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Inelastic momentum

  1. Jun 2, 2009 #1
    1. The problem statement, all variables and given/known data
    A car of mass 600kg traveling at 0.8c hits another, stationary car of 800kg. The two cars stick together. What is the mass and speed of the coupled cars?


    2. Relevant equations



    3. The attempt at a solution
    I found the momentum of the first car as p=gamma*mu = (600kg*0.8c)/sqrroot(1-0.8^2) = 2.4*10^11 kg.m/s

    The momentum of car 2 is zero.

    Then according to p=mu, i thought p(inital)=p(final)=m(final)*u(final)
    This would mean i would find the final velocity, u = p(initial)/m(final) = 2.4*10^11/(800+600)

    However i don't think this is right? If anyone could help me it would be great, thanks.
     
  2. jcsd
  3. Jun 2, 2009 #2

    Hao

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    Unfortunately, the only way to solve this is to use the relativistic equations for conservation of mass-energy and conservation of momentum, and solve algebracially for v and m (They will be complicated).

    Mass-Energy: [tex]\gamma_1 m_1 c^2 + m_2 c^2 = \gamma_3 m_3 c^2[/tex]
    Momentum: γ1 m1 v1 = γ3 m3 v3

    There are a few ways to simplify the calculations, but I don't think you have encountered them before:
    1) 4-vectors will find you [tex]m_3[/tex] directly.
    2) Using [tex]E^2 + p^2 c^2 = m^2 c^4[/tex] will find you [tex]m_3[/tex] directly as well.
    3) Transforming to the Zero-Momentum Frame will make the calculation trivial, but finding the Zero-Momentum Frame may be difficult.

    An intuitive way to approach this question is that the collision is inelastic, so some kinetic energy is converted to mass. Hence, the rest mass of m3 is not just m1+m2. Furthermore, there is a net momentum, so m3 must be moving. Hence, the gamma factor for m3 is not 1.
     
    Last edited: Jun 2, 2009
  4. Jun 2, 2009 #3

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi Skullmonkee! Hi Hao! :smile:

    Have a mu: µ and a gamma: γ

    and try using the X2 and X2 tags just above the Reply box :wink:
     
  5. Jun 2, 2009 #4
    Using four-momentum:
    (c=1, kg=1, b=beta, y=gamma)
    (P = four-momentum)
    (p = momentum)

    P1 = (p1, 0, 0, iE1)
    P2 = (0, 0, 0, im2)

    then using the information from the question:

    b1 = 0.8 -> y1 = 5/3 -> E1 = y1*m1 = 1000 -> p1 = sqrt(E1^2 - m1^2) = 800
    m2 = 800

    we get the four-momenta:

    P1 = (800, 0, 0, i1000)
    P2 = (0, 0, 0, i800)

    conservation of four-momentum:

    P3 = P1 + P2

    P3 = (800, 0, 0, i1800)

    m3 = sqrt(E3^2 - p3^2) = 1612 -> y3 = E3/m3 = 1.116 -> b3 = 0.44

    hope that helps :)


    ---
     
    Last edited by a moderator: Aug 6, 2009
  6. Jun 2, 2009 #5
    Thank you everyone. I think im getting there.

    b.szczesny: thanks for the great help but can i just ask why E1=y1*m1 = 1000 and not E1=y1*m1*c^2?

    I know that for the four-momenta ic can be written as i(E/c) but dosn't that result in i1000c not i1000?
    Im still learning 4-vectors
     
  7. Jun 2, 2009 #6

    Cyosis

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    Homework Helper

    At the start of his post he defined c=1. Therefore i1000c=i1000. That said he does use an antiquated version of four momentum. I personally find it easiest to use the invariance of the innerproduct of a 4-momentum vector. Let [itex]p_\mu=(E/c,\vec{p}), p_\mu p^\mu=-m^2c^2[/itex]

    In your case [itex]p_\mu=((\gamma_1 m_1 c+m_2c,\gamma_1 m_1 v,0,0)[/itex]. We know that the 4-momentum before the collision needs to be equal to the 4-momentum after the collision. Therefore [itex]p_\mu p^\mu=-(\gamma_1 m_1 c+m_2c)^2+(\gamma_1 m_1 v)^2=-m_3^2 c^2[/itex]. Now knowing that energy and momentum after the collision needs to be the same as prior to the collision you can calculate the speed of the composite object through [itex]\beta=pc/E[/itex]
     
    Last edited: Jun 2, 2009
  8. Jun 2, 2009 #7
    thanks for all the help. i learned a lot.
     
  9. Jun 2, 2009 #8
    Note that in classical physics you could not use conservation of energy. But in relativity the energy is conserved, because the kinetic energy that is dissipated in the form of heat, is accounted for by the rest mass.

    I agree with Cyosis that using invariance equations are simplest. You can simplify the calculations a bit more by using c = 1 units. If P is the total four-momentum, then we have in c = 1 units:

    P^2 = m^2

    where m is the mass after the collision and P the total momentum. We can write:

    P = P1 + P2

    P1 = m1 gamma1 (1, v1)

    P2 = m2(1,0)

    So:

    m^2 = (m1 gamma1 + m2)^2 - m1^2 gamma1^2 v1^2


    To find the speed, you use can use that v= momentum/energy (or that energy = gamma m and then obtain v from gamma):

    v = m1 gamma1 v1/[m1 gamma1 v1 + m2]
     
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