Elastic and Inelastic Collision in Two Dimensions

Firstly a note in order to avoid any misunderstandings: the exact kinematics of a particle collision is rarely of interest in plasma physics as it is impractical to track a large number of particles individually. I have derived the relationships below actually in a different context but could not find a better place to put them. Also, note that for the determination of the 'impact angle', I used here the hard sphere model which obviously would only be an approximation for charged particles (strictly speaking the Rutherford scattering formula would have to be used for this). Apart from this, the solution below is a completely general and exact description of a 2D collision event (and in any case it provides exact conservation of momentum and energy).

The relationship between the velocities of masses m1 and m2 before the collision (unprimed) and after the collision (primed) is given by the conservation equations for momentum and energy:

(1)       m1.vx,1 = m1.vx,1' + m2.Δvx,2'

(2)       m1.vy,1 = m1.vy,1' + m2.Δvx,2'.tan(θ)

(3)       m1/2.(vx,12+vy,12) = m1/2.(vx,1'2+vy,1'2) + m2/2.Δvx,2'2.(1+tan2(θ))  

For the sake of simplicity it has been assumed here that mass 2 is initially resting i.e. vx,2=0, vy,2=0 and vx,2'=Δvx,2' (this does not affect the general validity of the final result as the assumption will be dropped again later by referring the velocities to mass 2 explicitly).
The angle θ in Eq.(2) (and (3)) is the angle of the final velocity vector of ball 2 with the x-axis. This will be determined separately later from the geometry of the collision (alternatively, one can also treat θ as a free parameter).

Solving now Eq.(1) for vx,1' and Eq.(2) for vy,1' yields

(4)       vx,1' = vx,1 - m2/m1.Δvx,2'

(5)       vy,1' = vy,1 - m2/m1.Δvx,2'.tan(θ)  

Inserting (4) and (5) into Eq.(3) results in a quadratic equation for Δvx,2' which (after some lengthy but straightforward algebraic manipulations) yields the solution:

(6)       Δvx,2' = 2[ vx,1 + tan(θ).vy,1 ] / [(1+tan2(θ)).(1+m2 /m1 )]   .

Referring now the initial velocities explicitly to ball 2 and noting that the angle θ is the sum of the relative velocity angle between ball 1 and 2

(7)       γv = arctan [ (vy,1-vy,2)/(vx,1-vx,2) ]

and the impact angle α (see below); one can finally write

(8)       Δvx,2' = 2[ vx,1 - vx,2 + a.(vy,1 - vy,2 ) ] / [(1+a2).(1+m2 /m1 )]   ,


(9)       a = tan(θ) = tan(γv+α)   .

The 'impact angle' α can vary between -90o and +90o (or -π/2 and π/2 when using radians) (0o corresponds to a head-on- and the extreme values to a grazing collision). (in most treatments of collision problems, the center-of -mass scattering angle Θ is used, which relates to α through α=(π-Θ)/2; however, as α is independent of the reference frame, it is a much better choice here).
The actual value of α depends on the exact coordinates of the particles, so if the latter are not known, one has to treat α as a free parameter and generate it for instance through a random number generator. (otherwise see below for the determination of α).

From Eqs.(2),(4),(5),(8),and (9) the velocity components after the collision are therefore:

(10)       vx,2' = vx,2 + Δvx,2'   ,
(11)       vy,2' = vy,2 + a .Δvx,2'   ,
(12)       vx,1' = vx,1 - m2/m1.Δvx,2'   ,
(13)       vy,1' = vy,1 - a. m2/m1.Δvx,2'   .

The 'impact angle' α in Eq.(9) can be determined from the collision geometry as shown in the illustration below:

Geometry of 2D collision
Geometry of 2D collision

If the coordinates of the balls (with radius r1 and r2) are x1,y1 and x2,y2, α is given by

(14)       α = arcsin [ d.sin(γx,yv) / (r1+r2) ] ,


(15)       d = √ [ (x2-x1)2 +(y2-y1)2 ] ,
(16)       γx,y= arctan [ (y2-y1)/((x2-x1) ] ,

This will give a value for α as long as the absolute value of d1,2.sin(γx,yv) ≤(r1+r2) (otherwise the arcsin- function above is not defined, in which case the balls would not collide anyway). Note that one also has to exclude the case where γx,y and γv differ by more then π/2 and less than 1.5*π as then the balls would move away from each other.

If one is tracking the balls for further collisions, it is also necessary to update the coordinates in addition to the velocities. This is achieved as follows: using the law of cosines for the collision triangle and solving the corresponding quadratic equation for the distance which ball 1 travels to the collision point, one obtains the time that elapses from the original coordinates to the collision as

(17)   t = { d.cos(γx,yv) ± √ [(r1+r2)2- (d.sin(γx,yv)) 2] } / √ [ (vx,1-vx,2)2 +(vy,1-vy,2)2 ] .

Before the first square root, the minus sign holds if cos(γx,yv)>0 and the plus sign otherwise.
Note again that a valid solution requires |d.sin(γx,yv)| ≤ (r1+r2) (otherwise the balls would not collide).
The new coordinates (the position of the center of the balls at the moment of collision) are therefore

(18)       x1' = x1 + vx,1.t
(19)       y1' = y1 + vy,1.t
(20)       x2' = x2 + vx,2.t
(21)       y2' = y2 + vy,2.t

Generalization to Inelastic Collisions

The generalization of the above formulae to inelastic collisions is ultimately simple:   we just have to refer the velocity components (Eq.(10)-(13)) to the center of mass reference frame, apply the restitution coefficient to these, and add again the center of mass velocity to return to the lab frame, i.e. with

(22)       vx,cm = ( m1.vx,1 + m2.vx,2 )/( m1+ m2)  
(23)       vy,cm = ( m1.vy,1 + m2.vy,2 )/( m1+ m2)  

we have

(24)       vx,1'' = (vx,1'-vx,cm).R + vx,cm 
(25)       vy,1'' = (vy,1'-vy,cm).R + vy,cm 
(26)       vx,2'' = (vx,2'-vx,cm).R + vx,cm 
(27)       vy,2'' = (vy,2'-vy,cm).R + vy,cm 

where R is the restitution coefficient (=1 for a perfectly elastic collision; =0 for a perfectly inelastic collission (balls stick together after the collision)).

I have tested the formulae for several cases with a Fortran Program (for those who can't use Fortran, I have also translated this into a C/C++ Version). Additionally, I have included there now also a simplified version which does not do the 'remote' collision detection illustrated above but just returns the new velocities assuming that the input coordinates are already those of the collision.

The elastic and inelastic collision in 3 dimensions can be derived in a similar way, with the only difference that now two 'impact angles' need to be defined to determine all the velocity components.

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Thomas Smid (M.Sc. Physics, Ph.D. Astronomy)