# Elastic Collision with 2 Carts

1. Oct 26, 2012

### SherBear

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

2. Oct 26, 2012

### SherBear

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 ?

3. Oct 26, 2012

### haruspex

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?

4. Oct 27, 2012

### SherBear

They are the same and I don't know

5. Oct 27, 2012

### 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

6. Oct 28, 2012

### haruspex

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.

7. Oct 29, 2012

### SherBear

typo vib=0 is Vao = 0, because initially it is at rest

8. Oct 29, 2012

### SherBear

I have it figured out, I will post what I have after school

9. Oct 29, 2012

### 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

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. Oct 29, 2012

### haruspex

It's not a theorem, it's a law.
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. Oct 30, 2012

### SherBear

.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. Oct 30, 2012

### haruspex

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. Oct 30, 2012

### SherBear

Ok I understand. Thank you for all your time and patience. =-)