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  • Discuss two dimensional collisions as an extension of one dimensional analysis.
  • Define point masses.
  • Derive an expression for conservation of momentum along x -axis and y -axis.
  • Describe elastic collisions of two objects with equal mass.
  • Determine the magnitude and direction of the final velocity given initial velocity, and scattering angle.

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 size 12{p} {} is conserved. The simplest collision is one in which one of the particles is initially at rest. (See [link] .) The best choice for a coordinate system is one with an axis parallel to the velocity of the incoming particle, as shown in [link] . Because momentum is conserved, the components of momentum along the x size 12{x} {} - and y size 12{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.)

A purple ball of mass m1 moves with velocity V 1 toward the right side along the X direction. The orange ball of mass m 2 is initially at rest. The total momentum is the momentum possessed by purple ball only. After collision purple ball moves with velocity v 1prime in the positive X Y plane making an angle theta 1 with the x axis and the orange ball moves in the X Y plane below the x axis making an angle theta 2 with the x axis. The total momentum would be the sum of the momentum of purple ball p1 prime and the orange ball p 2 prime. In two-dimensional collision too the momentum before and after collision remains the same.
A two-dimensional collision with the coordinate system chosen so that m 2 size 12{m rSub { size 8{2} } } {} is initially at rest and v 1 size 12{v rSub { size 8{1} } } {} is parallel to the x size 12{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 size 12{x} {} -axis, the equation for conservation of momentum is

p 1 x + p 2 x = p 1 x + p 2 x .

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

m 1 v 1 x + m 2 v 2 x = m 1 v 1 x + m 2 v 2 x .

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Practice Key Terms 1

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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