In the previous two sections, we considered only one-dimensional collisions; during such collisions, the incoming and outgoing velocities are all along the same line. But what about collisions, such as those between billiard balls, in which objects scatter to the side? These are two-dimensional collisions, and we shall see that their study is an extension of the one-dimensional analysis already presented. The approach taken (similar to the approach in discussing two-dimensional kinematics and dynamics) is to choose a convenient coordinate system and resolve the motion into components along perpendicular axes. Resolving the motion yields a pair of one-dimensional problems to be solved simultaneously.

One complication arising in two-dimensional collisions is that the objects might rotate before or after their collision. For example, if two ice skaters hook arms as they pass by one another, they will spin in circles. We will not consider such rotation until later, and so for now we arrange things so that no rotation is possible. To avoid rotation, we consider only the scattering of point masses—that is, structureless particles that cannot rotate or spin.

We start by assuming that F_net=0, so that momentum p is conserved. The simplest collision is one in which one of the particles is initially at rest. (See Figure 1.) The best choice for a coordinate system is one with an axis parallel to the velocity of the incoming particle, as shown in Figure 1. Because momentum is conserved, the components of momentum along the x– and y-axes (p_x and p_y)  will also be conserved, but with the chosen coordinate system, p_y is initially zero and p_x is the momentum of the incoming particle. Both facts simplify the analysis. (Even with the simplifying assumptions of point masses, one particle initially at rest, and a convenient coordinate system, we still gain new insights into nature from the analysis of two-dimensional collisions.)

Figure 1. A two-dimensional collision with the coordinate system chosen so that m₂ is initially at rest and v₁ is parallel to the x-axis. This coordinate system is sometimes called the laboratory coordinate system, because many scattering experiments have a target that is stationary in the laboratory, while particles are scattered from it to determine the particles that make-up the target and how they are bound together. The particles may not be observed directly, but their initial and final velocities are.

Along the x-axis, the equation for conservation of momentum is

p_1x + p_2x = p′_1x + p′_2x.

Where the subscripts denote the particles and axes and the primes denote the situation after the collision. In terms of masses and velocities, this equation is

m₁v_1x + m₂v_2x = m₂v_2x = m₁v′_1x + m₂v′_2x.

But because particle 2 is initially at rest, this equation becomes

m₁v_1x = m₁v′_1x + m₂v′_2x.

The components of the velocities along the x-axis have the form v cos θ. Because particle 1 initially moves along the x-axis, we find v_1x = v₁.

Conservation of momentum along the x-axis gives the following equation:

m₁v₁ = m₁v′₁ cos θ₁ + m₂v′₂ cos θ₂, where θ₁ and θ₂ are as shown in Figure 1.

Along the y-axis, the equation for conservation of momentum is

p_1y + p_2y = p′_1y + p′_2y or m₁v_1y + m₂v_2y = m₁v′_1y + m₂v′_2y.

But v_1y is zero, because particle 1 initially moves along the x-axis. Because particle 2 is initially at rest, v_2y is also zero. The equation for conservation of momentum along the y-axis becomes

0 = m₁v′_1y + m₂v′_2y.

The components of the velocities along the y-axis have the form v sin θ.

Thus, conservation of momentum along the y-axis gives the following equation:

0 = m₁v′₁ sin θ₁ + m₂v′₂ sin θ₂.

The equations of conservation of momentum along the x-axis and y-axis are very useful in analyzing two-dimensional collisions of particles, where one is originally stationary (a common laboratory situation). But two equations can only be used to find two unknowns, and so other data may be necessary when collision experiments are used to explore nature at the subatomic level.

Figure 2. A collision taking place in a dark room is explored in Example 1. The incoming object m1 is scattered by an initially stationary object. Only the stationary object’s mass m2 is known. By measuring the angle and speed at which m1 emerges from the room, it is possible to calculate the magnitude and direction of the initially stationary object’s velocity after the collision.

Elastic Collisions of Two Objects with Equal Mass

Some interesting situations arise when the two colliding objects have equal mass and the collision is elastic. This situation is nearly the case with colliding billiard balls, and precisely the case with some subatomic particle collisions. We can thus get a mental image of a collision of subatomic particles by thinking about billiards (or pool). (Refer to Figure 1 for masses and angles.) First, an elastic collision conserves internal kinetic energy. Again, let us assume object 2 (m₂) is initially at rest. Then, the internal kinetic energy before and after the collision of two objects that have equal masses is

[latex]\frac{1}{2}{mv_1}^2=\frac{1}{2}{mv′_{1}}^2+\frac{1}{2}{mv′_{2}}^{2}\\[/latex].

Because the masses are equal, m₁ = m₂ = m. Algebraic manipulation (left to the reader) of conservation of momentum in the x– and y-directions can show that

[latex]\frac{1}{2}mv_1^2=\frac{1}{2}{mv′_{1}}^2+\frac{1}{2}{mv′_{2}}^{2}+mv′_{1}v′_{2}\cos\left(\theta_{1}-\theta_{2}\right)\\[/latex].

(Remember that θ₂ is negative here.) The two preceding equations can both be true only if

mv′₁v′₂ cos(θ₁ − θ₂) = 0.

There are three ways that this term can be zero. They are

• v′₁ = 0: head-on collision; incoming ball stops
• v′₂ = 0: no collision; incoming ball continues unaffected
• cos(θ₁ − θ₂)=0: angle of separation (θ₁ − θ₂) is 90º after the collision

All three of these ways are familiar occurrences in billiards and pool, although most of us try to avoid the second. If you play enough pool, you will notice that the angle between the balls is very close to 90º after the collision, although it will vary from this value if a great deal of spin is placed on the ball. (Large spin carries in extra energy and a quantity called angular momentum, which must also be conserved.) The assumption that the scattering of billiard balls is elastic is reasonable based on the correctness of the three results it produces. This assumption also implies that, to a good approximation, momentum is conserved for the two-ball system in billiards and pool. The problems below explore these and other characteristics of two-dimensional collisions.

Section Summary

• The approach to two-dimensional collisions is to choose a convenient coordinate system and break the motion into components along perpendicular axes. Choose a coordinate system with the x-axis parallel to the velocity of the incoming particle.
• Two-dimensional collisions of point masses where mass 2 is initially at rest conserve momentum along the initial direction of mass 1 (the x-axis), stated by m₁v₁ = m₁v′₁ cos θ₁ + m₂v′₂ cos θ₂ and along the direction perpendicular to the initial direction (the y-axis) stated by 0 = m₁v′_1y + m₂v′_2y.
• The internal kinetic before and after the collision of two objects that have equal masses is

[latex]\frac{1}{2}{mv_{1}}^{2}=\frac{1}{2}{mv′_{1}}^{2}+\frac{1}{2}{mv′_{2}}^{2}+{\text{mv}′}_{1}{v′}_{2}\text{cos}\left({\theta }_{1}-{\theta }_{2}\right)\\[/latex].

• Point masses are structureless particles that cannot spin.

Conceptual Questions

1. Figure 3 shows a cube at rest and a small object heading toward it. (a) Describe the directions (angle θ₁) at which the small object can emerge after colliding elastically with the cube. How does θ₁ depend on b, the so-called impact parameter? Ignore any effects that might be due to rotation after the collision, and assume that the cube is much more massive than the small object. (b) Answer the same questions if the small object instead collides with a massive sphere.

Figure 3. A small object approaches a collision with a much more massive cube, after which its velocity has the direction θ₁. The angles at which the small object can be scattered are determined by the shape of the object it strikes and the impact parameter b.

Glossary

point masses: structureless particles with no rotation or spin