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: M _{1}V_{1i} + M_{2}V_{2i }= M_{1}V_{1f}
+ M_{2}V_{2f } 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 V_{2i } = 0 and V_{1f}
= V_{2f } = V_{f} = common velocity of
the two entangled (stuck together) objects. Thus we have:M _{1}V_{1i} _{ }= (M_{1} + M_{2})
V_{f} . I gave an example with something like M_{1}
= 1.0 kg , M_{2} = 3.0 kg and V_{1i}
= 8.0 m/s. This gives: 8.0 kg m/s = (4.0 kg)V_{f} yielding
V_{f} = 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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