2. Drops of rain fall perpendicular to the roof of a parked car during a rainsto

Question

2. Drops of rain fall perpendicular to the roof of a parked car during a rainstorm. The drops strike the roof with a speed of 18 m/s, and the mass of rain per second striking the roof is 0.069 kg/s. (a) Assuming the drops come to rest after striking the roof, find the average force exerted by the rain on the roof. magnitude N direction (b) If hailstones having the same mass as the raindrops fall on the roof at the same rate and with the same speed, how would the average force on the roof compare to that found in part (a)? (Assume the hailstones bounce back up off the roof.) The magnitude would be the same as in part (a). The magnitude would be four times that in part (a). The magnitude would be half that in part (a). The magnitude would be double that in part (a).

Explanation / Answer

a)as we know force =mass *a

from the units we can find out  if we multiply 18m/s x 0.067kg/s then the unit will come as a average force

so F=18m/s x 0.067kg/s

F=1.206 Kgm/s^2

F=1.206 N

b) The hail particle's initial momentum will be
p=m*v
gets reversed when it bounces, so it becomes
p=-m*v
The change in momentum of the hail particle is
P=2*m*v
So, the force experienced by the roof is now double than in the case of raindrop, because the momentum that the roof absorbs from every hail particle is double the momentum it absorbs from every droplet.

answer is D

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