How Fast Must a Pion Travel to Cover 15 Meters Before Decaying?

[Eich]

How fast must a pion be moving in order to travel 15 m before it decays?
The avg lifetime, at rest, is 2.6 x 10^-8 s.

The answer is supposed to be 0.89c

This sounds so easy. Like, just plugging in the formula but I'm not getting it. I haven't done relativity in so long. I think I must've forgotten more than I thought. I can't even find my notes now. Can someone please tell me how to get to the answer?

 

[Yapper]

We arent even touching on relativity in my AP Physics BC class so I can't help you

 

[Eich]

Thanks yapper for looking at the problem, at least.

Anyone else?

 

[Yapper]

Try the college boards

 

[vincentchan]

the answer is not .89c, \gamma = 1/ \sqrt{1-.89^2}~2 it must travel at much higher speed, at leat .99c up (Sorry, i don't have a calculator handy and can't give you the exact answer), your answer is propabaly right, don't alway trust the textbook answer...

 

[learningphysics]

Find a formula for the time in the stationary frame in terms of the velocity v. Then use that time in the time dilation formula. T'=T(1-v^2/c^2)^(1/2). Then solve for v.

Remember T' is the time in the moving frame (in the pion's frame).

I get 2.66*10^-8m/s which is approximately 0.89c

 

[Curious3141]

Let the pion's resting lifespan be \tau. The distance it is required to travel from the perspective of a resting observer is d, and the pion's velocity relative to the resting observer is v. c is the speed of light.

From the frame of the pion, it will always "measure" its lifespan as \tau. This is the proper time interval. The resting observer will measure this time interval as \gamma \tau where \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}.

The observer measures the velocity of the pion as v, where :

\frac{d}{\gamma \tau} = v

\frac{d}{\tau} = \frac{v}{\sqrt{1 - \frac{v^2}{c^2}}}

Put the relevant values in and solve for v and the answer is v = 0.89c

 

[Eich]

Awesome. Thanks!
I got the same answer as vincentchan. And other people did too so I just left it at that. I'm so grateful it's been proved otherwise.

V.v.v. thankful.

 

[vincentchan]

I get 2.66*10^-8m/s which is approximately 0.89c

just want to point out...
2.66*10^-8m/s is 0.89*10^-16c

 

[vincentchan]

Let the pion's resting lifespan be \tau. The distance it is required to travel from the perspective of a resting observer is d, and the pion's velocity relative to the resting observer is v. c is the speed of light.

From the frame of the pion, it will always "measure" its lifespan as \tau. This is the proper time interval. The resting observer will measure this time interval as \gamma \tau where \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}.

The observer measures the velocity of the pion as v, where :

\frac{d}{\gamma \tau} = v

\frac{d}{\tau} = \frac{v}{\sqrt{1 - \frac{v^2}{c^2}}}

Put the relevant values in and solve for v and the answer is v = 0.89c

y don't you simply use t = \gamma \tau
and again, v is not .89c

 

[Curious3141]

y don't you simply use t = \gamma \tau
and again, v is not .89c

Show us your reasoning.

 

[vincentchan]

oh, sorry, I thought its say the particle travel 15 mins... I didn't realize it is 15 meters, you are right, and so do the answer