/*! 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 Body A of mass \(M\) has an orig... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 58

Body A of mass \(M\) has an original velocity of \(6.0 \mathrm{m} / \mathrm{s}\) in the \(+x\)-direction toward a stationary body (body \(\mathrm{B}\) ) of the same mass. After the collision, body A has velocity components of $1.0 \mathrm{m} / \mathrm{s}\( in the \)+x\( -direction and \)2.0 \mathrm{m} / \mathrm{s}$ in the + y-direction. What is the magnitude of body B's velocity after the collision?

Short Answer

Expert verified
Answer: The magnitude of the velocity of body B after the collision is approximately 5.39 m/s.

Step by step solution

01

Set up the momentum conservation equations for x and y directions

We know that momentum is conserved in each direction during the collision. Let's set up the momentum equations for the x and y directions separately: In x-direction: Before collision: \(M \cdot v_{Ax}=M \cdot v_{Ax(init)} = M\cdot6.0\) After collision: \(M (v_{Ax(final)}+v_{Bx}) = M(1.0+v_{Bx})\) In y-direction: Before collision: \(M \cdot 0 = 0\) After collision: \(M (v_{Ay(final)}+v_{By}) = M(2.0+v_{By})\)
02

Solve the momentum conservation equations

We can now solve these momentum conservation equations for the final velocity components of body B, \(v_{Bx}\) and \(v_{By}\). Solve for \(v_{Bx}\): \(M \cdot 6.0 = M(1.0+v_{Bx}) \implies 6.0 = 1.0+v_{Bx} \implies v_{Bx} = 5.0 \: m/s\) Solve for \(v_{By}\): \(M \cdot 0 = M(2.0+v_{By}) \implies 0 = 2.0+v_{By} \implies v_{By} = -2.0 \: m/s\)
03

Find the magnitude of the final velocity of body B

Now that we have found the x and y components of the final velocity of body B, we can find the magnitude of its final velocity using the Pythagorean theorem: \(V_B = \sqrt{v_{Bx}^2 + v_{By}^2} = \sqrt{(5.0)^2 + (-2.0)^2} = \sqrt{25+4} = \sqrt{29} \approx 5.39 \, m/s\) The magnitude of the velocity of body B after the collision is approximately \(5.39 \, m/s\).

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 91影视!

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

Particle A is at the origin and has a mass of \(30.0 \mathrm{g} .\) Particle B has a mass of \(10.0 \mathrm{g} .\) Where must particle \(\mathrm{B}\) be located if the coordinates of the \(\mathrm{cm}\) are \((x, y)=(2.0 \mathrm{cm},\) $5.0 \mathrm{cm}) ?$
A helium atom (mass \(4.00 \mathrm{u}\) ) moving at $618 \mathrm{m} / \mathrm{s}$ to the right collides with an oxygen molecule (mass 32.0 u) moving in the same direction at \(412 \mathrm{m} / \mathrm{s}\). After the collision, the oxygen molecule moves at \(456 \mathrm{m} / \mathrm{s}\) to the right. What is the velocity of the helium atom after the collision?
In a circus trapeze act, two acrobats actually fly through the air and grab on to one another, then together grab a swinging bar. One acrobat, with a mass of \(60 \mathrm{kg},\) is moving at \(3.0 \mathrm{m} / \mathrm{s}\) at an angle of \(10^{\circ}\) above the horizontal and the other, with a mass of $80 \mathrm{kg},\( is approaching her with a speed of \)2.0 \mathrm{m} / \mathrm{s}$ at an angle of \(20^{\circ}\) above the horizontal. What is the direction and speed of the acrobats right after they grab on to each other? Let the positive \(x\) -axis be in the horizontal direction and assume the first acrobat has positive velocity components in the positive \(x\)- and \(y\)-directions.
An object located at the origin and having mass \(M\) explodes into three pieces having masses \(M / 4, M / 3,\) and \(5 M / 12 .\) The pieces scatter on a horizontal frictionless xy-plane. The piece with mass \(M / 4\) flies away with velocity \(5.0 \mathrm{m} / \mathrm{s}\) at \(37^{\circ}\) above the \(x\)-axis. The piece with mass \(M / 3\) has velocity \(4.0 \mathrm{m} / \mathrm{s}\) directed at an angle of \(45^{\circ}\) above the \(-x\)-axis. (a) What are the velocity components of the third piece? (b) Describe the motion of the CM of the system after the explosion.
A block of wood of mass \(0.95 \mathrm{kg}\) is initially at rest. A bullet of mass \(0.050 \mathrm{kg}\) traveling at \(100.0 \mathrm{m} / \mathrm{s}\) strikes the block and becomes embedded in it. With what speed do the block of wood and the bullet move just after the collision?
See all solutions

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Read Explanation

Translational Dynamics

Read Explanation

Medical Physics

Read Explanation

Linear Momentum

Read Explanation

Fields in Physics

Read Explanation

Space 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.