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Help in momentum

  1. Oct 18, 2003 #1
    help in momentum!!!!

    PLEASE HELP FAST!

    i have 1 question on a homework.. its killing me.. its dues in a couple of hours!! someone HELP!

    its 2 questions.. but i jus need the second one.. coz the seconds a part of the first..

    HELP!!

    1. A 4.73 g particle is moving at 3.2 m/s toward a stationary 10.2 g particle.

    With what speed does the heavier parti_ cle approach the center of mass of the two particles? Answer in units of m/s.

    2. What is the magnitude of the momentum of the lighter particle,relative to the center of mass? Answer in units of N s.

    If you guys dont help me out.. im gonna die

    HELP!! FAST!! I HAVE FEW HOURS LEFT!!

    --------------------------------------------------

    see i got the first part right.. its like..

    momentum = total mass * velocity .. ie m1v1+m2v2/(m1+m2)..

    but what about the 2nd part??

    I DONT GET IT
     
    Last edited by a moderator: Oct 18, 2003
  2. jcsd
  3. Oct 19, 2003 #2
    Re: help in momentum!!!!


    You should've asked sooner since all the smart people are
    asleep by now. You're liable to get bad advice this time
    of the morning. BTW, are you in a different country? In the
    U.S. there is no school tommorrow.
     
  4. Oct 19, 2003 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You got the first part right?
    "see i got the first part right.. its like..

    momentum = total mass * velocity .. ie m1v1+m2v2/(m1+m2).."

    The problem did not ask for momentum, it asked for "speed relative to the center of mass". What does that have to do with this?

    Do you know how to find center of mass? It is the point at which the system would balance. In particular, is x1 and x2 are measured from the center of mass to m1 and m2 respectively, then m1x1= m2x2.
    In particular, if we set up a "coordinate system" with origin at m2 (which is very nicely sitting still) with positive value extending toward m1, then x2= x, the coordinate of the center of mass. x1= d-x where d is the distance from m1 to m2. We then have m1(d-x)= m2x so that m1d- m1x= m2x and (m1+m2)x= m1d so x= m1d/(m1+ m2) (you probably have a formula much like that in your book. Since we are asked for the speed at which partical m1 is moving toward the center of mass, we differentiate (or divide by time to get speed if you haven't taken calculus) to get x/t= (m1/(m1+m2))d/t (m1 and m2 aren't changing- the only thing that gives "speed" is the fact that d, the distance between m1 and m2, is changing.
    The problem tells you that m1= 4.73 g, m2= 10.2 g and d/t, the speed at which the two particles are moving toward each other, is
    3.2 m/s. Plug in those values to find x/t, the speed with which EACH particle is approaching their mutual center of mass.

    Once you have found that, since momentum= mass times speed, just multiply the speed you just found by the mass of the lighter particle, 4.73 grams.

    But the problem asks you to answer in "N-s" or Newton-seconds(surely you didn't mean Newtons PLURAL- that's a measure of force, not momentum) so you will need to convert 4.73 grams to kilograms- A "Newton-second" is defined as the momentum of a one KILOGRAM mass moving at 1 meter/second.
     
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