1. You have 2 balls moving across a flat plain, sliding not rolling. They are both the same mass and have the same velocity. What would be the most primitive formula for calculating the resulting vectors from the collision?

2.

3. Originally Posted by shawngoldw
You have 2 balls moving across a flat plain, sliding not rolling. They are both the same mass and have the same velocity. What would be the most primitive formula for calculating the resulting vectors from the collision?
m1u1+m2u2=m1v1+m2v2

e=v2-v1/u1-u2

m1 & m2 are mass
u1, u2, v1and v2 are velocity before collision and after collision.

4. thanks, what's e?

5. I suspect basim is thinking of a 1D case. His first equation expresses conservation of momentum. The second equation defines the coefficient of restitution.

I didn't understand the original question as a 1D problem (why would you talk about vectors?). Is it meant to be 1D or 2D?

Anyway, the extension from 1D to 2D is really simple. The 1D equations apply on the velocity components normal to the impact surface. The tangential velocity components remain unchanged as long as you don't have friction.

6. im thinkin about 2d. I'm not so much interested in velocity, mainly direction.

7. Assumptions: No rolling, no friction, perfectly elastic (no kinetic energy loss on impact), both balls have the same mass, spherical (circular) shape.

Given: initial positions and initial velocities (2D-vectors u1, v1) of each ball.

1. Determine the positions of both balls at impact. Define the direction of the line through both centers of mass: the "impact line".

2. Decompose each of the initial velocity vectors in two components: The component parallel to the impact line, and the component normal (perpendicular) to the impact line. For example, the initial velocity of ball 1 should be composed of

u1 = u1p +u1n (2D vector sum)

where component vector u1p is parallel to the impact line, and u1n normal to it. Likewise, for the second ball:

v1 = v1p +v1n

3. Obtain the velocity vectors after impact. Under the given assumptions, this is very easy to do: Both balls retain their normal components and exchange their parallel components, i.e. after the impact

Ball 1: u2 = v1p +u1n
Ball 2: v2 = u1p +v1n

You're done!

The only part that involves any appreciable mathematical effort is step 2. Knowledge of some vector algebra will help you do the decomposition (you could call it a coordinate transformation).

Some secondary effects we neglected according to the above assumptions:

Rolling balls largely retain their rotational energy through the impact, which will slightly alter the path.

Friction between the balls causes (slight) changes in the velocity components normal to the impact line.

Imperfect elasticity causes (slight) changes in the parallel components, as some kinetic energy will be transformed to heat.

8. Thanks, that is exactly what I'm looking for! That should work perfect!

edit* I spent some time thinking about it and isn't something missing from that formula? Take 2 situations, both with the same equations of the paths. In the first one your equation works. But in the second one you move one of the balls an inch back. Now the equations are still the same but they will take a path that doesn't fit your equation because they will contact at different parts of the balls.

Not sure if i was clear there, ill clarify if you don't understand.

9. Not sure if I understand the question. You know the positions of the balls at the time of impact, and their velocities just before impact. Those conditions uniquely describe the situation, and those are the only conditions that enter the calculation. There is no ambiguity. If you move one ball back 1 inch, you need to re-evaluate the new impact position and the situation will be different. How is this reflected in the equations? Through the definition of the impact line (i.e. the vector between the two centers of mass). This line will be oriented differently.

10. I'm not too good with circles thats why i asked. So ill try this and see if it works. Thanks.

11. Haven't read the posts before this one so this may have already been covered.

In any of these types of questions, use conservation of energy/momentum.

i.e. find an expression for enrgy/momemtum before, find one for after then equate the two.

12. Haven't read the posts before this one so this may have already been covered.

In any of these types of questions, use conservation of energy/momentum.

i.e. find an expression for enrgy/momemtum before, find one for after then equate the two.

Yes, that's the short and general answer, and in essence that's what we did. The long answer, catered to the specific problem, was given to show how simple the solution becomes after all assumptions have been incorporated, and to provide an algorithmic recipe. This solution is also well-known enough to make re-inventing the wheel by first principles unnecessary. The full derivation is of secondary interest.

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