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Homework Help: Calculate KINETIC ENERGY of proton traveling close to speed of light

  1. Feb 2, 2010 #1
    1. The problem statement, all variables and given/known data

    A proton is traveling with a speed of v = 2.990x108 m/s. Calculate the (relativistically correct) value of the kinetic energy for this proton. Assume that the speed of light is c = 3.000 x 108 m/s and that the mass of the proton is 1.673 x 10-27 kg.

    Note: Do your calculations to 4 significant figures. Then round off to give an answer good to 3 significant figures. Be sure to include the correct abbreviation for the SI unit.


    2. Relevant equations

    K = 1/2 mv2

    Limit is K = 1/2mC2

    3. The attempt at a solution

    K=1/2(1.673x10-27)(2.990x108)
    K=2.50x10-19J

    Is this correct? I feel like I'm missing something here...
     
  2. jcsd
  3. Feb 2, 2010 #2
    O should I be using the equation:

    K =( g - 1 )MC2
    or
    K= (1/2)gMV2
    ?

    What is the correct formula for calculating kinetic energy when a particle's speed approaches the speed of light??
     
  4. Feb 3, 2010 #3
    you should use the relativistic equations to calculate the kinetic energy .. I suggest you to use :

    Ek = gama*m*c^2 - mc^2 , where gama= 1/sqrt(1-(v/c)^2 ) ..
     
  5. Feb 3, 2010 #4
    when a particle moves with very high velocity ( close to the speed of light )
    then the moving mass of the particle is given as :

    M = m/sqrt(1-(v^2/c^2))

    here,
    m is ths rest mass of the particle = 1.673X10^(-27)kg
    v is the velocity of the particle = 2.990X10^(8)m/s
    c is the speed of light = 3.00X10^(8)m/s

    using above formula the mass of the proton moving with the speed close to that of light can be calculated :

    M = 6.480X10^(-26)kg

    then we can obtain the kinectic energy of the proton as:

    KE = (1/2)M(v^2)

    KE = 2.914X10^(-19) joule

    please correct if m wrong!!
     
    Last edited: Feb 3, 2010
  6. Feb 3, 2010 #5
    I dont recall doing this when I used to take that course .. I just substitute in the formula I wrote
     
    Last edited: Feb 3, 2010
  7. Feb 3, 2010 #6

    vela

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    That's wrong. You should abandon the notion of relativistic mass because it leads to errors like this.

    In special relativity, the energy of an object of rest mass m is given by

    [tex]E=\frac{mc^2}{\sqrt{1-(v/c)^2}}[/tex]

    When v=0, you get E=mc2. This is the rest energy of the object. The object's kinetic energy is the energy in excess of the rest energy, so it's given by

    [tex]K=E-mc^2[/tex]

    If you express the total energy E as a series in (v/c)2, you get

    [tex]E=mc^2\left[1+\frac{1}{2}\left(\frac{v}{c}\right)^2+...\right]\cong mc^2+\frac{1}{2}mv^2+...[/tex]

    When the speed of the object is much smaller than the speed of light, you can neglect the higher-order terms and recover the Newtonian expression for kinetic energy

    [tex]K=\frac{1}{2}mv^2[/tex]

    To get this result, we assumed v<<c, so this formula is only good when the speeds are not a significant fraction of the speed of light. In particular, you can not use it in this problem.
     
  8. Feb 3, 2010 #7
    then what should be the kinetic energy for such problem!!
     
  9. Feb 3, 2010 #8

    vela

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    I clearly said what it was in my previous post.
     
  10. Feb 3, 2010 #9
    ohh ya ..thanx
     
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