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Finding momentum of a two-block system

  1. Apr 8, 2013 #1
    1. The problem statement, all variables and given/known data
    A system consists of a 3 kg block moving with velocity ‹ 14, 5, 0 › m/s and a 7 kg block moving with velocity ‹ −4, 6, 0 › m/s.

    (a) What is the momentum of this two-block system?

    (b) Next, due to interactions between the two blocks, each of their velocities change, but the two-block system is nearly isolated from the surroundings. Now what is the momentum of the two-block system?


    2. Relevant equations
    p(sys)=M(total)V(Centre of mass)


    3. The attempt at a solution
    I broke up the momentum into separate vector directions.
    x=(3+7)(14+ -4)
    x= 21*10
    x=210

    y=(3+7)(6+5)
    y= 21*11
    y=231

    There is no velocity in the z direction, therefore momentum = <210,231,0>
    And for part b, if the system is isolated, then even if the velocity changes, the total momentum of the system should remain constant. Therefore, the momentum should still be <210,231,0>

    I was sure my understanding was right. I am not sure why this answer is wrong. Is it a fundamental flaw in my understanding?

    [EDIT]: from information gained from this website, http://en.wikibooks.org/wiki/FHSST_Physics/Momentum/System

    I am given to understand that the total momentum in a system in the sum of the momentums in the system. In this case, the correct answer would be to find the vector momentum of each object and add them together.

    My question: how is the the total momentum? Is it because it takes into account the momentum of all objects inside the system, and as some parts of the momentum are negative or positive, it cancels those parts and leaves a weighted average of the momentums in the system? Thanks!
     
    Last edited: Apr 8, 2013
  2. jcsd
  3. Apr 8, 2013 #2

    haruspex

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    That's not the right equation. Figure out the contribution of each to the x component of the momentum separately, then add them together.
    3+7=21?!
    Exactly so.
     
  4. Apr 8, 2013 #3
    Thank you!

    I didn't even notice that. Just shows that I need to take more care in my problems when working. Thanks very much.
     
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