Relative Motion: Collision of Two Spaceships on Earth

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Homework Help Overview

The problem involves two spaceships moving towards each other at relativistic speeds, with an observer on Earth noting their initial positions and velocities. The main questions pertain to the timing of their collision from both the Earth frame and the frames of the individual spaceships, as well as the relative velocities between the two ships.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the setup of the problem, including the use of constant velocities and the calculation of collision time. There is an attempt to calculate relative velocities using classical addition, which raises questions about the validity of this approach given the relativistic context.

Discussion Status

Some participants are exploring the calculations and questioning the assumptions made regarding the addition of velocities. There is recognition that the speeds involved are significant fractions of the speed of light, which may affect the calculations. A hint for further exploration has been requested.

Contextual Notes

Participants note the need to consider relativistic effects when calculating velocities and collision times, as the speeds are comparable to the speed of light. There is an emphasis on ensuring that the calculations adhere to the principles of relativity.

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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?

Homework Equations





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?

A hint to go on, please?
 
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The velocity of the spaceship 2 to the frame reference of the spaceship 1 is Uba=Ua+Ub=1.4c
The velocity of the spaceship 1 to the frame reference of the spaceship 2 is also Uab=Ua+Ub=1.4c

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

t=AB/Uab=4,2.10^8/(1,4.3.10^8)=1 sec

Something is wrong. It cannot be the same.
 
You don't consider the fact that the speeds are comparable to the speed of light.
 
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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