More Problems

Problem 1:

Consider a hoop of mass m and radius r rolling without slipping down an incline.
Assume that at t = 0 the hoop is at x = 0 and θ = 0.  Find the equations of motion for x and θ and the forces of constraint using the method of Lagrange multipliers.

Solution:

• Concepts:
  Lagrange's Equations, Lagrange multipliers
  Reasoning:
  Assume you have chosen coordinates for a system that are not independent, but are connected by m equations of constraints of the form
  Σ_ka_lk dq_k + a_lt dt =  0,  l = 1, ..., m.
  Then the equations of motion are be obtained from
  d/dt(∂L/∂(dq_k/dt)) -  ∂L/∂q_k = ∑_lλ_la_lk,  Σ_k a_lk dq_k + a_lt dt =  0.
• Details of the calculation:
  L = T - U.
  T = (m/2)(dx/dt)² + (mr²/2)(dθ/dt)², U = -mgxsinΦ.
  L = (m/2)(dx/dt)² + (mr²/2)(dθ/dt)² + mgxsinΦ.
  Constraint: dθ = dx/r,  dx - rdθ  = 0.
  We have two coordinates, x and θ, and one equation of constraint, dx - rdθ  = 0, which must be cast into the form λ∑_ka_1k dq_k = 0,  λ(a_x dx + a_θ dθ) = 0.  Therefore a_x = 1, a_θ  = -r.
  There is only one equation of constraint, so we can drop the index l.
  ∂L/∂v = mdx/dt,  (d/dt)∂L/∂v = md²x/dt², ∂L/∂x = mgsinΦ.
  ∂L/∂ω = mr²dθ/dt,  (d/dt)∂L/∂ω = mr²d²θ/dt²,  ∂L/∂θ = 0.
  md²x/dt² - mgsinΦ = λ,  mr²d²θ/dt² = -rλ,   d²x/dt² = rd²θ/dt².
  3 equations, 3 unknowns.
  d²x/dt² = (g/2)sinΦ, d²θ/dt² = (g/(2r))sinΦ,  λ = -(mg/2)sinΦ.
  Here λ = F_x is the force of constraint associated with the coordinate x.  It is the static frictional force of constraint between the hoop and the incline.
  The generalized force of constraint associated with the coordinate θ is -rλ =  r(mg/2)sinΦ.  It is the torque acting on the hoop, pointing into the page.

Problem 2:

Small spheres of radius "r" are incident with velocity v on a stationary hard sphere of radius "R" and scatter elastically.  What is the total scattering cross section?

Solution:

• Concepts:
  Definition of the scattering cross section, geometry
• Reasoning:
  The concept of cross section, as its name suggests, is that of effective area for collision.  The cross section of a spherical target of radius R for point particles is σ = πR².  If the scattering particles have radius r, then the cross section is σ = π(r+R)². 
  

Problem 3:

(a)  Derive the relationship between the impact parameter b and the scattering angle θ for Rutherford scattering of a projectile of mass m by a fixed target particle of mass M. 
(b)  If 4 MeV alpha particles are incident on a gold foil (Z = 79), calculate the impact parameter that would give a deflection of 10 degrees.
(c)  Explain what modifications of this calculation must be made if the gold atom is allowed to recoil during the collision.

Solution:

• Concepts:
  Motion in a central potential, elastic scattering
• Reasoning:
  For the case of the repulsive Coulomb potential U(r) = α/r, we are asked to find the relationship between the impact parameter and the scattering angle.
• Details of the calculation:
  In a central potential the motion is in a plane and M and E are constant.
  E = ½m(dr/dt)² + b²E/r² + U(r),  dr/dt = ±[(2/m)(E(1 - b²/r²) - U(r))]^½.
  From
  dr/dt = ±[(2/m)(E(1 - b²/r²) - U(r))]^½,  d/dt = [b(2mE)^½/(mr²)] d/dφ,
  we find the equation for the trajectory.
  dφ = [b(2mE)^½/(mr²)]dr [(2/m)(E(1 - b²/r²) - U(r))]^-½.
  φ(u) = b∫₀^udu'/[1 + b²u'² - U(u')/E]^½,
  with u = 1/r and φ(z = -∞) = 0.
  
  Let θ be the angle between the incident and the scattered direction and φ₀ be the angle between r(z = -∞) and r_min.  Then
  φ₀ = b∫₀^umaxdu'/[1 - b²u'² - U(u')/E]^½,
  and
  E = M²/(2mr²_min) + U(r_min) = b²E/r²_min + U(r_min) =  b²Eu²_max + U(u_max)
  determines u_max.  We have
  θ = π - 2φ₀ for a repulsive potential,
  θ = 2φ₀ - π for an attractive potential, (or θ = π - 2φ₀, θ < 0).

  If U(r) = α/r,  U(u) = αu,  U(u_max) = αu_max, then
  u_max = -α/(2b²E) + [α²/(4b⁴E²) + 1/b²]^½ and
  φ₀ = b∫₀^umaxdu'/[1 - b²u'² - αu'/E]^½.

  Let bu' = x, αu'/E = 2cx, c = α/(2bE).
  Then b*u_max = -c + (c² + 1)^½ and
  φ₀ = b∫₀^umaxdu'/[1 - b²u'² - αu'/E]^½ = ∫₀^b*umaxdx/[1 - x² - 2cx]^½
  = ∫₀^b*umaxdx/[1 + c² - (x + c)²]^½ = (1 + c² )^-½∫₀^b*umaxdx/[1 - (x + c)²/(1 + c² )]^½
  = ∫_ymin^ymaxdy/[1 - y²]^½),
  with y = (x + c)/(1 + c²)^½,  dy = dx (1 + c²)^½,  y_min = c/(1 + c²)^½,  y_max = 1.
  Since |y| ≤ 1, let y = cosβ,  dy = -sinβdβ. Then
  φ₀ = ∫_ymin^ymaxdy/[1 - y²]^½ = ∫₀^acos(ymin)dβ =  cos^-1(y_min) = cos^-1(c/(1 + c²)^½).
  cosφ₀ = c/(1 + c²)^½.
  1/cos²φ₀ = tan² + 1.  tan²φ₀ = 1/c²,   cotφ₀ = c = α/(2bE).
  cot(θ/2) = cot(π/2 - φ₀) = tanφ₀ = 1/cotφ₀.
  b = [α/(2E)] cot(θ/2)
  is the relationship between the impact parameter b and the scattering angle θ for Rutherford scattering.

  (b)  E = 4 MeV = 4*10⁶ eV(1.6*10^-19 J/eV) = 6.4*10^-13 J.
  α = 1/(4πε₀)(2)(79)(1.6*10^-19 C)² 
  =  (9.0*10⁹ Nm²/C²)(158)(2.56*10^-38 C²)
  =  3640*10^-29 Nm² = 3.6*10^-26 Jm.
  θ = π - 2Φ₀ = 10^o.
  cot 5^o  = 11.43.
  b = (3.6*10^-26 Jm)(11.43)/2(6.4*10^-13 J)
   = 3.2*10^-13  m.
  
  (c)  If the target is allowed to recoil, then we solve for the motion of the CM and the motion about the CM, i.e. the relative motion.  The CM moves with constant velocity V = m v₁/(m + M).  The problem of two interacting particles in their CM frame is equivalent to the problem of a fictitious particle of reduced mass μ moving in a central potential U(r).  The reduced mass of the alpha particle-gold nucleus system is μ = mM/(m + M).  If the energy of the alpha particle in the lab frame is E, then the energy of the fictitious particle is E' = ME/(M + m).  Given a scattering angle θ in the lab frame we need to find the corresponding scattering angle θ' in the CM frame. 
  
  v'sinθ' = vsinθ,  v² = v'² + V² + 2v'Vcosθ'.
  Then we use 
  b = (α/2E')cot(θ'/2)
  to find the impact parameter b for θ'.