1) Subatomic particles called muons can be created in the upper atmosphere by collisions of cosmic rays (energetic particles coming from ast

Question

1) Subatomic particles called muons can be created in the upper atmosphere by collisions of cosmic rays (energetic particles coming from astrophysical sources). As we shall see in a few lectures, muons quickly decay (in 2.2 microseconds on average) into other particles. Consider a muon that was created at a height the same as the top of a nearby mountain, traveled straight down towards the Earth through the atmosphere at 0.9 times the speed of light, and then decayed at a point the same height as that of a tree it reaches, after exactly 2.2 microseconds in the muon frame.

A) Which direction is the tree moving in the muon rest frame?

A. Up
B. Down
C. Left
D. Right
E. None of above

B) How far in meters did the muon travel in the muon’s rest frame?
C) How far in meters did the tree travel in the muon’s rest frame?
D) What is the distance in meters from the top of the mountain to the tree in the muon’s rest frame?
E) How fast in meters per second is the muon moving in the rest frame of someone standing on the mountain?
F) In which direction is the muon moving in the rest frame of someone standing on the mountain?

A. Up
B. Down
C. Left
D. Right
E. None of above

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Ben Gia 2 months 2021-07-27T17:33:59+00:00 1 Answers 1 views 0

Answers ( )

    0
    2021-07-27T17:35:19+00:00

    Answer and explanation:

    A.

    Muon travelled straight down towards the earth. Therefore the tree moves up in the rest frame of muon (option a)

    B.

    In muon rest frame it travels Zero meters

    C.

    Distance, d = Velocity, v * Time, s

    where, v = 0.9c = 0.9 \times 8 \times 10^8 , s = 2.2 \mu s

    d = 0.9 \times 3 \times 10^8 \times 2.2 \times 10^{-6}\\\\d = 594m

    D.

    Distance from the top of the mountain to the tree is the same as the distance travelled by the tree in the muons rest frame

    that is same as in part C which is 594m

    E.

    Using lorentz contraction

    In the rest frame of someone standing on the mountain

    the distance is given by

    d' = \frac{d}{\gamma} = d\sqrt{1 - \frac{v^2}{c^2}}, where, \frac{1}{\gamma}= \sqrt{1 - \frac{v^2}{c^2}}

    d' = 594\sqrt{1 - \frac{(0.9c)^2}{c^2}}

    d' = 594\sqrt{1 - 0.81}

    d' = 594 \times 0.4359

    d’ = 258.92m

    F.

    in the rest frame of someone standing on the mountain,

    muon moves straight down

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