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Homework Help: Special Theory of Relativity

  1. Oct 31, 2007 #1
    Thanks for taking a look. The following question is that is that from a grade 12 academic physics course.

    1. The problem statement, all variables and given/known data

    Scientist Ludwig von Drake, while in his laboratory, measures the half-life of some radioactive material which is in a bomb, approaching with speed v. Donald Duck, who is riding on the bomb, also measures the half-life. His answer is a factor of 2 smaller than Ludwig's. What is the value of v, expressed as a fraction of c?

    Answer: .87

    2. Relevant equations

    Special relativity equation: t=t0/square root (1-v2/c2) Note: 2 = squared
    Other: I know there is at least one more I must use but for the life of me don't know what it is.

    3. The attempt at a solution

    First we set V=Drakes. If this is so than Donald's equation must be equal to:

    t=[t0/square root (1-v2/c2)]/2

    Therefore to= 2t[square root (1-v2/c2)]

    The problem is in equating the equations. They end up cancelling out because one is a direct derivative of the other. This leads me to believe I need at least 1 more equation.

    It must also be noted that while von Drake may use the special equation of relativity Donald duck cannot (at least this is what I figure). I say this because Donald duck is viewing the half-life from the bomb at rest, meaning he would be more in the realm of inertial frame of reference. What equation i now use knowing that I have no idea.

    Thanks in Advance guys.
  2. jcsd
  3. Oct 31, 2007 #2


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    t=t0/square root (1-v2/c2) (this is the time dilation formula)

    t is Drake's time

    t0 is the proper time which is Donald's time.

    we are given that t = 2t0.

    so you have equations:

    t=t0/square root (1-v2/c2)
    t = 2t0

    so solve these 2 equations.
  4. Oct 31, 2007 #3
    Donal Duck is measuring the proper time interval [tex]t_0[/tex], so the interval measured by Drake is [tex]t=t_o\lambda[/tex]. We're given that [tex]t=2t_o[/tex]. Dividing both equations we get that [tex]\lambda=(1-\frac{v^2}{c^2})^{-1/2}=2[/tex]. Solving the equation yields the solution.
  5. Nov 1, 2007 #4
    Thanks guys. My solution set is similar to both of yours.

    I used 2T= To/sqreroot(1-v2/c2) and 2T = To/x where x=.5

    This yields sqreroot(1-v2/c2) = .5

    Through squaring both sides and moving the variables/numbers around you obtain


    Simply square root that to get the answer which is .866c=v

    Once again thanks for your help!
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