7.4

Collisions in two Dimensions

The collisions discussed so far have been “head-on,” or one-dimensional, because the velocities of the objects all point along a single line before and after contact. Collisions often occur, however, in two or three dimensions. Figure 7-15 shows a two-dimensional case in which two balls collide on a horizontal frictionless table.

Figure 7-15 (a) Top view of two balls colliding on a horizontal frictionless table. (b) This part of the drawing shows the x and y components of the velocity of ball 1 after the collision.

For the system consisting of the two balls, the external forces include the weights of the balls and the corresponding normal forces produced by the table. Since each weight is balanced by a normal force, the sum of the external forces is zero, and the total momentum of the system is conserved, as Equation 7.7b indicates. Momentum is a vector quantity, however, and in two dimensions the x and y components of the total momentum are conserved separately. In other words, Equation 7.7b is equivalent to the following two equations:

| (7.9a)

| (7.9b)

These equations are written for a system that contains two objects. If a system contains more than two objects, a mass-times-velocity term must be included for each additional object on either side of Equations 7.9a and 7.9b. Example 9 shows how to deal with a two-dimensional collision when the total linear momentum of the system is conserved.

Problem solving insight Momentum, being a vector quantity, has a magnitude and a direction. In two dimensions, take into account the direction by using vector components and assigning a plus or minus sign to each component.

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 9 | | | A Collision in Two Dimensions

For the data given in Figure 7-15a, use momentum conservation to determine the magnitude and direction of the final velocity of ball 1 after the collision. Reasoning We can obtain the magnitude and direction of the final velocity of ball 1 from the x and y components of this velocity. These components can be determined by using the momentum-conservation principle in its component form, as expressed in Equations 7.9a and 7.9b.
Knowns and Unknowns The following data are given in Figure 7-15a:

Description | Symbol | Value | Comment
Mass of ball 1 | m1 | 0.150 kg |
Magnitude of initial velocity of ball 1 | v01 | 0.900 m/s | Before collision occurs.
Directional angle of initial velocity of ball 1 |  | 50.0° | See Figure 7-15a.
Mass of ball 2 | m2 | 0.260 kg |
Magnitude of initial velocity of ball 2 | v02 | 0.540 m/s | Before collision occurs.
Magnitude of final velocity of ball 2 | vf2 | 0.700 m/s | After collision occurs.
Directional angle of final velocity of ball 2 |  | 35.0° | See Figure 7-15a.
Unknown Variables
Magnitude of final velocity of ball 1 | vf1 | ? | After collision occurs.
Directional angle of final velocity of ball 1 | θ | ? | After collision occurs. See Figure 7-15a.

Modeling the Problem

| Step 1 Vector Components
| Figure 7-15b shows the final velocity vf1 of ball 1 oriented at an angle above the +x axis. Also shown are the components vf1x and vf1y. The components are perpendicular, so that we can use the Pythagorean theorem and trigonometry and write Equations 1a and 1b at the right.

| (1a)

| (1b)

| Step 2 Conservation of Linear Momentum
| Momentum is a vector quantity, so the conservation principle applies separately to the components of the momentum. Applying momentum conservation (Equation 7.9a) to the x-direction we find that x Component

Applying momentum conservation (Equation 7.9b) to the y-direction we find that
y Component

These equations can be solved to obtain values for the components vf1x and vf1y:

The substitution of these values into Equations 1a and 1b is shown at the right.

Solution Algebraically combining the results of each step, we find that
Related Homework: Problems 30, 36

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