When two objects collide and one of the masses is much larger (maybe a factor of 100 or so), we can avoid parts of the calculation from the previous section. For instance, one thing we know for sure is that the more massive object's velocity in E is approximately equal to the velocity of the center of mass C in E. This is easy to see if you look back at the definition of the velocity of the center of mass expression which is just a weighted average of terms. [br][br]I originally used an example of a course grade with weighted course components to describe the equation. In this case it's as if an exam worth 200 points and maybe a two point attendance quiz were your only two course scores. If you got 90% on the exam and 50% on the quiz, what is your course percentage? I hope you see without calculating that it's almost exactly 90%. [br][br]The equivalent here is that [math]\vec{v}_{C,E}\approx \vec{v}_{1,E}[/math] if object one is more massive. Making this assumption will save you calculation time. Making this assumption will also make both the initial and final velocities of the more massive object zero in the center of mass frame. This is just because the object itself is assumed to be the center of mass and it can't be moving with respect to itself! [br][br]I want you to understand that the reality is that the more massive object will have a very small, but non-zero velocity in C. It must have a non-zero velocity since otherwise the total momentum in C could not be zero. In most cases, we needn't worry about the tiny velocity, however.

If a collision takes place in which the motion of both objects stays on a line, we may simplify the equations from the previous section. It would be a useful exercise to prove to yourself that the equations in C will reduce to:[br][br][center][math]\vec{v}_{1,C}_f=-e\vec{v}_{1,C}_i \\[br]\vec{v}_{2,C}_f=-e\vec{v}_{2,C}_i [/math][/center]What this implies is that when viewed from the center of mass frame, collisions simply lead to the reversal of the motion and a reduction in speed by a factor of e. If the collision does not dissipate any energy (e=1), then the collision will be a perfect reversal as if a video was played in reverse. In fact there is a deep connection between elastic systems and time-reversal symmetry. The short version is that if energy is not dissipated, there is no way to distinguish time going "forward" vs "backwards". Nature, on the other hand, always tends in any but the simplest systems to dissipate energy. This is equivalent to the notion of the increase in entropy of systems as you may have discussed in chemistry or elsewhere.

A ping pong (really called table tennis) paddle is roughly 50-100 times as massive as the ball. Use the discussions above to find the final velocities in E for a paddle p and ball b that start with velocities given below, assuming that the collision is linear. The velocities below represent a paddle that's moving at speed v in the x direction, and a ball moving at twice that speed in the negative x direction.[br][center][br] [math]\vec{v}_{p,E}_i=v\hat{i} \\[br]\vec{v}_{b,E}_i=-2v\hat{i}. [/math][/center]

Using the subscript 'p' for paddle and 'b' for ball, we may assume since the paddle's mass is much larger that: [br][br][center][math]\vec{v}_{C,E}=\vec{v}_{p,E}_i=v\hat{i} \\[br]\text{and therefore } \\[br]\vec{v}_{p,C}_i=\vec{0}. [/math][/center][br][br]We still need to find the ball's initial velocity in C:[br][math]\vec{v}_{b,C}_i=\vec{v}_{b,E}_i+\vec{v}_{E,C} \\[br]\vec{v}_{b,C}_i=\vec{v}_{b,E}_i-\vec{v}_{C,E} \\[br]\vec{v}_{b,C}_i=-2v\hat{i}-v\hat{i}=-3v\hat{i}. [/math][br][br]The result above means that if you had an action camera strapped to the paddle, that it would look like a ball is traveling toward it (in the -x direction) at a speed 3v. [br][br]To find the final velocities of the two objects in C, simply multiply by -e. Therefore: [br][math] [br]\vec{v}_{b,C}_f=-e\vec{v}_{b,C}_i = -e(-3v\hat{i})=3ve\hat{i} \\[br]\vec{v}_{p,C}_f=-e\vec{v}_{p,C}_i =\vec{0} [/math][br][br]The last step is to convert back to earth frame E:[br][math][br]\vec{v}_{b,E}_f=\vec{v}_{b,C}_f+\vec{v}_{C,E} \\[br] \vec{v}_{b,E}_f = 3ev\hat{i}+v\hat{i}=(3e+1)v\hat{i} \\[br]\text{For the paddle we find } \\[br]\vec{v}_{p,E}_f=\vec{v}_{p,C}_f+\vec{v}_{C,E} [br]\vec{v}_{p,E}_f=\vec{0}+v\hat{i}=v\hat{i}.[/math][br][br]What this tells us is that the paddle moves right through the collision without change in velocity and that the ball leaves with a speed that at maximum in a perfectly elastic collision would approach 4v, but realistically with e=0.8 would give us 3.4v.[br][br]Recognize that in reality there will be a very slight decrease in speed of the paddle, but that it is assumed negligibly small here. Correspondingly, the speed of the ball would be every so slightly less than 3.4v in reality. If you want to know for sure what those values are, then use the full-blown versions of the equations from the previous section of the text and don't make the assumption about the paddle's mass being much larger.