Question

During a hailstorm, hailstones with an average mass of 2g and a speed of 15 m/s strike a window pane at a 45^o angle. The area of the window is 0.5 m² and the hailstones hit it at a rate of 30 per second. What average pressure do they exert on the window? How does this compare to the pressure of the atmosphere?

Short answer

Pressure exerted on the window is 2.54 N/m^2.

Step-by-step solution

Given information

Average mass of hail = 2g

speed of 15 m/s

Angle of fall = 45^o

Area of the window = 0.5 m² a

Rate of hail strike = 30 per second.

Explanation

From the pressure force relationship we can calculate average force on window by

$\Delta F_x=\frac{\Delta P_x}{\Delta t}.....................................\left(1\right)$

Change in momentum can be defined as

ΔPx = mΔVx

So we get

role="math" localid="1650443757487" $\Delta F_x=\frac{m\Delta v_x}{\Delta t}..............................\left(2\right)$

We can calculate average pressure as

role="math" localid="1650443764918" $P=\frac{\Delta F_x}{A}.................................\left(3\right)$

As the hail strikes at an angle of 45^o. So we have to find the horizontal component as it will only add to pressure

$\Delta v_x=v\cos \theta -(-v\cos \theta )=2v\cos \theta$

Substitute the given values, we get

$$\begin{gathered} \Delta v_x=2\times (15m/s)\times \cos 45 \\ \Delta v_x=(30m/s)\times \frac{1}{\sqrt{2}} \\ \Delta v_x=21.21\mathrm{m}/\mathrm{s} \end{gathered}$$

Now calculate force by substituting values in equation (2)

$$\begin{gathered} \Delta F_x=\frac{(30\times 2\times {10}^{-3}kg)\times (21.21m/s)}{1s} \\ \Delta F_x=1.27\mathrm{N} \end{gathered}$$

Now calculate average pressure by substituting values in equation(3)

$$\begin{gathered} P=\frac{1.27N}{0.5m^2} \\ P=2.54{\mathrm{Nm}}^{-2} \end{gathered}$$

The value of atm pressure is 101325 N/m²

The pressure on the window is much less than that of atm. pressure.