When finding the force of a falling raindrop, how to account for not knowing the collision time? So my question is to find the force of impact on the ground of a raindrop of mass $m$ that falls at velocity $v$ right as it hits the ground. So what I did was find the momentum of the object, which is $mv$, then divided by time $t$ to get the force. However, this is where I'm stuck, because it seems like $t$ is very small and unknown.
 A: You are right to suspect you can't find the force knowing only $m$ and $v$. In fact, dimensionally there is no force you can cook up out of them, so it's clear we are going to need to introduce some extra physics. That is, not every sphere of mass $m$ and speed $v$ hitting a wall experiences the same force. The difference between different spheres is the material they are made of so we're looking for a material property.
The simplest model I can think of to give at least an answer is this: assume we know also the density $\rho$ of the water and that as the raindrop lands it completely falls apart into some sort of puddle of negligible height. Then we can say the floor has done work on the raindrop to bring it to rest equal to:
$W_{floor} = Fr = \frac{1}{2}mv^2 $
We can use the relation $m = \frac{4}{3}\pi r^3$ to get rid of one of $r,m,\rho$. It's much easier to leave in terms of $m,r$ only but I suppose an engineer might prefer $m,\rho$ so:
$F = \frac{m}{2r}v^2 = (\frac{\pi}{6}\rho m^2)^{1/3}v^2$
(A more complicated model may want to take account of the fact that some work also needs to go into overcoming any cohesive energy the raindrop has, the fact the force isn't uniform so the peak force will be higher than this, and maybe some fluid mechanics effects I haven't thought of.)
