Question

We have a tunnel, 800 feet long in its rest frame, with doors on each end which can be used to seal the tunnel. The train is 1,000 feet long in its own rest frame, as shown in the illustration. You can close the doors simultaneously any time you want in the tunnel frame.

Find the required minimum speed of the train so that the front side of the train crashes only through the right door. The left door must not touch the train.

Answer 1

Is it just a simple case of applying the length contraction formula?
So we want
800=1000/gamma
So
\[\frac{1}{\sqrt{1-\frac {v_{min} ^2} {c^2}}}=\gamma = 5/4\]
\[\Rightarrow v_{min} =c \sqrt{1-0.8 ^2}=0.6c \ ms^{-1}\]

Answer 2

ooops, inches per sec, not m/s :P

Answer 3

*feet

Answer 4

Awesome! that will do haha !

Answer 5

yay ^_^

Answer 6

I have another question if you have time @Bobo-i-bo
But this is a real question thats troubling me from couple of hours

Answer 7

Sounds awesome :D :Bring it on, haha

Answer 8

you familiar with lorentz transformation equations right ?

Answer 9

let me copy paste them here maybe

Answer 10

\(x' = \gamma(x-vt)\)
\(t' = \gamma(t-vx/c^2)\)

\(x = \gamma(x'+vt')\)
\(\color{red}{t = \gamma(t'+vx'/c^2)}\)

Answer 11

The highlighted red equation tells me that the \(x_2'\gt x_1' \implies t_2\gt t_1\).

That is, as viewed from S frame,
the clock located on the right side in S' frame runs "ahead" of the clock located located on the left side in S' frame.

Answer 12

Is my interpretation correct ?
Below youtube video says the exact opposite of my interpretation :
https://youtu.be/ev9zrt__lec?t=432

Answer 13

In that video the spaceships are going right

It says all the clocks run at same rate and the left clock runs ahead of the right clock

Answer 14

From lorentz transformation I interpret the exact opposite !

Answer 15

see what im confused about ?

Answer 16

I do, let me get my head around it

Answer 17

Thanks :)

Answer 18

I think it's this: So say it is 12:00. Then you see 12:00 at t_2 at a later time than t_1, therefore t_1 is running ahead

Answer 19

Yes? possibly? maybe? :P

Answer 20

Thats what video says, but I still don't get how it follows from lorentz transformation...

Answer 21

\(\color{red}{t = \gamma(t'+vx'/c^2)} \)

suppose the observer in S frame looks at the clocks in S' frame at \(t' = 0\).

the clock at \(x'=0\) shows the time \(\gamma v*0/c^2\)
the clock at \(x'=1\) shows the time \(\gamma v*1/c^2\)
the clock at \(x'=2\) shows the time \(\gamma v*2/c^2\)

the clocks location on the right side are running ahead ?

Answer 22

*
the clocks located to the right are running ahead ?

Answer 23

No. So in frame S, you SEE the clock say 12:00 (or t'=0) at x'=0 at time gamma*v*0/c^2. You see the clock say 12:00 at x'=0 at time gamma*v*1/c^2. You do not see the clock saying gamma*v*0/c^2 when looking at x'=0 from frame S, instead you are seeing the time 12:00 at time gamma*v*0/c^2.

Answer 24

I believe that's where your confusion lies. I don't know if i'm explaining very well though '~'

Answer 25

Wow! \(t\) is not what clocks in S' show. \(t\) is the time coordinate in S frame. the larger the value of \(t\), the longer the observer in S frame waits ! Beautiful xD thanks a lot !

Answer 26

Exactly! Np :)