Newton's cradle: why not twice the balls at half the speed? For "Sinterklaas" (Dutch holiday) I got a Newton's cradle. I understand the basics of it, but a couple things bother me.
If I drop one ball, one bounces off. If I drop two, two bounce off. Why, if I drop one, don't two bounce off at half the speed? And the other way around, if I drop two, why doesn't one bounce off at twice the speed?
What would happen if I'd stick two together, and drop one on the other side? Would the two then go with half the speed? Is this the same as a ball that's twice as heavy as the others?
And for bonus points: would there be a way to make one ball shoot off at a higher speed as whatever I drop?
If possible, please eli5.
 A: Newton's cradle relies on a special sort of collision. This is when a sphere makes a head-on elastic collision with another sphere of the same mass. It's easy to show from conservation of kinetic energy and momentum (and even more easily by viewing the collision in a frame of reference in which the spheres are moving towards each other with equal speeds) that the first sphere stops and the second moves off with the velocity that the first sphere had! [Note that macroscopic bodies can't collide perfectly elastically, that is with no loss of total KE, but steel spheres probably retain >90% of their KE.]
Newton's cradle is easier to analyse if you imagine that the spheres aren't quite touching each other. Then when the first sphere hits the second, the second will move off with roughly the same velocity that the first one had, and will hit the third and so on. The last sphere doesn't have anything to hit – until it has swung out and returned. So we have a succession of collisions of the sort discussed in the first paragraph. 
You should be able to figure out the rest from here. 
A: 
If I drop one ball, one bounces off. If I drop two, two bounce off.
  Why, if I drop one, don't two bounce off at half the speed?

Because you have to conserve momentum and energy, P=mv and E=1/2mv^2.
So if you have 1 ball raised to a height so it's going 1 speed, you get P=1 and E=1/2. 
So now let's see what happens if you want two balls at half the speed, P=(2)(.5)=1, E=1/2(2)(.5)^2 =.25 <- does not balance.
Play with some examples, you'll quickly see there's no other way it could work that balances. The v on one side vs. the v^2 on the other is the gotcha. One more for Mr. Newton!
