chapter 9 · 40 ¥ elastic collisions, 3 ¥ for v2i = 0, eqs. 9-67 & 9-68 indicate (pp. 221-222) :...

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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


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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


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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


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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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• 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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• 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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• 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)


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• Elastic collisions

• Both p & KE are conserved

(Eqs. 9-63 &9-64, p. 221with v2i # 0)


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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)


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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


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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(&)


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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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• 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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• 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)

chapter 9 · 40 ¥ elastic collisions, 3 ¥ for v2i = 0, eqs. 9-67 & 9-68 indicate (pp. 221-222) :...

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