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Special Relativity -- Dynamics -- Energy

  1. Dec 8, 2015 #1
    1. The problem statement, all variables and given/known data
    Antihydrogen is the only antimatter element that has been produced in the laboratory, albeit just a few atoms at a time. Each antihydrogen atom consists of a positron in orbit around an antiproton and has the same atomic mass as hydrogen. If an antihydrogen atom collides with a hydrogen atom, they annihilate each other and create gamma radiation.

    If this energy could be harnessed in a matter-antimatter automobile engine, how far could a car travel on 10 mg each of antihydrogen and hydrogen? At highway speeds, a typical automobile expends about 2.5×10^3 J per meter.

    2. Relevant equations
    E = mc^2

    3. The attempt at a solution
    I know the answer is 9.00 * 10^8 m, but I haven't been able to work out an answer that is even close to this.

    I've solved the minimum amount of energy that is released in this process (3.01 * 10^-10 J). I'm not sure if this is relevant, but I solved it if necessary.
     
    Last edited: Dec 8, 2015
  2. jcsd
  3. Dec 8, 2015 #2
    Hi Barry:

    I think you omitted an important equation: E = mc2.

    Hope this helps.

    Regards,
    Buzz
     
  4. Dec 8, 2015 #3
    Okay, so if i use E = mc^2 with m = 10 mg then i get 9 * 10^12 J. Where do I go from here?
     
  5. Dec 8, 2015 #4
    Hi Barry:

    Look at the end of 1. Problem Statement. Also be careful about your units.

    Regards,
    Buzz
     
  6. Dec 8, 2015 #5
    I realize that you should be able to take E / 2.5 * 10^3 J/m , but this is incorrect.
     
  7. Dec 8, 2015 #6
    Hi Barry:

    What units are you using for c and m?

    Regards,
    Buzz
     
  8. Dec 8, 2015 #7
    c: m/s
    m: kg
     
  9. Dec 8, 2015 #8
    Hi Barry:

    What values are you using for c and m?

    Regards,
    Buzz
     
  10. Dec 8, 2015 #9

    c: 3.00 * 10^8 m/s
    m: 10 * 10^-5 kg
     
  11. Dec 8, 2015 #10
    Hi Barry:

    How are you converting 10 mg into kgs?

    Regards,
    Buzz
     
  12. Dec 8, 2015 #11
    Hi Barry:

    Also, what is the TOTAL input mass, hydrogen and anti-hydrogen?

    Regards,
    Buzz
     
  13. Dec 8, 2015 #12

    This is exactly what I'm doing:

    [(20 * 10^-6)(3 * 10^8)^2] / (2.5 * 10^3) = 720000000

    which is incorrect.
     
  14. Dec 8, 2015 #13
    i figured it out. thank you.
     
  15. Dec 8, 2015 #14
    Hi Barry:

    OK. You get the same answer I do, which is not the same as the answer you say you know is correct: 9.00 * 10^8 m. Where did this "correct" answer come from?

    Regards,
    Buzz
     
  16. Dec 8, 2015 #15
    Hi Barry:

    Glad to have been of some help.

    Regards,
    Buzz
     
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