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F = dp/dt ?

  1. Jul 19, 2007 #1
    Hello Everyone,

    I was working on a problem which involved throwing snowballs into a car to propel it forward. The mass of the car would be increasing, since the snowballs stayed inside the car. The official solution acknowledged the changing mass, but used the equation:

    m * dv/dt = dp/dt

    I would have used:

    dp/dt = d(mv)/dt = dm/dt * v + m * dv/dt

    I could be misunderstanding the solution, and I don't understand infinitesimals very much, so help would be appreciated.

    Thanks!
     
    Last edited: Jul 19, 2007
  2. jcsd
  3. Jul 19, 2007 #2

    berkeman

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    Staff: Mentor

    Thread moved to the Homework Help forums. Do you have more information on your question, EFuzzy?
     
  4. Jul 19, 2007 #3
    I'm sorry, I was using a beta version browser, which didn't work correctly, and I wasn't able to complete my post. I've edited it now.
     
  5. Jul 19, 2007 #4

    berkeman

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    Staff: Mentor

    You are on the right track. For the snowball, what is dm/dt? Granted, it splatters a bit when hitting the car, but assume it doesn't break apart.
     
  6. Jul 19, 2007 #5

    nrqed

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    Science Advisor
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    Gold Member


    You are correct.

    This said, it is possible that the dm/dt v term is much smaller than the m dv/dt term and this *might* explain why they drop it. But to be sure, you would have to post their complete solution and all the numbers involved. But strictly speaking, you are correct.
     
  7. Jul 19, 2007 #6
    The problem is this:

    The effective mass rate, from the car's perspective, is [tex]\frac{\sigma * (u-v)}{u}[/tex].
    So dp/dt = [tex]\frac{\sigma * (u-v)^2}{u}[/tex], since (u-v) is the speed of the snowball from the car's perspective.
    Also dm/dt = [tex]\frac{\sigma * (u-v)}{u}[/tex]

    Thus dp/dt = dm/dt * (u-v). Up to here, I agree with the official solution. However, depending on the expansion for dp/dt, we get different answers.
     
  8. Jul 19, 2007 #7
    To me it looks like the official solution is right. Equation
    [tex]
    F=\frac{dp}{dt}
    [/tex]
    is not so clear when objects are breaking into smaller pieces, or getting together to form bigger objects.

    Consider a following example. A truck is being driven with constant velocity, and precisly with the same velocity with the wind (ok this is slightly theoretical), and it is raining. Now the rain drops do not exert any force on the truck, that would accelerate or slow it down. However, the mass of the truck is increasing, because more and more water is coming in its back. In this situation, the momentum of the truck is increasing, because velocity remains constant but mass increases, but how is it possible, when rain drops are not causing an accelerating force on it?

    The lesson out of this is, that if smaller objects are getting together to form bigger objects, you should not interpret them as objects whose mass is changing. At least not so seriously, that you would use
    [tex]
    F=\frac{dm}{dt}v + m\frac{dv}{dt}
    [/tex]
    on them.

    EDIT: Or more precisly, I'm not sure what happens in the official solution, or in your solution, but it looked like this point I made, needed attention. It is surely possible work that problem out somehow avoiding mistakes anyway.
     
    Last edited: Jul 19, 2007
  9. Jul 19, 2007 #8
    Thanks for all of the responses.

    I'm not sure what the full implications of the raindrop problem are, but it seems that especially in that situation dm/dt * v + m * dv/dt gives the right answer, since it reduces to dm/dt *v, whereas the m*dv/dt gives an answer of 0.
     
    Last edited: Jul 19, 2007
  10. Jul 19, 2007 #9
    No, [itex]F=(dm/dt)v[/itex] would not be right, because there is no force at all. The mass is increasing without accelerating force. That is what I meant with this "driving with the wind" situation. Rain drops just land on the truck, without causing any horizontal force.
     
  11. Jul 19, 2007 #10
    Oh, OK, I see what you were trying to say then.

    Is there any time we would want to use the dm/dt *v + m *dv/dt expansion then? Because the only way mass could change is if objects combined or broke up.
     
  12. Jul 19, 2007 #11
    A good question! I think I don't know any other example than the relativistic particles. My physics teacher long time ago explained that you should use that with rockets, that are losing fuel while accelerating, but I'm not sure if that was correct. The acceleration of rocket is anyway solved better using conservation of momentum than forces, and I've actually never put any effort into checking how forces would work.

    I don't know at the moment how you original problem should be solved, and cannot say much to it (because I haven't thought about it), but my message was, that be careful with these situations where objects are sticking together to form new objects, because they are tricky. The conservation of momentum is something you can trust always, at least. Forces are more difficult.
     
  13. Jul 19, 2007 #12
    There are two approaches to this problem: you can consider the force acting on the car alone (as the official solution did), or you can consider the force acting on the combined car + snowballs-it's-carrying object (as you did, EFuzzy).

    In the first case, the car's mass is constant, so you don't have to worry about that expansion thingy. The change in momentum of the car has to counteract that of the snowballs, and:

    [tex]\frac{dp}{dt} = (u - v) \frac{dm}{dt}[/tex].

    (Note that dm/dt is the rate at which snowballs pile up in the back of the car.) Hence we have:

    [tex]m \frac{dv}{dt} = (u - v) \frac{dm}{dt}[/tex].

    Using the second method, the car absorbs the momentum of incoming snowballs and also gains mass so:

    [tex]\frac{dp}{dt} = u \frac{dm}{dt}[/tex].

    And so:

    [tex]m \frac{dv}{dt} + v \frac{dm}{dt} = u \frac{dm}{dt}[/tex].

    But these statements are equivalent! Both methods should yield the same velocity function for the car.
     
  14. Jul 19, 2007 #13
    EFuzzy, well my point was to show how it is possible to do things wrong with the dp/dt expression, as a warning, but now durt showed how to do things right with it, so there you have it.

    durt, I think it is not so clear how much the quantity dp/dt behaves as a traditional force in the second way you show. Consider the rain example, which is now the same thing as setting u-v=0. According to this convention, the truck would experience some non-zero force, altough rain drops land on it without relative horizontal velocity component, and it is not being accelerated.
     
  15. Jul 19, 2007 #14
    Thank you all for your responses! I've got a good idea now of how the problem works now.

    I think what's happening with the "non-traditional" force is that force is best defined as dp/dt, with f=ma only applying when mass is constant. Thus force can maybe either increase mass or increase speed.
     
  16. Jul 20, 2007 #15
    No!

    No, never treat mass as variable when using Newton's second law! This is an example of sophisticated foolishness inherent in an unfortunately large number of elementary texts. Kleppner and Kolenkow give a full argument against this practice in their Introduction to Mechanics. The main point is that N2L applies to particles, and by extension to clearly defined systems of particles. But if you consider the system to have variable mass because matter is accumulating/going out, your system is not clearly defined as a collection of certain particles anymore and N2L does not apply!

    Using N2L like this can give you very wrong answers unless you are lucky. Always take systems whose mass is conserved and apply conservation of linear momentum to get the answer. Sometimes using dp/dt = m dv/dt + v dm/dt may be temptingly simpler, but it is dangorous!

    In non-relativistic mechanics, it is perfectly general to write F = m dv/dt instead of F = dp/dt. Force never ever changes mass, only velocity. Don't make this mistake.

    Molu
     
  17. Jul 20, 2007 #16
    OK. Hold on. I was confused. :redface: This time time I decided to actually do the problem.

    First, the incoming snowballs impart a momentum change not only to the car, but also to the snowballs it's already carrying. So you can't say that the change in momentum of the car (not including the snowballs it's carrying) has the same magnitude as that of the incoming snowballs. That probably doesn't make much sense to you readers... Anyway, we're going to treat the car as a object with increasing mass. Yes, that's right, loom91. If you see another way to solve the problem, please do tell.

    Let the car's mass as a function of time be [tex]m[/tex], and let its velocity be [tex]v[/tex]. Applying conservation of momentum to an incoming snowball of mass [tex]\Delta m[/tex] and the car, we find that:

    [tex]\Delta m (u-v) = m \Delta v[/tex].

    Separating variables and integrating, we get mass as a function of velocity:

    [tex]m = \frac{M u}{u - v}[/tex].

    Now, recall that:

    [tex]\frac{dm}{dt} = \frac{u-v}{u} \sigma[/tex].

    Now equate this with the derivative of [tex]m[/tex] with respect to [tex]t[/tex], solve for [tex]v[/tex], and you might get:

    [tex]v = u \left( 1 - \sqrt{\frac{1}{1+ \frac{2 \sigma}{M} t}} \right)[/tex].

    Is this the correct answer?
     
  18. Jul 20, 2007 #17
    Sorry to jump in, but I'd like to give my two cents...

    Initial momentum:
    [tex]Mv + u\Delta m[/tex]
    [tex]Mv + u\alpha\Delta t[/tex]

    Final momentum:
    [tex](M + \alpha\Delta t)(v + \Delta v)[/tex]

    Equating and simplifying gives:

    [tex]\alpha(u-v)\Delta t = M\Delta v[/tex]

    In differential terms:

    [tex]\frac{\alpha}{M} dt = \frac{dv}{u-v}[/tex]

    Which leads to the solution:

    [tex]v(t) = u(1 - e^{\frac{-\alpha t}{M}})[/tex]
     
  19. Jul 20, 2007 #18
    This is same thing that durt wrote earlier:
    but with different notation. You have [tex]\alpha = \frac{dm}{dt}[/tex] and [tex]M=m[/tex].

    The solution
    looks correct to me. I think durt did something wrong in his second post. I'm not sure what, but it is more complicated, so change of mistake is greater.
     
  20. Jul 20, 2007 #19
    Nope, you understand differentiation perfectly, however:

    Mass, in Newtonian/Galilean mechanics, is conserved. Meaning, it does not change. What is the derivative of something that does not change?
     
  21. Jul 20, 2007 #20

    I might be wrong, but I don't think that the mass is conserved in this problem, as more mass is being introduced from the outside.

    There might be snow in the car, but I think internal changes/forces cannot affect the system(so car wouldn't move in this case). Am I right?
     
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