Acompletely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axis. just before the collision, one ball of mass 3.0 kg is moving upward at 20 m/s, and the other ball, of mass 2.0 kg, is moving downward at 14 m/s. how high do the combined two balls of putty rise above the collision point? (neglect air drag.)

h = 2.087 m

Explanation:

Given

m₁ = 3 kg

v₁ = 20 m/s

m₂ = 2 kg

v₂ = - 14 m/s

In a completely inelastic collision the colliding objects stick together after the collision and move together as a single object.

In the given problem, lets assume that the balls of putty are initially moving along the  y  axis, upward direction being the positive  y  direction. And the collision occurs at the origin of the coordinate system.

We can apply the equation

vs = (m₁*v₁ + m₂*v₂) / (m₁ + m₂)  

⇒   vs = (3 kg*20 m/s + 2 kg*(- 14 m/s)) / (3 kg + 2 kg)  

⇒   vs = 6.4 m/s (↑)

To calculate the maximum height  h  attained by the combined system of two balls of putty after the the collision, we use the expression for linear motion under gravity:

vf² = vi² - 2*g*h

where

vf = 0 m/s  

g = 9.81 m/s²

vi = vs = 6.4 m/s

finally we get h:

h = vi² / (2*g)

⇒   h = (6.4 m/s)² / (2*9.81 m/s²) = 2.087 m

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2