SOLUTION: Two billiard balls collide. Ball 1 moves with a velocity of 4 m/s, and ball 2 is at rest. After the collision, ball 1 comes to a complete stop. What is the velocity of ball 2 after

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Question 1198185: Two billiard balls collide. Ball 1 moves with a velocity of 4 m/s, and ball 2 is at rest. After the collision, ball 1 comes to a complete stop. What is the velocity of ball 2 after the collision? Is this collision elastic or inelastic? The mass of each ball is 0.16 kg.
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Refer to this page
https://openstax.org/books/physics/pages/8-3-elastic-and-inelastic-collisions
Quote: "In an elastic collision, the objects separate after impact and don’t lose any of their kinetic energy."
also
Quote: "An inelastic collision is one in which objects stick together after impact, and kinetic energy is not conserved."

We can see that since the billiard balls don't stick together, and instead separate, this means the collision is elastic.

m1 = mass of the first ball
vi1 = initial velocity of the first ball
vf1 = final velocity of the first ball

m2 = mass of the second ball
vi2 = initial velocity of the second ball
vf2 = final velocity of the second ball

Due to the elastic collision, we can use this conservation of momentum formula
m1*vi1 + m2*vi2 = m1*vf1 + m2*vf2
an example problem can be found here
https://www.ck12.org/physics/Elastic-and-Inelastic-Collisions/lesson/Elastic-and-Inelastic-Collisions-PHYS/

We have these values
m1 = m2 = 0.16 kg
vi1 = 4 m/s
vf1 = 0 m/s (since the 1st ball comes to a stop)
vi2 = 0 m/s (the 2nd ball starts at rest)
vf2 = x = unknown

Let's determine x
m1*vi1 + m2*vi2 = m1*vf1 + m2*vf2
0.16*4 + 0.16*0 = 0.16*0 + 0.16*x
0.16*4 = 0.16*x
x = 4
It's not particularly interesting since each ball has the same mass, so we expect the amount of speed going in is the amount coming out (assuming no kinetic energy is lost to heat)
I suggest you try different mass values for each ball to see how the vf2 value will change.

Further reading
https://www.khanacademy.org/science/physics/linear-momentum/elastic-and-inelastic-collisions/a/what-are-elastic-and-inelastic-collisions

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Answers:
Velocity of the 2nd ball, after the collision, is 4 m/s
The collision is elastic.

Answer by ikleyn(52831) About Me  (Show Source):
You can put this solution on YOUR website!
.
Two billiard balls collide. Ball 1 moves with a velocity of 4 m/s, and ball 2 is at rest.
After the collision, ball 1 comes to a complete stop. What is the velocity of ball 2 after the collision?
Is this collision elastic or inelastic? The mass of each ball is 0.16 kg.
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The logic and explanations in the post by @math_tutor2020 are not perfectly correct,
so I came to explain everything in a right way.

In particular, tutor @math_tutor2020 writes quoting

    "An inelastic collision is one in which objects stick together after impact, 
     and kinetic energy is not conserved."

And then @math_tutor2020 makes a conclusion

    We can see that since the billiard balls don't stick together, 
    and instead separate, this means the collision is elastic.


        It is NOT CORRECT. By the definition, a collision of two bodies is called inelastic collision
        if two bodies remain separate, but kinetic energy is not conserved.


The "sticking together" condition determines OTHER type of collision, called
"plastic collision", or, in other terminology, "perfectly plastic collision"

But "inelastic collision" means loosing of kinetic energy, only: it is transformed
into other types of energy (heat energy or interior energy).


            The solution to the given problem is as follows:

Two bals have the same mass of 0.16 kg.
Before collision, ball 1 moves with a velocity of 4 m/s; ball 2 is at rest.
After collision, ball 1 comes to a complete stop.


To determine the speed of the second ball, use the momentum conservation law,
which is held at any collision (elastic, inelastic, plastic).


it says  "momentum before the collision = momentum after the collision",  or,  for given conditions, 


    m%5B1%5D%2Av%5B1%5D = m%5B2%5D%2Av%5B2%5D,  or   0.16%2A4 = 0.16%2Av%5B2%5D,  which implies  v%5B2%5D = 4 m/s.


So, the balls "exchanged" their velocities after the collision.

It makes it obvious that the kinetic energy is conserved;  so, the collision is ELASTIC.

Solved, with complete explanations.

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This topic on collisions requires full and completely clear understanding, if a person claims to be an expert.