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 speed of 15 m/s. They move along nearby lines and pass each other without colliding. At the end of 2.0 s, the height above Earth's surface of the center of mass of the two-ball system is:

Answer 1

The center of mass of the two-ball system is 7.05 m above ground.

Step-by-step explanation:

Motion of 0.50 kg ball:

Initial speed, u = 0 m/s

Time = 2 s

Acceleration = 9.81 m/s²

Initial height = 25 m

Substituting in equation s = ut + 0.5 at²

s = 0 x 2 + 0.5 x 9.81 x 2² = 19.62 m

Height above ground = 25 - 19.62 = 5.38 m

Motion of 0.25 kg ball:

Initial speed, u = 15 m/s

Time = 2 s

Acceleration = -9.81 m/s²

Substituting in equation s = ut + 0.5 at²

s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m

Height above ground = 10.38 m

We have equation for center of gravity

`\bar{x}=(m_1x_1+m_2x_2)/(m_1+m_2)`

m₁ = 0.50 kg

x₁ = 5.38 m

m₂ = 0.25 kg

x₂ = 10.38 m

Substituting

`\bar{x}=(0.50* 5.38+0.25* 10.38)/(0.50+0.25)=7.05m`

The center of mass of the two-ball system is 7.05 m above ground.