202104080551 Solution to 1973-AL-AMATH-I-5

A particle of mass $m$ moving in a straight line in a medium with a velocity $v$ is subjected to a force $mk(v^3+a^2v)$ in the direction opposite to the velocity, where $k$ and $a$ are constants. Show that, at any subsequent time $t$ , the velocity $v$ is related to the initial velocity $v_0$ by

$\displaystyle{\frac{v}{(v^2+a^2)^{1/2}}=\frac{v_0}{(v_0^2+a^2)^{1/2}}e^{-a^2kt}}$ .

Show that for any value of the initial velocity, no matter how large, the particle will never travel a distance greater than $\pi /(2ka)$ .

Background.

Newton's second law:

$\begin{aligned} F = -mk(v^3+a^2v) & = ma \\ -mk(\dot{x}^3+a^2\dot{x}) & = m\ddot{x} \\ \frac{\mathrm{d}\dot{x}}{\mathrm{d}t} & = -k(\dot{x}^3+a^2\dot{x}) \\ \frac{\mathrm{d}\dot{x}}{\dot{x}^3+a^2\dot{x}} & = -k\,\mathrm{d}t \\ \dots \textrm{integrating from }& (v=v_0,\, t=0)\textrm{ to }(v=v,\, t=t) \\ \int_{v_0}^{v} \frac{1}{\dot{x}(\dot{x}^2+a^2)}\,\mathrm{d}\dot{x} & = \int_{0}^{t}-k\,\mathrm{d}t \end{aligned}$

(to be continued)

Roughwork.

Let . Then,

$\begin{aligned} 1 & \equiv A(u^2+a^2)+Bu^2+Cu \\ 1 & \equiv (A+B)u^2+Cu+Aa^2 \\ A+B & = 0 \\ C & = 0 \\ A & = \frac{1}{a^2} \\ \Rightarrow B & = -\frac{1}{a^2} \end{aligned}$
Thus,
$\displaystyle{\int \frac{1}{u(u^2+a^2)}\,\mathrm{d}u = \int \bigg(\frac{1}{a^2u} - \frac{u}{a^2(u^2+a^2)}\bigg) \,\mathrm{d}u}$

(continue)

$\begin{aligned} & \quad \int_{v_0}^{v} \frac{1}{\dot{x}(\dot{x}^2+a^2)}\,\mathrm{d}\dot{x} \\ & = \int_{v_0}^{v} \bigg(\frac{1}{a^2\dot{x}} - \frac{\dot{x}}{a^2(\dot{x}^2+a^2)}\bigg) \,\mathrm{d}\dot{x} \\ & = \int_{v_0}^v\frac{1}{a^2\dot{x}}\,\mathrm{d}\dot{x} - \int_{v_0}^v \frac{\dot{x}}{a^2(\dot{x}^2+a^2)}\,\mathrm{d}\dot{x} \\ \end{aligned}$

The first term is

$\begin{aligned} \int_{v_0}^v\frac{1}{a^2\dot{x}}\,\mathrm{d}\dot{x} & = \bigg[ \frac{\ln \dot{x}}{a^2} \bigg]\bigg|_{v_0}^{v} \\ & = \bigg( \frac{\ln v}{a^2} \bigg) - \bigg( \frac{\ln v_0}{a^2} \bigg) \\ & = \frac{1}{a^2}\ln \bigg( \frac{v}{v_0} \bigg) \end{aligned}$

and the second term is

$\begin{aligned} &\quad -\int_{v_0}^v \frac{\dot{x}}{a^2(\dot{x}^2+a^2)}\,\mathrm{d}\dot{x}\\ \dots & \textrm{ let } \dot{x}=a\tan\theta \enspace\dots \\ \dots & \textrm{ let } \theta = \tan^{-1}\bigg(\frac{v}{a}\bigg) \textrm{ and } \theta_0 = \tan^{-1}\bigg(\frac{v_0}{a}\bigg) \\ & = -\int_{\theta_0}^{\theta} \frac{a\tan\theta}{a^2(a^2\tan^2\theta +a^2)}\cdot (a\sec^2\theta)\,\mathrm{d}\theta \\ & = -\int_{\theta_0}^{\theta} \frac{\tan\theta}{a(a^2\sec^2\theta)}\cdot (a\sec^2\theta )\,\mathrm{d}\theta \\ & = -\int_{\theta_0}^{\theta} \frac{\tan\theta}{a^2}\,\mathrm{d}\theta \\ & = -\bigg[ -\frac{\ln |\cos\theta |}{a^2} \bigg]\bigg|_{\theta_0}^{\theta} \\ & = \frac{1}{a^2} \bigg[\ln \bigg| \frac{a}{\sqrt{\dot{x}^2+a^2}} \bigg|\bigg]\bigg|_{v_0}^{v} \\ & = \frac{1}{a^2} \bigg( \ln\bigg| \frac{a}{\sqrt{v^2+a^2}} \bigg| -\ln\bigg| \frac{a}{\sqrt{v_0^2+a^2}} \bigg| \bigg) \\ & = \frac{1}{a^2} \ln \bigg( \frac{\sqrt{v_0^2+a^2}}{\sqrt{v^2+a^2}} \bigg) \\ \end{aligned}$

Back to an earlier line

$\displaystyle{\int_{v_0}^{v} \frac{1}{\dot{x}(\dot{x}^2+a^2)}\,\mathrm{d}\dot{x} = \int_{0}^{t}-k\,\mathrm{d}t}$

we have

$\begin{aligned} \frac{1}{a^2}\ln\bigg( \frac{v}{v_0} \bigg) + \frac{1}{a^2}\ln \bigg( \frac{\sqrt{v_0^2+a^2}}{\sqrt{v^2+a^2}}\bigg) & = -kt \\ \ln\bigg( \frac{v}{v_0} \bigg) + \ln \bigg( \frac{\sqrt{v_0^2+a^2}}{\sqrt{v^2+a^2}}\bigg) & = -a^2kt \\ \ln\bigg( \frac{v}{v_0}\frac{\sqrt{v_0^2+a^2}}{\sqrt{v^2+a^2}} \bigg) & =-a^2kt \\ \frac{v}{v_0}\frac{\sqrt{v_0^2+a^2}}{\sqrt{v^2+a^2}} & = e^{-a^2kt} \\ \end{aligned}$

or,

$\boxed{\displaystyle{\frac{v}{\sqrt{v^2+a^2}}=\frac{v_0}{\sqrt{v_0^2+a^2}}e^{-a^2kt}}}$

as desired.

It remains to be shown that for any value of the initial velocity, no matter how large, the particle will never travel a distance greater than $\pi /(2ka)$ .

(to be continued)

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