Newton’s Cradle is a desktop toy consisting of five metal balls in a line hanging from a frame. You pull one back and let it go. It hits the other four, leaving three motionless (more or less), while the fourth goes swinging out from the opposite end. All in all, a very cool way to demonstrate conservation of momentum. OK, here’s where I’m confused. No matter how hard you swing the first ball, only one will react from the other side. If you swing back two balls, however, two balls will react on the opposite side. How does it know to do that? Momentum is mass times velocity. What’s the difference between one ball and high velocity and two balls at low velocity?
This is caused by conservation of energy. The system has to conserve both momentum (mass x velocity) and energy (1/2 mass x velocity squared, assuming very little energy is lost as heat). So if you swing two balls and only one on the other side reacts, but with twice the velocity, then momentum is, as you noted, conserved. However, the kinetic energy would be doubled without any other energy being freed up.
Momentum has direction. To properly analyze momentum you need to consider both its magnitude and its direction. In combination this gives you a momentum vector.
Conservation of energy shows why the momentum is preserved in a given direction. However, the reason the toy ‘knows’ that you’re dropping two balls in opposite direction as opposed to one ball in a given direction is that momentum is preserved according to direction.
In the two ball case, the momentum that the left ball has going left, is transferred through the many balls to the right side and is taken up by the rightmost ball. The momentum that the right ball has going right is likewise transferred through the many balls to the left side and is taken up by the left ball. You can think of this as pulses of kinetic energy travelling through each other, if you like.
Perhaps I should field this one.
Momentum must be conserved. But that’s not the whole story.
Consider only momentum. Consider the five ball case that you raised. All five balls have equal masses (m). The initial velocity of ball 1 is v[sub]1i[/sub]. The initial velocities of balls 2-5 are 0. Assume also that all velocities are directed along the line joining the (initially stationary) balls. This last is an oversimplification, but it won’t change our result.
The final velocities of all five balls is unknown. It must obey the conservation of momentum:
mv[sub]1i[/sub]=mv[sub]1f[/sub]+mv[sub]2f[/sub]+mv[sub]3f[/sub]+mv[sub]4f[/sub]+mv[sub]5f[/sub]
We know the “right” solution, that v[sub]1i[/sub] equals v[sub]5f[/sub], and all other velocities are zero. But that’s not the only solution. You correctly point out that
v[sub]1i[/sub]=(v[sub]4f[/sub]+v[sub]5f[/sub])/2
is another solution, where v[sub]4f[/sub]=v[sub]5f[/sub] (i.e. two balls moving at half the initial velocity of ball 1 each).
But wait! We have to preserve kinetic energy, too.
That is:
mv[sub]1i[/sub][sup]2[/sup]=mv[sub]1f[/sub][sup]2[/sup]+mv[sub]2f[/sub][sup]2[/sup]+mv[sub]3f[/sub][sup]2[/sup]+mv[sub]4f[/sub][sup]2[/sup]+mv[sub]5f[/sub][sup]2[/sup]
Now, we can’t have the solution you proposed:
v[sub]1i[/sub][sup]2[/sup]=(v[sub]4f[/sub][sup]2[/sup]+v[sub]5f[/sub][sup]2[/sup])/2
Doesn’t work when v[sub]4f[/sub]=v[sub]5f[/sub] (that implies that 1=sqrt(2), which it don’t).
There are possible solutions for all the v[sub]nf[/sub]'s that satisfy the two equations above, but there are other properties of Newton’s cradle. For instance, before the collision:
v[sub]1i[/sub]<=…<=v[sub]ni[/sub], and vsubi[/sub]=…=v[sub]5i[/sub]
and after
v[sub]1f[/sub]<=v[sub]2f[/sub]<=v[sub]3f[/sub]<=v[sub]4f[/sub]<=v[sub]5f[/sub]
That is, a ball cannot move faster than one it is “pushing”, or slower than one that is “pushing” it.
Basically, the only solution that conserves both momentum and energy, and that doesn’t have a ball “passing” another on the way in (before the collision) or on the way out (after the collision) (i.e. all the balls remain in line with each other) is the one that you observe.
kg m²/s²
Assuming all five spheres have the same mass and diameter, the case of one ball moving to the right at 1 m/s, colliding, and resulting in 2 balls moving to the right at 2/3 m/s, and one moving to the left a 1/3 m/s, not only conserves momentum and energy, it is physically possible. More detail on how to do this can be found in an old thread on this board (Hint: the spheres can be same mass and diameter yet not be identical).
Saying that the balls at the other end magically know they should conserve momentum and energy is inaccurate. What really happens is that where the first ball or group of balls makes contact, a compression wave radiates in opposite directions from the point of impact. Those waves travel through the balls, until they reach the ends of the row, at which point they are reflected back where they came from. If all the balls are identical, those waves eventually meet again and superimpose in such a way that the same number of balls continue on from the other end. I left a detailed description of how these waves interact in the old thread. Sorry, but this board is too slow for me to search for the old thread.
Here’s the other thread: Gizmo with Five Swinging Balls
OK, I’m more or less satisfied with the kinetic energy explanation. I was thinking that KE didn’t have to be preserved because of the clicking sound for one thing, but really that just skews the results slightly from the ideal. So .5m * 2v has the same momentum, but different kinetic energy. Got it.
Now let’s play “what if”. What if the one end ball was twice as heavy as the other four? Would swinging it kick up two balls on the opposite side? I think so. Now for the hard part, what if you swing one ball on the opposite side? It can’t kick up the heavy ball at half the velocity since that would violate the KE rule. I’m thinking some combination of the heavy ball reacting and the light ball recoiling is the answer, but I’m not sure.