# Minimum possible kinetic energy in an interaction

1. Nov 17, 2012

### Neutrinogun

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
M1V1o + M2V2o = M1V1f + M2V2f
KE = (.5)(m)(v2)
Kinetic energy is minimized in a completely inelastic collision.

3. 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.)

Thanks.

2. Nov 17, 2012

### Staff: Mentor

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.)

3. Nov 17, 2012

### Neutrinogun

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