned}\mathrm{d}\mathbf{s}\times \hat{\mathbf{r}} & = \begin{array}{|ccc|}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\r & 0 & r\sec^2\theta\\\cos\theta & 0 & \sin\theta \\\end{array} \, \cdot\, \mathrm{d}\theta \\\end{aligned}$

$\begin{aligned}B & = \frac{\mu_0I}{4\pi r^2}\int_{\theta =-\frac{\pi}{2}}^{\theta =\frac{\pi}{2}} (-r\sin\theta +r\cos\theta )\,\mathrm{d}\theta \\& = \frac{\mu_0I}{4\pi r}(2+0) \\& = \frac{\mu_0I}{2\pi r} \\\end{aligned}$

Derive Fleming’s left hand rule and right hand rule by vector analysis. Read More

$\begin{aligned}\mathbf{F} & = q(\mathbf{E}+\mathbf{v}\times\mathbf{B}) \\\frac{1}{q}\begin{pmatrix}F_x \\ F_y \\ F_z \end{pmatrix} & = \begin{pmatrix} E_x \\ E_y \\ E_z \end{pmatrix} + \begin{pmatrix} v_yB_z-v_zB_y \\ v_zB_x - v_xB_z \\ v_xB_y-v_yB_x \end{pmatrix}\end{aligned}$

(C004) Observation.

Similarities in formulae for finding the equivalent. Read More

Current $I$ in series:

$I_{\textrm{eq}}=I_1=I_2=\cdots =I_n$

Current $I$ in parallel:

$I_{\textrm{eq}} = I_1+I_2+\cdots +I_n$

Voltage $V$ in series:

$V_{\textrm{eq}}=V_1+V_2+\cdots +V_n$

Voltage $V$ in parallel:

$V_{\textrm{eq}}=V_1=V_2=\cdots =V_n$

Resistance $R$ in series:

$R_{\textrm{eq}}=R_1+R_2+\cdots +R_n$

Resistance $R$ in parallel:

$\displaystyle{\frac{1}{R_{\textrm{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots +\frac{1}{R_n}}$

Capacitance $C$ in series:

$\displaystyle{\frac{1}{C_{\textrm{eq}}}=\frac{1}{C_1}+\frac{1}{C_2}+\cdots +\frac{1}{C_n}}$

Capacitance $C$ in parallel:

$C_{\textrm{eq}}=C_1+C_2+\cdots +C_3$

Inductance $L$ in series:

$L_{\textrm{eq}}=L_1+L_2+\cdots +L_n$

Inductance $L$ in parallel:

$\displaystyle{\frac{1}{L_{\textrm{eq}}}=\frac{1}{L_1}+\frac{1}{L_2}+\cdots +\frac{1}{L_n}}$

Similarities in terms of inverse square law. Read More

Newton’s law of gravitational force :

$\boxed{\displaystyle{F_G=G\frac{Mm}{r^2}}}$

Gravitational field strength:

$\boxed{\displaystyle{g=\frac{GM}{r^2}}}$

Coulomb’s law of electrostatic force:

$\boxed{\displaystyle{F_E=k\frac{Qq}{r^2}}}$

Electric field strength:

$\boxed{\displaystyle{E=\frac{q}{4\pi\varepsilon_0r^2}}}$

Biot–Savart law of magnetic field:

$\boxed{\displaystyle{\mathbf{B}=\oint_{\Sigma}\frac{\mu_0I}{4\pi}\frac{\mathrm{d}\mathbf{s}\times\hat{\mathbf{r}}}{r^2}}}$

Intensity:

$\boxed{\displaystyle{I=\frac{P}{4\pi r^2}}}$

Correspondence between intensive (/intrinsic) and extensive (/extrinsic) properties. Read More

$\boxed{\textrm{Resistivity }\displaystyle{\rho=\frac{RA}{l}}}$ vs $\boxed{\textrm{Resistance }\displaystyle{R=\frac{\rho l}{A}}}$ .

$\boxed{\textrm{Specific heat capacity }\displaystyle{c=\frac{Q}{m\Delta T}=\frac{C}{m}}}$ vs $\boxed{\textrm{Heat capacity }\displaystyle{C=\frac{Q}{\Delta T}=mc}}$

Comparison between categories. Read More

$\begin{array}{|c|c|}\hline\textrm{transverse wave} & \textrm{longitudinal wave} \\\hline\textrm{vibration} \perp \textrm{propagation} & \textrm{vibration} \parallel \textrm{propagation}\\\textrm{energy}\parallel \textrm{propagation} & \textrm{energy}\parallel \textrm{propagation}\\\textrm{crest/trough} & \textrm{compression/rarefaction} \\\hline\end{array}$

$\begin{array}{|c|c|}\hline\textrm{natural light} & \textrm{laser beam} \\\hline\textrm{continuous spectrum} & \textrm{monochromatic} \\\textrm{incoherent} & \textrm{coherent} \\\textrm{divergent} & \textrm{unidirectional} \\\textrm{not intense} & \textrm{very intense} \\\hline\end{array}$

$\begin{array}{|c|c|}\hline\textrm{particle} & \textrm{antiparticle} \\\hline\textrm{electron }e^- & \textrm{positron }e^+ \\\textrm{proton }p^+ & \textrm{antiproton }p^- \\\textrm{neutrino }\nu & \textrm{antineutrino }\bar{\nu} \\\hline\end{array}$

Perspective from macroscopic to microscopic. Read More

$\begin{array}{|c|c|}\hline\textrm{macroscopic} & \textrm{microscopic} \\PV=nRT=NkT & PV=\frac{1}{3}Nm\overline{v^2} \\\hline\end{array}$

Differences between categories. Read More

Between centre of mass (CM) and centre of gravity (CG):

$\boxed{\displaystyle{\mathbf{r}_{\textrm{CM}}=\frac{1}{M}\int\mathbf{r}\,\mathrm{d}m}}$

which depends $\textrm{\scriptsize{ONLY}}$ on mass distribution;

$\boxed{\displaystyle{\mathbf{r}_{\textrm{CG}}=\frac{1}{M}\int\mathbf{r}\cdot g(r)\,\hat{\mathbf{k}}\,\mathrm{d}m}}$

which depends $\textrm{\scriptsize{EXTRA}}$ on gravity variation.

Between mass and weight:

Mass $m$ of a body is a constant of matter whereas its weight $W=mg$ is a variable of gravity $g$ .

Between conservative and nonconservative force.

(C005) Formulation.

Ex. 1, Gravity varying with position. Read More

Assumed that the Earth is a perfect sphere of mass $M$ , radius $R$ , and volume $\displaystyle{V=\frac{4}{3}\pi R^3}$ having uniform mass density $\displaystyle{\rho =\frac{M}{V}=\textrm{Const.}}$ The weight $m\mathbf{g}$ of a test particle $m$ at some height $x$ above sea level is provided by the Earth’s gravitational force $\mathbf{F}_G$ with gravitational constant $G$ and $x$ -dependent gravity $g=g(x)$ , i.e.,

$\begin{aligned}F_G & = mg \\G\frac{Mm}{(R+x)^2} & = mg \\g = g(x) & = G\frac{M}{(R+x)^2}\end{aligned}$

For $g(x) |\, x\in\{ -R, 0, \infty\}$ , please explain the physical meanings of $g(-R)$ , $g(0)$ , and $g(\infty)$ .

Ex. 2, Projectile in parabolic equation. Read More

Let an object be projected from an angle $\alpha$ with the level, at an initial speed $v$ , and subjected to gravity $g$ . Begin with

$\begin{aligned}\frac{v_y}{v_x} & = \tan\alpha \\x(t) & = (v_x)t \\y(t) & = (v_y)t+\frac{1}{2}(-g)t^2 \\\end{aligned}$

and

$\begin{aligned}v_x & = v\cos\alpha \\v_y & = v\sin\alpha \\v & = \sqrt{v_x^2+v_y^2} \\\end{aligned}$

one may write time $t(x)$ in terms of $x$

$\displaystyle{t(x)=\frac{x}{v_x}=\frac{x\tan\alpha}{v_y}}$

hence

$\begin{aligned}y(t(x)) & = (v_y)\bigg(\frac{x\tan\alpha}{v_y}\bigg) + \frac{1}{2}(-g)\bigg(\frac{x}{v_x}\bigg)^2 \\\end{aligned}$

and thus

$\boxed{\displaystyle{y(x)=\Big( -\frac{g}{2v^2\cos^2\alpha}\Big)x^2+(\tan\alpha )x}}$

Ex. 3, Object and thin lens. Read More

Given for convex (/converging) lens or concave (/diverging) lens

$\displaystyle{\frac{1}{f}=\frac{1}{u}+\frac{1}{v}}$ .

When $f$ is a constant, write $u(v)$ in terms of $v$

$\displaystyle{u(v)=\frac{fv}{v-f}}$

and write $v(u)$ in terms of $u$

$\displaystyle{v(u)=\frac{fu}{u-f}}$

s.t. $u\circ v = v\circ u = \textrm{id}$

For positive and negative focal length $\pm f$ , find $v(u)$ for $u\to\infty$ , $u>2f$ , $u=2f$ , $f<u<2f$ , $u=f$ , and $u<f$ and hence find whether the image is real or virtual.

(C006) Visualization.

Number line for i. scale, ii. range, and iii. arrow. Read More

Vector Addition by Parallelogram and Head-to-tail Method. Read More

The relationship between displacement $s$ graph, velocity $v$ graph, and acceleration $a$ graph, of independent variable time $t$ . Read More

- ① implies ④, which implies ⑨.
- ② implies ⑤, ⑥, or ⑦.
- ⑤ implies ⑩.
- ⑥ implies ⑪.
- ⑦ implies ⑫.
- ③ implies ⑧, which implies ⑬.

Exercises. (drawing graphs)

The graphs of isobaric process, isochoric/isovolumetric process, isothermal process, and adiabatic process. Read More

1^st law of thermodynamic

$\boxed{\displaystyle{\Delta Q=\Delta U+\Delta W}}$

Isobaric process.

$\Delta P=0\qquad \displaystyle{\frac{V}{T}=\textrm{Const.}}$

Isochoric/Isovolumetric process.

$\Delta V=0 \qquad \Delta W=0 \qquad \displaystyle{\frac{P}{T}=\textrm{Const.}}$

Isothermal process.

$\Delta T=0 \qquad \Delta U=0 \qquad PV=\textrm{Const.}$

Adiabatic process.

$\Delta Q=0$

The graphs of average velocity $v_{\textrm{avg}}$ , instantaneous velocity $v$ , average acceleration $a_{\textrm{avg}}$ , and instantaneous acceleration $a$ .

Field representation by field lines. E.g., electric field, magnetic field, and gravitational field.

July 8, 2022October 24, 2022

202207081631 Statics Figures (Elementary) Q1

In a lift, the panel screen displays signs on an array of seven bars of light-emitting diodes (LEDs):

such that the numeric digits $0$ , $1$ , $2$ , $3$ , $4$ , $5$ , $6$ , $7$ , $8$ and $9$ when lit up are as follow:

For each configuration of lit-up cells, assuming that the mass distribution is uniform and the plate thickness negligible, find the centre of mass (CM), and hence the centre of gravity (CG).

Intuition.

Try arguing $\textrm{\scriptsize{NOT}}$ by symmetry in mathematics, $\textrm{\scriptsize{BUT}}$ by moment in physics.

We state without proof the centre of mass (CM) in each cell is as shown below:

For $N$ particles in 2-D:

$\begin{aligned} x_{\textrm{CM}} & = \frac{m_1x_1+m_2x_2+\cdots +m_Nx_N}{m_1+m_2+\cdots +m_N} = \frac{\sum_{i=1}^{N}m_ix_i}{M} \\ y_{\textrm{CM}} & = \frac{m_1y_1+m_2y_2+\cdots +m_Ny_N}{m_1+m_2+\cdots +m_N} = \frac{\sum_{i=1}^{N}m_iy_i}{M} \\ \end{aligned}$

Define the center at the origin $O(0,0)$ , and let $\mathbf{r}_1$ , $\mathbf{r}_2$ , $\mathbf{r}_3$ , $\mathbf{r}_4$ , $\mathbf{r}_5$ , and $\mathbf{r}_6$ be the vectors pointing to the CM of each cell:

such that

$\begin{aligned} \mathbf{r}_1 & = 2\,\hat{\mathbf{j}} \\ \mathbf{r}_2 & = -1\,\hat{\mathbf{i}} + 1\,\hat{\mathbf{j}} \\ \mathbf{r}_3 & = 1\,\hat{\mathbf{i}} + 1\,\hat{\mathbf{j}} \\ \mathbf{r}_4 & = -1\,\hat{\mathbf{i}} -1\,\hat{\mathbf{j}} \\ \mathbf{r}_5 & = -2\,\hat{\mathbf{j}} \\ \mathbf{r}_6 & = 1\,\hat{\mathbf{i}} -1\,\hat{\mathbf{j}} \\ \end{aligned}$

To illustrate how to get the CM for the number signs, we first begin with $1$ :

$\mathbf{1}_{\textrm{CM}} = (x_{\textrm{CM,1}},y_{\textrm{CM,1}})$ .

$\begin{aligned} \mathbf{1}_{\textrm{CM}} & = \frac{m\mathbf{r}_3+m\mathbf{r}_6}{m+m} \\ & = \frac{1}{2}\mathbf{r_3} + \frac{1}{2}\mathbf{r_6} \\ & = \frac{1}{2}(1,1) + \frac{1}{2}(1,-1) \\ & = (1,0) \\ \end{aligned}$

The remaining are left the reader as an exercise.

July 7, 2022October 24, 2022

202207071711 Solution to 1976-CE-AMATH-I-XX

July 7, 2022October 24, 2022

202207071209 Exercise 45W (Q1)

Separate $46$ into four parts such that the second part is $1$ more than twice the first, the third is twice the fourth, and the fourth is $3$ more than the first.

_{J. R. Lux & R. S. Pieters. (1969). Exercises in Elementary Algebra}

Roughwork.

Let $a$ , $b$ , $c$ , and $d$ be positive real numbers ( $\in\mathbb{R}^{+}$ ) such that $0<a,b,c,d<46$ .

Given that

$\begin{aligned} 46 & = a+b+c+d \\ 1 & = b - 2a \\ 2d & = c \\ 3 & = d-a \\ \end{aligned}$

or,

$\begin{aligned} \begin{bmatrix} 1 & 1 & 1 & 1 \\ -2 & 1 & 0 & 0 \\ 0 & 0 & 1 & -2 \\ -1 & 0 & 0 & 1\end{bmatrix} \begin{bmatrix} a \\ b \\ c\\ d \end{bmatrix} & = \begin{bmatrix} 46 \\ 1 \\ 0 \\ 3 \end{bmatrix} \\ \end{aligned}$

or,

$\begin{aligned} \begin{bmatrix} a \\ b \\ c\\ d \end{bmatrix} & = \begin{bmatrix} 1 & 1 & 1 & 1 \\ -2 & 1 & 0 & 0 \\ 0 & 0 & 1 & -2 \\ -1 & 0 & 0 & 1\end{bmatrix}^{-1}\begin{bmatrix} 46 \\ 1 \\ 0 \\ 3 \end{bmatrix} \\ \end{aligned}$

To find the inverse, begin with

$\begin{aligned} & \boxed{\begin{array}{cccc|cccc} 1 & 1& 1 & 1 & 1 & 0 & 0 & 0 \\ -2 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & -2 & 0 & 0 & 1 & 0 \\ -1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ \end{array} }\\ R1' & \Longleftarrow R1-R2-R3-3R4 \\ \begin{bmatrix} 6 \\ 0 \\ 0 \\ 0\\ \end{bmatrix}^{\mathrm{T}} & = \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}^{\mathrm{T}} - \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix}^{\mathrm{T}} - \begin{bmatrix} 0 \\ 0 \\ 1 \\ -2 \end{bmatrix}^{\mathrm{T}} - 3\begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \end{bmatrix}^{\mathrm{T}} \\ \begin{bmatrix} 1 \\ -1 \\ -1 \\ -3 \\ \end{bmatrix}^{\mathrm{T}} & = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}^{\mathrm{T}} - \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}^{\mathrm{T}} - \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}^{\mathrm{T}} - 3\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}^{\mathrm{T}} \\ \end{aligned}$

proceed with

$\begin{aligned} & \boxed{\begin{array}{cccc|cccc} 6 & 0& 0 & 0 & 1 & -1 & -1 & -3 \\ -2 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & -2 & 0 & 0 & 1 & 0 \\ -1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ \end{array} }\\ R2' & \Longleftarrow R2+\frac{R1}{3} \\ \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \\ \end{bmatrix}^{\mathrm{T}} & = \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0\end{bmatrix}^{\mathrm{T}} + \frac{1}{3}\begin{bmatrix} 6 \\ 0 \\ 0 \\ 0 \end{bmatrix}^{\mathrm{T}}\\ \begin{bmatrix} 1/3 \\ 2/3 \\ -1/3 \\ -1 \end{bmatrix}^{\mathrm{T}} & = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}^{\mathrm{T}} +\frac{1}{3}\begin{bmatrix} 1 \\ -1 \\ -1 \\ -3 \end{bmatrix}^{\mathrm{T}} \\ \end{aligned}$

proceed with

$\begin{aligned} & \boxed{\begin{array}{cccc|cccc} 6 & 0& 0 & 0 & 1 & -1 & -1 & -3 \\ 0 & 1 & 0 & 0 & 1/3 & 2/3 & -1/3 & -1 \\ 0 & 0 & 1 & -2 & 0 & 0 & 1 & 0 \\ -1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ \end{array} }\\ R4' & \Longleftarrow 6R4+R1 \\ \begin{bmatrix} 0 \\ 0 \\ 0 \\ 6 \\ \end{bmatrix}^{\mathrm{T}} & = 6\begin{bmatrix} -1 \\ 0 \\ 0 \\ 1\end{bmatrix}^{\mathrm{T}} + \begin{bmatrix} 6 \\ 0 \\ 0 \\ 0 \end{bmatrix}^{\mathrm{T}}\\ \begin{bmatrix} 1 \\ -1 \\ -1 \\ 3 \end{bmatrix}^{\mathrm{T}} & =6\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}^{\mathrm{T}}+\begin{bmatrix} 1 \\ -1 \\ -1 \\ -3 \end{bmatrix}^{\mathrm{T}} \\ \end{aligned}$

proceed with

$\begin{aligned} & \boxed{\begin{array}{cccc|cccc} 6 & 0& 0 & 0 & 1 & -1 & -1 & -3 \\ 0 & 1 & 0 & 0 & 1/3& 2/3 & -1/3 & 0 \\ 0 & 0 & 1 & -2 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 6 & 1 & -1 & -1 & 3 \\ \end{array} }\\ R3' & \Longleftarrow R3+\frac{1}{3}R4 \\ \begin{bmatrix} 0\\0 \\1 \\0 \\ \end{bmatrix}^{\mathrm{T}} & = \begin{bmatrix} 0 \\ 0 \\ 1 \\ -2\end{bmatrix}^{\mathrm{T}} + \frac{1}{3}\begin{bmatrix} 0 \\ 0 \\ 0 \\ 6 \end{bmatrix}^{\mathrm{T}}\\ \begin{bmatrix} 1/3\\ -1/3 \\2/3 \\ 1 \end{bmatrix}^{\mathrm{T}} & =\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}^{\mathrm{T}}+\frac{1}{3}\begin{bmatrix} 1 \\ -1 \\ -1 \\ 3 \end{bmatrix}^{\mathrm{T}} \\ \end{aligned}$

proceed with

$\begin{aligned} & \boxed{\begin{array}{cccc|cccc} 6 & 0& 0 & 0 & 1 & -1 & -1 & -3 \\ 0 & 1 & 0 & 0 & 1/3& 2/3 & -1/3 & 0 \\ 0 & 0 & 1 & 0 & 1/3 & -1/3 & 2/3 & 1 \\ 0 & 0 & 0 & 6 & 1 & -1 & -1 & 3 \\ \end{array} }\\ R1',R4' & \Longleftarrow \frac{1}{6}\cdot R1, R4 \\ & \boxed{\begin{array}{cccc|cccc} 1 & 0& 0 & 0 & 1/6 & -1/6 & -1/6 & -1/2 \\ 0 & 1 & 0 & 0 & 1/3& 2/3 & -1/3 & 0 \\ 0 & 0 & 1 & 0 & 1/3 & -1/3 & 2/3 & 1 \\ 0 & 0 & 0 & 1 & 1/6 & -1/6 & -1/6 & 1/2 \\ \end{array} }\\ \end{aligned}$

Hence

$\begin{aligned} \begin{bmatrix} 1 & 1 & 1 & 1 \\ -2 & 1 & 0 & 0 \\ 0 & 0 & 1 & -2 \\ -1 & 0 & 0 & 1\end{bmatrix}^{-1} & = \begin{bmatrix} 1/6 & -1/6 & -1/6 & -1/2 \\ 1/3 & 2/3 & -1/3 & 0 \\ 1/3 & -1/3 & 2/3 & 1 \\ 1/6 & -1/6 & -1/6 & 1/2 \end{bmatrix} \\ \end{aligned}$

and thus

$\begin{aligned} \begin{bmatrix} a \\ b \\ c\\ d \end{bmatrix} & = \begin{bmatrix} 1/6 & -1/6 & -1/6 & -1/2 \\ 1/3 & 2/3 & -1/3 & 0 \\ 1/3 & -1/3 & 2/3 & 1 \\ 1/6 & -1/6 & -1/6 & 1/2 \end{bmatrix}\begin{bmatrix} 46 \\ 1 \\ 0 \\ 3 \end{bmatrix} \\ \end{aligned}$

$\begin{aligned} a & = \bigg( \frac{1}{6} \bigg) (46) + \bigg(-\frac{1}{6} \bigg) (1) + \bigg(-\frac{1}{6} \bigg) (0) + \bigg(-\frac{1}{2} \bigg) (3) \\ & =\,\dots \\ & = 6 \\ \end{aligned}$

$\begin{aligned} \because & \enspace 3 = d-a\\ \therefore & \enspace 3=d-(6) \\ \therefore & \enspace d=9 \\ \end{aligned}$

$\begin{aligned} \because &\enspace 2d = c\\ \therefore &\enspace 2(9) = c\\ \therefore &\enspace c=18 \\ \end{aligned}$

$\begin{aligned} \because &\enspace 1=b-2a\\ \therefore &\enspace 1=b-2(6)\\ \therefore &\enspace b=13 \\ \end{aligned}$

$\boxed{\begin{pmatrix} a & b & c & d \end{pmatrix} = \begin{pmatrix} 6 & 13 & 18 & 9\\ \end{pmatrix}}$

$\therefore$ The first number is $6$ , the second $13$ , the third $18$ , and the fourth and last $9$ .

July 7, 2022October 24, 2022

202207071143 Exercise 18.1 (Q1)

The Sun is a yellow star and emits most of its radiation in the yellow portion of the spectrum. If the sun’s radiation peaks at a frequency of $5.20\times 10^{14}\,\mathrm{Hz}$ , how much energy is emitted by one photon of this visible yellow light?

_{J. B., Hickman. (2002). Problem-Solving Exercises in Physics}

Roughwork.

In joules,

$\begin{aligned} E&=hf \\ & = (6.63\times 10^{-34}\,\mathrm{J\cdot s})(5.20\times 10^{14}\,\mathrm{Hz}) \\ & = 3.4476\times 10^{-19}\,\mathrm{J} \\ \end{aligned}$

In electron volts,

$\begin{aligned} E & = \frac{3.4476\times 10^{-19}\,\mathrm{J}}{1.60\times 10^{-19}\,\mathrm{J/eV}} \\ & = 2.15475\,\mathrm{eV} \\ \end{aligned}$

July 6, 2022October 24, 2022

202207061152 Solution to 1971-CE-AMATH-I-XX

If $k$ is an integer, simplify

$\displaystyle{\cos\bigg[ \frac{1}{2}(4k+1)\pi +\theta \bigg]\tan\bigg[\frac{1}{2}(4k+3)\pi -\theta \bigg]}$ .

Roughwork.

First,

$\begin{aligned} & \quad \cos\bigg[ \frac{1}{2}(4k+1)\pi +\theta \bigg]\tan\bigg[\frac{1}{2}(4k+3)\pi -\theta \bigg] \\ & = \cos\bigg( 2k\pi+\frac{\pi}{2}+\theta \bigg) \tan \bigg( 2k\pi+\frac{3\pi}{2}-\theta \bigg) \\ \dots\textrm{as } &\textrm{are}\cos (\angle ),\tan (\angle )\textrm{ shifted by full periods, \textit{i.e.}, }k\cdot 2\pi\,\dots \\ & = \cos\bigg(\frac{\pi}{2}+\theta\bigg) \tan \bigg(\frac{3\pi}{2}-\theta \bigg) \\ \end{aligned}$

Second,

recall that

$\begin{aligned} \textrm{(in radians)} & \quad\textrm{(in degrees)} \\ \pi & = 180^\circ \\ \frac{\pi}{2} & = 90^\circ \\ \frac{3\pi}{2} & = 270^\circ\textrm{ \scriptsize{OR} }-90^\circ = -\frac{\pi}{2}\\ \end{aligned}$

Hence,

$\begin{aligned} & \quad \cos\bigg(\frac{\pi}{2}+\theta\bigg) \tan \bigg(\frac{3\pi}{2}-\theta \bigg) \\ & = \cos\bigg(\frac{\pi}{2}+\theta\bigg) \tan \bigg(-\frac{\pi}{2}-\theta \bigg) \\ & = \cos\bigg(\frac{\pi}{2}+\theta\bigg) \tan \bigg[ -\bigg(\frac{\pi}{2}+\theta \bigg)\bigg] \\ \dots &\textrm{ as }\tan(-\theta )=-\tan\theta \enspace\dots \\ & = -\cos\bigg(\frac{\pi}{2}+\theta\bigg) \tan \bigg(\frac{\pi}{2}+\theta\bigg) \\ \end{aligned}$ .

Recall that

$\begin{aligned} \because \enspace & \tan\theta = \frac{\sin\theta}{\cos\theta} \\ \therefore \enspace & \cos\theta\tan\theta = \sin\theta \\ \end{aligned}$ .

Hence,

$\begin{aligned} &\quad -\cos\bigg( \frac{\pi}{2}+\theta\bigg)\tan\bigg( \frac{\pi}{2}+\theta\bigg) \\ & = -\sin\bigg( \frac{\pi}{2}+\theta\bigg) \\ \end{aligned}$

Third,

recall that

$\displaystyle{\sin \bigg(\theta\pm\frac{\pi}{2}\bigg)}=\pm\cos\theta$ .

$\begin{aligned} &\quad -\sin \bigg( \frac{\pi}{2}+\theta\bigg) \\ & = -\cos\theta \\ \end{aligned}$

Answer.

$\displaystyle{\cos\bigg[ \frac{1}{2}(4k+1)\pi +\theta \bigg]\tan\bigg[\frac{1}{2}(4k+3)\pi -\theta \bigg]}=-\cos\theta$ .

July 5, 2022October 24, 2022

202207051339 Solution to 1974-CE-AMATH-I-XX

(a) In the figure below, $ABCDEF$ is a regular hexagon. Which one of the $6$ vectors $\overrightarrow{AD}$ , $\overrightarrow{DA}$ , $\overrightarrow{FC}$ , $\overrightarrow{CF}$ , $\overrightarrow{EB}$ , $\overrightarrow{BE}$ is equal to $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{DC}$ ?

(b) $\mathbf{u}$ and $\mathbf{v}$ are unit vectors making an angle $60^\circ$ with each other as shown in the figure below. $AB$ is a vector making an angle $30^\circ$ with $\mathbf{u}$ and $|\overrightarrow{AB}|=2$ . Express $\overrightarrow{AB}$ in terms of $\mathbf{u}$ and $\mathbf{v}$ .

(c) In the figure below, $\angle B=90^\circ$ and $m(\overrightarrow{AB})=10$ . Calculate $\overrightarrow{AB}\cdot\overrightarrow{AC}$ .

(a)

$\begin{aligned} \mathbf{AD} & = -\mathbf{DA} \\ |\mathbf{AD}| & = |\mathbf{DA}| \\ \mathbf{FC} & = -\mathbf{CF} \\ |\mathbf{FC}| & = |\mathbf{CF}| \\ \mathbf{EB} & = -\mathbf{BE} \\ |\mathbf{EB}| & = |\mathbf{BE}| \\ \end{aligned}$

Ans. $\mathbf{FC}$ $\textrm{\scriptsize{OR}}$ $\overrightarrow{FC}$ by inspection.

Working.

$\begin{aligned} &\quad \mathbf{AB} + \mathbf{BC} + \mathbf{DC} \\ & = (\mathbf{AB} + \mathbf{BC}) + \mathbf{DC} \\ & = \mathbf{AC} + \mathbf{DC} \\ \dots & \textrm{ as }\mathbf{AC}=\mathbf{FD} \enspace\dots \\ & = \mathbf{FD} + \mathbf{DC} \\ & = \mathbf{FC} \\ \end{aligned}$

(b)

Write, in Cartesian components of unit vectors $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$ , the following:

$\begin{aligned} \mathbf{AB} & = |\mathbf{AB}|\cos 30^\circ\,\hat{\mathbf{i}} + |\mathbf{AB}|\sin 30^\circ\,\hat{\mathbf{j}} \\ & = 2\cos 30^\circ\,\hat{\mathbf{i}} + 2\sin 30^\circ\,\hat{\mathbf{j}} \\ & = \sqrt{3}\,\hat{\mathbf{i}} + 1\,\hat{\mathbf{j}} \\ \end{aligned}$

and similarly,

$\begin{aligned} \mathbf{u} & = |\mathbf{u}|\,\hat{\mathbf{i}} \\ \mathbf{v} & = |\mathbf{v}|\cos 60^\circ\,\hat{\mathbf{i}} + |\mathbf{v}|\sin 60^\circ\,\hat{\mathbf{j}}\\ & \\ \because\enspace & \mathbf{u}, \mathbf{v}\textrm{ are unit vectors} \\ \therefore\enspace & |\mathbf{u}|=|\mathbf{v}|=1 \\ & \\ \mathbf{u} & = \hat{\mathbf{i}} \\ \mathbf{v} & = \frac{1}{2}\,\hat{\mathbf{i}} + \frac{\sqrt{3}}{2} \hat{\mathbf{j}} \\ \end{aligned}$

$\begin{aligned} \begin{bmatrix} \mathbf{u} \\ \mathbf{v} \end{bmatrix} & = \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} \begin{bmatrix} \hat{\mathbf{i}} \\ \hat{\mathbf{j}} \end{bmatrix} \\ \begin{bmatrix} \hat{\mathbf{i}} \\ \hat{\mathbf{j}} \end{bmatrix} & = \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}^{-1}\begin{bmatrix} \mathbf{u} \\ \mathbf{v} \end{bmatrix} \\ \end{aligned}$

Lemma. (Inversion of 2-by-2 matrices)

For $2\times 2$ matrices, inversion can be done as follows:

$\begin{aligned} \mathbf{A}^{-1} & = \begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} \\ & = \frac{1}{\mathrm{det}\mathbf{A}}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \\ & = \frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \\ \end{aligned}$

_{Wikipedia on Invertible matrix}

Then,

$\begin{bmatrix} 1 & 0 \\ \frac{1}{2} & \frac{\sqrt{3}}{2}\end{bmatrix}^{-1} = \begin{bmatrix} 1 & 0 \\ -\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}} \end{bmatrix}$

so that

$\begin{bmatrix} \hat{\mathbf{i}} \\ \hat{\mathbf{j}} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}} \end{bmatrix}\begin{bmatrix}\mathbf{u} \\ \mathbf{v}\end{bmatrix}$

Thus

$\begin{aligned} \mathbf{AB} & =\begin{pmatrix} \sqrt{3} & 1\end{pmatrix}\begin{bmatrix}\hat{\mathbf{i}} \\ \hat{\mathbf{j}} \end{bmatrix} \\ & = \begin{pmatrix} \sqrt{3} & 1\end{pmatrix} \begin{bmatrix} 1 & 0 \\ -\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}} \end{bmatrix} \begin{bmatrix}\mathbf{u} \\ \mathbf{v}\end{bmatrix} \\ & = \begin{bmatrix} \frac{2}{\sqrt{3}} & \frac{2}{\sqrt{3}} \end{bmatrix}\begin{bmatrix}\mathbf{u} \\ \mathbf{v}\end{bmatrix} \\ &=\frac{2}{\sqrt{3}}\,\mathbf{u}+\frac{2}{\sqrt{3}}\,\mathbf{v} \\ \end{aligned}$

There might be some mistake if conceptually.

(c)

I don’t know how. Skip it.