Question

In a 30.0-s interval, 500 hailstones strike a glass window with an area of 0.600m2 at an angle of 45.0°to the window surface. Each hailstone has a mass of 5.00g and a
speed of 8.00 m/s. If the collisions are elastic, what are the average force and pressure on the window?

Answer 1

or one hailstone we have;
Force = Mass X acceleration = 0.005kg x 9.8.} This is when the hailstone is not inclined at an angle.
When the hailstone is inclined at an angle of 45, then the component of force along the glass window will be F =0.005kg x 9.8 x sin45= 0.005kg x 9.8 x 0.707= 0.0346N.
Therefore, total force for the 500 hailstones would be 500x0.0346N=17.32N
This force is acting on an area equal to 0.600m2
Pressure = Force per unit area = 17.32N/0.600m2 = 28.9Pa

Answer 2

(a) The average force exerted by the hailstone is 0.667 N.

(b) The pressure exerted by the force on the window is 1.1 N/m².

Average force exerted by the forces

The average force exerted by the hailstone is calculated by applying Newton's second law of motion as follows;

$F = ma\\\\F = \frac{mv}{t} \\\\F = \frac{500 (0.005) \times 8}{30} \\\\F = 0.667 \ N$

Pressure on the window

The pressure exerted by the force on the window is calculated as follows;

$P = \frac{F}{A} \\\\P = \frac{0.667}{0.6} \\\\P = 1.11 \ N/m^2$

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