Special Relativity and Relative motion

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Homework Help Overview

The problem involves two spaceships moving towards each other at relativistic speeds, as observed from Earth. The original poster seeks to determine the time of collision in the Earth frame and the relative velocities of the spaceships from each other's frames of reference, utilizing concepts from special relativity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the setup of the problem, including the distances and velocities involved. There is an exploration of the use of Lorentz transformations to find relative velocities and the timing of the collision from different reference frames.

Discussion Status

Some participants confirm the correctness of initial calculations and engage in further exploration of the Lorentz transformation equations. There is ongoing dialogue about the timing of the collision from different frames, with suggestions to focus on time rather than distance for certain calculations.

Contextual Notes

Participants note the need to clarify the use of Lorentz transformations and the distinction between the questions of "when" and "where" regarding the collision event.

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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So far, correct.
 
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.
 
Numerically, yes. Opp signs because in opp directions. But that shouldn't bother you.
 
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.
 
Thank you very much for your time.
 

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