Question

In a Broadway performance, an 80.0-kg actor swings from a 3.75-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

Explanation:

It is given that,

Mass of actor, m₁ = 80 kg

Length of cable, h₁ = 3.75 m

Mass of costar, m₂ = 55 kg

Using the conservation of energy to find the speed of the performer at the bottom of the arc as :

Potential energy at start = kinetic energy at end

$mgh_1=\dfrac{1}{2}mv_1^2$

$v_1=\sqrt{2gh_1}$

$v_1=\sqrt{2\times 9.8\times 3.75}$

$v_1=8.57\ m/s$

Let v₂ is its speed after the collision. Now using the conservation of momentum as :

$m_1v_1=(m_1+m_2)v_2$

$v_2=\dfrac{m_1v_1}{(m_1+m_2)}$

$v_2=\dfrac{80\times 8.57}{80+55}$

$v_2=5.07\ m/s$

Let h₂ is the maximum height they reach after their upward swing. Again using the conservation of energy as :

$\dfrac{1}{2}mv_2^2=mgh_2$

$h_2=\dfrac{v_2^2}{2g}$

$h_2=\dfrac{(5.07)^2}{2\times 9.8}$

$h_2=1.31\ m$

Hence, this is the required solution.