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Momentum problem. Urgent, can help me pls?

  1. Mar 13, 2009 #1
    The problem statement, all variables and given/known data

    In a shipping company distribution center, an open cart of mass 51.0kg is rolling to the left at a speed of 5.50m/s (see the figure). You can ignore friction between the cart and the floor. A 10.0kg package slides down a chute that is inclined at 37degrees from the horizontal and leaves the end of the chute with a speed of 2.60m/s . The package lands in the cart and they roll off together.



    (a) If the lower end of the chute is a vertical distance of 4.00 above the bottom of the cart, what is the speed of the package just before it lands in the cart?
    ans: 9.23m/s

    (b) What is the final speed of the cart?
    ans: 4.26m/s


    The attempt at a solution

    I think this is a prefectly inelastic collision when the package falls into the cart.
    For first part, is it use v^2 = u^2 + 2as?
    where u = 2.6m/s and a = -9.8 and s = -4?
    i get the answer 9.23, but i not sure if this is correct cos i din take into acct that the package falls from a slope at an angle. does the angle matter?


    For 2nd part, i duno how to do. I tried using conservation of momentum but i cant get the answer.
    my eqn is: Mp * Vpi + McVci = (Mp + Mc)Vf
    and i get Vf as 5.02m/s

    Pls help me.. Thanks!
     

    Attached Files:

  2. jcsd
  3. Mar 13, 2009 #2

    Doc Al

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    Staff: Mentor

    Your thinking is probably a bit off, but lucky for you you get the right answer anyway. To do it right, using acceleration, you'd need the component of the initial velocity in the vertical direction. But then you'd be adding back the horizontal component later to find the total velocity.

    Even better would be to use conservation of energy. Then it would be clear that the angle doesn't matter (for this part).

    Hint: During the collision of package and cart, momentum (a vector) is only conserved in one direction. (Which direction?)
     
  4. Mar 13, 2009 #3
    oh.. we need to add back the horizontal part! it din occur to me that we need to add that back. hmm. how do we know if we can use energy to solve a qns?
     
  5. Mar 13, 2009 #4
    during collision, momentum is conserved in x direction?
     
    Last edited: Mar 13, 2009
  6. Mar 13, 2009 #5

    LowlyPion

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    Homework Helper

    Yes.
     
  7. Mar 13, 2009 #6

    Doc Al

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    Just try it and see! In this case, mechanical energy is conserved (in part A) so it can be used.

    Right! Now you need the speed of the package in that direction to analyze the collision.
     
  8. Mar 13, 2009 #7
    so its Mp*Vpxi + Mc*Vcxi = (Mc + Mp)*Vxf right?

    i get 4.94m/s, but answer is 4.26m/s..
     
  9. Mar 13, 2009 #8

    Doc Al

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    Right.
    What were your values for Vpxi & Vcxi?
     
  10. Mar 13, 2009 #9
    my Vpxi is 2.6cos 37
    Vcxi is 5.5

    correct?
     
  11. Mar 13, 2009 #10

    Doc Al

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    Correct. Different signs, of course.

    Recheck your arithmetic: 2.6 cos 37 = ?
     
  12. Mar 13, 2009 #11
    we must include the signs?? means vcxi is -5.5??
    2.6cos 37 is 2.0764
     
  13. Mar 13, 2009 #12
    means even for momentum qns, we need to consider the signs for velocity?

    can i ask sth? like when we calculating potential energy, the gravt acceleration, do we need to consider the -ve sign if we take upward as positive?
     
  14. Mar 13, 2009 #13

    LowlyPion

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    Absolutely. Signs matter.

    And that's the difference in your answer that you are looking for.
     
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