Drag models approach different terminal speeds · source reference

Printed question

A metal ball is released from the surface of the liquid in a tall cylinder. The ball falls vertically through the liquid and the distance, $x$ cm, that the ball has fallen in time $t$ seconds is measured. The speed of the ball at time $t$ seconds is $v$ cm s⁻¹. The ball is released in a manner such that $x=0$ and $v=0$ when $t=0$.

(a) The motion of the ball is modelled by the differential equation $\frac{d^2x}{dt^2}+\frac12\frac{dx}{dt}-10=0$. It is given that $v=\frac{dx}{dt}$.

(i) Show that the differential equation can be written as $\frac{dv}{dt}=10-\frac12v$. [1]

(ii) Using the differential equation in (a)(i), find $v$ in terms of $t$. Hence find $x$ in terms of $t$. [6]

(b) The metal ball is now released in another tall cylinder filled with a different liquid. However, for this liquid, the motion of the ball is modelled by $\frac{dv}{dt}=10-k^2v^2$, where $k$ is a positive constant. It is given that $x=0$ and $v=0$ when $t=0$.

(i) Find $v$ in terms of $t$ and $k$. [5]

(ii) When the ball falls through this liquid, its speed will approach its “terminal speed” which is the speed it will attain after a long time. Find the ball’s terminal speed in terms of $k$. You must show sufficient working to justify your answer. [2]

[2024/TJC/Prelim/I/11]

Printed working

Linear drag (a)

Given $x''+x'/2-10=0$ (1), $v=x'$, so $v'=x''$. Substitute: $v'+v/2-10=0\Rightarrow v'=10-v/2$.

First printed separation. $v'=(20-v)/2$; $\int dv/(20-v)=\int\frac12dt$; $-\ln|20-v|=t/2+c$; $|20-v|=e^{-t/2-c}$; $20-v=\pm e^{-t/2-c}=Ae^{-t/2}$, $A=\pm e^{-c}$.

Alternative separation. $\int dv/(10-v/2)=\int1dt$; $-2\ln|20-v|=t+c$; $\ln|20-v|=-t/2-c/2$; $|20-v|=e^{-t/2-c/2}$; $20-v=\pm e^{-t/2}e^{-c/2}=Ae^{-t/2}$, $A=\pm e^{-c/2}$.

At $t=0,v=0$, $20=A$. Therefore $v=20-20e^{-t/2}$.

$x'=20-20e^{-t/2}$; $x=\int(20-20e^{-t/2})dt=20t-20[e^{-t/2}/(-1/2)]+d=20t+40e^{-t/2}+d$.

At $t=0,x=0$, $0=40+d\Rightarrow d=-40$; $x=20t+40e^{-t/2}-40$.

Quadratic drag (b)(i)

$v'=10-k^2v^2$; $\int dv/[(\sqrt{10})^2-(kv)^2]=\int1dt$.

$\frac1k\int k\,dv/[(\sqrt{10})^2-(kv)^2]=t+f$.

$\frac1{2k\sqrt{10}}\ln\left|\frac{\sqrt{10}+kv}{\sqrt{10}-kv}\right|=t+f$.

$\ln| (\sqrt{10}+kv)/(\sqrt{10}-kv)|=2k\sqrt{10}t+2k\sqrt{10}f$.

$(\sqrt{10}+kv)/(\sqrt{10}-kv)=\pm e^{2k\sqrt{10}f}e^{2k\sqrt{10}t}=Be^{2k\sqrt{10}t}$; $B=\pm e^{2k\sqrt{10}f}$.

At $t=0,v=0$, $\sqrt{10}/\sqrt{10}=1=B$.

$\sqrt{10}+kv=e^{2k\sqrt{10}t}(\sqrt{10}-kv)$; $kv(1+e^{2k\sqrt{10}t})=\sqrt{10}(e^{2k\sqrt{10}t}-1)$.

$v=\frac{\sqrt{10}}k\frac{e^{2k\sqrt{10}t}-1}{e^{2k\sqrt{10}t}+1}$.

Four printed terminal-speed methods (b)(ii)

1. As $t\to\infty$, $e^{2k\sqrt{10}t}\to\infty$, the quotient tends to 1, so $v\to\sqrt{10}/k$ cm s⁻¹.
2. Multiply numerator and denominator by $e^{-2k\sqrt{10}t}$: $v=(\sqrt{10}/k)(1-e^{-2k\sqrt{10}t})/(1+e^{-2k\sqrt{10}t})\to(\sqrt{10}/k)(1-0)/(1+0)=\sqrt{10}/k$.
3. Long division: $v=(\sqrt{10}/k)[1-2/(e^{2k\sqrt{10}t}+1)]\to\sqrt{10}/k$.
4. At terminal speed, $v'=10-k^2v^2=0\Rightarrow v^2=10/k^2\Rightarrow v=\pm\sqrt{10}/k$. Since speed $v>0$, terminal speed is $\sqrt{10}/k$ cm s⁻¹.

Printed answer key

(a)(ii) $v=20-20e^{-t/2}$; $x=20t+40e^{-t/2}-40$. (b)(i) $v=(\sqrt{10}/k)(e^{2k\sqrt{10}t}-1)/(e^{2k\sqrt{10}t}+1)$. (ii) $\sqrt{10}/k$ cm s⁻¹.

Teacher note 1

Given $dv/dt=f(v)$, integrate $\int dv/f(v)=\int1dt$.

Need to put in the modulus sign when integration results in a logarithm, since its argument may be negative. In removing the modulus sign, remember to put in “$\pm$”. In (a) $20-v$ may be negative; in (b) $(\sqrt{10}+kv)/(\sqrt{10}-kv)$ may be negative.

Mathematical review

(a)(i) Since $v=x'$, $v'=x''=10-v/2$.

(a)(ii) Separate: $-2\ln|20-v|=t+C$, giving $v=20+Ae^{-t/2}$. Initial rest gives $A=-20$. Thus $v=20(1-e^{-t/2})$ and $x=20t+40e^{-t/2}-40$.

(b)(i) Put $L=\sqrt{10}/k$. Separation gives $\frac1{2k\sqrt{10}}\ln[(\sqrt{10}+kv)/(\sqrt{10}-kv)]=t$ on $0\le v<L$. Hence $v=(\sqrt{10}/k)(e^{2k\sqrt{10}t}-1)/(e^{2k\sqrt{10}t}+1)$. Equivalently $v=L\tanh(k\sqrt{10}t)$, whose derivative is $10\operatorname{sech}^2(k\sqrt{10}t)=10-k^2v^2$.

(b)(ii) Divide numerator and denominator by the exponential. The ratio tends to 1, so terminal speed is $\sqrt{10}/k$ cm/s. The positive equilibrium is approached from below; the negative equilibrium is incompatible with the falling-from-rest speed trajectory.

Provenance

TJC JC2 2025 H2 Maths (9758) MSM. Native transcription checked against original visible question, working and sidebars. Question page(s): 2. Working page(s): 5, 6, 7, 8. Original records remain private for QA; this native article has no PDF or raster worksheet dependency.