Speed of Proton at 750keV Kinetic Energy

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SUMMARY

The speed of a proton accelerated to a kinetic energy of 750 keV can be calculated using the relativistic kinetic energy equation K=(gamma-1)mc^2. The user converted the kinetic energy to joules, resulting in K = 1.202E-13 J. However, the calculation led to an undefined result when attempting to isolate the speed variable v, indicating a miscalculation in the manipulation of the gamma factor. The discussion emphasizes the importance of maintaining symbolic representation throughout derivations to facilitate error tracing.

PREREQUISITES
  • Understanding of special relativity concepts
  • Familiarity with the kinetic energy equation K=(gamma-1)mc^2
  • Knowledge of the mass-energy equivalence E0=mc^2
  • Ability to manipulate algebraic expressions involving square roots
NEXT STEPS
  • Review the derivation of the gamma factor in special relativity
  • Practice solving problems involving relativistic kinetic energy
  • Learn about dimensional analysis to verify equations
  • Explore the implications of relativistic speeds on particle physics
USEFUL FOR

Physics students, educators, and anyone studying special relativity or particle acceleration will benefit from this discussion.

dancergirlie
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Homework Statement



What is the speed of a proton after being accelerated to a kinetic energy of 750keV?

Homework Equations



E=K+E0
E0=mc^2
K=(gamma-1)mc^2
gamma=1/(sqroot(1-v^2/c^2))

The Attempt at a Solution



Alright, so to find the speed, I used the equation:

K=(gamma-1)mc^2
which i simplified to:
1/(gamma-1)=mc^2/K

I converted K into joules which is equal to 1.202E-13 J
and what I ended up getting was that
sqroot(1-v^2/c^2) - 1=1250.42
but if i continue on with that I will need to take the square root of a negative number because i will bring v^2/c^2 to the right side and then I would have to do 1-1250.42 and take the square root of that which is undefined. Any help would be great!
 
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Do yourself a favor and use symbols throughout your derivation. Plug in at the very end. It would be easier to trace your error(s) if you did. As it is, I cannot figure out what that 1250.42 means or where it came from. If you have gotten to the stage of special relativity you should be able to handle expressions symbolically. Doing so enables you to find your mistakes much more easily through dimensional analysis.
 

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