Question

At the same instant that a 0.50-kg ball is dropped from 25 m above Earth, a second ball, with a mass of 0.25 kg, is thrown straight upward from Earth's surface with an initial velocity of 19.6 m/s. They move along nearby lines and pass each other without colliding. When the second ball is at its highest point, what is the velocity of the center of mass of the two-ball system

Answer 1

The velocity of the center of mass of the two-ball system is 13.1 m/s.

Step-by-step explanation:

Given;

mass of the first ball, m₁ = 0.5 kg

mass of the second ball, m₂ = 0.25 kg

initial velocity of the second ball, u₂ = 19.6 m/s

At the highest point the velocity of the second ball, v₂ = 0

The highest point reached by the second ball is calculated as;

v₂² = u₂² - 2gh

0 = u₂² - 2gh

2gh = u₂²

h = u₂² / 2g

h = (19.6²) / (2 x 9.8)

h = 19.6 m

The final velocity of the first ball when it had traveled 19.6 m down;

v₁² = u₁² + 2gh

v₁² = 0 + 2gh

v₁ = √2gh

v₁ = √(2 x 9.8 x 19.6)

v₁ = 19.6 m/s

The velocity of the center of mass of the two-ball system is calculated as;

`v = (m_1v_1 \ + \ m_2v_2)/(m_1 \ + \ m_2) \\\\v = (0.5* 19.6 \ + \ 0.25* 0)/(0.5 \ + \ 0.25) \\\\v = 13.1 \ m/s`