Relativity Question Regarding Muons in movement

[HarleyM]

Homework Statement

A muon has a lifetime of 2.2x10^-6 s when at rest, after which time it decays into other particles.

a)ignore any effects of relativity discussed in this lesson, if the muon was moving at 0.99c how far would it travel before decaying into other particles, according to Newtonian mechanics? *this line confuses me, I hope i obeyed them in my answer.. if anyone can let me know that would be great!

b) how long would the muon last according to an observer in the Earth's frame of reference who viewed the muon moving at 0.99c?

c) How far would the muon actually travel, when viewed moving at 0.99c?

d) Compare the distances travelled. Explain why this type of evidence is excellent support for the theory of relativity.

Homework Equations

Δt_m = Δt_s/√(1-v²/c²)

The Attempt at a Solution

a) 0.99(3x10^8) = 2.97x10^8

2.97x10^8 m/s * 2.2x10^-6 s = 653.4 m (not sure whether this is according to Newtonian mechanics?)

b) Δt_m = Δt_s/√(1-v²/c²)
Δt_m = 2.2x10^-6/√(1-(0.99c)²/c²)
Δt_m = 2.2x10^-6/√(1-0.99)
Δt_m = 2.2x10^-5

c) 2.97x10⁸ m/s * 2.2x10^-5s
= 6534 m

D) this is excellent supportive evidence for the theory of relativity because it displays as matter approaches the speed of light significant time dilation occurs and the muon lasts much longer than it would with Earth as its reference frame.Does everything look logical ?

Thanks!

 

[alexg]

But if this is to be according to Newtonian mechanics, there is no time dilation, and equation 2 is not correct.

 

[bm0p700f]

Well I think your method is fine but your half life is off. You forgot to square the 0.99!

 

[HarleyM]

But if this is to be according to Newtonian mechanics, there is no time dilation, and equation 2 is not correct.

only the first question is according to Newtonian mechanics.. by second equation do you mean the Δt_m?

all other questions use relativity unless otherwise stated I believe!

 

[HarleyM]

Well I think your method is fine but your half life is off. You forgot to square the 0.99!

in the Δt_m equation you mean? The C² cancels out so I am left with 1-0.99

at least I think..


thanks for the input

 

[cupid.callin]

in the Δt_m equation you mean? The C² cancels out so I am left with 1-0.99

at least I think..


thanks for the input

it would be 0.99² = 0.98

 

[HarleyM]

it would be 0.99² = 0.98

Thats not the method the book shows, they show both c² cancelling out so I am going with that method, simply because if they say its wrong I'll show them the book and say the book is also wrong

 

[cupid.callin]

Thats not the method the book shows, they show both c² cancelling out so I am going with that method, simply because if they say its wrong I'll show them the book and say the book is also wrong

the author might have missed is.
\Large{\frac{v^2}{c^2}} and if v=.99c \Large{\frac{(0.99c)^2}{c^2} = \frac{0.99^2 c^2}{c^2}} = 0.99^2

 

[HarleyM]

the author might have missed is.
\Large{\frac{v^2}{c^2}} and if v=.99c \Large{\frac{(0.99c)^2}{c^2} = \frac{0.99^2 c^2}{c^2}} = 0.99^2

I went back and looked over it all, they didn't do it for one example but all others so it threw me for a loop , I changed my answer thank you very much both of you for pointing that out!