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B.7.14 Conservation of energy
Potential plus kinetic energy of a closed system is conserved ( Eq. B.128 ) . This is useful, for example, when
solving for an object's current height (current potential energy) when its initial height (initial potential
energy), initial velocity (initial kinetic energy), and current velocity (current kinetic energy) are known.
U a þ K a ¼ U b þ K b
(B.128)
B.7.15 Conservation of momentum
In a closed system, total momentum (mass times velocity) is conserved. This means that it does not
change ( Eq. B.129 ) and that the momenta of all objects in a closed system always sum to the same
amount ( Eq. B.130 ). This is useful, for example, when solving for velocities after a collision when
the velocities before the collision are known.
d
dt ððm 1 v 1 þ m 2 v 2 þ ...þ m n v n ÞÞ ¼
0
(B.129)
m 1 v 1 þ m 2 v 2 þ ...þ m n v n ¼ constant
(B.130)
B.7.16 Oscillatory motion
In some systems, the stability of an object is subject to a linear restoring force, F. The force is linearly pro-
portional (using k as the constant of proportionality) to the distance x the object has been displaced from its
equilibriumposition ( Eq. B.131 ) . Associatedwith this force is the potential energy ( Eq. B.132 ) thatresults
from the object's displacement. These systems have the property that, if they are disturbed from equilib-
rium, the restoring force that acts on them tends tomove themback into equilibrium. When disturbed from
equilibrium, they tend to overshoot that point when they return, due to inertia. Then the restoring force acts
in the opposite direction, trying to return the system to equilibrium. The result is that the system oscillates
back and forth like a mass on the end of a spring or the weight at the end of a pendulum.
F ¼kx
(B.131)
1
2 kx
2
U ¼
(B.132)
Combining the basic equations of motion (force equals mass times acceleration) with
Equation B.131 , one can derive the differential equation for oscillatory motion. This equation is sat-
isfied by the displacement function x ( t )( Eq. B.133 ) .
F ¼ ma
F ¼kx
2
d
x
dt
a ¼
(B.133)
2
m d
x
dt ¼kx
d
2
x
dt ¼kx=m
 
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