Two colliding spaceships in Special Relativity

[shyguy79]

1. Homework Statement

---------> A. B <----------
v. v
------------------------------------ this is static frame of ref (D) the space station

Where v=0.97c

In the static frame of reference, ship A has the same length as ship B and are traveling towards each other. Calculate the speed of ship A relative to ship B according to an observer on the space station. Express your answer as a multiple of vA.

Use your answer as part of a short argument to show that the time required for the whole of ship A to pass through the whole of ship B is ΔtA, as observed from the space station.

2. Homework Equations
v'= u - v / 1-vu/c^2

3. The Attempt at a Solution
I think that v' = 0.97c - (-0.97c) / 1 - 0.97c*-0.97c/c^2 because they are traveling towards each other and comes out approximately as 2.9995e8 ms^-1... Or am I barking up the wrong tree and how would you express as a multiple of vA?

How would I work out the 2nd part though... I'm stumped?

Help much appreciated... Thanks for reading

 

[vela]

1. Homework Statement

---------> A. B <----------
v. v
------------------------------------ this is static frame of ref (D) the space station

Where v=0.97c

In the static frame of reference, ship A has the same length as ship B and are traveling towards each other. Calculate the speed of ship A relative to ship B according to an observer on the space station. Express your answer as a multiple of vA.

Use your answer as part of a short argument to show that the time required for the whole of ship A to pass through the whole of ship B is ΔtA, as observed from the space station.

2. Homework Equations
v'= u - v / 1-vu/c^2

3. The Attempt at a Solution
I think that v' = 0.97c - (-0.97c) / 1 - 0.97c*-0.97c/c^2 because they are traveling towards each other and comes out approximately as 2.9995e8 ms^-1... Or am I barking up the wrong tree and how would you express as a multiple of vA?

You should really get in the habit of using parentheses. What you've written means
$$v' = 0.97 c - \frac{-0.97c}{1} - \frac{0.97c\times (-0.97c)}{c^2}$$ In any case, the velocity-addition formula doesn't apply here. What you found was the speed of A an observer at rest with respect to B would see. The question is asking you for the speed of A relative to B an observer in the rest frame of the space station would see. Hint: the answer will be greater than c.

How would I work out the 2nd part though... I'm stumped?

Help much appreciated... Thanks for reading

Say the tips of the ships meet at t=0 and x=0. Some time later, their tails will both be at x=0. How long does that take?

A ------>.<------ B  at t=0
B <------.------> A  at t=?

 

[shyguy79]

You should really get in the habit of using parentheses. What you've written means
$$v' = 0.97 c - \frac{-0.97c}{1} - \frac{0.97c\times (-0.97c)}{c^2}$$ In any case, the velocity-addition formula doesn't apply here. What you found was the speed of A an observer at rest with respect to B would see. The question is asking you for the speed of A relative to B an observer in the rest frame of the space station would see. Hint: the answer will be greater than c.

I thought nothing could travel faster than c?

 

[vela]

Nothing is traveling faster than c.

 

[shyguy79]

...it's just seen from the space station that the speed is faster than c?

 

[vela]

In the space station's rest frame, the relative speed of the ships is faster than c. This just means the distance between the ships is decreasing at a rate higher than c. This is the similar to sending two photons off in opposite directions. Both photons travel with speed c relative to you, but the distance between them increases at a rate of 2c.

 

[shyguy79]

Ahhhh! Gotcha... so what I've worked out would be from the v' = 0.97c - (-0.97c) / 1 - 0.97c*-0.97c/c^2 is the speed in the frame of reference of one of the spaceships?

btw... thank you very much for your help!