Question

During a heavy rain, hailstones of average size 1.0 cm in diameter fall with an average speed of 20 m/s. Suppose 2000 hailstones strike every square meter of a 10 m × 10 m roof perpendicularly in one second and assume that the hailstones do not rebound. Calculate the average force exerted by the falling hailstones on the roof. Density of a hailstone is 900 kg/m³.

Answer 1

It is given that:
Diameter of hailstone =  1 cm =  0.01 m
⇒ Radius of hailstone, r =  0.005 m 

Average speed of hailstone =  20 m/s
Density of hailstone =  900 kg/m3 = 0.9 g/cm3
\[\text{ Volume of the hailstones is given as, }\] 
\[V = \frac{4}{3}\pi r^3 \]
\[ \Rightarrow V = \frac{4}{3}\pi(0 . 005 )^3 = 5 . 235 \times {10}^{- 7} m^3 \]
\[\text{ Mass = volume }\times \text{ density } = 5 . 235 \times {10}^{- 7} \times 900\]
\[ = 4 . 711 \times {10}^{- 4} \text{ kg }\]
\[ \therefore \text{ Mass of 2000 hailstone } = 2000 \times 4 . 711 \times {10}^{- 4} = 0 . 9422\]
\[\text{ Rate of change of momentum } = 0 . 9422 \times 20 \approx 19 \text{N/ m}^2 \]

∴ The total force exerted on the roof = 19 × 100 = 1900 N