What if a particle travelled at c/3

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In summary, a particle is created at t = 100ns, travels with velocity Ux = -c/3, and decays at t = 300ns. To an observer moving at -3/5 c along the x-x' axis, the particle's velocity during its trip is Ux' = 7/9 c. The distance between the positions of creation and decay can be calculated using D = Vt = 7c/9 x 250ns. The time elapsed between creation and decay is 250ns. The fact that both observers in different frames get the same answers for distance and time does not contradict relativity, as the effects of time dilation and length contraction occur for both observers due to their
  • #1
stunner5000pt
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A particle is observed to be created at t = 100ns , travel with speed Ux = -c/3 and then decay at t = 300ns. To an observer who was moving at speed of -3/5 c along the common x - x' axis:

A) waht was the velocity of the particle during its trip?

Ux' = Ux - V / 1 - VUx/c^2

however this time they are moving in the same direction so the numerator and denominator becomes plus signs and i get 13/18 c

B) what distance between the position where the particle was created and the position at which it decayed?

Ok since moving clocks move slower, the observer moving will see the 200ns actually take 200 times gamma = 250ns to complete

Since D = Vt = 13c / 18 times 250ns

C) How much time elapsed between the instant it was creaed and the instant it decayed?
I think i answered it in B = 250ns

D) How can it be that the observers in two different frames namely S and S' get the same answers for parts b and c? Desnt this contradict relaitivity? (i.e. what happened to length contraction and time dilation?)
I don't understand what frame is S and S' here but I would think since in either frame the observer thinks that his frame is at rest but the world moves past him in one direction or the other. So at relative speed length contraction and tiem dilation occur for both sides - an observer would se time somehwhere else move slower, while the observer from somewhere else would see the original observer's time move slower too.

pease tell me if i am correct in what i have done so far...

your assitance is greatly appreciated
 
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  • #2
Stunner,

Please stop posting links to your threads in other Forums. It is considered spam. The same goes for posting to your own threads over and over to illicit a response. I know your assigment is due tomorrow, but you can't expect instant service here. If you are having trouble with your homework, then be responsible, plan accordingly, and post your questions before it gets to this.

stunner5000pt said:
A) waht was the velocity of the particle during its trip?

Ux' = Ux - V / 1 - VUx/c^2

however this time they are moving in the same direction so the numerator and denominator becomes plus signs and i get 13/18 c

Check your math. I get Ux'=c/3.

B) what distance between the position where the particle was created and the position at which it decayed?

Ok since moving clocks move slower, the observer moving will see the 200ns actually take 200 times gamma = 250ns to complete

Since D = Vt = 13c / 18 times 250ns

You'll have to fix this, based on the error in Part A.

As for the last question, ponder the new result for Part A. Since relativistic effects arise due to the relative motion of two bodies, does it not stand to reason that the observed times are the same, given that the observed particle speeds are the same?
 
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  • #3
i'm sorry about that ... i should have planned earlier... alas

thank you for answering though

i got the answer for the time using 200 times gamma which is 200 x 1.25 = 250 ns but i guess tha would be wrong??
 
  • #4
I just edited the end of my post, so that it's clearer. Try the problem again?
 
  • #5
Tom Mattson said:
I just edited the end of my post, so that it's clearer. Try the problem again?

i checked my math and i get 7/8 c

perhaps the signs which i explained earlier are wrong?

Ux' = Ux - V / 1 - VUx/c^2

however they are moving in the same direction so the numerator and denominator becomes plus signs and i get 14/18 c = 7 /9 c
 

1. What is the significance of a particle travelling at c/3?

The speed of light, denoted by c, is a fundamental constant in the universe and is considered to be the fastest possible speed for anything with mass. Therefore, a particle travelling at c/3 is moving at one-third of the speed of light, which is still incredibly fast and has significant implications in the field of physics.

2. How does the speed of a particle affect its mass?

According to Einstein's theory of relativity, as a particle approaches the speed of light, its mass increases significantly. This means that a particle travelling at c/3 would have a greater mass than the same particle at rest.

3. Can a particle actually travel at c/3?

Based on our current understanding of physics, it is not possible for a particle with mass to reach the speed of light. This is because as the particle's speed approaches c, its mass and energy also approach infinity, making it impossible to accelerate any further.

4. How does the concept of time dilation apply to a particle travelling at c/3?

According to the theory of relativity, time appears to slow down for objects moving at high speeds. This means that for a particle travelling at c/3, time would appear to pass slower for the particle than for an observer at rest. This phenomenon is known as time dilation.

5. What are the practical applications of understanding the speed of particles?

The study of particle speed and relativity has led to many groundbreaking discoveries and technological advancements. It has helped us understand the fundamental laws of the universe and has led to developments in fields such as space travel, nuclear energy, and telecommunications.

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