# Collision: Final velocities of two particles, calculated from impulse

1. Aug 17, 2010

### Icebone1000

Hi, I dont know if this is the right place to ask this..

As the title says, I want know the final velocities of two partcles after collision, but I want to get those from the impulse, not from the final velocities equations you see arround there...

The final step would be that:
Vf = particle initial velocity + impulse/mass
( I mean, is where I want to get to solve the final velocity )

So considering two particles:
A( m = 5, V = 1 ), B( m = 1, V = -5 )

How do I calculate?

That what I was doing, but I cant get reasonable results..:

relative Closing velocity = 1 -(-5) = 6;
considering COR = 1, Closing vel after collision = -6;(-1*6)(coefficient of restitution, 1 = no velocity lost )
impulse = 6*( -6 - 6) = -72 ;(impulse = m*( V1 - V0),the change in momentum thing, I guess the mass here is the system total mass..but I guess those values are absurds..)

So the velocity of A(from impulse):
1 + (-72/5) = 1 - 14.4 = -13.4;//v0 + impulse/mass
B:
-5 + (72/1) = 67

What am I missing...?

2. Aug 18, 2010

### AJ Bentley

Why not do it the way everyone else does?

It's pretty much the same thing, Impulse is just the integral of force over time - so it's just the change in momentum.
The same force is applied to each object over the same length of time, so the impulse for each, and hence the change in momentum is the same for each.
But that's just a re-statement of conservation of momentum. So why not just accept that fact and apply the usual principles of conservation of momentum and energy?

3. Aug 21, 2010

### LawlQuals

You only need two equations here to solve for your two final velocities.

Let subscripts $$A$$ and $$B$$ denote each particle, and the superscripts $$+$$ and $$-$$ mark pre and post impact paramters.

It is important to remember when equations are applicable, and to consider each case carefully. The coefficient of restitution (COR), $$e$$ is applicable along the line of impact (the line that traces out each contact force during the collision), here this is a horizontal line (if we choose). So you have the relation, with $$e=1$$

$$v_B^+ - v_A^+ = v_A^- - v_B^-$$

You are given $$v_B^-$$ and $$v_A^-$$, thus we only need one more equation. Consider the momentum-impulse principle along the line of impact, for the *system* ($$m_A$$ and $$m_B$$).

$$(m_Av_A^- + m_Bv_B^-) + \int_{t^-}^{t^+} Fdt = (m_Av_A^+ + m_Bv_B^+)$$

where $$F$$ represents the contact forces during the interval $$t\in (t^-,t^+)$$. Given that we are evaluating the system, can you make any statements about the value of the integral? A free body diagram of the system at the moment of contact will be instructive. If you can validly make an insight here, you will have two equations and two unknowns, enabling a solution. This is not guesswork, as your reluctance seemed to indicate, can you see *why* I have chosen to analyze the system as opposed to just one mass at a time with the momentum-impulse system?

4. Aug 23, 2010

### Icebone1000

My problem is not really get a answer, since I alredy have a working solution..

What Im looking for is the logic of impulse, and why my method for getting the final velocities from impulse are wrong(since Im doing this on the logic way I understand), looking at my example, could anyone calculate the final velicities in the way Im looking for?

LawlQuals, I dont get much of what you say on the last paragraph..but in the way Im trying to calculate it start on gathering system info, not per particle info(relative velocity, COR, impulse), and then finnally applying the correct impulse for each particle.

(to make more clear ever, you can calculate the velocity of a particle after an impulse by simply adding the velocity gathered from the impulse formula(i = m.v so v = i/m ) plus current velocity, pretty obvious:
particle A( mass 2, vel 5) ->apply impulse 20 (so 20/2 = 10) -> particle vel = 15
So what Im trying to do is get the impulse relative to the colliding particles(gathered trough the relative velocity) so I can add the velocity resulted from that impulse to each particle current velocity..)

To me looks like a simple issue, but no one seems to understand what Im trying to do and keep pushing me the final velocities formulas, Im probaly missing some simple concept in the way..

5. Aug 23, 2010

### LawlQuals

I will look at your work later when I get more time. I had skimmed your work, but regrettably I did not spend much time understanding it because it was dense (all numbers, not formatted, no units), and inconvenient for me to read (sorry). So, I just suggested a general strategy to aid you to help see if that could steer you in the proper direction. You did clearly ask "what [are you] missing?" though, so I should have addressed the question more directly.

As a general rule, you *always* use the impulse-momentum principle in impact problems, that is what I was getting at in the final paragraph of my previous post. In many cases, the impulse principle collapses to conservation of momentum, but if and only if you can substantiate it. That is to say, the hint in the final paragraph I was making was that $$\vec{F} = 0$$, but why? If this were true, then the relation I wrote down would simply be a statement of momentum conservation.

The point is, it is a common mistake that I see (I am a TA), students abusing momentum conservation. One must be careful about claiming any conservation laws are true. If they really are true, then they can be shown to be true. To be safe, we always start with the moment-impulse principle. If it happens to be the case where we can collapse the principle into a momentum conservation relation (like we can here), then we should figure out when and how that happens (here, only along the line of impact, and only for the system of both particles). Momentum is not conserved for each individual particle though, that one must be careful to not claim being true. In the end, perhaps I was driving a point that was irrelevant to your confusion, but perhaps not.

I will look at your work in a bit. Thanks for your patience.

6. Aug 25, 2010

### Icebone1000

I thank you for the patience.

Ill post the link of two articles that explains what Im trying to do..Those articles use directions(not 1D like my example)..so I guess on 1D it becomes what Im trying to do(the contact normal can be discarded..I guess)

http://chrishecker.com/images/e/e7/Gdmphys3.pdf
http://www-2.cs.cmu.edu/afs/cs/user/baraff/www/pbm/rigid2.pdf

Those articles have status on game physics programming topic, as I have found. Could anyone compare it to what I do on my example?

¦D I guess Im starting to be annoying..

7. Aug 29, 2010

### LawlQuals

Hi there, I am really sorry but I have been super busy this past week, and will continue to be so until Friday. Just skimming your answer, it appears you wrote something like m_total*(difference in velocities), but there is no such term that is characterized by both masses. The particles move separately (so you must model your equations with the masses as separate, described by separate velocities), if they moved together you would use the sum of both masses. It physically distinct to write the mass in your equations as being that of both. I will respond more thoroughly after Friday, and sorry again for so many rain checks.

8. Aug 29, 2010

### Icebone1000

I get it now, thanks for the help!

#1#Impulse = (Vab_1 - Vab_0) / (1/ma + 1/mb)

#2#This is writed mostly times as -(1+COR).Vab_0 / (1/ma + 1/mb)..

Simple maths, if you put Vab_0 in evidence on the first numerator, you get the second numerator, since Vab_1 = -COR.Vab_0