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Conservation of Momentum Question

  1. Apr 14, 2013 #1
    Hello, I've recently came across a question that I have two solutions for. Though I don't see how either of them can be wrong, I'm getting two different answers :( Can someone explain me why this is happening (maybe I'm doing something wrong?) and which solution is better to use.

    1. The problem statement, all variables and given/known data

    During a free dance program in figure skating, Phrank (m = 71kg) glides at a 2.1m/s to a stationary Phyllis (m= 52kg) and hangs on. How far will the pair slide after the "collision" if coefficient of kinetic friction (μk) between their skates and the ice is 0.052?

    2. Relevant equations

    m1v1 1+m2v2 2=(m1+m2)vf (Inelastic collision, since Phrank grabs on)

    Solution 1:
    F⋅Δt=Δp =m(vf−vi) (Impulse)
    Δd = 1/2(vi+vf)Δt

    Solution 2:
    W=F⋅d
    Wnet=ΔK

    3. The attempt at a solution

    Solution 1:

    Use the inelastic collision formula to solve for the final velocity (vf).
    m1v1 + m2v2 = (m1+m2)vf
    (71kg)(2.1m/s) + (52kg)(0) = (71kg + 52kg)vf
    vf = 1.2122m/s

    Plug that into the impulse forumla.
    F⋅Δt=Δp =m(vf−vi)
    Δt = Δp/F
    Δt = m(vf - vi)/μk⋅m⋅g (masses cancel out)
    Δt = vf - vi/μk⋅g
    Δt = (0 - 2.1m/s)/0.052⋅9.81 m/s2
    Δt = 2.3763s

    Plug the time into the the second kinematics equation.
    Δd = 1/2(vi+vf)Δt
    Δd = 1/2(2.1m/s + 1.2122m/s)(2.3763s)
    Δd = 1.44 m

    Solution 2:

    Use the inelastic collision formula to solve for the final velocity (vf).
    m1v1 + m2v2 = (m1+m2)vf
    (71kg)(2.1m/s) + (52kg)(0) = (71kg + 52kg)vf
    vf = 1.2122m/s

    Combine the two work equations.
    W=F⋅d
    Wnet=ΔK
    F⋅d=ΔK
    Δd = ΔK/F
    Δd = 1/2(m1 + m2)vf2 - 1/2(m1)(v1)2 / μk⋅m⋅g

    Its messy, so I'll skip plugging in all the values.

    Δd = -66.185/-62.74476
    Δd = 1.0548m

    Both solutions seem reasonable to me, but why am I getting two different answers? Which solution is more accurate?
     
  2. jcsd
  3. Apr 14, 2013 #2

    gneill

    User Avatar

    Staff: Mentor

    Hi Ketchup1, Welcome to Physics Forums.

    I think something's gone wrong in your second solution. To find the distance traveled via the work done keep in mind that the initial velocity is that of the post-collision pair, and the final velocity is zero (they come to rest when their traveling is done).
     
  4. Apr 14, 2013 #3
    Oh! That works out! I didn't think of that for some reason
    Thank you very much :D
    The answer works out perfectly.
     
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