Complete question is;
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 12 m/s. How high do the combined two balls of putty rise above the collision point? (Neglect air drag.)
Answer:
2.64 m
Step-by-step explanation:
We are told the collision between the two balls is completely inelastic. This means that they will stick together after collision.
Since momentum is conserved and they stick together after collision, the formula for inelastic collision is;
m1•v1 + m2•v2 = (m1 + m2)v
Where v is the velocity after Collision.
We are given;
m1 = 3 kg
v1 = 20 m/s
m2 = 2 kg
v2 = -12 m/s (negative because it's moving downwards)
Thus, plugging into the equation, we have;
(3 × 20) + (2 × -12) = (3 + 2)v
60 - 24 = 5v
36 = 5v
v = 36/5
v = 7.2 m/s
To find the height that the combined two balls of putty rise above the collision point, we will use the formula;
v² = u² + 2gh
After the collision, initial velocity is the one they moved with after colliding. Thus, u = 7.2 m/s and final velocity is v = 0 m/s
Making h the subject and Plugging in relevant values, we have;
h = (v² - u²)/2g
Since gravity is acting in a direction opposite to the motion, then g = -9.81 m/s²
Thus;
h = (0² - 7.2²)/(2 × -9.81)
h = -51.84/-19.62
h = 2.64 m