GR8677 #20


Problem



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 restmass 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, longdivision by hand yields approximately .
The author of this site prefers to use fractions instead of decimals. Thus . 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:
, or perhaps a more elaborate list of correlations.
If one knew that beforehand, then one would immediately arrive at choice (E).


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Comments 
ernest21 20190810 03:09:34  Its a good approach to solve problems @ home,interesting. wolfgang krauser   fredluis 20190809 04:13:03  Both get me closer to 2 compared to the other given answers, but nothing I can think of can give me an exact answer of two. As for the answer explanation here, I am just fairly uncomfortable with the idea that adding two different units together can get the correct answer. tree removal   joshuaprice153 20190808 07:33:21  Pretty insightful post. Never thought that it was this simple after all. I had spent a good deal of my time looking for someone to explain this subject clearly and youâ€™re the only one that ever did that. Keep it up. house painting   JoshWasHere 20140820 08:12:31  In working this out, I approximated = =1.9 which makes it easier to work with. Then in working out the algebra, I approximated = which made taking the square root extremely simple. is nearly .85 c so without too many assumptions or generous approximations, I got nearly the right answer in a short amount of time.   dak213 20140630 17:58:42  I think they forgot to add the 1 in the equation. I got then solving for I got exactly 0.85c E
dak213 20140630 18:00:13 
Ah, I see this issue has been addressed. I should have read the comments at the bottom.

  deniskrasnov 20130119 00:08:42  i have a suspicion that ETS only wants you to remember Pythagorean triple 3:4:5
wich translates into:
beta = 0.8 => gamma = 5/3 = 1.66...
beta = 0.6 => gamma = 5/4 = 1.25
and thus, dealing with similar problem one should first check if one of these is enough to identify the correct answer quickly
the first one is sufficient in this case
gamma = 938/494 > 850/500 > 1.7 > 1.66.. => beta > 0.8   oliTUTilo 20121101 16:43:56  If we algebraically solve for v (or ) instead of , as Yosun does, it seems easier to approximate the answer.
is approximately , so is approximately , which is 1.7 something divided by 2, or approximately 0.85. v is , so badabingbadaboom.
In general, I'd algebraically solve for the variable in question first, gaining the greatest transparency, but redmomatt's method is even a bit faster here.
  redmomatt 20111014 08:59:20  Why can't we do this....?
(E)   checkyoself 20111005 10:46:45  Approximating is fine. You end up with which you know must be greater than 0.75c so the answer must be E.   archard 20100525 18:33:08  Instead of doing the long division by hand, you could just say that gamma is less than 2. Solving the inequality you get that v > . E is the only answer that satisfies this.
Crand0r 20101112 08:49:13 
Actually, you should get that . All of the answers satisfy this, but you also know that , which leaves only E.

  lamejiaa 20090830 09:09:49  The result is good but (19/10)(19/10) = 361/100
walczyk 20110406 17:37:34 
^^this is true. you can check this quickly by . After a couple steps you will end up with . Do a little math and you'll get Phew, all these relativistic problems really have you scratching away with your pencil, but the most time consuming part of this problem will be the long division, good luckk.

  jmason86 20090816 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 20090925 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.

  nz_gre 20070924 18:12:08  why are we neglecting the fact that
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 20071005 13:31:04 
The expression for is in fact equivalent to the expresssion
as follows (here is relativistic mass, is rest mass)
Here I will use
QED

FutureDrSteve 20111029 13:44:19 
Another consideration with your approach is that you still need to calculate because once you get down to
in order to solve for v, you need to use:
p =
Since you need the other approach to solve for , you might as well stick with that method to save time and effort

FutureDrSteve 20111029 13:56:04 
Correction: You don't NEED to solve for using the other method. You can use the definition of to solve for v, but using the simpler approach will reduce the risk of making a careless mistake, plus you will save lots of time, which, as far as I'm concerned, is the cardinal rule of the PGRE.

ryanjsfx 20141020 11:32:20 
I thought using wasn't so bad. Observe (with c set to 1):rnrn
is about 1.7 and 1.7 divided by 2 is about 0.84
Done!

  leftynm 20061030 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 20080322 13:10:35 
How is that not, basically, what she did? Just in more detail.

  radicaltyro 20061021 12:59:15  1.9^2 = 3.61 = 361/100, not 262/100
achca 20120626 03:55:13 
absolutely right!

 

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