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Impulse momentum

  1. Apr 25, 2014 #1
    1. The problem statement, all variables and given/known data
    https://scontent-a-sjc.xx.fbcdn.net/hphotos-frc1/l/t1.0-9/10155336_1407954212814131_2465716609293795371_n.jpg


    2. Relevant equations
    ∫Fdt = mvf-mvi



    3. The attempt at a solution

    I have tried integrating kxcdt from 0 - .01 and havent had any luck.

    I am not quite sure how to go about solving this problem...
     
  2. jcsd
  3. Apr 25, 2014 #2

    PhanthomJay

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    I don't know where you are getting the t= .01 s from. You are looking for the momentum change of m1 between its start point and the point where it leaves the spring. Try using conservation of energy to solve for the speed of m1 as it leaves the spring.
     
  4. Apr 25, 2014 #3
    thank you!

    this is what I did:

    .5Kxc2=.5m1v12

    v1=sqrt(kxc2/m1) = 1.02 m/s

    then

    ∫Fdt = m1v1= 1.02(2.4) = 2.448Ns

    Its not the exact same answer as the paper gives though,
    any thoughts on that?
     
  5. Apr 25, 2014 #4

    BiGyElLoWhAt

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    This looks like a 3 step problem to me. One with no integrals.

    Having an inelastic collision means CoE is no good during the collision, but it says nothing about afterwards. You need both conservation of momentum and energy for this one.
     
  6. Apr 26, 2014 #5

    rude man

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    I haven't worked out the m2 part but I think m2 does not collide with m1 until after m1 is released from the spring. Proceeding with this assumption, which should of course be verified:

    No matter, the problem certainly does require an integration. You will need to solve ∫F dt.
    Hint: relate v[x(t)] to x(t), the position of m1. Then use a chain rule to solve the integral.

    EDIT: thanks folks for reminding me that there is a hard way and an easy way. I certainly took the hard way!
     
    Last edited: Apr 26, 2014
  7. Apr 26, 2014 #6

    ehild

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    Do not round too early, too much. v1 is inaccurate.

    ehild
     
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