
In an elastic collision, kinetic energy is conserved as well as momentum. An example of a perfectly elastic collision occurs when an ideal (frictionless) rubber ball bounces off a floor and reaches the same height from which it was initially dropped. A nearlyelastic collision occurs in billiards when a fastmoving cue ball strikes another ball, causing the cue ball to stop in place and the target ball to move off in the same direction. Real collisions are rarely perfectly elastic however, the amount of kinetic energy lost may be so small that it is often a good approximation to assume perfect elasticity.

If you’ve ever played with a device like the one shown above, you’ve probably asked yourself: “How can it know how many spheres I’ve lifted?” Newton’s cradle, as it’s called, is a row of little pendulums. Each is an identical metal sphere that hangs from two threads or pieces of fishing line. Lift a single sphere at one end of the row and let it drop: One will pop up at the other end. Lift and drop two, and two will leap from the other end. What’s more, the ones that rise up will fall back toward the center and strike the remaining spheres. The cycle will then repeat itself a dozen or more times!
Newton’s cradle is an example of collisions at work. Your first thought might well be, “well, momentum must be conserved, so one ball in means one ball out, both traveling at the same speed.” And you’d be right: Momentum is conserved in Newton’s cradle. But is this the only way for this system to conserve momentum? Couldn’t two balls pop out from the other end, each with half the speed of the one that dropped in?
Yes, but there’s a catch! A pair of spheres at half the speed has the same momentum as one sphere at full speed. The total kinetic energy of the pair, however, would only be half as much as the single sphere’s. The rest of the kinetic energy would be missing! The collisions in a welldesigned Newton’s cradle are nearly elastic, and elastic collisions conserve kinetic energy. There is only one possible outcome that will conserve both momentum and kinetic energy; if one ball swings down at full speed, one and only one ball must pop out, and at full speed.

(11.4)  $$\frac{1}{2}{m}_{1}{v}_{{}_{i1}}^{2}+\frac{1}{2}{m}_{2}{v}_{{}_{i2}}^{2}=\frac{1}{2}{m}_{1}{v}_{{}_{f1}}^{2}+\frac{1}{2}{m}_{2}{v}_{{}_{f2}}^{2}$$
  Conservation of energy for elastic collisions


Elastic collision problems typically involve two equations: conservation of momentum and conservation of kinetic energy. The momentum equation involves the masses and velocities before and after the collision. The energy equation involves the masses and the velocities squared before and after the collision. The squared velocities make the algebra of solving momentum problems a little more challenging. In problems involving two and three dimensions, momentum must be conserved separately in each direction. Kinetic energy is a scalar however, and there is typically only one kinetic energy equation.

What can make an elastic collision problem even more difficult is that the conservation of kinetic energy equation is quadratic in speed; i.e., the speed variables are all squared. In algebra class you learned that if you have a quadratic equation of the form
then the two solutions (or roots) of the equation are
$$x=\frac{b\pm \sqrt{{b}^{2}4ac}}{2a}$$

More difficult elastic collision problems involving these quadratic equations will be covered in more depth in later chapters in the book.

A 0.16 kg cue ball traveling at 4 m/s strikes a stationary 0.16 kg eight ball. After the collision, the cue ball travels at 0.2 m/s while the eight ball travels at 3.8 m/s. Is this an elastic collision? Why or why not?
Asked: 
whether it is an elastic collision; i.e., is kinetic energy conserved? 
Given: 
masses of the balls, m_{1} = m_{2} = 0.16 kg;
initial speed v_{i1} = 4 m/s of the cue ball;
initial speed v_{i2} = 0 m/s of the eightball;
final speed v_{f1} = 0.2 m/s of the cue ball;
final speed v_{f1} = 3.8 m/s of the eight ball 
Relationships: 
conservation of energy for two objects in an elastic collision: $$\frac{1}{2}{m}_{1}{v}_{{}_{i1}}^{2}+\frac{1}{2}{m}_{2}{v}_{{}_{i2}}^{2}=\frac{1}{2}{m}_{1}{v}_{{}_{f1}}^{2}+\frac{1}{2}{m}_{2}{v}_{{}_{f2}}^{2}$$


Solution: 
Compare the kinetic energies of the balls before and after the collision:
$$\begin{array}{ccc}\frac{1}{2}(0.16\text{kg}){(4\text{m/s})}^{2}+\frac{1}{2}(0.16\text{kg}){(0)}^{2}& \stackrel{?}{=}& \frac{1}{2}(0.16\text{kg}){(0.2\text{m/s})}^{2}+\frac{1}{2}(0.16\text{kg}){(3.8\text{m/s})}^{2}\\ 1.28\text{J}+0\text{J}& \ne & \text{0}\text{.003J}+1.155\text{J}\end{array}$$


Answer: 
The collision is not elastic because some kinetic energy is lost. (Since only a small amount of the kinetic energy is lost, the collision is “nearly elastic.”) 

In elastic collisions, which of the following quantities are conserved?
 velocity
 momentum
 kinetic energy
 I only
 I and III only
 II and III only
 I, II, and III

The correct answer is c, both momentum and kinetic energy. Momentum is conserved in all collisions, and what is special about elastic collisions is that kinetic energy is also conserved.

 
