Special Relativity Spaceship Question

[Lucille]

Homework Statement

bob and anna are in two space ships of proper length 100m; anna's ship passes bob's and is moving in the -x dir at v=0.75c. the front of anna's ship and the tail of bob's ship coincide at x=x'=t=t'=0. bob sits at the window in his ship and sees a clock inside anna's ship
a) what position is the window in bob's space ship
b) what time does this clock in anna's ship read

Homework Equations

x'=gamma(x-vt)
x=gamma(x'+vt')
t'=gamma(-v/c^2*x+t)
t=gamma(v?c^2*x+t')
delta t = gamma delta to
L = Lo/gamma

The Attempt at a Solution

a)

so if bob sees the clock at the end of anna's ship, then:

L=Lo/gamma = sqrt(1-v^2/c^2)*Lo = sqrt(1-0.75^2)*(100)=66.1m <-- position of bob's window in his spaceship (distance from the end of his spaceship to the window)

b)

i tried :

x=100m
v=0.75c

t'=x'/v = 4.45 * 10^-7 s

But I'm not sure if this is correct...

 

[Chestermiller]

One thing that isn't clear: Bob is situated at the front of his space ship?

Chet

 

[Lucille]

The question doesn't state that. It's asking where the window is - I'm guessing that's the point at which he can see Anna's clock, and her window is at the end of her spaceship.

 

[Chestermiller]

The question doesn't state that. It's asking where the window is - I'm guessing that's the point at which he can see Anna's clock, and her window is at the end of her spaceship.

So, under this interpretation, you are looking at the event at which x' = L₀ and t = 0. In part a, you showed that $x=L_0/\gamma$. Now all you need to do is substitute this into the equation for t'.

Chet

 

[HalfLostAndSearching]

Your approach for part a) is correct. For part b), you are on the right track, but there are a few errors in your calculations.

First, the equation you are using, t'=x'/v, is for time dilation, not for finding the time on a clock in a moving frame. The correct equation is t' = gamma(t-vx/c^2), where t is the time on the clock in the stationary frame (in this case, Bob's frame) and x is the distance between the clock and the observer (again, in this case, Bob). So, using the values given in the problem, we have:

t' = gamma(t-vx/c^2) = gamma(0-0.75*100/299792458^2) = 4.45 * 10^-7 s

This is the correct time on Anna's clock as observed by Bob. Note that this is the same value you calculated, but with a different equation.