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v rn = [ v 1− v 2− ] • n
v rn = [(71 m/s) i + (0 m/s) j ] • [ (0.875) i − (0.484) j ]
v rn = 62.1 m/s
The velocity components of the bat and ball in the normal direction are:
v 1n− = v 1− n = 27.1 m/s
v 2n− = v 2− n = −35.0 m/s
Applying the principle of conservation of momentum in the normal direction and
solving for v 1n+ yields:
m 1 v 1n− + m 2 v 2n− = m 1 v 1n+ + m 2 v 2n+
(1.02 kg) (27.1 m/s) + (0.15 kg) (−35.2 m/s) =
=(1.02 kg) v 1n+ + (0.15 kg) v 2n+
v 1n+ = 21.92 m/s − (0.14 m/s) v 2n+
As in the previous example, applying the formula for coefficient of restitution with the
preceding formula for v 1n+ yields:
e = (−v 1n+ + v 2n+ ) / (v 1n− − v 2n− )
0.46 = [−21.92 m/s + (0.14 m/s) v 2n+ + v 2n+ ] / [27.1 m/s + 35.2
m/s]
v 2n+ = 44.4 m/s and v 1n+ = 15.7 m/s
Here again, since this impact is frictionless, each object retains its original tangential
velocity component. For the bat, this component is 15 m/s, while for the ball it's −19.3
m/s. Converting these normal and tangential components to x - y coordinates yields the
following bat and ball velocities for the instant just after impact:
v 1+ = 21.0 m/s i − 5.5 m/s j
v 2+ = 30 m/s i + 38.7 m/s j
Both of these examples illustrate fundamental impact analysis using the classical ap‐
proach. They also share an important assumption—that the impacts are frictionless. In
reality, you know that billiard balls and baseballs and bats collide with friction; other‐
wise, you would not be able to apply English in billiards or create lift-generating spin
on baseballs. Later in this chapter we'll discuss how to include friction in your impact
analysis.

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