### Sample 2-D Collision

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 p1ix = 0.35*2.5
p1ix = 0.875 kg m/s

p2ix = 0

p1iy = 0

p2iy = 0

After p1fx = 0.35*1.7*cos(30)
p1fx= 0.515 kg m/s

p2fx = ?

p1fy = 0.35*1.7*sin(30)
p1fy = 0.2975 kg m/s

p2fy = ?

Appliyng conservation of momentum in the x-direction:

p1ix + p2ix = p1fx + p2fx

p2fx = p1ix - p1fx

p2fx = 0.875 - 0.515 = 0.36 kg m/s

Appliyng conservation of momentum in the y-direction:

p1iy + p2iy = p1fy + p2fy

p2fy = -p1fy = -0.2975 kg m/s

The total momentum of the second puck after the collision is found from:

p2f2 = p2fx2 + p2fy2

p2f = 0.467 kg m/s
v2f =
 p2f m2
=
 0.467 0.51
= 0.916 m/s

The direction of puck 2 after the collision can be found from:
tan(θ) =
 p2fy p2fx

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

θ = 39.6 degrees