# Two colliding spaceships in Special Relativity

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 travelling towards eachother. 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 travelling towards eachother 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

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vela
Staff Emeritus
Homework Helper
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 travelling towards eachother. 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 travelling towards eachother 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?

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

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
Staff Emeritus
Homework Helper
Nothing is traveling faster than c.

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

vela
Staff Emeritus