Lecture Corrections

#1 Chapter 6,  Lecture Date: June 25, 2009. See the discussion of inelastic collisions revolving around  Figure 6.14. In the absence of external forces and in the presence  only of mutually canceling  internal forces, we have conservation of momentum: M1V1i + M2V2i = M1V1f + M2V2f    where the symbol V represent the velocity, 1 or 2 each mass, i  = initial (before the collision) and f = final ( after) .  Let the positive direction be right. When V is positive, the object moves to the right as in the case when   V = + 5.0 m/s.  When V is negative, the object moves to the left,  i.e. when V = - 4.6 m/s.  Now let V2i = 0 and  V1f  = V2f   = Vf  = common velocity of the two entangled (stuck together) objects. Thus we have:   M1V1i  = (M1 + M2) Vf  .  I gave an example with something like M1 = 1.0 kg ,  M2  = 3.0 kg  and V1i  = 8.0 m/s. This gives: 8.0 kg m/s = (4.0 kg)Vf  yielding Vf =   2.0 m/s. We  correct   the expression for the final common velocity, tying it  to the final (not the initial) velocity of each object.

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