- #1
luckis11
- 272
- 2
https://en.wikipedia.org/wiki/Elastic_collision
μα+mβ=μx+my,
μα^2+mβ^2=μx^2+my^2
I want x in relation of all variables except y, therefore I need to replace-eliminate y:
μα+mβ=μx+my =>y=(μα+mβ-μx)/m
μα^2+mβ^2=μx^2+my^2=>y=((μα^2+mβ^2-μx^2)/m)^0.5
and it is eliminated if I equate these two parts of the two equalities with which y is equal to:
((μα+mβ-μx)/m)^2=(μα^2+mβ^2-μx^2)/m=>
(μα+mβ-μx)^2/m=μα^2+mβ^2-μx^2,
A=μα, Β=mβ, C=μx,
(A+B-C)^2=Α^2+ΑΒ-ΑC+AB+B^2-BC-AC-BC+C^2= Α^2+2ΑΒ-2ΑC+B^2-2BC+C^2=>
μ^2α^2+2μαmβ-2μαμx+m^2β^2-2mβμx+μ^2x^2=mμα^2+m^2β^2-mμx^2
so far wolframalfa answers x=(-m α + 2 m β + α μ)/(m + μ) which is the solution according to theory. But the last relation is equivalent with the trionym:
μ^2x^2+mμx^2-2μαμx-2mβμx+μ^2α^2+2μαmβ-mμα^2=0
A=μ^2+mμ, Β= -2αμ^2-2mβμ, C=μ^2α^2+2μαmβ-mμα^2,
Αx^2+Bx+C=0=>x=(-B+-(B^2-4AC)^0.5)/(2A)=>
x=(-(-2αμ^2-2mβμ)+-((-2αμ^2-2mβμ)^2-4(μ^2+mμ)(μ^2α^2+2μαmβ-mμα^2))^0.5)/
(2(μ^2+mμ))
And the latter equation is what also wolframalfa answers now!
And not only this does not seem how it can be factorized, but replacing arithmetic values:
(-m α + 2 m β + α μ)/(m + μ), m=2, α=3, β=5, μ=7=>35/9
(2*3*7^2+2*5*7*2+((-2*3*7^2-2*5*7*2)^2-4(7^2+7*2)(3^2*7^2-3^2*7*2+2*3*5*7*2))^0.5)/(2(2^2+2*7))=245/18≠35/9
(2*3*7^2+2*5*7*2-((-2*3*7^2-2*5*7*2)^2-4(7^2+7*2)(3^2*7^2-3^2*7*2+2*3*5*7*2))^0.5)/(2(2^2+2*7))=21/2≠35/9
whereas:
(7^2)x^2+2*7*x^2-2*49*3x-2*2*5*7*x+49*3^2+2*7*3*2*5-2*7*3^2=0=>x=35/9
Where is the mistake?
μα+mβ=μx+my,
μα^2+mβ^2=μx^2+my^2
I want x in relation of all variables except y, therefore I need to replace-eliminate y:
μα+mβ=μx+my =>y=(μα+mβ-μx)/m
μα^2+mβ^2=μx^2+my^2=>y=((μα^2+mβ^2-μx^2)/m)^0.5
and it is eliminated if I equate these two parts of the two equalities with which y is equal to:
((μα+mβ-μx)/m)^2=(μα^2+mβ^2-μx^2)/m=>
(μα+mβ-μx)^2/m=μα^2+mβ^2-μx^2,
A=μα, Β=mβ, C=μx,
(A+B-C)^2=Α^2+ΑΒ-ΑC+AB+B^2-BC-AC-BC+C^2= Α^2+2ΑΒ-2ΑC+B^2-2BC+C^2=>
μ^2α^2+2μαmβ-2μαμx+m^2β^2-2mβμx+μ^2x^2=mμα^2+m^2β^2-mμx^2
so far wolframalfa answers x=(-m α + 2 m β + α μ)/(m + μ) which is the solution according to theory. But the last relation is equivalent with the trionym:
μ^2x^2+mμx^2-2μαμx-2mβμx+μ^2α^2+2μαmβ-mμα^2=0
A=μ^2+mμ, Β= -2αμ^2-2mβμ, C=μ^2α^2+2μαmβ-mμα^2,
Αx^2+Bx+C=0=>x=(-B+-(B^2-4AC)^0.5)/(2A)=>
x=(-(-2αμ^2-2mβμ)+-((-2αμ^2-2mβμ)^2-4(μ^2+mμ)(μ^2α^2+2μαmβ-mμα^2))^0.5)/
(2(μ^2+mμ))
And the latter equation is what also wolframalfa answers now!
And not only this does not seem how it can be factorized, but replacing arithmetic values:
(-m α + 2 m β + α μ)/(m + μ), m=2, α=3, β=5, μ=7=>35/9
(2*3*7^2+2*5*7*2+((-2*3*7^2-2*5*7*2)^2-4(7^2+7*2)(3^2*7^2-3^2*7*2+2*3*5*7*2))^0.5)/(2(2^2+2*7))=245/18≠35/9
(2*3*7^2+2*5*7*2-((-2*3*7^2-2*5*7*2)^2-4(7^2+7*2)(3^2*7^2-3^2*7*2+2*3*5*7*2))^0.5)/(2(2^2+2*7))=21/2≠35/9
whereas:
(7^2)x^2+2*7*x^2-2*49*3x-2*2*5*7*x+49*3^2+2*7*3*2*5-2*7*3^2=0=>x=35/9
Where is the mistake?