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Inelastic Collision Mass Ratio

  1. Oct 21, 2007 #1
    1. The problem statement, all variables and given/known data
    George of the Jungle, with mass m, swings on a light vine hanging from a
    stationary branch. A second vine of equal length hangs from the same point,
    and a gorilla of larger mass M swings in the oposite direction on it.
    Both vines are horizontal when the primates start from rest at the same
    moment. George and the ape meet at the lowest point in their swings.
    Each is afraid that one vine will break, so they grab each other and hang on.
    They swing upward together, reaching a point where the vines make an
    angle of 35.0° with the vertical.
    Find the value of the ratio m/M.

    2. Relevant equations
    Angular Momentum - L = mvr
    Potential Energy - U = mvh
    Kinetic Energy - K = 1/2 mv^2

    3. The attempt at a solution
    When both the man and ape start from rest the potential energy is
    U = mgh, where h = r-rcosθ, r being the length of the vine
    at 0°, cos 0° = 1, r-r =0
    so U = mg

    The knetic energy at the bottom given by K = 1/2 mv^2
    setting an equality of K=U
    1/2 mv^2 = mg
    v= sqrt 2g
    This would apply to both the man and the ape.
    vM = vm = vf - right?

    The potential energy of the man/ape system would be
    U = (M+m) g r(1-cos 35°)

    The kinetic energy of the man/ape system would be
    1/2v^2(M-m)
    So
    1/2v^2(M-m) = (M+m) g r(1-cos 35°)
    would be one equation for energy
    and
    Mvr + mvr = (M+m)vr
    would equate angular momentum

    ... How do I continue to get a ratio of m/M?
     
  2. jcsd
  3. Oct 22, 2007 #2
    Would it be proper to say the man lost his kinetic energy after the
    collision and write the final energy equation as:
    1/2(M+m)vF^2 = (M+m)gh where h= r(1-cos35)
    and vF = sqrt 2gr(1-cos35)

    then vM would be sqrt 2g

    Um = m(vm)gh, where h = (r-rcos180), cos 180 =-1, r-(-r) r+r = 2r
    1/2m(vm)^2 = m(vm)g2r
    vm would be sqrt 2g4r

    Do I then plug the 3 velocities into the first momentum equation:
    M vM + m vm = (M+m) vF

    I still don't think enough will cancel out to me a ratio of m/M.
     
  4. Oct 24, 2007 #3
    If I go back to my original assumption that both man and ape have the same initial
    velocity and discount that one is at 0 degrees and the other is at 180 degrees
    I can write their energy formulas as:
    mgR = 1/2mv^2 and MgR = 1/2Mv^2
    v = sqrt 2gR
    Then if I write the momentum formula as:
    (M-m)v = (M+m)v2
    and write the collision energy formula as
    1/2(M+m)v2^2 = (M+m) gR(1-cos35)
    v2 = sqrt 2gR(1-cos35)
    Sub back into the momentum formula with the two
    velocities:
    (M-m)sqrt 2gR = (M+m) sqrt 2gr (1-cos35)
    From there it is pretty straightforward.
     
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