Domino waves

Let's consider the energy in two rows.

Besides the initial push, the kinetic energy in two rows are all coming from the gravitational potential energy of the dominoes when they are falling down. The different work done by the two initial push will be transferred to the later dominoes with some certain fraction in each collision. And after a great amount of dominoes, this part of energy will become negligible. The gravitational potential energy, which are the same in two rows, will be the main cause of the motion of, say, the 1000th domino.

Therefore, we can estimate that the time taken for two rows to fall down will be approximately the same.

While the force on the initial domino B is 3 times the force on the initial domino A, the dominoes within each chain are not touching, so the force on each domino in chain B does not remain at triple the force on chain A. They will finish at just about the same time. Try this out for yourself!

Run a mental simulation or write a code in python with the mechanics of dominoes falling, you will see that whilst B gets a minuscule head-start they both equalize exponentially since the initial force doesn't translate to every domino tile equally. So whilst there is there that row B will finish by a 0.66 secs faster, it will never finish 2/3 or 1/3 of the time A finishes, so the logical conclusion is to answer the one option that comes closest to the prediction and eliminate those who are off by a large margin, so we are left with the answer of A and B finishing in "about the same" time. (note the key word "about" and not not necessarily "the same")

If the dominos were evenly spaced in an exactly straight row in gravity-free space, and if you strike the first one with some velocity $v$ , thus imparting the system with a certain amount of energy $E$ , then by Conservation of Energy, the system will only contain constant energy $E$ throughout, and so at the end of the row of domino, the last one will fly off with the same velocity $v$ . The faster $v$ is, the faster the "wave" will get to the other end. It might seem that, therefore, $F_B$ would finish quicker.

However, on the ground with gravity, we have to consider the gravitational potential energy of falling dominos, and it's not necessarily a fact that all of the energy of one falling domino is transferred to the next. After all, one domino could miss the next. But assuming that the wave is asymptotically moving forward with constant speed, if the instantaneous velocity of the tip of one standing domino just struck by the previous one is $v$ , then it's the same for the next standing one when struck. Given $(v, h, d, s, g)$ , or standing tip velocity, height of domino, thickness of domino, space between dominos, and gravitational constant, one can compute the tip velocity of one falling domino at the moment of its impact on the next standing domino, which determines its standing tip velocity, which is the same $v$ . Solving for $v$ in terms of $(h, d, s, g)$ involves elliptic integrals, but regardless of the exact answer, the initial velocity at the start doesn't have a bearing on this long term asymptotic wave speed. Regardless of how the first domino is struck, the long term wave speed of falling dominos depends only on the dimensions and spacing of the dominos and gravitational constant.

Striking the first domino such its initial instantaneous tip speed is less than the asymptotic wave speed can be replicated by having a few more dominos before the first one, such that the "new first one" is barely tipped over, so that $F_B$ becomes like $F_A$ in the long run, which is why "Row A and Row B take about the same amount of time to finish falling".

The force of the first domino doesen't really have a measurable impact on the entire system thus If you consider the first domino to be the trigger for starting a system of 999 dominos the main diffrence between row A and B finishing is likely to be around 0.66 seconds.

An easier way to think of this is that while Force B is greater and takes less time to push the first domino, the speed at which the dominoes fall is a wave and equivalent to the speed of the wave propagation through the system since each domino isn't touching at t=0.

Since there are 1000 domino's in each row, it doesn't really matter how strong the first domino is pushed. In the end they are all pushed down (or falling) by gravity with always the same force. Pushing the first stone is only necessary to introduce the chain reaction. For example starting a fire with a high powered laser or a classical lighter has no influence on how long it takes to burn an haystack, as long it was hot enough to get the process started.

Let n = the number of dominos, m = mass of each domino.

Kinetic energy gained in each falling domino equals to its loss in potential energy. $KE_{total} = KE_{initial} + nPE$

Using $KE = 0.5mv^2$ , the speed at which all the dominos fall is:

$v = \sqrt{\frac{2KE_{total}}{nm}} = \sqrt{\frac{2(KE_{initial}+ nPE)}{nm}} = \sqrt{\frac{2KE_{initial}}{nm} + \frac{2PE}{m}}$

As n increases to a large number, the inital KE term decreases and the PE term dominates.

1000 is pretty large, so it is reasonable to assume that the two rows fall at the same speed, and hence about the same time.

violets are blue, roses are red, from the second domino speed remain the same till the end.

If we look at the system's center of mass for both cases we can assume that the initial and final statements are the same for both rows. Since there's only horizonal velocity at the beginning, the time of the free falling depends only on the gravity acceleration , which is the same for them both. Thus the time from initial to final situation is the same.

Row A starts with V1, Row B starts with V2.

V1 < V2.

The dominos take T time and, and have a constant and identical acceleration A.

So as time progresses:

The speed of Row A, becomes: Va = V1 + A•T

And the speed of Row B, becomes: Vb = V2 + A•T

As T goes to infinity, T→∞, the speeds become identical:

Va = 0 + A•T = Vb

Additionally, any distance D made between the two rows, will become insignificant as well as T→ ∞:

Xb = V2•T + ½AT²

Xa = D + Xb = D + V2•T + ½AT²

T→∞:

Xb = 0 + ½AT² = ½AT²

Xa = 0 + Xb = 0 + 0 + ½AT² = ½AT² = Xb

What is the meaning of this conclusion? The meaning is that, the distance passed by both rows with respect to time, will become closer and closer to each other(in terms of ratio).

This may help!

Based on life experience, it seems that the speed of domino is the same.