Force of water drops (momentum)

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1. The problem statement, all variables and given/known data

Large limestone caverns have been formed by dripping water. If water droplets of 0.0305 mL fall from a height of 5.17 m at a rate of 14 droplets per minute, what is the average force exerted on the limestone floor by the droplets of water during a 1.00-min period? (Assume the water does not accumulate on the floor)

2. Relevant equations

p = m*v

3. The attempt at a solution

I think I've tried everything for this problem. - I tried following this but it doesn't give me the right answer. So I calculate momentum but then how do I find impulse if I don't know the length of time for the impact? I'm really confused.


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Force = Rate of change of momentum.

o What is the momentum of a single drop just before it hits the limestone floor?
o Following the first question, the change in momentum of a single drop as it hits the limestone floor is what?
o There are 14 drops per minute, so the total change of momentum over a minute is what?
o What is the average total change of momentum per second?

[Edit: By the way, the link you pointed to describes an "Nmgt" term in the impulse equation. But the author of the post seems to have put that there to account for the gravitational force due to the accumulation of water on floor (i.e. Fg = Nmg, where N is the total number of drops and m is the mass of a single drop). But that doesn't apply to your particular problem, since your problem states, "assume the water does not accumulate on the floor."]
Last edited:
Yup I got it! Thank you! I think I was over thinking it :(

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