Kinetic Energy in Reference Frames

[aclark609]

K'_sys=1/2M_totalV_cm²+1/2[itex]\mu[/itex]V_reli²

I understand this equation represents the total kinetic energy in a reference frame. What I'm not getting out of this is the overall concept. I understand that the first part of the equation is supposed to represent the total kinetic energy required to conserve momentum in the system, but I don't know how.

In other words, how is taking the kinetic energy of the total mass of the system using the center of mass velocity equal to the energy needed to conserve momentum. Why would you use the total mass and the center of mass velocity? Perhaps I need to understand the concept of center of mass velocity a little better and it's purpose.

Same for the second part. Why would you use mu to find the kinetic energy that can be converted? Is it not possible to do it any other way? Perhaps this is the easiest?

 

[Simon Bridge]

K'_sys=1/2M_totalV_cm²+1/2[itex]\mu[/itex]V_reli²

I understand this equation represents the total kinetic energy in a reference frame. What I'm not getting out of this is the overall concept. I understand that the first part of the equation is supposed to represent the total kinetic energy required to conserve momentum in the system, but I don't know how.

How would you do it?

In other words, how is taking the kinetic energy of the total mass of the system using the center of mass velocity equal to the energy needed to conserve momentum. Why would you use the total mass and the center of mass velocity? Perhaps I need to understand the concept of center of mass velocity a little better and it's purpose.

Yes - it is not "center of mass velocity" but "the velocity of the center of mass" and it is frame-dependent.

Same for the second part. Why would you use mu to find the kinetic energy that can be converted? Is it not possible to do it any other way? Perhaps this is the easiest?

It is possible to do it in many different ways. This way distinguishes between different forms the energy takes so it makes the math easier.