Discussion
A raindrop falls under gravity and captures water molecules from atmosphere. Its mass changes at the rate \(\lambda m(t)\), where \(\lambda\) is a positive constant and m(t) is the instantaneous mass. Assume that acceleration due to gravity is constant and water molecules are at rest with respect to earth before capture. Which of the following statements is correct?
Answer: C
The answer is the speed approaches a constant value when \(\lambda t \gg 1\).
The captured molecules start at rest, so Newton's law for variable mass reads
\[ \begin{aligned} \frac{d(mv)}{dt} &= mg \\ m\frac{dv}{dt} + \lambda m v &= mg \\ v(t) &= \frac{g}{\lambda}\left(1 - e^{-\lambda t}\right) \end{aligned} \]
For \(\lambda t \gg 1\) the exponential dies away and the drop approaches the terminal speed \(g/\lambda\). For \(\lambda t \ll 1\) the speed is about \(gt\), growing linearly, not constant. The speed never grows exponentially; the exponential term decays.
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