What is the speed of each particle relative to the other?

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Homework Help Overview

The discussion revolves around determining the speed of two particles shot out simultaneously at right angles with a given velocity v, specifically focusing on their relative speeds as perceived by each other. The context involves concepts from classical mechanics and potentially relativistic physics.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the calculation of relative velocity using vector analysis, questioning the assumptions behind their calculations. Some discuss the implications of relativistic effects when considering speeds close to the speed of light.

Discussion Status

The discussion is active, with participants providing various insights into the relative velocity calculations. There is an exploration of both classical and relativistic frameworks, with some guidance offered regarding the use of vector subtraction and the relativistic addition formula.

Contextual Notes

There are mentions of different frames of reference and the need for clarity on whether relativistic effects are significant in the problem context. Some participants express uncertainty about the necessity of the relativistic addition formula in their calculations.

pivoxa15
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Two particles are shot out simultaneously from a given point with velocity v. They start at right angles to each other. What is the speed of each particle relative to the other?

I got one particle thinks the other one is traveling at v\sqrt2 whereas they are traveling at v and vice versa. But I think I calculated it from a frame that is at rest relative to both particles and imagined what each particle would think if they had information in my frame.
 
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pivoxa15 said:
Two particles are shot out simultaneously from a given point with velocity v. They start at right angles to each other. What is the speed of each particle relative to the other?

I got one particle thinks the other one is traveling at v\sqrt2 whereas they are traveling at v and vice versa. But I think I calculated it from a frame that is at rest relative to both particles and imagined what each particle would think if they had information in my frame.
If you stick to a vector analysis, it is not so confusing. The relative velocity of a to b is the velocity of a relative to the origin - velocity of b relative to the origin (or + the velocity of the origin relative to b). So just subtract the velocity vectors.

The relative velocity, therefore is \vec v_a + (-\vec v_b). Relative speed is the length of that vector, which is the hypotenuse of a right triangle with sides of length v: v_{rel}\sqrt{2v^2} = \sqrt{2}v.

AM
 
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So veclocity of a to b is usually just \vec v_a -\vec v_b.

So if we look at another example, ship a is traveling at 4c/5. ship b is traveling at 3c/5 in 1D motion. The relative velocity of ship a to b or what ship b thinks a is traveling is v_a -v_b = c/5.
 
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If the speeds are relativistic and in different directions, you need to use the vector form of the relativistic "velocity addition" formula. See the section "The velocity addition formula for non-parallel velocities" on the following page from the Usenet Physics FAQ:

http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html
 
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But if I am just doing velocity addition for different frames (not events in frame), i.e know what the other frame is traveling relative to me, do I still need the relativistic addition formula? If I do apply it than I could pretend the 'event' is a person standing still in the other frame. So the velocity of that person is 0 in their frame S'. I get the speed I see them is v the relative velocity of the frames. This value can be calculated with the method I suggested in #3?
 
pivoxa15 said:
But if I am just doing velocity addition for different frames (not events in frame), i.e know what the other frame is traveling relative to me, do I still need the relativistic addition formula?
Yes. The relativistic addition formula tells you how velocities transform from one frame to another, which is what you want.
 

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