What if a particle travelled at c/3

[stunner5000pt]

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) what 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

 

[quantumdude]

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.

A) what 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 U_x'=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?

 

[stunner5000pt]

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??

 

[quantumdude]

I just edited the end of my post, so that it's clearer. Try the problem again?

 

[stunner5000pt]

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