Patrick Godon

March 31, 1999 11:23 
Re: physics, Dynamics
Hi there, the impulse received by the ground during the collision of the drop is equal to the change of momentum. Assuming that the drop transfers all its momentum to the ground, the problem becomes fairly easy (otherwise you have to calculate what part of the momentum is transfered to heating the water, splaching, etc..). The impulse is then just equal to the momentum of the drop, namely: J=mv, where m is the mass of the drop (tens of miligrams) time the velocity of the drop (depending on the wind, etc.. say a few meter/second). For more than one drop, you just need to add as a function of time, and J/t gives you an estimate of the force applied to the ground. I hope this helps, Patrick.
