a. 37500; 299,792,457.9 m/s
The formula for the length contraction for a particle moving close to the speed of light is:
[tex]L' = \frac{L}{\gamma}[/tex]
where
L' is the length observed by the particle moving
L is the length observed by an observer at rest
In this problem, we have
[tex]L = 3 km = 3000 m[/tex] is the length of the SLAC measured by an observer at rest
[tex]L' = 8cm = 0.08 m[/tex] is the length measured by the electrons moving
Substituting into the formula, we find the gamma factor
[tex]\gamma = \frac{L}{L'}=\frac{3000 m}{0.08 m}=37,500[/tex]
The formula for the gamma factor is
[tex]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
where v is the electron's speed and c is the speed of light. Re-arranging the equation, we find v:
[tex]1-\frac{v^2}{c^2}=\frac{1}{\gamma^2}\\v=c \sqrt{ 1-\frac{1}{\gamma^2}}=(299 792 458 m/s)\sqrt{1-\frac{1}{(37500)^2}}=299,792,457.9 m/s[/tex]
b. [tex]1.0\cdot 10^{-5}s[/tex]
For an observer at rest in the laboratory, the electron is moving at a speed of
v = 299,792,457.9 m/s
and it covers a total distance of
L = 3000 m
which is the length of the SLAC measured by the observer. Therefore, the time it takes for the electron to travel down the tube is
[tex]t=\frac{L}{v}=\frac{3000 m}{299,792,457.9 m/s}=1.0\cdot 10^{-5}s[/tex]
c. [tex]2.67\cdot 10^{-10}s[/tex]
From the electron's point of view, the length of the SLAC is actually contracted, so the electron "sees" a total distance to cover of
[tex]L' = 0.08 m[/tex]
And this means that the total time of travel of the electron, in its frame of reference will be shorter; in particular it is given by the formula:
[tex]t' = \frac{t}{\gamma}[/tex]
where
[tex]t=1.0\cdot 10^{-5}s[/tex] is the time measured by the observer at rest
[tex]\gamma=37500[/tex]
Substituting,
[tex]t' = \frac{1.0\cdot 10^{-5}s}{37500}=2.67\cdot 10^{-10}s[/tex]
