Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A 75.19 -kg fisherman is sitting in his 28.09 -kg fishing boat along with his \(13.63-\mathrm{kg}\) tackle box. The boat and its cargo are at rest near a dock. He throws the tackle box toward the dock with a speed of \(2.911 \mathrm{~m} / \mathrm{s}\) relative to the dock. What is the recoil speed of the fisherman and his boat?

Short Answer

Expert verified
Answer: The recoil speed of the fisherman and his boat is 0.384 m/s.

Step by step solution

01

Calculate the Initial Momentum

Initially, the fisherman, boat, and tackle box are all at rest. So, the initial total momentum is 0.
02

Set up the Conservation of Momentum Equation

Let v be the recoil speed of the fisherman and his boat. (initial momentum) = (final momentum) 0 = (m1 + m2) * v + m3 * v1
03

Plug in the Values

Now, we will plug in the values of the given data into the conservation of momentum equation. 0 = (75.19 kg + 28.09 kg) * v + (13.63 kg)(2.911 m/s)
04

Solve for the Recoil Speed v

Rearrange the equation to solve for v: v = -(13.63 kg)(2.911 m/s) / (75.19 kg + 28.09 kg) Now, do the arithmetic: v = -(39.66 kg m/s) / (103.28 kg) v = -0.384 m/s
05

Write the Final Answer

Since we got a negative value for v, this indicates that the recoil speed of the fisherman and his boat is 0.384 m/s in the opposite direction of the thrown tackle box which is relative to the dock. Thus, the recoil speed of the fisherman and his boat is 0.384 m/s.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Momentum
Momentum is a fundamental concept in physics, representing the quantity of motion an object has. It is a vector quantity, meaning it has both magnitude and direction, and is defined as the product of an object's mass and its velocity. The formula for linear momentum (\textbf{p}) is expressed as \( \textbf{p} = m \times \textbf{v} \), where \( m \) is the mass and \( \textbf{v} \) is the velocity of the object.

In the given problem, understanding momentum helps to analyze the motion of the fisherman and the boat after the tackle box is thrown. Initially, the system is at rest; hence the total momentum is zero. The act of throwing the tackle box changes the distribution of momentum within the system, illustrating the principle of conservation of momentum: the total momentum before the event equals the total momentum after the event as long as no external forces are acting on the system.

In problems involving conservation of momentum, it is crucial to pay attention to the direction of the velocity vectors. Negative or positive signs in the computed velocities indicate the direction of motion relative to a chosen frame of reference, which in this case, is the dock.
Recoil Speed
Recoil speed refers to the velocity gained by an object moving in the opposite direction of another object being ejected or thrown away. This concept is commonly encountered in examples involving firearms, rockets, or any scenario where an object is expelled from a system, leading the remaining system to move in the converse direction due to the conservation of momentum.

When the fisherman throws the tackle box, the boat and the fisherman gain a velocity in the opposite direction. This velocity is the recoil speed. The magnitude of the recoil speed can be calculated using the conservation of momentum principle. The actual value of the recoil speed will depend on the mass of the objects involved as well as the speed of the expelled object. In the problem's context, the negative sign obtained in the final computation conveys that the fisherman and his boat move in the direction opposite to that of the thrown tackle box, aligning with the concept of recoil.
Physics Problem Solving

Identifying the Relevant Concept

The first step in solving any physics problem is to understand the concept involved. In the case of the fisherman and his boat, the concept is the conservation of momentum. Recognizing this allows you to set up the necessary equations properly.

Setting Up Equations

Once the concept is identified, the next step is translating the physical situation into a mathematical model — in this case, applying the conservation of momentum equation. The signs used in the equation reflect direction and should align with the chosen frame of reference.

Solving and Interpreting the Results

Substitute the given values into the equations and solve for the unknown. After obtaining the numerical answer, interpret it in the context of the question. A negative sign in velocity, for instance, may indicate direction rather than implying anything about the speed's magnitude.

Following these structured steps in physics problem-solving enhances understanding and accuracy.
Kinematics
Kinematics is the branch of mechanics that deals with the motion of objects without considering the forces causing the motion. It involves the description of motion in terms of displacement, velocity, and acceleration. In the context of the fisherman problem, understanding kinematics is vital as it deals with the velocities of moving objects. The primary kinematic equation relates to the velocity and the movement of the fisherman and his boat after throwing the tackle box.

Although the problem at hand does not delve into acceleration and displacement, kinematic principles allow us to understand how objects move and interact with each other after a particular kinematic event, such as the tackle box being thrown in this scenario. The problem leverages kinematic concepts to deduce the final velocities of the involved objects, which is a core aspect of kinematic analysis.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A cart running on a frictionless air track is propelled by a stream of water expelled by a gas-powered pressure washer stationed on the cart. There is a \(1.00-\mathrm{m}^{3}\) water tank on the cart to provide the water for the pressure washer. The mass of the cart, including the operator riding it, the pressure washer with its fuel, and the empty water tank, is \(400 . \mathrm{kg} .\) The water can be directed, by switching a valve, either backward or forward. In both directions, the pressure washer ejects \(200 . \mathrm{L}\) of water per \(\mathrm{min}\) with a muzzle velocity of \(25.0 \mathrm{~m} / \mathrm{s}\) a) If the cart starts from rest, after what time should the valve be switched from backward (forward thrust) to forward (backward thrust) for the cart to end up at rest? b) What is the mass of the cart at that time, and what is its velocity? (Hint: It is safe to neglect the decrease in mass due to the gas consumption of the gas-powered pressure washer!) c) What is the thrust of this "rocket"? d) What is the acceleration of the cart immediately before the valve is switched?

A motorcycle stunt rider plans to start from one end of a railroad flatcar, accelerate toward the other end of the car, and jump from the flatcar to a platform. The motorcycle and rider have a mass of \(350 . \mathrm{kg}\) and a length of \(2.00 \mathrm{~m}\). The flatcar has a mass of \(1500 \mathrm{~kg}\) and a length of \(20.0 \mathrm{~m}\). Assume that there is negligible friction between the flatcar's wheels and the rails and that the motorcycle and rider can move through the air with negligible resistance. The flatcar is initially touching the platform. The promoters of the event have asked you how far the flatcar will be from the platform when the stunt rider reaches the end of the flatcar. What is your answer?

The center of mass of the Sun and Jupiter is located a) exactly at the center of the Sun. b) near the center of the Sun. c) exactly at the center of Jupiter. d) near the center of Jupiter. e) halfway between the Sun and Jupiter.

A proton with mass \(1.673 \cdot 10^{-27} \mathrm{~kg}\) is moving with a speed of \(1.823 \cdot 10^{6} \mathrm{~m} / \mathrm{s}\) toward an alpha particle with mass \(6.645 \cdot 10^{-27} \mathrm{~kg}\), which is at rest. What is the speed of the center of mass of the proton-alpha particle system?

Young acrobats are standing still on a circular horizontal platform suspended at the center. The origin of the two-dimensional Cartesian coordinate system is assumed to be at the center of the platform. A \(30.0-\mathrm{kg}\) acrobat is located at \((3.00 \mathrm{~m}, 4.00 \mathrm{~m})\), and a 40.0 -kg acrobat is located at \((-2.00 \mathrm{~m},-2.00 \mathrm{~m})\). Assuming that the acrobats stand still in their positions, where must a 20.0 -kg acrobat be located so that the center of mass of the system consisting of the three acrobats is at the origin and the platform is balanced?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free