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Rain Drop

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abyss43 | 14:28 Wed 22nd Oct 2008 | Science
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How could I as accurately as possible calculate the impulse exterted on a surface by a falling raindrop assuming its impact velocity is 10m/s. (assume water is fresh water at 10 Degress celcius) and impact surface is rigid. Dynamic viscosity water spread etc should be taken into account.
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Sorry I forgot mass of rain drop is 3 grams
14:45 Wed 22nd Oct 2008
Avatar Image MatydaLover
The force F on the surface will be F = ma ... a being the deceleration v/t. This gives the impulse of the force as Ft = mv ... ... which looks right because the impulse is related to the change in momentum.

It does not matter how long the raindrop takes to stop ... at the end its momentum is zero, so the impluse is mv. with your numbers that is 0.003 kg x 10 m/s 0.03 N (I think it is Newtons).
15:53 Wed 22nd Oct 2008
Avatar Image MatydaLover
... I meant Ns, Newton seconds.
15:55 Wed 22nd Oct 2008

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