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GR8677 #20
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Problem
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Special Relativity }Rest Energy
If one remembers the formulae from special relativity, arithmetic would be the hardest part of the problem.
The problem is to solve  , where  and  .
The rest-mass of both the kaon and the proton are given in the problem. Thus, the equation reduces to  .
Now, because the range of velocities vary significantly between  and  , one can't directly approximate that as 2. Boo. So, long-division by hand yields approximately  .
The author of this site prefers to use fractions instead of decimals. Thus ^2= 1+\frac{81}{100} +\frac{180}{100} =\frac{262}{100}) . Express  in terms of  to get  .
 is about 64, so the velocity has to be greater than 0.8c. The only choice left is (E).
If one has the time, one might want to memorize the following:
&=&1.005\\
\gamma(\beta=0.25)&=&1.033\\
\gamma(\beta=0.5)&=&1.155\\
\gamma(\beta=0.75)&=&1.51\\
\gamma(\beta=0.9)&=&2.29<br />
\end{eqnarray})
, or perhaps a more elaborate list of  correlations.
If one knew that before-hand, then one would immediately arrive at choice (E).
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Alternate Solutions |
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Comments |
lamejiaa
2009-08-30 09:09:49 | The result is good but (19/10)(19/10) = 361/100 |  | jmason86
2009-08-16 16:22:01 | If you DO use  then you get roughly  . Not bad, I say. If the listed velocities were closer together, we might have a problem but they are pretty well spaced. Since it takes about 20 seconds to solve through with , I suggest using this, and if the answer came out too far from those listed, to go back and solve for 1.x. If you did have to do this, at least you already solved for v and you could just plug back in without deriving it again.
wittensdog
2009-09-25 17:26:02 |
I strongly agree - whenever you see spaced out answers, that's always your clue to approximate.
One thing I always remember is that when gamma is 2, the speed is sqrt(3)/2 , which is something like 0.866 or something like that. A value of 2 for gamma comes up in a lot of places, such as when rest energy and kinetic energy are equal, so it's a good thing to remember. Seeing that the ratio is almost two, that the answers are well spaced, and that there's something in the ball park of 86% of c, you should immediately go for it.
Seriously, NEVER do math if you see spaced out answers. One big example I can think of is the drift velocity problem, where each of the answers were spaced out by like 7 orders of magnitude. There was a factor of 1.6(pi) or something like that, but that's certainly not enough to swing 7 OM! Of course it's not quite as dramatic here, but still, no one wants to do long division on the physics GRE.
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| | nz_gre
2007-09-24 18:12:08 | why are we neglecting the fact that
^{2} + m_{0}c^{2})
And then finding the momentum velocity?
Surely, if the total kaon energy is equal to the proton rest mass then
938 = (pc) + (494) and we go from there?
bkardon
2007-10-05 13:31:04 |
The expression for  is in fact equivalent to the expresssion
as follows (here  is relativistic mass,  is rest mass)
^2)})
^2) = m^2_0 c^4)
^2 = m^2_0 c^4)
Here I will use 
QED
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| | leftynm
2006-10-30 17:05:27 | This is much simpler than that.
Once you have gamma*m_k = m_p, plug in gamma = [1-(v/c)^2)]^-1, and solve for v/c. You get v/c = sqrt(1 - (m_k/m_p)^2). If you allow m_k/m_p = 1/2, which it just about is, then you get v/c = sqrt(3)/2. From seeing this so much in trig, I know sqrt(3)/2 = 0.866. Then v/c is about 0.866c. So choice E is very close.
blah22
2008-03-22 13:10:35 |
How is that not, basically, what she did? Just in more detail.
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| | radicaltyro
2006-10-21 12:59:15 | 1.9^2 = 3.61 = 361/100, not 262/100 |  |
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