December 2021 – Physicspupil

December 10, 2021October 24, 2022

202112100933 Kinematics graphs (Elementary) Q2

This post is depreciated as it is misleading the reader about the speed of train.

The number $038$ should be the “mission order” of the train which indicates to stationed staff of its running railway, service time, and need of assistance if any (MTR Academy, 2017).

An MTR train enters the station at a speed of $38$ kilometres an hour, i.e., $38\,\mathrm{km/h}$ .

_{Retrieved image from http://mtr.hk365day.com/}

If the subway is $100\,\mathrm{m}$ long and the train terminates at the stop with constant deceleration, i.e., $\mathbf{a}=-a\,\hat{\mathbf{i}}\quad (a=\textrm{Const.}>0)$ ,

_{Modified figures retrieved from https://www.shutterstock.com/}

what is the time required for the train to come to a full stop?

Background. (Equations of linear motion in uniform acceleration)

$\begin{cases} \enspace & v = u + at \\ \enspace & s = \displaystyle{\frac{(u+v)}{2}t} \\ \enspace & s = \displaystyle{ut+\frac{1}{2}at^2} \\ \enspace & v^2 = u^2 + 2as \\ \end{cases}$

Solution.

Take the rightward to be positive direction.

Provided that the initial velocity $\mathbf{u}$ is

$\begin{aligned} \mathbf{u} & =+38\,(\mathrm{km\, h^{-1}})\,\hat{\mathbf{i}} \\ & = +38\times\frac{1000}{60\times 60}\,(\mathrm{m\, s^{-1}})\,\hat{\mathbf{i}} \\ & = +10.5556\,(\mathrm{m\, s^{-1}})\,\hat{\mathbf{i}}\quad (4\,\mathrm{d.p.}) \\ \end{aligned}$

the final velocity $\mathbf{v}$ is

$\mathbf{0}$ , or simply put, $0$ ;

and the displacement $\mathbf{s}$ for the duration is

$\begin{aligned} \mathbf{s} & =s\,\hat{\mathbf{i}} \\ & =+100\,\hat{\mathbf{i}}\\ \end{aligned}$ ,

so, out of five variables:

$a$ , $s$ , $t$ , $u$ , and $v$ ,

we already know three exactly:

$\mathbf{s}$ (of magnitude $s$ );
$\mathbf{u}$ (of magnitude $u$ );
$\mathbf{v}$ (of magnitude $v$ ).

If the first step were to solve for only one unknown in the equations of motion i – iv below,

i. $v=u+at$ is $\textrm{\scriptsize{NOT}}$ solvable for there are two unknowns $a$ and $t$ ;

ii. $s=\frac{(u+v)}{2}t$ solvable for there is $\textrm{\scriptsize{ONLY}}$ one unknown $t$ ;

iii. $s=ut+\frac{1}{2}at^2$ $\textrm{\scriptsize{NOT}}$ solvable for there are two unknowns $a$ and $t$ ;

iv. $v^2=u^2+2as$ solvable for there is $\textrm{\scriptsize{ONLY}}$ one unknown $a$ .

thus, we should pick equation ii. to calculate the unknown $t$ .

That said, solving for time $t$ ,

$\begin{aligned} s & = \frac{(u+v)}{2}t \\ 100 & = \frac{(10.5556+0)}{2}t \\ t & = 18.9\,\mathrm{s}\quad \textrm{(3 s.f.)} \\ \end{aligned}$

Afterword.

$\dagger$ If you wish to know about the rate of deceleration $-a$ , you can use equation iv., yet this is left the reader.

$\ddagger$ The time $t$ might seem longer than expected, because normally the deceleration of train is non-constant.

December 4, 2021June 6, 2023

202112041209 Homework 1 (Q5)

Suppose that light of intensity $10^{-9}\,\mathrm{W/m^2}$ normally shines on a metal surface. The metal is made up of a simple cubic lattice of atoms with lattice spacing $0.3\,\mathrm{nm}$ . Each atom has one free electron. The binding energy at the metal surface is $8\,\mathrm{eV}$ . Suppose further that the light is uniformly distributed over the surface and all its energy is absorbed by the surface electrons.

(a) If the incident radiation were well described by classical physics, how long would one have to wait after switching on the light source until an electron gains enough energy to be released as a photoelectron?

(b) In reality, how long is this time duration? Explain briefly.

Reading Comprehension.

Highlighting some keyword(s) will help doing the question:

(S1) […] light […] normally shines on a […] surface […] ;

(S2) […] metal is made up of a simple cubic lattice of atoms […] ;

(S3) […] Each atom has one free electron […] .

Solution.

(The solution below is based on the manuscript of 2015-2016 PHYS2265 Modern Physics Homework 1 Solution.)

(a)

There are eight atoms each lattice and one electron each atom. Hence,

$\begin{aligned} t & = \frac{8 \times (1.602\times 10^{-19})}{10^{-9} \times (0.3\times 10^{-9})^2} \\ & = 1.424\times 10^{10}\,\mathrm{s} \\ \end{aligned}$

(b)

No electron can be released as a photoelectron, for no photon has energy greater than $8\,\mathrm{eV}$ . (Why?)

December 3, 2021November 15, 2022

202112031054 Kinematics graphs (Elementary) Q1

The graph below illustrates three paths in Red ( $R$ ), Green ( $G$ ), and Blue ( $B$ ).

For a person walking along paths $R$ , $G$ , and $B$ at a constant speed $2\,\mathrm{m\, s^{-1}}$ , find, in each path,

(a) the distance travelled;
(b) the time needed from start to finish; and
(c) the displacement and velocity on the journey.

Solution.

(a)

Along path $R$ , the walking distance $d$ is

$\begin{aligned} \textrm{Distance }d & = \bigg(\frac{1}{2}\bigg) \big(\pi (90-50)\big) + \bigg(\frac{1}{2}\bigg) \big( \pi (50-30)\big) \\ & = \bigg(\frac{1}{2}\bigg) (40\pi ) + \bigg(\frac{1}{2}\bigg) (20\pi ) \\ & = 20\pi + 10\pi \\ & = 30\pi\,\mathrm{m} \\ \end{aligned}$

Along path $G$ , the walking distance $d$ is

$\begin{aligned} \textrm{Distance }d & = \sqrt{(30-0)^2+(60-20)^2} + \sqrt{(30-0)^2+(100-60)^2} \\ & = \sqrt{900+1600} + \sqrt{900+1600} \\ & = \sqrt{2500} + \sqrt{2500} \\ & = 50+50 \\ & = 100\,\mathrm{m} \end{aligned}$

Along path $B$ , the walking distance $d$ is

$\begin{aligned} \textrm{Distance }d & = (20-0) + (20-0) + (50-20) + (50-20) \\ & \qquad\quad + (100-50) + (100-50) \\ & = 20+20+30+30+50+50 \\ & = 200\,\mathrm{m} \\ \end{aligned}$

(b)

Along path $R$ , the time $t$ needed is

$\begin{aligned} \textrm{Time }t & = \frac{30\pi\,\mathrm{m}}{2\,\mathrm{m\, s^{-1}}} \\ & = 47.1\,\mathrm{s}\\ \end{aligned}$

Along path $G$ , the time $t$ needed is

$\begin{aligned} \textrm{Time }t & = \frac{100\,\mathrm{m}}{2\,\mathrm{m\, s^{-1}}} \\ & = 50\,\mathrm{s}\\ \end{aligned}$

Along path $B$ , the time $t$ needed is

$\begin{aligned} \textrm{Time }t & = \frac{200\,\mathrm{m}}{2\,\mathrm{m\, s^{-1}}} \\ & = 100\,\mathrm{s}\\ \end{aligned}$

(c)

Read the following graph, and you shall see each and every displacement in dashed lines.

For path $R$ , the displacement $\mathbf{s}$ travelled is

$\begin{aligned} \textrm{Displacement }\mathbf{s} & = - s\,\hat{\mathbf{i}} \\ & = - (90-30)\,\hat{\mathbf{i}} \\ & = - 60\,\mathrm{m}\,\hat{\mathbf{i}} \\ \end{aligned}$

and the velocity $\mathbf{v}$ is

$\begin{aligned} \textrm{Velocity }\mathbf{v} & = \frac{\mathbf{s}}{t} \\ & = \frac{- 60\,\mathrm{m}\,\hat{\mathbf{i}}}{47.1\,\mathrm{s}} \\ & = -1.27\,\mathrm{m\, s^{-1}}\,\hat{\mathbf{i}} \\ \end{aligned}$

For path $G$ , the displacement $\mathbf{s}$ travelled is

$\begin{aligned} \textrm{Displacement }\mathbf{s} & = s\,\hat{\mathbf{j}} \\ & = (100-20)\,\hat{\mathbf{j}} \\ & = 80\,\mathrm{m}\,\hat{\mathbf{j}} \\ \end{aligned}$

and the velocity $\mathbf{v}$ is

$\begin{aligned} \textrm{Velocity }\mathbf{v} & = \frac{\mathbf{s}}{t} \\ & = \frac{80\,\mathrm{m}\,\hat{\mathbf{j}}}{50\,\mathrm{s}} \\ & = +1.6\,\mathrm{m\, s^{-1}}\,\hat{\mathbf{j}} \\ \end{aligned}$

For path $B$ , the displacement $\mathbf{s}$ travelled is

$\begin{aligned} \textrm{Displacement }\mathbf{s} & = 100\,\mathrm{m}\,\hat{\mathbf{i}} + 100\,\mathrm{m}\,\hat{\mathbf{j}} \\ \end{aligned}$

or, the magnitude of displacement is

$\begin{aligned} s & = \sqrt{(100)^2+(100)^2} \\ & = 100\sqrt{2}\,\mathrm{m} \\ \end{aligned}$

such that

$\mathbf{s} = s\cos 45^\circ\,\hat{\mathbf{i}} + s\sin 45^\circ\,\hat{\mathbf{j}}$

the velocity $\mathbf{v}$ is

$\begin{aligned} \textrm{Velocity }\mathbf{v} & = \frac{\mathbf{s}}{t} \\ & = \frac{100\,\mathrm{m}\,\hat{\mathbf{i}} + 100\,\mathrm{m}\,\hat{\mathbf{j}}}{100\,\mathrm{s}} \\ & = +1\,\mathrm{m\, s^{-1}}\,\hat{\mathbf{i}} + 1\,\mathrm{m\, s^{-1}}\,\hat{\mathbf{j}}\\ \end{aligned}$

December 2, 2021October 24, 2022

202112021444 Solution to 1979-CE-AMATH-I-1

Let $\displaystyle{y=x+\frac{1}{x^2}}$ . Find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ from first principles.

Solution.

From first principles,

$\begin{aligned} y:=f(x) & =x+\frac{1}{x^2} \\ f(x+\Delta x) & =(x+\Delta x) + \frac{1}{(x+\Delta x)^2} \\ f(x+\Delta x)-f(x) & = \Delta x + \frac{1}{(x+\Delta x)^2} - \frac{1}{x^2} \\ & = \Delta x + \frac{x^2-(x+\Delta x)^2}{x^2(x+\Delta x)^2} \\ & = \Delta x - \frac{2x\Delta x+(\Delta x)^2}{x^2(x+\Delta x)^2} \\ \frac{f(x+\Delta x)-f(x)}{\Delta x} & = 1 - \frac{2x + \Delta x}{x^2(x+\Delta x)^2} \\ \lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} & = 1 - \frac{2x + 0}{x^2(x+0)^2} \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = 1 - \frac{2}{x^3} \\ \end{aligned}$

Playground.

To make fun of calculus, do let

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

where $f(x,y)$ is a degree $3$ polynomial in two variables $x$ and $y$ .

The set of solutions to $f(x,y)=0$ is

$\begin{cases} \enspace x \neq 0 \\ \enspace y = \frac{x^3+1}{x^2} \end{cases}$

Computing $\partial_xf$ and $\partial_yf$ as follow:

$\begin{aligned} \partial_xf & = \frac{\partial}{\partial x}f(x,y) \\ & = 3x^2-2xy \\ \end{aligned}$

and

$\begin{aligned} \partial_yf & = \frac{\partial}{\partial y}f(x,y) \\ & = -x^2 \\ \end{aligned}$ ,

we oversee the relation

$\mathrm{d}f = (\partial_xf)(\Delta x)+(\partial_yf)(\Delta y)$ .

Definition. (Singularity and Smoothness)

A point $p=(a,b)$ on a curve $\mathcal{C}=\{ (x,y)\in\mathbb{C}^2:f(x,y)=0\}$ is said to be singular if

$\begin{aligned} \frac{\partial f}{\partial x}(a,b) & = 0 \\ \frac{\partial f}{\partial y}(a,b) & = 0 \\ \end{aligned}$ .

A point that is not singular is called smooth. If there is at least one singular point on $\mathcal{C}$ , then curve $\mathcal{C}$ is called a singular curve. If there are no singular points on $\mathcal{C}$ , the curve $\mathcal{C}$ is called a smooth curve.

_{C.f. Definition 1.9.1, T. Carrity, et al., Algebraic Geometry: A Problem Solving Approach}

That said, in our scenario

$f(x,y)=x^3-x^2y+1$ ,

the set of candidates for singularity

$\{ (a,b)\in\mathbb{C}^2\enspace \mathrm{s.t.}\enspace f(a,b)=0\textrm{ and }\partial_xf(a,b)=\partial_yf(a,b)=0\}$ .

is empty.

$\therefore$ The curve $\mathcal{C}=\{ (x,y)\in\mathbb{C}^2:x^3-x^2y+1=0\}$ is smooth, thus everywhere differentiable.

Wait, the function in question should be the curve

$F(x)=\displaystyle{x+\frac{1}{x^2}}$ ;

still, for better or worse, $F(x)$ is $\textrm{\scriptsize{NOT}}$ a differentiable function because the derivative does $\textrm{\scriptsize{NOT}}$ exist at $x=0$ , for that point is a discontinuity.

December 2, 2021October 24, 2022

202112021138 SQL Table 004

Background. (Equilibrium)

Equilibrium is established by the exchange of energy, volume, or particle number between different systems or subsystems.

_{Section 2.2. Arovas. (2019). Thermodynamics and Statistical Mechanics}

CREATE TABLE SQL_Table_004
    (ID int,
    Object_of_exchange varchar(255),
    Constancy varchar(255)
    Subject_of_equilibrium varchar(255)
    );
INSERT INTO SQL_Table_004
    (Object_of_exchange, Constancy, Subject_of_equilibrium)
    VALUES
    ('energy', 'temperature (T)', 'thermal')
    ('volume', 'momentum/temperature (p/T)', 'mechanical')
    ('particle', 'chemical potential/temperature (\mu /T)', 'chemical')

CREATE TABLE SQL_Table_004

(ID int,

Object_of_exchange varchar(255),

Constancy varchar(255)

Subject_of_equilibrium varchar(255)

);

INSERT INTO SQL_Table_004

(Object_of_exchange, Constancy, Subject_of_equilibrium)

VALUES

('energy', 'temperature (T)', 'thermal')

('volume', 'momentum/temperature (p/T)', 'mechanical')

('particle', 'chemical potential/temperature (\mu /T)', 'chemical')

Output.
+------+---------------------+-------------------------------+--------------------------+
|  ID  | Object_of_exchange  |  Constancy                    |  Subject_of_equilibrium  |
+------+---------------------+-------------------------------+--------------------------+
|  1   |  energy             |  temperature (T)              |  thermal                 |
|  2   |  volume             |  momentum/temperature (p/T)   |  mechanical              |
|  3   |  particle           |  chemical potential/          |  chemical                |
|      |                     |    temperature (\mu /T)       |                          |
+------+---------------------+-------------------------------+--------------------------+

Output.

+------+---------------------+-------------------------------+--------------------------+

+------+---------------------+-------------------------------+--------------------------+

+------+---------------------+-------------------------------+--------------------------+

December 2, 2021December 12, 2022

202112021037 Exercise 1.4 Lagrangian actions

For a free particle an appropriate Lagrangian is

Eq. (1.8):

$\mathcal{L}(t,x,v)=\displaystyle{\frac{1}{2}mv^2}$ .

Suppose that $x$ is the constant-velocity straight-line path of a free particle, such that $x_a=x(t_a)$ and $x_b=x(t_b)$ . Show that the action on the solution path is

Eq. (1.9):

$\displaystyle{\frac{m}{2}\frac{(x_b-x_a)^2}{t_b-t_a}}$ .

_{Extracted from Structure and Interpretation of Classical Mechanics, SICP, 2e}

Background. (Lagrangians and Lagrangian actions)

The function $\mathcal{L}$ is called a Lagrangian for the system, and the resulting action,

Eq. (1.4):

$S[q](t_1,t_2)=\displaystyle{\int_{t_1}^{t_2}\mathcal{L}\circ\Gamma [q]}$ ,

is called the Lagrangian action. For Lagrangians that depend only on time, positions, and velocities the action can also be written

Eq. (1.5):

$S[q](t_1,t_2)=\displaystyle{\int_{t_1}^{t_2}\mathcal{L}(t,q(t),\mathrm{D}q(t))\,\mathrm{d}t}$ .

_{(Section 1.3 The Principle of Stationary Action)}

Working. (roughly)

$\begin{aligned} S & = \int_{t_a}^{t_b}\frac{1}{2}mv^2\,\mathrm{d}t \\ & = \frac{m}{2}\int_{t_a}^{t_b}(\mathbf{v}\cdot\mathbf{v})\,\mathrm{d}t \\ & = \frac{m}{2}\int_{t_a}^{t_b}\bigg( \frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}\cdot\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}\bigg) \,\mathrm{d}t \\ & = \frac{m}{2}\int_{x_a,t_a}^{x_b,t_b}\frac{(\mathrm{d}x)^2}{(\mathrm{d}t)} \\ & = \frac{m}{2}\frac{(x_b-x_a)^2}{t_b-t_a} \\ \end{aligned}$