/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 In a \(30.0-\) interval, 500 hai... [FREE SOLUTION] | 91Ó°ÊÓ

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In a \(30.0-\) interval, 500 hailstones strike a glass window of area \(0.600 \mathrm{m}^{2}\) at an angle of \(45.0^{\circ}\) to the window surface. Each hailstone has a mass of \(5.00 \mathrm{g}\) and moves with a speed of \(8.00 \mathrm{m} / \mathrm{s} .\) Assuming the collisions are elastic, find the average force and pressure on the window.

Short Answer

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Step by step solution

01

Calculate momentum change of one hailstone

Firstly, calculate the momentum change for one hailstone. The momentum is given as mass multiplied by velocity. However, since the collision is elastic and at an angle, only the component of velocity perpendicular to the window will reverse. We decompose the velocity vector into the perpendicular component using the cosine of the angle provided. Therefore, the change in momentum \( \Delta p \) for a single hailstone will be twice the mass multiplied by the component of velocity perpendicular to the window surface,i.e., \( \Delta p = 2mV_{\perp} \) where \( V_{\perp} = V \cos(\theta) \). Substituting the given values: \( m = 5.00 g = 5.00 × 10^{-3} kg \), \( V = 8.00 m/s \), and \( \theta = 45.0^{\circ} = 45.0 \pi / 180 rad \), we get the final result.
02

Calculate the average force

Next, calculate the average force exerted by all hailstones per unit time. The force can be obtained from the relation \( F = \Delta p / \Delta t \) where \( \Delta p \) is the change in momentum and \( \Delta t \) is the time interval. Considering that there are 500 hailstones hitting the window in a duration of \( 30.0s \), the total change in momentum becomes \( 500 \Delta p \). Substituting this into the equation gives us the average force.
03

Calculate the average pressure

Then, we calculate the average pressure which is given by the force divided by the area over which the force is distributed. Therefore, we have \( P = F/A \), where \( A = 0.600 m^2 \) is the window area provided in the exercise. Substituting \( F \) from the previous step we can calculate the average pressure.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Elastic Collision
An elastic collision is a fundamental concept in physics where two or more bodies collide and bounce off each other without losing their kinetic energy. It's an ideal scenario, often used in problems to simplify calculations, as it assumes no energy is lost to sound, heat, or deformation during the collision. In practical terms, elastic collisions don't perfectly occur in the real world, but some events, like the collision of billiard balls or hailstones hitting a surface, can come close.

In the context of the given problem, when hailstones hit the window, they are assumed to undergo elastic collisions. This means that the speed of each hailstone perpendicular to the window just before and just after the collision remains the same, but the direction reverses. Hence, the fact that the problem specifies an elastic collision allows us to calculate the change in momentum accurately using this assumption.
Momentum Change
Momentum change is a key topic in physics, particularly when examining the motion of objects after they interact. It is defined as the difference in an object's momentum before and after an event, such as a collision. Momentum itself is a product of an object's mass and velocity. It's a vector quantity, meaning it has both magnitude and direction.

In our hailstone problem, the momentum change for one hailstone arises when it hits the window and subsequently bounces off. Since the collision is elastic and the hailstones strike at a 45-degree angle, only the perpendicular component of their velocity changes. This change in momentum is crucial for calculating the force exerted by the hailstones on the window, as force is directly related to the rate of change of momentum over time.
Average Force Calculation
Average force calculation involves finding the constant force that would produce the same effect on an object's motion as the actual force exerted over some time interval. This concept is deeply connected to Newton's second law of motion, which states that the force acting on an object is equal to the rate of change of its momentum.

For the hailstone problem, we find the average force by dividing the total momentum change by the time interval during which the collisions occur. Since we are dealing with multiple hailstones striking the window, we must consider the momentum change for each hailstone and then add them up to find the total momentum change. This total momentum change, divided by the time over which the hailstones strike the window, gives us the average force exerted by the hailstones on the window.
Pressure Calculation
Pressure calculation is another important concept in many fields, such as physics, engineering, and meteorology. It is defined as the force applied perpendicular to the surface of an object per unit area. In simple terms, pressure explains how concentrated a force is over a particular area and is measured in pascals (Pa) in the International System of Units.

In the textbook problem, once we have the average force exerted by the hailstones, calculating the pressure on the window involves dividing that force by the area of the glass window. The larger the area over which the force is spread, the smaller the pressure. This calculation is relevant, as it can determine if a material will withstand the force applied to it without breaking, crucial for understanding the impacts of hailstones on structures.

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Most popular questions from this chapter

A 1.00 -mol sample of an ideal monatomic gas is at an initial temperature of \(300 \mathrm{K}\). The gas undergoes an isovolumetric process acquiring \(500 \mathrm{J}\) of energy by heat. It then undergoes an isobaric process losing this same amount of energy by heat. Determine (a) the new temperature of the gas and (b) the work done on the gas.

A gas is at \(0^{\circ} \mathrm{C}\). If we wish to double the rms speed of its molecules, to what temperature must the gas be brought?

In a period of \(1.00 \mathrm{s}, 5.00 \times 10^{23}\) nitrogen molecules strike a wall with an area of \(8.00 \mathrm{cm}^{2} .\) If the molecules move with a speed of \(300 \mathrm{m} / \mathrm{s}\) and strike the wall head-on in elastic collisions, what is the pressure exerted on the wall? (The mass of one \(\mathrm{N}_{2}\) molecule is \(4.68 \times 10^{-26} \mathrm{kg} .\) )

(a) If it has enough kinetic energy, a molecule at the surface of the Earth can "escape the Earth's gravitation," in the sense that it can continue to move away from the Earth forever, as discussed in Section \(13.7 .\) Using the principle of conservation of energy, show that the minimum kinetic energy needed for "escape" is \(m g R_{E},\) where \(m\) is the mass of the molecule, \(g\) is the free-fall acceleration at the surface, and \(R_{E}\) is the radius of the Earth. (b) Calculate the temperature for which the minimum escape kinetic energy is ten times the average kinetic energy of an oxygen molecule.

At what temperature would the average speed of helium atoms equal (a) the escape speed from Earth, \(1.12 \times 10^{4} \mathrm{m} / \mathrm{s}\) and \((\mathrm{b})\) the escape speed from the Moon, \(2.37 \times 10^{3} \mathrm{m} / \mathrm{s} ?\) (See Chapter 13 for a discussion of escape speed, and note that the mass of a helium atom is \(\left.6.64 \times 10^{-27} \mathrm{kg} .\right)\)

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