Minimum possible kinetic energy in an interaction

[Neutrinogun]

Homework Statement

Object A of mass 10 kg moving at 10 m/s [E] interacts with (but does not touch) Object B of mass 8 kg moving at 12 m/s [N]. What is the minimum possible kinetic energy of the system after the collision?

Homework Equations

M₁V_1o + M₂V_2o = M₁V_1f + M₂V_2f
KE = (.5)(m)(v²)
Kinetic energy is minimized in a completely inelastic collision.

The Attempt at a Solution

10(10) + 8(12) = (10 + 8)v
v = 10.89 m/s
KE = (.5)(10)(10.89)² + (.5)(8)(10.89)²
KE = 1067 J
Correct answer: 534 J (which is one-half of the answer I got.)

Help please?
Thanks.

 

[Doc Al]

Realize that momentum is a vector and must be added as such. (One momentum points east and the other points north. Find their correct vector sum.)

 

[Neutrinogun]

Oh...Thanks! Previous problems like this had the second object initially at rest, so I used that way without thinking.