Momentum and collisions question .help please

2 objects called A and B are weighted.3KG and 2KG.A has 2m/s velocity and B has 6m/s before the collision and they collide face to face.after the collision ,object A moves to opposite direction with a speed of 4m/s.What is the velocity of B after the collision?

answer has to be 15m/s.But I never get it

By the law of conservation of momentum,
momentum before collision = momentum after collision
(3kg * 2m/s) + (2kg * 6m/s) = (3kg * 4m/s) + (2kg * v)
6kgm/s + 12kgm/s = 12kgm/s + 2kg * v
6kgm/s = 2kg * v
v = 3m/s

The answer cannot be 15m/s. Assuming there are no external forces, the momentum in the system (i.e. A and B) must remain the same before and after the collision. If it were 15m/s, the momentum would have increased by 24kgm/s after the collision.

Hmmm. I keep doing (3KG)*2 + (2KG)*6 = (3KG)*4 + (2KG)*?
and ? = 3 m/s. Weird.

Are you sure the answer's right? Collectively, they were only going 18 m/s.



Okay, I got it. Velocity is a vector, which means that when A is going forwards, it's positive, and when A is moving in the opposite direction (like in the problem above, after B bumps it), A becomes negative.

So the equation become (3)(2) + (2)(6) = -(3)(4) + (2)(?)
then ? = 15 m/s. yay!

This question is supposed to be asked for physics guys!!