Conservation of momentum in Newton’s cradle


Imagine a Newton's cradle with 5 balls with mass of each ball is $m$. In a case where two balls are dropped against three balls, if we write an equation considering that momentum is conserved,
$$ 2mu=2mv_1+3mv_2$$
where $u$ is the initial velocity of the two balls dropped, $v_1$ is the final velocity of the two balls dropped, $v_2$ is the velocity of the three balls that were stationary.

And an equation for the kinetic energy conserved (assuming the collision is elastic),
And solving the two equations, $v_1$ doesn't equate to $0$ and $v_2$ doesn't equate to $u$. But in a Newton's cradle, a number of balls equal to the dropped number of balls always should go up from the other side with the same velocity. My question is why can't we use conservation laws considering the collision as between two systems, one with mass $2m$ and the other with mass $3m$?

Best Answer

You are right, conservation of energy and 1D momentum only give 2 equations, but there are 5 unknowns (the final velocities of all 5 balls), so there are many possible outcomes for a given initial condition. The one with two balls coming out and the two that came in coming to rest seems like it "ought" to be the right one because it's nice and symmetrical, and indeed it's (at least approximately) what we see if we do the experiment, but how does Nature know to pick that one? It turns out, for reasons that are not at all obvious, that you can treat the overall collision as a set of pairwise collisions happening in sequence. It helps to imagine that there's a small gap between the balls, even though in practice they are usually hanging in contact. If we number the balls so that 1 and 2 come in with velocity u and 3, 4, and 5 are initially at rest, first 2 collides with 3, then 3 collides with 4 while 1 collides with 2, then 4 collides with 5 while 2 collides with 3, then 3 collides with 4. The net result is that 1, 2, and 3 are at rest and 4 and 5 exit with velocity u.

Here are the balls' velocities at each step: \begin{eqnarray*} (u, u, 0, 0, 0)\\ (u, 0, u, 0, 0)\\ (0, u, 0, u, 0)\\ (0, 0, u, 0, u)\\ (0, 0, 0, u, u) \end{eqnarray*}

The transmission of the momentum from the left end to the right end happens at the speed of sound in the balls (because it is a compression wave), which for the usual metal balls is so fast as to appear instantaneous (but see this video to see what happens with some other materials).

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