Hard Sphere Collision

1D Collision

Momentum and kinetic energy conserved.

$m_{1} v_{1i} + m_{2} v_{2i} = m_{1} v_{1f} + m_{2} v_{2f}$
$m_{1} (v_{1i} - v_{1f}) = m_{2} (v_{2f} - v_{2i})$

$m_{1} v_{1i}^{2} + m_{2} v_{2i}^{2} = m_{1} v_{1f}^{2} + m_{2} v_{2f}^{2}$
$m_{1} (v_{1i}^{2} - v_{1f}^{2}) = m_{2} (v_{2f}^{2} - v_{2i}^{2})$
$m_{1} (v_{1i} - v_{1f}) (v_{1i} + v_{1f}) = m_{2} (v_{2f} - v_{2i}) (v_{2f} + v_{2i})$

Solving for $v_{1f}$ and $v_{2f}$ . There is a trick : express in terms of differences and divide one equation by the other. This gives two linear equations to solve, avoiding the quadratic.

$m_{1} v_{1i} + m_{2} v_{2i} = m_{1} v_{1f} + m_{2} v_{2f}$
$(v_{1i} + v_{1f}) = (v_{2f} + v_{2i})$

Solve for $v_{1f}$ .

$m_{1} v_{1i} + m_{2} v_{2i} = m_{1} v_{1f} + m_{2} v_{2f}$
$m_{2} v_{1i} + m_{2} v_{1f} = m_{2} v_{2f} + m_{2} v_{2i}$
Subtract.

solution
$v_{1f} (m_{1} + m_{2}) = 2 m_{2} v_{2i} + (m_{1} - m_{2}) v_{1i}$
$v_{2f} (m_{1} + m_{2}) = 2 m_{1} v_{1i} + (m_{1} - m_{2}) v_{2i}$

2D Collision

Poor mans collision detection was implemented. This name is negative and refers to the fact that collision is detected after the event as opposed to before. If two particles(circles or spheres) are intersecting then a collision has occurred.

One response is to go back in time and recalculate collision when an intersection occurs. Another is to continue iterating the system forward.

The major sticking point of the model was when two or more particles are trapped and cling together. eg particles are somehow already intersecting reversing back in time and going forward leaves some still together in an endless binding or loop.

The resolution of this was to force the particles apart and in the face of uncertainty keep the system incrementing. See void d2particle::collisionPoint ( d2particle & p, double const h ) routine.

Maths of Exact Collision

Geometry has been used to determine the collision exactly. The conservation of momentum and kinetic energy are not enough to solve the problem. Perfect spheres geometry is used for simplicity.

$Δt$ step simulation. Increment two particles. If their centers are too close then a collision has occurred and solve the collision exactly.

Since we can solve the intersection exactly this is not the most general but is easily visualized and may result in reduced evaluation of the quadratic equation when solving for line intersection.

Vectors are used for simplicity.

$x_{1} (t) = x_{1i} + t v_{1}$
$x_{2} (t) = x_{2i} + t v_{2}$

$(x_{1i0} + t v_{10}, x_{1i1} + t v_{11})$
$(x_{2i0} + t v_{20}, x_{2i1} + t v_{21})$

Their centers are the sum of their radius es apart at the point of collision.

$(x_{1i0} + t v_{10} - (x_{2i0} + t v_{20}))^{2} + (x_{1i1} + t v_{11} - (x_{2i1} + t v_{21}))^{2} = (r_{1} + r_{2})^{2}$

$z_{1} = x_{1} (t) + (- x_{1} (t) + x_{2} (t)) \frac{r_{1}}{r_{1} + r_{2}}$

$x_{1} (t_{1}) = x_{2} (t_{2})$
$x_{1i0} + t_{1} v_{10} = x_{2i0} + t_{2} v_{10}$
$x_{1i1} + t_{1} v_{11} = x_{2i1} + t_{2} v_{11}$
Solve for $t_{1}$

$z_{2} = x_{1} (t_{1})$