Question about collisions in 2-dimensions

[Blairmc2]

Q.
A 0.150 kg billiard ball (A) is rolling toward a stationary billiard ball (B) at 10.0 m/s. After the collision, ball A rolls off at 7.7 m/s at an andle of 40.0 degrees clockwise from its original direction. What is the speed and direction of Ball B after the collision?

Known:
M of A = .150 kg
M of B = .150 kg
Vi of B = 10.0 m/s
Vf of A = 7.7 m/s[70degrees clockwise]

Implied:
Vi of A = 0 m/s
Ff = 0N

Unknown:
Vi of B = ?

I started out using the law of conservation of energy but I messed up somewhere early and got fully lost...I know you need to use this method to conserve the momentum of each object but I just can't remember how, can someone start please?

 

[dicerandom]

This is a purely conservation of momentum problem. Pick your x and y axes, initially you have some amount of momentum in x and y due to the motion of ball A and in the final situation you have the same amount of momentum in x and y due to the combined motions of A and B.

 

[phucnv87]

The conservation of momentum

$$m\vec{v_{iA}}=m\vec{v_{fA}}+m\vec{v_{fB}}$$[/color]

 

[assyrian_77]

Well, using conservation of momentum should do it. Remember, though, that you should treat the vector components separately, i.e. work it out for a chosen x-direction and y-direction. Easiest would be to have ball A roll in the x-direction.EDIT: Wow, the answers keep rolling in.