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Relativistic Mass Exam Question

  1. Nov 10, 2018 #1
    2018-11-10 (5).png

    2. Relevant equations
    Relativistic mass equation given in the question

    The first part of the question:
    I understand 200 MeV is the energy lost as it initially moves with 200MeV however is brought to rest and thus this total kinetic energy must have been transferred from the 'two equal parts'
    For the worked solution below,
    2018-11-10 (4).png
    Q : I am confused on how the total energy E=mc^2 = 200 MeV as implied in the work solution.
    3. I thought
    Ek = mc^2 - moc^2

    does the worked solution instead mean that
    delta theta = m - mo
    and so delta theta = 200MeV/ c^2

    The second Part of the question: What speed did the two masses have?
    2018-11-10 (3).png
    I see the worked solution rearranges the relativistic mass equation to make v the subject, it does this by using the approximation m = mo as delta m is so small
    Q : however I am confused why they plug in 200*10^6 eV for delta m in the 2nd to last line of working shown.

    Attached Files:

    Last edited by a moderator: Nov 10, 2018
  2. jcsd
  3. Nov 10, 2018 #2


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    Homework Helper
    Gold Member

    If m and mo refer to the relativistic and rest masses of one of the fragments, then Δm would equal ½ ⋅ 200 Mev/c2. But then,
    mo½ (2.21 x 105 Mev/c2). If you use these values for Δm and mo in the expression
    then you can check that this gives the same result as the expression given in the solutions:
    upload_2018-11-10_18-18-16.png since the ½ factors cancel.

    Note, however, that evaluation does not give upload_2018-11-10_18-20-14.png . This is not the correct answer.
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