Collision Geometry

Kvant 1970, No. 5 p. 18-25 Tag 2O2Y

Rough automatic draft, low quality: expect mistakes and awkward wording. It will be replaced by a better translation. Original →

GEOMETRY
COLLISIONS

Y. A. SMORODINSKY,
E A SURKOV

A LITTLE INTRODUCTION.

One of the problems that one often encounters in modern physics is the problem of elastic scattering. It is usually put like this: it is necessary to determine the velocities (in magnitude and direction) of two parts after they collided and scattered (as they say, scattered), if the directions of their motion and energy before the collision were known.

A collision is called elastic if it is not accompanied by the release of heat, and we can assume that the sums of the kinetic energies of the part before and after the collision are equal. In addition, particles obey the law of conservation of momentum.

When two particles - gas molecules or two nucleons - collide with each other, the result of the collision can be different. If the particles fly at a sufficiently large distance from each other,

Above: photograph of an off-center impact of a moving ball on an equally stationary one. The balls are separated at right angles. The stationary ball moves to the right after colliding. The illumination is strokosoppic, the flash frequency is 25 per second.

then the directions of their movement may hardly change and the path of the particles will only deviate slightly from the rectilinear one; in other cases, when particles collide, so to speak, head-on, the direction of their movement can change greatly. If we know the initial state of the particles (vectors of their velocities), then the velocities of the particles after the collision are determined if we set the scattering angle - the angle at which one will deviate

Bottom: successive collisions of a proton with other protons, initially at rest. The photograph was taken in a bubble chamber with liquid hydrogen. Traces of flying protons are visible in the form of a chain of tiny bubbles.

from particles upon collision (we will define this concept more precisely below).

Although finding the connection between particle velocities and the scattering angle is not a difficult task in principle, nevertheless, the calculations must be carried out in the correct sequence - otherwise it is quite easy to get confused in them.

A very visual and convenient geometric method for solving the collision problem is based on the fact that

Each collision process can be associated with a quadrilateral composed of velocity vectors. Using this method, the determination of angles and velocities can be reduced to the solution of triangles known from trigonometry. This method is also convenient because it can be generalized to more complex processes, for example, nuclear reactions, and even to solving problems in the mechanics of the theory of relativity. Here we will talk only about the simplest tasks.

In order to use the geometric method, it is necessary to introduce the concept of velocity space. Since velocities are vectors, the operations that can be performed with it are very similar to the operations that are performed with displacement (distance) vectors. Therefore, just as all possible displacements form (or, if you like, fill) a common space (we will call it coordinate space), so all possible velocities form a similar geometric object - velocity space. It is in the velocity space that we will construct and solve triangles.

It is most convenient to talk about the properties of velocity space based on an analogy with coordinate space. Therefore, we will first fill in some properties of the latter.

COORDINATE SPACE.

Consider a two-dimensional space, for example the floor of a room, extended in all directions to infinity.

If there are several points (objects on the floor) on the plane, then in order to talk about the position of these points, we need to select a cutoff body - one of the points, for example $O$ (Fig. 1). then the position of any point $A$ relative to $O$ is given by the radius vector $\vec{OA}$ , looking to this point from point $O$ , or, as you can also say, the displacement vector from point $O$ to point $A$ .

If we wanted to tell someone how to find the point $A$ , then we would have to choose, in addition to the cutting body $O$ , another point $C$ (Fig. 2), in order to say what angle $\alpha$ is made by the vector $\vec{OA}$ with the direction $OC$ . This angle, together with the length of the radius vector $\vec{OA}$ , determines the point $A$ .

One could do it differently: draw through the point $O$ two mutually perpendicular lines $OX$ and $OY$ (coordinate axes), one of which passes through the point $C$ , and specify two numbers $A_x$ and $A_y$ - projections of the vector $\vec{OA}$ on the coordinate axes.

In the future, we will consider only those quantities that do not depend on the coordinate system -

distances between points and angles between straight lines. Therefore, we will not need to introduce a coordinate system.

If we choose not the point $O$ as the body of the report, but some other point of the plane, for example, the point $B$ (Fig. 1), then the position of the point $A$ relative to $B$ will be determined by the radius vector $\vec{BA}$ , equal to the sum of the vectors $\vec{BO}$ and $\vec{OA}$ . In other words, if we move to a new body of the report - point $B$ , which is specified by the vector $\vec{OB}$ , then in order to find the radius vector $\vec{BA}$ , which determines the position of the body $A$ relative to $B$ , we need to go to the vector $\vec{OA}$ , which determines the position of the body $A$ relative to the first initial body of the report $O$ , add vector $\vec{BO} = -\vec{OB}$ .

Obviously correct addition of vectors is depicted graphically somewhat differently. Added vectors are drawn as coming from one point, that is, the construction vector $\vec{BD}$ , the parallel vector $\vec{OA}$ and equal in length (Fig. 3). The vector $\vec{BA}$ is considered the sum of $\vec{BO} + \vec{BD}$ . But you can add vectors as shown in figure 1. In what follows we will need

This is exactly the method for adding velocities.

Do such pins go horizontally or vertically, defining the relative positions of the body lying on the plane. Now we will be interested in the relative velocities of a body moving uniformly and rectilinearly along a plane.

SPACE OF SPEED

Let’s put a point on the plane, denote it $V_0$ , and draw the velocity vectors of any bodies moving relative to the reporting body $O$ so that the origins of these vectors are at point $V_0$ . Let $V_A$ be the point at which the end of the body velocity vector $A$ (Fig. 4) comes. Since speed is a vector of displacement per unit time, velocity vectors add up in exactly the same way as displacement vectors. Therefore, if as the body of the report we choose not the point $O$ , but some other body, for example $B$ , whose velocity vector relative to $O$ is the vector $\vec{V}_0\vec{V}_B$ , then the velocity vector of the body $A$ relative to $B$ can be found by using the rule of vector addition, which we talked about when we considered coordinate space. The speed of the body $A$ relative to $B$ is equal to the sum of the vectors $\vec{V}_B\vec{V}_O$ and $\vec{V}_O\vec{V}_A$ , that is, the vector

ru $V_B V_A$ . Essentially we just need to check the vector from point $V_B$ to point $V_A$ .

It is clear that the speed of any body $C$ on our plane must correspond to a certain point $V_c$ , which is the end of the vector $\vec{V}_o V_c$ - the speed of body $C$ relative to body $O$ . A vector drawn from point $V_c$ to any other point in the plane, for example, at $V_A$ , is represented by the velocity vector of body $A$ relative to body $C$ .

This allows us to represent the velocities of bodies not only as vectors, but also as points on a plane in velocity space. Any point on this plane means the speed of the body, and the vector going from one point to another is the vector of the relative speed of the body. It is clear what the meaning of the point $V_o$ with which we started is. This is the speed of the body $O$ .

Of course, using velocity space, we do not distinguish between bodies located at different points in coordinate space, but having the same velocities. All of them correspond to the same point in velocity space. Therefore, by figure 4 we cannot say whether the body $C$ will reach the body $O$ or move away from it. This is determined generally by the condition of the problem that we are solving and by the positions of the bodies in coordinate space.

KINEMATIC DIAGRAM

Let us now move on to solving the scattering problem. Let a particle with mass $m_1$ and speed $v_1$ collide with another particle having mass $m_2$ and speed $v_2$ , and let after the collision the velocities of the corresponding particles be $v_1'$ and $v_2'$ ). The task is to find the velocities

*) We believe that the particles move like billiard balls: the impact itself lasts a very short time, and the rest of the time the particles move as if they were free. Then the expressions “before the event” and “after the event” have a completely understandable meaning.

$v_1'$ and $v_2'$ , if the vectors $v_1$ and $v_2$ and the particle masses are known.

The velocities $v_1'$ and $v_2'$ cannot, of course, have arbitrary values. Firstly, they must satisfy the law of conservation of momentum (momentum):

$m_1 v_1' + m_2 v_2' = m_1 v_1' + m_2 v_2' ,$

(1) secondly, they must satisfy the law of conservation of energy:

$\frac{m_1 v_1^2}{2} + \frac{m_2 v_2^2}{2} = \frac{m_1 v_1'^2}{2} + \frac{m_2 v_2'^2}{2} .$

In the case when the particles before and after the impact move in one straight line (such an impact is called central), these equations are solved quite simply. Instead of the vector equation (1), we can write exactly the same scalar equation for the lengths of the vectors. Then we get two scalar equations with two unknowns: $v_1'$ and $v_2'$ . But more often you have to deal with an off-center impact, in which particles are separated at different angles. Having projected the particle velocity vectors into two mutually perpendicular directions, one of which is conveniently chosen to be parallel to the velocity of any of the particles before scattering, together with equation (1) we can write two exact scalar equations for the vector projections. Therefore, the conservation laws in the case of an off-center impact should be useful for the particles after scattering and the two angles that these vectors make with the direction we have chosen. If one of the quantities is specified, then the equation can be solved.

The collided part in any reporting system. Let’s take, for example, one in which particle 1 rummaged around before impact (such a system is called a laboratory system). It contains $v_1 = 0$ , and the law of conservation of momentum looks like this:

$m_2 v_2 = m_1 v_1' + m_2 v_2' .$

This means that three vectors

$m_2v_2, m_1v_1'$ and $m_2v_2'$ form a triangle. Any relative particle velocities (before and after collision) lie in the plane of this triangle. Therefore, it is enough for us to consider only the two-dimensional velocity space.

Let us mark $v_1, v_2, v_1'$ and $v_2'$ on the velocity plane. We will simply denote the corresponding points by the numbers 1, 2, 1’ and 2’ (Fig. 5). Let us prove that these points lie at the vertices of an equal trapezoid. To do this, we introduce a new reporting system: center of mass systems. It is selected so that the impulses of the colliding particles in it are equal in size and opposite in direction. For particles 1 and 2, point $C$ —the velocity of the center of mass—lies on a straight line connecting these points and divides this straight line into segments inversely proportional to the masses of the particles. If we denote the particle velocities in the center of mass system by $u_1$ and $u_2$ , then $m_1u_1 = m_2u_2$ .

This equality is similar to the lever equilibrium rule. The point that corresponds to a light particle is located further from the center. It is clear that the total momentum of the particle in the center of mass system is zero. Due to the law of conservation of momentum in this system, the sum of the particle impulses after calculation will also be equal to zero. This means that the point $C$ lies on the line connecting these points and divides this line into segments inversely proportional to the masses of the particles.

It is not difficult to show that, due to the law of conservation of energy in the center of mass system, the magnitudes of the impulses of each of the particles do not change, but only their directions change. To do this, the energies of the particles must be expressed through the magnitudes of their impulses and, writing the law of conservation of energy in the form

$\frac{(m_1u_1)^2}{2m_1} + \frac{(m_2u_2)^2}{2m_2} = \frac{(m_1u_1')^2}{2m_1} + \frac{(m_2u_2')^2}{2m_2},$

replace in this equation the impulses of particle 2 with impulses of particle 1 equal in magnitude or vice versa. In the first case we get that $m_1u_1' = m_1u_1$ or $u_1' = u_1$ , and in the second that $m_2u_2' = m_2u_2$ or $u_2' = u_2$ . And this means that at figure 5 the result is an equal trapezoid.

If the masses of both particles are equal, then this trapezoid turns into a rectangle. (Since $m_1 = m_2$ , then $u_1 = u_2 = u_1' = u_2'$ .)

MULTIPLE TASKS

Let’s look at several problems that can be solved using kinematic diagrams.

Nanhem with the case of non-central impact of particles with equal masses (Fig. 6). If point 1 is before the interpretation

buy, then the angle 212’ is the angle by which the trajectory of particle 2 deviates during scattering. Particle 1 flies off in direction 1’. The angle 211’ is called the recoil angle. The diagram shows that in the coordinate system in which one of the particles (particle 1) was bought before the collision, particles 1 and 2 after the impact scatter at right angles in directions 11’ and 12'.

Let’s consider two coordinate systems: α is the scattering angle in the center of mass system, β is the scattering angle in the system where particle 1 was purchased before impact. The diagram shows that α = 2β, since α is an exterior angle not adjacent to the angle β in the equilateral triangle 12’C. The scattering angle changed in the laboratory system is half the scattering angle in the center of mass system.

If we rotate segments 12 and 12’ around point C - it is more convenient, however, to leave segment 12 motionless and rotate only segment 12’ - then we will describe all possible cases of scattering at given energies (or, what is the same, at given velocities) of particles before scattering, that is, all possible cases of collisions.

In this case, we will see that the scattering angle α in the center of mass system changes by 360°: 180° clockwise and 180° counterclockwise. Since in the laboratory system the scattering angle β is half the scattering angle β is half the scattering angle β: v₂’ = v₂cosβ, where v₂ is the initial velocity of particle 2 in this system.

Let us now consider the problem of the collision of a part with unequal masses. To increase all possible cases of scattering, we will again rotate the segment 12’ around point C. Particle 1 is lighter, then point 1 describes a larger circle than

point $2'$ (Fig. 7, a). (Since $m_1u_1 = m_2u_2$ , then at $m_1 < m_2$ , $u_1 > u_2$ .) From diagram 7, a it is clear that in the reference frame in which the heavy part 2 was buried before scattering, the light part 1 can scatter at any angle (scattering angle - $121'$ ).

The situation is different if the lighter part 1 is dug. Since the point $2'$ lies on a circle of small radius, it is clear that the segment $C2'$ is necessarily located in the angle at which this circle is visible from point 1 (Fig. 7, b). This means that the maximum scattering angle of particle 2 in the coordinate system where particle 1 was buried before scattering (angle $212'$ ) is equal to the angle at which the irrigation of the circle is visible from point 1, resulting from the expression of the segment $2'C$ around point $C$ . From figure 8 (this is an enlarged part figure 7, b) we find that

$\sin \alpha_{\max} = \frac{u_2'}{u_1} = \frac{u_2}{u_1} = \frac{m_1}{m_2}.$

If the heavy part scatters on the light part, then it cannot deviate by too large an angle. The maximum angle is determined by the mass ratio.

Thus, you need to learn to read diagrams, and then move on to simple trigonometry, which gives an answer to any question. We can now not solve equations expressing the laws of conservation of energy and momentum each time, but immediately draw a kinematic diagram in the velocity space (for which these laws are satisfied automatically) and then consider it from the observable point of view, moving at different speeds. On figure 9 the same collision process is depicted from the point of view of three different frames of reference: a) in the center of mass frame, b) in the LABorator frame, in which part 1 was dug before scattering, b) in an arbitrary frame of reference O.

Solve the following problems yourself.