Question

Tarzan, who has a mass of 75 kg, holds onto the end of a vine that is at a 12 angle from the vertical. He steps off his branch and, just at the bottom of his swing, he grabs onto his chimp friend Cheetah, whose mass is 35 kg. Part A What is the maximum angle the rope reaches as Tarzan swings to the other sido? Express your answer in degrees. θ ΑΣΦ 0, Submit Request Answer

Answer 1

Consider that the length of the vine is R.
Initial potential energy with respect to the bottom, PE = (M * g * R) - (M * g * R * cos$\theta$i)
Where M is Tarzon's mass, $\theta$ i = 12 degrees
PE = M * g * R (1 - cos$\theta$i)
At the bottom, the kinetic energy, KE = 1/2 * M * Vi²
Where Vi is the velocity at the bottom

Using conservation of energy,
1/2 * M * Vi² = M * g * R (1 - cos$\theta$i)
Vi² = 2 * g * R * (1 - cos$\theta$i)
= 2 * g * R * (1 - cos(12))
Vi² = 2 * g * R * 0.022 ...(1)
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Consider that Tarzan made a collision with chimp and the final velocity of Tarzan+chimp is Vf.
Using conservation of momentum,
M * Vi = (m + M) * Vf
Where m is the mass of the chimp
Vf = [M / (m + M)] * Vi
= [75 / (75 + 35)] * Vi
= 0.682 * Vi ...(2)
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The kinetic energy of Tarzan+chimp just after the collision,
Eb = 1/2 * (m + M) * Vf²
Potential energy of Tarzan+chimp at the maximum angle,
Et = (m + M) * g * R * (1 - cos$\theta$f)

Using conservation of energy, Eb = Et
1/2 * (m + M) * Vf² = (m + M) * g * R * (1 - cos$\theta$f)
Vf² = 2 * g * R * (1 - cos$\theta$f)
Substituting (2),
(0.682 * Vi)² = 2 * g * R * (1 - cos$\theta$f)
0.465 * Vi² = 2 * g * R * (1 - cos$\theta$f)
Substituting (1),
0.465 * [2 * g * R * 0.022] = 2 * g * R * (1 - cos$\theta$f)
0.010 = (1 - cos$\theta$f)
cos$\theta$f = 0.9898
$\theta$ f = cos^-1(0.9898)
= 8.2 degrees