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Conservation of Linear Momentum

  1. Sep 1, 2013 #1
    In my Classical Dynamics book, I am reading about the topic alluded to in the title of this thread.
    Here is an excerpt that is provided me with confusion:

    "The total linear momentum [itex]\vec{p}[/itex] of a particle is conserve when the total force on it is zero

    Note that this result is derived from the vector equation [itex]\vec{p} = \vec(0}[/itex], and therefore applies for each component of the linear momentum. To state the result in other terms, we let [itex]\vec{s}[/itex] be some constant vector such that [itex]\vec{F} \cdot \vec{s} = \vec{0}[/itex]indepent of time. Then [itex]\vec{p} \cdot {s} = \vec{F} \cdot \vec{s} = \vec{0}[/itex] or, integrating with respect to time [itex]\vec{p} \cdot \vec{s} = c[/itex] which states that the component of linear momentum in the direction in which in the force vanishes is constant in time

    The part in bold is particularly confusing. Could someone help, please?
     
  2. jcsd
  3. Sep 1, 2013 #2

    BruceW

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    yeah, that is very weird. It looks almost like they are using ##\vec{s}## as an arbitrary vector. So in other words, imagine ##\vec{s}## is any arbitrary (but constant with time) vector, then ##\vec{F} \cdot \vec{s} = 0## Yeah, also, it should be 0 (a scalar) not a vector, since the dot product of two vectors is a scalar.

    I think there is something in the definition of a linear vector space that says if the dot product of a vector ##\vec{F}## with any arbitrary vector in the vector space ##\vec{s}## is zero, then the vector ##\vec{F}## must be the zero vector. (In other words, it is just another way of saying that ##\vec{F}## is the zero vector).
     
  4. Sep 1, 2013 #3
    What it is trying to say (not too well) is that the change in the component of momentum in the direction perpendicular to the applied force is zero.
     
  5. Sep 1, 2013 #4

    D H

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    You misread. Your book (apparently Marion & Thornton) says
    "To state the result in other terms, we let ##\vec s## be some constant vector such that ##\vec F \cdot \vec s = 0## independent of time. Then ##\dot{\vec p} \cdot \vec s=\vec F \cdot \vec s = 0## or, integrating with respect to time ##\vec p \cdot \vec s = c## which states that the component of linear momentum in the direction in which in the force vanishes is constant in time."​

    You missed the derivative of momentum, ##\dot{\vec p}##.
     
  6. Sep 1, 2013 #5

    BruceW

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    Ah, yeah. It could mean that. So if we choose a specific ##\vec{s}## for which ##\vec{F} \cdot \vec{s}=0## then this means for that specific vector ##\vec{s}##, we have: ##\vec{p} \cdot \vec{s} = c## (constant with time). So for example, if there are zero forces in the 'x' direction, then the momentum in the 'x' direction is constant.
     
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