2D Elastic Ball Collision Physics

2d physics collision

0WJYxW9FMN · Feb 4, 2016 · Viewed 10k times · Source

I am making a program that involves elastic ball physics. I have worked out all of the maths for collision against walls and stationary objects, but I cannot figure out what happens when two moving balls collide. I have mass and velocity (x and y velocity to be exact, but velocity of each ball and their direction will do) and would like the formulae for those. Remember - this is a perfectly elastic collision - so no spinning balls, etc.

Answer

This wikipedia article provides a formula to compute velocities after collision between two particles :

$\mathbf{v'_1}=\mathbf{v_1}-\frac{2m_2}{m_1+m_2}\frac{\left \langle \mathbf{v_1}-\mathbf{v_2}| \mathbf{x_1}-\mathbf{x_2}\right \rangle }{{\left \| \mathbf{x_1}-\mathbf{x_2} \right \|}^{2}} (\mathbf{x_1}-\mathbf{x_2})$

$\mathbf{v'_2}=\mathbf{v_2}-\frac{2m_1}{m_1+m_2}\frac{\left \langle \mathbf{v_2}-\mathbf{v_1}| \mathbf{x_2}-\mathbf{x_1}\right \rangle }{{\left \| \mathbf{x_2}-\mathbf{x_1} \right \|}^{2}} (\mathbf{x_2}-\mathbf{x_1})$

There are many reasons to use this formula :

you just need the velocity vectors of your balls before collision, their mass and their position,
you don't need to define angles of deviation,
the operations are simple (just dot product required),
the vectors can be expressed in any coordinates system.

There is no proof in the wikipedia article so I provide it below.

Definition of the problem

For each ball we define :

mi the mass
vi the vector of velocity before collision
v'i the vector of velocity after collision
Oi the point of center
xi the vector of Oi position

The unit vector n is normal to the surfaces of balls at the point of contact.

$\mathbf{n}=\frac{\mathbf{O_1O_2}}{\left \| \mathbf{O_1O_2}\right \|}=\frac{\mathbf{x_2 - x_1}}{\left \| \mathbf{x_2 - x_1}\right \|}$

The unit vector t is tangent to the surfaces of balls at the point of contact.

Physics law to use

The conservation of the total momentum is expressed by :

$m_1\mathbf{v'_1}+m_2\mathbf{v'_2}=m_1\mathbf{v_1}+m_2\mathbf{v_2}$

The conservation of total kinetic energy is expressed by :

$\frac{m_1 {v'}_1^{2}}{2}+\frac{m_2 {v'}_2^{2}}{2}=\frac{m_1 v_1^{2}}{2}+\frac{m_2 v_2^{2}}{2}$

As there is no force applied in the tangential direction, the tangential components of velocities are unchanged after collision :

$\left\{\begin{matrix} \left \langle \mathbf{v'_1}|\mathbf{t} \right \rangle=\left \langle \mathbf{v_1}|\mathbf{t} \right \rangle \\ \left \langle \mathbf{v'_2}|\mathbf{t} \right \rangle=\left \langle \mathbf{v_2}|\mathbf{t} \right \rangle \end{matrix}\right.$

Proof

The tangential components of velocities are unchanged. So we can rewrite the conservation laws with normal components and we have a 1D problem now :

$\left\{\begin{matrix} m_1 {\left \langle \mathbf{v'_1}|\mathbf{n} \right \rangle}+m_2 {\left \langle \mathbf{v'_2}|\mathbf{n} \right \rangle} =m_1 {\left \langle \mathbf{v_1}|\mathbf{n} \right \rangle}+m_2 {\left \langle \mathbf{v_2}|\mathbf{n} \right \rangle} \\ m_1 {\left \langle \mathbf{v'_1}|\mathbf{n} \right \rangle}^{2}+m_2 {\left \langle \mathbf{v'_2}|\mathbf{n} \right \rangle}^{2}=m_1 {\left \langle \mathbf{v_1}|\mathbf{n} \right \rangle}^{2}+m_2 {\left \langle \mathbf{v_2}|\mathbf{n} \right \rangle}^{2} \end{matrix}\right.$

The conservation of kinetic energy can be factorized then simplified with the conservation of momentum :

$m_1 {\left \langle \mathbf{v'_1}|\mathbf{n} \right \rangle}^{2}-m_1 {\left \langle \mathbf{v_1}|\mathbf{n} \right \rangle}^{2}=m_2 {\left \langle \mathbf{v_2}|\mathbf{n} \right \rangle}^{2}-m_2 {\left \langle \mathbf{v'_2}|\mathbf{n} \right \rangle}^{2}$

$\Rightarrow {\left \langle \mathbf{v'_1}|\mathbf{n} \right \rangle}+{\left \langle \mathbf{v_1}|\mathbf{n} \right \rangle}={\left \langle \mathbf{v_2}|\mathbf{n} \right \rangle}+{\left \langle \mathbf{v'_2}|\mathbf{n} \right \rangle}$

We combine this last expression with the conservation of momentum and we get the normal component of v'1 :

$\left \langle \mathbf{v'_1}| \mathbf{n}\right \rangle=\frac{(m_1-m_2)\left \langle \mathbf{v_1}| \mathbf{n}\right \rangle + 2 m_2\left \langle \mathbf{v_2}| \mathbf{n}\right \rangle}{m_1+m_2} =\left \langle \mathbf{v_1}| \mathbf{n}\right \rangle-\frac{ 2 m_2\left \langle \mathbf{v_1-v_2}| \mathbf{n}\right \rangle}{m_1+m_2}$

Finally, we find the formula of the wikipedia article for v'1 :

$\mathbf{v'_1}=\left \langle \mathbf{v'_1}| \mathbf{n}\right \rangle\mathbf{n}+\left \langle \mathbf{v'_1}| \mathbf{t}\right \rangle\mathbf{t}=\mathbf{v_1}-\frac{2m_2}{m_1+m_2}\frac{\left \langle \mathbf{v_1}-\mathbf{v_2}| \mathbf{x_1}-\mathbf{x_2}\right \rangle }{{\left \| \mathbf{x_1}-\mathbf{x_2} \right \|}^{2}} (\mathbf{x_1}-\mathbf{x_2})$

The formula of v'2 is symmetrical.

2D Elastic Ball Collision Physics

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