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GR9677 #48
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Alternate Solutions |
Mexicana
2007-10-03 18:06:00 | The second easiest way to solve this, after 'banana's solution' (which is reeeally nice btw!) is just to simply solve for  , getting ^2)}} ) and then just do the ratio  , which is reeally easy to calculate since the lifetime is given in units of  and  .... Hence, the nice ratio of  exact. |  |
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Comments |
shafatmubin
2009-11-03 13:55:26 | For the physics GRE, consider memorising the gamma-v relativistic table:
v = 0.999c -> gamma ~23
v = 0.995c -> gamma ~10
v = 0.99c -> gamma~7
v = 0.98c -> gamma~6
v = 0.97c -> gamma~5
v = 0.96c -> gamma~4
v = 0.95c -> gamma~3
v = 0.9c -> gamma~2.2
v = 0.85 -> gamma~2
v = 0.8c -> gamma = 5/3=1.67
I spent a while on Mathematica to figure these out. But it was worth the effort - I could guess the answer by looking at the gamma required (gamma=7/3 -> v~0.9 i.e v=Sqrt(4/5) )
jack238
2010-04-04 10:08:06 |
How could you quickly tell that gamma is 7/3?
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|  | ams379
2009-10-07 10:46:45 | I found an easier way to do this problem.
Just use L=L_o/
where L=vt=v( )= 
then just plug in the values:
(2/ ( )()= 15 \sqrt{5}/c^2})
15/=15/ | | ams379
2009-10-07 10:29:28 | I found an easier way to do this problem. Just use the length contraction equation. L=Lo Gamma
Where L=vt =(2.5x10^-8)v=15
and just plug in the numbers.
(2.5x10^-8)(2/(3x10^8)=15
15/=15/ | | shartacus
2008-08-06 12:52:58 | I think we can save ourselves some hairy arithmetic by using a little logic: Choices (D) and (E) are obviously not correct and can be eliminated.
Assume for the moment that the mesons are traveling at about c. Then, ignoring relativity for the moment, the time for the beam to reach the detector is 
Notice that this is  . This means that , the relativistic factor, must be at least 2, since time in the frame of the mesons must be slowed by at least a factor of two for half of them to reach the detector. Gamma stays close to 1 until v is about .8c or .9c, where it increases asymptotically. Thus, choices (A) and (B) are too low. Only (C) remains.
In a nutshell, relativistic effects are large, so pick the choice that is closest to c without being greater than or equal to c.
Hope you guys followed my logic. This is the messed up way I think of things. | | Mexicana
2007-10-03 18:06:00 | The second easiest way to solve this, after 'banana's solution' (which is reeeally nice btw!) is just to simply solve for , getting and then just do the ratio , which is reeally easy to calculate since the lifetime is given in units of and .... Hence, the nice ratio of exact.
HaveSpaceSuit
2008-10-16 21:13:37 |
Mexicana, I think that there is a unit discrepancy in your solution. It looks like it is unit-less. I think that factor of c in the numerator is not supposed to be there. Correct me if I'm wrong.
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| | mm
2007-09-22 23:16:31 | Plug each choice into
^2}})
, and you can reach the answer.
| | banana
2005-12-07 18:17:12 | Much easier: The interval is the same in all frames. Thus: 
Gaffer
2007-10-27 09:47:59 |
Great soln! The math is easier if you think of it like:
The RHside
^2 +15^2 = 15^2(1/4+1) = t^2)
Then  (2/sqrt{5}) )
Basically, don't multiply big numbers unless you have too. This saves time.
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jw111
2008-11-05 12:53:37 |
Similar way(c=1)
the time for travel is
t = 15/v
and t is also the half life seen on the ground
t = 7.5r
so 15/v=7.5r -> 2/v = r
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| | mg
2005-11-27 16:05:44 | The arithmetic simplifies considerably if one realizes that  =2c. Then one has v=2c^2}) | |
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