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At one instant, the center of mass of a system of two

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 56E Chapter 8

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 56E

At one instant, the center of mass of a system of two particles is located on the x-axis at x = 2.0 m and has a velocity of One of the particles is at the origin. The other particle has a mass of 0.10 kg and is at rest on the x-axis at x = 8.0 m. (a) What is the mass of the particle at the origin? (b) Calculate the total momentum of this system. (c) What is the velocity of the particle at the origin?

Step-by-Step Solution:

Solution 56E Step 1 : It is given Center of velocity v = 5.0 m/s c Position of center of mass x =c2.0 m Velocity of particle 1 v =1 Mass of particle 1 m =1 Velocity of particle 2 v = 0 2 Mass of particle m =20.10 kg By the law of conservation of momentum , momentum at the center of mass has to be equal to the sum of momentum of each particle That is m c c m v + m v 1 1 2 2 But m =cm + m1and v2= 5m/s c Substituting values we get (m 1 m ) 2 5m/s = m v + 1 110 kg × 0) (m + 0.10 kg) × 5m/s = m v 1 1 1 (5m 1 0.50 kg m/s) = m v 1 1 Consider particles m × (x x ) = m × (x x ) c 1 c 1 2 2 Substituting values we get m 1 (2.0 m 0 ) = 0.10 kg × (8.0 m 2.0 m ) m 1 (2.0 m 0 ) = 0.10 kg × (6.0 m ) m = 0.60 /2.0 m 1 m 1 0.3 kg Hence the particle 1 is found to be 0.3 kg

Step 2 of 2

Chapter 8, Problem 56E is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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At one instant, the center of mass of a system of two