Elastic collisions

Problem:

A 3 kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 60^o with the surface.  It bounces off with the same speed and angle.  If the ball is in contact with the wall for 0.2 s, what is the average force exerted on the ball by the wall?

Solution:

• Concepts:
  Impulse: dp = Fdt,  ∆p = F_avg∆t
• Reasoning:
  In the elastic collision with the "infinitely" massive wall, the ball receives an impulse ∆p.  Enough information is given to calculate ∆p.  Since the contact time is given, F_avg can be calculated.
• Details of the calculation:
  The balls initial momentum is
  p_i = p_xii + p_yij
  = (3 kg 10 m/s)sin60^oi + (3 kg 10 m/s)cos60^oj.
  Its final momentum is
  p_f = p_xfi + p_yfj
  = -(3 kg 10 m/s)sin60^oi + (3 kg 10 m/s)cos60^oj.
  ∆p = p_f - p_i= -2(30 kgm/s)sin60^oi = -(51.96 kgm/s)i.
  ∆p = F_avg ∆t.
  F_avg = -(51.96 kgm/s)i/(0.2s) = -259.8 Ni.

Problem:

A block of mass m₁ and initial velocity v₁ collides head-on with a stationary block of mass m₂.  The mass m₂ compresses a spring of spring constant k.  Neglect friction and assume the collision is elastic.
(a)  What is the velocity v₂' of m₂ just after the collision?
(b)  What is the maximum compression of the spring?

Solution:

• Concepts:
  Elastic collisions, energy and momentum conservation
• Reasoning:
  In elastic collisions energy and momentum are conserved.
• Details of the calculation:
  (a)  Let v₁'and v₂' be the velocities of m₁ and m₂ just after the collision.
  Momentum conservation: m₁v₁ = m₁ v₁' + m₂ v₂'.
  Energy conservation:  m₁v₁² = m₁ v₁'² + m₂ v₂'².
  v₂' = 2v₁/(1 + m₂/m₁).
  (b)  ½m₂v₂'² = ½kd², where d is the maximum compression of the spring
  d = (m₂/k)^½2v₁/(1 + m₂/m₁).

Problem:

A (non-relativistic) neutron in a reactor makes an elastic head-on collision with the nucleus of a carbon atom initially at rest.
(a) What fraction of the neutron's kinetic energy is transferred to the carbon nucleus?
(b) If the initial kinetic energy of the neutron is 1.6*10^-13 J, find its final kinetic energy and the kinetic energy of the carbon nucleus after the collision.
(The mass of the carbon nucleus is about 12 times the mass of the neutron.)

Solution:

• Concepts:
  Elastic collisions, momentum conservation, energy conservation
• Reasoning:
  The collision of the two particles is elastic.
  Momentum conservation:
  (i) m₁v_1i = m₁v_1f + m₂v_2f.
  Energy conservation:
  (ii) ½m₁v_1i² = ½m₁v_1f² + ½m₂v_2f².
• Details of the calculation:
  (a)  We can solve this system of two equations for the ratio v_2f/v_1i.  We obtain
  v_2f² = (m₁/m₂)(v_1i² - v_1f²)   from (ii),
  v_1f = v_1i - (m₂/m₁)v_2f    from (i).
  We can therefore write
  v_1f² = v_1i² + (m₂/m₁)²v_2f² - 2(m₂/m₁)v_1iv_2f.
  v_2f² = (m₁/m₂)( 2(m₂/m₁)v_1iv_2f - (m₂/m₁)²v_2f²) = 2v_1iv_2f - (m₂/m₁)v_2f².
  (1 + (m₂/m₁))v_2f = 2v_1i.
  v_2f = 2m₁v_1i/(m₁+m₂).
  v_2f/v_1i = 2m₁/(m₁+m₂) = 2/13 = 0.15.
  The fraction of the kinetic energy transferred to the carbon nucleus is
  (m₂/m₁)(v_2f/v_1i)² = 12(0.15)² = 0.284.

  (b)  The initial kinetic energy of the of the neutron is 160 fJ and the final kinetic energy of the carbon nucleus is 45.4 fJ. (1 femtoJoule = 1 fJ = 10^-15 J.)  The final kinetic energy of the neutron is (160 - 45.4) fJ = 114.6 fJ.

Problem:

A non-relativistic neutron with mass m and kinetic energy T scatters elastically off a nucleus with mass M.  What is the maximum kinetic energy that can be transferred to the nucleus in one collision if
(a)  M is initially at rest,
(b)  or M is initially allowed to move?

Solution:

• Concepts:
  Elastic collisions, energy and momentum conservation
• Reasoning:
  The maximum kinetic energy is transferred in a head-on collision.  We use energy and momentum conservation.
• Details of the calculation:
  (a)  If M is at rest and m = m₁, M = m₂.
  Momentum conservation:
  (i) m₁v_1i = m₁v_1f + m₂v_2f.
  Energy conservation:
  (ii) ½m₁v_1i² = ½m₁v_1f² + ½m₂v_2f²
  v_2f² = (m₁/m₂)(v_1i² - v_1f²)   from (ii),
  v_1f = v_1i - (m₂/m₁)v_2f    from (i).
  We can therefore write
  v_1f² = v_1i² + (m₂/m₁)²v_2f² - 2(m₂/m₁)v_1iv_2f.
  v_2f² = (m₁/m₂)( 2(m₂/m₁)v_1iv_2f - (m₂/m₁)²v_2f²) = 2v_1iv_2f - (m₂/m₁)v_2f².
  (1 + (m₂/m₁))v_2f  = 2v_1i.
  v_2f  = 2m₁v_1i/(m₁+m₂).
  The maximum amount of kinetic energy transferred is ½Mv_2f² = [4mM/(M+m)²]T.
  (b)  If M is allowed to move, then the maximum amount of kinetic energy transferred is T, and m is at rest after the collision.
  (For this to happen M must initially moves with speed ½(1 - m/M)v into the same direction as m.)

Problem:

Two perfectly elastic balls, the larger of mass M and the smaller of mass m, with m << M, are dropped from a height h >> radius of either ball above a solid surface.  Mass M rebounds elastically from the surface and mass m rebounds elastically from M.  Find the height H that the small ball reaches in terms of h.  You need only find an approximate result in the limit that m/M --> 0.  The general result depends upon the ratio m/M.

Solution:

• Concepts:
  Elastic collisions, frame transformations
• Reasoning:
  When an object of mass m₁ << m₂ and velocity v collides head on with an object of mass m₂ at rest, then m₁ rebounds with a velocity nearly equal to -v.
• Details of the calculation:
  Assume M collides with velocity v = -(2gh)^½ j = -vj with the ground.  It rebounds with velocity vj.  In a frame moving with velocity vj equal to the velocity of M right after the collision with the ground, m approaches with velocity -2vj and rebounds with velocity 2vj.  In the lab frame m moves with velocity 3vj = (2gh')^½ j right after its collision with M, and therefore rises to a height of h' = 9h.

Problem:

An elastic ball is dropped on a long inclined plane.  It bounces, hits the plane again, bounces, and so on.  Let us label the distance between the points of the first and the second hit d₁₂ and the distance between the points of the second and the third hit d₂₃.  Find the ratio d₁₂/d₂₃.

Solution:

• Concepts:
  Elastic collisions, motion with constant acceleration
• Reasoning:
  The component of the acceleration perpendicular to the inclined plane and the component of the acceleration parallel to the inclined plane are constant.  The balls motion perpendicular to the inclined plane is a series of bounces from some initial height above the plane to the plane and back to the initial height.  The time interval between successive contacts with the plane is constant.  But the ball also accelerates in the direction parallel to the plane, so the distance along the plane between successive contacts with the plane increases.
• Details of the calculation:
  Assume the inclined plane makes an angle θ with the horizontal.  Orient the axes of your coordinate system as shown in the figure.  The x-axis makes an angle θ with the horizontal and the y-axis is perpendicular to the inclined plane.  The ball is dropped vertically and just after the first bounce its velocity vector makes an angle of 90^o - θ with the x-axis. Let x = 0, y = 0, t = 0 be the coordinates of the ball at the instant the ball makes contact with the inclined plane the first time.
  In the time interval between the first and second bounce the equations giving the position and velocity of the ball as a function of time are:
  x(t) = v sinθ t  + ½ g sinθ t²,    y(t) = v cosθ t  - ½ g cosθ t²
  v_x(t) = v sinθ   + g sinθ t,    v_y(t) = v cosθ   - g cosθ t
  The ball bounces for the second time when y(t) = 0, t = t_a = 2v/g.
  x(t_a) = (2v²/g) sinθ  + (2v²/g) sinθ = (4v²/g) sinθ
  v_x(t_a) = v sinθ  + 2v sinθ = 3v sinθ,  v_y(t_a) = v cosθ - 2v cosθ = -v cosθ
  Just after the second bounce, the velocity of the ball is v_x = 3v sinθ, v_y = v cosθ.
  Let t' = t - t_a.  Between the second and the third bounce the equations giving the position and velocity of the ball as a function of time are:
  x(t) - x(t_a) = 3v sinθ t' + ½ g sinθ t'²,    y(t) = v cosθ t' - ½ g cosθ t'²
  The ball bounces for the third time when y(t) = 0, t' = t_b = 2v/g.
  x(t_a + t_b) - x(t_a) = (6v²/g)sinθ  + (2v²/g) sinθ = (8v²/g) sinθ
  Therefore d₁₂/d₂₃ = x(t_a)/( x(t_a + t_b) - x(t_a)) = ½.

Problem:

Three elastic spheres of equal size are suspended on light strings as shown; the spheres nearly touch each other.  The mass M of the middle sphere is unknown; the masses of the other two spheres are 4m and m.  The sphere of mass 4m is pulled sideways until it is elevated a distance h from its equilibrium position and then released.  What must the mass of the middle sphere be in order for the sphere of mass m to rise to a maximum possible elevation after the first collision with the middle sphere?  What is that maximum elevation H?

Solution:

• Concepts:
  Elastic collisions, energy and momentum conservation
• Reasoning:
  In elastic collisions mechanical energy and momentum are conserved.
• Details of the calculation:
  For a collision between a sphere of mass m₁ and initial velocity v_1i and a stationary sphere of mass m₂, energy and momentum conservation yield v_2f = 2m₁v_1i/(m₁ + m₂).
  Let mass 4m move with speed v = (2gh)^1/2 just before the first collision.
  Here we have for the collision between 4m and M: v_2f = 8mv/(4m + M).
  Now v_2f becomes v_1i for the collision between M and m.
  We have for this collision: v_f = 16mMv/[(m + M)(4m + M)], where v_f is the speed of mass m immediately after the collision.
  We want to maximize v_f with respect to M.
  v_f = 16mMv/(M² + 5Mm + 4m²),
  dv_f/dM = 0 -->  16m(M² + 5Mm + 4m²) - 16mM(2M + 5m) = 0.
  16M² = 64m², M = 2m. 
  This is the only physical value for M for which dv_f/dM = 0.
  v_f --> 0 as M --> 0 or infinity, we have a maximum.
  v_fmax = (16/9)v = (16/9)(2gh)^1/2.
  H = v_fmax²/2g = (16/9)²h = 3.16 h.