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Conservation of momentum - Vertical spring

  1. Mar 6, 2017 #1
    1. The problem statement, all variables and given/known data
    A block of mass 200g is suspended through a vertical spring. The spring is stretched by 1.0 cm when the block is in equilibrium. A particle of mass 120g is dropped on the block from a height of 45 cm. The particle sticks to the block after the impact. Find the maximum extension of the spring. Take g=10 m/s^2.

    2. Relevant equations
    Initial Momentum of system = Final Momentum of system
    Ui+Ki=Uf+Kf

    3. The attempt at a solution
    Since the spring is initially in equilibrium,
    kxi = Mg
    using this we can find the spring constant.
    Also, at the point of maximum elongation,
    kxf=(M+m)g
    using this we can find the maximum elongation.
    The answer I get is 1.6 cm which is wrong (the correct answer is 6.1 cm). When I looked at the solution, I saw that conservation of momentum was applied first and then conservation of energy. But, I don't see why the momentum can be conserved. There is a net external force on the system because of gravity.
    Could someone explain to me what is wrong with my approach and also why momentum conservation can be applied in this case?
    Here is a diagram of this problem I drew
    IMG_20170306_203905396.jpg
     
  2. jcsd
  3. Mar 6, 2017 #2
    You have an inelastic collision when the particle strikes the block so conservation of energy does not apply. Momentum is conserved during the collision, however.

    If you drop a weight from some height onto a spring, do you think it will directly go to the equilibrium position?
     
  4. Mar 6, 2017 #3
    No. Not directly but only after the kinetic energy of the system becomes zero.
     
  5. Mar 6, 2017 #4

    BvU

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    At the moment of collision, the net external force is zero: gravity is canceled by the force the ceiling exerts. But you are correct: the conservation of momentum is indeed not applicable over a time interval > 0. Only at the moment of collision, which is probably what they do in the book solution (that I can't see) to calculate the kinetic energy of the block + particle after the inelastic collision

    [edit]Tom's faster, but we agree.
     
  6. Mar 6, 2017 #5
    Is there any mention of damping?
     
  7. Mar 6, 2017 #6
    That is
    No. I've typed the entire question as it is.
     
  8. Mar 6, 2017 #7

    BvU

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    Animated-mass-spring.gif
    Kinetic energy zero doesn't mean equilibrium position !
     
  9. Mar 6, 2017 #8
    Have you ever stepped onto one of the old mechanical scales? Don't you notice that the reading does not go immediately to the final weight. It tends to oscillate above and below the final weight that it settles on. (Due to friction, it eventually stops oscillating.) However, in your problem, since there is no mention of damping, I think it's safe to assume that there is no damping.
     
  10. Mar 6, 2017 #9
    But why not? Wouldn't K.E. be zero when the spring is at maximum elongation?
     
  11. Mar 6, 2017 #10

    BvU

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    It would. But you answered that in response to a question on equilibrium position. Which is probably what you calculated in post #1.
     
  12. Mar 6, 2017 #11
    So since there is no damping, can energy be conserved?
     
  13. Mar 6, 2017 #12

    BvU

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    Not in the collision. Afterwards, yes.
     
  14. Mar 6, 2017 #13
    Once you use conservation of momentum to figure out the velocity of the particle/block combination, energy will be conserved forever after that point.

    @BvU: Great picture above.
     
  15. Mar 6, 2017 #14
    Also, could anyone please explain what was wrong with my approach?
     
  16. Mar 6, 2017 #15

    BvU

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    Picture stolen from Wiki, I'm afraid - no credit to me.
     
  17. Mar 6, 2017 #16

    BvU

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    That's the equilibrium position.

    Show your work in detail.
     
  18. Mar 6, 2017 #17
    If you set the particle on the block very lightly - as opposed to dropping it from some height - don't you think the maximum extension would be different?
     
  19. Mar 6, 2017 #18
    Yes! If it is dropped from a height the extension will be more as compared to when it is placed lightly.
     
  20. Mar 6, 2017 #19
    Yes, so you need another tool other than F=kx.
     
  21. Mar 6, 2017 #20
    Also, here is a link to the solution I referred to
     
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