November 2022 – Physicspupil

November 25, 2022November 26, 2022

202211251510 Solution to 1982-AL-AMATH-I-3

A rocket fully loaded with fuel has total mass $M$ including mass $F$ of fuel. The rocket is fired vertically upwards at $t=0$ . During its journey, the fuel is allowed to burn at a constant rate $b$ such that the relative backward velocity of the exhaust gases is $u$ .

(a) If $m(t)$ is the mass of the rocket plus fuel and $v(t)$ its velocity at time $t$ , show that, if air resistance is neglected, the equation of motion is

$\displaystyle{-mg=m\frac{\mathrm{d}v}{\mathrm{d}t}+u\frac{\mathrm{d}m}{\mathrm{d}t}}$ .

(b) Show that, the speed of the rocket at time $\displaystyle{t\leqslant \frac{F}{b}}$ is given by

$\displaystyle{-gt-u\ln \bigg( 1-\frac{b}{M}t\bigg)}$ .

(c) The height $H$ which is reached by the rocket at the instant when the fuel is all burnt depends on the rate of burning $b$ . Determine the rate of burning $b_0$ such that the height reached will be maximal. (Hint: For the stationary value to be maximal, first show that

$\displaystyle{\frac{\mathrm{d}^2H}{\mathrm{d}b^2}=-\frac{gF^2}{b_0^4}}$

at $b=b_0$ )

Roughwork.

(a) Jot $m(t)\big|_{t=0}=M$ . Take $\big\uparrow \textrm{(+ve)}$ . By Newton’s 2^nd law,

$\begin{aligned} \textrm{Net }F & =\frac{\mathrm{d}p}{\mathrm{d}t} \\ -mg & = \frac{\mathrm{d}}{\mathrm{d}t}\big( mv\big) \\ -mg & = m\frac{\mathrm{d}v}{\mathrm{d}t} + u\frac{\mathrm{d}m}{\mathrm{d}t} \\ \end{aligned}$

(b)

$\begin{aligned} \int (-g)\,\mathrm{d}t & = \int\bigg(\frac{\mathrm{d}v}{\mathrm{d}t}\bigg)\,\mathrm{d}t+\int\bigg(\frac{u}{m}\frac{\mathrm{d}m}{\mathrm{d}t}\bigg)\,\mathrm{d}t \\ -gt & = \int\mathrm{d}v + u\int\frac{\mathrm{d}m}{m}\\ \end{aligned}$

and the result follows.

(c) From

$\begin{aligned} v(t) & = -gt-u\ln \bigg( 1-\frac{b}{M}t\bigg) \\ H(t) & =\int_0^t v(t')\,\mathrm{d}t' \\ & = \int_0^t \bigg[ -gt'-u\ln \bigg( 1-\frac{b}{M}t'\bigg) \bigg]\,\mathrm{d}t' \\ & = -\frac{g}{2}t'^2\bigg|_0^t +\frac{uM}{b}\int_{t'=0}^{t'=t} \ln T\,\mathrm{d}T \\ \dots\enspace & \textrm{as }\int \ln x\,\mathrm{d}x=x\ln x-x+C\enspace \dots \\ \end{aligned}$

This problem is not to be attempted.

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 24, 2022December 2, 2022

202211241139 Solution to 1970-HL-PHY-I-4

Two spherical copper bulbs $A$ and $B$ are interconnected through a fine tube. The bulbs are filled with one mole of an ideal gas at temperature $27\,^\circ\mathrm{C}$ . The gas pressure is found to be one atmosphere, and the volume of bulb $B$ is four times that of bulb $A$ . Now bulb $B$ is immersed in a bath at constant temperature $327\,^\circ\mathrm{C}$ , while bulb $A$ is maintained at its original temperature. Find the final pressure inside the bulbs, assuming that the volume of the fine tube can be neglected. The coefficient of linear expansion of copper is given as $2\times 10^{-5}$ per degree Celsius.

Roughwork.

For isotropic materials the volumetric thermal expansion coefficient is three times the linear coefficient:

$\alpha_V=3\alpha_L$

_{Wikipedia on Thermal Expansion}

Initially,

$\begin{aligned} P(V_A+V_B) & = nRT \\ (1.01\times 10^5)(5V_A) & = (1)(8.31)(300) \\ V_A & = 4940\,\mathrm{cm^3} \\ V_B & = 19760\,\mathrm{cm^3}\\ \end{aligned}$

finally,

$\begin{aligned} P_AV_A & =n_ART_A \\ P_BV_V & =n_BRT_B \\ \end{aligned}$

where

$\begin{aligned} n_A+n_B & =1 \\ V_A & = 0.00494 \\ V_B & = 0.01976 (1+(3)(2\times 10^{-5})(300))\\ T_A & = 300 \\ T_B & = 600 \\ \end{aligned}$

The remaining are left as an exercise to the reader.

November 24, 2022November 24, 2022

202211241033 Solution to 1970-HL-PHY-I-1

Show that, when an object of mass $m$ is weighed by a spring-balance suspended from the ceiling of a lift moving upwards with acceleration $a$ , the reading $W'$ in the spring-balance is given by $m(g+a)$ , where $g$ is the acceleration due to gravity. $W'$ is called the apparent weight of the object, and $mg$ the true weight.

It is known that a passenger in a lift will experience discomfort if his appparent weight exceeds his true weight by more than one sixth, or falls short of his true weight by more than one eighth. A lift in a hotel is designed to transport passengers from the ground floor direct to the top floor in the shortest possible time without causing discomfort. Find this time if the distance between the ground floor and the top floor is $100$ metres.

Roughwork.

By Newton’s 2^nd law,

$\begin{aligned} F_{\textrm{net}}& = ma \\ W'-mg & = ma \\ W' & = m(g+a) \\ \end{aligned}$

By boundary conditions,

$\begin{aligned} & \quad \enspace \begin{cases} \displaystyle{\frac{W'-W}{W}\leqslant \frac{1}{6} }\\ \displaystyle{\frac{W-W'}{W}\leqslant \frac{1}{8} }\\ \end{cases} \\ & \Rightarrow \frac{8W}{9} \leqslant W' \leqslant \frac{7W}{6} \\ & \Rightarrow \frac{8}{9}mg\leqslant m(g+a)\leqslant \frac{7}{6}mg \\ & \Rightarrow -\frac{g}{9} \leqslant a\leqslant \frac{g}{6} \\ & \Rightarrow |a|\leqslant \frac{g}{9} \\ \end{aligned}$

Take upward positive and let $g=10\,\mathrm{m\,s^{-2}}$ ,

$\begin{aligned} s & = ut+\frac{1}{2}at^2 \\ (100) & = (0)t + \frac{1}{2}\bigg(\frac{10}{9}\bigg) t^2 \\ t & = 6\sqrt{5}\,\mathrm{s} \\ \end{aligned}$

This problem is not to be attempted.

November 24, 2022November 24, 2022

202211240932 Solution to 1965-HL-PHY-I-2

A $1.0\,\mathrm{kg}$ object rests on a table $1.0\,\mathrm{m}$ above the floor, and the floor is $4.0\,\mathrm{m}$ above the street.

(a) What is the potential energy of the object with reference to i. the table top, ii. the floor, and iii. the street?
(b) Now the object is put on the floor. Find the change in potential energy for this decrease of $1.0\,\mathrm{m}$ in the height of the object when the reference level is i. the table top, ii. the floor, and iii. the street?
(c) Let the object fall freely from the original position on the table. Find its speed and kinetic energy when it reaches the floor.

Roughwork.

(a) Take $g=10\,\mathrm{m\,s^{-2}}$ ,

i. $\textrm{PE}=mgh=(1)(10)(0)=0$
ii. $\textrm{PE}=mgh=(1)(10)(1)=10\,\mathrm{J}$
iii. $\textrm{PE}=mgh=(1)(10)(1+4)=50\,\mathrm{J}$

(b)

i. $h_{\textrm{table}}=0$ :

$\begin{aligned} \Delta\textrm{PE} & = \textrm{PE}_{\textrm{final}} - \textrm{PE}_{\textrm{initial}} \\ & = mgh_f - mgh_i \\ & = (1)(10)(-1) - (1)(10)(0) \\ & = -10\,\mathrm{J} \\ \end{aligned}$

ii. $h_{\textrm{floor}}=0$ :

$\begin{aligned} \Delta\textrm{PE} & = \textrm{PE}_{\textrm{final}} - \textrm{PE}_{\textrm{initial}} \\ & = mgh_f - mgh_i \\ & = (1)(10)(0) - (1)(10)(1) \\ & = -10\,\mathrm{J} \\ \end{aligned}$

iii. $h_{\textrm{street}}=0$ :

$\begin{aligned} \Delta\textrm{PE} & = \textrm{PE}_{\textrm{final}} - \textrm{PE}_{\textrm{initial}} \\ & = mgh_f - mgh_i \\ & = (1)(10)(4) - (1)(10)(1+4) \\ & = -10\,\mathrm{J} \\ \end{aligned}$

(c) Take downward positive,

$\begin{aligned} s & = ut+\frac{1}{2}at^2 \\ (1) & = (0)t + \frac{1}{2}(10)t^2 \\ t & = \frac{\sqrt{5}}{5}\,\mathrm{s} \\ \end{aligned}$

$\begin{aligned} v & = u + at \\ & = (0) + (10)\bigg(\frac{\sqrt{5}}{5}\bigg) \\ & = 2\sqrt{5}\,\mathrm{m\,s^{-1}} \\ \end{aligned}$

$\begin{aligned} \textrm{KE} & =\frac{1}{2}mv^2} \\ & = \frac{1}{2}(1)(2\sqrt{5})^2 \\ & = 10\,\mathrm{J} \\ & = -\Delta\textrm{PE} \\ \end{aligned}$

This problem is not to be attempted.

November 23, 2022November 23, 2022

202211231608 Solution to 1982-CE-AMATH-I-5

Show that the tangents to the curve

$y=x^3-9x^2+30x+4$

cannot be parallel to the $x$ -axis.

Roughwork.

The slope $y'(x)$ of tangent to the curve is

$\begin{aligned} y'(x) & = \frac{\mathrm{d}y}{\mathrm{d}x} \\ & = \frac{\mathrm{d}}{\mathrm{d}x}(x^3-9x^2+30x+4) \\ & = 3x^2-18x+30 \\ \end{aligned}$

Assume to the contrary that such tangent as $y'(x)=0$ exists. Then

$\begin{aligned} 3x^2-18x+30 & = 0 \\ \frac{-(-18)\pm\sqrt{(-18)^2-4(3)(30)}}{2(3)} & = x \\ \end{aligned}$

where the discriminant

$\Delta = -36<0$ ,

implies there are no real roots. Contradiction, which was to be demonstrated.

November 23, 2022November 23, 2022

202211231536 Solution to 1986-CE-AMATH-I-4

Find the equation of the tangent to the curve

$x^2+xy+y^2=7$

at the point $(2,1)$ .

Roughwork.

$\begin{aligned} x^2+xy+y^2-7 & = 0 \\ 2x\,\mathrm{d}x+x\,\mathrm{d}y+y\,\mathrm{d}x + 2y\,\mathrm{d}y & = 0 \\ 2x + x\frac{\mathrm{d}y}{\mathrm{d}x}+y+2y\frac{\mathrm{d}y}{\mathrm{d}x} & = 0 \\ -\frac{2x+y}{x+2y} & = y'(x) \\ \end{aligned}$

The slope of tangent at $(x=2,y=1)$ is

$y'(x)|_{(2,1)}=\displaystyle{-\frac{2(2)+(1)}{(2)+2(1)}=-\frac{5}{4}}$ .

and its equation

$\begin{aligned} & \quad\enspace \frac{y-1}{x-2} = -\frac{5}{4} \\ & \Rightarrow 5x + 4y = 14 \\ \end{aligned}$

November 23, 2022November 23, 2022

202211231154 Solution to 1969-HL-PHY-I-2

An aeroplane flying with a uniform speed $v$ in a small circle of radius $r$ at a constant altitude $h$ releases a body. Assuming that air resistance can be neglected, find the angle of inclination between the ground and the line of sight of the pilot when he observes the body touching the ground.

Roughwork.

The angular speed $\omega$ of the aeroplane is

$\displaystyle{\omega =\frac{v}{r}}$

and its position in Cartesian coordinate system

$(x,y,z)=(r\cos (\omega t),r\sin (\omega t),h)$ .

WLOG assume the top view be

so that

$s=\sqrt{(vt-r\sin \omega t)^2+(r-r\cos \omega t)^2}$ .

Since the altitude $h$ is related to the time of flight $t$ by

$\displaystyle{h =\frac{1}{2}gt^2}$ ,

the angle of inclination $\alpha$ is in such

$\displaystyle{\tan\alpha =\frac{h}{s}}$

as to be desired.

November 22, 2022November 22, 2022

202211221451 Problem 4.10

A sphere of radius $R$ carries a polarization

$\mathbf{P}(\mathbf{r})=k\mathbf{r}$ ,

where $k$ is a constant and $\mathbf{r}$ the vector from the centre.
(a) Calculate the bound charges $\sigma_b$ and $\rho_b$ .
(b) Find the field inside and outside the sphere.

_{Extracted from David J. Griffiths. (1999). Introduction to Electrodynamics.}

Roughwork.

The surface charge is defined by Eq. (4.11):

$\sigma_b\equiv \mathbf{P}\cdot\hat{\mathbf{n}}$

whereas the volume charge by Eq. (4.12):

$\rho_b\equiv -\nabla \cdot \mathbf{P}$ .

_{Text on pp. 167-168}

On the surface $r=|\mathbf{r}|=R$ :

$\begin{aligned} \sigma_b & = P(R\,\hat{\mathbf{r}})\cdot\hat{\mathbf{r}} \\ & = k(R\,\hat{\mathbf{r}})\cdot\hat{\mathbf{r}} \\ & = kR \end{aligned}$

Under the surface $\mathbf{r}=\mathbf{r}(x,y,z)$ :

$\begin{aligned} \rho_b & = -\nabla\cdot\mathbf{P} \\ & = -\bigg(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\bigg)\cdot k(x,y,z) \\ & = -k\bigg(\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial z}{\partial z}\bigg) \\ & = -3k \end{aligned}$

Notice $\displaystyle{(-3k)\bigg(\frac{4\pi R^3}{3}\bigg) +(kR)(4\pi R^2)=0}$ , so the sphere has zero net charge, i.e., $Q_\textrm{net}=0$ , and thus produces no electric field outside, i.e., $\mathbf{E}(\mathbf{r}:r>R)=\mathbf{0}$ .

Eq. (4.13):

$V(\mathbf{r})=\displaystyle{\frac{1}{4\pi\epsilon_0}\iint_S\frac{\sigma_b}{r}\,\mathrm{d}a'+\frac{1}{4\pi\epsilon_0}\iiint_V\frac{\rho_b}{r}\,\mathrm{d}\tau'}$

and the rest is left as an exercise to the reader.