Momentum and conservation of energy

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Homework Help Overview

The discussion revolves around the concepts of momentum and conservation of energy in the context of elastic collisions between two objects. Participants are examining the relationships between momentum and kinetic energy, questioning the validity of certain equations derived from these principles.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationships between momentum and kinetic energy, questioning the validity of derived equations. Some express confusion over apparent contradictions in the conservation principles.

Discussion Status

The discussion is ongoing, with participants offering insights into the relationships between the equations of momentum and kinetic energy. Some guidance has been provided regarding the constraints of elastic collisions and the implications of selecting final velocities.

Contextual Notes

Participants are working under the assumption of elastic collisions, where both momentum and kinetic energy are conserved. There is an exploration of specific values for masses and velocities, which may not fully align with the principles discussed.

Werg22
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I two object collide, and there is no loss of kinetic energy, the following relationship is true:

[tex]{v'_1}{m'} + {v_1}{m}= {v'_2}{m'} + {v_2}{m}[/tex]

But also by conservation of energy,[tex]{v'_1^{2}}{m'}/2 + {v_1^{2}}{m}/2= {v'_1^{2}}{m'}/2 + {v_1^{2}}{m}/2[/tex][tex]{v'_1^{2}}{m'} + {v_1^{2}}{m} = {v'_2^{2}}{m'} + {v_2^{2}}{m}[/tex]

That means for any two mass, and velocities that share the relationship[tex]{v'_1}{m'} + {v_1}{m}= {v'_2}{m'} + {v_2}{m}[/tex]

they also share this relationship

[tex]{v'_1^{2}}{m'} + {v_1^{2}}{m} = {v'_2^{2}}{m'} + {v_2^{2}}{m}[/tex]

which is obviously not true. Why is there a paradox?
 
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Why do you think it is not true?
 
If we start with the following:

[tex]{v'_1}{m'} + {v_1}{m}= {v'_2}{m'} + {v_2}{m}[/tex]

[tex]{v'_1}{m'} - {v'_2}{m'} = {v_2}{m} - {v_1}{m}[/tex][tex]m'(v'_1 - v'_2) = m(v_2 - v_1)[/tex]Now with this relationship[tex]{v'_1^{2}}{m'} + {v_1^{2}}{m} = {v'_2^{2}}{m'} + {v_2^{2}}{m}[/tex][tex]m'(v'_1^{2} - v'_2^{2}) = m(v_2^{2} - v_1^{2})[/tex]Let m' = 4, m = 4, v'_1 = 4, v_1 = -3[tex]4(4 - v'_2) = 4(v_2 + 3)[/tex]

[tex]4 - v'_2 = v_2 + 4[/tex]

Let v'_2 be -2, so that v_2 = 3

[tex]m'(v'_1^{2} - v'_2^{2}) = m(v_2^{2} - v_1^{2})[/tex]

[tex]4(4^{2} - (-2)^{2}) = 4(3^{2} - ((-3)^{2})[/tex]

[tex]12 = 0[/tex]

So it is not true
 
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Werg22 said:
If we start with the following:

[tex]{v'_1}{m'} + {v_1}{m}= {v'_2}{m'} + {v_2}{m}[/tex]

[tex]{v'_1}{m'} - {v'_2}{m'} = {v_2}{m} - {v_1}{m}[/tex]


[tex]m'(v'_1 - v'_2) = m(v_2 - v_1)[/tex]


Now with this relationship


[tex]{v'_1^{2}}{m'} + {v_1^{2}}{m} = {v'_2^{2}}{m'} + {v_2^{2}}{m}[/tex]


[tex]m'(v'_1^{2} - v'_2^{2}) = m(v_2^{2} - v_1^{2})[/tex]


Let m' = 4, m = 4, v'_1 = 4, v_1 = -3


[tex]4(4 - v'_2) = 4(v_2 + 3)[/tex]

[tex]4 - v'_2 = v_2 + 4[/tex]

Let v'_2 be -2, so that v_2 = 3

No, you can't do that. If you have an elastic collision between two bodies, you can't select the final speed of either. Given your initial information:
both masses= 4, v'_1= 4, v_1= -3,
Then:
conservation of momentum: 4(4)+4(-3)= 4v'_2+ 4(v_2)
conservation of kinetic energy: 4(16)+ 4(9)= 4(v'_2)^2+ 4(v_2)^2.
Of course, you can just cancel all of the masses (4s):
v'_2+ v_2= 1 and v'_2^2+ v_2^2= 25.

Now, you have two equations in two unknowns.
 
Humm yes... they are consequent one of another so this makes more sense. Thank you!
 

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