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Two particles collide and form a new particle

  1. Jun 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Two particles of the same rest mass m have the same kinetic energy of 2.00*mc^2. They move in opposite directions and collide to form a new particle with mass M.
    a) What is the momentum of the new particle?
    b) What is its kinetic energy?
    c) Find the rest mass M of the new particle in terms of m.


    2. Relevant equations
    momentum=sqrt(total e.^2-rest e.^2)/c or p=mv

    I don't know if I can use total e.=k.e.+rest e. because it is a strange kinetic energy (2.00*mc^2) and I don't know how the new particle is formed...do you have add the two momenta from the two particles to get the momentum of the newly formed particle??? I don't know how to start...Thanks for your help and time
     
  2. jcsd
  3. Jun 9, 2009 #2

    Cyosis

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    This problem requires the use of special relativity, therefore [itex]p=\gamma m v[/itex]. Use both conservation of momentum and conservation of energy.

    There is nothing strange about the kinetic energy, and yes you've to add the two momenta, however keep in mind you're adding vectors here so don't forget the directions.
     
  4. Jun 9, 2009 #3
    This means that I first have to find out the momenta of the two particles by calculating first the velocity (2.00*mc^2=0.5*mass which I don't know*v^2) What do I have to insert in the 2.00*mc^2 part? Which nr. for mc^2 because I don't know if it is an electron, photon, proton etc. And also the mass...? When I would have the speed in the end I can calculate p=1/sqrt(1-v^2)*m*v. Then if I have a result x I can calculate the momentum of the new particle: x-x (x+(-x)) and this is the new momentum?!
     
  5. Jun 9, 2009 #4

    Cyosis

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    Well you could calculate the speed in terms of m, but that is not necessary. The problem states that they have the same rest mass and the same kinetic energy, but travel in opposite direction. So how do the two momenta relate to each other?
     
  6. Jun 9, 2009 #5
    They are the same except that one of them is negative...But what can I do know if I have this in "words" and not in numbers? Thx
     
  7. Jun 9, 2009 #6
    This doesn't make sense. If the two particles have the same rest mass and the kinetic energy, then their momenta are exact opposites (since it's a head on collision). It sounds like they are moving exactly into each other, meaning that the angle of seperation of the momentum vectors would sum to zero. The net momentum would be zero. Am I missing anything?

    Secondly, if you did have this exact scenario I describe, in order to conserve momentum, two particles would be shot out, not one. The two resulting particles would speed away from each other as to conserve net momentum of zero.
     
  8. Jun 10, 2009 #7

    Cyosis

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    The exercise states that they combine to form a new particle. If two cars speed towards each other with the same speed and collide head on, you would expect them to somewhat merge and not just bounce off each other and reverse directions. If they fully merge (a totally inelastic collision) I would also expect the wreck to be stationary. Momentum is conserved this way just fine.

    Well lets put it into equation form [itex]\vec{p_i}=\vec{p_f} \Rightarrow \vec{p_1}+\vec{p_2}=\vec{p_f}[/itex]. What do you know about [itex]\vec{p_f}[/itex] now?
     
    Last edited: Jun 10, 2009
  9. Jun 10, 2009 #8
    That pf is equal to p1+p2 but if I don't have p1 and p2 how can I calculate it? I have an exam tomorrow, can you just tell me how I have to calculate it..just give me the formulas and I'll put in the numbers so that I also did something :) Thanks, i'm really desperate
     
  10. Jun 10, 2009 #9
    Since when can a particle be completely stationary? If it has mass, then it's moving, otherwise, you would be the Nobel prize winner for finding absolute zero.
     
  11. Jun 10, 2009 #10

    Cyosis

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    Care to elaborate? If you mean that there is no absolute rest frame sure, but a particle can be at rest just fine with respect to a certain inertial frame.

    Yes and you already stated in a previous post that p1 and p2 are equal and opposite so what is pf?
     
  12. Jun 10, 2009 #11

    HallsofIvy

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    The total momentum, before and after the collision, is 0. Which tells you that their final velocity is 0. Now, what does conservation of mass-energy tell you about their total mass, M, after the collision?
     
  13. Jun 10, 2009 #12
    Okay, let's do some math.

    KE= 2mc^2
    E =ymc^2 (y = gamma = 1/[tex]\sqrt{1- v^2/c^2}[/tex]
    2= y = 1/[tex]\sqrt{1- v^2/c^2}[/tex] solve for v:

    sqrt{1- v^2/c^2}[/tex] (2) = 1
    (1- v^2/c^2) (4) = 1
    4(\frac{c^2-v^2}{c^2}) = 1
    \frac{4(c^2) - 4(v^2)}{c^2} = 1
    4(c^2) - 4(v^2) = c^2
    4(v^2) = 3(c^2)
    4/3 (v^2) = c^2
    1.3333 (v^2) = c^2
    1.1547v = c
    c/1.1547 = v
    v = 259,627,884.5 m/s

    Now we know the velocity of the original particles.
    But they are going head on into each other, so it doesn't end up helping that much, because the problem is unsolvable.
     
  14. Jun 10, 2009 #13
    Exactly. Exactly what I was saying above.
     
  15. Jun 10, 2009 #14

    Cyosis

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    This is not correct. You equated the kinetic energy to the total energy from the looks of it. Kinetic energy is total energy- rest energy. In formula form [itex]K=(\gamma-1)mc^2[/itex]. Secondly you don't have to calculate the velocities of the particles at all to solve this problem.
     
  16. Jun 10, 2009 #15
    But in order to have the same velocity measured for both particles that are going in complete opposite directions (in order to collide head on), you need to be in a stationary reference frame. Look at the other posts I and HallsofIvy have posted. Momentum must equal zero.
     
  17. Jun 10, 2009 #16
    My bad on the calculation part; I won't redo it since I now know that it was useless. I was trying to get the velocity to find the momentum somehow, but then again, you don't have and mass given to you, so it wouldn't have helped that much anyway I guess.
     
  18. Jun 10, 2009 #17

    Cyosis

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    And what would be problematic about being in a reference frame where you see both speeds as equal and opposite?

    Where have I ever said momentum does not equal zero. The problem with your statement is that from this you conclude there must be two particles flying in opposite directions, this is not correct.
     
  19. Jun 10, 2009 #18
    When two particles collide in a head on collision, the net momentum is zero. If one particle spewed out, then the momentum would be non-zero, since that particle has either energy or both energy and mass. This cause a problem, because momentum was not conserved.

    If you had two particles spew out, then they could be going in complete opposite directions. This means that even though the particles have either energy or both energy and mass, momentum is still conserved, because the net momentum would be zero.
     
  20. Jun 10, 2009 #19

    Cyosis

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    Spewed out? You pretend that there has to be a moving particle after the collision. According to the problem statement there is a single particle after the collision. Momentum is obviously conserved meaning that this single particle has zero momentum. There is no violation of conservation of momentum anywhere to be seen yet we only have one particle after the collision.

    Anyway if you're having problems with it I can solve it for you in a PM, this topic is to guide the OP not spew out a complete solution.
     
  21. Jun 10, 2009 #20
    I'm thinking that the OP's problem has a fallacy. If the momentum of the particle created after the collision has zero momentum, then that means one of two things. Either the particle has 0 velocity, or it has zero mass. If it has zero mass, then it must have eneryg, otherwise it wouldn't be a particle. In that case, you would have to use E=pc to find the momentum, which still comes out non-zero.

    If I'm right, then the OP is right to be confused about a problem like this, because it would have no solution. If I'm wrong, and there is one, and you said you can solve it, send me a personal message here so that the answer is not given away to the OP.
     
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