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How exactly does momentum work?

  1. Dec 18, 2013 #1
    How exactly does momentum work? I'll use an arrow flying off a bow as an example. Does momentum increase the second it is released or does it actually increase as it flies through the air after a certain range? If because of the second part, is it due to friction from the air?
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  3. Dec 18, 2013 #2


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    Momentum is a function of an objects mass times its velocity. In other words, a moving object simply has momentum. It gains momentum the moment it accelerates, which is the moment the string is let go. As it accelerates, its velocity and momentum both increase. It has nothing to do with friction.
  4. Dec 18, 2013 #3


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    For the specific case of linear momentum, you can look at Newton's second law, which states that ##\dot{\boldsymbol p} = {\boldsymbol F}##, where ##\dot{\boldsymbol p}## is the continuous time derivative of the linear momentum ##{\boldsymbol p}## and ##{\boldsymbol F}(t)## is a continuous and generally time-variant forcing function.

    This means that, for a given time ##t##, the system has a momentum ##{\boldsymbol p}(t)## at that time instance. Indeed, you can go further and use the momentum as a state of the system, e.g. you work with linear and angular momenta (and their time-derivatives) as your variables instead of traditional velocity and acceleration kinematic variables.

    Momenta are defined quantities; for linear momentum we have the familiar[tex]{\boldsymbol p} \equiv m{\boldsymbol v}[/tex]and, in one case, we have the angular momentum about a fixed point ##o##[tex]{\boldsymbol h}_o \equiv {\boldsymbol r} \times m{\boldsymbol v}.[/tex]We can also define absolute and relative angular momentums about an arbitrary point. (Note that in physics ##{\boldsymbol L}## is generally used as the symbol for angular momentum.) These quantities all give instantaneous momentums, e.g. the momentum has a certain value for a given system configuration at that time instance.

    For the second part of your question, let's look at the example of air resistance with a point mass model. Suppose, for simplicity, that the only force on the system is air resistance, and let it be modeled as ##{\boldsymbol f} = -cv{\boldsymbol v}##, where ##c## is assumed constant (generally it is a function of atmospheric density, projected area, and a dimensionless drag coefficient), and ##v = \|{\boldsymbol v}\|_2##. This then gives us a differential equation[tex]\dot{\boldsymbol p} = -cv{\boldsymbol v}[/tex]which, if we know the initial state of the system, we can integrate numerically and observe time histories of the system states. (If we use the momentum as our state, we can obtain ##{\boldsymbol v}## at each time step by inverting the momentum-velocity relationship.)
    Last edited: Dec 18, 2013
  5. Dec 19, 2013 #4
    Thank you.
  6. Dec 19, 2013 #5


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    Momentum only changes whilst a force is being applied. That's Newton's First Law of motion and is what you should apply in these sorts of classical problems.
    The arrow accelerates whilst the bow string is actually pushing it. (It gains Momentum from the Bow.)
    It then loses Momentum to the air particles (friction slows it down) on its way to the target. On impact, it will share its momentum with the target (and the ground that the target is standing on.)
    Momentum (Mass times Velocity) is one of the fundamental quantities in Science and it is a term that is often mis-used. This often gets in the way of understanding. Also, 'Inertia' is a word which is often used as a synonym for Momentum but 'Inertia' can also refer to Mass - so it is even more confusing when people try to explain things using that very loosely defined term. Momentum is very tightly defined, though.
  7. Dec 19, 2013 #6
    Thank you.
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