/*! 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 153 A 1-oz bullet is traveling with ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 153

A 1-oz bullet is traveling with a velocity of 1400 ft/s when it impacts and becomes embedded in a 5-lb wooden block. The block can move vertically without friction. Determine (a) the velocity of the bullet and block immediately after the impact, (b) the horizontal and vertical components of the impulse exerted by the block on the bullet.

Short Answer

Expert verified
(a) Use conservation of momentum to find final velocity. (b) Impulse's horizontal component equals total impulse; vertical is zero.

Step by step solution

01

Understand the problem

We have a bullet with a mass of 1 oz traveling at 1400 ft/s that embeds into a 5 lb wooden block. We need to determine the velocity of the bullet and block system immediately after impact and the components of the impulse exerted.
02

Convert units

First, we convert the mass of the bullet from ounces to pounds. Since there are 16 oz in a pound, the bullet's mass is \( \frac{1}{16} \) lb. Then convert these masses into slugs, knowing 1 lb = 1/32.2 slugs.
03

Calculate initial momentum

We use the formula for momentum \( p = mv \). Calculate the initial momentum of the bullet: \( p_{ ext{bullet}} = \left( \frac{1}{16} \times \frac{1}{32.2} \right) \times 1400 \text{ ft/s} \).
04

Apply conservation of momentum

Since the block is initially at rest, its initial momentum is zero. Using the conservation of momentum principle: \( p_{ ext{initial}} = p_{ ext{final}} \). Thus, \( (m_{ ext{bullet}} \cdot v_{ ext{bullet}}) = (m_{ ext{bullet}} + m_{ ext{block}}) \cdot v_{ ext{final}} \).
05

Solve for final velocity

Plug in the numbers: \( \left( \frac{1}{16} \times \frac{1}{32.2} \times 1400 \right) = \left( \frac{1}{16} \times \frac{1}{32.2} + \frac{5}{32.2} \right) \cdot v_{ ext{final}} \). Solve for \( v_{\text{final}} \).
06

Calculate the impulse

Impulse is given by the change in momentum. Compute the initial and final momentum of the bullet and use: \( J = p_{ ext{final}} - p_{ ext{initial}} \). Since we know the initial and final momenta, calculate the impulse's magnitude.
07

Determine components of impulse

Since the impulse is exerted in the direction of travel, the horizontal component equals the total impulse and the vertical component is zero because there is no initial vertical motion.

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.

Conservation of Momentum
The principle of conservation of momentum states that the total momentum of a closed system remains constant if no external forces are acting on it. This is an important concept in solving physics problems involving collisions. For our exercise, this means:
  • Before the collision, the bullet is moving, while the block is stationary. Thus, the bullet has initial momentum, and the block has zero momentum.
  • During the collision, the bullet embeds into the block, forming a single system.
  • Since there are no external horizontal forces acting, the total horizontal momentum before and after the collision is the same.
This principle allows us to set up the equation:\[ m_{\text{bullet}} \cdot v_{\text{bullet}} = (m_{\text{bullet}} + m_{\text{block}}) \cdot v_{\text{final}}, \]where all terms contribute to ensuring momentum is conserved in the horizontal direction.
Unit Conversion
In physics, using consistent units is crucial for accurate calculations. Here, we need to convert the mass of objects from one system to another:
  • The bullet's mass is given as 1 oz. We know there are 16 ounces in a pound, so the bullet's mass can be converted to pounds: \(\frac{1\, \text{oz}}{16} = 0.0625 \text{ lb}\).
  • For gravitational calculations, the unit often used is the slug, where 1 lb is equal to \(1/32.2\) slugs. Thus, the bullet's mass in slugs is \(0.0625 \div 32.2\).
This consistent unit conversion is necessary to ensure the accuracy of momentum calculations, involving both the bullet and the block.
Velocity Calculation
Determining the velocity of the combined bullet-block system after impact requires careful application of the conservation of momentum:
  • The initial momentum of the system is solely due to the moving bullet: \( p_{\text{bullet}} = m_{\text{bullet}} \times v_{\text{bullet}} \).
  • The final momentum of the system includes both the bullet and block moving together: \( (m_{\text{bullet}} + m_{\text{block}}) \times v_{\text{final}} \).
  • Equating initial and final momentum, solve for \( v_{\text{final}} \):\[ v_{\text{final}} = \frac{m_{\text{bullet}} \times v_{\text{bullet}}}{m_{\text{bullet}} + m_{\text{block}}} \].
This formula plugs in the known values, enabling students to effectively calculate the new velocity after impact.
Impulse Components
Impulse is a measure of the change in momentum and has both magnitude and direction. It is represented as:\[ J = \Delta p = p_{\text{final}} - p_{\text{initial}} \].Breaking down impulse in the given scenario:
  • The impulse exerted on the bullet by the block equals the change in momentum, calculated using the initial separate and final combined velocities.
  • The horizontal component of the impulse corresponds exactly to this change, as the problem specifies no vertical force components due to the horizontal motion.
  • The vertical impulse component is zero, aligning with the absence of initial vertical motion or external vertical forces acting on the combined system.
Understanding these components helps grasp how impulse affects motion directionally, critical in dynamics scenarios.

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

One of the requirements for tennis balls to be used in official competition is that, when dropped onto a rigid surface from a height of 100 in., the height of the first bounce of the ball must be in the range 53 in. \(\leq h \leq 58\) in. Determine the range of the coefficients of restitution of the tennis balls satisfying this requirement.

A satellite describes an elliptic orbit of minimum altitude 606 km above the surface of the earth. The semimajor and semiminor axes are 17 440 km and 13 950 km, respectively. Knowing that the speed of the satellite at point C is 4.78 km/s, determine (a) the speed at point A, the perigee, (b) the speed at point B, the apogee.

Rockfalls can cause major damage to roads and infrastructure. To design mitigation bridges and barriers, engineers use the coefficient of restitution to model the behavior of the rocks. Rock \(A\) falls a distance of \(20 \mathrm{m}\) before striking an incline with a slope of \(\alpha=40^{\circ}\). Knowing that the coefficient of restitution between rock \(A\) and the incline is \(0.2,\) determine the velocity of the rock after the impact.

Two disks sliding on a frictionless horizontal plane with opposite velocities of the same magnitude \(v_{0}\) hit each other squarely. Disk \(A\) is known to have a weight of 6 lb and is observed to have zero velocity after impact. Determine \((a)\) the weight of disk \(B,\) knowing that the coefficient of restitution between the two disks is \(0.5,\) (b) the range of possible values of the weight of disk \(B\) if the coefficient of restitution between the two disks is unknown.

Steep safety ramps are built beside mountain highways to enable vehicles with defective brakes to stop. A 10 -ton truck enters a \(15^{\circ}\) ramp at a high speed \(v_{0}=108 \mathrm{ft} / \mathrm{s}\) and travels for \(6 \mathrm{s}\) before its speed is reduced to \(36 \mathrm{ft} / \mathrm{s}\). Assuming constant deceleration, determine \((a)\) the magnitude of the braking force, \((b)\) the additional time required for the truck to stop. Neglect air resistance and rolling resistance.
See all solutions

Recommended explanations on Physics Textbooks

Force

Read Explanation

Energy Physics

Read Explanation

Linear Momentum

Read Explanation

Famous Physicists

Read Explanation

Particle Model of Matter

Read Explanation

Engineering Physics

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.