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Partical Mass - Energy - Quantum Dynamics - Distance

  1. Sep 5, 2009 #1
    1. The problem statement, all variables and given/known data

    This is the question.

    As an appreciation for his excellent contributions towards making Universiti Sains Malaysia an APEX university, Bala is given the opportunity to do an experiment at the Stanford particle accelerator, USA. During one of his experiments, he detected the creation of an  particle. From his calculation, the  particle has a rest energy 1672 MeV and total energy 2330 MeV.

    (a) He wonders: is the  particle a particle without mass? Help him by explaining your answer.

    (b) If the  particle decays and leaves a track 24 mm long, what is the

    (i) speed of the particle while making the imprint on the track?

    (ii) momentum of the particle?

    (c) He wrote a note in his laboratory book:

    mean lifetime of  particle = 8.2 x 1011 s

    Is this value correct? Justify your answer.

    2. Relevant equations
    This is the formula that i thought

    E = mc2

    E = (1/2)mu

    p = mu

    v = d/t

    a = v/t

    3. The attempt at a solution

    i've 1st and 2nd attempt at this cases. For question 1)b)i)

    -- get the Ekinetic with Etotal = Erest + Ekinetic

    -- then, get the mass using E=mc2.

    -- then, i get the velocity using Ekinetic = (1/2)mv

    -- then, i get the time using v = d/t.

    -- the data is : Ekinetic = 658 MeV, Mass =7.31 x 10 power of -17 g, V = 1.8 x 10 power of 17 m/s, t = 1.33 x 10 power of -19 s.

    -- lastly using the a = v/t is 1.35 x 10 power of 36 m/s

    for 1)b)ii)

    using the p=mu i get the momentum 13.158

    for 1)c) lifetime using lifetime = P/triangle(t)

    = then using smooth T = 1/r

    i get 1.33 x 10 power of -19s.



    LASTLY, i think this is not the best solution for the question, i think i miss something? i wonder if my answer right or wrong/// anybody please>>>?
     
  2. jcsd
  3. Sep 6, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi mahdzer! Welcome to PF! :smile:
    No, Ekinetic = mc2(1/√(1 - u2/c2) - 1) [or approx (1/2)mu2]
     
  4. Sep 6, 2009 #3
    No, Ekinetic = mc2(1/√(1 - u2/c2) - 1) [or approx (1/2)mu2]

    this is the formula of partical without mass isn't?

    but how can i prove this partical is partical without mass?
     
  5. Sep 6, 2009 #4
  6. Sep 7, 2009 #5

    tiny-tim

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    (btw, it's spelled "particle" :wink:)

    No, Ekinetic = mc2(1/√(1 - u2/c2) - 1) is the standard formula for KE for mass m.

    (for a particle without mass, the KE is the same as the total energy, since obviously there is no rest mass: in other words, m = 0)

    Where are you copying your formulas from? :confused:
    I don't follow you :redface: … which post are you referring to?
     
  7. Sep 7, 2009 #6
    so that for partical without mass, the KE is the same as total energy, Wallah! i think i can get the idea right here... and this formula just a basic formula, from the book i read. Nevertheless, our modul is too huge, yet i can't find the Partical Without Mass explanation in detail... thanks anyway... i think my solution would be for the Q1 is -

    = no, the partical that Bala examined is a Mass because the KE is not the same as total E, from the formula

    Ekinetic = mc2(1/√(1 - u2/c2) - 1) - then i think is this answer the Q1a)

    - i'm sorry i just start 1st semester Bachelor at Malaysia Science University for this Q, and i've only have Multimedia IT Background b4. So, i'm lack of Physics Knowlege, thanks 4 helping me... i just learn little basics Physics in form 4 and form5 at Secondary School, its long2 time ago.

    p:s/ how do i the symbol c2 right and get in the keyboard rite for c the power of 2?
     
  8. Sep 7, 2009 #7

    tiny-tim

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    Hi mahdzer! :smile:

    Just remember the formulas energy = m/√(1 - v2/c2), momentum = (v/c) times energy …

    so, obviously, if v/c = 1, then momentum = energy (and m = 0).

    btw, looking at the question again, i see that the question doesn't mention kinetic energy … you introduced kinetic energy for some reason.

    Kinetic energy is not a helpful idea in relativity: it's much better only to use rest energy (m) and total energy, as the question does. :wink:

    (to produce c2, either [noparse]type "c2", or click the [/noparse]QUOTE button at the bottom of any post, to take you to the reply page, where you'll see a lot of icons above the reply field, and clicking on the X2 icon will raise whatever you type next … eg c2n :wink:)​
     
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