May be you know about famous problem for its answer.

Place two billiard balls ($B_1$ and $B_2$) on the half-line $[0,\infty)$: the $B_1$ is at rest, the $B_2$ is to the right of it and is moving the left. At $x=0$ there is an elastic wall. Assume perfectly elastic collisions: the kinetic energy is preserved. How many collisions will happen depends on the masses of the ball, or rather their ratio.

For more details see Here.

Let me change condition of this problem a bit.

Place two billiard balls ($B_1$ and $B_2$). Both are at rest. Next, place one more ball $B_0$ between $B_1$ and $B_2$ which moves to the ball $B_2$. Every collision is elastic: kinetic energy is preserved. The same question.

We can begin with simple case when masses of $B_1$ and $B_2$ are equal to $M$ and mass of $B_0$ is equal to $m$.

I tried to solve this and my result is a system of recurrences which I am not able to solve. I will post my solution below.

  • $\begingroup$ So, what is your question? $\endgroup$
    – Jon Custer
    Commented Jan 12, 2017 at 21:07
  • $\begingroup$ How many collisions will happen? $\endgroup$
    Commented Jan 12, 2017 at 21:35
  • $\begingroup$ The second problem is without a wall, right? $\endgroup$ Commented Jan 12, 2017 at 22:02
  • $\begingroup$ Yes. Just three balls. $\endgroup$
    Commented Jan 12, 2017 at 22:13
  • $\begingroup$ Your title says 2 balls, but now you are asking about 3 balls. ... This paper might be of some help. $\endgroup$ Commented Jan 13, 2017 at 2:18

1 Answer 1


Let us introduce some notations.


$M, m$ for mass of $B_1$ ($B_2$) and $B_0$ respectively.
$K$ for $\frac{M}{m}$
$V$ for velocity of $B_0$
$W$ for velocity of $B_1$ or $B_2$
$V_i$ for velocity of $B_0$ after $i$th collision.

Next, it easy to get some helpful results. But firstly note that answer is not depends on initial value $V_0$ of $B_0$ therefore without loss of generality we can assume $V_0 = 1$. So, see, fig. $$W_1 = \frac{2}{1+K},\quad\text{i.e. velocity of $B_2$ after 1st collision}$$ $$V_1 = \frac{1-K}{1+K},\quad\text{i.e. velosity of $B_0$ after 1st collision}$$ $$W_2 = \frac{2}{1+K}\cdot V_1,\quad\text{i.e. velocity of $B_1$ after 2nd collision}$$ $$V_2 = \Bigl(\frac{1-K}{1+K}\Bigr)^2,\quad\text{i.e. velocity of $B_0$ after 2nd collision}$$ And so on. You get the point. We can write it in general: \begin{align} V_{i+1} &= (V_i - W_{i-1})\frac{1-K}{1+K} + W_{i-1}\\ W_{i+1} &= (V_i - W_{i-1})\frac{2}{1+K} + W_{i-1} \end{align} with initial conditions $$V_0 = 1, W_{-1} = W_0 = 0$$ That's all. Obviously, that if for some $i$ it is fulfilled that $$|V_i| \leq |W_{i-2}|$$ then process terminated. But how to know the first such $i$. I solved these recurrences numerically and this what I have got. enter image description here Where Theory is $$f(K) = \frac{\pi\sqrt{K}}{2\sqrt{2}}.$$ Where this function comes from? In paper mentioned by @sammygebril in comments similar problem solved. That result is two times bigger, i.e. $$f(K) = \frac{\pi\sqrt{K}}{\sqrt{2}},$$ for $K\gg1$.

  • $\begingroup$ If $K$>1, I'd rewrite the equations with $K-1$ numerators. Also in the first of your general equations, shouldn't it be $V_{i-1}$ at the end? $\endgroup$ Commented Jan 12, 2017 at 22:11
  • $\begingroup$ It does not matter because $V_i$ changes its sign from $i$ to $i$. $\endgroup$
    Commented Jan 12, 2017 at 22:16
  • $\begingroup$ Sorry. I did not see your 2nd sentence in the first time. I think it shouldn't because it is convenient to look at collision from reference system where one of he balls $B_1$ or $B_2$ is at rest. That is why $W$ appears in both recurrences. $\endgroup$
    Commented Jan 12, 2017 at 22:27

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