In 2-dimensional situations where momentum is conserved, the conservation law must be applied along each axis independently.  Here's an example of how one would write the conservation equations for such a situation.  The diagram below shows two hockey pucks on a level air table before and after a collision.  Puck 2 is initially at rest.  After the collision, Puck 1 moves off at an angle α above the +x-axis, and Puck 2 moves off at an angle b below the +x-axis.  Suppose we want to find the final velocities given the masses and initial velocities.

Here's how we would go about setting up conservation equations for this situation.
 

	The x-component of the total momentum of the system of the 2 pucks is conserved.
	The total x-momentum is expressed as the sum of the x-momenta of the individual objects.
	The definition of momentum is applied.
	Velocities are substituted.  The final velocities are the product of a magnitude and the cosine of the angle.
	Note that both x-components will be positive in this case, since cos(-ß) = cosß.
	The y-component of the total momentum of the system of the 2 pucks is conserved.
	The total y-momentum is expressed as the sum of the y-momenta of the individual objects.
	The definition of momentum is applied.
	Velocities are substituted.  The final velocities are the product of a magnitude and the sine of the angle.
	One y-component will be negative, since sin(-ß) = -sinß.

We end up with two equations in 4 unknowns:  v_1f, v_2f, α, and ß.  Two more independent equations would be required in order to find a unique solution for the magnitude and direction of the final velocities.  If the collision were elastic, a third equation would come from the application of conservation of kinetic energy:  .   Note that this equation is expressed in terms of magnitudes of the velocities rather than components.  That's because energy is a scalar.  The fourth equation would require knowing the nature of the forces that the pucks exert on each other.  In principle this is possible but in practice it's very difficult.  Usually, a problem will give you additional information in order to simplify the situation.  An example of a 2-dimensional inelastic collision is given next.

Example 4.  A 900-kg car traveling east at 15 m/s collides with a 750-kg car traveling north at 20 m/s. The cars stick together. With what velocity does the wreckage move just after the collision?
 

Given: v1ix = 15 m/s; v1iy = 0 v2ix = 0; v2iy = 20 m/s m1 =900 kg m2 = 750 kg Goal: Find System:  cars External forces: Gravity and normal are external forces, but the net, external force is 0. We're ignoring friction. Initial state:  cars before collision Final state:  cars just after collision		Since there is just one object to deal with after the collision, there are only 2 unknowns:  the magnitude and direction of the final velocity of the combined cars.  Thus, the 2 conservation of momentum equations will be sufficient to solve the problem. We choose x- and y-axes so that one of the cars has no y-component of momentum and the other has no x-component.  Because of our choice, we can see that car 1's initial momentum will be the x-component of the final momentum and car 2's momentum will be the y-component of the final momentum.
Solution:
(1)	Apply conservation of momentum along the x-axis.
(2)	Apply conservation of momentum along the y-axis.
	Divide equation (2) by equation (1).  This will eliminate the unknown vf.  Solve for the angle.
	Solve equation (1) for vf and substitute.
	As a check, we get the same result using equation (2).