What is the Final Velocity of Cart A After an Elastic Collision?

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Homework Help Overview

The discussion revolves around an elastic collision between two carts, A and B, with given masses and initial velocities. The original poster seeks to determine the final velocity of cart A after the collision, using the conservation of momentum principle.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of the conservation of momentum equation, questioning variable definitions and consistency in the values used. There are attempts to clarify the relationship between initial and final velocities, as well as the direction of motion post-collision.

Discussion Status

The discussion includes various interpretations of the problem setup, with some participants providing insights into the implications of mass and direction on the outcome of the collision. There is ongoing clarification regarding variable names and their meanings, but no consensus has been reached on the final velocity of cart A.

Contextual Notes

Participants note potential confusion arising from variable naming conventions and the need for consistent definitions. There is also a mention of assumptions regarding directionality in the context of the collision, which may affect the interpretation of the results.

SherBear
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Homework Statement


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


Homework 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


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
 
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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 ?
 
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?
 
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
 
SherBear said:
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 theorem

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 ?
 
  • #10
SherBear said:
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 theorem
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!
 
  • #11
.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.
 
  • #12
SherBear said:
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.
Whether you should have made the .3m/s negative depends on the information provided. You were given two speeds with the same sign. If no direction was specified you should assume they were in the same direction, so if you flip the sign of one you must flip that of the other.
But to answer your question, it depends on the relative masses. If momentum and energy are both conserved, and one object starts at rest:
- if the rest object is the lighter (the case we have here), they will move off in the same direction.
- if the rest object is heavier, the lighter object will bounce back
- if equal in mass, the impacting object will come to rest.
 
  • #13
Ok I understand. Thank you for all your time and patience. =-)
 

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