/*! 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 36 At the instant a \(3.0 \mathrm{~... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 36

At the instant a \(3.0 \mathrm{~kg}\) particle has a velocity of \(6.0 \mathrm{~m} / \mathrm{s}\) in the negative \(y\) direction, a \(4.0 \mathrm{~kg}\) particle has a velocity of \(7.0 \mathrm{~m} / \mathrm{s}\) in the positive \(x\) direction. What is the speed of the center of mass of the two-particle system?

Short Answer

Expert verified
The speed of the center of mass of the two-particle system is approximately 4.75 m/s.

Step by step solution

01

Calculate the total mass

First, add the masses of the two particles to find the total mass of the system. Let the masses of the two particles be \(m_1\) and \(m_2\). Thus, the total mass \(M = m_1 + m_2 = 3.0 \, kg + 4.0 \, kg = 7.0 \, kg\).
02

Calculate the momentum of each particle

Next, calculate the momentum of each particle. The momentum of a particle is given by \(p = m \, v\), where \(m\) is the mass and \(v\) is the velocity. For the first particle (particle 1): \(\vec{p}_1 = (3.0 \, kg)(6.0 \, m/s) \, (-\hat{j}) = -18.0 kg \, m/s \, \hat{j}\). For the second particle (particle 2): \(\vec{p}_2 = (4.0 \, kg)(7.0 \, m/s) \, \hat{i} = 28.0 \, kg \, m/s \, \hat{i}\).
03

Calculate the total momentum

Add the momentum vectors of the two particles to get the total momentum of the system. Total momentum \(\vec{P} = \vec{p}_1 + \vec{p}_2 = -18.0 \, kg \, m/s \, \hat{j} + 28.0 \, kg \, m/s \, \hat{i}\).
04

Calculate the velocity of the center of mass

The velocity of the center of mass \(\vec{v}_{cm}\) is given by \(\vec{v}_{cm} = \frac{\vec{P}}{M}\). Substituting the total momentum and total mass: \(\vec{v}_{cm} = \frac{-18.0 \, kg \, m/s \, \hat{j} + 28.0 \, kg \, m/s \, \hat{i}}{7.0 \, kg} = -2.57 \, m/s \, \hat{j} + 4.0 \, m/s \, \hat{i}\).
05

Calculate the speed of the center of mass

The speed of the center of mass is the magnitude of the velocity vector \(\vec{v}_{cm}\). The magnitude of a vector \(\vec{v} = v_x \hat{i} + v_y \hat{j}\) is given by \(|\vec{v}| = \sqrt{v_x^2 + v_y^2}\). Substituting the components: \(|\vec{v}_{cm}| = \sqrt{(-2.57 \, m/s)^2 + (4.0 \, m/s)^2} = \sqrt{6.60 + 16.00} = \sqrt{22.60} \, m/s 鈮� 4.75 \, m/s\).

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!

Key Concepts

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

Momentum
Momentum is a measure of the motion of an object and is defined by the product of its mass and velocity. It's a vector quantity, meaning it has both a magnitude and a direction. Mathematically, momentum \( \vec{p} \) is given by the formula: \[ \vec{p} = m \vec{v} \] where \( m \) is the mass of the object and \( \vec{v} \) is its velocity. In this exercise:
  • For particle 1: \( \vec{p}_1 = (3.0 \ \text{kg})(6.0 \ \text{m/s}) \ (- \hat{j}) = -18.0 \ \text{kg m/s} \ \hat{j} \)

  • For particle 2: \( \vec{p}_2 = (4.0 \ \text{kg})(7.0 \ \text{m/s}) \ \hat{i} = 28.0 \ \text{kg m/s} \ \hat{i} \)

By adding these momenta, we determined the total momentum of the system, which is crucial for finding the center of mass's motion.
Velocity
Velocity describes how fast an object moves and in which direction. It's different from speed, which only measures how fast something moves irrespective of direction. Velocity is also a vector quantity. In our calculation:
  • Particle 1 has a velocity of \( -6.0 \ \text{m/s} \) in the \(- \hat{j}\) or negative y-direction.
  • Particle 2 has a velocity of \(+ 7.0 \ \text{m/s} \) in the \( \hat{i}\) or positive x-direction.
We combined these velocities to find the velocity of the center of mass: \[ \vec{v}_{cm} = \frac{-18.0 \ \text{kg m/s} \hat{j} + 28.0 \ \text{kg m/s} \hat{i}}{7.0 \ \text{kg}} = -2.57 \ \text{m/s} \hat{j} + 4.0 \text{m/s} \hat{i} \]
Mass
Mass is the amount of matter in an object and is a scalar quantity (it only has magnitude, not direction). In the problem:
  • Particle 1 has a mass of \( 3.0 \ \text{kg} \).
  • Particle 2 has a mass of \( 4.0 \ \text{kg} \).
Adding these together gives the total mass of the system, which is \( 7.0 \ \text{kg} \). This total mass helped us find the velocity of the center of mass.
Vector Magnitude
Vector magnitude is a measure of the length or size of a vector. It's calculated using the Pythagorean theorem for two-dimensional vectors. If a vector \( \vec{v} \) has components \( v_x \) and \( v_y \), the magnitude \( |\vec{v}| \) is: \[ |\vec{v}| = \sqrt{v_x^2 + v_y^2} \] In our exercise, the center of mass velocity components are:
  • \( v_{cmx} = 4.0 \ \text{m/s} \)
  • \( v_{cmy} = -2.57 \ \text{m/s} \)
Plugging these values into the formula: \[ |\vec{v}_{cm}| = \sqrt{4.0^2 + (-2.57)^2} \approx \sqrt{22.60} \ \text{m/s} \approx 4.75 \ \text{m/s} \] This gives the speed of the center of mass.

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

An old Chrysler with mass \(2400 \mathrm{~kg}\) is moving along a straight stretch of road at \(80 \mathrm{~km} / \mathrm{h}\). It is followed by a Ford with mass \(1600 \mathrm{~kg}\) moving at 60 \(\mathrm{km} / \mathrm{h} .\) How fast is the center of mass of the two cars moving?

A big olive \((m=0.50 \mathrm{~kg})\) lies at the origin and a big Brazil nut \((M=1.5 \mathrm{~kg})\) lies at the point \((1.0,2.0) \mathrm{m}\) in an \(x y\) plane. At \(t_{1}=0\), a force \(\vec{F}_{o}=(2.0 \mathrm{~N}) \hat{\mathrm{i}}+(3.0 \mathrm{~N}) \hat{\mathrm{j}}\) begins to act on the olive, and a force \(\vec{F}_{n}=(-3.0 \mathrm{~N}) \hat{\mathrm{i}}+(-2.0 \mathrm{~N}) \hat{\mathrm{j}}\) begins to act on the nut. In unit-vector notation, what is the displacement of the center of mass of the olive-nut system at \(t_{2}=4.0 \mathrm{~s}\), with respect to its position at \(t_{1}=0\) ?

At \(t_{1}=0\), a \(1.0 \mathrm{~kg}\) jelly jar is projected vertically upward from the base of a 50 -m-tall building with an initial velocity of \(40 \mathrm{~m} / \mathrm{s}\). At the same instant and directly overhead, a \(2.0 \mathrm{~kg}\) peanut butter jar is dropped from rest from the top of the building. How far above ground level is the center of mass of the two-jar system at \(t_{2}=3.0 \mathrm{~s}\) ?

A stone is dropped at \(t_{1}=0 .\) A second stone, with twice the mass of the first, is dropped from the same point at \(t_{2}=100 \mathrm{~ms}\). (a) How far below the release point is the center of mass of the two stones at \(t_{3}=300 \mathrm{~ms}\) ? (Neither stone has yet reached the ground.) (b) How fast is the center of mass of the twostone system moving at that time?

A \(4.0 \mathrm{~kg}\) particle-like object is located at \(x=0, y=2.0 \mathrm{~m} ;\) a \(3.0 \mathrm{~kg}\) particle-like object is located at \(x=\) \(3.0 \mathrm{~m}, y=1.0 \mathrm{~m}\). At what (a) \(x\) and (b) \(y\) coordinates must a \(2.0 \mathrm{~kg}\) particle-like object be placed for the center of mass of the threeparticle system to be located at the origin?
See all solutions

Recommended explanations on Physics Textbooks

Kinematics Physics

Read Explanation

Rotational Dynamics

Read Explanation

Medical Physics

Read Explanation

Particle Model of Matter

Read Explanation

Turning Points in Physics

Read Explanation

Oscillations

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.