Past exam of the mathematics course of the University of Cambridge / 2001 / ia / Paper 4 / 10A (Dynamics)

Let $m=4\pi \rho a^3/3$ be the drop's mass, with constant mass density $\rho$. Its volume-growth rule gives $4\pi a^2\dot{a}=c\pi a^{2}v$, hence $\dot{a}=cv/4$ and $\dot{m}=\rho c\pi a^{2}v$. The incoming mist is stationary, so the variable-mass system momentum balance isExpanding the derivative and substituting the accretion rate gives the accreting raindrop with quadratic drag equationsThe extra $c$ contribution is the momentum needed to accelerate collected matter from rest; omitting it would treat the growing drop as a constant-mass particle.

Set $q=v^2/a$ and $B=7c/8+3k/4$. The chain rule givesTo justify the limiting value as well as its stability sign, use $c>0$, as appropriate to volume growth. Divide by $\dot{a}=cv/4$ and integrate the resulting first-order equation:Thus $q$ remains between two positive constants, its initial value and $g/B$. Since $d\sqrt{a}/dt=c\sqrt{q}/8$, the radius is defined for all increasing time and tends to infinity. The correction vanishes, proving the speed-radius attractor of an accreting raindrop:The strict inequality uses $k>0$. The growing radius prevents a finite terminal velocity; instead the acceleration approaches this positive constant.