Question

Consider a system consisting of three particles: m1 = 4 kg, vector v1 = < 10, -5, 15 > m/s, m2 = 9 kg, vector v2 = < -15, 5, -3 > m/s, m3 = 3 kg, vector v3 = < -24, 36, 18 > m/s.
a) what is the total momentum of this system?
b) what is the velocity of the center of mass of this system?
c) what is the total kinetic energy of this system?
d) what is the translational kinetic energy of this system?
e) what is the kinetic energy of this system relative to the center of mass?
one way to calculate krel is to calculate the velocity of each particle relative to the center of mass, by subtracting the center-of-mass velocity from the particle's actual velocity to get the particle's velocity relative to the center of mass, then calculating the corresponding kinetic energy, then adding up the three relative kinetic energies. however, there is a much simpler way to determine the specified quantity, without having to do all those calculations; think about what you know about the relationships among the various kinds of kinetic energy in a multiparticle system.

Answer 1

The answers to your questions are as listed below

A) The total momentum of the system is :  < - 167, 133, 87 >   kg m/s

B) Velocity of the center of mass of this system : < - 10.44 , 8.31, 5.44 > m/s

C) Total kinetic energy of the system : 5159.5 J  

D) Translational kinetic energy of the system :  1660.84 J

E) kinetic energy relative to the center of mass : 3498.66 J

Given data :

m₁ = 4 kg ,   V₁ = < 10, -5, 15 > m/s

m₂ = 9 kg,   V₂ = < -15, 5, -3 > m/s

m₃ = 3 kg    V₃ = < -24, 36, 18 > m/s.

A) Determine the total momentum of the system

Total momentum = Σ mv

= < 4*10 + 9*(-15) + 3*(-24), 4*(-5) + 9*5 + 3*36, 4*15 + 9 (-3) + 3*18 >

= < - 167, 133, 87 >   kg m/s

B) Velocity at the center of mass of this system

Velocity = < - 167 / 16 , 133/ 16, 87/ 16 >

              =  < - 10.44 , 8.31, 5.44 > m/s

C) Determine the total kinetic energy of the system

Total Kinetic energy = 1/2 m₁v₁² + 1/2 m₂v₂²  + 1/2 m₃v₃²  --- ( 1 )

where : v² = v₁² + v₂² + v₃²

therefore

Total kinetic energy = 700 + 1165.5 + 3294

                                  = 5159.5 J  

D) Determine the translational kinetic energy of the system

Translational kinetic energy ( Kcm ) = 1/2 ( m₁ + m₂ + m₃ ) * Velocity²

                                                           = 1/2 ( 16 ) * (  - 10.44² + 8.31² + 5.44² )

Hence translational kinetic energy = 1660.84 J

E) The Kinetic energy of the system relative to the center of mass

Krel = Ktotal - Kcm

       = 5159.5 - 1660.84

       = 3498.66 J

Hence we can conclude that the answers to your questions are as listed above.

Learn more about Translational kinetic energy :brainly.com/question/25959744