So the CB only accelerates to 1.5 x the stick speed because the stick slowing to half its speed during contact is as if a stick half its weight (1.5 x CB weight) hit the CB?
pj <- dumb it down for me
chgo
The easy way to get to roughly the answer without going into actually solving equations is to change the frame of reference of the stick and ball to one where they are coming at each other to start with and the total momentum is zero. After the collision, the total momentum must still be zero, and if no energy is lost the two objects must have perfectly reflected at the contact point. That makes the analysis really simple, and gives the 2:1 speed multiplier for very light targets that I mentioned above.
I'm afraid that changing the frame of reference loses a lot of people. And, you have to get them to agree to conservation of momentum and energy. Also, you have to introduce the idea that momentum has a direction, and that momentum is the product of mass and velocity, with velocity having a direction. So, it's only a slightly dumber way to look at things.
The actual equation of the speed of the ball relative to the speed of the stick, which has been posted here before and is in Byrne's book and is on Dr. Dave's website and is in every high school physics text book, is:
VelocityBall/VelocityStick = 2 MassStick/(MassStick+MassBall)
This can be checked for reasonableness by looking at extreme cases. If the ball is very, very heavy, it's speed will almost be zero because its huge mass dominates the denominator on the right side. If the masses are equal (as with direct ball-ball collisions) the speeds are the same. If the ball is very light, like the ping pong ball and the locomotive, the ball speed is twice the locomotive speed.