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Momentum=Newton second?

  1. Dec 14, 2009 #1
    Hello!
    if we write momentum of Ns, we drive 1kg * m / s and if we write Newton's second law of unity that is the same, (in N.s=newton*seconds), So 1Kg*m/s^2 *s=1Kg*m/s which is the correct unit for momentum . It does not mean that momentum is the same as Newton's second law?
     
    Last edited: Dec 14, 2009
  2. jcsd
  3. Dec 14, 2009 #2
    it's the same thing when you want to work, therefore, energy. it can express in different ways but ultimately it is the same;
    [tex]mgh=energy\\ [/tex]

    [tex]\frac{mv^2}{2}=energy\\ [/tex]
    [tex]F\cdot \Delta s=energy(=work)[/tex]

    so what is the difference between Newton's second law and the momentum?
     
  4. Dec 14, 2009 #3
    The difference between Newton's second law and change in momentum over time, more precisely the time derivative of momentum, is very subtle and negligible in most cases. In fact, Newton's second law is only generally correct. Force is always defined as a time derivative of momentum, and if the mass constant then we get Newton's second law.

    [tex]F=\frac{dp}{dt}=m\frac{dv}{dt}=ma[/tex]

    If mass isn't constant then we've got a product rule.

    [tex]F=\frac{dm}{dt}+\frac{dv}{dt}[/tex]
     
  5. Dec 14, 2009 #4
    but the mass at fault is always constant?
    why the thing we write as a product, the mass will never be different
     
  6. Dec 14, 2009 #5
    Sometimes the mass is different, some simple examples are of a rocket or a raindrop. From special relativity we also know that the mass isn't constant, but this factor is really negligible for real world applications. For most purposes, the mass is the same, and thus change in momentum over change in time is the same as force.
     
  7. Dec 14, 2009 #6
    but I can not understand the difference between kinetic energy and momentum
     
  8. Dec 14, 2009 #7
    Kinetic energy and momentum are very different. You could relate them by

    [tex]KE=\frac{p^2}{2m}[/tex]

    Maybe you could post a more specific question about what you do not understand between kinetic energy and momentum.
     
  9. Dec 14, 2009 #8
    I can not understand the properties of momentum has. what is momentum? what distinguishes rörlsemängd from Newton's second law?
     
  10. Dec 14, 2009 #9
    I guess there are a lot of answers to "what is momentum?" At the basic physics level, I think of momentum as a quantity that tells me how hard it will be to change the path of the object, such as how much an object with lots of momentum can alter another object. If a car is sliding quickly on ice, an object with lots of momentum, will I, an object with little momentum, be able to stop the car by walking into it? No.

    Momentum is intuitive, just find your intuition for it.
     
  11. Dec 14, 2009 #10
    Can you mention other things, then I will understand it better. thanks
     
  12. Dec 14, 2009 #11
    Surely you mean

    [tex]\mathbf{F}=\mathbf{v}\frac{dm}{dt}+m\frac{d\mathbf{v}}{dt}[/tex]

    Momentum is exactly what its formula tells you: [itex]\mathbf{p}=m\mathbf{v}[/itex]. It is the mass times the velocity, thus it is a vector quantity that describes, as Isaac Newton put it, "the quantity of motion."
     
  13. Dec 14, 2009 #12
    Now I do not understand. why should it be this:
    [tex]F\cdot \Delta t=v\cdot dm+m\cdot dv[/tex]
    How can we have in the:
    constant v multiply variable with m + constant m multiply variable with v=[tex]F\cdot \Delta t[/tex]
     
    Last edited: Dec 14, 2009
  14. Dec 14, 2009 #13

    Think of it as this (this is called the product rule, Leibniz's law, or derivation): By definition, the differential of two functions (we'll use mass and velocity here) is given by:

    [tex]
    d\left(m\cdot v\right)=(m+dm)\cdot(v+dv)-m\cdot v
    [/tex]

    so just expanding the product,

    [tex]
    d\left(m\cdot v\right)=m\cdot v+m\cdot dv+dm\cdot v+dm\cdot dv-m\cdot v
    [/tex]

    The first and last terms cancel. Now the argument is that since [itex]dm[/itex] and [itex]dv[/itex] are such infinitesimal amounts, then when we multiply them together, we get an even smaller amount and we can thus ignore it. So then

    [tex]
    d\left(m\cdot v\right)=m\cdot dv+dm\cdot v
    [/tex]

    Since the left hand side of the above equation is equal to the infinitesimal change in impulse ([itex]d(Ft)=F\,dt[/itex] for a constant Force) then it must be that

    [tex]
    F\,\Delta t=m\Delta v+v\Delta m[/tex]
     
    Last edited: Dec 14, 2009
  15. Dec 15, 2009 #14
    Yes, of course. The troubles of doing math without actually writing it...:redface:
     
  16. Dec 15, 2009 #15
    in the end could be Please use this formula? one can say that this formula is useful when we watch photons hitting against a free agent?
     
  17. Dec 15, 2009 #16
    Usually this formula would only be good for cases of http://www.relativitycalculator.com/mass_variable.shtml" [Broken] to move forwards. The two links are pretty in-depth views (in English) of applications of the variable-mass impulse relation I derived above.


    In most cases of first year physics (and in some cases too, second year physics), mass is considered constant and the [itex]v\,\Delta m[/itex] term can be neglected because [itex]\Delta m=0[/itex]. In the cases where mass can change (low-speed, inelastic collisions), you would not use this equation but probably conservation of momentum:

    [tex]
    v_{1,initial}m_1+v_{2,initial}m_2=(m_1+m_2)v_{final}
    [/tex]
     
    Last edited by a moderator: May 4, 2017
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