• P14 A group of particles of total mass 35 kg has a total kinetic energy of 340 J. The kinetic energy relative to the center of mass is 85 J. What is the speed of the center of mass?
• P14 A group of particles of total mass 35 kg has a total kinetic energy of 340 J. The kinetic energy relative to the center of mass is 85 J. What is the speed of the center of mass?
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Problem P14:**
A group of particles of total mass \(35 \, \text{kg}\) has a total kinetic energy of \(340 \, \text{J}\). The kinetic energy relative to the center of mass is \(85 \, \text{J}\). What is the speed of the center of mass?
**Explanation:**
To find the speed of the center of mass, use the formula for kinetic energy \((KE)\):
\[ KE = \frac{1}{2}mv^2 \]
* Let the total kinetic energy of the system be \( KE_{\text{total}} = 340 \, \text{J} \).
* The kinetic energy relative to the center of mass is \( KE_{\text{relative}} = 85 \, \text{J} \).
The kinetic energy of the center of mass \((KE_{\text{CM}})\) is:
\[ KE_{\text{CM}} = KE_{\text{total}} - KE_{\text{relative}} \]
\[ KE_{\text{CM}} = 340 \, \text{J} - 85 \, \text{J} = 255 \, \text{J} \]
Using the kinetic energy formula:
\[ 255 = \frac{1}{2}(35)v^2 \]
Solve for \( v \) (speed of the center of mass):
\[ 255 = 17.5v^2 \]
\[ v^2 = \frac{255}{17.5} \]
\[ v^2 = 14.57 \]
\[ v = \sqrt{14.57} \]
\[ v \approx 3.82 \, \text{m/s} \]
Thus, the speed of the center of mass is approximately \(3.82 \, \text{m/s}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9e3774d6-7392-46bd-8e3a-4f3b8babb74d%2Fa1a3927f-1714-4e79-8409-843fa763ddd6%2F09uco57_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem P14:**
A group of particles of total mass \(35 \, \text{kg}\) has a total kinetic energy of \(340 \, \text{J}\). The kinetic energy relative to the center of mass is \(85 \, \text{J}\). What is the speed of the center of mass?
**Explanation:**
To find the speed of the center of mass, use the formula for kinetic energy \((KE)\):
\[ KE = \frac{1}{2}mv^2 \]
* Let the total kinetic energy of the system be \( KE_{\text{total}} = 340 \, \text{J} \).
* The kinetic energy relative to the center of mass is \( KE_{\text{relative}} = 85 \, \text{J} \).
The kinetic energy of the center of mass \((KE_{\text{CM}})\) is:
\[ KE_{\text{CM}} = KE_{\text{total}} - KE_{\text{relative}} \]
\[ KE_{\text{CM}} = 340 \, \text{J} - 85 \, \text{J} = 255 \, \text{J} \]
Using the kinetic energy formula:
\[ 255 = \frac{1}{2}(35)v^2 \]
Solve for \( v \) (speed of the center of mass):
\[ 255 = 17.5v^2 \]
\[ v^2 = \frac{255}{17.5} \]
\[ v^2 = 14.57 \]
\[ v = \sqrt{14.57} \]
\[ v \approx 3.82 \, \text{m/s} \]
Thus, the speed of the center of mass is approximately \(3.82 \, \text{m/s}\).
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