What Is the Speed of the 0.300 kg Puck After Collision?

[_mae]

Excuse the Title =X

Homework Statement

A 0.300 kg puck, initially at rest on a horizontal, frictionless surface, is struck by a 0.200 kg puck moving initially along the x-axis with a speed of 2.00 m/s. After the collision, the 0.200 kg puck has a speed of 1.00 m/s at an angle of \Theta= 50.0° to the positive x axis.

http://img265.imageshack.us/img265/5695/811vx2.gif

Determine the speed of the 0.300 kg puck after the collision.​

Homework Equations

m1v1i + m2v2i = (m1 + m2)vf

The Attempt at a Solution

total momentum before collision = 0
total momentum after collision = 0.3 * v1f *sin 50 - 0.2 * v2f *sin φ
from the law of conservation of momentum ,
0.3 * v1f * sin 50 - 0.2 * v2f * sin φ = 0
0.3 * v1f * sin 50 = 0.2 * v2f * sin φ
where v2f = 1 m / s
φ = 90 - 50 = 40

so: (0.3)(v1f)(sin50) - (0.2)(1)(sin40) = 0.5596 m/s ??

 

[rl.bhat]

total momentum before collision = 0 This statement is wrong, because 0.2 kg mass is moving with a velocity 2 m/s.
total momentum after collision = 0.3 * v1f *sin 50 - 0.2 * v2f *sin φ This is only y-components. What about x-components?