Question Part Points Submissions Used 1 2 3 –/1 1/1 1/1 0/5 1/5 1/5 Total 2/3 Dr

Question

Question Part
Points
Submissions Used
1 2 3
–/1 1/1 1/1
0/5 1/5 1/5
Total
2/3

Drops of rain fall perpendicular to the roof of a parked car during a rainstorm. The drops strike the roof with a speed of 14 m/s, and the mass of rain per second striking the roof is 0.065 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 Correct: Your answer is correct.

(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 four times that in part (a). The magnitude would be the same as in part (a). The magnitude would be double that in part (a). The magnitude would be half that in part (a). Correct: Your answer is correct.

Explanation / Answer

a. The change of velocity of raindrops is 14m/s and the mass hitting the roof is .065kg/s.
So rate of change of momentum=14 x .065 kg m /s2=.91 N. Force is vertically downward.

b. As the Hailstones have negligible mass compared to the car, they will bounce back with the same velocity they strike the car with. So, change of momentum will be double.

force=2 x .91 N =1.82 N.

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