Two billiard balls with the same mass undergo a perfectly elastic head-on collision. if one ball's initial speed was 2 m/s, and the other's was 3.6 m/s in the opposite direction, what will be their speeds and directions after the collision

The ball with an initial velocity of 2 m/s rebounds at 3.6 m/s
 The ball with an initial velocity of 3.6 m/s rebounds at 2 m/s

   There are two principles involved here
 Conservation of momentum and conservation of energy.
  I'll use the following variables
 a0, a1 = velocity of ball a (before and after collision)
  b0, b1 = velocity of ball b (before and after collision)
 m = mass of each ball.

   For conservation of momentum, we can create this equation:
 m*a0 + m*b0 = m*a1 + m*b1
 divide both sides by m and we get:  
a0 + b0 = a1 + b1

   For conservation of energy, we can create this equation:
 0.5m(a0)^2 + 0.5m(b0)^2 = 0.5m(a1)^2 + 0.5m(b1)^2
 Once again, divide both sides by 0.5m to simplify
 a0^2 + b0^2 = a1^2 + b1^2

   Now let's get rid of a0 and b0 by assigned their initial values. a0 will be 2, and b0 will be -3.6 since it's moving in the opposite direction.
 a0 + b0 = a1 + b1
 2 - 3.6 = a1 + b1
 -1.6 = a1 + b1
 a1 + b1 = -1.6

   a0^2 + b0^2 = a1^2 + b1^2
 2^2 + -3.6^2 = a1^2 + b1^2
 4 + 12.96 = a1^2 + b1^2
 16.96 = a1^2 + b1^2
 a1^2 + b1^2 = 16.96

   The equation a1^2 + b1^2 = 16.96 describes a circle centered at the origin with a radius of sqrt(16.96). The equation a1 + b1 = -1.6 describes a line with slope -1 that intersects the circle at two points. Those points being (a1,b1) = (-3.6, 2) or (2, -3.6). This is not a surprise given the conservation of energy and momentum. We can't use the solution of (2, -3.6) since those were the initial values and that would imply the 2 billiard balls passing through each other which is physically impossible. So the correct solution is (-3.6, 2) which indicates that the ball going 2 m/s initially rebounds in the opposite direction at 3.6 m/s and the ball originally going 3.6 m/s rebounds in the opposite direction at 2 m/s.

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