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Newton's First Law

  1. Apr 30, 2006 #1
    I read the following question in a Yr11 High School Physics Test.

    A trolley is travelling along at unifrom velocity when a weight is dropped onto it which stays there.

    The velocity of the trolley will...

    A) Decrease
    B) Increase
    C) Stay the same

    ???

    The answer says "decrease" because of law of conservation of momentum.

    But this doesnt seem right to me because it seems to disobey Newtons first Law which states that an object wont change its motion unless it is acted upon by a net force. Where is the net force??

    Can anyone help explain?

    Thanks
     
  2. jcsd
  3. Apr 30, 2006 #2

    Hootenanny

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    Do you know the mathematical equation which represents Newton's first law?

    ~H
     
  4. Apr 30, 2006 #3
    Well sorta...

    F(net)=ma

    describes both Newton's 1st and 2nd laws. Right? How does this answer the question?
     
  5. Apr 30, 2006 #4

    Hootenanny

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    You are correct, m represents mass in newton's second law. What happens to the mass of the object (in this case a trolley) if a weight is dropped into it.

    ~H
     
    Last edited: Apr 30, 2006
  6. Apr 30, 2006 #5
    m increases,
    but a=0 (because a=F/m, F=0).

    How does this show that the trolley decelerates exactly?
     
  7. Apr 30, 2006 #6

    Hootenanny

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    As your in year 11, I'll assume you have done anything on friction ([itex]F = \mu mg[/itex]) for example. Think of it this way, if your pushing a trolley around tesco, when the trolley is full, you need to push harder than if it was empty to keep it going at a constane speed.

    ~H
     
    Last edited: Apr 30, 2006
  8. Apr 30, 2006 #7
    So conservation of momentum doesn't have anything to do with it?

    What if it was on a frictionless surface and there was no force from friction, would conservation of momentum be disobeyed?
     
  9. Apr 30, 2006 #8

    Hootenanny

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    Yes, conservation of momentum has everything to do with it; I apologise, its been a while since High school and I can't remember whether momentum eas covered at GCSE. (And thinking about Newton's laws through me a bit :grumpy: ). Can you write an equation for momentum representing this collision?

    ~H
     
  10. Apr 30, 2006 #9
    Law of Conservation of Momentum:
    For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.

    or,

    Sum(m1v1) = Sum(m2v2)


    But the system in the question not an isolated system. Trolleys run on surfaces, and surfaces apply reaction forces.
     
  11. Apr 30, 2006 #10

    Hootenanny

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    Yes but going back to the question, "A trolley is travelling along at unifrom velocity", this means there are no net external forces and therefore, momentum is conserved.

    ~H
     
  12. Apr 30, 2006 #11
    Ok so if conservation is suppose to apply then how is momentum conserved if the trolley slows down?

    When you add m1v1's (of trolley and dropping mass) you get a vector sum which is in a diagonal direction. However the final monetum in the trolley example is horizontal (appearing to disobey the conservation law).
     
  13. Apr 30, 2006 #12

    Hootenanny

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    Just consider the mometum in the x direction (horizontal plane). Also, initally before it is dropped, the weight is at rest, and hence has no momentum.

    ~H
     
    Last edited: Apr 30, 2006
  14. Apr 30, 2006 #13
    This comes back to the earlier question...

    If the experiment were done on a frictionless surface, where would the net force come from to create the deceleration of the trolley ?

    This is assuming that deceleration does occur in accordance with your explanation using the law of conservation of momentum in the x-direction?
     
  15. Apr 30, 2006 #14

    arildno

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    Remember that the trolley accelerates the falling mass up to its own horizontal velocity so that they become co-moving in the horizontal direction. Hence, by Newton's third law, the falling mass exerts a retarding force on the trolley of equal magnitude.
    Thus, the resultant common velocity is lower than the trolley's original one.

    This, of course, is in accordance with conservation of momentum.
     
  16. Apr 30, 2006 #15

    Hootenanny

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    Conservation of momentum is implied in by newton's first law. Newton's second law as F = ma only applies for constant mass. When mass is not constant you need to apply newton's law in its differential form;

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

    As, momentum is conserved dp = 0 and hence F = 0, thus no net external forces act.

    ~H
     
  17. Apr 30, 2006 #16

    arildno

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    If the surface of the trolley were frictionless, then the falling mass wouldn't attain the trolley's velocity by any means, i.e, would eventually slide off it. No change would be observed in the trolly's velocity.
     
  18. Apr 30, 2006 #17

    arildno

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    There doesn't exist a single example in classical mechanics where a material system's mass change over time. Conservation of mass holds absolutely.

    For geometric systems (which does not contain the same material particles over time) , the correct version of Newton's 2.law is:
    F=dp/dt+M, where dp/dt is the rate of change of momentum inside the system and M is the momentum flux out of the system.
    F is the net external force acting upon the coinciding material system.
     
  19. Apr 30, 2006 #18
    I haven't heard of this before, is there a reference?
     
  20. Apr 30, 2006 #19
    Thank you very much arildno, that explains a lot.
     
  21. Apr 30, 2006 #20

    Hootenanny

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    Thank's for that aldrino, haven't seen that before. I'm goin to look it up now.

    ~H
     
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