Conservation of momentum problem?

[I Like Pi]

Homework Statement

Two billiard balls collide on a horizontal table (assume no friction)). Billiard ball A has a mass of 0.175 kg,travelling at 1.20 m/s [E 40 deg S]. Billiard ball B has a mass of 0.225 kg, traveling at 0.68 m/s [E]. The velocity of billiard ball B after collision is 0.93 m/s [E 23 deg S]. What is the velocity of billiard A after the collision and what is the percentage of kinetic energy lost in the collision?

Homework Equations

P = m*v

The Attempt at a Solution

I believe this is a conservation of momentum problem?

therefore: $$P_{i}=P_{f}$$
$$(mv_{A})_{i}+(mv_{B})_{f}=(mv_{A})_{f}+(mv_{B})_{f}$$
$$.175*1.20+.225*.68=.175*.93+.225*v_{A}_{f}$$
$$v_{A}_{f} = .89 m/s$$

What I am unsure of is how i would find the direction of that velocity... I know it's vector problem, but how would i go about solving this, preferably algebraically?

EDIT: never mind, i found out how to do it using the vertical components

And to solve for % of kinetic energy lost, would I need to do it separately for each ball? using $$\frac{1}{2}mv^{2}$$?

Thanks in advance.

 

[chaoseverlasting]

Yes, you would need to calculate the kinetic energy of each ball individually.

 

[I Like Pi]

Thanks, but I realized I did the above wrong.. I need to find the x and y component of Ball A after the collision... so what I did was:
mv_aix + mv_bix = mv_afx + mv_bfx
= .175(1.20sin40) + .225(.68) = .175(v_afx) + .225(.93sin23)
v_afx = 1.18

mv_aiy + mv_biy = mv_afy + mv_bfy
= .175(1.20cos40) + .225(0) = .175(v_afy) + .225(.93cos23)
v_afy = -0.18

though I don't know if i am right as i also get a negative...

EDIT: sorry, I need to interchange the trig signs for both the components... DOI!

Now how would I go about with the kinetic energy?