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Conservation of energy with kinetic energy and spring

  1. Nov 14, 2012 #1
    1. The problem statement, all variables and given/known data
    A dynamics cart 1 ahs a mass of 1.8 kg and is moving with a velocity of 4.0 m/s
    along a frictionless track. Dynamics cart 2 has a mass of 2.2 kg and is moving at 6.0 m/s
    . The carts collide in a head-on elastic collision cushioned by a spring with spring constant k = 8.0 x 104 N/m.

    Determine the compression of the spring in cm, during the collision when cart 2 is moving at 4.0m/s


    and

    Calculate the maximum compression of the spring in cm


    3. The attempt at a solution

    so the givens are these:
    m1 = 1.8 kg
    m2 = 2.2 kg
    v1=4 m/s

    v2 = 6 m/s

    K = 80000N/m


    a)

    Ek = 1/2m2v2`
    Ek = 1/2(2.2kg)(4m/s
    )^2
    Ek = 17.6 J

    17.6 J = 1/2kx^2
    17.6 J = 1/2(80000N/m)x^2
    0.020976176 m = x

    b)

    I am not sure what to do for this particular part?

    v1`=(m1-m2)/(m1+m2)v1 + (m1xm2)/(m1+m2)v2
    v1`=0.4 m/s
    +5.94 m/s

    v1`= 6.34 m/s


    v2` = (2m1)/(m1+m2)v1 + (m2-m1)/(m1+m2)v2
    v2` = 3.6 m/s
    + 0.6 m/s

    v2` = 3 m/s


    Ektot = 1/2m1v1^2 + 1/2m2v2^2
    Ektot = 14.4 J +39.6 J
    Ektot = 54 J

    Ektot` = 1/2m1v1`^2 + 1/2m2v2`^2
    Ektot` = 36.17604 J + 9.9 J
    Ektot` = 46.07604 - but they are not the same since its elastic they are supposed to be the same​
     
    Last edited: Nov 14, 2012
  2. jcsd
  3. Nov 14, 2012 #2

    SammyS

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    For part a), which is incorrect:

    What is the KE of cart 1, when cart 2 is moving at 4.0m/s
    ? You need to include this in your total KE.​
     
  4. Nov 14, 2012 #3


    would it be Ek = 1/2m1v1^2
    Ek = 1/2(1.8kg)(4m\s)^2
    Ek= 14.4 J

    so then itll be

    Ektot = 14.4 J + 17.6 J
    Ektot = 32 J

    32 J = 1/2(80000N/m)x^2
    0.02828 m = x

    how about this?​
     
  5. Nov 14, 2012 #4

    SammyS

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    Incorrect.

    Use conservation of momentum to find the velocity of Cart 1, when Cart 2 is moving at 4.0m/s
     
  6. Nov 14, 2012 #5


    Okay .. but my other tutor said you can't use momentum because the energy is being conserved into the spring not the balls itself...? But if you suggest it i can try it that way...

    It would be m1v1 + m2v2 = m1v1` + m2v2`

    (1.8 kg)(4 m/s
    ) + (2.2 kg)(6 m/s
    ) = (1.8 kg)v1` + (2.2 kg)(4m/s
    )
    7.2 kgm/s
    + 13.2kgm\s
    = (1.8kg)v1` + 8.8kgm/s

    6 kgm/s
    - 8.8kgm\s
    = (1.8kg)v1`
    2.8 kgm\s
    / 1.8 kg = v1`
    1.556 m/s
    = v1`

    Ektot= 1/2m1v1`^2 + 1/2m2v2`^2
    Ektot = 2.1790 J + 17.6 J
    Ektot = 19.779 J

    Ektot = 1/2kx^2
    19.779 J / 40000 N/m = x^2
    0.022236 m = x​
     
  7. Nov 14, 2012 #6


    but it doesn't say that the m1 would also be changing in the velocity.... then why do we need to find the v1` ?​
     
  8. Nov 14, 2012 #7

    SammyS

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    ... because the velocity of m1 does change! Conservation of momentum tells you that, without it being stated in the problem .
     
  9. Nov 14, 2012 #8
    Oh i see what you are talking about ..... so how is my part a calculation i just did?
     
  10. Nov 14, 2012 #9
    ANY BODY?!:confused:
     
  11. Nov 14, 2012 #10

    SammyS

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    Patience !

    You need to quit prematurely bumping your posts !
     
  12. Nov 14, 2012 #11

    SammyS

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    Why are you equating the Potential Energy of the spring to the total KE of the two carts?​
     
  13. Nov 14, 2012 #12
    Well the amount of energy that the carts have would be transferring into the spring wouldn't it?
     
  14. Nov 14, 2012 #13
    Well when i come to think of it...
    Ektot = 19.779 J is the kinetic energy when they have come together compressing the spring..
    but how much did they have the initial energy....
    To do this i did this:

    Ektot(initial) = 1/2m1v1^2 + 1/2m2v2^2
    14.4 J + 39.6 J = 54 J

    And if i subtract the 54 J - 19.779 J = 34.221 thats the energy lost in the spring correct?

    so then i can use that ...

    34.221 J = 1/2(80000N/m)x^2
    0.029249 m = x
     
  15. Nov 14, 2012 #14

    SammyS

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    The amount of (kinetic) energy the carts lose will be transferred to PE of the spring. The Total energy is conserved.
     
  16. Nov 14, 2012 #15
    Yes that is correct similarely what i have done like this..

    Ektot(initial) = 1/2m1v1^2 + 1/2m2v2^2
    14.4 J + 39.6 J = 54 J

    And if i subtract the 54 J - 19.779 J = 34.221 thats the energy lost in the spring correct?

    so then i can use that ...

    34.221 J = 1/2(80000N/m)x^2
    0.029249 m = x

    im i correct on this calculation?
     
  17. Nov 14, 2012 #16

    SammyS

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    Yes. now convert that to cm.
     
  18. Nov 15, 2012 #17
    Okay but now what about the the sexond part where you need to find the maximum spring compression ? i showed my work above... but the energy before and after is different.. is that the way u should do it?
     
  19. Nov 15, 2012 #18

    SammyS

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    Where did you get those equations for v1' & v2'? For what conditions are those equations true? I suspect that they're for the velocities resulting from an elastic collision. As you point out, there's something wrong because there is kinetic energy lost.

    While you have an error somewhere when you used these. They have nothing to do with solving part b).

    At the instant the spring is at maximum compression, the two carts have the same velocity, v1 = v2 . Why is that the case?​
     
    Last edited: Nov 15, 2012
  20. Nov 15, 2012 #19
    well those equations are true for two moving object.. and i found the mistake where i was wrong.... for the v1` . the second mart is (2m1/m1+m2)v2

    i used these to find the kinetic energy i just wanted to see if they equal the kinetic energy before and after.. and its true... so now to find the maximum compression.. i need to find the velocity of them in the middle.... right?
     
  21. Nov 15, 2012 #20

    SammyS

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    In the middle of what ?

    Use conservation of momentum to find v2 & v2, when they're equal. Of course that's when v1 = v2 = vcenter of mass. And, I asked earlier, if you know why that gives the distance of closest approach.
     
    Last edited: Nov 15, 2012
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