Conference rusure::math

I'm sorry to disappoint you, but there is no answer.  Collisions in space are
not predetermined even if a great deal is assumed.
                                      OB   Moving particle A hits still
                                     /     particle B.  B takes an upward
                                    /      path, and A goes down at an angle.
                                   / b)    You can't assume a head-on collision.
    OA--------------------------->OB-----  All the mv_ (mass times velocity 
                                   \ a)    vector) comes from A.
                                    \      m(A) is the mass of A.
                                     \     v_(A) is the velocity vector of A.
                                      OA   The y components must cancel.
All the momentum (mv, or dp/dt) comes from A.
m(A)v_(A)[initial] = (m(A)v_(A)[final] * cos a) + (m(B)v_(B)[final] * cos b)
All the energy comes from A.
.5*m(A)(v_(A)[initial]**2)=1/2m(A)(v_(A)[final]**2) + 1/2m(B)(v_(B)[final]**2)
p_ = mv_
p_(A)[final] * sin a = p_(B)[final] * sin b

Three equations and four unknowns.  The four unknowns are the angles and the 
magnitudes of the velocities.

In one dimension a great deal is taken for granted.

O----------------------------------------->0--------------------->

vAf = final velocity of A 
vAf = ((mA-mB)/(mA+mB))vAi   +   (2mB/(mA+mB))vBi
vBi = velocity of b, initially 
vBf = (2mA/(mA+mB))vAi  +  ((mA - mB)/(mA + mB))vBi
if you think (mA-mB)/(mA+mB) resembles the expression for rho, the
reflection coefficient in a terminated line, you're right.

anyway, if you assume head-on collisions, you still have problems.

0        O
 \      /
  \    /
   \  /
    \/
     O????
We can assume the s-direction is head-on and use the 1-D equations.
However, what are the y up-down velocities?
If you're writing a video game, the best bet is guess at one of the
angles in the first set of equations above.  No one will know the 
difference.  Most people don't even expect a rubber ball, falling from
their hand as they walk, to continue moving with them until it hits
the ground; they expect it to fall straight down relative to the ground.
yrs,
Tom

I must correct and amplify.  I was watching Mechanical Universe on tv on
saturday.  it's the only TV program (except homework hotline on channel
58 in Los Angeles) that does real algebra and calculus with the intent of
teaching it to the viewer.  Anyway, they showed that two colliding particles
without spin must leave at right angles to one another.  This includes the
case of one ball completely stopping while the other balls leaves in the
same path as the first ball.  So, the dot product of the momenta is zero.
after the collision.  You still don't know where they'll point, but you know
they'll leave at right angled paths to one another.
Tom

Here's the demonstration done on Mechanical Universe from Cal Tech's Prof.
Goodstein:

Two balls, (1) and (2) of mass m. m(1)=m(2)= m

v^ is velocity = speed with direction.
m is a scalar.  p^ is a vector.

Definition: Momentum p^ = m * v^

Conditions: ball 1 is moving in a straight line.  Ball 2 is at rest.
Ball 1 hits ball 2.  What are the resulting momenta?
                                o
                               /
                              /
                             /
o-->------------------------o
                             \
                              \
                               \
                                o

Before collision: 
p^(1) = m * v^(1) 
p^(2) = m * v^(2) = m(2) * 0 = 0
Law: Momentum is conserved in a system.
p^(system) = p^(1)

After collision:
p^(1)' = m * v^(1)' 
p^(2)' = m * v^(2)'
p^(system)' = p^(1)' + p^(2)

but p^(ball) = m * v^(ball), so

p^(system)' = m * v^(1)' + m * v^(2)'

but p^(system)'=p^(system)=m*v^(1)

mv^(1) = mv^(1)' + mv^(2)'

Law: The total energy of the system is conserved

K(system)' = K(system), the energy of the system

If K= 1/2 * m v**2, and p = mv, then
K = mvmv/2m = (p^*p^)/2m   (p^*p^ is a dot product)

K(system)=K(1)'+K(2)'

K(system)= p^(1)**2/2m because ball one before the 
			collision had all the system energy
So:
p^(1)**2/2m = p^(1)'**2/2m + p^(2)'**2/2m

multiply by 2m
p^(1)**2 = p^(1)'**2 + p^(2)'**2

Gosh, it's the Pythagorean Theorem!  and m = m = m, so divide by m
v^(1)**2 = v^(1)'**2 + v^(2)'**2

Well, if the velocities follow Pythagoras, and the initial velocity is the
hypotenuse, then the resulting velocities are are at right angles to one 
another.

Tom

    re .1:
    
    Oh but I believe there IS an answer. Even in three space. I refer
    you to the book _Dynamics_, J. L. Meriam. Wiley and Sons Inc. 1975
    
    Library of Congress # TA352.M45 
    
                                                   
                  /
                 (  ___
                  ) ///
                 /
    

    Notes #175.2-3 discuss two-particle (or two-body) elastic collisions.
    They claim to "prove" that two particles hitting each other will
    bounce off at 90 degree angles with respect to each other.  I assume
    this means that if I drop a superball on the ground that either
    
    	1. the superball will bounce off parallel to the ground, or
  
    	2. the earth will start moving slowly sideways,
    
    or some linear combination of the two.  My superballs always bounced
    straight up!
    
    I think that there is a flaw in the analysis.  It may be that they 
    neglect the kinetic energy of the center of gravity of system:

                            /m1V1+m2V2\ 2
    	KE(cg) = 0.5(m1+m2)|-----------|
                            \  m1+m2  /
    
    Am I solving a different problem from these guys, or is the "Cal-Tech
    professor" full of prunes?
    
    Yours in confusion,
    John

    P.S. - given 2 velocities, I can always pick a coordinate system in
    which they are perpendicular (that is, V1 X V2 = {0,0,0} ).  Does this
    have something to do with it?

    Re .5:
    
    The mass of your superball does not equal the mass of the Earth.  .3
    assumes it does. 
    
    
    				-- edp