C.A.P.P.S Collective: March 2012

Two Dimensional Collision Physics ... by Brandon Williams

Let me start off by saying that the above title is a mighty long one and can sound intimidating. But, something I hope would have come out by now in my writings is that almost every area of math can be broken down into something very simple. But the above says 'physics!' Yes, but physics is mostly math and most certainly math in this case.

Next, let me tell you what this tutorial is going to be over. I am not going to be talking about two dimensional collisions in general, but instead one type: two circles colliding. This is one of the easiest cases to look at. Even though it is somewhat easy there are some very hard parts to it. In particular, if we were to take into account the "spin" of the circles and look at how that effects their velocities after collision we would be getting into something far beyond the scope of this tutorial. I will only be dealing with two circles colliding and how their velocities change afterward. I have already addressed this problem on a few of the Flash boards but it was such a sparse explanation I thought I would write about it.

I would hate to do this but I am going to assume that you have had a physics class at some point in your life and assume you are familiar with terms. It would not even need to be an in depth physics class. An introduction would do. If you have not then you may want to read this again in a few weeks because I have tutorials planned on mechanics (dynamics in particular) and kinematics. Those should bring you up to date. Well, let's start.

Common sense should tell you that whenever you come to a problem like this you should look at a simplified situation and see if you can extrapolate that to something bigger. A simplified version of two circles colliding in two dimensions would be two circles colliding in one dimension. The velocities of two objects after collision can be described with Newton's Conservation of Momentum. Momentum is the measure of an object's motion, denoted with a P, and is the product of the object's mass and velocity. Momentum is also a vector with the same direction as the object's velocity and a magnitude of the object's mass times the object's velocity (like I mentioned before). The equation for momentum:

      P = mv

The unit of momentum is kilogram * meter / second (kg*m/s).

Newton's Conservation of Momentum states that the sum of momentum before collision is equal to that afterwards. So if two objects collide and one object has momentum of 10 kg*m/s and the other has 2 kg*m/s their total momentum is 12 kg*m/s. When the two objects "bounce" off each other their momentum will change but the sum of their momentum will not. In mathematical terms:

      P_{inital_sum} = P_{final_sum}


      m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

      m₁ = mass of one object, m₂ = mass of other object
      v_1i = initial velocity of one object
      v_2i = initial velocity of other object
      v_1f = final velocity of one object
      v_2f = final velocity of other object

Now in the case of two balls colliding they should bounce off right? This is called an elastic collision. Here is a prime example of people giving others misconceptions about math and physics by using words out of context. The word "elastic" and "elasticity" seems to be a popular one and is usually used wrongly. Technically an elastic collision is one in which kinetic energy is conserved as well as momentum. Kinetic energy is measured in joules and is the amount of energy that is produced from the motion of the object. It is given with this equation:

      KE = 1/2mv²

If kinetic energy is conserved then, like momentum, the sum of the two objects' kinetic energy on collision is equal to that afterwards. In mathematical terms:

      KE_{inital_sum} = KE_{final_sum}

      1/2m₁v_1i² + 1/2m₂v_2i² = 1/2m₁v_1f² + 1/2m₂v_2f²

      m₁ = mass of one object, m₂ = mass of other object
      v_1i = initial velocity of one object
      v_2i = initial velocity of other object
      v_1f = final velocity of one object
      v_2f = final velocity of other object

Now let's examine what we have so far. Two equations that describe the collisions: one in terms of momentum and the other in terms of kinetic energy. Even though the equations state two very different things they use the same variables. Which variables do we know and which do we want to find? You should now the mass of the objects. That should be a given because you can make up anything you want. You should also know the velocities of the two objects because that is what describes the object's movement in the first place. The only variables you should not know (which are the ones you want to find) are v_1f and v_2f which are the velocities of the two objects after the collision. So we have two equations and two unknowns.

If you have had an algebra class (which I am hoping you have) then you probably noticed that the entire time you are always "hunting" for a variable. For example, you are given the equation [ 5x - 4 = 2x + 3 ] and you were asked to solve for "x". This time, however, we have two equations and two unknowns. This is known as a system of equations. You should have also been taught how to solve a system of equations. Here is an example just to refresh your memory:

       Solve for (x,y):

            3x + 5y = 7
            8x - 2y = -20

       1.)  You are going to solve the first equation for "x".
            To do this you are going to start by subtracting "5y" from
            both sides:

            3x = 7 - 5y

       2.)  Next you are going to divide both sides by 3 to get "x"
            by itself on the left side:

            x = 7/3 - 5/3y

       3.)  Now that we have this equation in terms of "x" we are going
            to plug it in place of "x" in the second equation.The reason
            we do not plug it into the original equation is because: 1.)
            you cannot solve an equation with two unknowns  2.) if you
            worked it out the "y's" would cancel out and you would be
            left with something like 7 = 7 which is not what you want.
            So, we plug it into the second equation:

            8 (7/3 - 5/3y) - 2y = -20

       4.)  The first thing you want to do is distribute the 8 into the
            parentheses:

            56/3 - 40/3y - 2y = -20

       5.)  As you can see we are getting some pretty nasty looking
            fractions.  Don't be scared by that though.  In this next
            step I am going to:  1.) subtract 56/3 from both sides  2.)
            combine the like terms on the left side of the equation:

            - 46/3y = -116/3

       6.)  And finally divide both sides by (-46/3) to leave the "y"
            by itself on the left:

            y = 348/138 = 58/23 »  2.521

       7.)  It's best to keep it in fraction form but I thought I should
            give you it in decimal so that you can check your work if
            you need to.  But, we are still not finished.  You have only
            solved for "y"!  What about "x"!  To solve for "x" simply
            plug "y" into one of the equations above and solve for "x".
            It does not even matter which one since you should get the
            same answer.  Just so you can check your work when I worked
            out the rest of the problem I got "x" equal to -129/69 or
            approximately 1.87

And there you go. Not one of the easiest examples because of the fractions, but you should be able to get the gist of it. Well, this is the same idea we are going to use for the two equations that represented the collision: conservation of momentum and conservation of kinetic energy. However I am not going to show you the work because it is long. Not necessarily hard but I assure you I would not be able to fit it in the margin of this document. But, I will still give you the equations just in case you are unable to work the problem out:

       v_i1(m₁-m₂) + v_i2(2m₂)
v_f1 = -------------------------
              m₁ + m₂

        v_i1(2m₁) + v_i2(m₁-m₂)
v_f2 = -------------------------
              m₁ + m₂

Anyway, just in case you lost touch with what we were trying to do after all of that, those last equations give the velocities of the two object after collision. And the good thing about the above equations is that they can be extrapolate to two dimensions with only a few new things going on. It is still much harder to visualize than the above put I have some pictures that should help.

First, a picture of some random ball-to-ball collision I thought of:

The white arrows are the circles' velocity vectors. Now what is one of the main differences between this collision and the one dimensional collision? Putting aside the fact that the balls' movements are in two dimensions (that's a little obvious), but something else pretty big is the fact that the balls' centers are not lined up anymore. In a one dimensional collision the balls hit dead on, but this time they are off. This new "thing" we are taking into account can be called the angle of collision. It is was makes this collision different from a one dimensional collision.

Anyway, in the case of a two dimensional collision you need to change the way you look at things. As the picture is now the balls' positions and and vectors are based on a regular Cartesian coordinate system with the x-axis being the horizontal and the y-axis being the vertical. The balls even have their own axes where their center is the origin. Here is a picture:

Well, like I said before, we are going to change that all around. Instead of looking at the collision like this in a regular coordinate system we are going to rotate around and look at the collision along the axis of collision.What is the axis of collision? It is an imaginary line which connects the two balls' centers. Like this:

Along this axis is where all your calculations will be based on. Then once you have everything calculated you rotate back to your regular coordinate system. But what "calculations" are going to be done on the axis? That is what I am getting into next.

The first thing you need to do is break the velocity vector up into its components (its x- and y-velocities) in the new coordinate system. The x- and y-velocities can also be thought of as the motion normal to the imaginary axis (x-velocity) and tangent to the axis (y-velocity). Yet another picture:

The blue lines are the x-velocities in the new coordinate system and the orange lines are the y-velocities.

So far we have gone through a lot of steps which you may not have even noticed. A collision took place, you get its angle of collision, you rotate your view and look at the situation along the axis of collision, and you have broken the objects' velocities down into its components along the x- and y-axis. All that is left is to calculate the new velocities (in the new coordinate system) and rotate back to your original coordinate system. This is the great part. You can use the conservation of momentum equations we derived above on the x-axis velocities, rotate back to your old coordinate system and you are ready to go. An object's y-velocity along the axis of collision does not change. Yes it is true. You get the components of the velocity vector in the new coordinate system, use the conservation of momentum equations on the x-axis velocity and rotate backwithout touching the y-velocity.

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