GREPhysics.NET: GR0177 #33

GR0177 #33

Problem

GREPhysics.NET Official Solution

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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).

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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.

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Comments

danielkwalsh
2010-09-05 18:25:18

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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.

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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.

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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

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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?

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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.

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