COLLISIONS AND SCATTERING OF PARTICLES

1. Central collisions of the balls.

Consider two spherical objects (balls) one of mass m₁ and the other of mass m₂. Let's set things up so these balls are approaching each other along the line joining their centers, a recipe for a head on collision. Let these balls not be rotating or vibrating. The motion is purely translation. Under these conditions we may choose a reference frame which is one dimensional, with the objects on the x-axis.

Applying the conservation of linear momentum to this situation we have,

m₁v_1i + m₂v_2i = m₁v₁ + m₂v₂

where v_i is the initial velocity of each object and v is the final velocity of each object. The conservation of kinetic energy gives us the equation

m₁v_1i² / 2 + m₂v_2i² / 2 = m₁v₁² / 2 + m₂v₂² / 2

These may be rewritten as follows:

m₁ (v_1i - v₁) = m₂ (v₂ - v_2i)

for the momentum equation and

m₁ (v_1i² - v₁²) = m₂ (v₂² - v_2i²)

for the energy.

As long as the difference between final and initial velocities is not zero for either object (meaning a collision actually happens), we may divide the second equation by the first one which yields

v_1i + v₁ = v₂ + v_2i or v_1i - v_2i = v₂ - v₁

In other words in a one dimensional elastic collision, the relative velocity of approach before the collision equals the relative velocity of separation after collision.

To get the final velocities in terms of the initial velocities and the masses, you would solve the last equation above for v₂ and plug that into the momentum equation and solve to get

v₁ = v_1i (m₁ - m₂) / (m₁ + m₂) + v_2i (2 m₂) / (m₁ + m₂)
v₂ = v_1i (2 m₁) / (m₁ + m₂) + v_2i (m₂ - m₁) / (m₂ + m₁)

So we have the basis for a model of a one dimensional elastic collision. For initial conditions v_1i and v_2i, if a collision happens, the final velocities depend on the masses as above.

Now if the two masses, m₁ and m₂ are equal we can replace both with simply m, like this.

v₁ = v_1i (m - m) / (m + m) + v_2i (2 m) / (m + m)

Likewise

v₂ = v_1i (2 m) / (m + m) + v_2i (m - m) / (m + m)

This reduces to the following expressions for final velocities v₁ and v₂.

v₁ = v_2i and v₂ = v_1i

In the case of equal mass collisions, the two objects just exchange velocities. Of course if the initial velocity of one of the balls were zero, then the one colliding with it would just stop and the hit ball would take off at the original speed of the incoming ball.

In general case of central and absolutely elastic collision of objects with different masses, one of which was still before collision (v_2i =0), the velocities of the objects after collision can be described by the equations:

v₁ = v_1i (m₁ - m₂) / (m₁ + m₂)
v₂ = v_1i (2 m₁) / (m₁ + m₂)

If the mass m₁ of incoming ball is greater than the mass m₂ of the still one, then after collision both velocities v₁ è v₂ will be positive and the balls will move in one direction, which coincides with the direction of initial motion of incoming ball.

If the mass m₁ of incoming ball is smaller than the mass m₂ of the still one, then after collision v₁ <0, v₂ >0 and the balls will move in opposite directions and since 2m₁>m₁-m₂ small ball will be reflected with the greater speed.

Next, we shall consider a case when a ball collides the chain of several identical balls, as shown in animation. In this case the incoming ball exchanges the velocity with the second ball, the second ball exchanges velocity with the third one, etc. As a result after collision all the balls except the last one will be in rest and last ball will be set into the motion with velocity equal to velocity of the incoming ball.

In practice the central collisions of identical balls can be observed with the aid of set-up shown in animation.  All the balls are suspended on long strings and the task is reduced to consideration of their paired collisions. Thus, all the system will behave as shown in animation, i.e. outermost balls will be pushed consequently with identical velocity and deviated by identical angles, while all the balls between them will be in rest. It is necessary to note, that above mentioned consideration is true only for the case of absolutely elastic central collision of the balls when there is no loss of energy. Actually, the total energy of the system will decrease due to air dumping, heating of the balls, excitation of acoustic waves, etc. Therefore, the amplitude of extreme balls deviation decreases with time and the central balls are set into the oscillatory motion.

In our consideration we suggested a series of two ball collisions, so that the last ball in the line leaves the group with approximately the same velocity as the ball which struck the group originally. If the contact between the balls is rigid, the incoming ball would see a single massive object comprised of several balls. It would then rebound from the collision in accordance with the rules developed above for collision between different masses. Our reasoning leads us to conclude that even though the balls in the line appear to be in contact, there must be initially a very soft reaction between adjacent balls.

Let us consider non-elastic collision in more details. During non-elastic collision the part of kinetic energy of incoming ball is lost in the form heating.  In the extreme case of absolutely non-elastic collision the incoming and still balls are stuck together, the kinetic energy of their relative motion vanishes and they continue the motion as single body. In many practical cases we deal with the partially elastic body when the attenuated flexible oscillation is excited in it after collision. Animation shows the collision of the elastic ball with the rigid wall. Such collision is accompanied with excitation of the flexural (deformation) oscillation of the elastic ball. In general case many modes of oscillation will be excited in a ball, but the one with the lowest frequency will dominate. The excited oscillation will attenuate with time and the energy of oscillation will be transformed into the heat.

Considering above the absolutely elastic collision in a chain of identical balls we concluded that all intermediate spheres remain in rest and only the extreme balls move.  Let us consider what will change if all the intermediate balls will be connected with springs.  Animation shows a case of two intermediate balls connected by a spring.  We see that intermediate balls are set into the oscillatory motion while their center of mass is practically still. Numerical simulation shows that the system will behavior by the similar way in case of three, four and more intermediate balls connected by springs.  The oscillation will attenuate with time and the system will remind the chain of free rigid balls considered above, but only partly.  The part of the energy of the system is gone away by the attenuated oscillation, so speed of the most right ball should be less than the speed of incoming ball. But, if we assume, that intermediate spheres are motionless, than the total impulse of the system will not be kept. Thus, the intermediate spheres have to move with small speed too. This conclusion coincide with the result of the numerical simulation.

v = (2M/m)(Lg)^1/2sin(j/2)

2. Collisions and scattering.

In our analysis of the swinging ball machine we discussed the elastic collision of two spherical objects when the collision was head on, that is the velocities of the two objects were along the line connecting the objects' centers. In order to understand the scattering of a group of light particles by a group of heavy ones, we must extend our analysis into more than one dimension.

Let us consider the case when the ball 1 collides the ball 2 without the friction. We shall suggest that the ball 2 of mass m₂ is in the rest before collision and the ball 1 of mass m₁ is moving with velocity v. The velocity v₁ of ball 1 and velocity v₂ of ball 2 after collision will depend upon the "aiming" distance d , which is equal to the distance between the center of the ball 2 and the line of the motion of the ball 1 before collision. The collision will happen if d < r₁ + r₂, where r₁ and r₂ are the radiuses of the ball 1 and ball 2 consequently. The force applied to ball 2 during collision from the side of the ball 1 is directed along the line joining the centers of the balls. So, after collision the ball 2 will move at angle q as shown in the figure.

(r₁ + r₂) sin q = d

During the collision the energy and momentum of the motion is constant:

From these equations we can find

Now we can move on to the situation illustrated. Here we have a number of light particles striking a number of heavier ones. There are differences between our analysis so far and this case. One is that multiple light particles may hit the same heavy one and the heavy particle is not so heavy that its reaction to collision may be neglected. This means that the heavy particle may not be at rest when the second and subsequent light particles hit it. Another difference is that a single light particle may rebound from one heavy particle and strike a second heavy particle. These differences enormously complicate the analysis. About all we can say at this point is that the scattering angles of the heavy particles will be in the range -p /2 to from +p /2 from the initial velocity vector of the light particles and this will bias the scattering of the group, introducing a preferential forward scattering. Some of the light particles may indeed be scattered into the back hemisphere, but more of them will be scattered into the forward hemisphere.

3. Rutherford's experiment

3.1. Scattering of a-particles by atoms of metal.

Existence in atom of heavy positively charged nucleus was discovered by English physicist E.Rutherford in 1906-1912 as a result of experiments with scattering of a-particles on the atoms of gold and other metals. These experiments showed that some of  a-particles (1 per 20000) passing through very thin metal layers were scattered by big angles, while the most of the particles were not deviated. Rutherford concluded that rare scattering of particles (which moves with velocity just 20 times lees that the velocity of light) at big angles may be explained by the presence in the matter the massive nucleus the size of which is much smaller as compared to the atom size. Animation shows the simulation of Rutherford's experiment. The beam of particles moving with about the same velocities collides several big balls simulating the atoms in metal. We see in animation that small part of the particles are scattered by big angles.

3.2. Differential section of scattering.  Rutherford's formula.

The scattering of particles by nucleus occurs due to Coulomb's electrostatic forces. Sighting parameter is the distance between the line of initial motion of particle and the center of the force center (nucleus). It is clear that not only the particles with sighting parameter smaller than the radius of nucleus will be scattered (see animation). Rutherford supposed that all mass of atom is concentrated in the positively charged small nucleus and obtained on this basis the expression for differential section of scattering :

 ds/dW = (Qe/2vp)²/(sin(j/2) )⁴

where Q is the charge of nucleus, v, p is the velocity and impulse of a-particle respectively, j  is the angle of scattering, ds is differential section of scattering which has dimensionality of area, dW is the spatial angle in which we observe the scattering. This expression is called Rutherford's formula. The calculated section for a-particles proved to be in very good correspondence with experimental results if we substitute in this formula Q= Ze, where Z is the atom number of the element.