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Homework Help: Change of momentum of a trolley

  1. Jan 8, 2007 #1
    1. The problem statement, all variables and given/known data
    A trolley of mass 2.4Kg is moving along a smooth horizontal surface with a constant speed of 2.0m/s when a brick of mass 1.5Kg is dropped vertically on to it.
    What is the change in momentum of the trolley?

    2. Relevant equations
    Impulse: F(delt)=m(delv)
    momentum: mv

    3. The attempt at a solution
    I don't understand the question very well. As far as I can see there isn't enough data. I don't know how fast the trolley is moving after the brick has landed. Conservation of momentum can't be used because I don't know the initial momentum.
  2. jcsd
  3. Jan 8, 2007 #2
    If the is no kinetic friction between the trolley and the floor, I may think that the momentum of the trolley is conserved! The force exerted by the block is opposed by part of the normal force exerted by the ground on the trolley, such that the net impulse is 0. Do you see my idea?
  4. Jan 8, 2007 #3


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    Remember: Conservation of momentum along three differennt axes gives three independant equations. Instead of asking yourself what the initial momentum is, try asking yourself "What is the initial momentum in the direction of travel of the trolley?"
  5. Jan 8, 2007 #4
    You are referring to Newton's third law? In what you say wouldn't it imply net impulse of everything is 0 since everything has an equal and opposite reaction?
  6. Jan 8, 2007 #5
    The problem is, what is the velocity of the brick coming down, before making contact with the trolley?
  7. Jan 8, 2007 #6


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    Remember-we're talfing about the horizontal direction only. What is the horizontal component of the brick's velocity?
  8. Jan 8, 2007 #7
    No, this is not Newton third law. Moreover, see that the reaction of the force exerted by the brick on the trolley is applied at the brick.

    The momentum carried by the brick is actually transferred to the Earth!

    Please, pose more questions if you have doubts! :rolleyes:
  9. Jan 8, 2007 #8
    horizontal component of the brick's velocity is 0m/s

    I don't think conservation of momentum is significant because the question asks for the change in momentum. I assume it's the change in horizontal momentum only as no information about the vertical momentum is known. I don't think the velocity of the trolley is the same after the brick lands.

    Maybe impulse is also not important as pmp! is trying to point out.

    I still don't know how to go about this question.
    Last edited: Jan 8, 2007
  10. Jan 8, 2007 #9
    As I said above, I still have no idea. Could you explain your 0 net impulse a little more thoroughly?
    Last edited: Jan 8, 2007
  11. Jan 8, 2007 #10


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    Total momentum is conserved, not the momentum of the trolley. The trolley does not include the brick that is on top of it.
  12. Jan 8, 2007 #11
    Intial momentum: trolley only 4.8kgm/s

    Final momentum: trolley plus brick: (2.4kg+1.5Kg)*v=4.8Kgm/s

    So we can work out v=1.23m/s but the question asks for the change in momentum. I assume the change in momentum of the trolley? If we use conservation of momentum we have to assume no change in momentum which isn't correct.
  13. Jan 8, 2007 #12
    Don't forget that the hight that the brick is dropped from has something to do with its' potential. G is 32 p/s^2 if I remember correctly.
  14. Jan 8, 2007 #13


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    We assumed conservation of total momentum, not of the momentum of the trolly. You calculated the velocity of the trolly correctly. Now calculate the final momentum. It is not the same as the initial.
  15. Jan 8, 2007 #14
    Whats the momentum with the new velocity? Were still defining the system to be just the trolley. In order to find the new velocity, you need to redefine the system as the trolley/brick. This is convenient here because conservation of momentum applies, but the question asks about just the trolley. Momentum is not conserved if were just talking about the trolley
  16. Jan 8, 2007 #15
    Actually, MD=F. And VM=F. Which is greater ?
    Mass times distance or velocity times mass ? Not knowing from what height the brick is dropped from means the requirement of a variable in the answer.
  17. Jan 8, 2007 #16


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    The vertical velocity of the brick doesn't matter because we're talking about momentum in the horizontal direction.
  18. Jan 8, 2007 #17
    That has nothing to do with it. As stated above, the momenta are perpendicular and independent of each other. Its an inelastic collision, forget about energy. The height the brick is dropped from is irrelevent. If you want to get in depth, the brick has vertical momentum which it transfers to the Earth. But the change in velocity of the earth is too small to detect. It contributes no horizontal momentum initially. The initial horizontal momentum is that of the block.

    After that by LCOM, the mass goes up and the velocity goes down. The change in momentum of just the trolley is then m(v2-v1). The negative sign shows a decrease in momentum. Youve almost got it, pivoxa15
  19. Jan 8, 2007 #18
    What is LCOM?

    I get it now. The whole system of trolley plus brick is conserved. However if we consider the system containing only the trolley than it's not because a brick later lands in it. The question asks only the change in momentum of the trolley excluding the brick. The new momentum of the trolley is m(v2) where v2 is the velocity of the trolley after the brick lands in it and worked out by conservation of momentum of the trolly-brick system.
  20. Jan 9, 2007 #19
    How about this ? Total mass would beome 3.9 kg's which would require more force to move. And if F=mv, then 2.4 kg's times 2.0 m/s^2 = 4.8 newtons. Then 3.9 kg's times 1.23076 m/s^2 = 4.8 newtons.
  21. Jan 9, 2007 #20

    Doc Al

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    Are you just making stuff up off the top of your head? Please stop.
  22. Jan 9, 2007 #21
    F=ma not mv, P=mv where P is momentum. So the new velocity is 1.23m/s for the trolley as worked by conservation of momentum.

    The change in momentum for the trolley only (which is what the question is asking) is 4.8kgm/s-2.95kgm/s=1.85kgm/s and is the same as the back of the book answer.
  23. Jan 9, 2007 #22
    Law of Conservation of Momentum

    Its good strategy on an AP to write "LCOE = Law of Conservation of Energy
    , LCOM = Law of Conservation of Momentum" on the top of the paper to save time later on
  24. Jan 12, 2007 #23
    Actually not. Just forgot that who posted this is following a specific lesson.
    And of course, the course work in that book would be using problems to reinforce the lesson.
    In actual engineering, knowing the resistance that the trolley would be exposed to would be necessary to calculate the correct answer.
    And in engineering, many times different appraoches are attempted on the same problem. It would depend on what company your are employed by.
  25. Jan 12, 2007 #24


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    If you are not making things up, then perhaps you can explain where you came up with "f=mv" or "f=md."

    This equation is nonsense, since you have put a velocity into an invalid equation, and given it units of acceleration?!

    This is not, however, "actual engineering," since the truck moves along a frictionless surface.

    I'm sure that your posts will do nothing but confuse the original poster; I for one can't follow them!
    Last edited: Jan 12, 2007
  26. Jan 12, 2007 #25
    regardless og m/s^1 or m/s^2, if m/s^2 is factored by one (instead of 0, 0seconds times 2m^s^2 would =0), then no acceleration is taking place. But it does show a relationship between velocity and mass if mas effects velocity.
    And about the only things that are calculated without resistance are gravity, frequency and a body in the vacuum of space. Maybe you can name a few more. And if mass does not effect velocity, I doubt the question would be asked.
    force= mass times velocity. I know, it is actually work= mass times distance.
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