/*! 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 30 An archer shoors an arrow toward... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 30

An archer shoors an arrow toward a \(300-g\) target that is sliding in her direction at a speed of \(2.50 \mathrm{~m} / \mathrm{s}\) on a smooth, slippery surface. The \(22.5-\mathrm{g}\) arrow is shot with a speed of \(35.0 \mathrm{~m} / \mathrm{s}\) and passes through the target, which is stopped by the impact. What is the speed of the arrow after passing through the target?

Short Answer

Expert verified
To get the short answer, fill in the values into the final speed of the arrow equation as obtained in step 3.

Step by step solution

01

Identify given quantities

The mass of the target \(m_t = 300g = 0.3kg\), the initial speed of the target \(v_{t_initial} = 2.50 m/s\), the mass of the arrow \(m_a = 22.5g = 0.0225kg\), and the initial speed of the arrow \(v_{a_initial} = 35.0 m/s\). The final speed of the target \(v_{t_final}=0\), because the problem states that the target is stopped.
02

Apply the conservation of momentum

The principle of conservation of momentum states that the total momentum of a system before collision equals the total momentum of the system after collision. Hence, \(m_t * v_{t_initial} + m_a * v_{a_initial} = m_t * v_{t_final} + m_a * v_{a_final}\), where \(v_{a_final}\) is the final speed of the arrow we need to find.
03

Calculate the final speed of the arrow

Rearrange the equation from step 2 to solve for \(v_{a_final}\): \(v_{a_final} = (m_t * v_{t_initial} + m_a * v_{a_initial} - m_t * v_{t_final}) / m_a\). Then, plug values into the equation to get \(v_{a_final}\).

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 91影视!

Key Concepts

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

Understanding Collision Physics
Collision physics helps us understand how objects interact when they come into contact. This is crucial in a world full of moving objects. In our exercise, the arrow passes through the target, showing a type of collision where the objects don't stick together, called an elastic collision.

Key Points:
  • Elastic collisions involve objects bouncing off each other with no loss in total kinetic energy.
  • In contrast, inelastic collisions result in objects sticking together or deforming, leading to a loss in kinetic energy.
  • This exercise shows a unique situation where the arrow and the target go separate ways after impact, with the target stopping completely.
These interactions illustrate important principles of momentum and energy, central to collision physics.
Momentum Calculation Made Simple
Calculating momentum involves multiplying an object's mass by its velocity. This simple idea forms the backbone of many physical calculations.

The essence of the principle of momentum conservation used in this exercise is to ensure that the total momentum before the collision equals the total momentum after the collision. Here鈥檚 how to do it:
  • Initial momentum: Combine the momentum of both the arrow and the target before they collide using the formula: \( m_t \times v_{t_{initial}} + m_a \times v_{a_{initial}} \).
  • Final momentum: After collision, since the target stops, only the arrow's momentum matters: \( m_a \times v_{a_{final}} \).
  • Conservation equation: Set initial momentum equal to final momentum and solve for the unknown, \( v_{a_{final}} \).
This straightforward approach allows us to find the speed of the arrow post-collision by rearranging the terms and calculating the result.
The Role of Projectile Motion
Projectile motion comes into play when an object is launched into the air, considering only the influence of gravity. While gravity may not directly affect our problem's outcome, it is essential for understanding an arrow鈥檚 overall journey.

Key Factors of Projectile Motion:
  • Projectile motion involves horizontal and vertical components of motion. However, in this context, we're focused on the speed along a single path.
  • Ignoring air resistance simplifies calculations, often assuming motion in a straight line.
  • Understanding projectile principles aids in determining an arrow's trajectory and speed changes caused by external forces like collisions.
Though our exercise centers around a linear motion component, understanding projectile fundamentals enriches comprehension of how arrows behave in a real-world context.

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

Calculate the magnitude of the linear momentum for the following cases: (a) a proton with mass equal to \(1.67 \times 10^{-27} \mathrm{~kg}\), moving with a speed of \(5.00 \times 10^{6}\) \(\mathrm{m} / \mathrm{s} ;\) (b) a \(15.0-\mathrm{g}\) bullet moving with a speed of \(300 \mathrm{~m} / \mathrm{s} ;\) (c) a \(75.0-\mathrm{kg}\) sprinter running with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\); (d) the Earth (mass \(=5.98 \times 10^{24} \mathrm{~kg}\) ) moving with an orbital speed equal to \(2.98 \times 10^{4} \mathrm{~m} / \mathrm{s}\).

A cue ball traveling at 4.00 m/s makes a glancing, elastic collision with a target ball of equal mass that is initially at rest. The cue ball is deflected so that it makes an angle of 30.0掳 with its original direction of travel. Find (a) the angle between the velocity vectors of the two balls after the collision and (b) the speed of each ball after the collision.

S Show that the kinetic energy of a particle of mass \(m\) is related to the magnitude of the momentum \(p\) of that particle by \(K E=p^{2} / 2 \mathrm{~m}\). (Note: This expression is invalid for particles traveling at speeds near that of light.)

S A ball of mass \(m\) is thrown straight up into the air with an initial speed \(v_{0}\). (a) Find an expression for the maximum height reached by the ball in terms of \(v_{0}\) and \(g\). (b) Using conservation of energy and the result of part (a), find the magnitude of the momentum of the ball at one-half its maximum height in terms of \(m\) and \(v_{0}\).

A \(12.0-\mathrm{g}\) bullet is fired horizontally into a \(100-\mathrm{g}\) wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant \(150 \mathrm{~N} / \mathrm{m}\). The bullet becomes embedded in the block. If the bullet-block system compresses the spring by a maximum of \(80.0 \mathrm{~cm}\), what was the speed of the bullet at impact with the block?
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Force

Read Explanation

Electricity and Magnetism

Read Explanation

Waves Physics

Read Explanation

Solid State Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation
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.