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Elastic collision with conservation of energy and momentum?

  • Thread starter zell_D
  • Start date
  • #1
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Homework Statement


a moving train collides with a train that is not moving, and the trains use their springy bumpers to bounce off of each other without damage. Assume diff masses for train A and B. Identify your isolated system, solve for the final velocity of EACH train in terms of INITIAL VELOCITY of the initially moving train and the 2 masses


Homework Equations


P=mv
Conservation of momentum: Pfinal=Pinitial
Assumption of no external forces, spring as part of the system. k not addressed

The Attempt at a Solution


pi=[mA](viA) (since B is at rest)
pf=[mA](vfA)+[mB](vfB)

pi=pf
[mA](viA)=[mA](vfA)+[mB](vfB)
since A's final velocity defined as in the negative direction in my system
[mA](viA)=-[mA](vfA)+[mB]()

vfA = [[mB]vfB-[mA](viA)]/mA
vfB = [[mA]vfA+[mA](viA)]/mB

my question this is the furthest i got to. I do not know if i am right or wrong. But it seems like thats not the furtherst i can solve this problem. since the final velocity variable still are present in both
 

Answers and Replies

  • #2
Doc Al
Mentor
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The Attempt at a Solution


pi=[mA](viA) (since B is at rest)
pf=[mA](vfA)+[mB](vfB)

pi=pf
[mA](viA)=[mA](vfA)+[mB](vfB)
Good.
since A's final velocity defined as in the negative direction in my system
[mA](viA)=-[mA](vfA)+[mB]()
OK, but no need for this step. It's OK to have negative values for V.

vfA = [[mB]vfB-[mA](viA)]/mA
vfB = [[mA]vfA+[mA](viA)]/mB
This won't help. What you need is a another independent equation to use in addition to conservation of momentum. Hint: What else is conserved? (See the title of this thread!)
 
  • #3
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energy is conserved....

so KEiA+KEiB=KEfA+KEfB?
KE initial for B = 0 so
KEiA=KEfA+KEfB

(1/2)mA[vi^2]=(1/2)mA[vfA^2]+(1/2)mB[vfB^2]

but wouldnt i end up with both final velocities again?
 
  • #4
Doc Al
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44,941
1,202
energy is conserved....

so KEiA+KEiB=KEfA+KEfB?
KE initial for B = 0 so
KEiA=KEfA+KEfB

(1/2)mA[vi^2]=(1/2)mA[vfA^2]+(1/2)mB[vfB^2]
Good.

but wouldnt i end up with both final velocities again?
You must combine both equations--solve them simultaneously. (This one and the one you had before from momentum conservation.)

Just take one of the final equations from your last post and substitute into this one. Then you can solve for both unknowns.
 
  • #5
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are the conservation of momentum equations right? since i made the velocity negative and all

oh and do i use FOIL since there are going to be squares lol havent done algebra in like forever -_-
 
  • #6
Doc Al
Mentor
44,941
1,202
are the conservation of momentum equations right? since i made the velocity negative and all
As long as you remember how you defined the sign.

oh and do i use FOIL since there are going to be squares lol havent done algebra in like forever -_-
Sounds like a plan to me.
 

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