Chapter 15 Collision Theory - MIT OpenCourseWare
Chapter 15 Collision Theory15.1 Introduction. 115.2 Reference Frames and Relative Velocities. 115.2.1 Relative Velocities . 315.2.2 Center-of-mass Reference Frame . 415.2.3 Kinetic Energy in the Center-of-Mass Reference Frame . 515.2.4 Change of Kinetic Energy and Relatively Inertial Reference Frames. 515.3 Characterizing Collisions . 715.4 One-Dimensional Collisions Between Two Objects. 715.4.1 One Dimensional Elastic Collision in Laboratory Reference Frame . 715.4.2 One-Dimensional Collision Between Two Objects – Center-of-MassReference Frame . 1115.5 Worked Examples. 12Example 15.1 Elastic One-Dimensional Collision Between Two Objects . 12Example 15.2 The Dissipation of Kinetic Energy in a Completely InelasticCollision Between Two Objects . 13Example 15.3 Bouncing Superballs . 1415.6 Two Dimensional Elastic Collisions . 1715.6.1 Two-dimensional Elastic Collision in Laboratory Reference Frame. 17Example 15.5 Elastic Two-dimensional collision of identical particles. 20Example 15.6 Two-dimensional elastic collision between particles of equal mass. 21Example 15.7 Two dimensional collision between particles of unequal mass . 2215.7 Two-Dimensional Collisions in Center-of-Mass Reference Frame . 2415.7.1 Two-Dimensional Collision in Center-of-Mass Reference Frame. 2415.7.2 Scattering in the Center-of-Mass Reference Frame . 25Example 15.8 Scattering in the Lab and CM Frames . 26
Chapter 15 Collision TheoryDespite my resistance to hyperbole, the LHC [Large Hadron Collider]belongs to a world that can only be described with superlatives. It is notmerely large: the LHC is the biggest machine ever built. It is not merelycold: the 1.9 kelvin (1.9 degrees Celsius above absolute zero) temperaturenecessary for the LHC’s supercomputing magnets to operate is the coldestextended region that we know of in the universe—even colder than outerspace. The magnetic field is not merely big: the superconducting dipolemagnets generating a magnetic field more than 100,000 times stronger thanthe Earth’s are the strongest magnets in industrial production ever made.And the extremes don’t end there. The vacuum inside the proton-containingtubes, a 10 trillionth of an atmosphere, is the most complete vacuum overthe largest region ever produced. The energy of the collisions are the highestever generated on Earth, allowing us to study the interactions that occurredin the early universe the furthest back in time.1Lisa Randall15.1 IntroductionWhen discussing conservation of momentum, we considered examples in which twoobjects collide and stick together, and either there are no external forces acting in somedirection (or the collision was nearly instantaneous) so the component of the momentumof the system along that direction is constant. We shall now study collisions betweenobjects in more detail. In particular we shall consider cases in which the objects do notstick together. The momentum along a certain direction may still be constant but themechanical energy of the system may change. We will begin our analysis by consideringtwo-particle collision. We introduce the concept of the relative velocity between twoparticles and show that it is independent of the choice of reference frame. We then showthat the change in kinetic energy only depends on the change of the square of the relativevelocity and therefore is also independent of the choice of reference frame. We will thenstudy one- and two-dimensional collisions with zero change in potential energy. Inparticular we will characterize the types of collisions by the change in kinetic energy andanalyze the possible outcomes of the collisions.15.2 Reference Frames and Relative Velocities We shall recall our definition of relative inertial reference frames. Let R be thevector from the origin of frame S to the origin of reference frame S′ . Denote the1Randall, Lisa, Knocking on Heaven's Door: How Physics and Scientific Thinking Illuminate the Universeand the Modern World, Ecco, 2011.15-1
position vector of the j th particle with respect to the origin of reference frame S by r jand similarly, denote the position vector of the j th particle with respect to the origin of reference frame S′ by r ′j (Figure 15.1).j th particlerjrjRSSFigure 15.1 Position vector of j th particle in two reference frames.The position vectors are related by r j r′j R .(15.2.1)The relative velocity (call this the boost velocity) between the two reference frames isgiven by dR.(15.2.2)V dtAssume the boost velocity between the two reference frames is constant. Then, therelative acceleration between the two reference frames is zero, dVA 0.dt(15.2.3)When Eq. (15.2.3) is satisfied, the reference frames S and S′ are called relativelyinertial reference frames.Suppose the j th particle in Figure 15.1 is moving; then observers in differentreference frames will measure different velocities. Denote the velocity of j th particle in frame S by v j dr j / dt , and the velocity of the same particle in frame S′ by v′j dr ′j / dt . Taking derivative, the velocities of the particles in two different referenceframes are related according to v j v′j V .(15.2.4)15-2
15.2.1 Relative VelocitiesConsider two particles of masses m1 and m2 interacting via some force (Figure 15.2).Figure 15.2 Two interacting particlesChoose a coordinate system (Figure 15.3) in which the position vector of body 1 is given by r1 and the position vector of body 2 is given by r2 . The relative position of body 1 with respect to body 2 is given by r1,2 r1 r2 .Figure 15.3 Coordinate system for two bodies. During the course of the interaction, body 1 is displaced by dr1 and body 2 is displaced by dr2 , so the relative displacement of the two bodies during the interaction is given by dr1,2 dr1 dr2 . The relative velocity between the particles is dr1,2 dr1 dr2 v1 v 2 .v1,2 dtdt dt(15.2.5)We shall now show that the relative velocity between the two particles is independent ofthe choice of reference frame providing that the reference frames are relatively inertial. ′ in reference frame S′ can be determined from using Eq.The relative velocity v12(15.2.4) to express Eq. (15.2.5) in terms of the velocities in the reference frame S′ , v1, 2 v1 v 2 (v1′ V) (v ′2 V) v1′ v′2 v1,′ 2(15.2.6)15-3
and is equal to the relative velocity in frame S .For a two-particle interaction, the relative velocity between the twovectors is independent of the choice of relatively inertial reference frames.15.2.2 Center-of-mass Reference Frame Let rcm be the vector from the origin of frame S to the center-of-mass of thesystem of particles, a point that we will choose as the origin of reference frame Scm ,called the center-of-mass reference frame. Denote the position vector of the j th particle with respect to origin of reference frame S by r j and similarly, denote the position vector of the j th particle with respect to origin of reference frame Scm by r′j (Figure15.4).j th particlerjrjrcmS!ScmFigure 15.4 Position vector of j th particle in the center-of-mass reference frame.The position vector of the j th particle in the center-of-mass frame is then given by r′j r j rcm .(15.2.7)The velocity of the j th particle in the center-of-mass reference frame is then given by v′j v j v cm .(15.2.8)There are many collision problems in which the center-of-mass reference frame is themost convenient reference frame to analyze the collision.Consider a system consisting of two particles, which we shall refer to as particle 1 andparticle 2. We can use Eq. (15.2.8) to determine the velocities of particles 1 and 2 in thecenter-of-mass, m2µ m1v1 m2 v 2v1′ v1 v cm v1 ( v1, v 2 ) v1, 2 .m1 m2m1m1 m2(15.2.9)15-4
where v12 v1 v 2 is the relative velocity of particle 1 with respect to particle 2 . Asimilar result holds for particle 2 : m1µ m1v1 m2 v 2 v′2 v 2 v cm v 2 ( v1 v 2 ) v1, 2 .m1 m2m2m1 m2(15.2.10)The momentum of the system the center-of-mass reference frame is zero as we expect, m1v1′ m2 v′2 µ v12 µ v12 0 .(15.2.11)15.2.3 Kinetic Energy in the Center-of-Mass Reference FrameThe kinetic energy in the center of mass reference frame is given byK cm 1 1 m v ′ v ′ m v′ v ′ .2 1 1 1 2 2 2 2(15.2.12)We now use Eqs. (15.2.9) and (15.2.10) to rewrite the kinetic energy in terms of the relative velocity v12′ v1′ v ′2 ,K cm 1 µ µ 1 µ µ vv vm m v2 1 m1 1, 2 m1 1, 2 2 2 m2 1, 2 m2 1, 2 1 11 1 µ 2 v1, 2 v1, 2 µv1, 2 22 m1 m2 2.(15.2.13)where we used the fact that we defined the reduced mass by1 11. µ m1 m2(15.2.14)15.2.4 Change of Kinetic Energy and Relatively Inertial Reference FramesThe kinetic energy of the two particles in reference frame S is given byKS 11m1v12 m2 v2 2 .22(15.2.15)We can take the scalar product of Eq. (15.2.8) to rewrite Eq. (15.2.15) as15-5
11 m1 (v1′ v cm ) (v1′ v cm ) m2 (v′2 v cm ) (v′2 v cm )22.1 2 12 12 m1v1′ m2 v2′ (m1 m2 )vcm (m1v1′ m2 v′2 ) v cm222KS (15.2.16)The last term is zero due to the fact that the momentum of the system in the center ofmass reference frame is zero (Eq. (15.2.11)). Therefore Eq. (15.2.16) becomesKS 111m1v1′ 2 m2 v2′ 2 (m1 m2 )vcm 2 .222(15.2.17)The first two terms correspond to the kinetic energy in the center of mass frame, thus thekinetic energies in the two reference frames are related by1K S K cm (m1 m2 )vcm 2 .2(15.2.18)We now use Eq. (15.2.13) to rewrite Eq. (15.2.18) asKS 11µv1, 2 2 (m1 m2 )vcm 222(15.2.19)Even though kinetic energy is a reference frame dependent quantity, because the secondterm in Eq. (15.2.19) is a constant, the change in kinetic energy in either reference frameis equal to1ΔK µ v1, 2 2 v1, 2 2 .(15.2.20)fi2(() ( ))This generalizes to any two relatively inertial reference frames because the relativevelocity is a reference frame independent quantity,the change in kinetic energy is independent of the choice of relativelyinertial reference frames.We showed in Appendix 13A that when two particles of masses m1 and m2 interact, thework done by the interaction force is equal toW ((1µ v1, 2 22) (v ) ) .2f1, 2i(15.2.21)Hence we explicitly verified that for our two-particle systemW ΔK sys .(15.2.22)15-6
15.3 Characterizing CollisionsIn a collision, the ratio of the magnitudes of the initial and final relative velocities iscalled the coefficient of restitution and denoted by the symbol e ,e vB.vA(15.2.23)If the magnitude of the relative velocity does not change during a collision, e 1, thenthe change in kinetic energy is zero, (Eq. (15.2.21)). Collisions in which there is nochange in kinetic energy are called elastic collisions,ΔK 0, elastic collision .(15.2.24)If the magnitude of the final relative velocity is less than the magnitude of the initialrelative velocity, e 1, then the change in kinetic energy is negative. Collisions in whichthe kinetic energy decreases are called inelastic collisions,ΔK 0, inelastic collision .(15.2.25)If the two objects stick together after the collision, then the relative final velocity is zero,e 0 . Such collisions are called totally inelastic. The change in kinetic energy can befound from Eq. (15.2.21),11 m1m2 2ΔK µ v 2A v , totally inelastic collision .22 m1 m2 A(15.2.26)If the magnitude of the final relative velocity is greater than the magnitude of the initialrelative velocity, e 1, then the change in kinetic energy is positive. Collisions in whichthe kinetic energy increases are called superelastic collisions,ΔK 0, superelastic collision .(15.2.27)15.4 One-Dimensional Collisions Between Two Objects15.4.1 One Dimensional Elastic Collision in Laboratory Reference FrameConsider a one-dimensional elastic collision between two objects moving in the x direction. One object, with mass m1 and initial x -component of the velocity v1x,i ,collides with an object of mass m2 and initial x -component of the velocity v2 x,i . Thescalar components v1x,i and v1x,i can be positive, negative or zero. No forces other thanthe interaction force between the objects act during the collision. After the collision, the15-7
final x -component of the velocities are v1x, f and v2 x, f . We call this reference frame the“laboratory reference frame”.Figure 15.5 One-dimensional elastic collision, laboratory reference frameFor the collision depicted in Figure 15.5, v1x,i 0 , v2 x,i 0 , v1x, f 0 , and v2 x, f 0 .Because there are no external forces in the x -direction, momentum is constant in the x direction. Equating the momentum components before and after the collision gives therelationm1v1x, i m2 v2 x, i m1v1x, f m2 v2 x, f .(15.3.1)Because the collision is elastic, kinetic energy is constant. Equating the kinetic energybefore and after the collision gives the relation111122m1v1x,i m2 v22x,i m1v1x, m2 v22x, ff2222(15.3.2)Rewrite these Eqs. (15.3.1) and (15.3.2) asm1 (v1x,i v1x, f ) m2 (v2 x, f v2 x,i )(15.3.3)22m1 (v1x,i v1x,) m2 (v22x, f v22x,i ) .f(15.3.4)Eq. (15.3.4) can be written asm1 (v1x,i v1x, f )(v1x,i v1x, f ) m2 (v2 x, f v2 x,i )(v2 x, f v2 x,i ) .(15.3.5)Divide Eq. (15.3.4) by Eq. (15.3.3), yieldingv1x,i v1x, f v2 x,i v2 x, f .(15.3.6)v1x,i v2 x,i v2 x, f v1x, f .(15.3.7)Eq. (15.3.6) may be rewritten asRecall that the relative velocity between the two objects is defined to be v rel v1,2 v1 v 2 .(15.3.8)15-8
where we used the superscript “rel” to remind ourselves that the velocity is a relativevelocity (and to simplify our notation). Thus vxrel,i v1x,i v2 x,i is the initial x -componentof the relative velocity, and vxrel, f v1x, f v2 x, f is the final x -component of the relativevelocity. Therefore Eq. (15.3.7) states that during the interaction the initial relativevelocity is equal to the negative of the final relative velocity v rel v relf , (1 dimensional energy-momentum prinicple) .i(15.3.9)Consequently the initial and final relative speeds are equal. We shall call this relationshipbetween the relative initial and final velocities the one-dimensional energy-momentumprinciple because we have combined these two principles to realize this result. Theenergy-momentum principle is independent of the masses of the colliding particles.Although we derived this result explicitly, we have already shown that the change inkinetic energy for a two-particle interaction (Eq. (15.2.20)), in our simplified notation isgiven by1ΔK µ ((v rel )2f (v rel )2i )(15.3.10)2Therefore for an elastic collision where ΔK 0 , the square of the relative speed remainsconstant(v rel )2f (v rel )2i .(15.3.11)For a one-dimensional collision, the magnitude of the relative speed remains constant butthe direction changes by 180 .We can now solve for the final x -component of the velocities, v1x, f and v2 x, f , asfollows. Eq. (15.3.7) may be rewritten asv2 x, f v1x, f v1x,i v2 x,i .(15.3.12)Now substitute Eq. (15.3.12) into Eq. (15.3.1) yieldingm1v1x,i m2 v2 x,i m1v1x, f m2 (v1x, f v1x,i v2 x,i ) .(15.3.13)Solving Eq. (15.3.13) for v1x, f involves some algebra and yieldsv1x, f m1 m22 m2v1x,i v .m1 m2m1 m2 2 x,i(15.3.14)15-9
To find v2 x, f , rewrite Eq. (15.3.7) asv1x, f v2 x, f v1x,i v2 x,i .(15.3.15)Now substitute Eq. (15.3.15) into Eq. (15.3.1) yieldingm1v1x,i m2 v2 x,i m1 (v2 x, f v1x,i v2 x,i )v1x, f m2 v2 x, f .(15.3.16)We can solve Eq. (15.3.16) for v2 x, f and determine thatv2 x, f v2 x,im2 m12 m1. v1x,im2 m1m2 m1(15.3.17)Consider what happens in the limits m1 m2 in Eq. (15.3.14). Thenv1x, f v1x,i 2mv ;m1 2 2 x,i(15.3.18)the more massive object’s velocity component is only slightly changed by an amountproportional to the less massive object’s x -component of momentum. Similarly, the lessmassive object’s final velocity approachesv2 x, f v2 x,i 2v1x,i v1x,i v1x,i v2 x,i .(15.3.19)relv2 x, f v1x,i v1x,i v2 x,i vx,i.(15.3.20)We can rewrite this asi.e. the less massive object “rebounds” with the same speed relative to the more massiveobject which barely changed its speed.If the objects are identical, or have the same mass, Eqs. (15.3.14) and (15.3.17) becomev1x, f v2 x,i , v2 x, f v1x,i ;(15.3.21)the objects have exchanged x -components of velocities, and unless we could somehowdistinguish the objects, we might not be able to tell if there was a collision at all.15-10
15.4.2 One-Dimensional Collision Between Two Objects – Center-of-Mass ReferenceFrameWe analyzed the one-dimensional elastic collision (Figure 15.5) in Section 15.4.1 in thelaboratory reference frame. Now let’s view the collision from the center-of-mass (CM)frame. The x -component of velocity of the center-of-mass isvx,cm m1 v1x,i m2 v2 x,im1 m2.(15.3.22)With respect to the center-of-mass, the x -components of the velocities of the objects arev1′x,i v1x,i vx,cm (v1x,i v2 x,i )v2′ x,i v2 x,i vx,cmm2m1 m2m1 (v2 x,i v1x,i ).m1 m2(15.3.23)In the CM frame the momentum of the system is zero before the collision and hence themomentum of the system is zero after the collision. For an elastic collision, the only wayfor both momentum and kinetic energy to be the same before and after the collision iseither the objects have the same velocity (a miss) or to reverse the direction of thevelocities as shown in Figure 15.6.Figure 15.6 One-dimensional elastic collision in center-of-mass reference frameIn the CM frame, the final x -components of the velocities arev1′x, f v1′x,i (v2 x,i v1x,i )v2′ x, fm2m1 m2m1 v2′ x,i (v2 x,i v1x,i ).m1 m2(15.3.24)The final x -components of the velocities in the “laboratory frame” are then given by15-11
v1x, f v1′x, f vx,cm (v2 x,i v1x,i ) v1x,im v m2 v2 x,im2 1 1x,im1 m2m1 m2(15.3.25)m1 m22 m2 v2 x,im1 m2m1 m2as in Eq. (15.3.14) and a similar calculation reproduces Eq. (15.3.17).15.5 Worked ExamplesExample 15.1 Elastic One-Dimensional Collision Between Two Objectsîîv1,i v1,x,i îv 2,i 012v1, f v1,x, f î1initial statem2 2m1v 2, f v2,x, f î final statem2 2m12Figure 15.7 Elastic collision between two non-identical cartsConsider the elastic collision of two carts along a track; the incident cart 1 has mass m1and moves with initial speed v1,i . The target cart has mass m2 2 m1 and is initially atrest, v2,i 0 , (Figure 15.7). Immediately after the collision, the incident cart has finalspeed v1, f and the target cart has final speed v2, f . Calculate the final x -component of thevelocities of the carts as a function of the initial speed v1,i .Solution The momentum flow diagram for the objects before (initial state) and after(final state) the collision are shown in Figure 15.7. We can immediately use our resultsabove with m2 2 m1 and v2,i 0 . The final x -component of velocity of cart 1 is givenby Eq. (15.3.14), where we use v1x,i v1,i15-12
1v1x, f v1,i .3(15.4.1)The final x -component of velocity of cart 2 is given by Eq. (15.3.17)v2 x, f 2v .3 1,i(15.4.2)Example 15.2 The Dissipation of Kinetic Energy in a Completely Inelastic CollisionBetween Two Objectsîîv1,iv 2,i 012initial statevffinal state12Figure 15.7b Inelastic collision between two non-identical cartsAn incident cart of mass m1 and initial speed v1, i collides completely inelastically with acart of mass m2 that is initially at rest (Figure 15.7b). There are no external forces actingon the objects in the direction of the collision. Find ΔK / K initial (K final K initial ) / K initial .Solution: In the absence of any net force on the system consisting of the two carts, themomentum after the collision will be the same as before the collision. After the collisionthe carts will move in the direction of the initial velocity of the incident cart with acommon speed v f found from applying the momentum conditionm1v1, i (m1 m2 )v f vf m1v1, i .m1 m2(15.4.3)The initial relative speed is virel v1, i . The final relative velocity is zero because the cartsstick together so using Eq. (15.2.26), the change in kinetic energy is1 m1m2 21ΔK µ (virel )2 v1, i .22 m1 m2(15.4.4)15-13
The ratio of the change in kinetic energy to the initial kinetic energy is thenΔK / K initial m2.m1 m2(15.4.5)As a check, we can calculate the change in kinetic energy via11ΔK (K f K i ) (m1 m2 )v 2f v1,2 i222 m1 2 1 21 (m1 m2 ) v1, i v1, i22 m1 m2 (15.4.6) m1 11 m1m2 2 1 m1v1,2 i v1, i . 2 m1 m2 m1 m2 2in agreement with Eq. (15.4.4).Example 15.3 Bouncing Superballs12gM 2 M1Figure 15.8b Two superballs droppingConsider two balls that are dropped from a height hi above the ground, one on top of theother (Figure 15.8). Ball 1 is on top and has mass M1 , and ball 2 is underneath and hasmass M 2 with M 2 M1 . Assume that there is no loss of kinetic energy during allcollisions. Ball 2 first collides with the ground and rebounds. Then, as ball 2 starts tomove upward, it collides with the ball 1 which is still moving downwards (figure belowleft). How high will ball 1 rebound in the air? Hint: consider this collision as seen by anobserver moving upward with the same speed as the ball 2 has after it collides withground. What speed does ball 1 have in this reference frame after it collides with the ball2?15-14
SolutionThe system consists of the two balls and the earth. There are five special states for thismotion shown in the figure below.part a)Initial State: the balls are released from rest at a height hi above the ground.State A: the balls just reach the ground with speed va 2ghi . This follows fromΔEmech 0 ΔK ΔU . Thus (1 / 2)mva2 0 mgΔh mghi va 2ghi .State B: immediately before the collision of the balls. Ball 2 has collided with the groundand reversed direction with the same speed, va , but ball 1 is still moving downward withspeed va .State C: immediately after the collision of the balls. Because we are assuming thatm2 m1 , ball 2 does not change its speed as a result of the collision so it is still movingupward with speed va . As a result of the collision, ball 1 moves upward with speed vb .Final State: ball 1 reaches a maximum height h f vb 2 / 2g above the ground. This againfollows from ΔK ΔU 0 (1 / 2)mvb2 mgΔh mgh f h f vb2 / 2g .Choice of Reference Frame:As indicated in the hint above, this collision is best analyzed from the reference frame ofan observer moving upward with speed va , the speed of ball 2 just after it rebounded with15-15
the ground. In this frame immediately, before the collision, ball 1 is moving downwardwith a speed vb′ that is twice the speed seen by an observer at rest on the ground (labreference frame).va′ 2 va(15.4.7)The mass of ball 2 is much larger than the mass of ball 1, m2 m1 . This enables us toconsider the collision (between States B and C) to be equivalent to ball 1 bouncing off ahard wall, while ball 2 experiences virtually no recoil. Hence ball 2 remains at rest in thereference frame moving upwards with speed va with respect to observer at rest onground. Before the collision, ball 1 has speed va′ 2 va . Since there is no loss of kineticenergy during the collision, the result of the collision is that ball 1 changes direction butmaintains the same speed,vb′ 2 va .(15.4.8)However, according to an observer at rest on the ground, after the collision ball 1 ismoving upwards with speedvb 2 va va 3va .(15.4.9)While rebounding, the mechanical energy of the smaller superball is constant (weconsider the smaller superball and the Earth as a system) hence between State C and theFinal State,ΔK ΔU 0 .(15.4.10)1ΔK m1 (3va )2 .2(15.4.11)ΔU m1 g h f .(15.4.12)The change in kinetic energy isThe change in potential energy isSo the condition that mechanical energy is constant (Equation (15.4.10)) is now1 m1 (3v1a )2 m1 g h f 0 .2(15.4.13)We can rewrite Equation (15.4.13) as1m1 g h f 9 m1 ( va )2 .2(15.4.14)15-16
Recall that we can also use the fact that the mechanical energy doesn’t change betweenthe Initial State and State A yielding an equation similar to Eq. (15.4.14),m1 g hi 1m ( v )2 .2 1 a(15.4.15)Now substitute the expression for the kinetic energy in Eq. (15.4.15) into Eq. (15.4.14)yieldingm1 g h f 9 m1 g hi .(15.4.16)Thus ball 1 reaches a maximum heighth f 9 hi .(15.4.17)15.6 Two Dimensional Elastic Collisions15.6.1 Two-dimensional Elastic Collision in Laboratory Reference FrameConsider the elastic collision between two particles in which we neglect any externalforces on the system consisting of the two particles. Particle 1 of mass m1 is initially moving with velocity v1, i and collides elastically with a particle 2 of mass m2 that isinitially at rest. We shall refer to the reference frame in which one particle is at rest, ‘thetarget’, as the laboratory reference frame. After the collision particle 1 moves with velocity v1, f and particle 2 moves with velocity v 2, f , (Figure 15.9). The angles θ1, fand θ 2, f that the particles make with the positive forward direction of particle 1 arecalled the laboratory scattering angles.v1, f11v1, i1, f222, fv 2, fFigure 15.9 Two-dimensional collision in laboratory reference frame Generally the initial velocity v1, i of particle 1 is known and we would like to determine the final velocities v1, f and v 2, f , which requires finding the magnitudes and directions15-17
of each of these vectors, v1, f , v2, f , θ1, f , and θ 2, f . These quantities are related by the twoequations describing the constancy of momentum, and the one equation describingconstancy of the kinetic energy. Therefore there is one degree of freedom that we mustspecify in order to determine the outcome of the collision. In what follows we shallexpress our results for v1, f , v2, f , and θ 2, f in terms of v1, i and θ1, f . m1v1,i m2 v 2,i in the initial state are givenThe components of the total momentum psysibysyspx,i m1v1,ip sys 0.y,i(15.5.1) m1v1, f m2 v 2, f in the final state are given byThe components of the momentum psysfpx,sysf m1 v1, f cosθ1, f m2 v2, f cosθ 2, fp sys m1 v1, f sin θ1, f m2 v2, f sin θ 2, f .y, f(15.5.2)There are no any external forces acting on the system, so each component of the totalmomentum remains constant during the collision,pxsys,i pxsys, f(15.5.3)p sys p sys.y,iy, f(15.5.4)m1 v1,i m1 v1, f cosθ1, f m2 v2, f cosθ 2, f ,(15.5.5)0 m1 v1, f sin θ1, f m2 v2, f sin θ 2, f .(15.5.6)Eqs. (15.5.3) and (15.5.4) becomeThe collision is elastic and therefore the system kinetic energy of is constantK i sys K f sys .(15.5.7)Using the given information, Eq. (15.5.7) becomes1112m1v1,i m1v1,2 f m2 v2,2 f .222(15.5.8)Rewrite the expressions in Eqs. (15.5.5) and (15.5.6) asm2 v2, f cosθ 2, f m1 (v1,i v1, f cosθ1, f ),(15.5.9)15-18
m2 v2, f sin θ 2, f m1v1, f sin θ1, f .(15.5.10)Square each of the expressions in Eqs. (15.5.9) and (15.5.10), add them together and usethe identity cos 2 θ sin 2 θ 1 yieldingv22, fm12 2 2 (v1,i 2v1,i v1, f cosθ1, f v1,2 f ) .m2(15.5.11)Substituting Eq. (15.5.11) into Eq. (15.5.8) yields111 m12 222m1v1,i m1v1, f (v1,i 2v1,i v1, f cosθ1, f v1,2 f ) .222 m2(15.5.12)Eq. (15.5.12) simplifies to m mm 2,0 1 1 v1,2 f 1 2v1,i v1, f cosθ1, f 1 1 v1,im2 m2 m2 (15.5.13)Let α m1 / m2 then Eq. (15.5.13) can be written as0 (1 α )v1,2 f 2α v1,i v1, f cosθ1, f (1 α )v1,2 i ,(15.5.14)The solution to this quadratic equation is given byv1, f (22α v1,i cosθ1, f α 2 v1,icos 2 θ1, f (1 α )v1,i(1 α ))1/2.(15.5.15)Divide the expressions in Eq. (15.5.9), yieldingv2, f sin θ 2, fv2, f cosθ 2, f v1, f sin θ1, fv1,i v1, f cosθ1, f.(15.5.16)Eq. (15.5.16) simplifies totan θ 2, f v1, f sin θ1, fv1,i v1, f cosθ1, f.(15.5.17)The relationship between the scattering angles in Eq. (15.5.17) is independent of themasses of the colliding particles. Thus the scattering angle for particle 2 is15-19
v1, f sin θ1, f θ 2, f tan 1 v1,i v1, f cosθ1, f (15.5.18)We can now use Eq. (15.5.10) to find an expression for the final velocity of particle 1v2, f v1, f sin θ1, f.α sin θ 2, f(15.5.19)Example 15.5 Elastic Two-dimensional collision of identical particlesv1, fˆjˆi11v1, i1, f2 302, f2v 2, fFigure 15.10 Momentum flow diagram for two-dimensional elastic collisionObject 1 with mass m1 is initially moving with a speed v1,i 3.0m s 1 and collideselastically with object 2 that has the same mass, m2 m1 , and is initially at rest. After thecollision, object 1 moves with an unknown speed v1, f at an angle θ1, f with respect to itsinitial direction of motion and object 2 moves with an
Chapter 15 Collision Theory Despite my resistance to hyperbole, the LHC [Large Hadron Collider] belongs
Part One: Heir of Ash Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26 Chapter 27 Chapter 28 Chapter 29 Chapter 30 .
TO KILL A MOCKINGBIRD. Contents Dedication Epigraph Part One Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Part Two Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18. Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26
DEDICATION PART ONE Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 PART TWO Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 .
5 1. Collision Physics – An Overview. 1.1 Outline. Collision physics includes ANY collision of a quantum particle with a target. Collision Particles may be: PHOTONS (eg from a Laser, Synchrotron source or FEL) ELECTRONS (usually of well defined momentum from an electron gun) IONS (usually from an ion source of well defin
Figure 2.3 Time histories of the girder-pier collision force As shown, for a minor β, the girder-pier collision force curve is close to a sinusoidal curve. And for a major β, the girder-pier collision force curve is close to a rectangular curve. The second model is based on the SDOF harmonic theory [8]. According to the theory, the axial .
236 Study Guide for An Introduction to Chemistry Section Goals and Introductions Section 14.1 Collision Theory: A Model for the Reaction Process Goals To describe a model, called collision theory, that helps us to visualize the process of many chemical reactions. To use collision theory to explain why not
About the husband’s secret. Dedication Epigraph Pandora Monday Chapter One Chapter Two Chapter Three Chapter Four Chapter Five Tuesday Chapter Six Chapter Seven. Chapter Eight Chapter Nine Chapter Ten Chapter Eleven Chapter Twelve Chapter Thirteen Chapter Fourteen Chapter Fifteen Chapter Sixteen Chapter Seventeen Chapter Eighteen
18.4 35 18.5 35 I Solutions to Applying the Concepts Questions II Answers to End-of-chapter Conceptual Questions Chapter 1 37 Chapter 2 38 Chapter 3 39 Chapter 4 40 Chapter 5 43 Chapter 6 45 Chapter 7 46 Chapter 8 47 Chapter 9 50 Chapter 10 52 Chapter 11 55 Chapter 12 56 Chapter 13 57 Chapter 14 61 Chapter 15 62 Chapter 16 63 Chapter 17 65 .
