Q. 1.21 - An Introduction to Thermal Physics

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?

Step-by-Step Solution

Answer

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

Step1: 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.

Step2: 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

$\Delta F_x=\frac{m\Delta v_x}{\Delta t}..............................\left(2\right)$

We can calculate average pressure as 

$P=\frac{\Delta F_x}{A}.................................\left(3\right)$

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 

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

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.