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A few general Qs about momentum and springs.

  1. Sep 24, 2003 #1
    if p is momentum then dp/dt = f right?

    I've heard that dp/dt can be written as f = m(dv/dt) + v(dm/dt)

    Can someone tell me what this actually means, because it would seem that m would be dependent on v and v dependent on m, so im not sure how this would work out.

    Also, can someone point tell me some important equations for springs other than f= -kx and e = .5kx^2
  2. jcsd
  3. Sep 24, 2003 #2
    "The time rate of change of the momentum of a particle is equal to the net force acting on the particle and is in the direction of that force" (quoted from Halliday, Resnick, Walker. "The Fundamentals of Physics", 6th ed. 2001)
    That is what they used to explain F(net)= dp/dt
    As far as, f = m(dv/dt) + v(dm/dt), I'm not sure.

    Somemore spring equations for ya:

    Work = (1/2)k*(x(initial)^2)-(1/2)k*(x(final)^2)
    Work = (1/2)k*(x^2) (work by a spring force with (x initial being 0))

    You can use work equations from above to solve different problems associated with energy in the system:

    K(final) - K(initial) = Work (k = (1/2)m*(v^2), of course

    Hope this helps.
  4. Sep 24, 2003 #3
    I hope that the following will help you.

    You have the equation f=dp/dt, but we know that p=mv so: f=d(mv)/dt.

    When m is constant then we have the famous f=mdv/dt we were using at school for years (a=dv/dt), but what happened when m is not constant. When m is not constant the derivative d(mv)/dt is equal to: f=m(dv/dt)+(dm/dt)v which is the full type of the second law of Newton.
    Last edited: Sep 28, 2003
  5. Sep 24, 2003 #4
    For the springs, how does one determine the period. Is there not a cosW somewhere?

    alright, i can see how u get the mv equation as a result of the chain rule, but i'm still not sure exactly how it applies. Here's one of the examples my TA used to try and explain this.

    A constant Force, F, is applied to a car with mass M containing within, some sand with mass m. (so mass total is M+m at the start). Concurrently, a hole opens up at the bottom of the car and it starts dropping sand at a rate of dm/dt. Find the speed of the car when all the sand is dropped.

    So basically we have F = dm/dt(which we know)v + m(dv/dt)

    he also pointed out that Pf - P0 = F (delta)t

    and that F =m(dv/dt)

    So (delta)t = m/(dm/dt)

    After this part i got confused - if dv/dt is changing based on mass and v is changing based on dv/dt, how can I do this. Is this a double integral?
  6. Sep 24, 2003 #5
    We have so far that F=m(dv/dt)+v(dm/dt) now multiply the last equation with dt (I know the mathematicians will loose their hair with this but as you know this work very well in physics) and you will get:
    Fdt=mdv+vdm and here is where you have the problem, divide with dm and you certainly will fell better when you see that:
    F(dt/dm)=m(dv/dm)+v . We know that dm/dt is constant so we can write dm/dt=-c (the - is because the mass of the sand is reducing). So we have:
    –F/c=m(dv/dm)+v and -(F/c+v)dm=mdv.
    I hope this will help.
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