Momentum and conservation of energy

AI Thread Summary
The discussion centers on the relationships between momentum and kinetic energy during collisions, specifically addressing a paradox when applying conservation laws. It highlights that while momentum conservation can be expressed as {v'_1}{m'} + {v_1}{m}= {v'_2}{m'} + {v_2}{m}, the corresponding kinetic energy equation does not hold true under certain conditions. Participants explore the implications of selecting final velocities in elastic collisions, emphasizing that both conservation laws must be satisfied simultaneously. The conversation concludes that the equations derived from momentum and energy conservation are interrelated, leading to a consistent solution when both are considered. The complexities of these relationships illustrate the nuances of collision physics.
Werg22
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I two object collide, and there is no loss of kinetic energy, the following relationship is true:

{v'_1}{m'} + {v_1}{m}= {v'_2}{m'} + {v_2}{m}

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

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

they also share this relationship

{v'_1^{2}}{m'} + {v_1^{2}}{m} = {v'_2^{2}}{m'} + {v_2^{2}}{m}

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:

{v'_1}{m'} + {v_1}{m}= {v'_2}{m'} + {v_2}{m}

{v'_1}{m'} - {v'_2}{m'} = {v_2}{m} - {v_1}{m}m'(v'_1 - v'_2) = m(v_2 - v_1)Now with this relationship{v'_1^{2}}{m'} + {v_1^{2}}{m} = {v'_2^{2}}{m'} + {v_2^{2}}{m}m'(v'_1^{2} - v'_2^{2}) = m(v_2^{2} - v_1^{2})Let m' = 4, m = 4, v'_1 = 4, v_1 = -34(4 - v'_2) = 4(v_2 + 3)

4 - v'_2 = v_2 + 4

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

m'(v'_1^{2} - v'_2^{2}) = m(v_2^{2} - v_1^{2})

4(4^{2} - (-2)^{2}) = 4(3^{2} - ((-3)^{2})

12 = 0

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

{v'_1}{m'} + {v_1}{m}= {v'_2}{m'} + {v_2}{m}

{v'_1}{m'} - {v'_2}{m'} = {v_2}{m} - {v_1}{m}


m'(v'_1 - v'_2) = m(v_2 - v_1)


Now with this relationship


{v'_1^{2}}{m'} + {v_1^{2}}{m} = {v'_2^{2}}{m'} + {v_2^{2}}{m}


m'(v'_1^{2} - v'_2^{2}) = m(v_2^{2} - v_1^{2})


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


4(4 - v'_2) = 4(v_2 + 3)

4 - v'_2 = v_2 + 4

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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