Question

Ask Your Teacher 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:

The maximum height is 1.41 m.

Explanation:

Given that,

Mass of actor = 83.0 kg

Radius = 3.90 m

Mass of costar = 55.0 kg

We need to calculate the velocity of swing

Use loss in gravitation energy = gain in kinetic energy

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

Put the value into the formula

$83.0\times9.8\times3.90=\dfrac{1}{2}\times83.0\times v^2$

$v^2=\dfrac{83.0\times9.8\times3.90\times2}{83.0}$

$v=\sqrt{\dfrac{83.0\times9.8\times3.90\times2}{83.0}}$

$v=8.74\ m/s$

We need to calculate the velocity of the system

Using conservation of momentum

$mv=(m_{1}+m_{2})v'$

Put the value into the formula

$83.0\times8.74=(83.0+55.0)\times v;$

$v'=\dfrac{83.0\times8.74}{138}$

$v'=5.25\ m/s$

We need to calculate the maximum height

Using formula of energy

Gain in gravitational energy = lost in kinetic energy

$mgh=\dfrac{1}{2}mv'^2$

$9.8\times h=\dfrac{1}{2}\times(5.25)^2$

$h=\dfrac{5.25^2}{2\times9.8}$

$h=1.41\ m$

Hence, The maximum height is 1.41 m.

Answer 2

Answer:

Explanation:

mass of actor, M = 83 kg

mass of costar, m = 55 kg

length, r = 3.9 m

let the speed of the actor is u before collision

By use of conservation of energy

mgh = 0.5 mu²

9.8 x 3.9 = 0.5 x u²

u = 8.74 m/s

Use conservation of momentum

M x u = ( M + m) x v

where v is the velocity after collision

83 x 8.74 = ( 83 + 55) x v

v = 5.26 m/s