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GR0877 Problem 24

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (B) To the observer, a meter stick of length l_0 which is passing by with a speed v = 0.8c is observed (sorry for tautology) to be contracted by a factor of \displaystyle \gamma = 1/\sqrt{1 - (0.8)^2} = \frac{5}{3}. Thus, the time it takes the stick to pass the observer is \displaystyle \Delta t = \frac{l_0}{\gamma v} = \frac{3}{5 \cdot 0.8 \cdot 3 \cdot 10^8} = 2.5 (ns).

Found a typo? Comment!

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  1. Someone Dummer Than You
    11.10.2011 at 09:49

    On the last line you are missing a 3 in the denominator from the speed of light, you have it as only 10^8m/s, as it is that expressions resolves to 7.5 ns. Also, you should be more clear/explicit about how you set up your equations. Don’t assume everyone is coming from the same set of background knowledge as you: for example, contracted by a factor of 5/3 doesn’t make (common) sense, how can a factor>1 contract (i.e. make smaller) anything? You should explicitly write out: l_rest=gamma*l_new

    • 11.10.2011 at 12:51

      Thanks for the typo! It has been fixed.
      As for “common sense”… 🙂 Well, look at the solution to the problem #23. There I’ve explicitly wrote what I (and other people) mean by saying that a stick is contracted by a factor of \gamma. I don’t think it’s worthwhile to mention this every time you’ve got such a factor in the solution.

  2. Someone Dummer Than You
    11.10.2011 at 22:41

    <DELETED by @physicsworks. For reasons see my comment >

    Thanks for helping with this one though, I was stymied for I don’t know how long because I couldn’t see the forest for the trees: ETS wanted time so I kept trying to use the time dilation equation to no avail….needed to to use l_new/velocity instead, as you’ve done above….

  3. 25.06.2016 at 09:51

    The meter stick has a length L_0=1m and its speed is v=0.8c. By using the law of relative length we can find the the relative length L and substitute it in the law of velocity to find the time.
    \displaystyle \L = 1/\sqrt{1 – frac{{0.8c}{c}}^2} = \cdot 0.6
    \displaystyle \t = \frac{L}{v} = \frac{0.6}{0.8 \cdot 3 \cdot 10^8} = 2.5n

    • 25.06.2016 at 09:59

      SORRY for the bad typing above.

      The meter stick has a length L0=1m and its speed is v=0.8c. By using the law of relative length we can find the the relative length L and substitute it in the law of normal velocity to find the time.

      L = L0*sqrt{1-(0.8c/c)^2}^0.5 = 1*sqrt{1-0.8^2}^0.5 = 0.6

      t = L/v = 0.6/0.8c = 0.6/(0.8*3*10^8) = 2.5ns

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