Which collision equations go with which situations?

[twenty5]

So yeah, I just want to get some things cleared up... with collision equations...

OK SO..

QUESTION:
which equations go with which situations?
+++++++++++++
Elastic collision:
----------------
1] V₁' = V₁ ( m₁ - m₂ ) / (m₁ + m₂ )

2] V₂' = 2m₁v₁ / (m₁ + m₂ )
1] and 2] are for ones where one object is stationary

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

++++++++++++++

Inelastic Collision:
-----------------

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂l

+++++++++++++++++

complete Inelastic Collisions:
----------------------------
m₁v₁ = (m₁ + m₂ ) V'

just want to get them straighten out for the test tomorrow ^^ and If I'm missing any... can you let me know thanks in advanced!

 

[rock.freak667]

So yeah, I just want to get some things cleared up... with collision equations...

OK SO..

QUESTION:
which equations go with which situations?
+++++++++++++
Elastic collision:
----------------
1] V₁' = V₁ ( m₁ - m₂ ) / (m₁ + m₂ )

2] V₂' = 2m₁v₁ / (m₁ + m₂ )
1] and 2] are for ones where one object is stationary

well for 1] and 2] let's rewrite these equations:

V_1'(m_1+m_2)=m_1v_1-m_2v_1

I am assuming the left side is what happened after collision.

So what does the term (m1+m2) symbolize for the bodies? and notice how on the right side the velocity is the same, what does the negative sign mean?

Also your equation for a completely inelastic collision is correct. But just know that the equation above it is for any collision in general i.e. it applies for both elastic and inelastic.

 

[twenty5]

well for 1] and 2] let's rewrite these equations:

V_1'(m_1+m_2)=m_1v_1-m_2v_1

I am assuming the left side is what happened after collision.

So what does the term (m1+m2) symbolize for the bodies? and notice how on the right side the velocity is the same, what does the negative sign mean?

uhm yup it's for ""after" collision ,and m1 + m2 is like when they are 1 mass I believe...so is everythiing else alrite? :) <br /> V_1single-quote(m_1+m_2)=m_1v_1-m_2v_1<br />

basically, on the right side , you can factor our v1 and then just divide by (m1 + m2) from the left side ;)

 

[rock.freak667]

uhm yup it's for ""after" collision ,and m1 + m2 is like when they are 1 mass I believe...so is everythiing else alrite? :)

Yep, meaning that the bodies stick together (so it is a completely inelastic collision) but when doing questions with momentum, you must always take into account direction.

So if we take +ve as moving to the right then -ve is to the left, right?
So from the right side of the equation m1 is moving to the right and m2 is moving to the left.

 

[twenty5]

Yep, meaning that the bodies stick together (so it is a completely inelastic collision) but when doing questions with momentum, you must always take into account direction.

So if we take +ve as moving to the right then -ve is to the left, right?
So from the right side of the equation m1 is moving to the right and m2 is moving to the left.

mmm kay

 

[rock.freak667]

mmm kay

so you should be good to go now.

 

[twenty5]

m₁v₁ = (m₁ + m₂ ) V'

for completely inelastic collision because 1 final velocity when they stick together at the end and... m1 + m2 same mass?

 

[twenty5]

phew okay thanks a whole bunch and another bunch :D

 

[rock.freak667]

m₁v₁ = (m₁ + m₂ ) V'

for completely inelastic collision because 1 final velocity when they stick together at the end and... m1 + m2 same mass?

Well the m1+m2 means the bodies stick together, so the new mass of the body is the sum of the masses m1 and m2