Question

In a Broadway performance, an 83.0-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

Answer:

$h = 2.821\,m$

Explanation:

The speed of the actor before the collision is found by means of the Principle of Energy Conservation:

$(83\,kg)\cdot(9.807\,\frac{m}{s})\cdot (3.90\,m) = \frac{1}{2}\cdot (83\,kg)\cdot v^{2}$

$v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (3.90\,m)}$

$v \approx 8.746\,\frac{m}{s}$

The speed after the inelastic collision is obtained by using the Principle of Momentum Conservation:

$(83\,kg)\cdot (8.746\,\frac{m}{s} )+(55\,kg)\cdot (0\,\frac{m}{s} ) = (83\,kg + 55\,kg)\cdot v$

$v = 5.260\,\frac{m}{s}$

Lastly, the maximum height is determined by using the Principle of Energy Conservation again:

$\frac{1}{2}\cdot (138\,kg)\cdot (5.260\,\frac{m}{s} )^{2} = (138\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot h$

$h = \frac{(5.260\,\frac{m}{s} )^{2}}{9.807\,\frac{m}{s^{2}} }$

$h = 2.821\,m$