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Linear momentum as a vector - help

  1. Sep 28, 2006 #1
    hi

    I'm really stuck and I hope you can help me.


    I know that linear momentum is a vector, but I'm still not sure how to add them together when :

    p1, p2, p ... linear momentum

    If two objects A1 ( p1 = 100) and A2 ( p2 = -100) represent an isolated system and are moving along ( parallel to ) x-axis ( A1 in positive and A2 in negative direction ), then total linear momentum p is zero. But what if A1 is still moving along the x axis, but this time function of object A1 is y(x) = 3, while before it was y(x) = 0. Then even though linear momentum vectors p1 and p2 are still parallel, they no longer lie on the same line. Can we still add vectors p1 and p2 as if they were lying on the same line? If yes, why? If not, why not?



    thank you
     
  2. jcsd
  3. Sep 28, 2006 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, remember that vectors can be "moved" (the mathematician in me cringes at that) and the total momentum of the system can be calculated as the sum of the two objects even if they are not at the same point (strictly speaking two distinct objects can never be at the same spot!).
     
  4. Sep 29, 2006 #3


    I know vectors can be 'moved', but why is this allowed with momentum vectors? Clearly, a system containing two objects with their veclocity vectors lying on same line differs ( somehow ) from system with velocity vectors parallel, but not lying on the same line. Why would we choose to neglect the fact that vectors ain't lying on the same line?
     
  5. Sep 29, 2006 #4

    Astronuc

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    Staff: Mentor

    We are not neglecting that fact.

    Momentum is proportional to velocity, which is a vector.

    Let [itex]\vec{v_1}\,=\,v_{1x}\hat{x}\,+\,\,v_{1y}\hat{y} [/itex],

    and [itex]\vec{v_2}\,=\,v_{2x}\hat{x}\,+\,\,v_{2y}\hat{y} [/itex]

    the

    [itex]\vec{v_1}\,+\,\vec{v_2}\,=\,(v_{1x}\,+\,v_{2x})\hat{x}\,+\,(v_{1y}\,+\,v_{2y})\hat{y}[/itex].

    If there is no velocity in the y direction, or if v1y = -v2y, then the sum is obviously zero, and there is not net momemtum in that direction. Adding these vectors does not involve the positions of the objects with those velocities.

    Now there is a property called angular momentum, which does involve the position or location of objects. Colinear particles with the same magnitude but opposite direction would have zero net linear momentum and zero net angular momentum, but non-colinear forces while having no net linear momentum would have a net angular momentum for the same velocity conditions.
     
    Last edited: Sep 29, 2006
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