Day: November 25, 2022

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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 2nd 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.

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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}