GREPhysics.NET: GR0177 #33

GR0177 #33

Problem

GREPhysics.NET Official Solution

This problem is still being typed.

Special Relativity $\Rightarrow$ }Frames

The distance the pion must travel is in the lab frame, and thus $L=30=vt$ .

The decay time is given in its proper time, and thus $t=\gamma t_0 \Rightarrow 30=v\gamma t_0$ . Solving for v, one finds that choice (D) works.

Alternate Solutions

danielkwalsh
2010-09-05 18:25:18

A quick trick: the equation $\frac{30m}{v}=\gamma t_0$ is correct, but for an extremely relativistic particle, as we clearly have here, the decay time in the lab frame can be well approximated as $\frac{30m}{c}$ , since $v\approx c$ . This simplifies the math a lot more in solving for $v$ . This gives $\beta=\frac{3\sqrt{11}}{10}$ , which is closest to choice (D).

Reply to this comment

Herminso
2009-09-21 13:42:50

Use $\triangle S^2=\triangle x^2-c^2\triangle t^2$ . In the own rest frame of the pion $\triangle x=0$ , and thus:

$\triangle S^2=-c^2\(10^{-8})^2=-9$ .

But in the Lab frame,

$\triangle S'^2=(30)^2-c^2\triangle t'^2$ .

Now, using the invariant $\triangle S^2=\triangle S'^2$ ,

$(30)^2-c^2\triangle t'^2=-9$ $\Rightarrow$ $\triangle t'=\sqrt{909/c^2}=sqrt{101}\times 10^{-8} s$

Thus, $v=\frac{\triangle x'}{\triangle t'}=\frac{30 m}{\sqrt{101}\times 10^{-8} s}\simeq2.98\times 10^8 m/s$ .

ramparts
2009-10-07 13:32:13

By far the best solution, because it's the most elegant and because it makes the algebra significantly easier without a calculator.

Reply to this comment

Comments

danielkwalsh
2010-09-05 18:25:18

Reply to this comment

Herminso
2009-09-21 13:42:50

hanin
2009-10-04 22:13:57

Why $\Delta x=0$ ?

kroner
2009-10-05 11:36:41

The pion's own rest frame is the frame where it doesn't move so $\Delta x = 0$ .

ramparts
2009-10-07 13:32:13

By far the best solution, because it's the most elegant and because it makes the algebra significantly easier without a calculator.

Reply to this comment

Ning Bao
2008-02-01 06:26:28

Pions don't reach speed of light (incidentally, this eliminates E). It must be going very close, however, to contract 30m to a little less than 3 meters to not violate this. This means D.

noether
2009-11-05 15:28:05

If 2.99*10^8 were listed as an answer, would you have chosen that?

gravity
2010-11-10 00:37:22

Yeah. This one was tricky. I did most of what everybody else did and ended with 30 c/(909) $^{1/2}$ which I figured looked more like more like c than it did 2.98 x 10^8.

Gah. I should have known better! At least it's not the real test.

Reply to this comment

StrangeQuark
2007-05-12 12:03:43

To make this faster (avoid the "messy" fraction)
Note to start that this is not a photon thus answer E is out.
now as above,
L=v $\gamma$ t
L $\sqrt{1-\frac{v^2 }{c^2}$ =v t
some simplification steps
$v^2$ = $\frac{L^2 c^2}{t^2 c^2+L^2}$
now plug and chug...
$v^2$ = $\frac{(30m)^2 ((3)(10^8)(\frac{m}{s}))^2}{(10^{-8}s)^2 ((3)(10^8)(\frac{m}{s}))^2+(30m)^2}$
Note now that we have a
$(10^{-8})(10^{8})$
in the first term in the denominator, leaving only
$3m^2+30m^2$
in the denominator,
but $3m^2\ll30m^2$
so we simplify
$v^2$ < $\frac{(30m)^2 ((3)(10^8)(\frac{m}{s}))^2}{(30m)^2}$
after canceling we see that
v< $(3)(10^8)\frac{m}{s}$
but only by a very small amount thus we have D

Reply to this comment

boundforthefloor
2006-11-26 05:25:22

Can anyone clarify this? I'm befuddled and can't find much that helps.

johnyp03
2006-11-29 17:48:43

So, you know L=30=vt in the lab frame. But, in the pion's frame, there is time dilation. So, t=t'=(gamma)t0 where t0=3*10^-8, gamma=1/(sqrt(1-v^2/c^2)). So:

30 = v(gamma)t0 = v*(10^-8)/(sqrt(1-v^2/c^2))

30*sqrt(1-v^2/c^2) = v*(10^-8)

900*(1-v^2/c^2) = v^2(10^-16)

900 = v^2*10^-16 + v^2*900/c^2

==================> 900/c^2=(900/9)*10^-16=10^-14

900 = v^2(10^-16 + 10^-14)

sqrt(900/(10^-16+10^-14)) = v

30/sqrt(1.01*10^-14) = v

30/1.005*10^-7 = v

v ~ 2.98*10^8

Hope this helps

boundforthefloor
2006-12-01 11:40:08

Thanks johnyp. The actual equation setup was killing me, now that I've seen the calcualtions it makes much more sense.

VanishingHitchwriter
2006-12-01 14:07:02

Surround your expressions with dollar signs for latex equations. Here's the guy's comment back again...

$t=t'=\gamma t_0$ where $t_0=3*10^{-8}$ , $\gamma=1/(\sqrt{1-v^2/c^2})$ . So:

$30 = v\gammat_0 = v*(10^{-8})/(\sqrt{1-v^2/c^2})$

$30*\sqrt{1-v^2/c^2} = v*(10^{-8})$

$900*(1-v^2/c^2) = v^2(10^{-16})$

$900 = v^2*10^{-16} + v^2*900/c^2$

==================> $900/c^2=(900/9)*10^{-16}=10^{-14}$

$900 = v^2(10^{-16} + 10^{-14})$

$\sqrt(900/(10^-16+10^-14)) = v$

$30/\sqrt(1.01*10^-14) = v$

$30/1.005*10^-7 = v$

$v ~ 2.98*10^8$

matweiss
2010-09-29 09:31:49

hey, I don't understand why t0 is 3 x 10^-8. it seems like I thought t in the frame of the particle was 10^-8 s and that (if you assume the particle to essentially be moving at light speed) then t in the lab frame equals 30/c= 1x 10^-7. any help?

Reply to this comment

senatez
2006-11-02 12:46:13

I ended up with a messy fraction 10c/10.1 which you estimate as .99c. This is about 2.98*10^8. I sure wish they would give easier arithmatic, it is waists a lot of time.

Reply to this comment

Post A Comment!

You are replying to:

To make this faster (avoid the "messy" fraction) Note to start that this is not a photon thus answer E is out. now as above, L=v $\gamma$ t L $\sqrt{1-\frac{v^2 }{c^2}$ =v t some simplification steps $v^2$ = $\frac{L^2 c^2}{t^2 c^2+L^2}$ now plug and chug... $v^2$ = $\frac{(30m)^2 ((3)(10^8)(\frac{m}{s}))^2}{(10^{-8}s)^2 ((3)(10^8)(\frac{m}{s}))^2+(30m)^2}$ Note now that we have a $(10^{-8})(10^{8})$ in the first term in the denominator, leaving only $3m^2+30m^2$ in the denominator, but $3m^2\ll30m^2$ so we simplify $v^2$ < $\frac{(30m)^2 ((3)(10^8)(\frac{m}{s}))^2}{(30m)^2}$ after canceling we see that v< $(3)(10^8)\frac{m}{s}$ but only by a very small amount thus we have D

Bare Basic LaTeX Rosetta Stone
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces $\alpha^2_0$ .
type this...	to get...
$\int_0^\infty$	$\int_0^\infty$
$\partial$	$\partial$
$\Rightarrow$	$\Rightarrow$
$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...

All-Solutions List | Latest Comments | Unanswered Questions (Cries for Help!) |::| About

This site is not affiliated with ETS, nor is it officially endorsed (yet) by ETS. This site is the work of an individual named Yosun Chang. The solutions, sans user comments and ETS questions, are © 2005 by Yosun Chang except where noted. All rights reserved. / The comments appending particular solutions are due to the respective users. / Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 1997 and © 2001 by ETS.

Bare Basic LaTeX Rosetta Stone