Problem 21 - An Introduction to Thermal Physics

Question

During a hailstorm, hailstones with an average mass of 2 g and a speed of \(15 \mathrm{m} / \mathrm{s}\) strike a window pane at a \(45^{\circ}\) angle. The area of the window is \(0.5 \mathrm{m}^{2}\) 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

The average pressure exerted by hailstones is 1.27 Pa, much less than atmospheric pressure.

Step 1: Identify the Variables

First, let's identify the given information:- Mass of each hailstone, \( m = 2 \, \text{g} = 0.002 \, \text{kg} \) (convert grams to kilograms)- Speed of each hailstone, \( v = 15 \, \text{m/s} \)- Angle of impact, \( \theta = 45^{\circ} \)- Window area, \( A = 0.5 \, \text{m}^2 \)- Rate of hailstones hitting the window, \( R = 30 \, \text{hailstones/second} \)

Step 2: Calculate the Momentum Change

The momentum of one hailstone is given by \( p = mv \).\[p = 0.002 \, \text{kg} \times 15 \, \text{m/s} = 0.03 \, \text{kg·m/s}\]Since the hailstones strike at an angle of \(45^{\circ}\), only the component of velocity normal (perpendicular) to the window contributes to the pressure. This component is \( v_{\text{normal}} = v \cos \theta \).\[v_{\text{normal}} = 15 \, \text{m/s} \times \cos(45^{\circ}) = 15 \, \text{m/s} \times \frac{1}{\sqrt{2}} \approx 10.61 \, \text{m/s}\]The normal change in momentum for each hailstone is:\[\Delta p_{\text{normal}} = 0.002 \, \text{kg} \times 10.61 \, \text{m/s} = 0.02122 \, \text{kg·m/s}\]

Step 3: Calculate the Force Exerted by Hailstones

The rate of momentum change (force) is given by the number of hailstones per second multiplied by the change in momentum per hailstone:\[F = R \cdot \Delta p_{\text{normal}} = 30 \, \text{hailstone/second} \times 0.02122 \, \text{kg·m/s} \approx 0.6366 \, \text{N}\]

Step 4: Calculate the Average Pressure on the Window

Pressure is defined as force per unit area. The average pressure (\( P \)) is thus:\[P = \frac{F}{A} = \frac{0.6366 \, \text{N}}{0.5 \, \text{m}^2} = 1.2732 \, \text{Pa}\]

Step 5: Compare with Atmospheric Pressure

The atmospheric pressure is approximately \( 101325 \, \text{Pa} \). Compare the calculated average pressure:\[\text{Atmospheric Pressure} = 101325 \, \text{Pa}\]\[\text{Hailstone Pressure} = 1.2732 \, \text{Pa}\]The pressure exerted by the hailstones is tiny compared to atmospheric pressure.

Key Concepts

Momentum

Momentum is a fundamental concept in physics. It describes the quantity of motion an object possesses and is the product of an object's mass and velocity, expressed as \( p = mv \). During a hailstorm, each hailstone carries momentum as it travels towards the window. By calculating the momentum of a single hailstone with mass \( 0.002 \, \text{kg} \) and velocity \( 15 \, \text{m/s} \), we find its momentum to be \( 0.03 \, \text{kg} \cdot \text{m/s} \). This number tells us how much motion energy the hailstone carries as it hits the window.

When an object like a hailstone strikes a surface, its momentum changes. Understanding this change is critical for evaluating the force exerted on impact. For surfaces like windows, the momentum's normal or perpendicular component is what affects the pressure exerted.

Force

Force is a measure of the interaction that changes the motion of an object. According to Newton's Second Law, force is the rate of change of momentum over time. Hence, when our hailstorm experiment is analyzed, we focus on how the momentum of hailstones changes with each impact.

The force exerted by the hailstones is calculated by multiplying the change in momentum of a single stone by the number of stones striking the window each second. In this case, each stone loses its normal momentum upon striking, imparting a force of approximately \( 0.6366 \, \text{N} \) during the continuous hitting sequence.

• Understanding force helps in predicting the effects such as deformation or potential damage to surfaces like glass windows.
• The force also directly relates to how much pressure is applied, pivotal in the subsequent pressure calculations.

Impact Angle

The angle at which a hailstone hits the window, known as the impact angle, significantly affects how much force the window experiences. For this problem, a 45-degree angle implies that only part of the stone's velocity contributes directly to the pressure on the window.

To evaluate this effect accurately, we calculate the velocity's normal component — the part of velocity perpendicular to the surface — using trigonometric functions like the cosine of the angle. With a 45-degree impact angle, the normal component is calculated as \( v \cos(\theta) \). This results in a reduced amount of velocity, diminishing the overall momentum change and thus the force exerted in the normal direction.

• Understanding the impact angle is vital in engineering and physical scenarios as it influences design choices and safety protocols.

Atmospheric Pressure

Atmospheric pressure is the force per unit area exerted by atmospheric gases on surfaces, often measured as \( 101325 \, \text{Pa} \). In comparison, the hailstones exert a much lesser pressure of around \( 1.2732 \, \text{Pa} \) on the window.

This comparison illustrates how small forces from hailstones are, compared to the continuous and much stronger push exerted by the atmospheric layer enveloping the Earth.

• Understanding atmospheric pressure is essential in various science and engineering fields, such as meteorology and aerodynamics.
• It also highlights how external factors like weather can have differing scales of impact, with varying consequences.