The simplest type of collision is one in which the involved objects stick together as the cars did in the earlier example. The reason that type is easy to calculate is that there is only one unknown variable, [math]\vec{v}_f.[/math][b]Collisions in which objects stick together after collision are called perfectly inelastic collisions[/b]. Among all collisions in nature, these lose the most energy to microscopic dissipative forms such as heat and vibrations.

The other end of the spectrum is the type of collision that loses none of the macroscopic kinetic energy to microscopic dissipative types. These do not exist in the world of our experience, but are roughly how very bouncy (elastic) balls collide off one another such as when billiards balls collide. Such collisions are much harder to deal with since there is an unknown velocity for each object after the collision. [br][br]Elastic collisions generally only occur in the microscopic realm of particle physics, but are not even guaranteed then as we shall see during third semester. If any energy is lost to sound (you can hear the collision), if the objects colliding vibrate (all macroscopic objects will), then the collision is not elastic. Many books will, however, approximate collisions between macroscopic objects as being elastic.

Collisions in which some of the kinetic energy of the incoming objects is dissipated as heat, sound or vibration are called inelastic. This is the type we mostly encounter in our experience of sports and life in general. The trouble with these is that we know that some balls bounce better than others. We therefore need a means of quantifying how "springy" a collision is. That's the point of the coefficient of restitution which we discuss next.

The extent to which collisions manage to conserve kinetic energy - from perfectly inelastic which dissipate the most kinetic energy to elastic which dissipate none - is quantified by a term called the coefficient of restitution, e. [b]This coefficient is defined as the ratio of the outgoing relative velocity of two objects over the incoming relative velocity. [/b] In other words, the ratio of how fast objects recede from one another after impact over how fast they were coming together before impact.[br][br]The easiest-to-imagine scenario is probably bouncing a ball on the ground. If a ball hits the ground while traveling 5.0m/s and leaves the ground traveling 4.0m/s, then [br][br][center][math]e=\frac{v_{recession}}{v_{approach}}=\frac{4.0m/s}{5.0m/s}=0.8.[/math][/center][br]This value is typical of sports balls that bounce well, like ping pong balls, for instance.

[i]The easiest way to find the coefficient of restitution of a ball such as a ping pong ball is to drop it from a known height and see how far it rises after bouncing. The height can't be too high (like a balcony) since otherwise energy is lost to air drag. At relatively slow speeds this loss is negligible, and the only reason for the rebound height to be lower than the drop height is that energy is lost during the bounce. The larger the energy loss, the lower the coefficient of restitution. [br][br]Suppose we drop a ping pong ball from a height of 1.00m and watch it rebound to a height of 72cm. Let's find the coefficient of restitution, e. First we find the speed at which the ball strikes the ground after being dropped from rest and from a height of 1.00m.[/i][br][br][center][math]\Delta K+\Delta U=0 \\[br]\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2+mg\Delta y=0 \\[br]\frac{1}{2}mv_f^2-0+mg\Delta y=0 \\[br]\frac{1}{2}v_f^2=-g\Delta y \\[br]v_f=\sqrt{-2g\Delta y} \\[br]v_f =\sqrt{-2(10m/s^2)(-1.00m)} \\[br]v_f = 4.47m/s \\ [/math][/center][br][br][i]Next we find how fast the ball must have left the ground such that it rises to a max height of 0.72m. Using the same equation but substituting different values, we find: [br][/i][br][center][math]\Delta K+\Delta U=0 \\[br]\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2+mg\Delta y=0 \\[br]0-\frac{1}{2}mv_i^2+mg\Delta y=0 \\[br]v_i = \sqrt{2g\Delta y} \\[br]v_i = 3.79m/s [/math][/center][br][i]The ratio of how fast ball left the ground to how fast it struck the ground is e, or:[/i][br][br][center][math][br]e=\frac{v_{recession}}{v_{approach}}=\frac{3.79m/s}{4.47m/s}=0.85 [/math][/center]

Notice that the ball leaves the ground slower than it struck the ground. This obviously means that the ball lost some kinetic energy during the impact. Since kinetic energy is given by [math]K=\tfrac{1}{2}mv^2[/math] we can find the fraction of remaining energy by taking the ratio of [math]\tfrac{K_f}{K_i}.[/math] This ratio turns out to be just [math]e^2=0.85^2=0.723.[/math] In other words, around 28% of the kinetic energy is lost during a single bounce of a ping pong ball. It is lost to sound energy, vibrations in the surface, and a tiny bit of heat production in both the ball and the surface.