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Constant momentum of an accelerated body

  1. Apr 3, 2014 #1
    Momentum of a body can be constant while it accelerates? I mean if velocity increases while mass decreases proportional.
    And if is true, what force produces acceleration?
     
  2. jcsd
  3. Apr 3, 2014 #2

    Dale

    Staff: Mentor

    That is an interesting idea.

    $$p=mv$$
    $$\frac{dp}{dt}=m\frac{dv}{dt}+v\frac{dm}{dt}$$

    So to keep momentum constant would require mass to change as:
    $$\frac{dm}{dt}=-\frac{ma}{v}$$
     
  4. Apr 3, 2014 #3
    Yes, but what produces that acceleration?
     
  5. Apr 3, 2014 #4

    Dale

    Staff: Mentor

    Any force would be fine. Have you studied Newton's 2nd law: ##\Sigma f = m a##
     
  6. Apr 3, 2014 #5

    D H

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    Staff Emeritus
    Science Advisor

    An external force could cause it, but an external force is not needed. That acceleration could come from a variable specific impulse rocket. Consider a rocket in deep space, far removed from any external forces. The rate at which the rocket's momentum changes is ##\dot p = \dot m (v-v_e)##. So all we need to do to keep the momentum constant is to keep increasing the exhaust velocity ##v_e## in tune with the rocket's velocity ##v##.
     
  7. Apr 3, 2014 #6

    A.T.

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    Science Advisor
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    This is kind of like shooting backwards while moving, so the projectile has zero final momentum. Therefore the empty cannon cannot have lost/gained any momentum compared to the loaded cannon. The cannon's gain in velocity is canceled by the loss of the mass of the projectile.

    https://www.youtube.com/watch?v=BLuI118nhzc
     
    Last edited: Apr 4, 2014
  8. Apr 3, 2014 #7
    In my opinion is a misconception that the initial momentum of a body can be constant while it accelerates.
    Let be initial momentun of the body [itex]p_1=m_1v_1[/itex]. The body splites in two bodies, with the momentum [itex]p_2[/itex] and [itex]p_3[/itex], so that [itex]p_2+p_3=p_1[/itex].
    If we put the condition [itex]p_1=p_2[/itex] = constant, then we got [itex]p_3=0[/itex].
    But we know to produce an acceleration we need a force [itex]F=\frac{dp}{dt}[/itex]. If [itex]dp=0[/itex] (like [itex]p_1=p_2[/itex]), there is no force and no acceleration.
     
  9. Apr 3, 2014 #8

    Dale

    Staff: Mentor

    It would be very unusual, but certainly not impossible provided the conditions above are met.

    Which is exactly what D H said in post 5 and A.T. said in post 6.

    This equation is only correct if ##\dot{m}=0##
     
  10. Apr 5, 2014 #9
    I think it is required to establish that the claim "momentum of a body can be constant while it accelerates" is valid for a system of bodies and not for the same body, because initial body splits in other bodies, as it accelerates.
     
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