Physics
posted by Fred .
A ball of mass 0.640 kg moving east (+x direction) with a speed of 3.10 m/s collides headon with a 0.780 kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision?
1. ball originally at rest
2. ball originally moving east
im alittle confused about this problem.
if you could give more details and give me the equations for me to start. that would be very helpful. much appreciated

Perfectly elastic collision assumes conservation of energy. Momentum is conserved for both elastic and inelastic collisions.
Let 1 and 2 denote each of the balls (1 moving, 2 initially stationary).
m1 and m2 their masses
v11 and v21 their initial velocities (v21=0)
v12 and v22 their final velocities.
Conservation of Kinetic energy equation:
(1/2)m1 v11² + (1/2)m2 v21²
= (1/2)m1 v12² + (1/2)m2 v22² ....(1)
Conservation of Momentum equation:
m1 v11 + m2 v21 = m1 v12 + m2 v22 ...(2)
All variables are known except for v12 and v22.
Solve for v12 and v22. 
wow, much help. thank you very much but i do have a quick question,
v21² AND v22² are both in the same equation, are you saying that im supposed to put v21² AND v22² on the same side? like this....
(1/2)m1 v11² + (1/2)m2 (1/2)m1 v12² + (1/2)m2
= v21² v22² ?????? 
(1/2)m1 v11² + (1/2)m2 v21²
This is the left hand side of the equation and represents the sum of kinetic energies of the two balls before the impact.
The right hand side of the equation:
(1/2)m1 v12² + (1/2)m2 v22²
represents the total kinetic energies after the elastic impact.
They are equated because energy is supposed to be conserved during an elastic impact.
The confusion probably arose because the equal sign was at the beginning of the next line.
By the way, you will probably find that v12 is negative, which physically means that ball 1 bumped backwards while pushing ball 2 forward.