Problem in Lorentz transformation

[bckcookie]

Homework Statement

Please see the attached file and advise me my solution is right or not! Thanks!

Homework Equations

~ d/t = v
~ Lorentz transformation : x'=(x-Vt)/(√(1-(V^2)/c^2), t' = [t-(V^2)/(c^2)]/√1-(V^2)/c^2),

The Attempt at a Solution

a) According to Cheryl, d/t =v, 15/(c/2) = 10^-7s

From Tom on ground,
i) when the light from the spark reaches mirror A
t1 = L/(c+V) = 15m/(3x10^8 + c/2 ) = 3.33 x 10^-8s

ii) when the light from spark reaches mirror B
t2 = L/(c-V) = 15m/(3x10^8 - c/2 ) = 10^-7s

iii) when the light from spark returns to Cheryl from mirror A
t2 = L/(c-V) = 15m/(3x10^8 - c/2 ) = 10^-7s

iv) when the light from spark returns to Cheryl from mirror B
t1 = L/(c+V) = 15m/(3x10^8 + c/2 ) = 3.33 x 10^-8s

b) How to apply the Lorentz transformation equation in the same four events take place?? please advise!

My approach:

t=10^-7s, d = 15m, v=c/2

x'=(x-Vt)/(√(1-(V^2)/c^2),
= 15-(c/2 x 10^-7)/(√(1-(V^2)/c^2) = 0

t' = [t-(V^2)/(c^2)]/√1-(V^2)/c^2),

= 8.66 x 10^-8s


c) They disagree that ligt from the spark reachs mirror A at the same time as mirror B, because when the light from the spark reaches mirror A of clock slow down and length contract. They also disagree the light returning from mirror A reaches Chery l at the same time as the light returning from mirror B, because the light returning from mirror B of clock slow down and length contract.

d) They agree it, because the clock is not slow down and length contract.

e) Using t' = t√1-v^2/c^2)

=10^-7s√1-c/v^2/c^2)

=8.66 x 10^-8s

"Moving clocks run slow"
ΔT = ΔT0/√1-v^2/c^2

 

[AvroArrow]

Therefore, Cheryl's clock is running slow by 1.14 x 10^-7s as compared to Tom's clock on the ground.