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Guide 9-2. Solving Conservation of Momentum Problems in Two Dimensions

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, Pucks 1 and 2 move off at angles α and β respectively. Both angles are measured counterclockwise with respect to 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. The magnitudes of the velocities are substituted. The sign of each velocity component is determined by the corresponding trig function. Note that both x-components will be positive in this case, since the cosines of Quadrant I and IV angles are positive. 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. The magnitudes of the velocities are substituted. The sign of each velocity component is determined by the corresponding trig function. In this case, sin(α) will be positive and sin(β) will be negative. This makes sense, because the components must add to 0.

We end up with two equations in 4 unknowns:  v1f, v2f, α, and β.  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).

Example 5:  Try this problem for practice. The magnitude and direction of the final velocity are given in the applet. Practice setting up the conservation of momentum problem in order to obtain those values.