December 28, 2022 – Physicspupil

December 28, 2022December 29, 2022

202212281204 Solution to 2015-DSE-PHY-IA-5

A constant net force acting on an object of mass $m_1$ produces an acceleration $a_1$ while the same force acting on another object of mass $m_2$ produces an acceleration $a_2$ . If this net force acts on an object of mass $(m_1+m_2)$ , what would be the acceleration produced?

Roughwork.

Write

$\begin{aligned} F_\textrm{net} & = m_1a_1 = m_2a_2 \\ & = (m_1+m_2)a_3 \\ \end{aligned}$

Provided are four options. Let’s check them one by one.

A. $a_3\stackrel{?}{=}a_1+a_2$

$\begin{aligned} (m_1+m_2)a_3 & = (m_1+m_2)(a_1+a_2) \\ & = (m_1a_1) + (m_2a_2) +m_1a_2+m_2a_1 \\ & = F_\textrm{net} + F_\textrm{net} +m_1a_2+m_2a_1 \\ & \gneq 2F_\textrm{net} \\ \therefore\enspace a_3 & \neq a_1+a_2 \\ \end{aligned}$

B. $a_3\stackrel{?}{=}\displaystyle{\frac{a_1+a_2}{2}}$

$\begin{aligned} (m_1+m_2)a_3 & = (m_1+m_2)\bigg(\frac{a_1+a_2}{2}\bigg) \\ & \stackrel{\textrm{(A)}}{=} F_\textrm{net} + \frac{m_1a_2+m_2a_1}{2} \\ & \gneq F_\textrm{net} \\ \therefore\enspace a_3 & \neq \frac{a_1+a_2}{2} \\ \end{aligned}$

C. $a_3\stackrel{?}{=}\displaystyle{\frac{a_1a_2}{a_1+a_2}}$

$\begin{aligned} (m_1+m_2)a_3 & = (m_1+m_2)\bigg(\frac{a_1a_2}{a_1+a_2}\bigg) \\ & = \frac{(m_1a_1)a_2+(m_2a_2)a_1}{a_1+a_2} \\ & = \frac{(F_\textrm{net})(a_1+a_2)}{a_1+a_2} \\ & = F_\textrm{net} \\ \therefore\enspace a_3 & = \frac{a_1a_2}{a_1+a_2} \\ \end{aligned}$

D. $a_3\stackrel{?}{=}\displaystyle{\frac{2a_1a_2}{a_1+a_2}}$ is so not to check.

Try-and-err was slower if steadier paced than fright-but-fight from head start,

$\begin{aligned} a_3 & = \frac{F_\textrm{net}}{m_1+m_2} \\ & = \bigg(\frac{m_1}{F_\textrm{net}}+\frac{m_2}{F_\textrm{net}}\bigg)^{-1} \\ & = \bigg(\frac{1}{a_1}+\frac{1}{a_2}\bigg)^{-1} \\ & = \bigg(\frac{a_1+a_2}{a_1a_2}\bigg)^{-1} \\ & = \frac{a_1a_2}{a_1+a_2} \\ \end{aligned}$

And the answer is C.

This problem is not to be attempted.

December 28, 2022January 6, 2023

202212281153 Solution to 1999-CE-AMATH-II-2

Find $\displaystyle{\int x(x+2)^{99}\,\mathrm{d}x}$ .

Roughwork.

$\begin{aligned} &\quad\enspace \int x(x+2)^{99}\,\mathrm{d}x \\ \dots\, &\textrm{let }u=x+2\textrm{ s.t. }\mathrm{d}u=\mathrm{d}x\,\dots \\ & = \int (u-2)(u)^{99}\,\mathrm{d}u \\ & = \int \big( u^{100}-2u^{99}\big)\,\mathrm{d}u \\ & = \frac{u^{101}}{101} - \frac{u^{100}}{50} + C\textrm{ for some constant} \\ & = \frac{(x+2)^{101}}{101} - \frac{(x+2)^{100}}{50} + C \\ \end{aligned}$

This problem is not to be attempted.

December 28, 2022December 28, 2022

202212281117 Problem 1.18

Show that an equation for the circle $C(z_0,r)$ is

$z\overline{z}-\overline{z}_0z-z_0\overline{z}+z_0\overline{z}_0=r^2$ .

_{Extracted from R. B. Ash & W. P. Novinger. (2004). Complex Variables.}

Roughwork.

Writing

$\begin{aligned} |z-z_0| & = r \\ |(a+bi)-(a_0+b_0i)| & = r \\ |(a-a_0)+(b-b_0)i| & = r \\ (a-a_0)^2 + (b-b_0)^2 & = r^2 \\ \end{aligned}$

and

$\begin{aligned} &\quad\enspace z\overline{z}-\overline{z}_0z-z_0\overline{z}+z_0\overline{z}_0 \\ & = (a+bi)(a-bi) - (a_0-b_0i)(a+bi) \\ & \qquad\qquad -(a_0+b_0i)(a-bi)+(a_0+b_0i)(a_0-b_0i) \\ & = \cdots \\ \end{aligned}$

is to let ends meet, without use of properties of conjugates:

$\begin{aligned} |\overline{z}| & = |z| \\ \mathrm{arg}\overline{z} & = -\mathrm{arg}z \\ \overline{z_1+z_2} & = \overline{z}_1+\overline{z}_2 \\ \overline{z_1-z_2} & = \overline{z}_1-\overline{z}_2 \\ \overline{z_1z_2} & = \overline{z}_1\overline{z}_2 \\ \mathrm{Re}z & = (z+\overline{z})/2 \\ \mathrm{Im}z & = (z-\overline{z})/2i \\ z\overline{z} & = |z|^2 \\ \end{aligned}$

(to be continued)