Relationship of multiple particles under collision Consider 3 particles. All 3 particles travel along the x-axis.


*

*The 1st particle possesses some mass, m, and its initial position is somewhere on the negative x-axis. It has some (positive) velocity v.

*The 2nd particle possesses some mass, A*m, and its initial position is at the origin of the x-axis (0). It has no velocity (at rest).

*The 3rd particle possesses some mass, B*m, and is initially situated somewhere on the positive x-axis. Like particle 2 it has no velocity (at rest).


Find the relation between A and B, under which the 1st and 2nd particles will collide with each other more than once.
note: it is assumed the collisions are elastic as well as there being no external forces considered. also the collisions are such that all particles remain on the x-axis
 A: Okay, let's first review the initial setup:


*

*1st particle: mass $m$ and initial velocity $v_m>0$ in +x direction

*2nd particle: mass $Am$ with assumption $A>0$, and initial velocity $v_A=0$

*3rd particle: mass $Bm$ with assumption $B>0$, and initial velocity $v_B=0$
Notation: after each collision, the new velocities will have a prime added to them, so after two collisions velocity of 2nd particle would be written $v_A''.$
From the setup, it is clear that the first collision takes place between 1st and 2nd masses, and knowing we have elastic collisions only, momentum and kinetic energy are both conserved. After the first collision, the 2nd mass moves(+x direction) towards the 3rd mass (still at rest), and they will collide, whereas after the 1st mass is going now in the -x direction (momentum conservation), so for the 2nd mass to collide with 1st again (more than one collision being the premise of the question), it has to catch up with the first mass, after its collision with $Bm$ i.e. the 3rd mass. Let's start solving:

*

*First collision momentum conservation gives: $$m \vec{v}_m=m\vec{v}_m'+Am \vec{v}_A'$$
kinetic energy conservation: $$\frac{1}{2}m v_m^2=\frac{1}{2}mv_m'^{2}+\frac{1}{2}Am v_A'^2$$
Coupling the two equations, we have for final velocities after collision: $$v_m'=\frac{(m-Am)v_m+2Am v_A}{m+Am}=\frac{(1-A)v_m}{1+A}$$
$$v_A'=\frac{2mv_m-(m-Am)v_A}{m+Am}=\frac{2v_m}{1+A}$$


*Second collision takes place between 2nd particle and the 3rd particle still at rest. Similarly as before equation of momentum conservation: $$Am \vec{v}_A'=Am\vec{v}_A''+Bm\vec{v}_B'$$
And the final velocities similar to before: $$v_A''=\frac{(A-B)v_A'}{A+B}$$
$$v_B'=\frac{2v_A'}{A+B}$$
where above are the simplified formulas knowing we can simplify the masses $m$ and knowing $v_B=0$ initially.

*

*Third collision: for it to take place between the 1st and 2nd masses, that now are both going in the -x direction after the collisions described above, it has to be able to catch up with $v_m'$ simply, so for them to collide again $A$ and $B$ need to satisfy the following: $$v_A''> v_m'$$
Lets start substituting:
$$\frac{A-B}{A+B}v_A'  > \frac{1-A}{1+A}v_m $$
$$\frac{A-B}{A+B}\frac{2v_m}{1+A}  > \frac{1-A}{1+A}v_m$$
Simplifying $v_m$, we can re-write the in-equation as(which is the first relation between $A$ and $B$):
$$A^2+A+AB>3B$$
Cases where the above holds (two cases to distinguish, in terms of A):

*

*$A>3$ where then we have $B>\frac{A^2+A}{3-A}$

*$A<3$ where then $B<\frac{A^2+A}{3-A}$

*Last possibility $A=3$
If either cases fulfilled, the first and second particles will collide more than once.
