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Really general question regarding basic Momentum

  1. Jun 24, 2011 #1
    My instruction material keeps talking about how "The Law of Conservation of Momentum" is only applicable in isolated systems, without friction or other dissipative forces.

    However, in the same breath it will go on to say that Momentum is always conserved in collisions.

    Would someone mind describing this in a little more depth?

    And, while I'll surely read any materials you link me to on the matter, my entire course is sans human contact, so communicating would be rad :).
     
  2. jcsd
  3. Jun 24, 2011 #2
    momentum is always conserved in collisions. we do not consider dissipative forces here only because when we calculate speed or velocity to compute momentum we actually consider its value just before and just after collision. here friction does not come into play. if we do consider the speed before or after a time interval before or after collision and use it to compute the momentum to verify or use the conservation of momentum principle there should not be any frictional loss, such that the speed remains same untill the collision takes place. that is why the term 'without friction' is used.
     
  4. Jun 24, 2011 #3

    tiny-tim

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    Hi AJKing! :smile:
    Yes, https://www.physicsforums.com/library.php?do=view_item&itemid=53"

    (and https://www.physicsforums.com/library.php?do=view_item&itemid=313")


    This is because "collision" is always understood to mean that the change can be assumed to happen in an infinitesimally short time, and so the https://www.physicsforums.com/library.php?do=view_item&itemid=75" by any ordinary force (such as friction or gravity) can be assumed to be zero.​

    eg when two bodies collide in mid-air, there are no external impulses (and only one external force, that of gravity) … external to the two-body system, that is … so momentum in all directions (and also angular momentum) is conserved.

    when one body hits a fixed object, and bounces off, or rotates around it, the impulse from the fixed object is external to the one-body system, so the momentum of the body is not conserved, although the angular momentum about the point of contact will be conserved, since the impulse has no moment about that point :wink:
     
    Last edited by a moderator: Apr 26, 2017
  5. Jun 24, 2011 #4
    Yeah , if external force is neglected then total momentum before collision is same as after the collision

    m1u1+m2u2=m1v1+m2v2

    Or p1=-p2
     
    Last edited by a moderator: Apr 26, 2017
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