Collisions are of importance in physics. Especially for small world such as atoms or nuclei, scattering is the crucial technique to investigate their nature. Before studying the scattering theory of quantum mechanics, we had better to get familiar with collisions in classical mechanics.
We have two main themes for studying the collision problems   . One is that we have to obtain velocities of the two particles after the collision from initial velocities by using conservation laws of momentum and energy. Since we have three equations from conservation law and four unknowns, one parameter out of four should be fixed according to the given collision problems. The other is that we have to study two ways of looking collisions, i.e. laboratory system and center-of-mass system. We need to convert between the two systems.
In order to achieve those two themes, the diagrammatic technique gives the powerful tool. For the collisions in one dimension, the mass-momentum diagram with mass m along the vertical axis and momentum p represented on the horizontal axis is the useful technique  . We obtain the whole story of the collision in one dimension from the single -diagram.
When we apply the mass-momentum diagram to the elastic collisions in two dimensions, three dimensional space is needed. However, in this case, the projection onto the -plane is sufficient  . For Newtonian mechanics  , we draw a circle in the two dimensional momentum space and also draw arrows of the momentum of the colliding particles into the circle. This is a way of looking collisions from laboratory system. In this article, we add one more circle. This is from the point of view of the center-of-mass system. These two circles give all information for two dimensional elastic collision problems.
This paper is organized in the following way. In Section 2, we recall two dimensional elastic collisions with equations. In Section 3, we show the diagrammatic approach for two dimensional elastic collision in order. First, we draw a circle for the center-of-mass system. Then we add to draw one more circle to obtain the momentum after the collision in the laboratory system. In Section 4, we investigate the special case where the target particle is at rest before the collision. Section 5 is devoted to a conclusion.
2. Elastic Collision between Two Particles in Two Dimensions
Let us recall the treatise of the two dimensional elastic collision with equations for later use. Figure 1 shows the collisions from the point of view in laboratory and center-of-mass systems and also show the notation which we use in this article. The projectile A has mass and velocity and the target B has mass and velocity before the collision. The momenta are given by and . They are known parameters or initial conditions of the collision. Usually, the velocities and are supposed to be in the
Figure 1. Left: Collisions in the laboratory system. Right:Collisions in the center-of-mass system.
same direction and are set along the x-axis in this article. The velocities after the collision are distinguished by the primes. And the asterisk is attached to the parameters in the center-of-mass system.
We have to obtain the four parameters after the collision in the laboratory system. However, we have only three equations, i.e. energy conservation and two components of momentum conservations. So, we have to fix one parameter out of four. We investigate the collision in the following way.
2.1. Velocities of the Center-of-Mass
We write down the conservation of momentum for two systems:
for laboratory system, (1)
for center-of-mass system. (2)
Let be the velocity of the center-of-mass. The relations between and the velocities of the particles in two systems are as follows:
Substituting Equation (3) into Equation (2), we obtain
which remains the same before and after collision.
2.2. Velocities and Momenta before Collision in Center-of-Mass System
Substituting Equation (4) into Equation (3), we obtain the velocities in the center-of-mass system
and the momenta
We clearly see that these expressions of the momentum satisfy the relation in Equation (2). Note that is a reduced mass and is a relative velocity before the collision.
2.3. Velocities and Momenta after Collision in Center-of-Mass System
We write energy conservations for two systems:
for laboratory system, (7)
for center-of-mass system. (8)
From the conservations of momentum in Equation (2), we obtain the following relations:
Substituting these relations into Equation (8), we obtain
This means that the velocities (and also momenta) of the two particles stay in magnitude before and after the collision in the center-of-mass system. Thus the collision simply rotates the velocities. However, the angle of the rotation cannot be determined from the conservations of momentum and energy because we have four unknowns and only three equations: the energy and the two components of momentum conservations. Namely, there is an infinite number of possible final states of outgoing particles in an elastic collision in two dimensions. Let be a unit vector in the direction of the velocity of the projectile A after the collision in the center-of-mass system. The scattering angle of the right figure in Figure 1 is related by . Accordingly, the velocities after the collision in the center-of-mass system are written by
and the momenta are given by
Again, these expressions of the momentum satisfy the relation in Equation (2).
2.4. Velocities and Momenta after Collision in Laboratory System
In order to return to the laboratory system, we must add Equation (11) to the velocity of the center-of-mass in Equation (4). We obtain velocities after the collision in the laboratory system,
Note that the sum of both sides clearly shows the momentum conservation of Equation (1) in the laboratory system.
Let be the -components of the momentum of the projectile A in Equation (14). We write down explicitly as follows:
where we note that the velocity has the x-component only due to Equation (4). From these equations and the relation , we obtain
This indicates the circle in momentum space, centered at with its radius . These quantities are uniquely determined by the initial condition of the collision.
Let us consider the case where the target particle is at rest before the collision. Setting in Equation (18), we obtain
Moreover, the case of which two particles have equal mass becomes more simple:
These equations of a circle will appear in the next sections.
3. Diagrammatic Technique
In this section, we deduce all relations, which we recalled in the former section, from the diagram in two dimensional momentum space.
3.1. Center-of-Mass System
Firstly, we draw a circle whose radius is in Equation (6), as depicted in Figure 2. This dashed circle centered at O describes the collision in the center-of-mass system. We draw dashed arrows of the momenta before the collision from Equation (6)
and after the collision from Equation (12)
Figure 2. Collision in the center-of-mass system. The figure shows the case that the projectile A has mass and velocity and the target B has mass and velocity . The vectors are the momenta of incident particles before the collision. The vector and are the momentum of outgoing particles after collision. The angle cannot be determined by the conservation laws only. We fix it according to the given collision problem.
where or is determined according to what we are asked in the collision problems. Since the scattering angle cannot be determined by the conservations of momentum and energy, the point C lies anywhere on this circle and the point D is opposite side against the point C on the circle.
3.2. Laboratory System
Next, as shown in Figure 3, we determine the point E on the px-axis so that . We draw another circle centered at E whose radius is the same as the dashed circle centered at O. This circle centered at E is appeared in the book of Landau and Lifshitz  , and its equation is written by Equation (18). This circle describes the laboratory system as explained below.
As shown in Figure 3, we draw a broken line from the point C in parallel to the px-axis until the broken line intersects with the circle centered at E. We call this point of intersection as F. Note that is always satisfied and it means . Then, the vector shows the momentum of the projectile A after the collision in the laboratory system. The tips of is always on the circle centered at E. The angle is the scattered angle of the projectile A in the laboratory system. We note that the angle in Figure 2 and the angle in Figure 3 are related each other. Once the is given by the collision problems, the is determined according to the prescription stated above. And the converse is also true. When the is given in the collision problem, we write down the vector at first. Then, we
Figure 3. Circle centered at E describes the collision in the laboratory system. The initial condition in Figure 2 shows that and . The vectors and indicate the momenta of the projectile and the target after the collision in the laboratory system.
draw a broken line from the point F to C. The vector and the angle are the momentum and the scattered angle of the projectile A in the center-of-mass system.
Next, we determine the point G on the px-axis so that . Then the vector shows the momentum of the target B after the collision. The angle is the scattered angle of the target B. The vector relation shows the momentum conservation before and after the collision.
In contrast with the equations in the previous section, what we need to do are the calculation of the radius of the circle and the lengths OE and EG. They are uniquely determined from the initial condition of the collision. We fix the angle or according to the given collision problems.
Let us consider the case where the target B is at rest before the collision. In that case, the point G definitely lies on the circle centered at E because of , i.e. , which is understood by Equations ((4) and (15)). The circle centered at E is described by Equation (19) and is depicted in Figure 4. The vector is equal to the momentum of the projectile A before the collision.
The point E lies inside or outside the dashed circle centered at O, according as and . The corresponding diagrams are shown in Figure 4 and Figure 5. It is evident from these figures that and can be expressed in terms of by
Figure 4. The initial conditions , and , show and . Then, and .
Figure 5. The initial conditions , , and show and . Then, and .
where we use the relation . It is also evident that since the triangle is an isosceles triangle, we obtain .
Applying the law of cosine to the triangle : , we obtain
The same application to the triangle gives
If , however, the projectile A can be deflected only through an angle not exceeding from its original direction, as shown in Figure 6. This maximum value of corresponds to the position F at which OF is a tangent to the circle centered at E. Evidently,
because EF = EG are both the radius of the circle.
The case becomes quite simple as shown in Figure 7. The circle centered at E is described by Equation (20). The point E lies on the dashed circle centered at O, i.e. OA = OE. The py-axis is a tangent to the circle centered at E. It is evident from the figure that . Further, Equations ((25) and (26)) become
Figure 6. The initial condition is as the same with Figure 5. If , the projectile A can be deflected only through an angle not exceeding from its original direction.
Figure 7. The initial conditions , and show and . Then, .
After the collision, the outgoing particles move at right angles to each other, that is .
We introduce two circles to analyze the two dimensional elastic collision in momentum space. One circle is for the center-of-mass system and the other is for the laboratory system. The relation between these two systems is clearly understood from these circles. Once we fix the scattered angle of projectile in one system, then we deduce all quantities, such as momenta and scattering angles of both particles in another system. The scattering problem in the special relativistic case is carried out in the next article.
The author thanks the anonymous reviewer for his helpful suggestions.