Elastic Collision between billiard balls

[testme]

Homework Statement

Two billiard balls of equal mass undergo a perfectly elastic head on collision. If one ball's initial speed was 2.00 m/s, and the other's was 3.00 m/s in the opposite direction, what will their speeds be after the collision?

Va = 2 m/s
Vb = 3 m/s

Homework Equations

MaVa + MbVb = MaVa' + MbVb'
1/2MaVa^2 + 1/2MbVb^2 = 1/2MaVa'^2 + 1/2MbVb'^2

The Attempt at a Solution

Let right be positive.

MaVa + MbVb = MaVa' + MbVb'
Va + Vb = Va' + Vb'
2 - 3 = Va' + Vb'
Va' = -1 - Vb'

1/2MaVa^2 + 1/2MbVb^2 = 1/2MaVa'^2 + 1/2MbVb'^2
Va^2 + Vb^2 = Va'^2 + Vb'^2
4 + 9 = (1-Vb')^2 + Vb'^2
0 = -13 + 1 + 2Vb' + Vb'^2 + Vb'^2
0 = 2Vb'^2 + 2Vb' - 12
0 = Vb'^2 + Vb' - 6
0 = (Vb' + 3)(Vb' - 2)
Vb' = -3 or Vb' = 2

I can figure this out, but I'm not sure which answer to choose and also for future questions how I know which to choose.

If Vb' = -3:

Va' = -1 - Vb'
Va' = -1 - (-3)
Va' = 2

If Vb' = 2:

Va' = -1 - Vb'
Va' = -1 - 2
Va' = -3

 

[Villyer]

Compare the final answers to your initial conditions.

What did you state Va and Vb to be?

 

[testme]

Well Va was 2m/s and Vb was -3 m/s

If they're colliding head on would they be pushed back directly in the opposite direction, making

Vb' = 2 m/s and Va' = -3 m/s?

 

[PeterO]

Homework Statement

Two billiard balls of equal mass undergo a perfectly elastic head on collision. If one ball's initial speed was 2.00 m/s, and the other's was 3.00 m/s in the opposite direction, what will their speeds be after the collision?

Va = 2 m/s
Vb = 3 m/s

Homework Equations

MaVa + MbVb = MaVa' + MbVb'
1/2MaVa^2 + 1/2MbVb^2 = 1/2MaVa'^2 + 1/2MbVb'^2

The Attempt at a Solution

Let right be positive.

MaVa + MbVb = MaVa' + MbVb'
Va + Vb = Va' + Vb'
2 - 3 = Va' + Vb'
Va' = -1 - Vb'

1/2MaVa^2 + 1/2MbVb^2 = 1/2MaVa'^2 + 1/2MbVb'^2
Va^2 + Vb^2 = Va'^2 + Vb'^2
4 + 9 = (1-Vb')^2 + Vb'^2
0 = -13 + 1 + 2Vb' + Vb'^2 + Vb'^2
0 = 2Vb'^2 + 2Vb' - 12
0 = Vb'^2 + Vb' - 6
0 = (Vb' + 3)(Vb' - 2)
Vb' = -3 or Vb' = 2

I can figure this out, but I'm not sure which answer to choose and also for future questions how I know which to choose.

If Vb' = -3:

Va' = -1 - Vb'
Va' = -1 - (-3)
Va' = 2

If Vb' = 2:

Va' = -1 - Vb'
Va' = -1 - 2
Va' = -3

Your duplicate answers arise since it was arbitrary which ball you chose as A and which as B - and which one was traveling in the positive direction.

Work the problem with Va = 2 and Vb = -3 initially, then Vb = 2 and Va = -3, then Va = -2 and Vb = 3 and finally Vb = -2 and Va = 3 and you will get seemingly different answers, which when interpretted to the situation mean the same thing.

It is a head on collision, so call the original motion Left and Right [one of those directions you will have to call one of those positive and the other negative for maths - it doesn't like left and right.

Once you interpret your answer you will find that the ball originally doing 2 will be traveling at 3 in the opposite direction to its original, and the ball doing 3 will be traveling at 2 in the opposite direction to its original, regardless of which ball you called A, and which direction you defined as positive.

 

[testme]

Okay, I got it, thanks, now I have another question :/
New topic ><

I don't find the questions hard just my answers aren't really matching up with teachers so I must be doing something wrong