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Confused about conservation of momentum

  1. May 24, 2009 #1
    I’m confused about conservation of momentum. I remember learning that the momentum before a collision is the same as after it:

    Pinitial = Pfinal .

    So I do that for a mass M1 colliding with another mass M2 that’s initially at rest,

    M1 x V1initial + 0 = M1 x V1final + M2 x V2final

    and I get an equation for the velocity V2final ,

    V2final = (M1/M2) x (V1initial - V1final) .

    Now let’s say that M2 is so large that in the limit it becomes infinite. In that case, V2final = 0, so that

    V1final = V1initial .

    But since M1 bounces off of M2 , shouldn't V1final and V1initial be pointing in opposite directions?
     
  2. jcsd
  3. May 24, 2009 #2

    atyy

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    Take M2 large but finite, then take M2 larger and larger.
     
  4. May 24, 2009 #3
    ??

    In the limit, unless V1initial - V1final is also "large but finite," and grows that way with increasing M2 , V2final = 0 .
     
  5. May 24, 2009 #4

    Doc Al

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    Staff: Mentor

    OK.

    OK.
    That doesn't follow. You cannot deduce anything about the difference between the initial and final speeds of M1, since you are essentially multiplying that difference by zero.

    Yes, if the collision is elastic. Use conservation of energy to deduce that the initial and final speeds are the same. And the fact that M1 doesn't pass through M2..
     
  6. May 24, 2009 #5
    I assume that what you are doing is to put the result [itex]v_{2final}=0[/itex] back into your first equation and then set [itex]m_2 v_{2final}=0[/itex] but that is not correct since you have taken the mass to be infinitely large you have something of the form [itex]\infty \times 0[/itex] which is not necessarily zero. In fact if you keep the expression for [itex]v_{2final}[/itex] and plug it into your first equation you get [itex]v_{1init}=v_{1init}[/itex] which is not very helpful :)

    The problem is that you have 2 unknown variables [itex]v_{1final}[/itex] and [itex]v_{2final}[/itex] and only one equation. You need to also consider conservation of energy in order to solve the equations.


    EDIT: Ooops Doc Al was faster...
     
  7. May 24, 2009 #6

    Dale

    Staff: Mentor

    Not necessarily. From the initial conditions there are two unknowns, the two final velocities. So you have one equation and two unknowns. You need another piece of information in order to have a fully-determined system. As Doc Al mentioned, the usual additional piece of information is either a specification of the collision as perfectly elastic (KE also conserved) or perfectly plastic (final velocities of each object are equal).
     
  8. May 24, 2009 #7
    You need to conserve momentum separately in every degree of freedom, so Px, Py, and Pz are separately conserved. Visualize an asteroid bouncing off the Earth. More than half the off-center asteroid impacts will result in a forward final asteroid velocity.
     
  9. May 24, 2009 #8

    atyy

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    magnitude of V2final=some smaller and smaller number, not zero.

    Actually, I'm not sure if it's a strict limit, or just an approximation. But you can't divide by infinity to get zero if you attempt to do it "strictly". So solve everything with finite M2, and only take the limit/approx (?) M2 >> M1 in the final step.
     
  10. May 24, 2009 #9

    Doc Al

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    True, but I suspect that the OP was thinking of a head-on collision in one dimension.
     
  11. May 24, 2009 #10
    It must be
    P(initial)+P(final)=o
    P(initial)=-P(final)
    M1.V(1initial)+M2.V(2initial)= -(M1.V(1final)+M2.V2(final))
    M1.V(1initial)+o = -(M1.V(1final)+M2.V(2final))
    M1.V(1initial)+M1.V(1final)= -M2.V(2final)
    M1/M2(V(1initial)+V(1final))= -V(2final)
    if M2 is infinitally large, V(2final)=o
    that means V(initial)+V(1final)=0
    V(initial) = -V(1final)
    Velocity in opposite direction
     
    Last edited: May 24, 2009
  12. May 24, 2009 #11
    V2 will be infinitely small, so result of momentum will be transfered to m1 in the opposite direction.

    V2final = (M1/M2) x (V1initial - V1final) .

    If m2 becomes infinite in this formula, then v2 final will be 0.

    And v1 final will be -

    v2final (M2/M1) - v1inital = -v1final

    That way v1final = infinite :P
     
  13. May 25, 2009 #12

    Doc Al

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    That's incorrect. Conservation of momentum says:
    P(initial) = P(final)

    This is an incorrect conclusion, as already pointed out several times in this thread.
     
  14. May 25, 2009 #13

    Doc Al

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    Wise guy. :smile: :tongue:
     
  15. May 25, 2009 #14
    Thanks for telling my math errors,I appologise.It was math deleimma in my mind at instant
     
  16. May 25, 2009 #15
    cool Doc Al..you are superb
     
  17. May 25, 2009 #16
    How correct of you...and dumb of me. (Thanks!)

    The "pass through" implication never occurred to me. Very nice!
     
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