Inelastic Collision Mass Ratio

[DkMuse]

Homework Statement

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.

Homework Equations

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

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?

 

[DkMuse]

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.

 

[DkMuse]

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.