17 done

1/16-disintegration-of-particles.md

@@ -1,130 +1,115 @@
 ---
-title: 16-disintegration-of-particles
+title: 16 Disintegration of particles
 ---
-IN many cases the laws of conservation of momentum and energy alone can
+
-be used to obtain important results concerning the properties of various mech-
+In many cases the laws of conservation of momentum and energy alone can be used to obtain important results concerning the properties of various mechanical processes. It should be noted that these properties are independent of the particular type of interaction between the particles involved.
-anical processes. It should be noted that these properties are independent of
+
-the particular type of interaction between the particles involved.
+Let us consider a "spontaneous" disintegration (that is, one not due to external forces) of a particle into two "constituent parts", i.e. into two other particles which move independently after the disintegration.
-Let us consider a "spontaneous" disintegration (that is, one not due to
+
-external forces) of a particle into two "constituent parts", i.e. into two other
+This process is most simply described in a frame of reference in which the particle is at rest before the disintegration. The law of conservation of momentum shows that the sum of the momenta of the two particles formed in the disintegration is then zero; that is, the particles move apart with equal and opposite momenta. The magnitude $p_0$ of either momentum is given by the law of conservation of energy:
-particles which move independently after the disintegration.
+
-This process is most simply described in a frame of reference in which the
+$$
-particle is at rest before the disintegration. The law of conservation of momen-
+E_i=
-tum shows that the sum of the momenta of the two particles formed in the
+E_{1i}+\frac{{p_0}^2}{2m_1}
-disintegration is then zero; that is, the particles move apart with equal and
++E_{2i}+\frac{{p_0}^2}{2m_2}
-opposite momenta. The magnitude Po of either momentum is given by the
+$$
-law of conservation of energy:
+
-here M1 and m2 are the masses of the particles, E1t and E2i their internal
+here $m_1$ and $m_2$ are the masses of the particles, $E_{1i}$ and $E_{2i}$ their internal energies, and $E_i$ the internal energy of the original particle. If $\epsilon$ is the "disintegration energy", i.e. the difference
-energies, and E the internal energy of the original particle. If € is the "dis-
+
-integration energy", i.e. the difference
+```load
-E= E:-E11-E2i,
+16.1
-(16.1)
+```
+
 which must obviously be positive, then
-(16.2)
+
-which determines Po; here m is the reduced mass of the two particles. The
+```load
-velocities are V10 = Po/m1, V20 = Po/m2.
+16.2
-Let us now change to a frame of reference in which the primary particle
+```
-moves with velocity V before the break-up. This frame is usually called the
+
-laboratory system, or L system, in contradistinction to the centre-of-mass
+which determines $p_0$; here $m$ is the reduced mass of the two particles. The velocities are $v_{10} = p_0/m_1, v_{20} = p_0/m_2$.
-system, or C system, in which the total momentum is zero. Let us consider
+
-one of the resulting particles, and let V and V0 be its velocities in the L and
+Let us now change to a frame of reference in which the primary particle moves with velocity $V$ before the break-up. This frame is usually called the *laboratory system*, or $L$ system, in contradistinction to the *centre-of-mass* system, or $C$ system, in which the total momentum is zero. Let us consider one of the resulting particles, and let $\v{v}$ and $\v{v}_0$ be its velocities in the $L$ and the $C$ system respectively. Evidently $\v{v} = \v{V}+\v{v}_0, or $\v{v}-\v{V} = \v{v}_0$, and so
-the C system respectively. Evidently V = V+vo, or V -V = V0, and SO
+
-(16.3)
+```load
-where 0 is the angle at which this particle moves relative to the direction of
+16.3
-the velocity V. This equation gives the velocity of the particle as a function
+```
-41
+
-42
+where $\theta$ is the angle at which this particle moves relative to the direction of the velocity $\v{V}$. This equation gives the velocity of the particle as a function of its direction of motion in the $L$ system. In `fig14` the velocity $\v{v}$ is represented by a vector drawn to any point on a circle[^\dagger] of radius $v_0$ from a point $A$ at a distance $V$ from the centre. The cases $V \lt v_0$ and $V\gt v_0$ are shown in `fig14a` and `fig14b` respectively. In the former case `\theta` can have any value, but in the latter case the particle can move only forwards, at an angle $\theta$ which does not exceed $\theta_\max$, given by
-Collisions Between Particles
+
-§16
+[^\dagger]: More precisely, to any point on a sphere of radius vo, of which `fig14` shows a diametral section.
-of its direction of motion in the L system. In Fig. 14 the velocity V is repre-
+
-sented by a vector drawn to any point on a circle+ of radius vo from a point
+```load
-A at a distance V from the centre. The cases V < vo and V>00 are shown
+16.4
-in Figs. 14a, b respectively. In the former case 0 can have any value, but in
+```
-the latter case the particle can move only forwards, at an angle 0 which does
+
-not exceed Omax, given by
+this is the direction of the tangent from the point $A$ to the circle.
-(16.4)
+
-this is the direction of the tangent from the point A to the circle.
+```fig
-C
+14
-V
+```
-V
+
-VO
+The relation between the angles $\theta$ and $\theta_0$ in the $L$ and $C$ systems is evidently
-VO
+
-max
+```load
-oo
+16.5
-oo
+```
-A
+
-V
+If this equation is solved for $\cos\theta_0$, we obtain
-A
+
-V
+```load
-(a) V<VO
+16.6
-)V>Vo
+```
-FIG. 14
+
-The relation between the angles 0 and Oo in the L and C systems is evi-
+For $v_0 > V$ the relation between $\theta_0$ and $\theta$ is one-to-one `fig14a`. The plus sign must be taken in `16.6`, so that $\theta_0 = 0$ when $\theta = 0$. If $v_0 \lt V$, however, the relation is not one-to-one: for each value of $\theta$ there are two values of $\theta_0$, which correspond to vectors $\v{v}_0$ drawn from the centre of the circle to the points $B$ and $C$ `fig14b`, and are given by the two signs in `16.6`.
-dently (Fig. 14)
+j
-tan
+In physical applications we are usually concerned with the disintegration of not one but many similar particles, and this raises the problem of the distribution of the resulting particles in direction, energy, etc. We shall assume that the primary particles are randomly oriented in space, i.e. isotropically on average.
-(16.5)
+
-If this equation is solved for cos Oo, we obtain
+In the $C$ system, this problem is very easily solved: every resulting particle (of a given kind) has the same energy, and their directions of motion are isotropically distributed. The latter fact depends on the assumption that the primary particles are randomly oriented, and can be expressed by saying that the fraction of particles entering a solid angle element $\dd{o}_0$ is proportional to $\dd{o}_0$, i.e. equal to $\dd{o}_0/4\pi$. The distribution with respect to the angle $\theta_0$ is obtained by putting $\dd{o}_0 = 2\pi \sin\theta_0\dd{o}_0$, i.e. the corresponding fraction is
-V
+
-0
+```load
-(16.6)
+16.7
-For vo > V the relation between Oo and 0 is one-to-one (Fig. 14a). The plus
+```
-sign must be taken in (16.6), so that Oo = 0 when 0 = 0. If vo < V, however,
+
-the relation is not one-to-one: for each value of 0 there are two values of Oo,
+The corresponding distributions in the $L$ system are obtained by an appropriate transformation. For example, let us calculate the kinetic energy distribution in the $L$ system. Squaring the equation $\v{v}=\v{v}_0+\v{V}$, we have $v^2={v_0}^2+V^2+2v_0V\cos\theta_0$, whence $\dd{(\cos\theta_0)} = \dd{(v^2)}/2v_0V$. Using the
-which correspond to vectors V0 drawn from the centre of the circle to the
+kinetic energy $T = \mfrac{1}{2}mv^2$, where $m$ is $m_1$ or $m_2$ depending on which kind of particle is under consideration, and substituting in `16.7`, we find the required distribution:
-points B and C (Fig. 14b), and are given by the two signs in (16.6).
+
-In physical applications we are usually concerned with the disintegration
+```load
-of not one but many similar particles, and this raises the problem of the
+16.8
-distribution of the resulting particles in direction, energy, etc. We shall
+```
-assume that the primary particles are randomly oriented in space, i.e. iso-
+
-tropically on average.
+The kinetic energy can take values between $T_\min = \mfrac{1}{2}m(v_0-V)^2$ and $T_\max=\mfrac{1}{2}m(v_0+V)^2$. The particles are, according to `16.8`, distributed uniformly over this range.
-t More precisely, to any point on a sphere of radius vo, of which Fig. 14 shows a diametral
+
-section.
+When a particle disintegrates into more than two parts, the laws of conservation of energy and momentum naturally allow considerably more freedom as regards the velocities and directions of motion of the resulting particles.  In particular, the energies of these particles in the $C$ system do not have determinate values. There is, however, an upper limit to the kinetic energy of any one of the resulting particles. To determine the limit, we consider the system formed by all these particles except the one concerned (whose
-§16
+mass is $m_1$, say), and denote the "internal energy" of that system by $E_i'$.  Then the kinetic energy of the particle $m_1$ is, by `16.1` and `16.2`, $T_{10} = {p_0}^2/2m_1 = (M-m_1)(E_i-E_{1i}-E_i')$ where $M$ is the mass of the primary particle. It is evident that $T_{10}$ has its greatest possible value when $E_i'$ is least. For this to be so, all the resulting particles except $m_1$ must be moving with the same velocity. Then $E_i'$ is simply the sum of their internal energies, and the difference $E_i-E_{1i}-E_i'$ is the disintegration energy $\epsilon$. Thus
-Disintegration of particles
+
-43
+```load
-In the C system, this problem is very easily solved: every resulting particle
+16.9
-(of a given kind) has the same energy, and their directions of motion are
+```
-isotropically distributed. The latter fact depends on the assumption that the
+
-primary particles are randomly oriented, and can be expressed by saying
+<!-- PROBLEMS -->
-that the fraction of particles entering a solid angle element doo is proportional
+<!-- PROBLEM 1. Find the relation between the angles 01, O2 (in the L system) after a disintegra- -->
-to doo, i.e. equal to doo/4rr. The distribution with respect to the angle Oo is
+<!-- tion into two particles. -->
-obtained by putting doo = 2m sin Oo doo, i.e. the corresponding fraction is
+<!-- SOLUTION. In the C system, the corresponding angles are related by 010 = n-020. Calling -->
-(16.7)
+<!-- 010 simply Oo and using formula (16.5) for each of the two particles, we can put -->
-The corresponding distributions in the L system are obtained by an
+<!-- 7+10 cos Oo = V10 sin Oo cot 01, V-V20 cos Oo = V20 sin Oo cot O2. From these two -->
-appropriate transformation. For example, let us calculate the kinetic energy
+<!-- equations we must eliminate Oo. To do so, we first solve for cos Oo and sin Oo, and then -->
-distribution in the L system. Squaring the equation V = V0 + V, we have
+<!-- form the sum of their squares, which is unity. Since V10/2.20 = m2/m1, we have finally, using -->
-2.2 = V02 + +200 cos Oo, whence d(cos 00) = d(v2)/200V. Using the
+<!-- `16.2`, -->
-kinetic energy T = 1mv2, where m is m1 or M2 depending on which kind of
+<!-- (m2/m1) sin202+(m1/m2) sin201-2 sin O1 sin O2 cos(A1+6 -->
-particle is under consideration, and substituting in (16.7), we find the re-
+<!-- PROBLEM 2. Find the angular distribution of the resulting particles in the L system. -->
-quired distribution:
+<!-- SOLUTION. When vo > V, we substitute `16.6`, with the plus sign of the radical, in `16.7`, -->
-(1/2mvov) dT.
+<!-- obtaining -->
-(16.8)
+<!-- (0  . -->
-The kinetic energy can take values between Tmin = 3m(e0-V)2 and
+<!-- When vo < V, both possible relations between Oo and 0 must be taken into account. Since, -->
-Tmax = 1m(vo+V)2. The particles are, according to (16.8), distributed
+<!-- when 0 increases, one value of Oo increases and the other decreases, the difference (not the -->
-uniformly over this range.
+<!-- sum) of the expressions d cos Oo with the two signs of the radical in `16.6` must be taken. -->
-When a particle disintegrates into more than two parts, the laws of con-
+<!-- The result is -->
-servation of energy and momentum naturally allow considerably more free-
+<!-- (0    max). -->
-dom as regards the velocities and directions of motion of the resulting particles.
+<!-- PROBLEM 3. Determine the range of possible values of the angle 0 between the directions -->
-In particular, the energies of these particles in the C system do not have
+<!-- of motion of the two resulting particles in the L system. -->
-determinate values. There is, however, an upper limit to the kinetic energy
+<!-- SOLUTION. The angle 0 = 01+02, where 01 and O2 are the angles defined by formula `16.5` -->
-of any one of the resulting particles. To determine the limit, we consider
+<!-- (see Problem 1), and it is simplest to calculate the tangent of 0. A consideration of the extrema -->
-the system formed by all these particles except the one concerned (whose
+<!-- of the resulting expression gives the following ranges of 0, depending on the relative magni- -->
-mass is M1, say), and denote the "internal energy" of that system by Ei'.
+<!-- tudes of V, V10 and V20 (for definiteness, we assume V20 > V10): 0 0 77 if V10 -->
-Then the kinetic energy of the particle M1 is, by (16.1) and (16.2),
+<!-- < 0 < TT if V V V10, 0 < Oo if V > U20. The value of Oo is given by -->
-T10 = 2/2m1 = (M-m1)(E:-E1-EiM where M is the mass of the
+<!-- sin = -->
-primary particle. It is evident that T10 has its greatest possible value
-when E/' is least. For this to be so, all the resulting particles except M1
-must be moving with the same velocity. Then Ei is simply the sum of their
-internal energies, and the difference E;-E-E; is the disintegration
-energy E. Thus
-T10,max = (M - M1) E M.
-(16.9)
-PROBLEMS
-PROBLEM 1. Find the relation between the angles 01, O2 (in the L system) after a disintegra-
-tion into two particles.
-SOLUTION. In the C system, the corresponding angles are related by 010 = n-020. Calling
-010 simply Oo and using formula (16.5) for each of the two particles, we can put
-7+10 cos Oo = V10 sin Oo cot 01, V-V20 cos Oo = V20 sin Oo cot O2. From these two
-equations we must eliminate Oo. To do so, we first solve for cos Oo and sin Oo, and then
-44
-Collisions Between Particles

1/17-elastic-collisions.md

@@ -1,165 +1,98 @@
 ---
-title: 17-elastic-collisions
+title: 17 Elastic collisions
 ---
-form the sum of their squares, which is unity. Since V10/2.20 = m2/m1, we have finally, using
+
-(16.2),
+A collision between two particles is said to be *elastic* if it involves no change in their internal state. Accordingly, when the law of conservation of energy is applied to such a collision, the internal energy of the particles may be neglected.
-(m2/m1) sin202+(m1/m2) sin201-2 sin O1 sin O2 cos(A1+6
+
-PROBLEM 2. Find the angular distribution of the resulting particles in the L system.
+The collision is most simply described in a frame of reference in which the centre of mass of the two particles is at rest (the $C$ system). As in `16`, we distinguish by the suffix $0$ the values of quantities in that system. The velocities of the particles before the collision are related to their velocities $\v{v}_1$ and $\v{v}_2$ in the laboratory system by $\v{v}_{10} = m_2\v{v}/(m_1+m_2), \v{v}_{20} = -m_1\v{v}/(m_1+m_2)$, where $\v{v} = \v{v}_1-\v{v}_2$; see `13.2`.
-SOLUTION. When vo > V, we substitute (16.6), with the plus sign of the radical, in (16.7),
+
-obtaining
+Because of the law of conservation of momentum, the momenta of the two particles remain equal and opposite after the collision, and are also unchanged in magnitude, by the law of conservation of energy. Thus, in the $C$ system the collision simply rotates the velocities, which remain opposite in direction and unchanged in magnitude. If we denote by $\v{n}_0$ a unit vector in the direction of the velocity of the particle $m_1$ after the collision, then the velocities of the two particles after the collision (distinguished by primes) are
-(0  .
+
-When vo < V, both possible relations between Oo and 0 must be taken into account. Since,
+```load
-when 0 increases, one value of Oo increases and the other decreases, the difference (not the
+17.1
-sum) of the expressions d cos Oo with the two signs of the radical in (16.6) must be taken.
+```
-The result is
+
-(0    max).
+In order to return to the $L$ system, we must add to these expressions the velocity $\v{V}$ of the centre of mass. The velocities in the $L$ system after the collision are therefore
-PROBLEM 3. Determine the range of possible values of the angle 0 between the directions
+
-of motion of the two resulting particles in the L system.
+```load
-SOLUTION. The angle 0 = 01+02, where 01 and O2 are the angles defined by formula (16.5)
+17.2
-(see Problem 1), and it is simplest to calculate the tangent of 0. A consideration of the extrema
+```
-of the resulting expression gives the following ranges of 0, depending on the relative magni-
+
-tudes of V, V10 and V20 (for definiteness, we assume V20 > V10): 0 0 77 if V10
+No further information about the collision can be obtained from the laws of conservation of momentum and energy. The direction of the vector $\v{n}_0$ depends on the law of interaction of the particles and on their relative position during the collision.
-< 0 < TT if V V V10, 0 < Oo if V > U20. The value of Oo is given by
+
-sin =
+The results obtained above may be interpreted geometrically. Here it is more convenient to use momenta instead of velocities. Multiplying equations `17.2` by $m_1$ and $m_2$ respectively, we obtain
-§17. Elastic collisions
+
-A collision between two particles is said to be elastic if it involves no change
+```load
-in their internal state. Accordingly, when the law of conservation of energy
+17.3
-is applied to such a collision, the internal energy of the particles may be
+```
-neglected.
+
-The collision is most simply described in a frame of reference in which the
+where $m = m_1m_2/(m_1+m_2)$ is the reduced mass. We draw a circle of radius $mv$ and use the construction shown in `fig15`. If the unit vector $\v{n}_0$ is along $OC$, the vectors $AC$ and $CB$ give the momenta $\v{p}_1'$ and $\v{p}_2'$ respectively.  When $\v{p}_1$ and $\v{p}_2$ are given, the radius of the circle and the points $A$ and $B$ are fixed, but the point $C$ may be anywhere on the circle.
-centre of mass of the two particles is at rest (the C system). As in $16, we
+
-distinguish by the suffix 0 the values of quantities in that system. The velo-
+```fig
-cities of the particles before the collision are related to their velocities V1 and
+15
-V2 in the laboratory system by V10 = M2V/(m1+m2), V20 = -m1V/(m1+m2),
+```
-where V = V1-V2; see (13.2).
+
-Because of the law of conservation of momentum, the momenta of the two
+Let us consider in more detail the case where one of the particles ($m_2$, say) is at rest before the collision. In that case the distance $OB = m_2p_1/(m_1+m_2) = mv$ is equal to the radius, i.e. $B$ lies on the circle. The vector $AB$ is equal to the momentum $\v{p}_1$ of the particle $m_1$ before the collision. The point A lies inside or outside the circle, according as $m_1\lt m_2$ or $m_1\gt m_2$. The corresponding diagrams are shown in `Figs. 16a, b`. The angles $\theta_1$ and $\theta_2$ in these diagrams are the angles between the directions of motion after the collision and the direction of impact (i.e. of $\v{p}_1$). The angle at the centre, denoted by $\chi$, which gives the direction of $\v{n}_0$, is the angle through which the direction of motion of $m_1$ is turned in the $C$ system. It is evident from the figure that $\theta_1$ and $\theta_2$ can be expressed in terms of $\chi$ by
-particles remain equal and opposite after the collision, and are also unchanged
+
-in magnitude, by the law of conservation of energy. Thus, in the C system
+```load
-the collision simply rotates the velocities, which remain opposite in direction
+17.4
-and unchanged in magnitude. If we denote by no a unit vector in the direc-
+```
-tion of the velocity of the particle M1 after the collision, then the velocities
+
-of the two particles after the collision (distinguished by primes) are
+```fig
-V10' m20120/(m1+m2), V20' = -mjono/(m1+m2).
+16
-(17.1)
+```
-§17
+
-Elastic collisions
+We may give also the formulae for the magnitudes of the velocities of the two particles after the collision, likewise expressed in terms of $\chi$:
-45
+
-In order to return to the L system, we must add to these expressions the
+```load
-velocity V of the centre of mass. The velocities in the L system after the
+17.5
-collision are therefore
+```
-V1' =
+
-(17.2)
+The sum $\theta_1 + \theta_2$ is the angle between the directions of motion of the particles after the collision. Evidently $\theta_1 + \theta_2 > \mfrac{1}{2}\pi$ if $m_1\lt m_2$, and $\theta_1+\theta_2\lt\mfrac{1}{2}\pi$ if $m_1\gt m_2$.
-V2' =
+
-No further information about the collision can be obtained from the laws
+When the two particles are moving afterwards in the same or in opposite directions (head-on collision), we have $\chi=\pi$, i.e. the point $C$ lies on the diameter through $A$, and is on $OA$ (`Fig. 16b` ; $\v{p}_1'$ and $\v{p}_2'$ in the same direction) or on $OA$ produced (`Fig. 16a`; $\v{p}_1'$ and $\v{p}_2'$ in opposite directions).
-of conservation of momentum and energy. The direction of the vector no
+
-depends on the law of interaction of the particles and on their relative position
-during the collision.
-The results obtained above may be interpreted geometrically. Here it is
-more convenient to use momenta instead of velocities. Multiplying equations
-(17.2) by M1 and M2 respectively, we obtain
-(17.3)
-P2' muno+m2(p1+p2)/(m1+m2)
-where m = m1m2/(m1+m2) is the reduced mass. We draw a circle of radius
-mv and use the construction shown in Fig. 15. If the unit vector no is along
-OC, the vectors AC and CB give the momenta P1' and P2' respectively.
-When p1 and P2 are given, the radius of the circle and the points A and B
-are fixed, but the point C may be anywhere on the circle.
-C
-p'
-no
-P'2
-B
-A
-FIG. 15
-Let us consider in more detail the case where one of the particles (m2, say) is
-at rest before the collision. In that case the distance OB = m2p1/(m1+m2) = mv
-is equal to the radius, i.e. B lies on the circle. The vector AB is equal to the
-momentum P1 of the particle M1 before the collision. The point A lies inside
-or outside the circle, according as M1 < M2 or M1 > M2. The corresponding
-46
-Collisions Between Particles
-§17
-diagrams are shown in Figs. 16a, b. The angles 01 and O2 in these diagrams
-are the angles between the directions of motion after the collision and the
-direction of impact (i.e. of P1). The angle at the centre, denoted by X, which
-gives the direction of no, is the angle through which the direction of motion
-of m1 is turned in the C system. It is evident from the figure that 01 and O2
-can be expressed in terms of X by
-(17.4)
-C
-p'
-P2
-pi
-P2
-0
-max
-10,
-X
-O2
-O2
-B
-B
-A
-0
-A
-Q
-0
-(a) m < m2
-(b) m, m m m
-AB=p : AO/OB= m/m2
-FIG. 16
-We may give also the formulae for the magnitudes of the velocities of the
-two particles after the collision, likewise expressed in terms of X:
-ib
-(17.5)
-The sum A1 + O2 is the angle between the directions of motion of the
-particles after the collision. Evidently 01 + O2 > 1/1 if M1 < M2, and 01+O2 < 1st
-if M1 > M2.
-When the two particles are moving afterwards in the same or in opposite
-directions (head-on collision), we have X=TT, i.e. the point C lies on the
-diameter through A, and is on OA (Fig. 16b ; P1' and P2' in the same direc-
-tion) or on OA produced (Fig. 16a; P1' and P2' in opposite directions).
 In this case the velocities after the collision are
-(17.6)
+
-This value of V2' has the greatest possible magnitude, and the maximum
+```load
-§17
+17.6
-Elastic collisions
+```
-47
+
-energy which can be acquired in the collision by a particle originally at rest
+This value of $\v{v}_2'$ has the greatest possible magnitude, and the maximum energy which can be acquired in the collision by a particle originally at rest is therefore
-is therefore
+
-(17.7)
+```load
-where E1 = 1M1U12 is the initial energy of the incident particle.
+17.7
-If M1 < M2, the velocity of M1 after the collision can have any direction.
+```
-If M1 > M2, however, this particle can be deflected only through an angle
+
-not exceeding Omax from its original direction; this maximum value of A1
+where $E_1 = \mfrac{1}{2}m_1{v_1}^2$ is the initial energy of the incident particle.
-corresponds to the position of C for which AC is a tangent to the circle
+
-(Fig. 16b). Evidently
+If $m_1\lt m_2$, the velocity of $m_1$ after the collision can have any direction.  If $m_1 > m_2$, however, this particle can be deflected only through an angle not exceeding $\theta_\max$ from its original direction; this maximum value of $\theta_1$ corresponds to the position of $C$ for which $AC$ is a tangent to the circle `Fig. 16b`. Evidently
-sin Omax = OC|OA = M2/M1.
+
-(17.8)
+```load
-The collision of two particles of equal mass, of which one is initially at
+17.8
-rest, is especially simple. In this case both B and A lie on the circle (Fig. 17).
+```
-C
+
-p'
+The collision of two particles of equal mass, of which one is initially at rest, is especially simple. In this case both $B$ and $A$ lie on the circle `Fig. 17`.
-P2
+
-Q2
+```fig
-B
+17
-A
+```
-0
+
-FIG. 17
 Then
-01=1x,
+
-A2 = 1(-x),
+```load
-(17.9)
+17.9
-12
+```
-=
+```load
-(17.10)
+17.10
+```
+
 After the collision the particles move at right angles to each other.
-PROBLEM
+
-Express the velocity of each particle after a collision between a moving particle (m1) and
+<!-- PROBLEM -->
-another at rest (m2) in terms of their directions of motion in the L system.
+<!-- Express the velocity of each particle after a collision between a moving particle (m1) and -->
-SOLUTION. From Fig. 16 we have P2' = 20B cos O2 or V2' = 2v(m/m2) cos O2. The momen-
+<!-- another at rest (m2) in terms of their directions of motion in the $L$ system. -->
-tum P1' = AC is given by OC2 = AO2tp12-2AO.p cos or
+<!-- SOLUTION. From Fig. 16 we have P2' = 20B cos O2 or V2' = 2v(m/m2) cos O2. The momen- -->
-Hence
+<!-- tum P1' = AC is given by OC2 = AO2tp12-2AO.p cos or -->
-for m1 > M2 the radical may have either sign, but for M2 > M1 it must be taken positive.
+<!-- Hence -->
-48
+<!-- for m1 > M2 the radical may have either sign, but for M2 > M1 it must be taken positive. -->
-Collisions Between Particles

1/equations/16.1.tex

@@ -0,0 +1 @@
+\epsilon=E_i-E_{1i}-E_{2i}

1/equations/16.2.tex

@@ -0,0 +1 @@
+\epsilon=\mfrac{1}{2}{p_0}^2\left(\frac{1}{m_1}+\frac{1}{m_2}\right)=\frac{{p_0}^2}{2m}

1/equations/16.3.tex

@@ -0,0 +1 @@
+v^2+V^2-2vV\cos\theta={v_0}^2

1/equations/16.4.tex

@@ -0,0 +1 @@
+\sin\theta_\max=v_0/V

1/equations/16.5.tex

@@ -0,0 +1 @@
+\tan\theta=v_0\sin\theta_0/(v_0\cos\theta_0+V)

1/equations/16.6.tex

@@ -0,0 +1 @@
+\cos\theta_0=-\frac{V}{v_0}\sin^2\theta\pm\cos\theta\sqrt{1-\frac{V^2}{{v_0}^2}\sin^2\theta}

1/equations/16.7.tex

@@ -0,0 +1 @@
+\mfrac{1}{2}\sin\theta_0\dd{\theta_0}

1/equations/16.8.tex

@@ -0,0 +1 @@
+(1/2mv_0V)\dd{T}

1/equations/16.9.tex

@@ -0,0 +1 @@
+T_{10,\max}=(M-m_1)\epsilon/M

1/equations/17.1.tex

@@ -0,0 +1,2 @@
+\v{v}_{10}'=m_2v\v{n}_0/(m_1+m_2),\qquad 
+\v{v}_{20}'=-m_1v\v{n}_0/(m_1+m_2)

1/equations/17.10.tex

@@ -0,0 +1,2 @@
+v_1'=v\cos\mfrac{1}{2}\chi,\qquad
+v_2'=v\sin\mfrac{1}{2}\chi

1/equations/17.2.tex

@@ -0,0 +1,4 @@
+\begin{align}
+	\v{v}_1' &= m_2v\v{n}_0/(m_1+m_2)+(m_1\v{v}_1+m_2\v{v}_2)/(m_1+m_2), \\
+	\v{v}_1' &= -m_1v\v{n}_0/(m_1+m_2)+(m_1\v{v}_1+m_2\v{v}_2)/(m_1+m_2).
+\end{align}

1/equations/17.3.tex

@@ -0,0 +1,4 @@
+\begin{align}
+	\v{p}_1' &= mv\v{n}_0+m_1(\v{p}_1+\v{p}_2)/(m_1+m_2), \\
+	\v{p}_2' &= -mv\v{n}_0+m_2(\v{p}_1+\v{p}_2)/(m_1+m_2),
+\end{align}

1/equations/17.4.tex

@@ -0,0 +1 @@
+\tan\theta_1=\frac{m_2\sin\chi}{m_1+m_2\cos\chi},\qquad \theta_2=\mfrac{1}{2}(\pi-\chi)

1/equations/17.5.tex

@@ -0,0 +1,2 @@
+v_1'=\frac{\sqrt({m_1}^2+{m_2}^2+2m_1m_2\cos\chi)}{m_1+m_2}v,\qquad
+v_2'=\frac{2m_1v}{m_1+m_2}\sin\mfrac{1}{2}\chi

1/equations/17.6.tex

@@ -0,0 +1,2 @@
+\v{v}_1'=\frac{m_1-m_2}{m_1+m_2}\v{v},\qquad
+\v{v}_2'=\frac{2m_1}{m_1+m_2}\v{v},

1/equations/17.7.tex

@@ -0,0 +1 @@
+{E_2}'_\max=\mfrac{1}{2}m_2{{v_2}'_\max}^2=\frac{4 m_1 m_2}{(m_1+m_2)^2}E_1,

1/equations/17.8.tex

@@ -0,0 +1 @@
+\sin\theta_\max=OC/OA=m_2/m_1

1/equations/17.9.tex

@@ -0,0 +1 @@
+\theta_1=\mfrac{1}{2}\chi,\qquad\theta_2=\mfrac{1}{2}(\pi-\chi),

1/index.md

@@ -24,14 +24,14 @@ III.  INTEGRATION OF THE EQUATIONS OF MOTION
 13. [The reduced mass](13-the-reduced-mass.html)
 14. [Motion in a central field](14-motion-in-a-central-field.html)
 15. [Kepler's problem](15-keplers-problem.html)
+IV.  COLLISION BETWEEN PARTICLES
+16. [Disintegration of particles](16-disintegration-of-particles.html)
+17. [Elastic collisions](17-elastic-collisions.html)
 
 <span style="background-color: yellow; color: white: width: 100%;">
 🚧 WORK IN PROGRESS BELOW THIS POINT 🚧
 </span>
 
-IV.  COLLISION BETWEEN PARTICLES
-16. [Disintegration of particles](16-disintegration-of-particles.html)
-17. [Elastic collisions](17-elastic-collisions.html)
 18. [Scattering](18-scattering.html)
 19. [Rutherford's formula](19-rutherfords-formula.html)
 20. [Small-angle scattering](20-small-angle-scattering.html)

tools/process-textract.sh

@@ -0,0 +1,8 @@
+#!/bin/bash
+
+FILE=$1
+sed -i 's/^(\([0-9]\+.[0-9]\+\))$/\n```load\n\1\n```\n/g' $FILE
+sed -i 's/(\([0-9]\+.[0-9]\+\))/`\1`/g' $FILE
+
+sed -i 's/ L / $L$ /g' $FILE
+sed -i 's/ C / $C$ /g' $FILE