What is the total kinetic energy of two protons in different reference frames?

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SUMMARY

The total kinetic energy of two protons, each moving at 0.5c in reference frame S', is calculated to be 290 MeV. The calculation involves using the relativistic kinetic energy formula, which accounts for the Lorentz factor (γ). The initial attempt using the classical kinetic energy formula (1/2 mv^2) yielded an incorrect result of 270 MeV, highlighting the necessity of applying relativistic principles for accurate results in high-speed scenarios.

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PsychonautQQ
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Homework Statement


In the reference frame S', two protons, each moving at .5c, approach each other head on. Calculate the total kinetic energy of the two protons in frame S', and calculate the total kinetic energy of the two protons as seen in reference frame S which is moving with one of the protons.






The Attempt at a Solution


So for the first part apparently the answer I'm looking for is 290MeV

v = .5c so γ = 1.1547

using 1/2 mv^2
m = 1.6726E-27
v = .5(3E8)

and then multiplying this times two for two protons and 1.15 from gamma, then converting that to eV by 1J = 6.24E18 eV I get 270MeV.. back of the book says 290 MeV.. what am I doing wrong ;-(?
 
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PsychonautQQ said:
using 1/2 mv^2

You'll need to use the relativistic formula for kinetic energy.
 
However we will not provide it. lol
 

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