## Elastic Collision with 2 Carts

1. The problem statement, all variables and given/known data
There are two carts, A and B, they hit each other and bounce off (Elastic)
Here is what is given: vi=0
Mass of A=.8kg I call it Ma
Mass of B=1.6 kg I call it Mb
Vbo=1.0 m/s
vib=0
vfb=.3 m/s

2. Relevant equations

conservation of Momentum I have p1 = p2 Ma Vai + Mb Vbi = Ma Vaf + Mbf Vbf The question is what is Vf on cart a? I do not know what to do after the conservation of momentum

3. The attempt at a solution
The question is what is Vf on cart a?

conservation of Momentum I have p1 = p2 Ma Vai + Mb Vbi = Ma Vaf + Mbf Vbf I do not know what to do after the conservation of momentum
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
 Is this correct? Ma Vao + Mb Vbo = MaVaf + Mbf Vbf .8 Vao + 1.6 Vbo = .8 Vaf + 1.6 Vbf .8 + 0 + 1.6 (1) = .8 Vaf + 1.6 (.3) 1.6 (.7) = .8 Vaf Vaf= sqrt 2.07 Vaf= 1.4 m/s ?
 Recognitions: Homework Help Science Advisor This is a bit confusing because you keep changing the names of the variables. In particular, what are Vbo and vib? Seem like they should be the same thing (original = initial?), but you quote different values for them. And why do you take sqrt at the end?

## Elastic Collision with 2 Carts

They are the same and I don't know
 Ma Vao + Mb Vbo = MaVaf + Mbf Vbf this is the same as Ma Vai + Mb Vbi = Ma Vaf + Mbf Vbf and I haven't changed the values, they are still Here is what is given: vi=0 Mass of A=.8kg I call it Ma Mass of B=1.6 kg I call it Mb Vbo=1.0 m/s vib=0 vfb=.3 m/s

Recognitions:
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 Quote by SherBear Ma Vao + Mb Vbo = MaVaf + Mbf Vbf this is the same as Ma Vai + Mb Vbi = Ma Vaf + Mbf Vbf and I haven't changed the values, they are still Here is what is given: vi=0 Mass of A=.8kg I call it Ma Mass of B=1.6 kg I call it Mb Vbo=1.0 m/s vib=0 vfb=.3 m/s
So Vbo and Vbi mean the same, but what is vib (which has a different value)? And is vfb the same as Vbf? If not, please define all the variables.
 typo vib=0 is Vao = 0, because initially it is at rest
 I have it figured out, I will post what I have after school
 Before the crash, Cart A is at rest and cart B is on the right side moving toward cart A. (moving toward cart A negatively) After the crash, Cart A would be moving to the left if cart B crashed into it, and cart B would bounce off and move the other way to the right? Taking the Conservation of Momentum Theorm Ma(Vao) + Mb (Vbo) = Ma(Vaf) + Mb(Vbf) .8kg + 0 + 1.6 kg (-1.0 m/s) = .8 (Vaf) + 1.6 kg (.3 m/s) Vaf = .8kg + 0 + 1.6 kg (-1.0 m/s) / .8 kg + 1.6 kg (.3 m/s) Vaf= -0.625 m/s ?

Recognitions:
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 Quote by SherBear Before the crash, Cart A is at rest and cart B is on the right side moving toward cart A. (moving toward cart A negatively) After the crash, Cart A would be moving to the left if cart B crashed into it, and cart B would bounce off and move the other way to the right? Taking the Conservation of Momentum Theorm
It's not a theorem, it's a law.
 Ma(Vao) + Mb (Vbo) = Ma(Vaf) + Mb(Vbf) .8kg + 0 + 1.6 kg (-1.0 m/s) = .8 (Vaf) + 1.6 kg (.3 m/s)
How did "Ma(Vao) " become ".8kg + 0"?
Also, you originally gave Vbo as 1m/s and Vbf as .3 m/s. Having arranged the model with B on the positive axis side of A, you've found it necessary to make Vbo negative, but you haven't changed the sign of Vbf. I think you'll find that will mean the system actually gained energy!
 .8kg(0) When two things collide and are elastic do they go separate ways if elastic or the same way? That's why I didn't make .3 m/s negative.

Recognitions:
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