Question

In a Broadway performance, an 85.0-kg actor swings from a R = 3.90-m-long cable that is horizontal when he starts. At the bottom of his arc, he picks up his 55.0-kg costar in an inelastic collision. What maximum height do they reach after their upward swing?

Answer 1

To solve this problem it is necessary to apply the concepts related to the conservation of the moment and the conservation of energy.

The conservation of energy implies that potential energy is transformed into kinetic energy or vice versa, depending on the process, in this way

$KE = PE$

$\frac{1}{2}mv^2 = mgh$

Where,

m = mass

v = velocity

h = height

g = gravitational acceleration

Re-arrange to find v,

$\frac{1}{2}mv^2 = mgh$

$\frac{1}{2}v^2 = gh$

$v = \sqrt{2gh_1}$

Replacing with our values we have that

$v = \sqrt{2gh_1}$

$v = \sqrt{2(9.8)(3.9)}$

$v = 8.74m/s$

At this point we can find the final speed through the conservation of the momentum, that is

$mv = (m+M)V$

Here,

m = mass of the first person

M = Mass of the second person

v = Initial velocity

V = Final Velocity

Replacing,

$mv = (m+M)V$

$(85)(8.74) = (85+55)V$

$V = \frac{(85)(8.74) }{(85+55)}$

$V = 5.30m/s$

Applying energy conservation again, but in this case using the values of the final state we have to

$\frac{1}{2}(m+M)V^2 = (m+M)gh_2$

$\frac{1}{2}V^2 = gh_2$

$h_2 = \frac{V^2}{2g}$

$h_2 = \frac{5.3^2}{2*9.8}$

$h_2 = 1.433m$

Therefore the maximum height that they reach after their upward swing is 1.433m