Drops of rain fail perpendicular to the roof of a parked car during a rainstorm.

Question

Drops of rain fail 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 0.074 kg/s. (a) Assuming the drops come to rest after striking the roof, find the average force exerted by the rain on the roof. (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 double that in part (a). The magnitude would be half that in part (a). The magnitude would be the same as in part (a). The magnitude would be one fourth times In part (a).

Explanation / Answer

(a) Let's take downward direction as negative.

Impulse on the rain drops by the roof = Change in momentum of the rain drops

=> Favg t = (m)(0 - v)

=> Favg = -(m/t)v = -0.074 * (-10) = 0.74 N

So, average force on the rain drops is upward, and by newton's third law, the rain drops apply equal force on the roof downward i.e. -1.74 N

(b) Favg t = (m)[v - (-v)]

=> Favg = 2(m/t)v = 2 * 0.074 * 10 = 1.48 N

So, the average force due to hail is -1.48 N. Hence, it is double in magnitude and in downward direction.

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