2D Momentum Question on pool balls

[SpyIsCake]

Homework Statement

Standard pool balls have a mass of 0.17kg. Before a collision, ball A is at rest and ball B is traveling at 4.5 m/s [N]. After the collision, ball B is traveling at 1.6 m/s [N20E]. What is the velocity of ball A?

Homework Equations

pinitial = pfinal
m1v1 + m2v2 = m1v1 + m2v2

The Attempt at a Solution

I believe that the initial momentum would be 0.17 * 4.5 m/s = 0.765kg m/s [N].
The momentum for B is 1.6 * 0.17 = 0.272
The initial momentum has to equal the final momentum, so:

0.765kg/m/s = 0.272 + A

I drew a triangle for B's angle.

For X component: 0.272 cos 20 deg = 0.255
For Y component: 0.272 sin 20 deg = 0.09

The X component will just be 0.255 [W] because the X component is 0.
The Y component will have to be the difference between 0.765 and 0.09302, because you need the north to have the same momentum. The answer to that is 0.7168.

Then I plug them all in and use Pythagorean theorem to find the momentum, which was 0.76 kg m/s.
I divide that by the mass of 0.17kg and got a velocity of 4.2 m/s.

But unfortunately, the answer is 3 m/s.

How?

 

[RUber]

The Y component will have to be the difference between 0.765 and 0.09302, because you need the north to have the same momentum. The answer to that is 0.7168.

The North components do not have to have the same momentum. The total momentum in the system cannot change.

 

[Andrew Mason]

Homework Statement

Standard pool balls have a mass of 0.17kg. Before a collision, ball A is at rest and ball B is traveling at 4.5 m/s [N]. After the collision, ball B is traveling at 1.6 m/s [N20E]. What is the velocity of ball A?

Homework Equations

pinitial = pfinal
m1v1 + m2v2 = m1v1 + m2v2

The Attempt at a Solution

I believe that the initial momentum would be 0.17 * 4.5 m/s = 0.765kg m/s [N].
The momentum for B is 1.6 * 0.17 = 0.272
The initial momentum has to equal the final momentum, so:

0.765kg/m/s = 0.272 + A

This last statement is not correct. The north component of B's momentum (i.e. mv(cos(20)) + the north component of A's momentum has to equal B's initial momentum. I think you are also assuming that this is an elastic collision. You cannot assume that.
AM

 

[haruspex]

You may have cos and sin switched over. N20E is 20 degrees E of N, so if you are taking the positive X axis as E then the X component will involve sin of 20 degrees.