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We present the usefulness of the diagrammatic approach for analyzing two dimensional elastic collision in momentum space. In the mechanics course, we have two major purposes of 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. The other is that we have to study two ways of looking collisions, i.e. laboratory system and center-of-mass system. For those two major purposes, we propose the diagrammatic technique. We draw two circles. One is for the center-of-mass system and the other is for the laboratory system. Drawing these two circles accomplish two major purposes. This diagrammatic technique can help us understand the collision problems quantitatively and qualitatively.

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 [

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 [

When we apply the mass-momentum diagram to the elastic collisions in two dimensions, three dimensional space ( m , p x , p y ) is needed. However, in this case, the projection onto the ( p x , p y ) -plane is sufficient [

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.

Let us recall the treatise of the two dimensional elastic collision with equations for later use.

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 ( v ′ A , v ′ B , θ , ϕ ) 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.

We write down the conservation of momentum for two systems:

m A v A + m B v B = m A v ′ A + m B v ′ B for laboratory system, (1)

m A v A ∗ + m B v B ∗ = m A v ′ A ∗ + m B v ′ B ∗ = 0 for center-of-mass system. (2)

Let V be the velocity of the center-of-mass. The relations between V and the velocities of the particles in two systems are as follows:

v A ∗ = v A − V , v B ∗ = v B − V . (3)

Substituting Equation (3) into Equation (2), we obtain

V = m A v A + m B v B m A + m B = p A + p B m A + m B , (4)

which remains the same before and after collision.

Substituting Equation (4) into Equation (3), we obtain the velocities in the center-of-mass system

v A ∗ = + m B m A + m B ( v A − v B ) , v B ∗ = − m A m A + m B ( v A − v B ) , (5)

and the momenta

p A ∗ = + m A m B m A + m B ( v A − v B ) , p B ∗ = − m A m B m A + m B ( v A − v B ) . (6)

We clearly see that these expressions of the momentum satisfy the relation in Equation (2). Note that m A m B m A + m B is a reduced mass and v A − v B ( v A > v B ) is a relative velocity before the collision.

We write energy conservations for two systems:

m A 2 ( v A ) 2 + m B 2 ( v B ) 2 = m A 2 ( v ′ A ) 2 + m B 2 ( v ′ B ) 2 for laboratory system, (7)

m A 2 ( v A ∗ ) 2 + m B 2 ( v B ∗ ) 2 = m A 2 ( v ′ A ∗ ) 2 + m B 2 ( v ′ B ∗ ) 2 for center-of-mass system. (8)

From the conservations of momentum in Equation (2), we obtain the following relations:

v B ∗ = − m A m B v A ∗ , v ′ B ∗ = − m A m B v ′ A ∗ . (9)

Substituting these relations into Equation (8), we obtain

v ′ A ∗ = v A ∗ , v ′ B ∗ = v B ∗ . (10)

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 n ∗ be a unit vector in the direction of the velocity v ′ A ∗ of the projectile A after the collision in the center-of-mass system. The scattering angle θ ∗ of the right figure in

v ′ A ∗ = + m B m A + m B | v A − v B | n ∗ , v ′ B ∗ = − m A m A + m B | v A − v B | n ∗ , (11)

and the momenta are given by

p ′ A ∗ = + m A m B m A + m B | v A − v B | n ∗ , p ′ B ∗ = − m A m B m A + m B | v A − v B | n ∗ . (12)

Again, these expressions of the momentum satisfy the relation in Equation (2).

In order to return to the laboratory system, we must add Equation (11) to the velocity of the center-of-mass V in Equation (4). We obtain velocities after the collision in the laboratory system,

v ′ A = v ′ A ∗ + V , v ′ B = v ′ B ∗ + V , (13)

and momenta

p ′ A = p ′ A ∗ + m A V = + m A m B m A + m B | v A − v B | n ∗ + m A m A + m B ( p A + p B ) , (14)

p ′ B = p ′ B ∗ + m B V = − m A m B m A + m B | v A − v B | n ∗ + m B m A + m B ( p A + p B ) . (15)

Note that the sum of both sides clearly shows the momentum conservation p ′ A + p ′ B = p A + p B of Equation (1) in the laboratory system.

Let p ′ A = ( p ′ A x , p ′ A y ) be the x , y -components of the momentum of the projectile A in Equation (14). We write down explicitly as follows:

p ′ A x = m A m B m A + m B | v A − v B | cos θ ∗ + m A V , (16)

p ′ A y = m A m B m A + m B | v A − v B | sin θ ∗ , (17)

where we note that the velocity V has the x-component only due to Equation (4). From these equations and the relation cos 2 θ ∗ + sin 2 θ ∗ = 1 , we obtain

( p ′ A x − m A V m A m B m A + m B | v A − v B | ) 2 + ( p ′ A y m A m B m A + m B | v A − v B | ) 2 = 1. (18)

This indicates the circle in momentum space, centered at ( m A V ,0 ) with its radius p A ∗ = p B ∗ = m A m B m A + m B | v A − v B | . 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 v B = 0 in Equation (18), we obtain

( p ′ A x − m A m A + m B p A m B m A + m B p A ) 2 + ( p ′ A y m B m A + m B p A ) 2 = 1. (19)

Moreover, the case of which two particles have equal mass m A = m B becomes more simple:

( p ′ A x − p A 2 p A 2 ) 2 + ( p ′ A y p A 2 ) 2 = 1. (20)

These equations of a circle will appear in the next sections.

In this section, we deduce all relations, which we recalled in the former section, from the diagram in two dimensional momentum space.

Firstly, we draw a circle whose radius is p A ∗ = p B ∗ in Equation (6), as depicted in

O A = p A ∗ = + m A m B m A + m B ( v A − v B ) , O B = p B ∗ = − p A ∗ , (21)

and after the collision from Equation (12)

O C = p ′ A ∗ = + m A m B m A + m B | v A − v B | n ∗ , O D = p ′ B ∗ = − p ′ A ∗ , (22)

where n ∗ = ( c o s θ ∗ , s i n θ ∗ ) or θ ∗ = ∠ C O A is determined according to what we are asked in the collision problems. Since the scattering angle θ ∗ 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.

Next, as shown in _{x}-axis so that O E = m A V . 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 [

As shown in _{x}-axis until the broken line intersects with the circle centered at E. We call this point of intersection as F. Note that O C = E F is always satisfied and it means ∠ C O A = ∠ F E A = θ ∗ . Then, the vector O F = p ′ A shows the momentum of the projectile A after the collision in the laboratory system. The tips of O F = p ′ A is always on the circle centered at E. The angle ∠ F O E = θ is the scattered angle of the projectile A in the laboratory system. We note that the angle θ ∗ in

draw a broken line from the point F to C. The vector O C = p ′ A ∗ and the angle θ ∗ 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 p_{x}-axis so that E G = m B V . Then the vector F G = O H = p ′ B shows the momentum of the target B after the collision. The angle ∠ G O H = ∠ F G O = ϕ is the scattered angle of the target B. The vector relation O G = O E + E G = O F + O H shows the momentum conservation p A + p B = p ′ A + p ′ B 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 θ or θ ∗ according to the given collision problems.

Let us consider the case where the target B is at rest v B = 0 before the collision. In that case, the point G definitely lies on the circle centered at E because of | O C | = | E F | = | E G | , i.e. p A ∗ = p B ∗ = m B V , which is understood by Equations ((4) and (15)). The circle centered at E is described by Equation (19) and is depicted in

The point E lies inside or outside the dashed circle centered at O, according as m A < m B and m A > m B . The corresponding diagrams are shown in

tan θ = E F sin θ ∗ O E + E F cos θ ∗ = m B V sin θ ∗ m A V + m B V cos θ ∗ = sin θ ∗ cos θ ∗ + m A m B , (23)

tan ϕ = E F sin θ ∗ E G − E F cos θ ∗ = sin θ ∗ 1 − cos θ ∗ = sin ϕ ∗ 1 + cos ϕ ∗ , (24)

where we use the relation θ ∗ + ϕ ∗ = π . It is also evident that since the triangle Δ E F G is an isosceles triangle, we obtain ϕ = π − θ ∗ 2 .

Applying the law of cosine to the triangle Δ O E F : O F 2 = O E 2 + E F 2 − 2 ⋅ O E ⋅ E F cos ( π − θ ∗ ) , we obtain

v ′ A = v A m A + m B m A 2 + m B 2 + 2 m A m B cos θ ∗ . (25)

The same application to the triangle Δ E F G gives

v ′ B = 2 m A v A m A + m B sin θ ∗ 2 . (26)

If m A > m B , however, the projectile A can be deflected only through an angle not exceeding θ max = ∠ F O A from its original direction, as shown in

s i n θ m a x = E F O E = E G O E = m B V m A V = m B m A , (27)

because EF = EG are both the radius of the circle.

The case m A = m B becomes quite simple as shown in _{y}-axis is a tangent to the circle centered at E. It is evident from the figure that θ ∗ = 2 θ . Further, Equations ((25) and (26)) become

v ′ A = v A cos θ ∗ 2 , v ′ B = v A sin θ ∗ 2 . (28)

After the collision, the outgoing particles move at right angles to each other, that is θ + ϕ = π 2 .

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.

The authors declare no conflicts of interest regarding the publication of this paper.

Ogura, A. (2018) Diagrammatic Approach for Investigating Two Dimensional Elastic Collisions in Momentum Space I: Newtonian Mechanics. World Journal of Mechanics, 8, 343-352. https://doi.org/10.4236/wjm.2018.89025