Two-dimensional Collision of Hard Balls

Drew Dolgert for Michael Fowler

August 18, 1998

We begin with two balls colliding, one moving, one at rest.

v⃗_{1} = v_{1}x̂  (1)
v⃗_{2} = 0  (2)

Immediately move to center of mass coordinates where

xcm = \frac{m_{1}x⃗_{1} + m_{2}x⃗_{2}}{m_{1} + m_{2}}  (3)

In these coordinates, the total linear momentum is zero.

m_{1}v⃗_{1} + m_{2}v⃗_{2} = 0  (4)
m_{1}v′_{1} + m_{2}v′_{2} = 0  (5)

The primes denote the velocities after the collision. The effect of this momentum conservation on the kinetic energy is that the particle velocities are related by

v⃗_{2} = −\frac{m_{1}}{m_{2}} v⃗_{1},  and  v′_{2} = −\frac{m_{1}}{m_{2}} v′_{1}.  (6)

Also, the speeds before and after the collision are unchanged

v′_{1} = v_{1},  and  v′_{2} = v_{2}.  (7)

Because v⃗_{1} and v⃗_{2} are so well connected, we can complete the calculation by studying v⃗_{1} alone.

Our main physical consideration is the moment of impact of the balls. If we define an impact parameter, B, as the ŷ distance between the centers of the two balls, we can define a single angle to describe the collision,

sin θ = \frac{B}{2r}  (8)

where r is the ball radius. For balls of differing radii, just use B/(r_{1} + r_{2}). During a head-on collison, θ = 0. A glancing blow of m_{1} above m_{2} is at θ = π/2 and below is at θ = −π/2.

At the moment of impact, the impulse of the impact is along the point of contact of the two balls. For the first ball, we can write the impulse as

m_{1}(v′_{1} − v⃗_{1}) = I(− cos θx̂ + sin θŷ).  (9)

The x̂ and ŷ components simplify to

m_{1}v′_{1x} − m_{1}v_{1} = −I cos θ  (10)
m_{1}v′_{1y} = I sin θ.  (11)

Making these a fraction gives us the needed angle,

tan θ = \frac{m_{1}v′_{1y}}{m_{1}v_{1} − m_{1}v′_{1x}} = \frac{sin ϕ}{1 − cos ϕ}.  (12)

In the last step, we replaced v_{1x} = v_{1} cos ϕ and v_{1y} = v_{1} sin ϕ. The angle at which m_{1} leaves the collision is now ϕ measured from the x̂ axis. We now solve this equation.

tan θ − cos ϕ tan θ = sin ϕ  (13)

Squaring both sides gives

tan^{2} θ(cos^{2} ϕ − 2 cos ϕ + 1) = sin^{2} ϕ = 1 − cos^{2} ϕ.  (14)
(tan^{2} θ + 1) cos^{2} ϕ − 2 tan^{2} θ cos ϕ + tan^{2} θ − 1 = 0  (15)

Using the quadratic formula, we find

cos ϕ = \frac{2 tan^{2} θ ± \sqrt{4 tan^{4} θ − 4(tan^{2} θ + 1)(tan^{2} θ − 1)}}{2(tan^{2} θ + 1)}.  (16)

Simplifying shows

cos ϕ = \frac{tan^{2} θ ± 1}{tan^{2} θ + 1}.  (17)

Clearly we take the negative root to find

cos ϕ = 2 sin^{2} θ − 1 = 1 − 2 cos^{2} θ.  (18)