Question

A 1.2kg coconut is hanging from a vine at a height of 2 meters above the ground. When the coconut is dropped it swings down and hits a .8kg lemur, who sticks to the coconut. The coconut then continues its arc up the other side.

a. What is the speed of the coconut before it hits the lemur?

b. What is the speed of the coconut and lemur after the collision?

c. If the lemur wants to reach a persimmon that is at a height of 1.2 m above the ground on the far side of the swing, will it be able to?

d. Through what maximum angle does the lemur swing?

Answer #1

here,

the mass of coconut , m1 = 1.2 kg

the mass of lemur , m2 = 0.8 kg

height , h1 = 2 m

a)

the speed of the coconut before it hits the lemur , u = sqrt(2 * g * h1)

u = sqrt(2*9.81*2) m/s

u = 6.26 m/s

b)

let the speed of the coconut and lemur after the collison be v

using conservation of momentum

m1 * u = (m1 + m2) * v

1.2 * 6.26 = (1.2 + 0.8 ) * v

v = 3.76 m/s

the speed of the coconut and lemur after the collison is 3.76 m/s

c)

the maximum height gained , h2 = u^2 /2g

h2 = 3.76^2 /(2*9.81) m = 0.72 m

as h2 < 1.2 m

the lemur will not be able to reach this height

d)

let the maximum angle be theta

h1 * ( 1 - cos(theta)) = h2

2 * ( 1 - cos(theta)) = 0.72

theta = 50.2 degree

the maximum angle is 50.2 degree

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