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Transcript
Lecture 28: More on Collisions • Last time we started discussing the fraction of the kinetic energy retained by m1 after an elastic collision in the LAB frame is: T1 v12 = 2 T0 u1 • To relate this to the LAB scattering angle, we use conservation of energy and momentum: p x = m1u1 = m1v1 cosψ + m2 v2 cos ζ p y = 0 = m1v1 sinψ − m2 v2 sin ζ 1 1 1 2 2 T = m1u1 = m1v1 + m2 v22 2 2 2 3 equations Treat u1/v1, v2, ψ and ζ as variables Can solve for u1/v1 as a function of ψ • From the energy equation, we find: m1 2 v2 = ( u1 − v12 ) m2 • Plugging this into the py equation gives: m1v1 sinψ = m2 sin ζ = m1 2 ( u1 − v12 ) sin ζ m2 m1v1 sinψ m1m2 ( u12 − v12 ) cos ζ = 1 − = 1 − cos2 ζ m v sin ψ = 2 2 m1m2 ( u1 − v1 ) 2 2 1 1 2 m2 ( u12 − v12 ) − m1 v12 sin 2 ψ m2 ( u12 − v12 ) • This means that: m2 v2 cos ζ = m2 2 2 2 2 m u v m v − − sin ψ ( ) m1 2 2 1 1 1 1 2 ( u1 − v1 ) m2 m2 ( u12 − v12 ) = m1 m2 ( u12 − v12 ) − m1 v12 sin 2 ψ • We can now insert this into the px equation to find: m1u1 = m1v1 cosψ + m1 m2 ( u12 − v12 ) − m1 v12 sin 2 ψ m1 ( u1 − v1 cosψ ) = m1 m2 ( u12 − v12 ) − m1 v12 sin 2 ψ m12 u12 − 2m12 v1 cosψ + m12 v12 cosψ = m1 m2 ( u12 − v12 ) − m1 v12 sin 2 ψ • Or: v12 ( m1 + m2 ) − 2m1u1v1 cosψ + u12 ( m1 − m2 ) = 0 v12 v1 m + m2 ) − 2m1 cosψ + m1 − m2 = 0 2 ( 1 u1 u1 from which we can find v1/u1 with the quadratic formula: 2 2 2 2 v1 2m1 cosψ ± 4m1 cos ψ − 4 ( m1 − m2 ) = u1 2 ( m1 + m2 ) 2 m1 m = cosψ ± cos2 ψ − 1 − 22 m1 + m2 m1 m1 m22 2 = cosψ ± − sin ψ 2 m1 + m2 m1 Inelastic Collisions • All collisions conserve energy (just like elastic collisions do) • But, if systems of particles collide, the internal energy of the systems may change – Therefore, the kinetic energy associated with the motion of the center of mass is not conserved • Such collisions are called inelastic – The extreme case is a collision between two objects that stick together after they collide (two blobs of clay or silly putty might behave this way) – These collisions are called totally inelastic • Most collisions lie somewhere between elastic and totally inelastic Quantifying Inelasticity • There are two ways to measure how inelastic a collision is 1. Measure the kinetic energy before and after the collision, and call the difference Q: T f − Ti = Q 1 2 2 1 2 2 1 2 2 1 2 2 Q + m1 u1 + m2 u2 = m1 v1 + m2 v2 2 2 2 2 The larger |Q| is, the more inelastic the collision – – If Q > 0, the final kinetic energy is greater than the initial. Such collisions are called exoergic – an explosion is one example If Q < 0, the final kinetic energy is less than the initial. These collisions are endoergic – a collision between blobs of clay falls into this category Coefficient of Restitution 2. Another way to think about inelasticity is to consider the relative velocity in a head-on collision. In the CM frame: Elastic collision m2 m1 u u2 1 v1= -u1 v2 = -u2 v1 − v2 =1 u1 − u2 • • Totally inelastic collision m1 u m2 u 1 2 m1 + m2 v1 = v2 = 0 v1 − v2 =0 u1 − u2 In general we define v1 − v2 = ε as the coefficient of u1 − u2 restitution For head-on collisions in non-CM reference frames, the velocity components normal to the collision plane enter the formula Impulse • Even though we may not know what forces act during a collision, we can determine something about those forces from Newton’s Second Law • During the collision: d ( mv ) F=p= dt Fdt = d ( mv ) t2 Fdt = ∆ ( mv ) = m∆v if m is constant t1 ∆(mv) is called the impulse, and givcn the • The quantity symbol P • The left-hand side is related to the time-average of the force acting during the collision, so we have: Favg ∆t = P Scattering Cross Sections • We’ve seen that if we know the initial velocities, and know the forces involved, we can exactly predict the results of the collision • In particle physics we want to reverse the process – i.e., we want to learn about the forces acting between particles by observing the results of a collision • There’s a catch, though: we can’t specify the initial state exactly – there’s a limit to how well we can aim the incoming particles – We can typically direct a beam of particles (say, electrons or protons) into a block of material, or at each other – In each case, we don’t know which two particles are going to collide, or the initial distance between them • In the CM frame, the collision looks like: θ T1 b T2 • The quantity b is called the “impact parameter”, and we usually don’t know its value for a collision – We typically do control the initial energies accurately, and can measure the scattering angle and final energies – We also know the intensity (number of particles per unit area) of the incoming beams • So we define a quantity in terms of the things we know or can measure: Collisions per particle that cause scattering into a solid angle element d Ω′ at angle θ σ (θ ) = Beam intensity • Note that the numerator has no dimension, while the denominator has dimensions of 1/area • So, σ(θ) has dimensions of area, which is why we give it the name “cross section” • As defined, σ(θ) relates to scattering into a differential element of solid angle, so it’s called the “differential cross section” • The equation for differential cross section is dN / d Ω′ ( ) σ θ = I dN ( ) σ θ d Ω′ = I • If the force between the colliding particles is symmetric about the collision axis (as all central forces are), we can integrate over φ to find: d Ω′ = 2π sin θ dθ • For particles of a given energy, the impact parameter determines the scattering angle • If collisions with impact parameter between b and b + db result in scattering into d Ω,′ we have: dN = IdA = I ⋅ 2π bdb • Relating the two expressions for dN gives: I ⋅ 2π bdb = I σ (θ ) d Ω′ = I σ (θ ) 2π sin θ dθ σ (θ ) = b db sin θ dθ • Actually, for most collision forces db/dθ is negative – the closer the objects come, the greater the scattering angle • Since the cross section should be a positive quantity, we write: b db σ (θ ) = sin θ dθ