Typesetting Exercises – Page 3

May 11, 2023May 11, 2023

202305111603 Exercise 3.1

Solve the following equations:

(a) $13x-4=3x+16$ .

(b) $\displaystyle{\frac{3x-9}{18}+\frac{x}{27}-\frac{2x-5}{4}=\frac{4}{3}-x}$ .

(c) $\displaystyle{\frac{3}{x-1}-\frac{2}{x+4}=\frac{4}{2-2x}}$ .

(d) $\displaystyle{\frac{2x}{x-3}-\frac{4x+1}{2x-1}=\frac{21}{2x^2-7x+3}}$ .

(e) $\displaystyle{\frac{1}{4}(3y-2)-\bigg[ y-\frac{1}{y}(7-3y)\bigg] =-\frac{1}{4}y-7}$ .

_{Extracted from K. L. Nielsen. (1958). College Mathematics.}

Roughwork.

(a)

$\begin{aligned} 13x-4 & = 3x+16 \\ 13x-3x & = 16+4 \\ 10x & = 20 \\ x & = 2 \\ \end{aligned}$

(b)

$\begin{aligned} \frac{3x-9}{18} + \frac{x}{27} - \frac{2x-5}{4} & = \frac{4}{3}-x \\ \frac{3x-9}{2\cdot 3^2} + \frac{x}{3^3} - \frac{2x-5}{2^2} & = \frac{4}{3}-x \\ 6(3x-9) + 4x-27(2x-5) & =36(4)-108x \\ 18x-54+4x-54x+135 & =144-108x \\ 18x+4x-54x+108x & = 144+54-135 \\ 76x & = 63 \\ x & = \frac{63}{76} \\ \end{aligned}$

(c)

$\begin{aligned} \frac{3}{x-1}-\frac{2}{x+4} & = \frac{4}{2-2x} \\ \frac{3}{x-1}-\frac{2}{x+4} & = -\frac{2}{x-1} \\ 3(x+4) - 2(x-1) & = -2(x+4) \\ 3x+12-2x+2 & = -2x-8 \\ 3x & = -22 \\ x & = -\frac{22}{3} \\ \end{aligned}$

(d) Not to be attempted.

(e) Not to be attempted.

January 17, 2023May 4, 2023

202301170910 Exercise 3.2

Solve these simultaneous equations:

1. $\begin{cases} x+y=5 \\ xy=6 \\ \end{cases}$

Roughwork. Read More

Substituting $x=5-y$ for $x$ :

$\begin{aligned} (5-y)y & = 6 \\ y^2-5y+6 & = 0 \\ (y-2)(y-3) & = 0 \\ y & = 2,3 \\ \begin{pmatrix}x \\ y \end{pmatrix} = \bigg\{ \begin{pmatrix} 2 \\ 3 \end{pmatrix},\begin{pmatrix} 3 \\ 2 \end{pmatrix} \bigg\} & \\ \end{aligned}$

2. $\begin{cases} x-2y=1\\ x^2+y^2=29 \\ \end{cases}$

Roughwork. Read More

Substituting $x=1+2y$ for $x$ :

$\begin{aligned} (1+2y)^2 + y^2 & =29 \\ (1+4y+4y^2)+y^2 & = 29 \\ 5y^2 +4y -28 & = 0 \\ (5y+14)(y-2) & = 0 \\ y & = -\frac{14}{5},2 \\ \begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}5\\2\end{pmatrix}, \begin{pmatrix}-23/5\\-14/5\end{pmatrix} \bigg\} & \\ \end{aligned}$

3. $\begin{cases} 2x+y=5 \\ x^2-xy=12 \\ \end{cases}$

Roughwork. Read More

Substituting $y=5-2x$ for $y$ :

$\begin{aligned} x^2-x(5-2x) & = 12 \\ 3x^2 -5x -12 & = 0 \\ (3x+4)(x-3) & = 0 \\ x & = -\frac{4}{3},3 \\ \begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}-4/3\\7/3\end{pmatrix},\begin{pmatrix}3\\-1\end{pmatrix}\bigg\} &\\ \end{aligned}$

4. $\begin{cases} 3x-y=7 \\ x^2-xy+y^2=7\\ \end{cases}$

Roughwork. Read More

Substituting $y=3x-7$ for $y$ :

$\begin{aligned} x^2-x(3x-7)+(3x-7)^2 & = 7 \\ 7x^2 -35x +42 & = 0 \\ (7x+7)(x-6) & = 0 \\ x & = -1,6 \\ \begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}-1\\-10\end{pmatrix} , \begin{pmatrix}6\\11\end{pmatrix}\bigg\} & \\ \end{aligned}$

5. $\begin{cases} 5x+y=9 \\ 3xy+y^2=-5 \\ \end{cases}$

Roughwork. Read More

Substituting $y=9-5x$ for $y$ :

$\begin{aligned} 3x(9-5x)+(9-5x)^2 & = -5 \\ 10x^2-63x +86 & = 0 \\ (10x-43)(x-2) & = 0 \\ x & = \frac{43}{10},2 \\ \begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}43/10\\-25/2\end{pmatrix} & , \begin{pmatrix}2\\-1\end{pmatrix}\bigg\} \\ \end{aligned}$

6. $\begin{cases} 3x+2y=13 \\ 3x^2+y^2=31 \\ \end{cases}$

Roughwork. Read More

Substituting $\displaystyle{y = \frac{13-3x}{2}}$ for $y$ :

$\begin{aligned} 3x^2 + \bigg(\frac{13-3x}{2}\bigg)^2 = 31 \\ 7x^2 - 26x+15 & = 0 \\ (7x-5)(x-3) & = 0 \\ x & = \frac{7}{5},3 \\ \begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}7/5\\22/5\end{pmatrix},\begin{pmatrix}3\\-13\end{pmatrix} \bigg\} & \\ \end{aligned}$

7. $\begin{cases} 2x-3y+11=0 \\ 2x^2-xy=36 \\ \end{cases}$

Roughwork. Read More

Substituting $\displaystyle{y=\frac{2x+11}{3}}$ for $y$ :

$\begin{aligned} 2x^2-x\bigg( \frac{2x+11}{3}\bigg) & =36 \\ 4x^2-11x-108 &= 0 \\ (4x-27)(x+4) & = 0 \\ x & = -4,\frac{27}{4} \\ \begin{pmatrix}x\\y\end{pmatrix} & = \bigg\{ \begin{pmatrix}-4\\1\end{pmatrix},\begin{pmatrix}27/4\\49/6\end{pmatrix}\bigg\} \\ \end{aligned}$

8. $\begin{cases} 2x-5y=1\\ x^2-xy+3y^2=9 \\ \end{cases}$

Roughwork. Read More

Substituting $\displaystyle{y=\frac{2x-1}{5}}$ for $y$ :

$\begin{aligned} x^2-x\bigg(\frac{2x-1}{5}\bigg)+3\bigg(\frac{2x-1}{5}\bigg)^2 & = 9 \\ 27x^2 -7x -222 & = 0 \\ (27x+74)(x-3) & = 0 \\ x & = -\frac{74}{27},3 \\ \begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}-74/27\\-35/27\end{pmatrix} , \begin{pmatrix}3\\1\end{pmatrix} \bigg\} & \\ \end{aligned}$

9. $\begin{cases} 2x+3y=7 \\ y^2=26-x^2 \\ \end{cases}$

Roughwork. Read More

Substituting $\displaystyle{x=\frac{7-3y}{2}}$ for $x$ :

$\begin{aligned} y^2 & = 26-\bigg(\frac{7-3y}{2}\bigg)^2 \\ 0 & = 13y^2 -42y -55 \\ 0 & = (13y-55)(y+1) \\ y & = -1,\frac{55}{13} \\ \begin{pmatrix}x\\y\end{pmatrix} & = \bigg\{ \begin{pmatrix}5\\-1\end{pmatrix} , \begin{pmatrix}74/26\\55/13\end{pmatrix}\bigg\} \\ \end{aligned}$

10. $\begin{cases} 5x+3y+7=0 \\ 3y^2=x^2-4y+3 \\ \end{cases}$

Roughwork. Read More

Substituting $\displaystyle{x=\frac{-7-3y}{5}}$ for $x$ :

$\begin{aligned} 3y^2 & = \bigg(\frac{-7-3y}{5}\bigg)^2-4y+3 \\ 0 & = 33y^2+29y-62 \\ 0 & = (33y+62)(y-1) \\ y & = 1,-\frac{62}{33}\\ \begin{pmatrix}x\\y\end{pmatrix} & = \bigg\{\begin{pmatrix} -2 \\1\end{pmatrix},\begin{pmatrix} -3/11\\-62/33\end{pmatrix}\bigg\} \\ \end{aligned}$

_{Extracted from A. Godman & J. F. Talbert. (1975). Additional Mathematics Pure and Applied in SI Units.}

This problem is not to be attempted.

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 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 $9$ .

May 31, 2022October 24, 2022

202205311653 Exercise 1W

When $a=1$ , $b=2$ , $c=5$ , $d=6$ , and $m=0$ , give the value of

1. $ad^2+5bc$ 2. $a^2bc^2-3ad+a^4$ 3. $mabc+3a^2b^2c^2$ 4. $b^2cd-abc^2$ 5. $a^4b^3c+bcd$ 6. $ma^3+4cd-ad$ 7. $\displaystyle{\frac{ab^2}{2}+5c}$ 8. $\displaystyle{\frac{md}{3}+\frac{4cd}{b}}$ 9. $\displaystyle{\frac{ac}{d}+a}$ 10. $\displaystyle{\frac{a^2}{c}+3a^2b}$ 11. $\displaystyle{\frac{ab^2}{a}-\frac{a^5d}{b}}$ 12. $\displaystyle{\frac{c+d}{d}+\frac{d-c}{a}}$ 13. $\displaystyle{\frac{d^2-a^2b^2}{a^2b^2}}$ 14. $\displaystyle{\frac{mcd+3abc}{d^2-4b^2}}$ 15. $\displaystyle{\frac{a^2d^2-9b^2}{abcd}}$

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

Solution.

1.

$\begin{aligned} ad^2+5bc & = (1)(6)^2+5(2)(5) \\ & = 36 + 50 \\ & = 86 \\ \end{aligned}$

2.

$\begin{aligned} a^2bc^2-3ad+a^4 & = (1)^2(2)(5)^2-3(1)(6)+(1)^4 \\ & = 50-18+1 \\ & = 33 \\ \end{aligned}$

3.

$\begin{aligned} mabc+3a^2b^2c^2 & = (0)(1)(2)(5)+3(1)^2(2)^2(5)^2 \\ & = 0 + 3(4)(25) \\ & = 300 \\ \end{aligned}$

4.

$\begin{aligned} b^2cd-abc^2 & = (2)^2(5)(6)-(1)(2)(5)^2 \\ & = (4)(5)(6)-(2)(25) \\ & = 120-50 \\ & = 70 \\ \end{aligned}$

5.

$\begin{aligned} a^4b^3c+bcd & = (1)^4(2)^3(5)+(2)(5)(6) \\ & = 8(5) + 60 \\ & = 40 + 60 \\ & = 100 \\ \end{aligned}$

6.

$\begin{aligned} ma^3+4cd-ad & = (0)(1)^3+4(5)(6)-(1)(6) \\ & = 0 + 120 - 6 \\ & = 114 \\ \end{aligned}$

7.

$\begin{aligned} \frac{ab^2}{2} + 5c & = \frac{(1)(2)^2}{2}+5(5) \\ & = \frac{4}{2} + 25 \\ & = 2 + 25 \\ & = 27 \\ \end{aligned}$

8.

$\begin{aligned} \frac{md}{3}+\frac{4cd}{b} & = \frac{(0)(6)}{3}+\frac{4(5)(6)}{(2)} \\ & = 0 + \frac{120}{2} \\ & = 60 \\ \end{aligned}$

9.

$\begin{aligned} \frac{ac}{d}+a & = \frac{(1)(5)}{(6)} + (1) \\ & = \frac{5}{6} + 1 \\ & = 1\frac{5}{6} \\ \end{aligned}$

10.

$\begin{aligned} \frac{a^2}{c}+3a^2b & = \frac{(1)^2}{(5)}+3(1)^2(2) \\ & = \frac{1}{5}+6 \\ & = 6\frac{1}{5} \\ \end{aligned}$

11.

$\begin{aligned} \frac{ab^2}{a}-\frac{a^5d}{b} & = \frac{(1)(2)^2}{(1)}-\frac{(1)^5(6)}{(2)} \\ & = 4 - 3 \\ & = 1 \\ \end{aligned}$

12.

$\begin{aligned} \frac{c+d}{d}+\frac{d-c}{a} & = \frac{(5)+(6)}{(6)} + \frac{(6)-(5)}{(1)} \\ & = \frac{11}{6} + \frac{1}{1} \\ & = 1\frac{5}{6} + 1 \\ & = 2\frac{5}{6} \\ \end{aligned}$

13.

$\begin{aligned} \frac{d^2-a^2b^2}{a^2b^2} & = \frac{(6)^2-(1)^2(2)^2}{(1)^2(2)^2} \\ & = \frac{36-(1)(4)}{(1)(4)} \\ & = \frac{32}{4} \\ & = 8 \\ \end{aligned}$

14.

$\begin{aligned} \frac{mcd+3abc}{d^2-4b^2} & = \frac{(0)(5)(6)+3(1)(2)(5)}{(6)^2-4(2)^2} \\ & = \frac{0+30}{36-4(4)} \\ & = \frac{30}{36-16} \\ & = \frac{30}{20} \\ & = \frac{3}{2} \\ & = 1\frac{1}{2} \\ \end{aligned}$

15.

$\begin{aligned} \frac{a^2d^2-9b^2}{abcd} & = \frac{(1)^2(6)^2-9(2)^2}{(1)(2)(5)(6)} \\ & = \frac{(1)(36)-9(4)}{60} \\ & = \frac{36-36}{60} \\ & = \frac{0}{60} \\ & = 0 \\ \end{aligned}$

January 25, 2022October 24, 2022

202201260719 Problem 1.57

Let $A=\{ x:x^2-3x+2=0\}$ , $B=\{ x:x^2\le 16\}$ . Determine whether or not $A\subset B$ .

_{Extracted from M. R. Spiegel. (1969). Schaum’s Outline of Theory and Problems of Real Variables}

Set $A$ and set $B$ are described according to property method; if described by the roster method:

$\begin{aligned} A & =\{ 1,2\} \\ B & = \{ -4,-3,-2,-1,0,1,2,3,4\} \\ \end{aligned}$

Obviously set $A$ is a subset of set $B$ .

Roughwork.

(set $A$ )

$\begin{aligned} x^2-3x+2 & = 0 \\ (x-2)(x-1) & = 0 \\ x & = 1,2 \\ \end{aligned}$

(set $B$ )

$\begin{aligned} x^2 & \le 16 \\ x^2 - 16 & \le 0 \\ (x-4)(x+4) & \le 0 \\ -4\le x & \le 4 \\ \end{aligned}$

January 25, 2022October 24, 2022

202201251933 Circumference 001

Draw a circle of radius $R$ around the center $(R,0)$ .

As the center of a circle is equidistant from all points on the circumference, using Pythagorean theorem, we have

$\begin{aligned} (x-R)^2+(y-0)^2 & = R^2 \\ y^2 & = R^2 - (x-R)^2 \\ y & = \pm\sqrt{2xR-x^2} \\ |y| & = \sqrt{2xR-x^2}\\ \end{aligned}$

Attempts. (reinventing the wheel)

Considering the differentials $\mathrm{d}x$ and $\mathrm{d}y$ , we have

$\mathrm{d}y =\displaystyle{\frac{R-x}{\sqrt{2xR-x^2}}\,\mathrm{d}x}$

Suppose I do not know the circumference is $2\pi R$ long. Let its unknown length be $s$ , and let it be partitioned into infinitesimal $\mathrm{d}s$ , such that

$s=\displaystyle{\int\mathrm{d}s}$

Assume we may write

$(\mathrm{d}s)^2=(\mathrm{d}x)^2+(\mathrm{d}y)^2$ .

Expand the right hand side as follows

$\begin{aligned} &\quad \textrm{RHS} \\ & = (\mathrm{d}x)^2+(\mathrm{d}y)^2 \\ & = (\mathrm{d}x)^2 + \bigg( \frac{R-x}{\sqrt{2xR-x^2}}\,\mathrm{d}x\bigg)^2 \\ & = \bigg(\frac{R^2}{2xR-x^2}\bigg)(\mathrm{d}x)^2 \\ \end{aligned}$

Then

$\begin{aligned} \mathrm{d}s & = \sqrt{\bigg( \frac{R^2}{2xR-x^2}\bigg)(\mathrm{d}x)^2} \\ s = \int\mathrm{d}s & = \int_{x=0}^{2R} \frac{R}{\sqrt{2xR-x^2}}\,\mathrm{d}x \\ \end{aligned}$

Suppose I evaluate the integral above $\textrm{\scriptsize{NOT}}$ by direct substitution $\textrm{\scriptsize{BUT}}$ by Riemann sum, so the definite integral due to Riemann is given by

$\displaystyle{\int_{a}^{b}f(x)\,\mathrm{d}x}=\lim_{n\to\infty}\sum_{i=1}^{n}f(x_{i}^{*})\, x$ .

For $i=0,1,2,\dots ,n$ , let $P=\{ x_i\}$ be a regular partition of $[0,2R]$ . Then

$x=\displaystyle{\frac{b-a}{n}=\frac{2R}{n}}$ .

By right-endpoint approximation for Riemann sums, for each interval $[x_{i-1},x_i]$ , we have

$\displaystyle{x_i=x_0+ix=0+i\bigg[\frac{2R}{n}\bigg]=\frac{2Ri}{n}}$ .

Let $f(x)\stackrel{\textrm{def}}{=}\displaystyle{\frac{R}{\sqrt{2xR-x^2}}}$ . Thus,

$\begin{aligned} f(x_i) & = \frac{R}{\sqrt{(2)\displaystyle{\bigg(\frac{2Ri}{n}\bigg)}(R)-\displaystyle{\bigg(\frac{2Ri}{n}\bigg)^2}}} \\ & = \dots \\ & = \frac{n}{2\sqrt{in-i^2}} \\ \end{aligned}$

Writing the Riemann sum in the form

$\begin{aligned} \sum_{i=1}^{n}f(x_i)\, x & = \sum_{i=1}^{n}\frac{n}{2\sqrt{in-i^2}}\bigg(\frac{2Ri}{n}\bigg) \\ & = R\cdot \sum_{i=1}^{n}\frac{1}{\sqrt{\frac{n}{i}-1}} \\ & = R\cdot g(n) \\ \end{aligned}$

Inspecting $R\cdot g(n)$ where

$g(n) = \displaystyle{ \sum_{i=1}^{n}\frac{1}{\sqrt{\frac{n}{i}-1}}}$

I guess, under correction, that $g(n)=2\pi$ .

(to be continued)

Solution. (arc-length parametrization)

Referring to the equation of locus on the very first line:

$(x-R)^2+(y-0)^2=R^2$ ,

then parameterizing $x(\theta )$ , $y(\theta )$ by $\theta$ ,

$\begin{aligned} x & = R+R\cos\theta \\ y & = R\sin\theta \\ \end{aligned}$

and computing the derivatives wrt $\theta$ :

$\begin{aligned} \frac{\mathrm{d}x}{\mathrm{d}\theta} & = -R\sin\theta \\ \frac{\mathrm{d}y}{\mathrm{d}\theta} & = R\cos\theta \\ \end{aligned}$

Note that

$\displaystyle{s=\int\mathrm{d}s}$

where

$\mathrm{d}s=\sqrt{\displaystyle{\bigg(\frac{\mathrm{d}x}{\mathrm{d}\theta}\bigg)^2+\bigg(\frac{\mathrm{d}y}{\mathrm{d}\theta}\bigg)^2}}\,\mathrm{d}\theta$ .

Then,

$\begin{aligned} s & = \int \mathrm{d}s \\ & = \int \sqrt{\bigg( \frac{\mathrm{d}x}{\mathrm{d}\theta}\bigg)^2+\bigg( \frac{\mathrm{d}y}{\mathrm{d}\theta}\bigg)^2}\,\mathrm{d}\theta \\ & = \int \sqrt{(-R\sin\theta )^2+(R\cos\theta )^2}\,\mathrm{d}\theta \\ & = \int \sqrt{R^2}\,\mathrm{d}\theta \\ & = \int_{\theta =0}^{2\pi} R\,\mathrm{d}\theta \\ & = 2\pi R \\ \end{aligned}$

(to be continued)

January 13, 2022July 25, 2022

202201131704 Practice in the Making

Let $x=2$ , $y=3$ , and $z=5$ . Then,

$\begin{aligned} &\quad 2x+3y+5z \\ &= 2(2)+3(3)+5(5) \\ &= 38 \\ & \\ &\quad 3x+5y+2z \\ & = 3(2)+5(3)+2(5) \\ & = 31 \\ & \\ & \quad 5x+2y+3z \\ & = 5(2)+2(3)+3(5) \\ & = 31 \\ \end{aligned}$

That is to say, I have made a set of three equations:

$\begin{cases} 2x+3y+5z =38 \\ 3x+5y+2z =31 \\ 5x+2y+3z =31 \\ \end{cases}$

in matrix form,

$\begin{bmatrix} 2 & 3 & 5 \\ 3 & 5 & 2 \\ 5 & 2 & 3 \\ \end{bmatrix}\begin{bmatrix} x \\ y \\ z\\ \end{bmatrix} = \begin{bmatrix} 38 \\ 31 \\ 31 \\ \end{bmatrix}$

complicating it,

$\begin{bmatrix} 2 & 0 & 3 & 0 & 5 \\ 0 & 0 & 0 & 0 & 0 \\ 3 & 0 & 5 & 0 & 2 \\ 0 & 0 & 0 & 0 & 0 \\ 5 & 0 & 2 & 0 & 3 \\ \end{bmatrix}\begin{bmatrix} x \\ 0 \\ y \\ 0 \\ z\\ \end{bmatrix} = \begin{bmatrix} 38 \\ 0 \\ 31 \\ 0 \\ 31 \\ \end{bmatrix}$

there should be no loss of information; instead, twenty $0$ ‘s are gained .

April 15, 2021April 16, 2021

202104150720 Matching Equal Angles

February 25, 2021August 18, 2025

202102250456 Remarks on “Conceptual Change”

$\textrm{\Huge{\S}}$ 0 Foreword

This article is sheer homework submitted for earning a grade. And it is not intended to survive academic rigor that it may have any contribution to the current literature.

$\textrm{\Huge{\S}}$ 1 Introduction

May the author introduce his readers to the history of conceptual change approach.

The term “conceptual change” originated from The Structure of Scientific Revolutions published by Thomas Kuhn in 1962. Kuhn, first a theoretical physicist and later a scholar in the history of science and philosophy, recorded the scientific revolutionary change that took place in physics at that time—from Newtonian mechanics to quantum mechanics. It was nevertheless the tension beneath that surface that loaned its name— the tension between the Received View $\dagger$ and the Theory Reduction $\ddagger$ (Vosniadou, 2007, pp. 1–2). The idea of “conceptual change” and “paradigm shift” transitioned from the philosophy of science into the science education circle equally appropriately, as did the idea of “gestalt switch” and Piaget’s stages of cognitive development from the study of psychology (Lillian Hoddeson, 2007, pp. 26–27).

(The Received View)

"... [t]he attempts by logical positivists and logical empiricists to treat scientific theories as sets of axioms that could be formulated in mathematical logic."

(The Theory Reduction)

"According to theory reduction, a theory that enjoys a high degree of confirmation cannot ever be disconfirmed, but can only be expanded to a theory with a wider scope, or absorbed into a more inclusive and comprehensive theory (see Suppe, 1977)."

Criticisms brought about to ”conceptual change approach” in teaching science are in many aspects, just to list a few:

“… [C]onceptual change is a slow and gradual process and not a dramatic, gestalt shift that happens over a short period of time.” (Caravita & Hallden, 1994; Vosniadou, 2003; Hatano & Inagaki, 1994; as cited in Vosniadou, 2007, pg. 3)
“… [S]cience learning does not require the replacement of ‘incorrect’ with ‘correct’ conceptions, but the ability on the part of the learner to take different points of view and understand when different conceptions are appropriate depending on the context of use.” (e.g., Pozo, Gomez, & Sanz, 1999; Spada, 1994; as cited in Vosniadou, 2007, pg. 3)
“… [C]ognitive conflict is not a successful instructional strategy for producing conceptual change, as students tend to patch up local inconsistencies in a superficial way.” (Chinn & Brewer, 1993; Smith, diSessa, & Roschelle, 1993; Vosniadou, 1999; as cited in Vosniadou, 2007, pg. 3)

Hoddeson (2007, pg.27) recalled how his PhD teacher Kuhn studied the “conceptual change” as in “Werner Heisenberg’s intellectual route to his formulation of quantum mechanics”. It can be seen that the term “conceptual change” was originally intended to:

$\checkmark$ : make reference to the leap of philosophy of science from the Received View to the Theory Reduction.

$\times$ : it is an intrinsic end, $\textrm{\scriptsize{NOT}}$ an instrumental end, i.e., this approach is descriptive of how science, in particular physics, evolves, but not prescriptive of how science education, inclusive of high school physics, should be attained or designed.

Von Aufschnaiter and Rogge (2010), arguing both theoretically with literature and empirically with their studies on pupils’ conceptual development in physics, posited that pupils are typically lacking in any explanatory conceptual understanding of the content knowledge, and a focus on missing conceptions is much more promising than on misconceptions.

The author couldn’t agree more with all their conclusions, and one of which he cannot overemphasize but ought to restate here is their lately approach—the importance of inclusion of pupils’ misconceptions to teaching is attached to the design of instruction rather than making them an explicit discussion and contrasting them with scientific concepts.

$\textrm{\Huge{\S}}$ 2 Research questions

Andrew A. diSessa (2017, pg. 10) defined “clinical interviewing” to be the technical version of “just talking with people”, and explained how clinical interviewing is connected to instruction. Namely, there are three ways of input.

Input to conjectures and expectations

“Knowing how people think within their zone of felt competence provides important, general and often very specific conjectures about how they can learn from instruction.”

Input to design

“… [S]uch research affects the very goals of instruction (what and when topics should be taught, and how they should be construed), in addition to instructional strategies.”

Input to observation

The clinical interviewer is able to observe “[b]oth the productive and sometimes less productive roles of pre-instructional knowledge”.

Research question

To what extent the argument proposed by Von Aufschnaiter and Rogge, namely, that missing conceptions should be focused prior to misconceptions, is confirmed by a clinical interview with a high school physics pupil?

$\textrm{\Huge{\S}}$ 3 Methodology

One interview was held with a Form 5 pupil, which lasted for eight minutes. A transcript of the interview is in the appendix.

The interviewee had not been taught uniform circular motion before. And the author found it good to have a closed-book talk on this topic with him.

During the interview, there is not any rigid interview protocol, as long as any digression from the subject might still maintain a good understanding between the interviewer and the interviewee.

Delimitation.

Notwithstanding that the interview questions should reveal equally likely misconceptions and missing conceptions, the interviewer must have confirmation bias towards a greater likelihood of missing conceptions.

One single interview alone can demonstrate but that at most one interviewee is able to get along with a pre-lesson discussion on expertise and argumentation. And without following up the interviewee’s learning progress, it cannot give evidence to whether such an axiomatic approach as concerns classical physics is a better alternative to the prevailing conceptual change approach.

Findings of interviews with more interviewees and on more diverse and deeper physics topics are in order for presenting themselves to the current literature.

$\textrm{\Huge{\S}}$ 4 Findings and discussion

Misconception 1: Impetus theory—there exists a force acting upon an object moving with constant velocity

Referring to the transcript from line 25 to 32. The interviewer prompted the interviewee to recall Newton’s first law of motion. The interviewee self-corrected his statement later on though, it might not be an over-extrapolation that he held the impetus theory which is common to physics beginners and laymen.

Such misconception is an instance of ontologically inappropriate miscategorisation of concepts.

Misconception 2: Vector vs. scalar—inability to distinguish speed and velocity.

Referring to the transcript from line 40 to 65. The interviewee was asked whether or not he could differentiate between speed and velocity. Although the reply being affirmative, it can be inferred that he had not mastered the notion of vectors, and could not tell speed and velocity apart.

When debriefed after the interview, he said that he had never studied vectors in mathematics, as the teaching was not then compulsory. What he could tell was by reciting his bookwork, that speed is the rate of change of distance, and that velocity the rate of change of displacement.

Missing conception 1: Centripetal force can be non-contact force.

Referring to the transcript from line 68 to 79. It can be extrapolated that a beginner, when first introduced to the centripetal force acting upon an object in circular motion, might be inclined to think there must be some tension between the centre and the orbiter, and dismiss the fact that centripetal force can account for non-contact force, let alone the fact that it is not a force a priori, but a posteriori the net force.

Missing conception 2: Force vs. pseudo force—distinguishing centripetal and centrifugal forces by inertial or non-inertial frame of reference.

Referring to the transcript from line 97 to 118. It can be observed, that in the scenario when a bus was taking a turn, the interviewee maintained that he should term the force centripetal, despite the fact that its pulling the passenger away from his seat should naturally be termed centrifugal. As an aside, the answer shall depend on which frame—the ground’s or the bus’—is taken reference to, and at this stage it is all too demanding of one high school pupil to tell the difference between inertial and non-inertial frames of reference.

It is worth mentioning, however, that the interviewee admitted that he opted for the term “centripetal” because he had had his supposition that “centrifugal force is a pseudo-force”, based on what disseminated from popular science.

This echoes with that of Ivarsson, Schoultz, and Säljö‘s sociocultural perspective, that human cognition is an interactive process in sociocultural context rather than a purely cognitve one, and “prior knowledge is neither an obstacle nor a prerequisite for conceptual change” (Mayer, 2002, pg. 106).

Misconception 3: The motion of an object is always in the same direction of the net force which it is applied.

Referring to the transcript from line 130 to 152. The interviewer asked the interviewee to consider the situation where there is a centripetal force acting on the satellite by the Earth but the satellite will not clash into the Earth. One reading between the lines could identify there might be one assumption held by the interviewee, that “the motion of an object is the same as that of its net force it experiences”.

The interviewee had previously been taught about electrostatics, and he should have rejected the wrong assumption above by giving a counterexample: the path of a charged particle in an E-field is not necessarily the same as the path of field lines.

Missing conception 3: Relating pre-conceptions to form new ones—from projectile motion to circular motion

Referring to the transcript from line 18 to 20. The interviewee was asked whether he could deduce from definition of uniform circular motion that there must exist a net force acting upon it, still less defining it the centripetal force. It was not until in halfway of the interview that he could understand the axiomatic deduction that come up to its existence.

Referring to the transcript from line 147 to 160. The interviewee could not account for the reason that in the presence of a centripetal force an orbiter and the centre are able to keep their distance. The failure might be mostly due to misconception 3, and partly attributed to an over-compartmentalisation of interrelated physics concepts. That is to say, should he relate an orbiter as a projectile with some initial velocity and a downward constant acceleration perpendicular to its motion, he might be given a clue.

$\textrm{\Huge{\S}}$ 5 Conclusions and implications on teaching

Conceptual change approach has had a long standing reputation for three decades in not only tertiary science teacher education, but teaching high school science also. Care and caution must be taken by educators, to cite Mayer (2002, pg. 127), that it is “… [n]ot a simple process of deletion or replacement of p-prims (i.e., phenomenological primitives), as in contrasting views of conceptual change, but rather a complex process of integration and reorganization”.

The author contends that a pre-instructional talk, chat, or, informal discussion, can help the teacher to design his instruction. Judging misconceptions is with the benefit of hindsight, as searching for missing conceptions is with the benefit of the doubt. Focusing on missing conceptions is indeed more promising than on misconceptions, at least in physics.

Last but not least, as food for thought, if what taught in high school physics is not only classical, but also non-classical (quantum), we have an excuse not to spare the big words—paradigm shift, gestalt switch, conceptual change, and what not. Otherwise, in the senior secondary school physics curriculum, as one in Asian countries, should we not found our education upon normal science, in Kuhn’s terms?

$\textrm{\Huge{\S}}$ 6 Reference

diSessa, A. A. (2017). Knowledge in pieces: an evolving framework for understanding knowing and learning. In Amin, T. G., & Levrini, O. (Eds.). (2017). Converging perspectives on conceptual change: Mapping an emerging paradigm in the learning sciences. Routledge.

Hoddeson, L. (2007). In the wake of Thomas Kuhn’s theory of scientific revolutions: The perspective of an historian of science. In Vosniadou, S., Baltas, A., & Vamvakoussi, X. (Eds.). (2007). Re-framing the conceptual change approach in learning and instruction. Elsevier Ltd.

Mayer, R. E. (2002). Understanding conceptual change: a commentary. In Limón, M. & Mason, L. (Eds.). (2002). Reconsidering conceptual change: issues in theory and practice. Kluwer Academic Publishers

Von Aufschnaiter, C., & Rogge, C. (2010). Misconceptions or missing conceptions?. Eurasia Journal of Mathematics, Science & Technology Education, 6(1).

Vosniadou, S. (2007). The conceptual change approach and its re-framing. In Vosniadou, S., Baltas, A., & Vamvakoussi, X. (Eds.). (2007). Re-framing the conceptual change approach in learning and instruction. Elsevier Ltd.

$\textrm{\Huge{\S}}$ 7 Appendix

Please turn this page over and see Transcript of Interview attached thereafter.

(Submitted on Saturday, 4 May, 2019)