George of the Jungle, with mass m. swings on a light vine hanging from a station

Question

George of the Jungle, with mass m. swings on a light vine hanging from a stationary tree branch. A second vine of equal length hangs from the same point, and a gorilla of larger mass M swings in the opposite direction on it. Both vines are horizontal when the primates start from rest at the same moment. (George and the gorilla meet at the lowest point of their swings. Each is afraid that one vine will break, so they grab each other and hang on. They swing upward together, reaching a poi??t where the vines make angle of 35.0 degree with the vertical. Find the value of the ratio m/M.

Explanation / Answer

Velocity just before collision
1/2mv2=mgh {not dependend on mass}

v = (2gh) for both man and gorilla

Let l be the lenght of the vine then h = l (initial position is horizontal)

v = (2gl).....1

Now by conservation of momentum

Mv - mv = (m+M) vf ; vf is there combined speed after they grab each other

vf = v(M-m)/(M+m)....2

now height achived = l*cos21.7

1/2(M+m)vf2 = (M+m)g *lcos21.7

using 1 and 2

1/2*v2 (M-m)2/(M+m)2=glcos21.7

=> 2gl*(M-m)2/(M+m)2= 2glcos21.7

=>(M-m)2/(M+m)2 = cos 21.7

=> (M-m)/(M+m) = cos21.7

using componendo and dividendo (Ratio Proportion)

2M/(-2m) = (cos21.7 + 1)/(cos21.7 - 1)

m/M = (1 - cos21.7)/(1 + cos21.7)

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.