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GRE Relativity Problem

  • #1
784
11

Homework Statement


http://grephysics.net/ans/8677/20

So to do this, I solved for the total rest energy of both particles.
The rest energy of Kaon + K = Rest energy of Proton
Rest energy of Proton - Rest Energy of Kaon = K
K = P^2/2m
((Rest Energy of Proton - Rest Energy of Kaon)*(2*Mass of Kaon))^1/2 = P

P = mv/(sqrt(1-v^2/c^2)

If I do this and do all the algebra correct and solve for v, will this method give me the correct answer? I got the wrong answer but I suck at numbers X_x.

I realize after looking at the answers on the website this is a poor way to do this problem with GRE time constraints, I just want to know if this thought process is flawed or not.
 

Answers and Replies

  • #2
19,921
4,096
If the rest energy of a particle is 494 MeV, and its total energy is 938 MeV, what is γ? Since [itex]γ=\frac{1}{\sqrt{1-\beta^2}}[/itex], square both sides and solve for β2 in terms of γ2. What is the value of β?
 
  • #3
784
11
Yeah, I see that this is the best way to do it. I was just wondering if my way works (even though it would take way longer).
 

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