The Law of Conservation of Momentum applies in two and three dimensions, too. To apply it in 2-D, split the momentum into x and y components and keep them separate. Write out two conservation of momentum equations, one for the x direction and one for the y direction.

For instance:

m₁v_1ix + m₂v_2ix = m₁v_1fx + m₂v_2fx

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	p1yi = 0 p2yi = 0
After	p1fx = 0.35*1.7*cos(30) p1fx= 0.515 kg m/s p2fx = ?	p1yf = 0.35*1.7*sin(30) p1yf = 0.2975 kg m/s p2yf = ?

Applying conservation of momentum in the x-direction:

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

Applying conservation of momentum in the y-direction:

p_1yi + p_2yi = p_1yf + p_2yf

p_2yf = -p_1yf = -0.2975 kg m/s

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

p_2f² = p_2fx² + p_2yf²

p_2f = 0.467 kg m/s

v_2f = p_2f / m₂ = 0.467 / 0.51 = 0.916 m/s

The direction of puck 2 after the collision can be found from:

tan(θ) = p_2yf / p_2fx

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

θ = 39.6 degrees