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I Momentum-Determining External Forces

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  1. Jan 31, 2017 #1
    I am studying momentum and I just want to check that I understand the idea correctly.

    Think theres a system.In this system theres two masses ##m_1## and ##m_2## moving with some velocity ##\vec v_1## and ##\vec v_2## and they exert a forces each other.Lets call the total force acting on ##m_1## is ##\vec F_1## and for ##m_2## is ##\vec F_2##.

    So If ##\vec F_1=\vec F^{ext}+\vec F_{21}##
    ##\vec F_2=\vec F^{ext}+\vec F_{12} ##

    Then the change in the total momentum;
    ##Δ\vec P=\vec {(F_{tot})^{ext}} Δt##

    If theres no exteral force then, ##Δ\vec P=0## .So If theres no external force then the change in momentum should be zero .If theres external force and its constant then the change in momentum will be constant but non-zero.##Δ\vec P≠0##

    If the change in momentum is zero(No external force) the center of mass of the system will be not moving.If theres external force it will accelerate by ##\frac {\vec {(F_{tot})^{ext}} } {M}=a_{com}## (##M## here total mass of the system)

    Just I am confused with the idea of external force...Think theres two object making projectile motion.and colliding in the air.Is the external force in here is gravity ? (No air drag)

    How can we determine the external forces in a given system ?

    Most collions happens in a small amount of time (##Δt## nearly ##0##), in this case can we say ##Δ\vec P=0## ? and If we can why ? (There will be external force)
     
    Last edited: Jan 31, 2017
  2. jcsd
  3. Jan 31, 2017 #2

    BvU

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    Think again: its momentum doesn't change. But it can well be nonzero.

    Correct. The reverse process: fireworks ! A rocket shoots off and explodes into a bunch of fireballs. Without air drag the center of mass simply continues the parabola the original rocket was following.

    For instance by using ##F = mg## when gravity is the only external force acting !
     
  4. Jan 31, 2017 #3
    So in any system The free body diagram will be showing us the external forces ?
     
  5. Jan 31, 2017 #4
    If we collide two objects in space.There would be no external force so the ##Δ\vec P=0## Even if theres external forces in small ##Δt## (the collusion time) from the equation of ##Δ\vec P=\vec {(F_{tot})^{ext}} Δt##

    ##Δ\vec P## should be zero
     
  6. Jan 31, 2017 #5
    ##Δ\vec V_{com}=0⇔Δ\vec P=0## ##Δ\vec V_{com}## can have any value bot the change must be zero
    Is this true ?
     
  7. Jan 31, 2017 #6

    BvU

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    Yes, that's about it. We restrict our considerations to constant mass cases, so ## {d\over dt} \left ( m\vec v \right ) = m {dv\over dt}##.
     
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