Public Examinations in Physics – Page 4

September 16, 2019October 24, 2022

201909160458 Solution to 2001-AL-PHY-IIA-5

(bad example)

The distance between $X$ and $Z$ is $s=ut+\frac{1}{2}at^2$ .

The mid-point $Y$ of $XZ$ is $\frac{1}{2}s$ away from $X$ after some time $t'<t$ .

$\begin{aligned} \frac{1}{2}s & =ut'+\frac{1}{2}at'^2 \\ \frac{1}{2}\bigg( ut+\frac{1}{2}at^2 \bigg) & =ut'+\frac{1}{2}at'^2 \\ 0 & = \bigg( \frac{1}{2}a\bigg) t'^2 + (u)t' + \bigg( -\frac{1}{2}ut-\frac{1}{4}at^2 \bigg) \\ t' & = \frac{-(u)\pm \sqrt{(u)^2-4\big( \frac{1}{2}a\big)\big( -\frac{1}{2}ut-\frac{1}{4}at^2\big)}}{2\big(\frac{1}{2}a\big)} \\ t' & = \frac{-u\pm \sqrt{u^2+aut+\frac{1}{2}a^2t^2}}{a} \\ \dots\textrm{ because } &v-u=at\enspace \dots \\ t' & = \frac{-u\pm \sqrt{u^2+u(v-u)+\frac{1}{2}(v-u)^2}}{a} \\ & = \frac{-u\pm \sqrt{\displaystyle{\frac{u^2+v^2}{2}}}}{a} \end{aligned}$

At mid-point $Y$ the speed $w$ should therefore be

$\begin{aligned} w & =u+at' \\ & = u + a \Bigg( \frac{-u\pm \sqrt{\frac{u^2+v^2}{2}}}{a} \Bigg) \\ & = \pm \sqrt{\displaystyle{\frac{u^2+v^2}{2}}} \end{aligned}$

And the answer is A.

Remark.

What above was done brutally and mechanically, without a sense of physical meaning.

July 28, 2019October 13, 2022

201907280623 Solution to 1982-AL-PHY-I-7

(Non-relativistic approach.)

Set up a 2-D Cartesian coordinate system, the origin being in the position of body $X$ at time $t=0$ , and at the point $(6,0)$ there being body $Y$ .

Then the position of body $X$ and of body $Y$ can each be given by a function of time $t$ :

$\begin{aligned} \mathbf{r}_X(t) & = 3t\, \hat{\mathbf{i}} \\ \mathbf{r}_Y(t) & =6\, \hat{\mathbf{i}} + 4t\, \hat{\mathbf{j}} \end{aligned}$

where $t\in [0,\infty )$ .

The separation $\mathbf{r}_{YX}$ of body $Y$ from body $X$ by time $t$ is:

$\mathbf{r}_{YX}(t)= \mathbf{r}_Y - \mathbf{r}_X =(6-3t)\, \hat{\mathbf{i}} + (4t)\, \hat{\mathbf{j}}$ .

The velocity $\mathbf{v}_{YX}$ of body $Y$ from body $X$ is:

$\begin{aligned} \mathbf{v}_{YX} & =\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}\Big(\mathbf{r}_{YX}\Big) \\ & = -3\, \hat{\mathbf{i}} + 4\, \hat{\mathbf{j}} \end{aligned}$

The magnitude $v_{YX}$ of the velocity is

$v_{YX}=|\mathbf{v}_{YX}|=\sqrt{(-3)^2+(4)^2}=5\quad (\mathrm{m\, s^{-1}})$ .

And the answer is B.

All above is overkill.

Notice and

July 27, 2019July 25, 2022

201907271536 Solution to 1991-CE-PHY-II-4

$T=\displaystyle{\frac{1}{f}}=1/5=0.2\,\mathrm{s}$ when $f=5\,\mathrm{Hz}$ .

Setup.

Given that from the figure.

Doing substitution for $s$ , $u$ , $a$ , and $t$ in

$s=ut+\displaystyle{\frac{1}{2}}at^2$ ,

I have some equations:

$\begin{aligned} 4 &= s_2 = u_1T + \frac{1}{2}aT^2 \\ 12 &= s_3 = u_1(2T) + \frac{1}{2}a(2T)^2 \\ 24 &= s_4 = u_1(3T) + \frac{1}{2}a(3T)^2 \\ \end{aligned}$

Let $a$ be the subject,

$\begin{aligned} a & = \frac{8}{T^2} - \frac{2u}{T} \\ a & = \frac{6}{T^2} - \frac{u}{T} \\ a & = \frac{16}{3T^2} - \frac{2u}{3T} \end{aligned}$

As is known $T=0.2$ ,

$\begin{aligned} a & = 200- 10u \\ a & = 150 - 5u\\ a & = \frac{400}{3} - \frac{10u}{3} \end{aligned}$

Solving,

$u=10\enspace (\mathrm{cm\, s^{-1}})$ and $a=100\enspace (\mathrm{cm})$ .

And the answer is D.

Roughwork.

Unnecessarily tedious,

,

be simple.

July 25, 2019January 19, 2023

201907251758 Solution to 1980-CE-PHY-II-5

The kinetic energy $E_\mathrm{k}$ of an object of mass $m$ and speed $v$ is given by the formula

$E_\mathrm{k}=\displaystyle{\frac{1}{2}}mv^2$ .

In the situation that the object is thrown upwards with initial speed $u$ , and subjected only to gravity $\mathbf{g}$ , it can be expected that after some time of flight $T$ , the object will return to its initial position, its downward speed in which is equal to the initial upward speed $u$ .

Define a piecewise scalar function $v(t)$ of time $t$ :

$v(t) = \begin{cases} u-gt & \quad \textrm{when }0\leq t\leq \displaystyle{\frac{T}{2}} \\ -u+gt & \quad \textrm{when } \displaystyle{\frac{T}{2}}\leq t\leq T \end{cases}$

or simply

$v:[0,T]\subset \mathbb{R} \rightarrow [0,u]\subset \mathbb{R}$ given by $t\mapsto \big|u-g(T-t)\big|$ .

Then

$\begin{aligned} v^2 & =(u-gt)^2\quad \big( =(-u+gt)^2\big) \\ & = u^2-2ugt+g^2t^2 \quad \big(\forall\, t\in [0,T] \big) \end{aligned}$ .

Thus the kinetic energy $E_\mathrm{k}(t)$ is

$\begin{aligned} E_\mathrm{k}(t) & =\displaystyle{\frac{1}{2}}m(u^2-2ugt+g^2t^2) \\ & = \bigg( \displaystyle{\frac{1}{2}}mg^2 \bigg) t^2 + ( -mug ) t + \bigg( \displaystyle{\frac{1}{2}}mu^2 \bigg) \end{aligned}$

where $m$ , $u$ , $g$ are constants.

The kinetic energy $E_\mathrm{k}(t)$ is set to zero at time $t'$ :

$\begin{aligned} t' & = \displaystyle{\frac{-(-mug)\pm\sqrt{\big( -mug\big)^2-4\big(\frac{1}{2}mg^2\big)\big(\frac{1}{2}mu^2\big)}}{2\big(\frac{1}{2}mg^2\big)}} \\ & = \displaystyle{\frac{mug\pm\sqrt{m^2u^2g^2-m^2u^2g^2}}{mg^2}} \\ & =\displaystyle{\frac{u}{g}} \end{aligned}$

Substituting $t'=\displaystyle{\frac{u}{g}}$ for $t$ in $v=u-gt$ :

$v=u-g\bigg( \displaystyle{\frac{u}{g}} \bigg) =0$ ,

as checked.

And the answer is B.

July 25, 2019January 19, 2023

201907251549 Solution to 1984-CE-PHY-II-2

From the graph, the magnitude of velocity $\mathbf{v}$ of an object can be given by a function $v(t)$ of time $t$ :

$v(t)=0.5t+2$ .

The magnitude of displacement $\mathbf{s}$ of the object from time $t=0$ to $t=T$ is a function of time $T$ , given by

$s(T)=\displaystyle{\int_0^T}v(t)\,\mathrm{d}t$ .

In the first 4 seconds ( $T=4$ ),

$\begin{aligned} s(4) & = \int_0^4 (0.5t+2) \,\mathrm{d}t \\ & = \big[0.25t^2+2t\big] \big|_0^4 \\ & = 0.25(4)^2+2(4) \\ & = 12 \end{aligned}$

And the answer is D.

March 31, 2019January 19, 2023

201903310628 Solution to 1980-AL-PHY-I-23

Faraday’s law:

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

where $\varepsilon$ is the e.m.f. induced, $N$ the number of turns in the coil, $\Phi$ the magnetic flux, also ( $\Phi =BA$ ) the product of magnetic flux density $B$ and area $A$ , $t$ the time, and the negative sign due to Lenz’s law.

The magnitude of the e.m.f. induced in the coil is therefore

$\varepsilon = \bigg| -N\displaystyle{\frac{\Delta \Phi}{\Delta t}} \bigg| = \bigg| -N \displaystyle{\frac{\Phi -0}{t}} \bigg| = N\Phi /t$ .

And the answer is C.

March 31, 2019January 19, 2023

201903310609 Solution to 1980-AL-PHY-I-14

By Snell’s law,

$\begin{aligned} n_1\sin\theta_1 & = n_2\sin\theta_2\\ n_1 \sin 45^\circ & = 1 \times \sin 90^\circ \\ n_1 & = \displaystyle{\frac{1}{\sin 45^\circ}} \\ & = \sqrt{2} \\ \end{aligned}$

And the answer is C.

March 31, 2019January 19, 2023

201903310557 Solution to 2004-AL-PHY-II-16

The original electric potential $V_B$ at point $B$ is due to charge $+Q$ at point $A$ :

$V_B =V_{BA} = k \displaystyle{\frac{(+Q)}{(2r)}}=k \displaystyle{\frac{Q}{2r}}$ .

The final electric potential $V_B$ is due to both charges $+Q$ and $-2Q$ at points $A$ and $C$ :

$V_B' = V_{BA} + V_{BC} = k \displaystyle{\frac{Q}{2r}} + k \displaystyle{\frac{(-2Q)}{(r)}}=-3 \bigg( k \displaystyle{\frac{Q}{2r}} \bigg) =-3V_B$ .

And the answer is B.

March 31, 2019January 19, 2023

201903310540 Solution to 1980-AL-PHY-I-2

The original gravitational force $F$ is

$F=G \displaystyle{\frac{(m)(m)}{(2r)^2}}=G \displaystyle{\frac{m^2}{4r^2}}$ .

The final gravitational force $F'$ is

$F'=G \displaystyle{\frac{(m)(m)}{(2r+4r)^2}}=G \displaystyle{\frac{m^2}{36r^2}}=\displaystyle{\frac{1}{9}}\bigg( G \displaystyle{\frac{m^2}{4r^2}} \bigg) = \displaystyle{\frac{1}{9}}F$ .

Thus $F:F'=9:1$ .

And the answer is B.

March 8, 2019January 19, 2023

201903080359 Solution to 2005-AL-PHY-IIA-12

We assume that the charges on a small sphere can be treated as one point charge.

At first, $P$ , $Q$ , and $R$ are of charge $\pm q$ , $\mp q$ , and $0$ .

Then, $P$ and $R$ are put in contact and share their charges evenly. They are now of charge $\pm \displaystyle{\frac{q}{2}}$ . We keep them separated afterwards.

Later on, $R$ and $Q$ are put in contact and share their charges evenly. Each of them is hence of charge:

$\displaystyle{\frac{\bigg( \pm \displaystyle{\frac{q}{2}}\bigg) +(\mp q)}{2}}=\mp\displaystyle{\frac{q}{4}}$ .

We keep them separated afterwards.

The initial magnitude $F$ of electrostatic force between $P$ and $Q$ is given by:

$F=\bigg| \displaystyle{k\frac{(\pm q)(\mp q)}{r^2}} \bigg|=k \displaystyle{\frac{q^2}{r^2}}$

The final magnitude $F'$ of electrostatic force between them is related to the initial $F$ by:

$F'=\bigg| \displaystyle{k\frac{ ( \pm \frac{q}{2}) (\mp \frac{q}{4})}{r^2}} \bigg|=\displaystyle{\frac{1}{8}}\bigg( k \displaystyle{\frac{q^2}{r^2}} \bigg) = \displaystyle{\frac{1}{8}}F$

$\begin{aligned} \quad & \quad & \textrm{Initially} & \quad & \rightarrow & \quad & P\textrm{ and }R\textrm{ in touch} &\quad & \rightarrow & \quad & R\textrm{ and }Q\textrm{ in touch}\\ P & & \pm q & & & & \pm \displaystyle{\frac{q}{2}} & & & & \boxed{\pm \displaystyle{\frac{q}{2}}} \\ Q & & \mp q & & & & \boxed{\mp q} & & & & \mp \displaystyle{\frac{q}{4}}\\ R & & 0 & & & & \pm \displaystyle{\frac{q}{2}} & & & & \mp \displaystyle{\frac{q}{4}} \end{aligned}$

And the answer is B.