No Slide Title

							Chapter 10
    The Take-Home Message on
           Chapter 10
   If two bodies are exerting STRONG forces on each
    other over a short period of time, then you have a
    collision problem.
   Linear Momentum is ALWAYS conserved.
   Kinetic Energy is NOT always conserved.
    – Kinetic Energy is conserved during an elastic collision.
    – Kinetic Energy is not necessarily conserved during an
      inelastic collision.
   Impulse-Linear Momentum Theorem
    – Relates strength and duration of collision force to
      change in momentum
    – Similar to Work-Kinetic Energy Theorem
               Equations
   Δp = pf - pi = J       (10-4)
 J = F Δt                 (10-8)
 F 
         n
            mv           (10-10)
        t
        (m1  m2 )
v1 f              v1i   (10-18)
        (m1  m2 )
           2m1
   2f              v1i
v                         (10-19)
        (m1  m2 )
            More Equations
   .                                        (10-28)

    .       2m1              (m2  m1 )
   v2 f               v1i              v2i (10-29)
           (m1  m2 )         (m1  m2 )
   .m v  (m  m )V                         (10-34)
      1      1   2


   . 1v1  m2v2  (m1  m2 )V
    m                                        (10-36)
               Last One
               P      m1v1i  m2v2i
   .
    vcm            
            m1  m2     m1  m2
                   (10-30)
            Problem Types
   Impulse and Linear Momentum
    – Series of Collisions
   Elastic Collision
    – 1-D
    – 2-D
    – Moving Target vs. Stationary Target
   Inelastic Collision
    – 1-D
    – 2-D
     How to Solve Them
   Impulse and Linear Momentum
    – Use Conservation of Momentum
    – Impulse-Linear Momentum Theorem
   Series of Collisions
    – Use Equation (10-10)
   Elastic Collisions
    – Use Conservation of Momentum
    – Use Conservation of Kinetic Energy
    – In Two Dimensions Treat Momentums as
      Vector Quantities (x,y), and Solve as Usual
                  Problem #1
   A 283g air-track glider
    moving at 69 m/s on a 24 m
    long air track collides
    elastically with a 467g glider
    at rest in the middle of the
    horizontal track.

    The end of the track over which the struck glider moves
    is not frictionless, and the glider moves with a
    coefficient of friction =0.02 with respect to the track.
    Will the glider reach the end of the track? Neglect the
    length of the glider.
          Problem #1 Solution
   m1 = 0.283 kg
   m2 = 0.467 kg
   Kinetic Energy Conserved (Elastic Collision)
   Momentum Conserved
   Ki=Kf
        1                    1               1
    Ki  (m1 )( v1i )  K f  (m1 )( v1 f )  (m2 )( v2 f )2
                     2                     2

        2                    2               2
    (0.283 kg)(0.69 m/s)2 = (0.283 kg)(v1f)2 + (0.467 kg)(v2f)2

            0.1445(v1f)2 + 0.2335(v2f)2 = 0.0674
    Problem #1 Solution (cont.)
   Pi = Pf
    m1v1i = m1v1f + m2v2f
    (0.283 kg)(0.69 m/s) = (0.283 kg)(-v1f) + (0.467 kg)(v2f)
          v1f = -0.69 m/s + (1.65)v2f


   Two Equations and Two Unknowns
    0.1415 (-0.69 m/s + 1.65v2f)2 + 0.2335(v2f)2 = 0.067
   Solve to find that v2f = 0.525 m/s
   Substitute this to Solve for Kinetic Energy of Struck Glider
    K = 1/2 m2(v2f)2 = 1/2 (0.467)(0.525)2
    KE of struck glider = 0.064 J
    Problem #1 Solution (cont.)
   Ffriction = μN = μm2g
             = (0.02)(0.467 kg)(9.8 m/s2)
             = 0.0915 N
   Work Done by Friction = (0.0915 N)(1.2 m) =
    0.11 J
   K < Wfriction
   Glider Does Not Reach the End of the Track
                     Problem #2
   Wile E. Coyote, having just
    passed Physics 207 at VCU,
    splurges on a 500 kg ACME
    rocket. He reads the rocket
    specs to discover a relative
    exhaust speed of 600 m/s and a
    fuel consumption rate of 2
    kg/s.
   He launches horizontally after the roadrunner and flies for twenty
    seconds before the road turns and the rocket does not work. He
    collides with the 4000 kg ACME delivery truck (initially
    stationary). Neglecting gravity and air resistance, what is the
    relative velocity of the truck just after the totally inelastic
    collision? Assume the super genius to have a mass of 25 kg.
           Problem #2 (cont.)

   Rocket Science is needed for this problem:
    – Use the second rocket equation to find the
      rocket speed at impact
                        mi
        vf - vi = u ln
                        mf
                               500kg  25kg
          vf = (600 m/s) ln
                                  485kg

          vf = 47.55 m/s
                 Problem #2 (cont.)
   Now solve for vtruck using conservation of momentum:
    – P i = Pf
    – mrocket ivrocket i + mtruck i vtruck i = mrocket f vrocket f + mtruck f vtruck f


                    mrocketi vrocketi       (485kg)( 47.55m / s )
    – vf =                                
                  mrocket f  mtruck f        485kg  4000kg

            = 5.14 m/s
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