Public Examinations in Physics – Page 2

January 4, 2023January 4, 2023

202301041207 Solution to 1969-HL-PHY-2

Derive the relation between Young’s modulus $Y$ and Hooke’s constant $k$ for a uniform steel wire of cross-section $A$ and length $L$ .

Two uniform wires of equal cross-section have the same Young’s modulus $Y$ but different Hooke’s constants $k_1$ and $k_2$ . If the two wires are connected in series to form a new wire, compare Young’s modulus of the connected wire with that of the original wires, and find Hooke’s constant of the connected wire in terms of $k_1$ and $k_2$ .

Roughwork.

$\displaystyle{\textrm{Young's modulus (}Y\textrm{)}=\frac{\textrm{Stress (}\sigma\textrm{)}}{\textrm{Strain (}\varepsilon\textrm{)}}}$

$Y \equiv \displaystyle{\frac{\sigma}{\varepsilon} = \frac{F/A}{\Delta l/L} = \frac{FL}{A\Delta l}\quad \textrm{\scriptsize{OR}}\quad \displaystyle{F = \frac{YA\Delta l}{L}}$

Lemma.

Hooke’s law for a stretched wire:

$\displaystyle{F=\bigg(\frac{YA}{L}\bigg)\Delta l=kx}$

such that

$\displaystyle{k\equiv \frac{YA}{L}}\qquad\because\enspace x\equiv \Delta l$ .

_{Wikipedia on Young’s modulus}

$\begin{aligned} \Delta l' & = \Delta l_1+\Delta l_2 \\ \frac{F}{A'}\frac{L'}{Y'} & = \frac{F}{A}\bigg(\frac{L_1}{Y_1}+\frac{L_2}{Y_2}\bigg) \\ \because &\enspace \begin{cases} A' = A \\ Y_1 =Y_2 =Y\\ L' =L_1+L_2 \\ \end{cases} \\ \therefore& \quad Y'=Y \\ \end{aligned}$

Hence

$\begin{aligned} k' & = \frac{Y'A'}{L'} \\ & = \frac{YA}{L_1+L_2} \\ & = \bigg(\frac{L_1}{YA}+\frac{L_2}{YA}\bigg)^{-1} \\ & = \bigg(\frac{1}{k_1}+\frac{1}{k_2}\bigg)^{-1} \\ & = \bigg(\frac{k_1+k_2}{k_1k_2}\bigg)^{-1} \\ & = \frac{k_1k_2}{k_1+k_2} \\ \end{aligned}$

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.

December 9, 2022December 21, 2022

202212091158 Solution to 1965-HL-PHY-3

(a) State the three different types of heat transfer, and give a simple example for each type of heat transfer.
(b) Calculate the amount of heat required to change $10\,\mathrm{g}$ of ice in a sealed container of volume $1\,\mathrm{L}$ and pressure at $1\,\mathrm{atm}$ at $-20\,^\circ\mathrm{C}$ to steam at $120\,^\circ\mathrm{C}$ . (Neglect the heat absorbed by the air and the container. The specific heat of ice and steam is $0.5\,\mathrm{cal\,g^{-1}\,^\circ C^{-1}}$ ). What is the pressure inside the container expressed in atmospheric pressure, when the $10\,\mathrm{g}$ of ice is completely changed to steam at $120\,^\circ\mathrm{C}$ ?

Roughwork.

(a) Skip it.

(b) Stuck. Any equation of state $f(P,V,T)=0$ as may well be applicable to solids and liquids as the ideal gas law $PV=nRT$ to gases, lacking the enthalpies of fusion and vaporisation unknown?

(to be continued)

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

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 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.