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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), $$\frac{1}{2}\left(2mu^2\right)=\frac{1}{2}\left(2mv_1^2\right)+\frac{1}{2}\left(3mv_2^2\right)$$ 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 $3m$?

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    $\begingroup$ Note that if the 1st 2 balls are dropped, then the last 2 balls are going to swing out together. So $v_1$ is not the velocity of the two balls that are dropped but the velocity of the two balls that fly off. $\endgroup$
    – jalex
    Commented May 13 at 18:42

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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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    $\begingroup$ The reasoning for why 2 balls come out with velocity $u$ is not all that complex - there are many other solutions which would conserve the momentum of the system, but they generally would not conserve the energy of the system, which must occur in an elastic collision. Conservation of energy and momentum are all you need to conclude that if 2 balls move in with velocity $u$, you can't have 4 balls come out with velocity $u/2$. I don't think your other comment is correct that it would be possible to have a different number of balls come out and still preserve both momentum and KE. $\endgroup$
    – Nuclear Hoagie
    Commented May 14 at 18:20
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    $\begingroup$ @NuclearHoagie Any combination of final velocities that satisfy $\sum v_i = 2u$ and $\sum v_i^2 = 2u^2$ will satisfy both momentum and energy conservation, including the one OP gave (which is what you would get, for example, if balls 1 and 2 were bound together and balls 3, 4, and 5 were also). $\endgroup$
    – pwf
    Commented May 15 at 0:40
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If you had an elastic collision between two solid masses, the moving one (from the right) of $2m$ and the stationary one of $3m$ you would get the solution you propose. The mass $3m$ block would move to the left at $\frac{4}{5}u$ and the $2m$ block would recoil to the right at $\frac{-1}{5}u$.
However, you do not have solid blocks. The interaction among the five balls is quite complex, depending on the separation of the various balls and the compresson waves that travel through them. see wiki Article. In this case the balance of momentum and kinetic energy is achieved when two blocks on the left move away with (close to) $u$.

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It's because your modeling of the problem is wrong. When you drop 2 balls, after the collision, only 2 balls will go up, not 3. Therefore, the equation you mentioned was wrong: $$2mu=2mv_1+3mv_2.$$ Actual equation would be: $$2mu=2mv_1+2mv_2 \\ mu=mv_1+mv_2 \\ u=v_1+v_2.$$

Also, right after the collision, the velocity of those 2 balls will be $0$ so, $$v_1=0 \\ u=0+v_2$$ therefore: $$u=v_2.$$

So, basically your modeling was wrong, that's it, and there is nothing wrong with conservation law.

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    $\begingroup$ You're missing the point. OP was asking how you know that only 2 balls come out, when there's mathematically a solution that conserves momentum and energy with 3 coming out. $\endgroup$
    – pwf
    Commented May 13 at 22:44
  • $\begingroup$ @pwf I don't think there is such a solution. If the 3 balls have the same momentum as the 2 balls, then they must have less KE (since momentum is linearly dependent on velocity, but KE depends on the square of the velocity). Conservation of energy requires the same number of balls to come out as went in. $\endgroup$
    – Nuclear Hoagie
    Commented May 14 at 19:22
  • $\begingroup$ @NuclearHoagie You're assuming balls 1 and 2 come to rest. The 3 balls don't have to have the same momentum as the 2 balls. Try $v_1 = v_2 = -\dfrac{1}{5}u$, $v_3 = v_4 = v_5 = \dfrac{4}{5}u$. $\endgroup$
    – pwf
    Commented May 15 at 0:53
  • $\begingroup$ @pwf That isn't a solution "with 3 coming out", that's a solution with 5 coming out at different speeds. "3 coming out" implies that the other 2 did not - i.e., that they are at rest. $\endgroup$
    – Nuclear Hoagie
    Commented May 15 at 13:21
  • $\begingroup$ @NuclearHoagie Excuse my sloppy language. By "3 coming out" I was referring to OP's scenario that allowed for 3 balls to emerge with one velocity and the other 2 to have another. Recognizing that the 2 must recoil, one might speak of the 3 as "coming out" and the 2 as "bouncing back". You are right that IF the 2 come to rest, then only 2 will advance, but the two coming to rest is not a necessary consequence of momentum and energy conservation alone. I read OP as asking why that has to be the outcome, and in fact it doesn't. $\endgroup$
    – pwf
    Commented May 15 at 17:55

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