Help with second part of question

[jlmac2001]

I'm stuck on the second question of the following problem:

The proper mean lifetime of a muon is 2.00 ms. Muons in a beam are traveling a .96c.

What is their mean lifetime as measured in the laboratory?

ans. 2.00ms/sqrt(1-.960^2)= 7.1428 ms

How far would they travel, on average, before they decay?

ans. I don't know how to do this part. Can someone help me?

Thanks.

Jessica

 

[HallsofIvy]

There are two ways to do this and they should give the same answer:

1) Ignore relativity (or think of this as "in the muon's frame of reverence). The muon is moving at .98 c and has a lifetime of 2 ms= 0.002 seconds. Convert .98 c into "meters per second" and multiply by 0.002.

2) You have already calculated that the muon will have a lifetime, measured in the lab frame of reference of 7.1428 ms= 0.0071428 seconds (although I'm a little suspicious of your accuracy!). This is, of course, because, in the lab frame, the muon is "aging" slower.
Multiply 0.0071428 by .98c and then, whatever length you get, apply the same tranformation to determine the shrinkage due to the motion. You should see the sqrt(1-c^2]) terms cancel to give the same answer as before.

 

[M98Ranger]

, to find the distance traveled before decay, we can use the formula d = vt, where d is the distance, v is the velocity (in this case, 0.96c), and t is the time (in this case, the mean lifetime of 7.1428 ms). Plugging in the values, we get d = (0.96c)(7.1428 ms) = 6.8574 m. Therefore, on average, the muons would travel 6.8574 meters before decaying. I hope this helps! If you have any other questions, please let me know.