9-4 Conservation of Linear Momentum
9-4 The Conservation of Linear
Momentum
—
In particular cases where there is no change in force on a
system, Newton’s 2nd Law can be manipulated to show a
conservation in momentum.
—
If ΣF=0, then the ∆p=0.
Writing the change of momentum in terms of its initial
and final values, we have: —
9-4 One Quick Reminder!
—  Remember
that both force and
momentum are _______ quantities.
—  Remember that _______ quantities can
have ____ ____________: an x and a ycomponent.
—  Finally, the momentum conservation
principle applies to each component
separately.
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Law of Conservation of Momentum
—  Total
momentum of a system remains the same
when the net external force acting on the system
is zero.
—  General formula for two objects:
m1v1i + m2v2i +… = m1v1f + m2v2f +…
Collisions in more detail
—  There
are two main types of collisions
◦
◦
—  Momentum
is ___________ conserved in
an ____________ collision, but
mechanical energy is not always conserved
—  If
kinetic energy is lost in a collision, where
does it go?
Inelastic Collisions
—  Total
kinetic energy of the system is not
conserved
—  Loses KE, but not max
◦  In various forms:
–  Deformation
–  Sound
–  Light
–  Heat
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9-5 Inelastic Collisions
—  In
any collision where the KE is not
conserved, it is referred to as an
inelastic collision.
—  In an inelastic collision, the momentum
of the system is conserved, but not its
KE.
—  Nearly all collisions in the world are a
type of inelastic collision.
9-5 Inelastic Collisions
—  In
the case where the two objects “stick
together” after the collision, it is said to
be completely inelastic or perfectly
inelastic. (Max KE lost)
—  In
the case where two objects “start
together” and push away from one
another, it is said to be a reverse
inelastic collision or explosion.
9-5 Inelastic Collisions
•  The first bullet passes
through the block and
maintains much of its original
momentum, but loses some KE
•  The second bullet, expands
as it enters the block of wood,
which prevents it from passing all the way through it.
Most of the momentum transfers and a max KE is lost.
•  The third bullet bounces off the block transferring “all
of its own momentum” and then borrowing some more
from the block. This has the most momentum
transferred to the block and loses no KE.
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Conservation of Linear Momentum
A honeybee with a mass of
0.150 g lands on one end
of a 4.75 g popsicle stick.
After sitting at rest for a
moment, it runs toward
the other end with a
velocity vb relative to the
still water. The stick
moves in the opposite
direction with a speed of
0.120 cm/s. What is the
velocity of the bee?
9-6 Elastic Collisions
—  In
an elastic collision, momentum and
KE are conserved.
—  Most collisions are typically inelastic
because with elastic collisions, KE is
conserved, meaning no deformations, no
sound, no heat.
Elastic collision problem
—
A 5.00-kg ball, moving to the
right at a velocity of +2.00 m/s
on a frictionless table, collides
head-on with a stationary 7.50kg ball. Find the final velocities
of the balls if the collision is
elastic.
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9-7 Center of Mass
—  The
center of mass of a system is the average
position of all the mass in the system
—  Mathematically, the
position:
mass-weighted average of
xcm =
—  This
equation can be generalized to any
number of particles and to two or more
dimensions
9-7 Center of Mass
—
—
To extend the definition further, think of a system in two
dimensions instead of just one.
In this case, the CM will have both an x and a ycomponent.
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Calculating the Center of Mass
—
The drawing shows a sulfur dioxide molecule. It consists of
two oxygen atoms and a sulfur atom. A sulfur atom is twice
as massive as an oxygen atom. Using this information and the
data provided in the drawing, find the center of mass of the
sulfur dioxide molecule.
Calculating the Center of Mass
XCM =
(32)(0) + (16)(−.124) + (16)(.124) 0
=
= 0 nm
(32 +16 +16)
64
YCM =
(32)(0) + (16)(.0715) + (16)(.0715) 2.288
=
= 0.0358nm
(32 +16 +16)
64
(0 nm, 0.358 nm)
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