I was watching an old episode of MythBusters, and the myth they tackled was the Seesaw Saga. It's about a sky diver hitting a seesaw and launching a girl on the other end high into the air. I immediately noticed that MythBusters had no idea how momentum transfer on a seesaw works, which isn't surprising, but then I quickly realized that my initial intuition about it was all wrong as well. I don't think of it as a hard problem, once you do it properly, but I found it interesting because of how wrong the intuition about it is. And having done a quick search for seesaw problems on this forum, I've found a lot of posts with similar mistakes! So I present the problem purely as a mental exercise. A massless, frictionless, perfectly rigid seesaw has a mass M placed on the end of one arm. Another body of mass 3M strikes the end of the other arm going 120mph straight down. 1) At what speed does the lighter body leave the seesaw? 2) Can different placement of the lighter body or the impact point increase the outgoing speed? If so, what is the optimal placement and maximum speed? Assume perfectly elastic collisions between seesaw and both bodies. Further assume that at the moment of impact the seesaw is horizontal and motionless for simplicity. Feel free to post your solutions. If nobody posts anything similar to my approach, I will post it later. In either case, I suspect there will be some disagreement and discussion, which will, hopefully, resolve to everyone's benefit.