GR9677 #48



Alternate Solutions 
tdl17 20190103 21:16:15  Since the particles only took one halflife, which is seconds, to travel 15m, it must be traveling close to the speed of light. \\\\r\\\\n\\\\r\\\\nAssume the speed of the particles is some fraction of the speed of light and let this fraction be , we now have . \\\\r\\\\n\\\\r\\\\nUsing , we can replace v with ac and from there, you can solve for a and that should get you to , which is choice C.   Mexicana 20071003 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.  

Comments 
tdl17 20190103 21:16:15  Since the particles only took one halflife, which is seconds, to travel 15m, it must be traveling close to the speed of light. \\\\r\\\\n\\\\r\\\\nAssume the speed of the particles is some fraction of the speed of light and let this fraction be , we now have . \\\\r\\\\n\\\\r\\\\nUsing , we can replace v with ac and from there, you can solve for a and that should get you to , which is choice C.   psitae 20161027 14:27:54  probably very close to the speed of light >   bcomnes 20111102 15:06:43  I found the solution confusing as written. This is my attempt to make it more clear to myself.rnrnGiven: rnDistance traveled 15m.rnHalf life of Pion is seconds.rnrnWhat we do:rnThe particles move relativistically, hinted at given the subject matter.rnrnSince the pion is moving, its clock runs slower than the lab clock. After traveling 15m, it has decayed less than another pion sample that is not moving, so it appears to have moved faster than the speed of light in order to traverse 15m and only decay 1 half life. The pion only decays for s since only one half of the pions show up. This is our proper time .rnrnDefine the speed of the pion in the lab frame. Notice that 15/2.5 = 6 and we get or rnrnWe want the speed of the pion that the pion actually feels (or in the pions frame), .rnrnRemembering our basic time dilation formula , all we need to do now is solve for v.rnrnPlug in: Skipping some algebra rnrnYou need to remember of course, but also you could remember where, is the speed of the particle that the rest frame sees and v is the actual speed of the moving particle as the particle experiences, that , or conversely define and remember that
bcomnes 20111102 15:08:31 
I found the solution confusing as written. This is my attempt to make it more clear to myself.
Given:
Distance traveled 15m.
Half life of Pion is seconds.
What we do:
The particles move relativistically, hinted at given the subject matter.
Since the pion is moving, its clock runs slower than the lab clock. After traveling 15m, it has decayed less than another pion sample that is not moving, so it appears to have moved faster than the speed of light in order to traverse 15m and only decay 1 half life. The pion only decays for s since only one half of the pions show up. This is our proper time .
Define the speed of the pion in the lab frame. Notice that 15/2.5 = 6 and we get or
We want the speed of the pion that the pion actually feels (or in the pions frame), .
Remembering our basic time dilation formula , all we need to do now is solve for v.
Plug in: Skipping some algebra
You need to remember of course, but also you could remember where, is the speed of the particle that the rest frame sees and v is the actual speed of the moving particle as the particle experiences, that , or conversely define and remember that

bcomnes 20111102 15:09:10 
Sorry to double post but the first post was totally garbled.

  CarlBrannen 20101007 14:50:48  The speeds are in multiples of c so use c=1. We have a messy half life time and a clean distance in the problem so work in meters. To convert the halflife from seconds to meters use to get 7.5 meters.
The equation for movement is distance = velocity x time, so we have . In the rest frame of the meson the time is 7.5 meters. But time is stretched by the gamma factor so we have:
then
then
so
  shafatmubin 20091103 13:55:26  For the physics GRE, consider memorising the gammav 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 20100404 10:08:06 
How could you quickly tell that gamma is 7/3?

  ams379 20091007 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}
15/=15/   ams379 20091007 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 20080806 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.
resinoth 20150918 19:24:11 
Exactly. The dilation factor is 2, which corresponds to about 0.85c. Just learn a simple table of these values.

  Mexicana 20071003 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 20081016 21:13:37 
Mexicana, I think that there is a unit discrepancy in your solution. It looks like it is unitless. I think that factor of c in the numerator is not supposed to be there. Correct me if I'm wrong.

grace 20101105 20:55:20 
I think you miss time c.

  mm 20070922 23:16:31  Plug each choice into
, and you can reach the answer.
  banana 20051207 18:17:12  Much easier: The interval is the same in all frames. Thus:
Gaffer 20071027 09:47:59 
Great soln! The math is easier if you think of it like:
The RHside
Then
Basically, don't multiply big numbers unless you have too. This saves time.

jw111 20081105 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

bac 20120330 12:02:39 
After you get to , here is how I was able to very quickly eyeball the answer. (A), (D), and (E) are obviously out.rnrnSquaring the answers (B) and (C), you get and , respectively. Which one does look more like? is closest to , so (C) is the correct answer.

  mg 20051127 16:05:44  The arithmetic simplifies considerably if one realizes that =2c. Then one has v=2c  

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