(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A primary cosmic ray (which comes from elsewhere in our Galaxy) hits the nucleus of an

atom in our atmosphere 38 km above the ground and creates an unstable secondary particle,

which then travels straight down toward the ground at a speed of 0.99400c. (The altitude and the speed given in the previous sentence are measured by a person at rest on the ground.)

a) As measured by a person on the ground, how long does it take this particle to reach the

ground?

b) Using time dilation, calculate the time it takes the particle to reach the ground, as

measured in the frame of the particle.

c) Using length contraction, calculate the distance from the point of the particle’s creation to

the ground as measured in the frame of the particle.

d) Using your answer to part (c), calculate the time to reach the ground, measured in the

particle’s frame.

2. Relevant equations

[tex] \Delta t = \frac{\Delta t_0}{\sqrt{1-{u^2/c^2}}} [/tex]

[tex] l = l_0 \sqrt{1-u^2/c^2} [/tex]

[tex] \gamma = \frac{1}{\sqrt{1-u^2/c^2}}[/tex]

3. The attempt at a solution

Part (a): [tex] \Delta t = \frac{l}{u} = \frac{38*10^3}{0.994*(3*10^8)} = 1.274*10^{-4} \, \mbox{seconds}[/tex]

Part (b): [tex] \Delta t_0 = \Delta t \sqrt{1-u^2/c^2} = 1.274*10^{-4} \sqrt{1-0.994^2} = 1.394*10^{-5} \, \mbox{seconds}[/tex]

Here is where I'm running into a problem, on part (c):

I think that for this part, you solve for [tex] l_0 [/tex], because that would give the proper length:

[tex] l_0 = \frac{l}{\sqrt{1-u^2/c^2}} = \frac{38*10^3}{\sqrt{1-0.994^2}} = 3.474*10^5 \, \mbox{meters} [/tex]

The solution to this problem, however, says:

[tex] l = \frac{l_0}{\gamma} = \frac{38*10^3}{9.142} = 4.156*10^3 \, \mbox{meters} [/tex]

This doesn't make any sense according to length contraction either, because this answer should be the longest possible distance, since this answer is the length in the frame of the object being measured. But this value is less than that from the observer on Earth.

Is there something that I'm missing here?

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# Homework Help: Calculating time/velocity for relativistic cosmic ray

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