Question

A completely 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.)

Answer 1

h = 2.087 m

Step-by-step 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