Amidst a heavy rain, hailstones having an average diameter of 1.0 cm descend at an average speed of 20 m/s . Assume 2000 hailstones impact each square meter of a 10 m 10 m roof perpendicularly every second, and that the hailstones do not rebound. Determine the average force exerted on the roof by the falling hailstones. The density of a hailstone is 900 kg/m ^3 .

Step-by-step solution

The volume of a single hailstone is given by: V = 4 3 r^3 = 4 3 (0.5 10⁻²)^3 = 6 10⁻⁶ m ^3 The mass of a single hailstone is: m = V = 900 6 10⁻⁶ = 150 10⁻⁶ kg The total number of hailstones hitting the roof per second is: N = n A = 2000 (10 10) = 2 10^5 s ⁻¹ The total mass of hailstones hitting the roof per second is: M t = N m = 2 10^5 150 10⁻⁶ = 30 kg/s Since the hailstones do not rebound, the average force exerted on the roof is equal to the rate of change of momentum: F = M t v = 30 20 = 600 N Substituting 3.14