p1i + p2i = p1f + p2f .
If the forces acting during the collision are conservative, then mechanical energy is conserved as well. If the objects moves along horizontal, only the kinetic energy, KE = ½ m v2 plays a role. Therefore
½ m1 v21i + ½ m2 v22i = ½ m1 v21f + ½ m2 v22f .
In this experiment, we will use an air track as an essentially frictionless surface. (An air track is an aluminum surface with many tiny holes punched regularly on it, such that when sufficient amount of air is blown through the holes continuously by means of an electric air blower, an object sliding on this thin cushion of air is considered to be moving without friction). Since all motions are one dimensional, the momenta can be added algebraically with appropriate positive or negative signs to incorporate the direction of motion. Furthermore, in order to simplify the analysis, the 2nd glider (m2) will be held to stay still (v2i = 0) before the collision, such that only the 1st glider (m1) will have an initial velocity v1i directed toward the stationary 2nd glider. For the Elastic Collision, the masses of the gliders are chosen to be almost exactly equal to each other (m1 ~ m2), such that after collision, both laws of conservation of momentum and kinetic energy give rise to: the 1st glider comes almost to a full stop (v1f ~ 0), and the 2nd glider takes off with the velocity almost equal to the initial velocity of the 1st glider (v2f ~ v1i). (See the following figures).
The instantaneous velocities v's will be determined in an approximate manner, by means of photogates, as v = delta x/ delta t with delta t a very small number. The numerator delta x is the width of a small flag mounted vertically at the top of the glider, whereas delta t is the time interval during which the flag blocks the photogate as the glider is sliding along the track.