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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


    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
    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))

    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


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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


    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


    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


    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


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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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