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## Homework Statement

A) at what velocity will be the KE of a spaceship 2/3 of its rest energy?

B) If the Spaceship travels at this velocity relative to the earth and its trip takes 10 years as measured in the earthbound system, how much time has elapsed on the spaceship clock?

C) What is the distance (in lightyears) it travels as seen from the earth and as seen from the spaceship?

## Homework Equations

m=m

_{o}/([tex]\sqrt{}1- (v^2/c^2)[/tex]

[tex]\Delta[/tex]t= [tex]\Delta[/tex]t

_{o}/([tex]\sqrt{}1- (v^2/c^2)[/tex]

L=L

_{o}[tex]\sqrt{}1-(v^2]/c^2)[/tex]

Sorry for some reason the subscript isn't working or I'm doing it completely wrong...

m

^{2}

## The Attempt at a Solution

A)

KE = (m-m

_{o})c^2 = 2/3m

_{o}c^2 .... m=1.67

_{o}(1+2/3 = 1.67)

1.67 = 1/([tex]\sqrt{}1- (v^2/c^2)[/tex]

2.7889 = 1/(1-(v^2/c^2))

1/2.7889 = (1-(v^2/c^2))

0.6414 = v^2/c^2

v= 0.80c

B)

[tex]\Delta[/tex]t= [tex]\Delta[/tex]t

_{o}/([tex]\sqrt{}1- (v^2/c^2)[/tex]

10 yrs = [tex]\Delta[/tex]t

_{o}/[tex]\sqrt{}1-.80^2[/tex]

[tex]\Delta[/tex]t

_{o}= 6 years

C)I didn't really know how to do this one so I just tried and equation:

L=L

_{o}[tex]\sqrt{}1-(v^2/c^2)[/tex]

L=10yrs[tex]\sqrt{}1-.80^2[/tex]

L=61 lightyears for the people on earth

L=L

_{o}[tex]\sqrt{}1-(v^2/c^2)[/tex]

10years = L

_{o}[tex]\sqrt{}1-.80^2[/tex]

L

_{o}16.67 Lightyears for the people in the spaceship

Unfortunately I don't have the answers to these so I cant check my own work, any help would be greatly appreciated..

Thanks