Drops of rain fall perpendicular to the roof of a parked car during a rainstorm?

Question

Drops of rain fall perpendicular to the roof of a parked car during a rainstorm. The drops strike the roof with a speed of 10 m/s, and the mass of rain per second striking the roof is 0.047 kg/s.

Assuming the drops come to rest after striking the roof, find the average force exerted by the rain on the roof.
magnitude ____________ N

Answer 1

we know F*t=m*(vf-vi)

so F=(m(vf-vi))/t

or to put it differently: F= (m/t)(vf-vi)

when we put it like this, the answer is easy!

F=(.047)(-10)

F=-.47



I hope this helps! Have fun in physics!

Answer 2

It's a question of impulse, using newtons 3rd law.



Impulse = Force x time



also



Impulse = mass x (initial velocity - final velocity)



Therefore



I = 0.047t x 10

I = 0.47t



The t is in there to take kg/s to kg (it will cancel out later, but it stops silly mistakes if you leave it out)



0.47t = I = F x t

The ts cancel



F = 0.47Newtons



Ta-dah. Hope this is well explained :)

Answer 3

It's +0.47 not negative.