Elastic Collision in Two Dimensions

Elastic collisions of two known masses in two dimensions conserve momentum in the two dimensions, and also kinetic energy. Therefore there are three equations which can be written. If the initial system is completely specified, then there are 4 known initial kinematic quanties, in addition to the two masses: magnitude and direction of the initial velocity of the first mass, and the magnitude and direction of initial velocity of the second mass.

There will be 4 final kinematic quantities: magnitude and direction of the final velocity of the first mass, and the magnitude and direction of the final velocity of the second mass. If just one of these final kinematic quantities is specified, then the other three final kinematic quantities can be derived from the conservation equations.

As an example of a two dimensional elastic collision, consider the collision of a first mass (Subscript[m, 1 ])traveling along the horizontal direction (x) with speed (Subscript[v, 1i ])and then striking a second mass (Subscript[m, 2])which is a rest. After the collision the first mass is seen to be traveling at an angle θ above the horizontal direction.  Calculate the magnitudes of the final velocities (Subscript[v, 1f])and (Subscript[v, 2f]) as well as the direction angle ϕ of the second mass.

The three conservation equations are:

(Subscript[m, 1]) (Subscript[v, 1i ])= (Subscript[m, 1 ])(Subscript[v, 1f  ])cos θ + (Subscript[m, 2 ])(Subscript[v, 2f  ])cos ϕ 
0= (Subscript[m, 1 ])(Subscript[v, 1f  ])sin θ - (Subscript[m, 2 ])(Subscript[v, 2f  ])sin ϕ 
(1/2)(Subscript[m, 1]) ((Subscript[v, 1i ])^2)= (1/2)(Subscript[m, 1 ])((Subscript[v, 1f  ])^2) + (1/2)(Subscript[m, 2 ])((Subscript[v, 2f  ])^2)

As an example, let's take the two masses to be proton masses, 1.67 x (10^(-27))kg, with an initial speed of the first proton mass to be 3.5 x (10^5) m/s, and the second proton is at rest.  After the collision the first proton is seen to have a velocity with an angle θ = (37^o) above the horizontal. What are the magnitudes of the two final velocities, and what is the direction angle ϕ (below the horizontal) of the second proton's velocity after the collision.

We can solve this with Mathematica after first assigning the known kinematic variables, including the conversion of the θ angle to radians:

mass1=1.67*10^-27;
mass2= mass1;
V1Initial=3.5*10^5;
θ=37.0*Pi/180;

Now we use the Mathematica simultaneous equation solver, and show the one correct solution:

Solve[{mass1*V1Initial== mass1*V1Final*Cos[θ] + mass2*V2Final*Cos[ϕ],
  mass1*V1Final*Sin[θ] - mass2*V2Final*Sin[ϕ]==0,
mass1*V1Initial*V1Initial==mass1*V1Final*V1Final + mass2*V2Final*V2Final},{V1Final, V2Final,ϕ}]

V1Final->279522,V2Final->210635,ϕ->0.92502

The ϕ angle in radians can be converted to 53 degrees.  It is a general result that when one mass collides with a second equal mass which is at rest, then the two masses have final velocities which are perpendicular to each other, that is 90 degrees apart.