Special Relativity and Relative motion

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olga11
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1. Homework Statement

An observer on Earth sees two spaceships moving in opposite directions and finally they collide. At t=0 the observer on Earth says that the spaceship 1 which moves to the right with Ua=0.8c is at the point A and the spaceship 2 which moves to the left with Ub=0.6c is at the point B. The distance AB=L=4,2.10^8 m.
When do the two spaceships collide to the Earth frame of reference?
What is the velocity of the spaceship 2 to the frame reference of the spaceship 1?
What is the velocity of the spaceship 1 to the frame reference of the spaceship 2?
When does the collision happen to the frame reference of the spaceship 1 and to the frame reference of the spaceship 2?

2. Homework Equations



3. The Attempt at a Solution

Let D be the point of the collision and AD=x, so DB=L-x
The velocity is constant.
spaceship 1: AD=x=0,8c.t
spaceship 1: DB=L-x=0,6c.t
By addition
L=AD+DB
L=(0,8c+0,6c).t
t=4,2.10^8/(1,4.3.10^8)
t=1sec
Is it ok so far?
 
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The velocity of the spaceship 2 to the frame reference of the spaceship 1 is the same as the velocity of the spaceship 1 to the frame reference of the spaceship 2.
 
Using Lorentz transformations

0,8c=(U21-0,6c)/(1-U21.0,6c/c^2)

U21=1,4c/1,48

Correct?
 
When does the collision happen to the frame reference of the spaceship 1 and to the frame reference of the spaceship 2?

Now we must use Lorentz transformation for the length?
 
olga11 said:
Using Lorentz transformations

0,8c=(U21-0,6c)/(1-U21.0,6c/c^2)

U21=1,4c/1,48

Correct?

(Is your keyboard very old, so you can't tell the , from the .? Just kidding...)

Yes. You could have got it also by (0.8+0.6)c/(1+0.8*0.6).
 
olga11 said:
When does the collision happen to the frame reference of the spaceship 1 and to the frame reference of the spaceship 2?

Now we must use Lorentz transformation for the length?

The question is "when", not where. So, use the formula for time.