Consider a cosmic ray colliding with a nucleus in the Earth\'s upper atmosphere

Question

Consider a cosmic ray colliding with a nucleus in the Earth's upper atmosphere that produces a muon which has a velocity v = 0.911c.

The muon then travels at constant velocity and lives 1.52 µs as measured in the muon's frame of reference. (You can imagine this as the muon's internal clock.)

(a)

How far (in km) does the muon in travel according to an Earth-bound observer?

______km

(b)

How far (in km) does it travel as viewed by an observer moving with it? Base your calculation on its velocity relative to the Earth and the time it lives (proper time).

______ km

(c)

Verify that these two distances are related through length contraction = 2.42.

Explanation / Answer

Here ,

v = 0.991c

a)

distance travelled by the muon = life time of muon * speed of muon

distance travelled by the muon = 1.52 *10^-6/sqrt(1 - 0.911^2) * (0.911 * 3 *10^8)

distance travelled by the muon = 1007.3 m

the distance travelled by the muon is 1007.3 m = 1.01 km

b)

for the observer moving with it

distance travelled by the muon = life time of muon * speed of muon

distance travelled by the muon = 1.52 *10^-6 * (0.911 * 3 *10^8)

distance travelled by the muon = 415 m

the distance travelled by the muon is 415 m = 0.42 km

c)

as 1.01/y = 1.1/2.42 = 0.42

hence , this is true

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