Chapter 15 Collision Theory

Transcription

1 Chapter 5 Collision Theory 5 Introduction 5 Reference Frames and Relative Velocities 5 Relative Velocities 3 5 Center-of-mass Reference Frame 4 53 Kinetic Energy in the Center-of-Mass Reference Frame 5 54 Change of Kinetic Energy and Relatively Inertial Reference Frames 5 53 Characterizing Collisions 7 54 One-Dimensional Collisions Between Two Objects 7 54 One Dimensional Elastic Collision in Laboratory Reference Frame 7 54 One-Dimensional Collision Between Two Objects Center-of-Mass Reference Frame 55 Worked Examples Example 5 Elastic One-Dimensional Collision Between Two Objects Example 5 The Dissipation of Kinetic Energy in a Completely Inelastic Collision Between Two Objects 3 Example 53 Bouncing Superballs 4 56 Two Dimensional Elastic Collisions 7 56 Two-dimensional Elastic Collision in Laboratory Reference Frame 7 Example 55 Elastic Two-dimensional collision of identical particles 0 Example 56 Two-dimensional elastic collision between particles of equal mass Example 57 Two dimensional collision between particles of unequal mass 57 Two-Dimensional Collisions in Center-of-Mass Reference Frame 4 57 Two-Dimensional Collision in Center-of-Mass Reference Frame 4 57 Scattering in the Center-of-Mass Reference Frame 5 Example 58 Scattering in the Lab and CM Frames 6

2 Chapter 5 Collision Theory Despite my resistance to hyperbole, the LHC [Large Hadron Collider] belongs to a world that can only be described with superlatives It is not merely large: the LHC is the biggest machine ever built It is not merely cold: the 9 kelvin (9 degrees Celsius above absolute zero) temperature necessary for the LHC s supercomputing magnets to operate is the coldest extended region that we know of in the universe even colder than outer space The magnetic field is not merely big: the superconducting dipole magnets generating a magnetic field more than 00,000 times stronger than the Earth s are the strongest magnets in industrial production ever made And the extremes don t end there The vacuum inside the proton-containing tubes, a 0 trillionth of an atmosphere, is the most complete vacuum over the largest region ever produced The energy of the collisions are the highest ever generated on Earth, allowing us to study the interactions that occurred in the early universe the furthest back in time 5 Introduction Lisa Randall When discussing conservation of momentum, we considered examples in which two objects collide and stick together, and either there are no external forces acting in some direction (or the collision was nearly instantaneous) so the component of the momentum of the system along that direction is constant We shall now study collisions between objects in more detail In particular we shall consider cases in which the objects do not stick together The momentum along a certain direction may still be constant but the mechanical energy of the system may change We will begin our analysis by considering two-particle collision We introduce the concept of the relative velocity between two particles and show that it is independent of the choice of reference frame We then show that the change in kinetic energy only depends on the change of the square of the relative velocity and therefore is also independent of the choice of reference frame We will then study one- and two-dimensional collisions with zero change in potential energy In particular we will characterize the types of collisions by the change in kinetic energy and analyze the possible outcomes of the collisions 5 Reference Frames and Relative Velocities We shall recall our definition of relative inertial reference frames Let R be the vector from the origin of frame S to the origin of reference frame S Denote the Randall, Lisa, Knocking on Heaven's Door: How Physics and Scientific Thinking Illuminate the Universe and the Modern World, Ecco, 0 5-

3 position vector of the j th particle with respect to the origin of reference frame S by r j and similarly, denote the position vector of the reference frame S by r j (Figure 5) j th particle with respect to the origin of j th particle S r j R r j S Figure 5 Position vector of j th particle in two reference frames The position vectors are related by r j = r j + R (5) The relative velocity (call this the boost velocity) between the two reference frames is given by V = dr (5) dt Assume the boost velocity between the two reference frames is constant Then, the relative acceleration between the two reference frames is zero, dv A = = 0 (53) dt When Eq (53) is satisfied, the reference frames S and S are called relatively inertial reference frames Suppose the j th particle in Figure 5 is moving; then observers in different reference frames will measure different velocities Denote the velocity of j th particle in frame S by v j = dr / dt, and the velocity of the same particle in frame S by j v j = dr j / dt Taking derivative, the velocities of the particles in two different reference frames are related according to v j = v j + V (54) 5-

4 5 Relative Velocities Consider two particles of masses and m interacting via some force (Figure 5) Figure 5 Two interacting particles Choose a coordinate system (Figure 53) in which the position vector of body is given by r and the position vector of body is given by r The relative position of body with respect to body is given by r, = r r Figure 53 Coordinate system for two bodies During the course of the interaction, body is displaced by dr and body is displaced by dr, so the relative displacement of the two bodies during the interaction is given by dr, = dr dr The relative velocity between the particles is dr, dr dr = = = v v (55) dt dt dt We shall now show that the relative velocity between the two particles is independent of the choice of reference frame providing that the reference frames are relatively inertial The relative velocity v in reference frame S can be determined from using Eq (54) to express Eq (55) in terms of the velocities in the reference frame S, = v v = (v + V) (v v + V) = v = (56) 5-3

5 and is equal to the relative velocity in frame S For a two-particle interaction, the relative velocity between the two vectors is independent of the choice of relatively inertial reference frames 5 Center-of-mass Reference Frame Let r be the vector from the origin of frame S to the center-of-mass of the system of particles, a point that we will choose as the origin of reference frame S, 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 S by r j 54) j th particle (Figure r j r j r S! S Figure 54 Position vector of j th particle in the center-of-mass reference frame The position vector of the th j particle in the center-of-mass frame is then given by r j = r j r (57) The velocity of the j th particle in the center-of-mass reference frame is then given by v j = v j v (58) There are many collision problems in which the center-of-mass reference frame is the most convenient reference frame to analyze the collision Consider a system consisting of two particles, which we shall refer to as particle and particle We can use Eq (58) to determine the velocities of particles and in the center-of-mass, m µ v v = v v v m = ( v +, v = v ) = m (59) 5-4

6 where v = v v is the relative velocity of particle with respect to particle A similar result holds for particle : v v = v m v + m v m = ( v + v ) = µ = v v, (50) The momentum of the system the center-of-mass reference frame is zero as we expect, v v = µ v µ v = 0 (5) 53 Kinetic Energy in the Center-of-Mass Reference Frame The kinetic energy in the center of mass reference frame is given by K = v v + m v v (5) We now use Eqs (59) and (50) to rewrite the kinetic energy in terms of the relative velocity v = v v, µ µ µ K = m v v µ v v m + m c,,,, m m (53) = µ v v + m = µ,, where we used the fact that we defined the reduced mass by + (54) µ m 54 Change of Kinetic Energy and Relatively Inertial Reference Frames The kinetic energy of the two particles in reference frame S is given by K S = v + m v (55) We can take the scalar product of Eq (58) to rewrite Eq (55) as 5-5

7 K S = (v + v ) (v + v ) (v + v ) (v + v ) (56) = m + + )v v m v ( + ( v v ) v The last term is zero due to the fact that the momentum of the system in the center of mass reference frame is zero (Eq (5)) Therefore Eq (56) becomes K = v + v S m + ( )v (57) The first two terms correspond to the kinetic energy in the center of mass frame, thus the kinetic energies in the two reference frames are related by K S = K + ( )v (58) We now use Eq (53) to rewrite Eq (58) as K S = µ + ( )v (59) Even though kinetic energy is a reference frame dependent quantity, because the second term in Eq (59) is a constant, the change in kinetic energy in either reference frame is equal to v v, f, i ΔK = µ(( ) ( ) ) (50) This generalizes to any two relatively inertial reference frames because the relative velocity is a reference frame independent quantity, the change in kinetic energy is independent of the choice of relatively inertial reference frames We showed in Appendix 3A that when two particles of masses and m interact, the work done by the interaction force is equal to v v, f, i W = µ(( ) ( ) ) (5) Hence we explicitly verified that for our two-particle system W = ΔK sys (5) 5-6

8 53 Characterizing Collisions In a collision, the ratio of the magnitudes of the initial and final relative velocities is called the coefficient of restitution and denoted by the symbol e, v B e = (53) v A If the magnitude of the relative velocity does not change during a collision, e =, then the change in kinetic energy is zero, (Eq (5)) Collisions in which there is no change in kinetic energy are called elastic collisions, ΔK = 0, elastic collision (54) If the magnitude of the final relative velocity is less than the magnitude of the initial relative velocity, e <, then the change in kinetic energy is negative Collisions in which the kinetic energy decreases are called inelastic collisions, ΔK < 0, inelastic collision (55) 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 be found from Eq (5), m ΔK = µ va = v A, totally inelastic collision (56) If the magnitude of the final relative velocity is greater than the magnitude of the initial relative velocity, e >, then the change in kinetic energy is positive Collisions in which the kinetic energy increases are called superelastic collisions, ΔK > 0, superelastic collision (57) 54 One-Dimensional Collisions Between Two Objects 54 One Dimensional Elastic Collision in Laboratory Reference Frame Consider a one-dimensional elastic collision between two objects moving in the x - direction One object, with mass and initial x -component of the velocity v x,i, collides with an object of mass m and initial x -component of the velocity v x,i The scalar components v x,i and v x,i can be positive, negative or zero No forces other than the interaction force between the objects act during the collision After the collision, the 5-7

9 final x -component of the velocities are v x, f laboratory reference frame and v x, f We call this reference frame the Figure 55 One-dimensional elastic collision, laboratory reference frame For the collision depicted in Figure 55, v x,i > 0, v x,i < 0, v x, f < 0, and v 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 the relation v x, i v x, i = v x, f v x, f (53) Because the collision is elastic, kinetic energy is constant Equating the kinetic energy before and after the collision gives the relation Rewrite these Eqs (53) and (53) as v x,i v x,i = v x, f v x, f (53) (v x,i v x, f ) = m (v x, f v x,i ) (533) (v x,i v x, f ) = m (v x, f v x,i ) (534) Eq (534) can be written as (v x,i v x, f )(v x,i + v x, f ) = m (v x, f v x,i )(v x, f + v x,i ) (535) Divide Eq (534) by Eq (533), yielding Eq (536) may be rewritten as v x,i + v x, f = v x,i + v x, f (536) v x,i v x,i = v x, f v x, f (537) Recall that the relative velocity between the two objects is defined to be v rel v v v (538), 5-8

10 where we used the superscript rel to remind ourselves that the velocity is a relative velocity (and to simplify our notation) Thus v x rel,i = v x,i v x,i is the initial x -component of the relative velocity, and v x rel, f = v x, f v x, f is the final x -component of the relative velocity Therefore Eq (537) states that during the interaction the initial relative velocity is equal to the negative of the final relative velocity rel vi = v rel f, ( dimensional energy-momentum prinicple) (539) Consequently the initial and final relative speeds are equal We shall call this relationship between the relative initial and final velocities the one-dimensional energy-momentum principle because we have combined these two principles to realize this result The energy-momentum principle is independent of the masses of the colliding particles Although we derived this result explicitly, we have already shown that the change in kinetic energy for a two-particle interaction (Eq (50)), in our simplified notation is given by rel rel ΔK = µ((v ) f (v ) i ) (530) Therefore for an elastic collision where ΔK = 0, the square of the relative speed remains constant rel (v ) = (v (53) f For a one-dimensional collision, the magnitude of the relative speed remains constant but the direction changes by 80 rel ) i We can now solve for the final x -component of the velocities, v x, f follows Eq (537) may be rewritten as and v x, f, as v x, f = v x, f + v x,i v x,i (53) Now substitute Eq (53) into Eq (53) yielding v x,i v x,i = v x, f (v x, f + v x,i v x,i ) (533) Solving Eq (533) for v x, f involves some algebra and yields m m vx, f = v x,i + v x,i (534) 5-9

11 To find v x, f, rewrite Eq (537) as v x, f = v x, f v x,i + v x,i (535) Now substitute Eq (535) into Eq (53) yielding v x,i v x,i = (v x, f v x,i + v x,i )v x, f v x, f (536) We can solve Eq (536) for v x, f and determine that m v x, f = v x,i + v x,i (537) m + m + Consider what happens in the limits >> m in Eq (534) Then v x, f v x,i + m v x,i ; (538) the more massive object s velocity component is only slightly changed by an amount proportional to the less massive object s x -component of momentum Similarly, the less massive object s final velocity approaches We can rewrite this as v x, f v x,i + v x,i = v x,i + v x,i v x,i (539) v x, f v x,i = v x,i v x,i = v rel (530) x,i ie the less massive object rebounds with the same speed relative to the more massive object which barely changed its speed If the objects are identical, or have the same mass, Eqs (534) and (537) become v x, f = v x,i, v x, f = v x,i ; (53) the objects have exchanged x -components of velocities, and unless we could somehow distinguish the objects, we might not be able to tell if there was a collision at all 5-0

12 54 One-Dimensional Collision Between Two Objects Center-of-Mass Reference Frame We analyzed the one-dimensional elastic collision (Figure 55) in Section 54 in the laboratory 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 is v x,i v x,i vx, = (53) With respect to the center-of-mass, the x -components of the velocities of the objects are x,i = v x,i v = (v x,i v x,i ) x, (533) v v m x,i = v x,i v = (v x,i v x,i ) x, In the CM frame the momentum of the system is zero before the collision and hence the momentum of the system is zero after the collision For an elastic collision, the only way for both momentum and kinetic energy to be the same before and after the collision is either the objects have the same velocity (a miss) or to reverse the direction of the velocities as shown in Figure 56 Figure 56 One-dimensional elastic collision in center-of-mass reference frame In the CM frame, the final x -components of the velocities are m v x, f = v x,i = (v x,i v x,i ) m (534) v x, f = v x,i = (v x,i v x,i ) The final x -components of the velocities in the laboratory frame are then given by 5-

13 v x, f = v x, f + v x, m v x,i v = x,i (vx,i v x,i ) + (535) m m = vx,i + v x,i m as in Eq (534) and a similar calculation reproduces Eq (537) 55 Worked Examples Example 5 Elastic One-Dimensional Collision Between Two Objects î i = x,i î v,i = 0 initial state m = v î î final state, f = x, f v, f = v,x, f î m = Figure 57 Elastic collision between two non-identical carts Consider the elastic collision of two carts along a track; the incident cart has mass and moves with initial speed i The target cart has mass m = and is initially at rest, v,i = 0, (Figure 57) Immediately after the collision, the incident cart has final speed and the target cart has final speed v, f Calculate the final x -component of the velocities of the carts as a function of the initial speed i Solution The momentum flow diagram for the objects before (initial state) and after (final state) the collision are shown in Figure 57 We can immediately use our results above with m = and v,i = 0 The final x -component of velocity of cart is given by Eq (534), where we use v x,i = i 5-

14 v x, f = 3 i (54) The final x -component of velocity of cart is given by Eq (537) v x, f = 3 i (54) Example 5 The Dissipation of Kinetic Energy in a Completely Inelastic Collision Between Two Objects î i v,i = 0 initial state î final state v f Figure 57b Inelastic collision between two non-identical carts An incident cart of mass and initial speed i collides completely inelastically with a cart of mass m that is initially at rest (Figure 57b) There are no external forces acting on 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, the momentum after the collision will be the same as before the collision After the collision the carts will move in the direction of the initial velocity of the incident cart with a common speed v f found from applying the momentum condition i = ( )v f (543) v f = i rel The initial relative speed is v i = i The final relative velocity is zero because the carts stick together so using Eq (56), the change in kinetic energy is rel ΔK = µ(vi ) = m v (544), i 5-3

15 The ratio of the change in kinetic energy to the initial kinetic energy is then ΔK / K initial = As a check, we can calculate the change in kinetic energy via m (545) ΔK = (K f K i ) = (m )v f v, i = (m m ) i v, i m = m i = m i (546) in agreement with Eq (544) Example 53 Bouncing Superballs g M >> M Figure 58b Two superballs dropping Consider two balls that are dropped from a height h i above the ground, one on top of the other (Figure 58) Ball is on top and has mass M, and ball is underneath and has mass M with M >> M Assume that there is no loss of kinetic energy during all collisions Ball first collides with the ground and rebounds Then, as ball `starts to move upward, it collides with the ball which is still moving downwards (figure below left) How high will ball rebound in the air? Hint: consider this collision as seen by an observer moving upward with the same speed as the ball has after it collides with ground What speed does ball have in this reference frame after it collides with the ball? 5-4

16 Solution The system consists of the two balls and the earth There are five special states for this motion shown in the figure below part a) Initial State: the balls are released from rest at a height h i above the ground State A: the balls just reach the ground with speed v a = gh i This follows from ΔE = 0 ΔK = ΔU Thus ( / )mv mech a 0 = mgδh = mgh i v a = gh i State B: immediately before the collision of the balls Ball has collided with the ground and reversed direction with the same speed, v a, but ball is still moving downward with speed v a State C: immediately after the collision of the balls Because we are assuming that m >>, ball does not change its speed as a result of the collision so it is still moving upward with speed v a As a result of the collision, ball moves upward with speed v b Final State: ball reaches a maximum height h f = v b / g above the ground This again follows from ΔK = ΔU 0 ( / )mv = mgδh = mgh h = v / g b f f b Choice of Reference Frame: As indicated in the hint above, this collision is best analyzed from the reference frame of an observer moving upward with speed v a, the speed of ball just after it rebounded with 5-5

17 the ground In this frame immediately, before the collision, ball is moving downward with a speed v b that is twice the speed seen by an observer at rest on the ground (lab reference frame) v a = v a (547) The mass of ball is much larger than the mass of ball, m >> This enables us to consider the collision (between States B and C) to be equivalent to ball bouncing off a hard wall, while ball experiences virtually no recoil Hence ball remains at rest in the reference frame moving upwards with speed v a with respect to observer at rest on ground Before the collision, ball has speed v a = v a Since there is no loss of kinetic energy during the collision, the result of the collision is that ball changes direction but maintains the same speed, v b a = v (548) However, according to an observer at rest on the ground, after the collision ball is moving upwards with speed v b = v a + v a = 3v a (549) While rebounding, the mechanical energy of the smaller superball is constant (we consider the smaller superball and the Earth as a system) hence between State C and the Final State, ΔK + ΔU = 0 (540) The change in kinetic energy is The change in potential energy is ΔK = m ) (3v a (54) ΔU = g h f (54) So the condition that mechanical energy is constant (Equation (540)) is now (3v a ) + g h f = 0 (543) We can rewrite Equation (543) as m g h ) f = 9 ( v a (544) 5-6

18 Recall that we can also use the fact that the mechanical energy doesn t change between the Initial State and State A yielding an equation similar to Eq (544), g h i = ( v a ) (545) Now substitute the expression for the kinetic energy in Eq (545) into Eq (544) yielding g h f = 9 g h i (546) Thus ball reaches a maximum height h f = 9 h i (547) 56 Two Dimensional Elastic Collisions 56 Two-dimensional Elastic Collision in Laboratory Reference Frame Consider the elastic collision between two particles in which we neglect any external forces on the system consisting of the two particles Particle of mass is initially moving with velocity i and collides elastically with a particle of mass m that is initially at rest We shall refer to the reference frame in which one particle is at rest, the target, as the laboratory reference frame After the collision particle moves with velocity and particle moves with velocity v, f, (Figure 59) The angles θ, f and θ, f that the particles make with the positive forward direction of particle are called the laboratory scattering angles i, f, f v, f Figure 59 Two-dimensional collision in laboratory reference frame Generally the initial velocity i of particle is known and we would like to determine the final velocities and, which requires finding the magnitudes and directions v, f 5-7

19 of each of these vectors,, v, f, θ, f, and θ, f These quantities are related by the two equations describing the constancy of momentum, and the one equation describing constancy of the kinetic energy Therefore there is one degree of freedom that we must specify in order to determine the outcome of the collision In what follows we shall express our results for, v, f, and θ, f in terms of i and θ, f The components of the total momentum p sys = m v,i i + m v,i in the initial state are given by p sys = m v x,i,i sys p y,i = 0 (55) The components of the momentum p sys f = m v, f v, f in the final state are given by sys p x, f = cosθ, f v, f cosθ, f (55) p sys = m v sinθ, f m sinθ, f y, f, f v, f There are no any external forces acting on the system, so each component of the total momentum remains constant during the collision, Eqs (553) and (554) become sys p x,i sys = p x, f (553) sys sys p = p (554) y,i y, f i = cosθ, f v, f cosθ, f, (555) 0 = sinθ, f m v, f sinθ, f (556) The collision is elastic and therefore the system kinetic energy of is constant sys sys K i = K f (557) Using the given information, Eq (557) becomes Rewrite the expressions in Eqs (555) and (556) as i = v, f (558) m v, f cosθ, f = (i cosθ, f ), (559) 5-8

20 m v, f sinθ, f = sinθ, f (550) Square each of the expressions in Eqs (559) and (550), add them together and use the identity cos θ + sin θ = yielding v, f (i + = i cosθ, f ) (55) m Substituting Eq (55) into Eq (558) yields i = + (i i cosθ, f + ) (55) m Eq (55) simplifies to 0 = + v v v cosθ, f,i, f, f v, (553),i m m m Let α = / m then Eq (553) can be written as 0 = (+ α ) The solution to this quadratic equation is given by αv v cosθ ( α )v, (554) f,i, f, f, i / αi cosθ, f ± (α i cos θ, f + ( α )i ) = (555) (+ α ) Divide the expressions in Eq (559), yielding Eq (556) simplifies to v, f sinθ, f v =, f sinθ, f (556) v, f cosθ, f i cosθ, f tanθ, f = sinθ, f i cosθ, f (557) The relationship between the scattering angles in Eq (557) is independent of the masses of the colliding particles Thus the scattering angle for particle is 5-9

21 sinθ, f = tan θ, f (558) cosθ, f i We can now use Eq (550) to find an expression for the final velocity of particle sinθ =, f v, f (559) α sinθ, f Example 55 Elastic Two-dimensional collision of identical particles i ĵ î = 30, f, f v, f Figure 50 Momentum flow diagram for two-dimensional elastic collision Object with mass is initially moving with a speed i = 30m s and collides elastically with object that has the same mass, m =, and is initially at rest After the collision, object moves with an unknown speed at an angle θ, f with respect to its initial direction of motion and object moves with an unknown speed v, f, at an unknown angle θ, f (as shown in the Figure 50) Find the final speeds of each of the objects and the angle θ, f Solution: Because the masses are equal, α = We are given that i = 30 m s and θ, f = 30 o Hence Eq (554) reduces to = i cosθ, f = (30 m s )cos30 = 6 m s (550) Substituting Eq (550) in Eq (557) yields 5-00

22 sinθ, f = tan θ, f i cosθ, f θ, f (6 m s )sin(30 ) = tan (55) 30 m s (6 m s )cos(30 ) = 60 The above results for and θ,f may be substituted into either of the expressions in Eq (559), or Eq (55), to find v, f = 5m s Eq (55) also has the solution v, f = 0, which would correspond to the incident particle missing the target completely Before going on, the fact that θ, f +θ, f = 90, that is, the objects move away from the collision point at right angles, is not a coincidence A vector derivation is presented in Example 56 We can see this result algebraically from the above result Substituting Eq (550) = i cosθ, f in Eq (557) yields cosθ, f sinθ, f tanθ, f = = cotθ, f = tan(90 θ, f ) ; (55) cosθ, f showing that θ, f +θ, f = 90, the angles θ, f and θ, f are complements Example 56 Two-dimensional elastic collision between particles of equal mass Show that the equal mass particles emerge from a two-dimensional elastic collision at right angles by making explicit use of the fact that momentum is a vector quantity i ĵ î, f, f v, f Figure 5 Elastic scattering of identical particles 5-

23 Solution: Choose a reference frame in which particle is initially at rest (Figure 5) There are no external forces acting on the two objects during the collision (the collision forces are all internal), therefore momentum is constant which becomes Eq (554) simplifies to sys sys p i = p f, (553) v = m v + m v (554), i, f, f v = v,i, f, f + v (555) Recall the vector identity that the square of the speed is given by the dot product v v = v With this identity in mind, we take the dot product of each side of Eq (555) with itself, v v = ( v + v ) ( + v ),i,i, f, f, f (556) = v v + v + v v, f This becomes v, f, f, f, f, f v +,i = v, f + v, f (557) Recall that kinetic energy is the same before and after an elastic collision, and the masses of the two objects are equal, so constancy of energy, (Eq (53)) simplifies to i = + v, f (558) Comparing Eq (557) to Eq (558), we see that v, f = 0 (559) The dot product of two nonzero vectors is zero when the two vectors are at right angles to each other justifying our claim that the collision particles emerge at right angles to each other Example 57 Two dimensional collision between particles of unequal mass Particle of mass, initially moving in the positive x -direction (to the right in the figure below) with speed i, collides with particle of mass m = m / 3, which is initially moving in the opposite direction (Figure 5) with an unknown speed v,i Assume that the total external force acting on the particles is zero Do not assume the collision is elastic After the collision, particle moves with speed = i / in the negative y -direction After the collision, particle moves with an unknown speed v, f, 5-

24 at an angle θ, f = 45 o with respect to the positive x -direction (i) Determine the initial speed v,i of particle and the final speed v, f of particle in terms of i (ii) Is the collision elastic? before i v,i after v, f = 45 ĵ î = m m = m / 3 = i / Figure 5 Two-dimensional collision between particles of unequal mass Solution: We choose as our system the two particles We are given that apply the two momentum conditions, = v /,i We Solve Eq (553) for v, f i ( / 3)v,i = ( / 3) v, f ( / ) (5530) 0 = ( / 3) v, f ( / ) (553) = 3 = 3 (553) v, f i Substitute Eq (553) into Eq (5530) and solve for v,i The initial kinetic energy is then v,i =(3/ )i (5533) 7 K i = i + ( / 3)v,i = i (5534) 8 The final kinetic energy is 3 7 K f = v, f = i + i = i (5535) Comparing our results, we see that kinetic energy is constant so the collision is elastic 5-3

25 57 Two-Dimensional Collisions in Center-of-Mass Reference Frame 57 Two-Dimensional Collision in Center-of-Mass Reference Frame Consider the elastic collision between two particles in the laboratory reference frame (Figure 59) Particle of mass is initially moving with velocity i and collides elastically with a particle of mass m that is initially at rest After the collision the particle moves with velocity v and particle moves with velocity v In section, f, f 57 we determined how to find, v, f, and θ, f in terms of i and θ, f We shall now analyze the collision in the center-of-mass reference frame, which is boosted form the laboratory frame by the velocity of center-of-mass given by v v =, i (5536) Because we assumed that there are no external forces acting on the system, the center-ofmass velocity remains constant during the interaction i v,i v, f Figure 53 Two-dimensional elastic collision in center-of-mass reference frame Recall the velocities of particles and in the center-of-mass frame are given by (Eq,(59) and (50)) In the center-of-mass reference frame the velocities of the two incoming particles are in opposite directions, as are the velocities of the two outgoing particles after the collision (Figure 53) The angle Θ between the incoming and outgoing velocities is called the center-of-mass scattering angle 5-4

26 57 Scattering in the Center-of-Mass Reference Frame Consider a collision between particle of mass and velocity i and particle of mass m at rest in the laboratory frame Particle is scattered elastically through a scattering angle Θ in the center-of-mass frame The center-of-mass velocity is given by v v =,i (5537) In the center-of-mass frame, the momentum of the system of two particles is zero Therefore 0 = i v,i = v, f (5538) m i = v (5539) m,i m = v (5540) m, f The energy condition in the center-of-mass frame is i v, i = v, (554) Substituting Eqs (5539) and (5540) into Eq (554) yields (we are only considering magnitudes) Therefore i = (554) v, i = v, (5543) Because the magnitude of the velocity of a particle in the center-of-mass reference frame is proportional to the relative velocity of the two particles, Eqs (554) and (5543) imply that the magnitude of the relative velocity also does not change v = v,, i,, f, (5544) verifying our earlier result that for an elastic collision the relative speed remains the same, (Eq (50)) However the direction of the relative velocity is rotated by the center-of-mass scattering angle Θ This generalizes the energy-momentum principle to two dimensions Recall that the relative velocity is independent of the reference frame, 5-5

27 v =, i v, i i v, i (5545) In the laboratory reference frame v,i = 0, hence the initial relative velocity is, = v =, and the velocities in the center-of-mass frame of the particles are then i,, i i v µ i = i (5546) m = µ (5547), i i m Therefore the magnitudes of the final velocities in the center-of-mass frame are µ µ µ = i =, = v = i,, i i (5548) µ µ µ = v, i =, = v = i,, i i m m m (5549) v, Example 58 Scattering in the Lab and CM Frames Particle of mass and velocity by a particle of mass m at rest in the laboratory i frame is scattered elastically through a scattering angle Θ in the center of mass frame, (Figure 54) Find (i) the scattering angle of the incoming particle in the laboratory frame, (ii) the magnitude of the final velocity of the incoming particle in the laboratory reference frame, and (iii) the fractional loss of kinetic energy of the incoming particle i ĵ î v, f,i, f v,i v, f v, f Figure 54 Scattering in the laboratory and center-of-mass reference frames 5-6

28 Solution: i) In order to determine the center-of-mass scattering angle we use the transformation law for velocities v = v, (5550) In Figure 55 we show the collision in the center-of-mass frame along with the laboratory frame final velocities and scattering angles, f v ĵ î i v, f v,i v, f v, f Figure 55 Final velocities of colliding particles Vector decomposition of Eq (5550) yields cosθ, i = cosθ v, (555) sinθ, i = sinθ (555) where we choose as our directions the horizontal and vertical Divide Eq (555) by (555) yields sinθ, i sinθ tanθ, i = = (5553) cosθ, i cosθ v Because i =, we can rewrite Eq (5553) as i sinθ tanθ, i = (5554) i cosθ v We now substitute Eqs (5548) and v / ( ) into Eq (5554) yielding = i 5-7

29 Thus in the laboratory frame particle scatters by an angle θ, i m sinθ tanθ, i = (5555) cosθ / m m sinθ = tan (5556) cosθ / m ii) We can calculate the square of the final velocity in the laboratory frame which becomes v, f = ( + v ) ( + v ) (5557) = + v, f v + v = v c, + v cosθ + v (5558) We use the fact that = i = (µ / ), i = (µ / ) i = (m / ) i to rewrite Eq (5558) as Thus m = i + m ) i cosθ + ( ( ( / m + m cosθ + m c ) = i m ) v, i (5559) (5560) (iii) The fractional change in the kinetic energy of particle in the laboratory frame is given by K, f K, i i m + m cosθ + m (cosθ ) = = = (556) K ) ), i i ( ( We can also determine the scattering angle Θ in the center-of-mass reference frame from the scattering angle θ, i of particle in the laboratory We now rewrite the momentum relations as In a similar fashion to the above argument, we have that cosθ, i + v = cosθ, (556) sinθ, i = sinθ (5563) 5-8

30 sinθ, f tanθ = (5564) cosθ, f + v Recall from our analysis of the collision in the laboratory frame that if we specify one of the four parameters, v, f, θ, f, or, then we can solve for the other three in terms of the initial parameters i and v, i With that caveat, we can use Eq (5564) to determine Θ 5-9

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