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Momentum / newtons 2nd

  1. Sep 3, 2005 #1
    I've a fairly good maths background, but I'm trying to improve my mechanics.
    Anyway, I read a version of newton's second law as follows (I've not got it to hand so this may not be exactly right)

    The change in momentum per unit time is proportional to the resultant force in that direction.

    Its the word propotional that's confusing me.
    If v = terminal velocity
    & u = initial velocity
    mass (resistance to change in velocity) is constant
    then change in momentum per unit time is (mv - mu) / t
    taking out a common factor of m gives m (v - u)/t
    as (v - u) / t = a then we have the change in momentum per unit time = to ma. Newton's second law F = ma

    Where does the 'proportional' into play? Shouldn't there be a constant of proportionality somewhere to make it an equation?

    Thanks
     
  2. jcsd
  3. Sep 3, 2005 #2

    LeonhardEuler

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    I would just point out that the equation as you have it is not exactly correct. The change in momentum per unit time is [tex]\frac{(mv-mu)}{\Delta t}[/tex] where [itex]\Delta t[/itex] is the change in time over the interval in which the velocity chnges from u to v. This statement doesn't make a whole lot of sense to me because if the amount of time is allowed to be any finite value, then the force could take an any number of values over that time. Hence the phrase "the resultant force" is ambiguous because it could mean any of the forces that were instananeously acting over the time interval. As far as the proportionality, that is also confusing because the derivative of the momentum is equal to the net force.
     
  4. Sep 4, 2005 #3
    Thanks. I'll get the book tomorrow and type it exactly.

    I think (though am not sure) the book uses t and not delta t because initially it considers uniform acceleration of a particle in a straight line and the SUVAT equations of motion. If the initial velocity is at t=0 then the change in t is just t.

    a = dv/dt (a is constant)

    dv = a dt

    v = at + d at t=0

    v=d (but this is just initial velocity u)

    v = at + u (latter substituted to get F = ma)

    The change in momentum (from t=0 to t=t), given uniform acceleration and a constant mass is

    mv - mu (impulse?)

    m(v -u) where v - u is the change in velocity.

    The impulse per unit time is

    (mv - mu) / t or (mv - mu) / (t - 0).

    If the mass of the particle varies then a differential equation needs to be formed. Is this done by assuming the mass remains constant over a small change in time? I'd like to see the working to set up the diff eqns for this in the case where
    1) the velocity is constant
    2) the velocity changes. For this do we assume that over a small change in time the change in velocity is small? Giving m(delta v)/delt t or ma in the limit as t tends to zero? If so, is there any justification for this assumption? Again how do we link this to the force side of the equation?

    I'm stuck trying to understand how to use newton's 2nd to fit this side to equal F. In particular, I don't understand why the law says is proportional to (definitely the words used) and not equal to the force. Second, I've seen force described as net force and wanted to understand where this fits in. I've been thinking it's because the wording used is 'resultant' force - the combined effect of several forces.

    It's really doing me in at the moment! Any more help would be greatly appreciated!
     
    Last edited: Sep 4, 2005
  5. Sep 4, 2005 #4
    It's only equal to instead of proportional because that is how we define our units. 1 newton is DEFINED to be the force that causes 1kg of mass to accelerate at 1 m/s^2. I could define a unit of force called a 'squiddelybob', and then say that F=Kma, where K is a constant. You can think about it being proportional, and the constant of proportionality is just equal to 1. If you were to measure mass in pounds instead of kilograms, then F=0.454ma.
     
  6. Sep 4, 2005 #5
    Ahhhh I seee!

    That makes sense! I reckon the more I think about physics the more I confuse myself. I'm with you on the squiddlebobs though after all, who is Newton anyway?

    Ok supposing that the mass varies. Do you consider small increments in time (delta t = dt) produce small changes in velocity (delta v) approximated by the differential dv

    Is this right? I've no physics books, just calculus.

    F is proportional to m(dv/dt)
    F = k m(dv/dt)

    Measuring in Newtons gives

    F = m (dv / dt)
    F dt = m dv

    Integral F dt = Integral m dv
    Definate integral from t0 to t1 of F dt = [mv] from t0 to t1
    Definate integral from t0 to t1 of F dt = mv(t1) - mv(t2)
    (As v is a function of t)

    I'm not convinced at all by this last working.

    What happens if we look at the small changes in mass delta m with a constant velocity?

    Can you recommend any decent books (preferably with pictures)?

    Cheers.
     
  7. Sep 4, 2005 #6
    Iff mass is a function of time, ie. mass varies, then

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

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

    [tex]v=\int \frac{F(t)}{m(t)}dt[/tex]

    As for textbooks, I don't know of any that deal with newtonian mechanics specifically, but if you are interested in physics in general I use Halliday, Resnick & Walker (which also seems to be the recommended text for many undergraduate physics courses around the world); It's expensive but covers everything from newton's laws, to electromagnetism, to quantum mechanics, and is easy to read. You'll need to learn some calculus as well. I use Stewart's "Calculus: concepts and contexts", which is again pretty expensive, and more in-depth than you would need, so someone else here can probably recommend something cheaper.
     
  8. Sep 5, 2005 #7

    HallsofIvy

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    That's why the word "proportional" is used- if the time interval is twice as long, the constant of proportionality would be twice as large. And, of course, this is a assuming a constant force.
     
  9. Sep 10, 2005 #8
    The James Stewart 5th edition calculus edition is great. I've ignored physics problems for years but began struggling with line integrals and realised that physics and maths were closely linked so I'd better brush up.
    I'll look into that physics book.

    I understand the variable seperable integration as you've shown it

    If m(t) = F'(t) then v = ln modulus m(t) + k or other methods

    but get confused when trying to derive it from small increments, in particular which is a function of time and which isn't.

    Displacement if changes (m) with time is
    Velocity (ms^-1) if changes with time is
    Acceleration (ms^-2)


    But is velocity and acceleration functions of time?
     
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