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Chapter 9
Center of Mass, LinearMomentum, & Collisions
Prof. Raymond Lee,revised 10-25-2010
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• Center of mass (CM)
• Center of mass (CM) of system orobject is a point that moves as if allsystem mass existed only there
• System moves as if an external F wereapplied to a single particle of mass M(total system mass) located at CM
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• CM coordinates
• Discrete-object CM coordinates (Eq. 9-5, p. 203):
where M is total system mass
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• CM position
• Locate CM using its position vector rCM:
• ri is position of ith particle & is defined by
(SJ 2008 Eq. 9.31, p. 246)
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• CM example
• Both masses on x-axis
• CM is on x-axis
• CM is closer to particlewith larger mass
(compare Fig. 9-2, p. 202)
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• CM for extended object
• Think of extended object as systemcomprised of many particles
• Since particle separation is small, considermass to be continuously distributed
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• Extended-object CM coordinates
• Extended- or continuous-object CMcoordinates (Eq. 9-9, p. 203):
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• Extended-object CM coordinates, 2
• CM position also given by rCM = (!r dm)/M(SJ 2008 Eq. 9.34, p. 246)
• Symmetric object’s CM is along an axis ofsymmetry & lies in a plane of symmetry(assumes uniform !)
• Center of gravity is point in object wherenet "mg acts (if g = constant over object,then CG coincides with CM)
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• CM example
• Consider anextended object as adistribution of smallmass elements, "m
• Then CM is locatedat position rCM
(SJ 2008 Fig. 9.15, p. 246)
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• Motion of system of particles
• Assume total system mass M = constant
• Describe system motion in terms ofsystem vCM & aCM
• Also describe system p & Newton’s 2ndlaw for system
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• Velocity of CM for system of particles is:
• System p is given by (Eq. 9-14, p. 207)
• Total linear p of system = total mass*vCM
(Eq. 9-17, p. 208)
v & p for system of particles
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• CM acceleration
• Find CM acceleration by differentiating vCMw.r.t. time:
(Eq. 9-18, p. 208)
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• Forces in system of particles
• CM acceleration is related to force F by
• If we add all internal forces, they cancelin pairs. Thus net F on system iscaused only by external F.
(Eq. 9-14, p. 207)
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• p for system of particles
• Total linear p of system of particles isconserved if no net external F acts onsystem
• M vCM = ptot = constant when #Fext = 0(Eq. 9-25, p. 211)
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• Newton’s 2nd law for asystem of particles
• If external F # 0, then net external F = totalsystem mass*CM acceleration:
• #Fext = M aCM (Eq. 9-14, p. 207)
• CM of a system of particles of combined massM moves just like a single particle of mass Mwould move under net external F
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• Motion of CM, example
• Projectile is fired into air &suddenly explodes
• Without explosion, projectilewould follow dotted line
• After explosion, fragments’ CMstill follows dotted line – sameparabolic path that projectilewould’ve followed withoutexplosion
(compare Fig. 9-5, p. 207)
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• Linear momentum
• Linear momentum of a particle (orobject modeled as a particle) of massm moving at velocity v is: p = m v (Eq. 9-22, p. 210)
• Use terms “momentum” & “linearmomentum” interchangeably here
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• Linear p, 2
• Linear momentum p = mv is vector quantitywith direction = v’s direction
• Dimensions of momentum are ML/T
• SI units of momentum are kg$m / s
• In component form, vector p is:
• px = mvx, py = mvy, pz = mvz (SJ 2008, p. 229)
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• Newton & p
• Newton called the product mv theparticle’s “quantity of motion”
• Newton’s 2nd law relates particle’smomentum to net force acting on it:
(above assumes constant m)
(Eq. 9-23, p. 210)
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• Newton’s 2nd law
• Time rate of change of particle’s p = net forceacting on particle• Newton presented 2nd law in this form (not as F=ma)
• It’s more general form than F=ma because it alsoaccounts for mass changes
• Applications to systems of particles areparticularly powerful
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• Impulse & p
• From N2L, Fnet = dp/dt (Eq. 9-27, p. 211)
• Solving for dp gives dp = #Fdt (Eq. 9-28, p. 212)
• Integrate to $ %p over interval %t
• Integral is called the impulse J of the force Facting on an object over "t
(Eqs. 9-30 & 9-31,
p. 212)J
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• Impulse-p theorem
• Eq. 9-31 is impulse-p theorem: Impulseof force F acting on particle = particle’s%p (equivalent to Newton’s 2nd law)
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• Impulse-p, 2
• Impulse is a vector quantity
• Impulse magnitude = areaunder force-time curve
• Impulse dimensions = M*L/T• Impulse isn’t a particle
property, but a property ofchange in particle’s p
(compare Fig. 9-9, p. 212)
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• Impulse-p, 3
• Impulse can also befound by using time-averaged force
• J = (#F)avg"t
• This $ same J as
does time-varying F
(compare Fig. 9-9, p. 212)
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• Impulse approximation
• Often 1 force acting on a particle » any otherforce acting on it
• Assume this if using impulse approximation,then call F the impulse force
• pf & pi represent momenta immediately< & > collision
• Assume particle moves negligibly duringcollision
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• Impulse approximation, 2
• p < & > collisionbetween car & wallcan be determined(p = m v)
• Find the impulse:• J = "p = pf – pi
• F = "p/"t
(SJ 2008 Ex. 9.3,pp. 233-234)
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• Conservation of linear p
• If & 2 particles in isolated systeminteract, total system p is constant• System p is conserved, not necessarily p
of individual particles
• Also, ptotal for isolated system = pinitial
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• Conservation of p, 2
• Conservation of p has several mathematicalforms• ptotal = p1 + p2 = constant (Eq. 9-42, p. 215)
• p1i + p2i = p1f + p2f (Eq. 9-43, p. 215)
• In component form, total p in each directionis conserved independently• pix = pfx piy = pfy piz = pfz
• Conservation of p can be applied to systemswith any number of particles
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• Conservation of p, 3
• Archer stands on africtionless surface (ice)
• Our approaches:• Newton’s 2nd law – lack
info about F or a
• Energy approach – lackinfo about work/energy
• Momentum – can usethis approach
(SJ 2008 Ex. 9.1, p. 230)
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• Conservation of p, 4
• Let system be the archer with bow (particle 1)& the arrow (particle 2)
• No external forces act in x-direction, sosystem is isolated in terms of p
• Total p < releasing arrow = 0
• Total p > releasing the arrow is p1f + p2f = 0
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• Conservation of p, 5
• Archer moves opposite arrow after itsrelease (consistent with Newton’s 3rd law)
• Since marcher » marrow, we know thatarcher’s a & v « arrow’s a & v
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• Collision characteristics
• Use term collision to describe 2 particlescoming very close to each other & interactingvia forces F
• Assume time interval between vf & vi is small
• Assume interaction F » any external F (thuscan use impulse approximation)
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• Collisions – Example 1
• Collisions can resultfrom direct contact
• Impulsive forcesmay vary in time incomplicated ways• This force is internal
to the system
• p is conserved(SJ 2008 Fig. 9.5, p. 235)
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• Collisions – Example 2
• Collision needn’t includeobjects’ physical contact
• Still can have forcesbetween particles
• Analyze this type ofcollision just as thosethat involve physicalcontact
(SJ 2008 Fig. 9.5, p. 235)
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• Types of collisions
• In an elastic collision, p & KE are conserved• Perfectly elastic collisions occur on a microscopic level
• In macroscopic collisions, only ~ elastic collisionsactually occur
• In an inelastic collision, KE is not conservedalthough p is
• In perfectly inelastic collisions, objects adhere toeach other afterward
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• Types of collisions, 2
• In an inelastic collision, some KE is lostbut objects don’t adhere
• Perfectly elastic & inelastic collisions arelimiting cases; most real collisions areintermediate cases
• p is conserved in all collisions
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• Perfectly inelastic collisions
• Since objects adhere >collision, they have same v
• m1v1i + m2v2i = (m1 + m2)vf(compare Eq. 9-53, p. 218)
(compare Fig. 9-15, p. 218)
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• Elastic collisions
• Both p & KE are conserved
(Eqs. 9-63 &9-64, p. 221with v2i # 0)
(compare Fig. 9-18, p. 221)
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• Elastic collisions, 2
• Typically, must solve for 2 unknowns & so need 2equations
• KE equation can be difficult to use since it uses v 2
• After some algebra, find that collision reverses theobjects’ relative velocities, or:
• v1i –v2i = –(v1f –v2f) (Eq. 9-74/Eq. 9-73, p. 222)• Use this result along with p conservation (Eq. 9-71,
p. 222), to calculate:
• 1-dimensional elastic collisions between 2 objectswith 1 stationary (Eqs. 9-67 & 9-68, p. 221)
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• Elastic collisions, 3
• For v2i = 0, Eqs. 9-67 & 9-68 indicate (pp. 221-222):
• If m1 = m2, then the particles exchange velocities(v1f = 0 = v2i & v2f = v1i)
• If massive particle 1 strikes head-on a very lightparticle 2 at rest (m1»m2), then m2 rebounds atv2f ' 2*v1i & m1 continues at v1f ' v1i.
• If very light particle 1 strikes head-on a massiveparticle 2 at rest (m1«m2), then m1’s velocity isreversed (v1f = –v1i ; ~ solid-wall collision) & v2f ' 0(m2 remains nearly at rest).
If both particles are moving, use Eqs. 9-75 & 9-76 (p. 222)
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• Ballistic pendulum
• Perfectly inelastic collision –bullet’s embedded in wood block
• Momentum equation will have 2unknowns, v1A & vB
• Use conservation of ME frompendulum $ v just after collision
• v1A = (m1+m2)*(2gh)1/2/m1
is bullet’s initial speed(compare Eq. 9-61, p. 220)
(compare Fig. 9-17, p. 220)
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• Ballistic pendulum, 2
• Multi-flash photo ofballistic pendulum
(SJ 2008 Ex. 9.6, p. 239)
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• 2-dimensional (2D) collisions
• Momentum is conserved in all directions
• Use subscripts for
• identifying the object
• indicating initial or final values
• the velocity components
• If collision is elastic, can also use conservation ofKE which leads to Eqs. 9-79 & 9-80 (p. 224)
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• 2D collision, example
• Particle 1 moves at velocity v1i& particle 2 is at rest
• In x-direction, initial px = m1v1i• In y-direction, initial py = 0
(compare Fig. 9-21, p. 224)
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• 2D collision, example
• After collision in x-direction:px = m1v1f cos(%) + m2v2f cos(&)
• After collision in y-direction:py = m1v1f sin(%) ' m2v2f sin(&)
(compare Fig. 9-21, p. 224)
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Problem-solving: 2D collisions
• Set up coordinate system & definevelocities w.r.t. system• Usually convenient to have x-axis coincide
with one initial velocity
• In coordinate-system sketch, draw &label all v & include all givens
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• Problem-solving: 2D collisions, 2
• Write terms for x- & y-components of eachobject’s p before & after collision
• Include appropriate signs for all v components
• Write terms for system’s ptotal in x-directionboth before & after collision. Then equatethese pre- & post-collision p. Repeat for y-direction ptotal.
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• Problem-solving: 2D collisions, 3
• If collision is inelastic, system KE isn’tconserved, & additional information islikely needed
• If collision is perfectly inelastic, then 2objects’ final vs are equal. Then solve pequations for unknowns.
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• Problem-solving: 2D collisions, 4
• If collision is elastic, system KE isconserved
• Equate KEtotal < collision to KEtotal >collision to get more info on relationshipbetween object vs
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• 2-dimensional collision
• < collision, car hastotal x-direction p;van has total y-direction p
• > collision, bothvehicles have px &py components
(SJ 2008 Ex. 9.8, p. 243)