Conservation of Linear momentum and Kinetic energy

AI Thread Summary
The discussion focuses on the conservation of linear momentum and kinetic energy in a system with two masses attached by a string, where one mass is removed at the equilibrium position. The initial kinetic energy is derived from both masses, but confusion arises when one mass is removed, leading to questions about the source of "extra" energy in the system. Participants clarify that conservation laws apply only to closed systems, and removing a mass alters the conditions, preventing the use of these laws. The importance of defining the scenario accurately is emphasized, as misunderstandings about the system's setup can lead to incorrect conclusions. Ultimately, the conversation highlights the necessity of maintaining a closed system to apply conservation principles effectively.
eonden
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Suppose that we are working with a horizontal string with two masses attached. There is no friction on the floor and we move those two attached masses away from the equilibrium position by an undefined distance.
Once the string with those two attached masses passes through the equilibrium possition, we remove one of the attached masses.
As there are no forces present in that moment (as we are in the equilibrium position), we should apply the conservation of the linear momentum such that:
(m1 + m2 ) * v_slow = m1 *v_fast (as we remove the m2 with no speed)
So, we move from an initial Kinetic energy = ((m1+m2)*v_slow^2)/2
To a final Kinetic energy = (m1*v_fast^2)/2 from which we can derive using the conservation of the linear momentum --> ((m1+m2)^2 * v_slow^2)/(2*m1) and as ((m1+m2)^2)/m1 > (m1+m2), we have essentialy given energy to the system.
Can someone explain me where does this "extra" energy come from?
Thanks
 
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eonden said:
Suppose that we are working with a horizontal string with two masses attached. There is no friction on the floor and we move those two attached masses away from the equilibrium position by an undefined distance.
What do you mean by "equilibrium position"? Please define the scenario you have in mind more completely.
 
Doc Al said:
What do you mean by "equilibrium position"? Please define the scenario you have in mind more completely.

https://ecourses.ou.edu/ebook/dynamics/ch10/sec101/media/d0121.gif

We start from an original position and move the masses away from the origin to the maximum amplitude of the Simple Harmonic Motion. Once we get to that position, we let it contract by the spring force.
Sorry for not including a simple diagram with the doubt.
 
If I understand you correctly, you want to remove part of the moving mass from the system at some point?
If so, then you can't use conservation laws - these work only for close systems.

In other words, if you magically remove one of the masses, the other one will not change velocity.
 
eonden said:
https://ecourses.ou.edu/ebook/dynamics/ch10/sec101/media/d0121.gif

We start from an original position and move the masses away from the origin to the maximum amplitude of the Simple Harmonic Motion. Once we get to that position, we let it contract by the spring force.
Sorry for not including a simple diagram with the doubt.
I see one mass attached to a wall by a spring. Your first post mentions two masses attached by a string. Are you changing the scenario?

You pull the mass from the equilibrium position and let it go. So what? (I don't see what this has to do with conservation of momentum.)

Please restate your question with reference to the diagram.
 
Bandersnatch said:
If I understand you correctly, you want to remove part of the moving mass from the system at some point?
If so, then you can't use conservation laws - these work only for close systems.

In other words, if you magically remove one of the masses, the other one will not change velocity.

Thanks, completely forgot that when talking about closed systems no matter can be exchanged.
Sorry for the stupid question.
 

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