Relativity Problem: Batman & Robin's Train Escape

[Aeighme]

Homework Statement

Holy Time Dilation! The Joker has captured Batman and Robin and has superglued them together to the roof of a train, which is moving at a steady velocity of magnitude v = 0.64c toward a low bridge. Confident that his diabolical plan will finish off the dynamic duo for good, the Joker returns to his seat in the train. Fortunately, Batman remembered to pack some super-bat-glue solvent in his utility belt, and is able to apply the solvent to the superglue. At the instant they are freed, the superheroes simultaneously set their stop watches to zero, and then by pre-arrangement they start running along the train in opposite directions at speed u = 0.74c, measured in the frame in which the train is at rest. They each run for a time Δt' = 19.3 ns, as measured on their own stop watches, and then they each jump from the train, just before the train passes under the bridge. They hit the ground with a sharp thud, and come instantly to a stop. Miraculously, neither member of the Dynamic Duo is hurt.
According to the Joker, who is at rest with respect to the train, how far apart are Batman and Robin when they jump?

Homework Equations

t=t¹*gamma
d=vt

The Attempt at a Solution

I tried finding the Joker's measured time and then calculated the distance..but it didn't work..nor did it seem right.

 

[Doc Al]

I tried finding the Joker's measured time and then calculated the distance..but it didn't work..nor did it seem right.

Sounds right to me. Show exactly what you did.

 

[Aeighme]

19.3e^-9s * (1/(1-.1²)^1/2= t


Then...

d=vt with v of joker wrt batman and time from above.

 

[Doc Al]

19.3e^-9s * (1/(1-.1²)^1/2= t

Where did you get a speed of 0.1c?

 

[Aeighme]

Where did you get a speed of 0.1c?

.74c -.64c

 

[Doc Al]

.74c -.64c

But Batman and Robin are moving at 0.74c with respect to the train, which is the Joker's frame. The speed of the train is irrelevant.

 

[Aeighme]

But Batman and Robin are moving at 0.74c with respect to the train, which is the Joker's frame. The speed of the train is irrelevant.

Ohhh. so if I use .74 instead, I should get the correct answer.

 

[Doc Al]

Give it a try.

 

[Aeighme]

19.3e^-9*(1-(.74)²)^-.5=t
t=2.86943e^-8 s
d=vt
d=.74*3e⁸*2.86943e^-8=6.37m

?

 

[Aeighme]

and then I double that answer to find the distance between batman and robin?
or do I find robin's distance a different way?

 

[Aeighme]

oh, got it. Just double d..and it works.

Thank you!

 

[Aeighme]

As measured on the ground, how far apart are the two points at which the pair hit the groundFor this next part, I now use the velocity wrt the ground..which would be .1c?

 

[Doc Al]

For this next part, I now use the velocity wrt the ground..

Yes. But you'll have to figure out the speeds of Batman and Robin with respect to the ground.

which would be .1c?

No, speeds don't simply add and subtract in special relativity the same as they do in Galilean/Newtonian relativity. You need to use the formula for relativistic addition of velocity.

Realize that since they move in opposite directions, they have different speeds with respect to the ground.

 

[Aeighme]

u=(u¹+v)/(1+(u¹/c²))


with u being the velocity I'm attempting to find, v being that of train and u¹ their speed wrt train?

 

[Aeighme]

Alrighty. I got the right answer.
Thank you again. This stuff is mind bending >.<

 

[Doc Al]

Excellent!