Showing Constant Acceleration for Falling Raindrop

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[SOLVED] Falling raindrop

Homework Statement


This problem is from Gregory. Yeah, classical mechanics is not being kind to me this week. I swear I'm not coming here for every single problem. :-p

A raindrop falls vertically through stationary mist, collecting mass as it falls. The raindrop remains spherical and the rate of mass accretion is proportional to its speed and the square of its radius. Show that, if the drop starts from rest with a negligible radius, then it has constant acceleration g/7.

m: the mass of the raindrop
r: the radius of the raindrop
v: the velocity of the raindrop

Homework Equations


Because the raindrop is a sphere: m = \rho 4/3 \pi r^3 for some constant \rho
m'(t) = k v r^2 for some constant k

The Attempt at a Solution


Plugging in our equation for m, we have the differential equation
m' = k v (\frac{3m}{4\pi \rho})^{2/3}
With initial conditions m = 0 (or close enough) and v = 0.
We need another differential equation, since both m and v are nonconstant.
Thinking about this intuitively, I don't see why the raindrop isn't just falling with acceleration g. I mean, they're just particles falling, with no other external force. But since I apparently can't assume that the acceleration is g, I see no logical reason to assume that the acceleration is even constant.
So I guess my question is, how can I find the second differential equation?
 
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I really, really don't want to sound impatient, because what you guys do is absolutely incredible. But, this homework is due on Tuesday, so I would really appreciate some help. Thanks!
 
In the case where the mass isn't constant you need to use F=(d/dt)(mv). That only simplifies to F=ma when the mass is constant. Does that help?
 
Use the superposition principle. Separate the action of gravity and raindrop growth. Gravity is easy to deal with. Now, what would happen to the raindrop if there gravity were not present? Conservation of momentum dictates that the raindrop slow down as it accumulates mass. The net behavior of the raindrop is the superposition of the constant acceleration due to gravity and the acceleration (deceleration) due to accumulating mass.
 
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> We need another differential equation, since both m and v are nonconstant.

Suppose the drop falls a dist of dx in time dt, and its mass changes from m to m+dm, and velo from v to v+dv. Then,

mv^2/2 + mg*dx = (m+dm)(v+dv)^2/2 =>
mg*dx = mv*dv + v^2*dm/2 =>
mgv = mv dv/dt + (v^2/2) dm/dt.

The v cancels out, giving you your 2nd diff eqn.
 
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