/*! 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 58 A sudden gust of wind exerts a f... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 58

A sudden gust of wind exerts a force of \(20 \mathrm{~N}\) for \(1.2 \mathrm{~s}\) on a bird that had been flying at \(5 \mathrm{~m} / \mathrm{s}\). As a result, the bird ends up moving in the opposite direction at \(7 \mathrm{~m} / \mathrm{s}\). What is the mass of the bird?

Short Answer

Expert verified
The mass of the bird is 1.6 kg

Step by step solution

01

Calculate initial and final momentum

Momentum is defined as the product of mass and velocity. We can start by calculating the initial momentum (Pi) which is just the bird鈥檚 mass times the bird鈥檚 initial speed. As we don't have the bird's mass, we can set it as \(m\): \(Pi = mv\). Next, calculate the bird's final momentum (Pf), once the gust of wind has affected its motion. The bird ends up moving in the opposite direction, thus its change in direction accounts for a negative speed: \(Pf = -m(v2)\) where \(v2 = 7 m/s\) is the final speed of the bird.
02

Calculate the impulse

Impulse (J) is defined as the force times the time it is exerted for, or the change in momentum. In formula terms this is: \(J = F鈭唗\). Thus, the impulse exerted by the wind is \(J = 20N * 1.2s = 24 Ns\). The impulse can also be presented as the difference between the final and initial momentum. Setting \(Pf - Pi = J\) and substitifying \(Pi\) and \(Pf\) from Step 1 gives: \(-m(v2) - (mv) = 24Ns\).
03

Solve for the mass

Finally, rearrange the equation from step 2 to solve for \(m\): \(m = -\frac{24 Ns}{v2 - v}\). Note the negative on both sides, which occurs because the direction of travel is considered negative due to the opposite direction after the wind gust.
04

Calculate the mass

Substitute the velocities \(v = 5 m/s\) and \(v2 = 7 m/s\) into the formula derived in step 3 to calculate the bird鈥檚 mass: \(m = -\frac{24 Ns}{-7 m/s - 5 m/s} = 1.6 kg\)

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Key Concepts

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

Impulse
Impulse is a core concept in physics that measures the change in momentum of an object when a force is applied over a period of time. In simple terms, it's like giving a push to the object. The formula for impulse is given by:
  • \( J = F \Delta t \)
where:
  • \( J \) is the impulse
  • \( F \) is the constant force applied
  • \( \Delta t \) is the time duration for which the force acts
Impulse can also be understood as the change in momentum, which is helpful in situations like collisions. In our exercise, a bird experiences a sudden impulse due to a gust of wind, changing its velocity as a result.
The impulse is calculated as \( 24 \text{ Ns} \), showing how this force, applied for 1.2 seconds, affects the bird's motion.
Force
Force is the interaction that causes an object to change its velocity or direction. It's one of the basic concepts in physics and can be felt when you push or pull on something. Force is measured in newtons (N) and is a vector, meaning it has both magnitude and direction.
In this exercise, a force of \( 20 \text{ N} \) acts on the bird, which is significant enough to change its velocity. The direction of the force is equally important, as it makes the bird move in the opposite direction. Understanding force helps us know how much energy is needed to change an object's state of motion.
Velocity
Velocity describes the speed of an object in a specified direction. It's different from speed because it includes the direction of motion, making it a vector quantity. Velocity can be positive or negative, indicating direction relative to a chosen reference point.
In the problem, the bird initially travels at \( 5 \text{ m/s} \) before the gust of wind changes its velocity to \( 7 \text{ m/s} \) in the opposite direction. This switch in direction is crucial and shows how velocity includes the concept of direction, not just how fast something is moving.
Newton's Second Law
Newton's Second Law of Motion is fundamental in understanding how motion changes due to applied forces. It states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. The law is usually expressed as:
  • \( F = ma \)
where:
  • \( F \) is the force
  • \( m \) is the mass
  • \( a \) is the acceleration
In our problem, Newton's Second Law helps us understand the bird's motion under the influence of the wind. The force resulted in a change of velocity (acceleration), and by rearranging the equations, we solved for the bird鈥檚 mass, arriving at \( 1.6 \text{ kg} \). This showcases how mass, force, and acceleration are deeply connected. Understanding this law provides a clearer picture of dynamics in everyday and extraordinary events.

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Most popular questions from this chapter

A friend suggests that if all the people in the United States dropped down from a 1 -m-high table at the same time, Earth would move in a noticeable way. To test the credibility of this proposal, (a) determine the momentum imparted to Earth by 300 million people, of an average mass of \(65 \mathrm{~kg}\), dropping from \(1 \mathrm{~m}\) above the surface. Assume no one bounces. (b) What change in Earth's speed would result? SSM

A large semitrailer truck and a small car have equal momentum. How do their speeds compare? A. The truck has a much higher speed than the car. B. The truck has only a slightly higher speed than the car. C. Both have the same speed. D. The truck has only a slightly lower speed than the car. E. The truck has a much lower speed than the car.

A \(0.10-\mathrm{kg}\) firecracker is hanging by a light string from a tree limb. The fuse is lit and the firecracker explodes into three pieces: a small piece \((0.01 \mathrm{~kg})\), a medium-sized piece \((0.03 \mathrm{~kg})\), and a large piece \((0.06 \mathrm{~kg})\). Assume the firecracker is at the origin of a coordinate system such that the \(+z\) axis points straight up, the \(+x\) axis points due east, and the \(+y\) axis points due north. The fragments fly off according to the following momentum vectors: Largest fragment: \(\quad \vec{p}_{L}=p_{\mathrm{Lx}} \hat{x}-p_{\mathrm{L}} \hat{z}\) Medium fragment: \(\bar{p}_{\mathrm{M}}=p_{\mathrm{M} y} \hat{y}+p_{\mathrm{M} z} \hat{z}\) Smallest fragment: \(\bar{p}_{5}=-p_{5 x} \hat{x}-p_{5 y} \hat{y}\) After the explosion, the component of velocity of the smallest piece in both the \(x\) and \(y\) directions is \(4 \mathrm{~m} / \mathrm{s}\). \(\left(v_{\mathrm{S} x}=v_{\mathrm{S} y}=4 \mathrm{~m} / \mathrm{s}\right.\). ) The velocity of the largest fragment is \(1 \mathrm{~m} / \mathrm{s}\) in the \(z\) direction \(\left(v_{\mathrm{Lz}}=1 \mathrm{~m} / \mathrm{s}\right)\). Determine the velocity vector of each fragment immediately after the explosion.

Consider a completely inelastic, head-on collision between two particles that have equal masses and equal speeds. Describe the velocities of the particles after the collision. A. The velocities of both particles are zero. B. Both of their velocities are reversed. C. One of the particles continues with the same velocity and the other comes to rest. D. One of the particles continues with the same velocity and the other reverses direction at twice the speed. E. More information is required to determine the final velocities.

A \(10,000-\mathrm{kg}\) train car moving due east at \(20 \mathrm{~m} / \mathrm{s}\) collides with and couples to a \(20,000-\mathrm{kg}\) train car that is initially at rest. Find the common velocity of the two-car train after the collision.
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