I. Elastic collisions of 2 particles II. Relate ψ and θ III. Relate ψ and ζ IV. Kinematics of elastic collisions


 Bernard Marshall
 13 days ago
 Views:
Transcription
1 I. Elastic collisions of particles II. Relate ψ and θ III. Relate ψ and ζ IV. Kinematics of elastic collisions 49
2 I. Elastic collisions of particles "Elastic": KE is conserved (as well as E tot and momentum p). i.e., the colliding objects are not deformed or reconfigured in a way that would change the potential energy. Issue: Many physical laws are expressed most simply in the CM frame. But measurements are made in the LAB frame. Goal: Relate lab system measurements (especially angles) to the CM reference frame. 50
3 Consider a system in the lab frame. Define the notation here: Before collision After collision moving m stationary u 1 u = 0 In the LAB frame: ψ ζ v v 1 u 1 = initial velocity of v 1 = final velocity of u = initial velocity of m v = final velocity of m (u = 0) v 0 ψ = deflection of ζ = deflection of m V = velocity of the CM in the LAB frame. It is constant, same before and after the collision. V 51
4 Same system but in the CM frame. Symbols for velocities have primes. Before collision After collision v 1 ' m θ u 1 ' u ' v ' Now, compared to V, m is moving leftward. By definition, V = 0 in the CM frame. So also, p tot = MV = 0. To maintain p tot = 0, the particles exit backtoback. u 1 ' = initial velocity of v 1 ' = final velocity of u ' = initial velocity of m v ' = final velocity of m θ = deflection angle of ( π θ ) = deflection angle of m 5
5 II. Relate ψ to θ Recall the definition of center of mass: R 1 M m ir i i For particles, R 1 M m 1r 1 r ( ) Take derivatives of both sides wrt time: R 1 ( M m 1r 1 r ) V = 1 ( M u 1 u ) but u = 0 So V = u 1 'Eq 0' 53
6 Relationships among the outgoing vectors: V v 1 ' v 1 θ ψ Notice: v 1 sinψ = v 1 'sinθ 'Eq 1' and: v 1 cosψ = v 1 'cosθ + V 'Eq ' Divide Eq 1 Eq : v 1 sinψ v 1 cosψ = tanψ = v 1 'sinθ v 1 'cosθ + V sinθ cosθ + V v 1 ' Now plug in V = u 1 54
7 tanψ = sinθ cosθ + u 1 v 1 ' 'Eq 3' Consider an elastic scatter: kinetic energy is conserved. In the CM: m u 1( ' 1 ) = ( v ' 1 ) So u 1 ' = v 1 ' from the definition of elastic 'Eq 4' But also u 1 ' = u 1 V = u 1 u 1 = u 1( ) u 1 from the definition CM u 1 ' = u 1 m 'Eq 5' Now compare Eq 4 to Eq 5. LHS's are equal, so RHS's must be equal: 55
8 v 1 ' = u 1m Plug this into Eq. 3 to get: tanψ = sinθ m u 1 1 m cosθ + 1 u 1 m tanψ = sinθ cosθ + m for elastic scatters 56
9 Notice special cases: 1) when << m, m 0, then tanψ sinθ cosθ = tanθ i.e., ψ θ for << m. ) when = m, tanψ = sinθ cosθ +1 = tanθ i.e., ψ = θ for = m 57
10 III. Relate ψ and ζ We apply an analysis similar to the one above. Notice: v V ζ θ v ' v 1 ' θ v sinζ = v 'sinθ 'Eq 6' and v cosζ = V v 'cosθ 'Eq 7' Divide Eq 6 Eq 7 : v sinζ v cosζ = v 'sinθ V v 'cosθ 58
11 sinθ tanζ = V v ' cosθ Find V v ' : ( u ' ) because KE is conserved, m u ' = v ' But u ' = V : So u ' = V V = v ' i.e, V v ' = 1. = m v ( ' ) if m is stationary in the LAB, it must move backward with magnitude V in the CM frame. Thus tanζ = sinθ 1 cosθ = cotθ = tan π θ 59
12 i.e., ζ = π θ ζ = π θ. Special case: when = m, ψ = θ [slide 56] θ = ψ Then ζ = π ψ ζ + ψ = π ζ +ψ = π 60
13 IV. Kinematics of elastic collisions Kinematics: relationships among energies of particles before and after To begin, combine 3 facts we derived previously: 1) elastic: u 1 ' = v 1 ' and u ' = v ' ) from the def. of CM (this was Eq 5): u 1 ' = u 1m 3) also from elastic, we just showed that v ' = V [slide 58] but V = u 1 [this was Eq 0] so v ' = u 1 We will use all of these. so also u ' = u 1 call this 'Eq B' 61
14 LAB frame CM frame Define: T 0 = total KE T 0 ' = total KE T 1 = KE of T 1 ' = KE of T = KE of m T ' = KE of m Goals: Express T o ' as a function of T 0 Express T 1 as a function of T 0 Express T as a function of T 0 Express T 1 ' as a function of T 0 Express T ' as a function of T 0 Begin: 6
15 T 0 = u 1 T 0 ' = 1 ( m u 1 1 ' u ' ) T 0 ' = 1 = 1 u 1 m u 1 m Subst. Eq 0 and Eq B: u 1 ( ) + m m 1 ( ) ( ) ( ) = 1 u 1 m T 0 m T 0 ' = T 0 63
16 T 1 ' = 1 v 1 ' = 1 m m 1u 1 m T 1 ' = T 0 Similarly, T ' = 1 m v ' Substitute v 1 ' = u 1 ' = u 1m = 1 m 1u 1 Substitute v ' = u ' = u 1 m ( ) T 1 ' = T 0 m ( ) 64
17 To find T 1 as a function of T 0, recall: T 1 = 1 v 1 1 T 0 u v 1 1 u 1 Recall from slide 53: v 1 ' θ ψ V ψ v 1 Notice v 1 ' = v 1 + V v 1 V cosψ So v 1 = v 1 V + v 1 V cosψ Then T 1 = v 1 T 0 u 1 V u 1 + v 1V cosψ u 1 Ter Term Term 3 65
18 Ter: Recall v 1 ' = u 1m So v 1 u = m 1 Term : Recall V = v ' = u ' = u 1 So V u = 1 66
19 Term 3: Recall v 1 'sinθ = v 1 sinψ So v 1 = v 1 'sinθ sinψ so v 1 V cosψ u 1 = V cosψ u 1 v 1 'sinθ sinψ = v 1 ' u 1 V u 1 sinθ tanψ Recall tanψ == sinθ So sinθ tanψ = cosθ + m cosθ + m [slide 55] = m cosθ + m Plug Terms 1,, and 3 into T 1 T 0 : 67
20 T 1 T 0 = = = = = = m 1 ( ) m 1 ( ) m 1 m 1 m ( ) 1 1 ( ) 1 + m ( ) cosθ + m m 1 + m cosθ + m 1 m m 1 + m cosθ + + m m + m m cosθ 1 0 ( ) ( + m ) m 1 cosθ ( m ( ) 1 ) m 1 cosθ ( ) 68
21 ( ) T 1 = 1 m 1 cosθ T 0 ( ) This is not ideal, because θ is measured in the CM frame T 1 T 0 = cosψ ± ( ) while T 1 and T 0 are defined in the LAB frame. To convert the angle to the LAB frame, use tanψ == m sinθ cosθ + m sin ψ [slide 55] to get: Similarly, T = 4m T 0 ( ) cos ζ 69
22 Special case of = m : T 1 T 0 cos ψ and T T 0 sin ψ Please read Thornton chapter 8, sections 8.4, 8.5, and 8.6 only. 70
23 I. Inelastic collision terminology II. Scattering angle 71
24 I. Inelastic collision terminology Inelastic: KE is not conserved. For a general collision, E final = E initial : v 1 + m v = m u m u + Q Subscripts "1" and "" refer to the participants in the collision. The "Q value" PE final PE initial Q = 0 elastic Q > 0 'exoergic' Q < 0 'endoergic' 7
25 Consider a headon collision (all motion is in 1dimension) Define ε v v 1 u u 1 The coefficient of restitution. These are magnitudes, not vectors. ε = 1 elastic [u 1 = v 1 and u = v ] ε 0 totally inelastic [v = v 1 ] If the collision is NOT headon, then ε applies only to the components in the 'headon' direction. 73
26 II. Scattering angle The path taken by a particle scattering from a force center is the same as a hyperbolic orbit, so equations that we use for centralforce orbital motion can be adapted for scattering problems. Thornton has a useful equation (8.17) that is not in Taylor. We will take a moment here to derive this equation of motion for the angle θ swept by an orbiting particle, as a function of its distance r from the force center. 74
27 The energy of an orbiting object E = µ r + +U 'Eq 1' (Taylor Eq. 8.35) µr Here = µr θ is angular momentum 'Eq ' (Taylor Eq. 8.3, Thornton Eq. 8.10) µ = reduced mass = m is called "centrifugal potential energy" (Taylor Eq. 8.8, Thornton Eq. 8.3) µr µ r is the translational KE U is the physical potential. Solve Eq 1 for r: r = ± µ E U µr 1 'Eq 3' Notice we can also invert Eq to get: θ = 'Eq 4' µr 75
28 Notice dθ = dθ dt dt dr dr = θ 1 r dr dθ = µr 1 ± 1 dr E U µ µr r θ = ± 1 dr 'Eq 5' µ E U µr 76
29 Now consider a particle of mass µ approaching a force center F (which may be another particle fixed in space). Particle "µ" has initial velocity v 0. If F is repulsive, particle "µ" will be deflected away: F v 0 µ Clarify the notation about the angles: 77
30 Compare the definitions of ellipse The set of points defined such that the sum of their distance from foci is constant: r'+ r = a r ' r hyperbola The set of points defined such that the difference of their distances from foci is constant: There are branches: r ' r = a and r' r = a a a r r ' Consider the physical situation in which the source of the potential is located at the focus indicated by. r = radial coordinate θ = angle subtended as particle travels. Eq 5 still holds. a 78
31 Thus for a hyperbola, as for an ellipse, we can show the relation between r and θ as: r θ r 1 The only adjustment we have to make is to note that traditionally, this angle is given a different symbol: Θ, so θ Θ. Thus for hyperbolas, Eq 5 becomes: r Θ = ± 1 dr 'Eq 6' µ E U µr 79
32 The symbol θ is instead used to indicate the angle through which the path is deflected: this would be the undeflected path θ asymptote of the deflected path so "θ" means different things for the ellipse and the hyperbola. 80
33 The problem is: angle Θ and angular momentum magnitude are not typically measured in an experiment, which makes it hard to use Eq. 6 in a scattering problem. We need to transform them into the usual observables. Impact parameter b is the distance by which the projectile would miss the force center if it did NOT deflect: Force center that is causing the deflection b Trajectory of the deflecting particle 81
34 Recall angular momentum L = r p. p = µv : reduced mass times velocity Recall that the force center defines the origin of r. Notice that if the particle DID NOT deflect, r would be to p when r = r min : at r = r min, r p = rp. On the undeflected trajectory, r min = b. Thus L = r p = bµv But also: kinetic energy T 0 ' = µv [Notice that the use of "µ" tells us that we are in the CM frame, and we use primes to indicate observables in that frame.] 8
35 So µv = T 0 ' ( ) = µ T 0 ' µ µv ( µv) = µt 0 ' µv = µt 0 ' ( ) Then = b µt 0 '. Plug this into Eq. 6: Θ = b µt 0 ' r 1 dr = µ E U b µt 0 ' µr ( b / r ) E T 0 ' U T 0 ' b r dr 83
36 Recall E is conserved. Also for elastic collisions, KE is conserved. Furthermore E = T 0 '+U 0 ' For r, U 0 ' 0 for classical fundamental forces. Then E = T 0 '. Thus set E T 0 ' Then : Θ = ( b / r ) = 1 for the asymptotic trajectory. 1 U T 0 ' b r dr 84
37 The maximum angle that a track can subtend is defined by: r b / r max ( ) Θ max = dr r min 1 U T 0 ' b r [Thornton calls this "ΔΘ"] r max = (asymptote) r min Redraw, labeling all angles: 85
38 β + γ +θ = π. By symmetry, β=γ. By parallel lines, Θ max = β. Thus Θ max +θ = π. Θ max θ β γ So θ = π Θ max Note: θ is the observable, measurable scattering angle. θ = π Θ max = π Call this 'Eq 7', but r min we need to find r min... ( b / r ) 1 U T 0 ' b r 86 dr
39 To find r min, we need to find the roots of 1 U T 0 ' b for the given U and T 0 '. r Why the roots? r = r min at the turning point ("apside") of the orbit. At the turning point, r = 0 instantaneously. Recall Eq 3 (slide 74): r = µ E U µr 1 These are different forms of the the same expression. 87
40 Procedure: Choose U and T 0 '. Set 1 U T 0 ' b r = 0. Solve for r = r min. Plug r min into the lower limit of the integral in Eq 7 and compute the integral (may be challenging). The answer has the form θ ( b,u,t 0 '). Invert the answer to get b( θ,u,t 0 '). Why do we want to know b? To predict the cross section, which depends functionally upon U. Then we measure the cross section and adjust the U function in the prediction, until the prediction matches the data. From this we infer the nature of the fundamental potential that produced the deflection. The next section explains cross section. 88
41 I. Cross sections II. Cross section formulas in the center of mass and in the lab frame III. Rutherford scattering Please read Thornton Chapter
42 I. Cross sections Consider a stream of particles directed at a target. Each particle's interaction with the target is one instance of a series of identical experiments. Different particles approach the target with slightly different and unknowable impact parameters b. We expect that deflection angle θ is related to: 1) impact parameter ) kinetic energy T 0 ', which can be controlled 3) details about the form of the potential U that is causing the deflection. We measure θ to infer U. To infer U, we have to eliminate b in favor of something that is actually measurable: cross section. 90
43 Consider N incident particles striking a target per unit area. Call N /area = intensity I. Suppose dn of them are scattered at angles ( θ,φ). Since those angles cannot be known with perfect precision, we must actually say that they are scattered into the ranges: ( θ to θ+dθ ) and ( φ to φ+dφ) The combination of these ranges is called solid angle dω' = sinθdθdφ. Define "cross section": dσ dn I dn is unitless. So the units of dσ are area. Normalize by solid angle: I has units of #/area "Differential cross section": σ ( θ ) dσ dω' = 1 I dn dω' 91
44 Notice that the particles that are scattered in range dθ were incident in range db: So dn = ( intensity of total area) ( fraction of total area contributing to the scatter ) = # area = I πbdb ( area covered by ring of width db) Notice also that scattering in angle φ is uniform, so dφ = π. Then dω' = dφ sinθdθ = π sinθdθ. Assemble σ ( θ ) 1 I dn dω' = 1 I πbdb I π sinθdθ = bdb sinθdθ 9
45 By convention, cross section is positive. But db dθ by convention, take the absolute value: σ ( θ ) b sinθ db dθ. is negative. So Now recall that we can predict θ from θ ( b) = ( b / r ) E T 0 ' U dr, for various possible potentials U. T 0 ' b r Invert it to get b( θ ). Take derivatives to get b sinθ Then also measure σ ( θ ) = 1 I db dθ. dn dω by counting deflected particles. Match the measurement to the predictions to infer the most correct form of 93 U.