Consider a collision between two pucks on a frictionless air hockey table. Puck 1 has a mass of 350 g and an initial velocity of 2.5 m/s in the +x direction. Puck 2 has a mass of 510 g and is initially at rest.
After the collision puck 1 has been deflected by 30 degrees and has a speed of 1.7 m/s. What is the speed and direction of puck 2 after the collision?
X  Y  

Before  p_{1ix} = 0.35*2.5 p_{1ix} = 0.875 kg m/s p_{2ix} = 0  p_{1iy} = 0 p_{2iy} = 0 
After  p_{1fx} = 0.35*1.7*cos(30) p_{1fx}= 0.515 kg m/s p_{2fx} = ?  p_{1fy} = 0.35*1.7*sin(30) p_{1fy} = 0.2975 kg m/s p_{2fy} = ? 
Appliyng conservation of momentum in the xdirection:
p_{1ix} + p_{2ix} = p_{1fx} + p_{2fx}
p_{2fx} = p_{1ix}  p_{1fx}
p_{2fx} = 0.875  0.515 = 0.36 kg m/s
Appliyng conservation of momentum in the ydirection:
p_{1iy} + p_{2iy} = p_{1fy} + p_{2fy}
p_{2fy} = p_{1fy} = 0.2975 kg m/s
The total momentum of the second puck after the collision is found from:
p_{2f}^{2} = p_{2fx}^{2} + p_{2fy}^{2}
p_{2f} = 0.467 kg m/s
v_{2f}  = 

= 

=  0.916 m/s 
The direction of puck 2 after the collision can be found from:
tan(θ)  = 

It's fine to just use the magnitudes  we know puck 2 has +x and y components.
θ = 39.6 degrees