Physics Challenge #134

Discussion in 'Archived Threads 2001-2004' started by Bill Catherall, Nov 4, 2002.

  1. Bill Catherall

    Bill Catherall Screenwriter

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    So I'm studying for the Physics GRE and I came upon a question in the sample test that reads:
    A lump of clay whose rest mass is 4 kilograms is traveling at three-fifths the speed of light when it collides head-on with an identical lump going the opposite direction at the same speed. If the two lumps stick together and no energy is radiated away, what is the mass of the composite lump?
    (A) 4 kg
    (B) 6.4 kg
    (C) 8 kg
    (D) 10 kg
    (E) 13.3 kg
    Naturally I'd think the answer was (C), but the answer key gives (D) as the correct answer. Why? Obviously it has something to do with relativity. What's interesting is that the speed of the second lump relative to the first lump is NOT 6/5c because it's impossible to exceed c. Right? But what I don't understand is how it affects mass. It looks like we would some how be converting energy to mass in this collision, since we can't create mass. So how do you calculate how much energy is converted to mass, and how much mass we get out of it?
     
  2. Will Pomeroy

    Will Pomeroy Stunt Coordinator

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    I'm trying to remember exactly how, but i know its the theory of relativity, E=mc^2 (didn't think anyone ever used that ever, did ya?), but i'm not sure how it works, i'm trying to remember two years ago, to grade 12 physics, but when you have two obects moving at different speed, it gets more complicated... when you have an object with a mass, say 1kg, then you know that E=mc^2, so you know that you have 1x(2.99e+8)^2)J of energy, so mass is dirrectly related to energy, and then converty the two energies back into mass, or something like that!

    Good luck!
     
  3. Holadem

    Holadem Lead Actor

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    The kinetic energy of each lump before the colision:
    K = (1/2)mV^2
    v = (3/5)C ==> K = (18/25)C^2
    K total = 2 x K
    K total = (36/25)C^2
    K total is not radiated so it's gotta go somewhere ==> mass.
    Now I would think m = (36/25) by virtue of E = mC^2, but it ain't working out for some reason, since that would give a weight gain of 1.44Kg. Obviously I am missing something. HELP!!! [​IMG]
    --
    Holadem
     
  4. Pascal A

    Pascal A Second Unit

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    Bill, the key here is that you are provided with a rest mass, not a relativistic mass. Since the bodies are travelling near the speed of light, you need to calculate the relativistic mass of the final lump.



    m_relativistic = gamma * m_rest

    gamma = 1 / sqrt[1 - (v^2/c^2)]
    = 1 / sqrt[1 - (3/5c)^2/c^2)]
    = 1 / sqrt[1 - (3/5)^2]]
    = 1.25

    Therefore,

    m_relativistic = gamma * m_rest
    = 1.25 * m1_rest + m2_rest
    = 1.25 * 2 * m_rest = 2.5 * 4 = 10 kg.
     
  5. Michael Varacin

    Michael Varacin Stunt Coordinator

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    OK...I'm no rocket scientist here, but fter the two lumps collide, they have a velocity of 0. So the mass would be 8 kg. The speed of light and the realtivity principle is a non issue in the problem.
     
  6. Pascal A

    Pascal A Second Unit

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    Actually, Michael, in the relative mass equation that I used in my solution, the equation does reduce to rest mass on a Newtonian level (v
     
  7. Bill Catherall

    Bill Catherall Screenwriter

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    Thanks Pascal! [​IMG] I had no idea how to calculate relativistic mass. The physics classes I took in college didn't cover much relativity. I was an engineering student (and graduated almost 5 years ago), and I'm looking at going back and getting a physics masters degree (acoustics [​IMG] ), so I have to take the physics GRE. Tough stuff...it's been a while, and much of it I never learned.
     
  8. Pascal A

    Pascal A Second Unit

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    No problem. [​IMG] I had to retake the Physics GRE earlier this year to get back to school to work on my PhD in Astrophysics (they wouldn't accept my last one from 1993 [​IMG]), so I was slowly re-living the nightmare. [​IMG]
    By the way, I was also engineering (ME) before going the Physics grad school route, and other than needing to take a couple of "refresher" type classes, I wasn't as lost as I thought I would be. I must admit, I'm really enjoying getting student discounts again. [​IMG]
     
  9. Bill Catherall

    Bill Catherall Screenwriter

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    Thanks for the encouraging words. I was ME also. I've already spoken to some of the profs at my alma mater. They seem to think I won't have any trouble getting in, regardless of how I do on this test. It's more of a formality. They did say that since I was in engineering there are a couple classes I'd have to take that are geared towards getting engineers up to speed with the rest of the physics students. After studying last night I was dreaming in calculus. [​IMG] [​IMG]
    At this point I'm more worried about financial aid and how I'm going to afford this. [​IMG]
     
  10. Ted Lee

    Ted Lee Lead Actor

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    man....i can't even balance my checkbook.
    in all seriousness...i wish i could do this kind of stuff. color me impressed! [​IMG]
     
  11. Danny R

    Danny R Supporting Actor

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    OK...I'm no rocket scientist here, but after the two lumps collide, they have a velocity of 0. So the mass would be 8 kg. The speed of light and the realtivity principle is a non issue in the problem.

    Not quite. Remember one key component of the problem: no energy is released. Since the speed at which the two objects are travelling involves a great amount of energy, and that energy is not released upon impact, it thus has to be converted to mass. So the extra 2 kilograms of rest mass is created just from the velocity the two objects had prior to impact.

    What I would like to know is what form the extra 2 kilograms would take.
     
  12. BrianW

    BrianW Cinematographer

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    Green Play-Doh.

    Or maybe blue... I'll check my Halliday & Resnick and get back to you.
     

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