Hong Kong Diploma of Secondary Education Examination (HKDSEE)

July 24, 2023July 24, 2023

202307241011 Solution to 2023-DSE-PHY-IA-25

Two cells of negligible internal resistance are connected to two resistors as shown.

Roughwork.

Label the nodes $a$ , $b$ , $c$ , and $d$ :

Identify potential differences $\Delta V_{ab}$ , $\Delta V_{cd}$ , $\Delta V_{ac}$ , and $\Delta V_{bd}$ :

Lemma. (Kirchhoff’s 2^nd law (/voltage law/loop rule))

The directed sum of the potential differences (voltages) around any closed loop is zero.

_{Wikipedia on Kirchhoff’s circuit laws}

$\begin{aligned} \Delta V_{ab}+\Delta V_{bd}+\Delta V_{dc}+\Delta V_{ca} & = 0 \\ (3)+((-I)(1))+(-(1.5))+(-(I)(2)) & = 0 \\ I & = +0.5\,\mathrm{A} \\ \end{aligned}$

This problem is not to be attempted.

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 23, 2022December 29, 2022

202212231640 Solution to 2016-DSE-PHY-IA-2

$0.3\,\mathrm{kg}$ of water at temperature $50\,^\circ\mathrm{C}$ is mixed with $0.2\,\mathrm{kg}$ of ice at temperature $0\,^\circ\mathrm{C}$ in an insulated container of negligible heat capacity. What is the final temperature of the mixture?

Given: specific heat capacity of water $=4200\,\mathrm{J\,kg^{-1}\,^\circ C^{-1}}$ ; specific latent heat of fusion of ice $3.34\times 10^5\,\mathrm{J\, kg^{-1}}$ .

Roughwork.

For all $0.3\,\mathrm{kg}$ liquid water, a temperature drop of $50\,\mathrm{C^\circ}$ will release

$\begin{aligned} \textrm{Sensible heat} & = (0.3)(4200)(50) \\ & = \textrm{63,000}\,\mathrm{J} \\ \end{aligned}$

and a phase transition of freezing will release further latent heat (of fusion)

$\begin{aligned} \textrm{Latent heat}& = ml_f \\ & = (0.3)(3.34\times 10^5) \\ & = \textrm{100,200}\,\mathrm{J} \\ \end{aligned}$

whereas for $0.2\,\mathrm{kg}$ of solid ice to melt all at once, latent heat (of liquidization)

$\begin{aligned} \textrm{Latent heat} & = ml_f \\ & = (0.2)(3.34\times 10^5) \\ & = \textrm{66,800}\,\mathrm{J} \\ \end{aligned}$

is to be absorbed.

Lemma.

Heat always moves from hotter objects to colder objects, unless energy in some form is supplied to reverse the direction of heat flow.

_{Wikipedia on Second law of thermodynamics}

By the inequalities

$0<\textrm{66,800}-\textrm{63,000}<\textrm{100,200}$

one will know.

December 20, 2022July 24, 2023

202212201748 Solution to 2020-DSE-PHY-IA-23

Three identical resistors, a battery of negligible internal resistance, and an ideal voltmeter are connected to form Circuits (a) and (b) respectively.

Given that the voltmeter reading is $8\,\mathrm{V}$ in Circuit (a), what is the voltmeter reading in Circuit (b)?

Roughwork.

Redrawing a labelled diagram for (a),

and noting that

$\begin{aligned} I_1=I_2 & = I/2 \\ I_3 & = 0 \\ I_4 & = I \\ R_\textrm{V}\parallel R & = R \\ R\parallel R & = R/2 \\ R_{\textrm{eq}} & = 3R/2 \\ \end{aligned}$

we are about to write

$\begin{aligned} I_4R & =8\,\mathrm{V}\textrm{ is given} \\ \mathcal{E} & =IR_{\textrm{eq}}\\ & = I\bigg(\frac{3R}{2}\bigg) \\ & = \frac{3}{2}(8) \\ & = 12\,\mathrm{V} \\ \end{aligned}$

and obtain that the battery has an emf $\mathcal{E}$ of $12\,\mathrm{V}$ . Likewise for (b),

where

$\begin{aligned} I_2 & = I_4 \\ I_3 & = 0 \\ R_\textrm{V}\parallel R & = R \\ R+R_\textrm{V}\parallel R & = 2R \\ R\parallel (R+R_\textrm{V}\parallel R) & = 2R/3 \\ 12= \mathcal{E} & = I\bigg(\frac{2R}{3}\bigg) \\ \Longrightarrow\enspace IR & = 18 \\ \end{aligned}$

$\begin{aligned} & \quad\enspace \begin{cases} I_1R =I_2(2R) \\ I_1+I_2 = 18/R \\ \end{cases} \\ & \Longrightarrow \begin{cases} I_1 = 12/R\\ I_2 = 6/R\\ \end{cases} \end{aligned}$

being now asked for $V=I_4R$ , the voltmeter reading, it is left to the reader.

November 25, 2022November 25, 2022

202211250902 Solution to 2018-DSE-PHY-IA-26

A mobile phone battery labelled $2800\,\mathrm{mA\,h}$ capacity is fully charged initially. What percentage of capacity remains after it has delivered a current of $200\,\mathrm{mA}$ for $3$ hours?

Background. (C-rate)

C-rate (in $\mathrm{h^{-1}}$ ) is defined the charging or discharging current (in $\mathrm{A}$ ) divided by the charge capacity (in $\mathrm{A\,h}$ ).

Background. (Capacity's)

Capacity $C$ rated in ampere hours:

$\begin{aligned} 1\,\mathrm{A\,h} & = 1\,\mathrm{A}\times 1\,\mathrm{h} \\ & = 1\cdot\mathrm{C\,s^{-1}}\times 3600\cdot\mathrm{s} \\ & = 3600\,\mathrm{C} \\ \end{aligned}$

is equivalent to a measure of charge as in coulometry; whereas in watt hours

$\begin{aligned} 1\,\mathrm{W\,h} & = 1\,\mathrm{W}\times 1\,\mathrm{h} \\ & = 1\cdot\mathrm{J\,s^{-1}}\times 3600\cdot\mathrm{s} \\ & = 3600\,\mathrm{J} \\ \end{aligned}$

a measure of energy as in voltammetry.

Background. (state of charge/depth of discharge)

$\textrm{SoC}+\textrm{DoD}=1$

Background. (full/nominal/cutoff voltage)

$\displaystyle{V_\textrm{nominal}=\frac{V_\textrm{full}-V_\textrm{cutoff}}{2}}$

Limited in my knowledge, may I write

$\begin{aligned} \textrm{SoC} & = \frac{2800-200\times 3}{2800} \times 100\% \\ & = 78.6\%\quad \textrm{(3 s.f.)} \\ \end{aligned}$

November 24, 2022November 24, 2022

202211241532 Solution to 2019-DSE-PHY-IA-6

A small ball after projection moves under the effect of gravity only. Its velocity at a certain instant is shown below. What is the speed of the ball $1\,\mathrm{s}$ before? Neglect air resistance. ( $g=9.81\,\mathrm{m\,s^{-2}}$ )

_{This figure is not one original but a modified.}

Roughwork.

Assume the angle of projection be $\theta (t_0) =\theta_0$ with speed $v(t_0)=v_0$ at launch time $t=t_0$ such that the resolved $x$ -, $y$ – components

$\begin{aligned} v_x(v_0,\theta_0 ) & = v_0\cos\theta_0 \\ v_y(v_0,\theta_0 ,t) & = v_0\sin\theta_0 -gt \\ \end{aligned}$

are combined to give the resultant speed at certain instant $t$ :

$\begin{aligned} v(v_0,\theta_0,t) & =\sqrt{\big(v_x(v_0,\theta_0)\big)^2+\big(v_y(v_0,\theta_0,t)\big)^2} \\ & = \sqrt{v_0^2-2v_0(\sin\theta_0)gt+g^2t^2} \\ \end{aligned}$

making with the level angle $\theta (t)$

$\begin{aligned} \theta (v_0,\theta_0,t) & = \arctan\bigg( \frac{v_y(v_0,\theta_0,t)}{v_x(v_0,\theta_0)}\bigg) \\ & = \arctan\bigg( \frac{v_0\sin\theta_0-gt}{v_0\cos\theta_0} \bigg) \\ \end{aligned}$

Now that for some instant $t=T$ we are given

$\begin{aligned} \theta (v_0,\theta_0,T) & =180^\circ -90^\circ -27^\circ =63^\circ \\ v(v_0,\theta_0,T) & =11\,\mathrm{m\,s^{-1}} \\ \end{aligned}$

recall

$\displaystyle{y(x) = \bigg(-\frac{g}{2v_0^2\cos^2\theta_0}\bigg) x^2+(\tan\theta_0)x}$

Equating

$\begin{aligned} \tan\theta \big|_{\theta =63^\circ} & = \frac{\mathrm{d}y}{\mathrm{d}x}\bigg|_{(x_0,y_0)} \\ & = \bigg(-\frac{g}{v_0^2\cos^2\theta_0}\bigg)x_0 +\tan\theta_0 \\ \end{aligned}$

Differentiating on $\theta (v_0,\theta_0,t)$ wrt time $t$ :

$\begin{aligned} \frac{\mathrm{d}\theta}{\mathrm{d}t} & = \frac{\mathrm{d}}{\mathrm{d}t}\bigg[\arctan \bigg(\frac{v_0\sin\theta_0-gt}{v_0\cos\theta_0}\bigg)\bigg] \\ & = \frac{1}{1+\Big(\frac{v_0\sin\theta_0-gt}{v_0\cos\theta_0}\Big)^2} \\ \theta'(t) & = \frac{v_0^2\cos^2\theta_0}{v_0^2-2v_0(\sin\theta_0)gt+g^2t^2} \\ \end{aligned}$

and also on $v(v_0,\theta_0,t)$ :

$\begin{aligned} \frac{\mathrm{d}v}{\mathrm{d}t} & = \frac{\mathrm{d}}{\mathrm{d}t}\bigg( \sqrt{v_0^2-2v_0\sin\theta_0gt+g^2t^2}\bigg) \\ v'(t) & = \frac{v_0(\sin\theta_0)g+g^2t}{\sqrt{v_0^2-2v_0(\sin\theta_0)gt+g^2t^2}} \\ & = \frac{v_0(\sin\theta_0)g+g^2t}{v_0\cos\theta_0} \cdot \sqrt{\theta'(t)} \\ \end{aligned}$

(to be continued)

November 25, 2021October 24, 2022

202111250910 Solution to 2020-DSE-PHY-1A-10

The diameter of Neptune is about $4$ times that of the Earth and its mass is about $17$ times that of the Earth. Estimate the acceleration due to gravity on Neptune’s surface.

Given: acceleration due to gravity on Earth’s surface $g=9.81\,\mathrm{m\, s^{-2}}$

Background.

From Newton’s law of gravitation, Eq. (B7):

$F=\displaystyle{\frac{Gm_{1}m_{2}}{r^2}}$

_{(pg. 15, List of data, formulae and relationship)}

Solution.

On the Earth, for some object of mass $m$ , we see that:

$\begin{aligned} F_{E} & = \frac{GmM_{E}}{r_{E}} \\ mg_{E} & = \frac{GmM_{E}}{r_{E}} \\ g_{E} & = \frac{GM_{E}}{r_{E}} \\ \end{aligned}$

Similarly, for some object of mass $m$ on Neptune, we have

$\begin{aligned} F_{N} & = \frac{GmM_{N}}{r_{N}} \\ mg_{N} & = \frac{GmM_{N}}{r_{N}} \\ g_{N} & = \frac{GM_{N}}{r_{N}} \\ \end{aligned}$

We compare $g_{N}$ to $g_{E}$ by the assumption:

$\begin{aligned} g_{N} & = \frac{GM_{N}}{r_{N}} \\ & = \frac{G(17M_E)}{(4r_{E})} \\ & = \frac{17}{4}\bigg(\frac{GM_{E}}{r_{E}}\bigg) \\ & = \frac{17}{4}g_{E} \\ \end{aligned}$

And the answer is D.

October 11, 2021October 25, 2022

202110111120 Solution to 2012-DSE-PHY-1A-5

There are two forces $\mathbf{F_1}$ and $\mathbf{F_2}$ , of constant magnitudes (i.e., $F_1=|\mathbf{F_1}|=\textrm{Const.}$ ; $F_2=|\mathbf{F_2}|=\textrm{Const.}$ ), acting at the same point. The angle $\theta$ between $\mathbf{F_1}$ and $\mathbf{F_2}$ increases from $0^\circ$ to $180^\circ$ .Apparently from the figure, $F_1>F_2$ .

Let the direction of $\mathbf{F_2}$ be fixed due east.

Then,

$\begin{aligned} \mathbf{F_1} & = F_1\cos\theta\,\hat{\mathbf{i}}+F_1\sin\theta\,\hat{\mathbf{j}} \\ \mathbf{F_2} & = F_2\,\hat{\mathbf{i}} + 0\,\hat{\mathbf{j}} \\ \mathbf{F_3} & = \mathbf{F_1} + \mathbf{F_2} \\ & = ( F_1\cos\theta + F_2 )\,\hat{\mathbf{i}} + F_1\sin\theta\,\hat{\mathbf{j}} \\ \end{aligned}$

$\begin{aligned} F_3 & = |\mathbf{F_3}| \\ & = \sqrt{(F_1\cos\theta + F_2)^2+(F_1\sin\theta )^2} \\ & = \sqrt{(F_1)^2\cos^2\theta + 2F_1F_2\cos\theta +(F_2)^2+(F_1)^2\sin^2\theta} \\ & = \sqrt{(F_1)^2+(F_2)^2+2F_1F_2\cos\theta} \\ \end{aligned}$

If $\theta =0^\circ$ , then $\cos\theta =1$ and $F_3=F_1+F_2$ ;

if $\theta =90^\circ$ , then $\cos\theta =0$ and $F_3=\sqrt{(F_1)^2+(F_2)^2}$ ;

if $\theta =180^\circ$ , then $\cos\theta =-1$ and $F_3=F_1-F_2$ .

By the triangle inequality,

$F_1+F_2>F_3=\sqrt{(F_1)^2+(F_2)^2}$ .

So the magnitude $F_3$ of the resultant force $\mathbf{F_3}$ decreases throughout.

(Countercheck)

Differentiating $F_3$ w.r.t. $\theta$ ,

$\begin{aligned} \quad \frac{\mathrm{d}}{\mathrm{d}\theta}(F_3) & = \frac{\mathrm{d}}{\mathrm{d}\theta} \sqrt{(F_1)^2+(F_2)^2+2F_1F_2\cos\theta} \\ & = \frac{-F_1F_2\sin\theta}{\sqrt{(F_1)^2+(F_2)^2+2F_1F_2\cos\theta}}\\ \end{aligned}$

As $\sin\theta \ge 0$ for $0\le \theta \le \pi$ , and $\sqrt{[\cdots ]}\ge 0$ , we have

$\displaystyle{\frac{\mathrm{d}F_3}{\mathrm{d}\theta}\le 0}$ .

And the answer is A.