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 0.044 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 half that in part (a). The magnitude would be double that in part (a). The magnitude would be the same as in part (a). The magnitude would be one fourth times in part (a). An object has a kinetic energy of 259 and a momentum of magnitude 29.5 kg middot m/s. (a) Find the speed of the object. m/s (b) Find the mass of the object.

Explanation / Answer

we know that

Impulse = Change in momentum = F*dt

F = CHange in momentum/dt

F = m*(dV)/dt

given that

m/dt = 0.044 kg/sec

So,

F = 0.044*(10 - 0) = 0.44 N

B.

Now in case of hailstones, hailstones bounces back with same, speed, So change in momentum will be

dP = m*dV = m*(V - (-V)) = 2*m*V

F = 2*m*V//dt

So magnitue of force would be doubled than that of part (a)

Correct option is B.

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