/*! 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 25 A hunter on a frozen, essentiall... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 25

A hunter on a frozen, essentially frictionless pond uses a rifle that shoots 4.20-g bullets at 965 m/s. The mass of the hunter (including his gun) is 72.5 kg, and the hunter holds tight to the gun after firing it. Find the recoil velocity of the hunter if he fires the rifle (a) horizontally and (b) at 56.0\(^\circ\) above the horizontal.

Short Answer

Expert verified
(a) -0.0560 m/s, (b) -0.0314 m/s.

Step by step solution

01

Understand the Law of Conservation of Momentum

Momentum is conserved in an isolated system, which means that the initial momentum before firing the rifle is equal to the final momentum after firing the rifle. Initially, the system is at rest, thus the total initial momentum is zero.
02

Express Conservation of Momentum Mathematically

The conservation of momentum can be expressed using the formula: \[ m_b \cdot v_b + m_h \cdot v_h = 0 \]where:- \( m_b \) is the mass of the bullet, 4.20 g (or 0.0042 kg).- \( v_b \) is the velocity of the bullet, 965 m/s.- \( m_h \) is the mass of the hunter and gun, 72.5 kg.- \( v_h \) is the recoil velocity of the hunter and gun.
03

Solve for Recoil Velocity Horizontally

Using the conservation equation:\[ 0.0042 \times 965 + 72.5 \times v_h = 0 \]Solve for \( v_h \):\[ v_h = -\frac{0.0042 \times 965}{72.5} \approx -0.0560 \text{ m/s} \]The negative sign indicates that the hunter moves in the opposite direction of the bullet.
04

Analyze the Recoil at an Angle

When the bullet is fired at an angle of 56.0掳 above the horizontal, only the horizontal component of the bullet's momentum affects the recoil velocity of the hunter.Calculate the horizontal component of the bullet's velocity: \[ v_{bx} = 965 \cdot \cos(56.0^\circ) \]
05

Solve for Recoil Velocity with Angle

Substitute the horizontal component of the bullet's velocity into the conservation equation:\[ 0.0042 \times (965 \cdot \cos(56.0^\circ)) + 72.5 \times v_h = 0 \]Solve for \( v_h \):\[ v_h = -\frac{0.0042 \times 965 \cdot \cos(56.0^\circ)}{72.5} \approx -0.0314 \text{ m/s} \]The negative sign once again indicates the hunter's recoil direction is opposite to the bullet's horizontal direction.

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.

Physics Problems
Physics problems are essential tools in understanding scientific concepts, such as momentum conservation. In our exercise, we explored the scenario of a hunter standing on a frictionless frozen pond.
The hunter fires a rifle, and this setting clear physics problems involve calculations of velocity and momentum.
By breaking down complex phenomena into simpler parts, physics problems help us delve into the detailed operational workings of physical laws.
  • Typically, physics problems require identifying known and unknown quantities.
  • They involve using formulas derived from laws or theories of nature, like the law of conservation of momentum in this case.
  • Comprehensive problem-solving in physics often involves analyzing different scenarios, such as movement in a horizontal direction or at an angle.
Tackling physics problems nurtures critical thinking and provides insights on applying theoretical knowledge to real-life situations.
Recoil Velocity
Recoil velocity refers to the speed at which an object moves backward in response to an action, like a gun firing a bullet. In our exercise, it's important to consider recoil velocity because it influences how entities respond when forces are applied.
Recoil velocity is evident when the hunter experiences backward momentum upon discharging a bullet forward. It's notable that the velocity is negative indicating opposite direction movement.
Recoil velocity depends on various factors:
  • The mass of both the bullet and the shooter (or object experiencing recoil) is key.
  • The speed at which the bullet is discharged.
  • The angle at which the bullet is fired, affecting how momentum divides between horizontal and vertical components.
Understanding recoil velocity is fundamental for designing stabilizing mechanisms for firearms and ensuring the safety of operators.
Momentum Conservation Formula
The momentum conservation formula is based on the principle that the total momentum of a closed system remains constant over time if it doesn't interact with external forces.
This principle is critical in solving the exercise since it explains how the hunter's system behaves post-shot.
The formula used was: \[ m_b \cdot v_b + m_h \cdot v_h = 0 \]
  • \(m_b\) is the mass of the bullet, and \(v_b\) is its velocity.
  • \(m_h\) is the mass of the hunter (including the gun), and \(v_h\) denotes the recoil velocity.
The momentum conservation principle applies well in our frictionless pond scenario since no external horizontal forces are in play.
By implementing this formula, we correctly predict how changes in variables like angle of firing impact the recoil velocity. Understanding this fundamental law allows solving numerous real-world and theoretical physics problems.

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

In a fireworks display, a rocket is launched from the ground with a speed of 18.0 m/s and a direction of 51.0\(^\circ\) above the horizontal. During the flight, the rocket explodes into two pieces of equal mass (see Fig. 8.32). (a) What horizontal distance from the launch point will the center of mass of the two pieces be after both have landed on the ground? (b) If one piece lands a horizontal distance of 26.0 m from the launch point, where does the other piece land?

The mass of a regulation tennis ball is 57 g (although it can vary slightly), and tests have shown that the ball is in contact with the tennis racket for 30 ms. (This number can also vary, depending on the racket and swing.) We shall assume a 30.0-ms contact time. The fastest-known served tennis ball was served by "Big Bill" Tilden in 1931, and its speed was measured to be 73 m/s. (a) What impulse and what force did Big Bill exert on the tennis ball in his record serve? (b) If Big Bill's opponent returned his serve with a speed of 55 m/s, what force and what impulse did he exert on the ball, assuming only horizontal motion?

Obviously, we can make rockets to go very fast, but what is a reasonable top speed? Assume that a rocket is fired from rest at a space station in deep space, where gravity is negligible. (a) If the rocket ejects gas at a relative speed of 2000 m/s and you want the rocket's speed eventually to be 1.00\(\times\) 10\(^{-3}c\), where \(c\) is the speed of light in vacuum, what fraction of the initial mass of the rocket and fuel is \(not\) fuel? (b) What is this fraction if the final speed is to be 3000 m/s?

To protect their young in the nest, peregrine falcons will fly into birds of prey (such as ravens) at high speed. In one such episode, a 600-g falcon flying at 20.0 m/s hit a 1.50-kg raven flying at 9.0 m/s. The falcon hit the raven at right angles to its original path and bounced back at 5.0 m/s. (These figures were estimated by the author as he watched this attack occur in northern New Mexico.) (a) By what angle did the falcon change the raven's direction of motion? (b) What was the raven's speed right after the collision?

One 110-kg football lineman is running to the right at 2.75 m/s while another 125-kg lineman is running directly toward him at 2.60 m/s. What are (a) the magnitude and direction of the net momentum of these two athletes, and (b) their total kinetic energy?
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Electricity and Magnetism

Read Explanation

Oscillations

Read Explanation

Force

Read Explanation

Wave Optics

Read Explanation

Magnetism and Electromagnetic Induction

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.