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Momentum and Impulse during a collision

  1. Oct 26, 2006 #1
    A 500 g cart is released from rest 1.0 m from the bottom of a frictionless, 30 degree ramp. The cart rolls down the ramp and bounces off a rubber block at the bottom. The figure shows the force during the collision.



    [​IMG]

    Here's what I did. Which is probably wrong for the very first part so I'll only show my work for first part:

    [tex]\[ v^2_f = v^2_i + 2ad \]
    \[ v^2_f = 0+ 2(0.5)(gsin30)(1)\]

    so \[v_f = 2.21 m/s \] [/tex]

    I took the area of the graph as [tex]\ [J=200\times26.7\times10^-^3\times0.5\][/tex]

    So can somebody please help with the first part, I don't understand.
     
    Last edited: Oct 26, 2006
  2. jcsd
  3. Oct 26, 2006 #2

    radou

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    Find the velocity of the cart at the bottom of the incline from energy conservation, i.e. mgh = 1/2 m v^2.
     
  4. Oct 26, 2006 #3

    OlderDan

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    The area of this graph has dimensions of Newtons*seconds, which is the same as the the dimensions of momentum. The area of the graph is called Impulse. How is that related to the momentum of the object that experiences this force? What is the formula for the area of a triangle? You can find the velocity before the bounce using your method, or you can look at energy conservation as radou suggested. The two are equivalent.

    I think this problem was done very recently in another thread.
     
    Last edited: Oct 26, 2006
  5. Oct 26, 2006 #4

    rsk

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    I think usually when you see a graph like that it's an impulse question - that is they want you to use Impulse ( Av Force x time) = change in momentum.

    So the area under the graph in this case will give you the chenge in momentum of the cart and from that you can calculate its final velocity.
     
  6. Oct 26, 2006 #5
    Okay thanks, I did that

    [tex] \[mgh = 0.5mv^2\]
    \[g\sin30 = 0.5v^2\]
    v= 3.13 m/s

    So

    \[J= p_f - p_1\]

    \[2.67 = 0.5(v_f + 3.13)\]

    so \[v_f= 2.21 m/s\]

    then
    \[mgh = 0.5mv^2\]

    \[gh = 0.5(2.12)^2\]

    so h= 0.229m!?!?!
    [/tex]
    Looks like something went horribly wrong somewhere.
    Durr... I took out my fractions because I can never do them.

    Can you tell me what is my problem?

    EDIT: OKAY I did not see the two last posts as it took me forever to write that out. I am bad at it. Also I know this is solved on another thread but it didn't help at all and NO ONE answered when I posted on it yesterday.

    Hi, also can someone anser my question why did I get a different answer when I did conservation of energy and kinematics. For that I am like 'wtf?'.
     
    Last edited: Oct 26, 2006
  7. Oct 26, 2006 #6

    rsk

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    What are you trying to find? You've confused me by ending up with h.

    What was the question asking you to do?
     
  8. Oct 26, 2006 #7

    radou

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    Where did the minus sign get lost? impulse = final momentum - initial momentum. Further on, does the length or the height of the incline equal 1? According to your calculation, it would be the length..
     
  9. Oct 26, 2006 #8
    Durr.... I am supposed to find how high the cart bounces back. Or maybe it is length.

    The 1m is apparently length.

    Edit: "After the cart bounces, how far does it roll back up the ramp?"

    Oh and I took the minus sign out because the final momentum is opposite the intial so the minus cancels?
     
    Last edited: Oct 26, 2006
  10. Oct 26, 2006 #9

    rsk

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    AH, Ok

    I agree with your velocities and the value for Impluse.

    I also agree with the 0.229 by conservation of energy.

    Why do you think this is wrong? - this is a height remember so you need to think about sines or cosines to get the lenght of track it rolls back up
     
  11. Oct 26, 2006 #10

    rsk

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    Just read your edit.

    Ok, so convert your 0.229 height to length - does that give you the right answer?
     
  12. Oct 26, 2006 #11
    Durr... I believe you two just gave me opposing information.

    and I calculate length to be l=0.459m but I lose marks for the wrong answer... durr.....

    EDIT: Durr.... the opposing info is gone and now I look like a crazy person. ;'( tear.
     
  13. Oct 26, 2006 #12

    rsk

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    Now you've really lost me - opposing information?

    When I said "why do you think this is wrong?" it was because I think it's right - but you seemed to think in Post 5 that it was horribly wrong.
     
  14. Oct 26, 2006 #13
    radou said something about not taking direction of momentum into account.

    I KNEW I LOOKED CRAZY :cry:

    Anyways, that's the right answer.

    Thanks so much. I am sastified customer.
     
  15. Oct 26, 2006 #14

    rsk

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    Ok, well done.
     
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