Kinetic Enrgey according to another viewer

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SUMMARY

The discussion focuses on calculating the kinetic energy T' of one particle as measured by another particle in a two-particle system, each with mass M. The user employs relativistic equations, specifically T=mc^2(r-1) and the velocity transformation formula u1=(u1'+u3)/(1+u1'u3), to derive T' but arrives at an incorrect result of T'=4/3M. The calculations involve the assumption of c=1 and the use of MeV units, indicating a context within relativistic physics.

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  • Understanding of relativistic kinetic energy equations
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2 particles with mass M and Kinetic Engergy T moving toward each other.
what is the kinetic energy T' as it measured by the other particle
given data: T=...
m=...

thanks
 
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Why you proposed this question?
 
blue_sky said:
Why you proposed this question?
training for test

that's what i did, could anyone tell what's my prob here:
for first particle
T=mc^2(r-1)
since we're working with Mev , c=1
T=m(r-1)
r=T/M+1
therefore
(1) V^2=T/(T+M)
(V is speed of particle one according to "the lab")

therefore the speed of particle2 is -V (simetry)
now we use additional formula
u1=(u1'+u3)/(1+u1'u3)
whille u1 is velocity of particle1 according to the lab (u1=V)
u1' is velocity of particle 1 according to particle 2
u3 is the velocity between lab to particle2 (u3=-V)
therefore
u1'=2V/(V^2+1)

now i look on kinetic energy of particle 1 according to particle 2:
T'=m(r-1)
while r=1/sqrt(1-u1'^2)
i assigned u1' & V and got
T'=4/3M
this isn't correct..
 

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