December 2022 – Page 4

December 7, 2022December 7, 2022

202212071606 Solution to 1976-AL-AMATH-I-8

During an epidemic in a country, the rate of spread of the disease is proportional to the number of people who are healthy; the rate of cure is proportional to the number currently sick. Let $P$ be the total population:

(a) Show that

$\displaystyle{\frac{\mathrm{d}W}{\mathrm{d}t}=kP-(k+k_1)W}$

where $W$ is the number of sick people; $k$ and $k_1$ are constants.
(b) Let the total population be growing at a steady rate, so that (independent of the epidemic)

$P=a+bt$

where $a$ and $b$ are positive. Show that the number of sick people is given by

$W=\alpha +\beta t+ce^{\mu t}$ .

Calculate $\alpha$ , $\beta$ , and $\mu$ in terms of the given constants $a$ , $b$ , $k$ , and $k_1$ .
What further information would you require to determine $c$ ?
(c) After a very long time, what is the proportion of people who will be in a state of sickness?

Roughwork.

$\begin{aligned} \frac{\mathrm{d}W}{\mathrm{d}t}+(k+k_1)W & = k(a+bt) \\ \dots\,\textrm{ by integrating factor } I=e^{\int (k+k_1)\,\mathrm{d}t} & =e^{(k+k_1)t}\,\dots \\ e^{(k+k_1)t}\bigg(\frac{\mathrm{d}W}{\mathrm{d}t}\bigg) + \Big( e^{(k+k_1)t} (k+k_1)\Big) W & = e^{(k+k_1)t} k(a+bt) \\ e^{(k+k_1)t}W & = \int e^{(k+k_1)t}k(a+bt)\,\mathrm{d}t \\ \end{aligned}$

$\begin{aligned} \textrm{RHS} & = \int e^{(k+k_1)t}k(a+bt)\,\mathrm{d}t \\ & = ka\int e^{(k+k_1)t}\,\mathrm{d}t + kb\int te^{(k+k_1)t}\,\mathrm{d}t \\ & = \frac{ka}{k+k_1}\Big( e^{(k+k_1)t}\Big) + \frac{kb}{(k+k_1)^2}\int t'e^{t'}\,\mathrm{d}t' \\ & = \frac{ka}{k+k_1}\Big( e^{(k+k_1)t}\Big) + \frac{kb}{(k+k_1)^2}\Big( e^{(k+k_1)t}((k+k_1)t-1)+C\Big) \\ W &= \textrm{RHS}\Big/ e^{(k+k_1)t} \\ & = \bigg(\frac{ka}{k+k_1}-\frac{kb}{(k+k_1)^2}\bigg) + \bigg(\frac{kb}{k+k_1}\bigg) t + \bigg(\frac{kbC}{(k+k_1)^2}\bigg) e^{(-(k+k_1))t} \\ & = \alpha +\beta t+ce^{\mu t} \\ \end{aligned}$

$\begin{aligned} \frac{W}{P} & = \frac{\alpha +\beta t+ce^{\mu t}}{a+bt} \\ \lim_{t\to \infty}\frac{W}{P} & = \lim_{t\to\infty}\frac{\beta t}{a+bt} \\ & = \frac{\beta}{b} \\ & = \frac{k}{k+k_1} \\ \end{aligned}$

This problem is not to be attempted.

December 7, 2022December 8, 2022

202212071217 Solution to 1971-HL-PHY-I-7

The figure below shows a three-dimensional network in the form of a pyramid, in which $A$ is the apex, and $BCDE$ the square base.

Each of the eight edges of the pyramid is a wire of resistance $1\,\mathrm{\Omega}$ . A $12\,\mathrm{V}$ battery with internal resistance of $0.1\,\mathrm{\Omega}$ is connected across $B$ and $D$ . Calculate

(a) the power input of the network,
(b) the terminal voltage across the battery when current flows through the network, and
(c) the potential at the points $B$ , $C$ , $D$ , and $E$ if the apex $A$ is earthed.

Roughwork.

Draw the circuit.

Label the potential.

Straighten the main.

Calculate the equivalent.

$\begin{aligned} R_{\textrm{eq}} & = \big( R\parallel R\parallel R\big) + \big( R\parallel R\parallel R\big) \\ & = \frac{1}{\frac{1}{R}+\frac{1}{R}+\frac{1}{R}}\times 2 \\ & = \frac{2R}{3} \\ & = \frac{2(1)}{3} \\ & = \frac{2}{3}\,\mathrm{\Omega} \\ \end{aligned}$

This problem is not to be attempted.

December 6, 2022December 6, 2022

202212061712 Solution to 1967-CE-AMATH-I-X

If the population of Hong Kong is $4$ million in $1967$ and it increases by $3\%$ each year, what will be the estimated population in $1987$ ? Give your answer correct to $4$ significant figures.

Roughwork.

$\begin{aligned} & \quad\enspace (4\times 10^6)(1+3\%)^{(1987-1967)} \\ & = (4\times 10^6)(1+0.03)^{20} \\ & = (4\times 10^6)\sum_{n=0}^{20}\begin{pmatrix}20\\ n\end{pmatrix}(1)^{20-n}(0.03)^n \\ & = (4\times 10^6)\sum_{n=0}^{20}\begin{pmatrix}20\\ n\end{pmatrix}(0.03)^n \\ & = \cdots \\ \end{aligned}$

This problem is not to be attempted.

December 6, 2022December 6, 2022

202212061213 Solution to 1976-HL-PHY-I-4

(a) A kettle of negligible heat capacity, full of liquid, was placed over a burner. It was found that the liquid, initially at a temperature of $297\,\mathrm{K}$ , reached the boiling point of $378\,\mathrm{K}$ in $9$ minutes, and after another $60$ minutes all the liquid in the kettle boiled away. Neglecting heat loss, calculate the latent heat of vaporisation of the liquid. Specific heat capacity of the liquid $=\textrm{4,000}\,\mathrm{J\,kg^{-1}\,K^{-1}}$ .

(b) Describe another method to determine the latent heat of vaporisation of a liquid.

Roughwork.

(a)

Let $P$ be the power of the burner, and $m$ the mass of the liquid. Assume no heat loss to the surroundings, then

$\begin{aligned} Pt = E & = Q = mc\Delta T \\ P\times 9(60) & = m\times 4000\times (378-297) \\ \frac{P}{m} & = 600 \\ \end{aligned}$

Suppose for the liquid the latent heat of vaporisation is $l_v$ , then

$\begin{aligned} Pt = E & = Q = ml_v \\ P\times 60(60) & = ml_v \\ l_v & = 3600\times\bigg(\frac{P}{m}\bigg) \\ & = 3600\times (600) \\ & = 2.16\times 10^6\,\mathrm{J\,kg^{-1}} \\ \end{aligned}$

(b) Left as an exercise to the reader.

December 5, 2022December 5, 2022

202212051136 Solution to 1978-AL-AMATH-II-3

A smooth homogeneous hollow right circular cylinder with open, flat ends stands freely on smooth horizontal ground so that its axis is vertical. Two spheres $A$ and $B$ of radii $a$ and $b$ ( $a<b$ ) and weights $W_a$ and $W_b$ respectively rest in equilibrium inside the cylinder as shown in the diagram. Suppose that the internal and external radii of the cylinder are $c$ and $d$ respectively, where $c<(a+b)$ .

i. Show that the vertical forces acting on the spheres reduce to a couple. Determine the moment of the couple.
ii. Determine the minimum weight of the cylinder such that it will not overturn.
iii. A third sphere $C$ , identical to $A$ , is then placed on top of $B$ in contact with the cylinder. Determine the minimum weight of the cylinder so that it will not overturn for the two possible equilibrium positions of $C$ .

You may assume that the cylinder is tall enough to hold all the spheres.

Roughwork.

WLOG reduce the problem from three-dimensional to two. Begin with three free-body diagrams as follow:

So many unknowns I don’t know how to get set.

(to be continued)

December 2, 2022December 5, 2022

202212021339 Solution to 1972-HL-PHY-I-2

A frictionless circular track of radius $15.3\,\mathrm{cm}$ is fixed vertically on the floor. A particle at the top of the track glides down from rest.

Find the height at which the particle will begin to leave the circular track. Calculate the horizontal and vertical components of the velocity with which the particle strikes the floor.

Roughwork.

Set up a coordinate system:

with an interface:

Write resolved $x$ -, $y$ -components of normal reaction $N$

$\begin{aligned} N_x & = N\cos\theta = mg\sin 2\theta \\ N_y & = N\sin\theta =mg\sin^2\theta \\ \end{aligned}$

For the particle to lose contact with the surface, necessarily there exists some largest possible angle $\theta\in [0,90^\circ )$ s.t.

$N_x(\theta )\textrm{ \scriptsize{OR} }N_y(\theta )=0$ ;

and sufficiently some smallest possible period $t\in \Big[0, \sqrt{\frac{2R}{g}}\Big)$ s.t.

$s_x^2(t)+s_y^2(t)>R^2$

hereby SUVAT equations of motion do $\textrm{\scriptsize{NOT}}$ apply because of non-uniform acceleration $a(\theta )$ depending on $\theta$ , e.g.,

$\begin{aligned} \textrm{Net }F_x = ma_x & = N_x \\ a_x(\theta ) & = g\sin 2\theta \\ \textrm{Net }F_y = ma_y & = W-N_y \\ a_y & = g-g\sin^2\theta \\ a_y(\theta ) & = g\cos^2\theta \\ \end{aligned}$

By quotient rule,

$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}\theta}\bigg(\frac{s_y(\theta)}{s_x(\theta )}\bigg) & = \frac{\mathrm{d}}{\mathrm{d}\theta} (\tan\theta )\\ \frac{s_x(\theta)v_y(\theta)-s_y(\theta)v_x(\theta)}{s_x^2(\theta)} & = \sec^2\theta \\ \end{aligned}$

Try considering the Lagrangian $\mathcal{L}=T-V$ by

$\begin{aligned} T & = \frac{1}{2}m(\dot{s_x}^2+\dot{s_y}^2) \\ V & = mgs_y \\ \end{aligned}$

will not work. Try instead mathematically, first by noting $s=\sqrt{s_x^2+s_y^2}$ and $\theta = s/R$ . On one hand, along $s$ we have one equation of motion with initial boundary conditions:

$\begin{aligned} 0 & =\frac{\mathrm{d}^2s}{\mathrm{d}t^2}-g\cos\bigg(\frac{s}{R}\bigg) \\ 0 & = \bigg[\frac{\mathrm{d}s}{\mathrm{d}t}\bigg]\bigg|_{t=0} \\ \frac{\pi R}{2} & = s(0) \\ \end{aligned}$

on the other hand, along line of action of the centripetal force,

$\begin{aligned} & F_\textrm{C} = \frac{m\dot{s}^2}{R} = mg\sin \bigg( \frac{s}{R} \bigg) \\ & \bigg(\frac{\mathrm{d}s}{\mathrm{d}t}\bigg)^2 - gR\sin\bigg(\frac{s}{R}\bigg) = 0 \\ \end{aligned}$

we have one another. Then,

$\begin{aligned} \frac{\mathrm{d}s}{\mathrm{d}t}\frac{\mathrm{d}^2s}{\mathrm{d}t^2}-g\frac{\mathrm{d}s}{\mathrm{d}t}\cos\bigg(\frac{s}{R}\bigg) & = 0 \\ \frac{\mathrm{d}s}{\mathrm{d}t}\frac{\mathrm{d}}{\mathrm{d}t}\bigg( \frac{\mathrm{d}s}{\mathrm{d}t}\bigg) -\frac{\mathrm{d}}{\mathrm{d}t}\bigg[ gR\sin\bigg(\frac{s}{R}\bigg)\bigg] & = 0 \\ \frac{\mathrm{d}}{\mathrm{d}t}\bigg[ \frac{1}{2}\bigg(\frac{\mathrm{d}s}{\mathrm{d}t}\bigg)^2 - gR\sin \bigg(\frac{s}{R}\bigg) \bigg] & = 0 \\ \end{aligned}$

This problem is not to be attempted.

(discontinued)

December 2, 2022December 2, 2022

202212020955 Solution to 1978-HL-PHY-II-4

A rectangular loop of length $0.2\,\mathrm{m}$ and width $0.1\,\mathrm{m}$ , carrying a steady current $I$ of $2\,\mathrm{A}$ is hinged along the $y$ -axis, and is situated in a uniform magnetic field $\mathbf{B}$ of $0.5\,\mathrm{T}$ parallel to the $x$ -axis.

If the plane of the loop makes an angle of $60^\circ$ with the $x$ – $y$ plane,

(a) calculate the force exerted by the magnetic field on each side of the loop, and
(b) calculate the torque required to hold the loop in this position.

Roughwork.

Force on a current-carrying conductor in a magnetic field is in magnitude

$\boxed{F=BIl\sin\theta}$

its direction to be determined by Fleming’s left hand rule.

(a) Have in mind a picture as viewing cross-sectionally:

and as down the top:

where

$\begin{aligned} F_1 = F_3 & = (0.5)(2)(0.2)\sin 90^\circ \\ F_2 = F_4 & = (0.5)(2)(0.1)\sin 60^\circ \\ \end{aligned}$

(b) This part is not to be attempted.

December 1, 2022December 2, 2022

202212011713 Solution to 1974-HL-PHY-I-2

A uniform ladder $6\,\mathrm{m}$ long and weighing $390\,\mathrm{N}$ rests with one end on the rough ground and the other end against a smooth wall. The ladder makes an angle of $60^\circ$ with the ground, and the coefficient of friction between the ladder and ground is $0.8$ .

(a) Draw a diagram to indicate the forces acting on the ladder.
(b) How far can a man weighing $980\,\mathrm{N}$ go up the ladder before the ladder begins to slip?

Roughwork.

(a)

Taking moment about the lower end of the ladder:

$\begin{aligned} \textrm{Torque}_\textrm{clockwise}\,(\tau_{\circlearrowright}) & = \textrm{Torque}_\textrm{anticlockwise}\,(\tau_{\circlearrowleft}) \\ (W\cos\theta )(d/2) & = (N_2\sin\theta )(d) \\ \end{aligned}$

whereas about its upper tip:

$\begin{aligned} (N_1\cos\theta )(d) & =(W\cos\theta )(d/2) + (f\sin\theta )(d) \\ \end{aligned}$

we have two equations.

(b)

This part is not to be attempted.