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Collision Question dealing with Spring Compression

  1. Mar 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Forum! I'm struggling on this one D:
    any help would be greatly appreciated!

    An elastic head-on collision, a 0.60kg cart moving 5.0m/s [W] collides with a 0.8-kg cart moving 2.0 m/s[E]
    The collision is cushioned by a spring (k=1200N/m)

    **Find the maximum compression of the spring
    I'm not sure if I did this right..^

    2. Relevant equations
    3. The attempt at a solution
    ** note that v prime (v’) is final velocity

    Required: maximum compression of the spring
    Given: k = 1200 N/m
    Analysis and Solution:
    Let [W] represent positive
    First we must find how fast the carts are moving at the point where the spring is at maximum compression. At this point, the two carts must have the same velocity; otherwise the spring would not be compressed to a maximum. To find the velocity, use conservation of momentum:

    PT = PT’
    m1v1 + m2v2 = (m1 + m2)v’
    (0.60)(5.0) + (0.80)( –2.0) = (0.60 + 0.80) v’

    Notice that v2 is negative because it is travelling [E] and we already decided [W] is positive

    (0.60)(5.0) + (0.80)( –2.0) = (0.60 + 0.80) v’
    3 + (–1.6) = (1.40) v’
    1.4 = (1.40) v’
    v’ = 1.0m/s
    Therefore, at maximum compression, the two carts are moving at 1.0m/s
    ****^this answer feels funny to me..

    Now use conservation of momentum to find the maximum compression of the spring:

    (1/2)m1v1^2 + (1/2)m2v2^2 = (1/2)(m1 + m2)v’^2 + (1/2)kx << could someone please explain this out in words how my equation makes sense.. Its from my book and looking at it is confusing me:yuck:
    Recall k = 1200N/m
    (1/2)m1v1^2 + (1/2)m2v2^2 = (1/2)(m1 + m2)v’^2 + (1/2)(1200N/m)x

    Solve for x (the maximum compression of the spring)

    (1/2)(0.60)(5.0)^2 + (1/2)(0.80)(2.0)^2 = (1/2)(0.60 + 0.80)(1.0)^2 + (1/2)(1200)x

    Notice that v2 is now positive despite the fact that it represents 2.0m/s [E].. and our forward direction was already decided to be [W]

    7.5 + 1.6 = 0.70 + 600x
    600x = 8.4
    x = 0.014
    Therefore, the spring compresses a maximum of 1.4cm or 0.014m at a spring constant of 1200 N/m
    ...^is this right :S


    back to the purple text.. I understand why I would make the 2.0m/s[E] negative by the first purple text..but then in the second part of my solution(where the second purple text is).. as per following the along with the book.. it instructs me to suddenly make this value positive..
    Can someone please explain why this makes sense D:
     
    Last edited: Mar 19, 2012
  2. jcsd
  3. Mar 19, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    Looks very reasonable. It also happens to be the speed of the center of mass of the system.
    It's saying that the kinetic energy before collision is equal to the kinetic energy when the masses are moving as one (same velocity), plus the energy stored in the spring. It's just a statement of conservation of energy.
    The energy of a spring is given by (1/2)kΔx2. Don't forget that the Δx is squared!

    Notice that v2 is now positive despite the fact that it represents 2.0m/s [E].. and our forward direction was already decided to be [W]

    7.5 + 1.6 = 0.70 + 600x
    600x = 8.4
    x = 0.014
    Therefore, the spring compresses a maximum of 1.4cm or 0.014m at a spring constant of 1200 N/m
    ...^is this right :S


    back to the purple text.. I understand why I would make the 2.0m/s[E] negative by the first purple text..but then in the second part of my solution(where the second purple text is).. as per following the along with the book.. it instructs me to suddenly make this value positive..
    Can someone please explain why this makes sense D:
    [/QUOTE]
    Squaring any value (except for imaginaries, of course) results in a positive result. So the sign of the velocities doesn't matter when you're working with kinetic energy.
     
  4. Mar 19, 2012 #3
    thanks so much gneill! That thread was up all day and you were the only one willing to take the time to the read it >.>

    and thanks for catching my mistake! I haven't handed it in yet thankfully!

    so my answer should be √0.014
    rounding to two sig digits:
    x = 0.12
     
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