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Collision Question (EASY?)

  • Thread starter twenty5
  • Start date
So yeah, I just wanna 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!
 
Last edited:

rock.freak667

Homework Helper
6,230
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So yeah, I just wanna 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:

[tex]V_1'(m_1+m_2)=m_1v_1-m_2v_1[/tex]

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

[tex]V_1'(m_1+m_2)=m_1v_1-m_2v_1[/tex]

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? :)


[tex]
V_1single-quote(m_1+m_2)=m_1v_1-m_2v_1
[/tex]

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

rock.freak667

Homework Helper
6,230
31
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.
 
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
 
m1v1 = (m1 + m2 ) V'

for completely inelastic collision because 1 final velocity when they stick together at the end and... m1 + m2 same mass?
 
phew okay thanks a whole bunch and another bunch :D
 

rock.freak667

Homework Helper
6,230
31
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
 

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