linear momentum and collisions chapter 9. i.linear momentum and its conservation ii.impulse and...

Linear momentum and Linear momentum and CollisionsCollisions

Chapter 9Chapter 9

I.I. Linear Momentum and its ConservationLinear Momentum and its Conservation

II.II. Impulse and MomentumImpulse and Momentum

III.III. Collisions in One DimensionCollisions in One Dimension

IV.IV. Collisions in Two DimensionsCollisions in Two Dimensions

V. V. The Center of Mass The Center of Mass

VI.VI. Motion of a System of ParticlesMotion of a System of Particles

VII.VII. Deformable SystemsDeformable Systems

VIII.VIII. Rocket PropulsionRocket Propulsion

Linear Momentum

The The linear momentumlinear momentum of a particle, or an of a particle, or an object that can be modeled as a particle of object that can be modeled as a particle of mass mass mm moving with a velocity moving with a velocity ,, is defined is defined to be the product of the mass and velocity:to be the product of the mass and velocity:

The terms The terms momentummomentum and and linear momentum linear momentum will be used interchangeably in the textwill be used interchangeably in the text

v

mp v

I. Linear momentumI. Linear momentum

vmp

The linear momentum of a particle is a vectorThe linear momentum of a particle is a vector p p defined as:defined as:

Momentum is a vector with magnitude equalMomentum is a vector with magnitude equal mvmv and has and has direction ofdirection of v v ..

The dimensions of momentum are ML/T SI unitSI unit of the momentum isof the momentum is kg-meter/secondkg-meter/second

Momentum can be expressed in component form:Momentum can be expressed in component form:

ppxx = = m vm vxx ppyy = = m vm vyy ppzz = = m vm vzz

Newton and MomentumNewton and Momentum

Newton’s Second Law can be used to Newton’s Second Law can be used to relate the momentum of a particle to the relate the momentum of a particle to the resultant force acting on itresultant force acting on it

with constant masswith constant mass

d md dm m

dt dt dt

vv pF a

Newton II lawNewton II law in terms of momentum:in terms of momentum:

amdt

vdm

dt

vmd

dt

pdFnet

)(

The time rate of change of the momentum of a particle is The time rate of change of the momentum of a particle is equal to the net force acting on the particle and is in the equal to the net force acting on the particle and is in the direction of the force.direction of the force.

Newton called the product Newton called the product mmvv the the quantity of motionquantity of motion of the of the particleparticle

System of particles:System of particles:

nnn vmvmvmvmppppP

....... 332211321

The total linear momentThe total linear moment PP is the vector sum of the is the vector sum of the individual particle’s linear momentum.individual particle’s linear momentum.

comvMP

The linear momentum of a system of particles is The linear momentum of a system of particles is equal to the product of the total mass equal to the product of the total mass M M of the of the system and the velocity of the center of mass.system and the velocity of the center of mass.

dt

PdFaM

dt

vdM

dt

Pdnetcom

com

Net external force acting on the system.Net external force acting on the system.

Conservation:Conservation:

If no external force acts on a closed, isolated system of If no external force acts on a closed, isolated system of particles, the total linear momentumparticles, the total linear momentum P P of the system cannot of the system cannot change.change.

ifnet PPdt

PdF

systemisolatedClosedconstP

0

),(

Closed:Closed: no matter passes through the systemno matter passes through the system boundary in any direction.boundary in any direction.

If the component of the net external force on a closed If the component of the net external force on a closed system is zero along an axis system is zero along an axis component of the linear component of the linear momentum along that axis cannot change.momentum along that axis cannot change.

The momentum is constant if no external forces act on a The momentum is constant if no external forces act on a closed particle system. Internal forces can change the linear closed particle system. Internal forces can change the linear momentum of portions of the system, but they cannot momentum of portions of the system, but they cannot change the total linear momentum of the entire system.change the total linear momentum of the entire system.

If no net external force acts on the system of particles the total If no net external force acts on the system of particles the total linear momentumlinear momentum PP of the system cannot change. of the system cannot change.

Each component of the linear momentum is conserved Each component of the linear momentum is conserved separately if the corresponding component of the net external separately if the corresponding component of the net external force is zero.force is zero.

Conservation of Linear MomentumConservation of Linear Momentum

Conservation of Linear MomentumConservation of Linear Momentum

• Whenever two or more particles in an Whenever two or more particles in an isolated system interact, the total isolated system interact, the total momentum of the system remains momentum of the system remains constantconstant– The momentum of the system is conserved, The momentum of the system is conserved,

not necessarily the momentum of an not necessarily the momentum of an individual particleindividual particle

– This also tells us that the total momentum of This also tells us that the total momentum of an isolated system equals its initial an isolated system equals its initial momentummomentum

Conservation of MomentumConservation of Momentum

• Conservation of momentum can be expressed Conservation of momentum can be expressed mathematically in various waysmathematically in various ways

• In component form, the total momenta in each direction In component form, the total momenta in each direction are independently conservedare independently conserved

ppixix = = ppfxfx ppiyiy = = ppfyfy ppiziz = = ppfzfz

• Conservation of momentum can be applied to systems Conservation of momentum can be applied to systems with any number of particleswith any number of particles

• This law is the mathematical representation of the This law is the mathematical representation of the momentum version of the isolated system modelmomentum version of the isolated system model

to ta l 1 2p = p + p = co n s ta n t

1 i 2 i 1 f 2 fp + p = p + p

Conservation of Momentum, Conservation of Momentum, Archer ExampleArcher Example

• The archer is standing on a The archer is standing on a frictionless surface (ice)frictionless surface (ice)

• Approaches:Approaches:– Newton’s Second Law – no, Newton’s Second Law – no,

no information about no information about FF or or aa– Energy approach – no, Energy approach – no,

no information about work or no information about work or energyenergy

– Momentum – yesMomentum – yes

Archer ExampleArcher Example• ConceptualizeConceptualize

– The arrow is fired one way and the archer recoils in The arrow is fired one way and the archer recoils in the opposite directionthe opposite direction

• CategorizeCategorize– MomentumMomentum

Let the system be the archer with bow (particle 1) Let the system be the archer with bow (particle 1) and the arrow (particle 2)and the arrow (particle 2)There are no external forces in the There are no external forces in the xx-direction, so it -direction, so it is isolated in terms of momentum in the is isolated in terms of momentum in the xx-direction-direction

• AnalyzeAnalyze– Total momentum before releasing the arrow is Total momentum before releasing the arrow is 00

Archer ExampleArcher Example

• Analyze.Analyze.– The total momentum after releasing the arrow isThe total momentum after releasing the arrow is

• FinalizeFinalize– The final velocity of the archer is negativeThe final velocity of the archer is negative

• Indicates he moves in a direction opposite the arrowIndicates he moves in a direction opposite the arrow• Archer has much higher mass than arrow, so velocity Archer has much higher mass than arrow, so velocity

is much loweris much lower

1 2 0f f p p

Impulse and MomentumImpulse and Momentum

• From Newton’s Second Law, From Newton’s Second Law,

• Solving for givesSolving for gives • Integrating to find the change in momentum Integrating to find the change in momentum

over some time intervalover some time interval

• The integral is called the The integral is called the impulse, ,impulse, , of the of the force acting on an object over force acting on an object over ΔΔtt

f

i

t

f i tdt p p p F I

d

dt

pF

dp

d dtp F

I

IV. Collision and impulseIV. Collision and impulse

Collision:Collision: isolated event in which two or more bodies exert isolated event in which two or more bodies exert relatively strong forces on each other for a relatively short time.relatively strong forces on each other for a relatively short time.

Single collisionSingle collision

pppdttFI

dttFpddttFpddt

pdF

if

t

t

t

t

p

p

f

i

f

i

f

i

)(

)()(

Measures the strength and duration of the collision Measures the strength and duration of the collision forceforce

Third law force pairThird law force pair

FFR R = - F= - FLL

Impulse:Impulse:

Impulse-linear momentum theorem Impulse-linear momentum theorem The change in the linear momentum of a body in a The change in the linear momentum of a body in a collision is equal to the impulse that acts on that body.collision is equal to the impulse that acts on that body.

Ippp if

Units:Units: kg m/skg m/s

zzizfz

yyiyfy

xxixfx

Ippp

Ippp

Ippp

tFI avg

FFavgavg such that:such that:

Area underArea under F(t)F(t) vsvs ΔΔtt curve curve = Area under= Area under F Favgavg vs vs tt

An estimated force-time curve for a baseball struck by a bat An estimated force-time curve for a baseball struck by a bat is shown in Figure. From this curve, determine (a) the is shown in Figure. From this curve, determine (a) the impulse delivered to the ball, (b) the average force exerted impulse delivered to the ball, (b) the average force exerted on the ball, and (c) the peak force exerted on the ball.on the ball, and (c) the peak force exerted on the ball.

More About ImpulseMore About Impulse• Impulse is a vector quantityImpulse is a vector quantity• The magnitude of the The magnitude of the

impulse is equal to the area impulse is equal to the area under the force-time curveunder the force-time curve– The force may vary with The force may vary with

timetime• Dimensions of impulse are Dimensions of impulse are

M L / TM L / T• Impulse is not a property of Impulse is not a property of

the particle, but a measure the particle, but a measure of the change in momentum of the change in momentum of the particleof the particle

ImpulseImpulse

• The impulse can also The impulse can also be found by using the be found by using the time averaged forcetime averaged force

• This would give the This would give the same impulse as the same impulse as the time-varying force doestime-varying force does

t I F

Impulse ApproximationImpulse Approximation

• In many cases, one force acting on a particle acts In many cases, one force acting on a particle acts for a short time, but is much greater than any for a short time, but is much greater than any other force presentother force present

• When using the Impulse Approximation, we will When using the Impulse Approximation, we will assume this is trueassume this is true– Especially useful in analyzing collisionsEspecially useful in analyzing collisions

• The force will be called the The force will be called the impulsive forceimpulsive force• The particle is assumed to move very little during The particle is assumed to move very little during

the collisionthe collision• represent the momenta represent the momenta immediatelyimmediately

before and after the collisionbefore and after the collisioni fandp p

Impulse-Momentum: Impulse-Momentum: Crash Test ExampleCrash Test Example

• CategorizeCategorize– Assume force exerted by Assume force exerted by

wall is large compared wall is large compared with other forceswith other forces

– Gravitational and normal Gravitational and normal forces are perpendicular forces are perpendicular and so do not effect the and so do not effect the horizontal momentumhorizontal momentum

– Can apply impulse Can apply impulse approximationapproximation

Crash Test ExampleCrash Test Example

• AnalyzeAnalyze– The momenta before and after the collision The momenta before and after the collision

between the car and the wall can be determined between the car and the wall can be determined – Find Find

• Initial momentumInitial momentum• Final momentumFinal momentum• ImpulseImpulse• Average forceAverage force

• FinalizeFinalize– Check signs on velocities to be sure they are Check signs on velocities to be sure they are

reasonablereasonable

Collisions – CharacteristicsCollisions – Characteristics • We use the term We use the term collisioncollision to represent an event to represent an event

during which two particles come close to each during which two particles come close to each other and interact by means of forcesother and interact by means of forces– May involve physical contact, but must be generalized May involve physical contact, but must be generalized

to include cases with interaction without physical to include cases with interaction without physical contactcontact

• The time interval during which the velocity The time interval during which the velocity changes from its initial to final values is assumed changes from its initial to final values is assumed to be shortto be short

• The interaction forces are assumed to be much The interaction forces are assumed to be much greater than any external forces presentgreater than any external forces present– This means the impulse approximation can be usedThis means the impulse approximation can be used

Collisions – Example 1Collisions – Example 1

• Collisions may be the Collisions may be the result of direct contactresult of direct contact

• The impulsive forces The impulsive forces may vary in time in may vary in time in complicated wayscomplicated ways– This force is internal to This force is internal to

the systemthe system

• Momentum is Momentum is conservedconserved

Collisions – Example 2Collisions – Example 2

• The collision need not The collision need not include physical include physical contact between the contact between the objectsobjects

• There are still forces There are still forces between the particlesbetween the particles

• This type of collision This type of collision can be analyzed in the can be analyzed in the same way as those same way as those that include physical that include physical contactcontact

Types of CollisionsTypes of Collisions

• In an elastic collision, momentum and kinetic energy are conserved– Perfectly elastic collisions occur on a microscopic

level– In macroscopic collisions, only approximately

elastic collisions actually occur• Generally some energy is lost to deformation,

sound, etc.

• In an inelastic collision, kinetic energy is not conserved, although momentum is still conserved– If the objects stick together after the collision, it is

a perfectly inelastic collision

CollisionsCollisions

• In an inelastic collision, some kinetic energy In an inelastic collision, some kinetic energy is lost, but the objects do not stick togetheris lost, but the objects do not stick together

• Elastic and perfectly inelastic collisions are Elastic and perfectly inelastic collisions are limiting cases, most actual collisions fall in limiting cases, most actual collisions fall in between these two types between these two types

• Momentum is conserved in all collisionsMomentum is conserved in all collisions

Perfectly Inelastic CollisionsPerfectly Inelastic Collisions

• Since the objects stick Since the objects stick together, they share together, they share the same velocity after the same velocity after the collisionthe collision

1 1 2 2 1 2i i fm m m m v v v

Elastic CollisionsElastic Collisions

• Both momentum and Both momentum and kinetic energy are kinetic energy are conservedconserved

1 1 2 2

1 1 2 2

2 21 1 2 2

2 21 1 2 2

1 1

2 21 1

2 2

i i

f f

i i

f f

m m

m m

m m

m m

v v

v v

v v

v v


• Typically, there are two unknowns to solve for and so Typically, there are two unknowns to solve for and so you need two equationsyou need two equations

• The kinetic energy equation can be difficult to useThe kinetic energy equation can be difficult to use• With some algebraic manipulation, a different With some algebraic manipulation, a different

equation can be usedequation can be used

vv11ii – v – v22ii = = vv11ff + + vv22ff

• This equation, along with conservation of momentum, This equation, along with conservation of momentum, can be used to solve for the two unknownscan be used to solve for the two unknowns– It can only be used with a one-dimensional, elastic It can only be used with a one-dimensional, elastic

collision between two objectscollision between two objects


• Example of some special casesExample of some special cases

mm11 = = mm22 – – the particles exchange velocitiesthe particles exchange velocities

– When a very heavy particle collides head-on with a very When a very heavy particle collides head-on with a very light one initially at rest, the heavy particle continues in light one initially at rest, the heavy particle continues in motion unaltered and the light particle rebounds with a motion unaltered and the light particle rebounds with a speed of about twice the initial speed of the heavy speed of about twice the initial speed of the heavy particleparticle

– When a very light particle collides head-on with a very When a very light particle collides head-on with a very heavy particle initially at rest, the light particle has its heavy particle initially at rest, the light particle has its velocity reversed and the heavy particle remains velocity reversed and the heavy particle remains approximately at restapproximately at rest

Problem-Solving Strategy: Problem-Solving Strategy: One-Dimensional CollisionsOne-Dimensional Collisions

• ConceptualizeConceptualize– Image the collision occurring in your mindImage the collision occurring in your mind– Draw simple diagrams of the particles before Draw simple diagrams of the particles before

and after the collisionand after the collision– Include appropriate velocity vectorsInclude appropriate velocity vectors

• CategorizeCategorize– Is the system of particles isolated?Is the system of particles isolated?– Is the collision elastic, inelastic or perfectly Is the collision elastic, inelastic or perfectly

inelastic?inelastic?

Problem-Solving Strategy: One-Problem-Solving Strategy: One-Dimensional CollisionsDimensional Collisions

• AnalyzeAnalyze– Set up the mathematical representation of Set up the mathematical representation of

the problemthe problem– Solve for the unknown(s)Solve for the unknown(s)

• FinalizeFinalize– Check to see if the answers are consistent Check to see if the answers are consistent

with the mental and pictorial with the mental and pictorial representationsrepresentations

– Check to be sure your results are realisticCheck to be sure your results are realistic

Example: Stress RelieverExample: Stress Reliever

• ConceptualizeConceptualize

– Imagine one ball coming in Imagine one ball coming in from the left and two balls from the left and two balls exiting from the rightexiting from the right

– Is this possible?Is this possible?

• CategorizeCategorize

– Due to shortness of time, the Due to shortness of time, the impulse approximation can impulse approximation can be usedbe used

– Isolated systemIsolated system

– Elastic collisionsElastic collisions


• AnalyzeAnalyze– Check to see if momentum is conservedCheck to see if momentum is conserved

• It isIt is

– Check to see if kinetic energy is conservedCheck to see if kinetic energy is conserved• It is notIt is not• Therefore, the collision couldn’t be elasticTherefore, the collision couldn’t be elastic

• FinalizeFinalize– Having two balls exit was not possible if only Having two balls exit was not possible if only

one ball is releasedone ball is released


• What collision What collision isis possiblepossible

• Need to conserve both Need to conserve both momentum and kinetic momentum and kinetic energyenergy– Only way to do so is Only way to do so is

with equal numbers of with equal numbers of balls released and balls released and exitingexiting

Collision Example – Ballistic PendulumCollision Example – Ballistic Pendulum• ConceptualizeConceptualize

– Observe diagramObserve diagram

• CategorizeCategorize– Isolated system of projectile and Isolated system of projectile and

blockblock– Perfectly inelastic collision – the Perfectly inelastic collision – the

bullet is embedded in the block of bullet is embedded in the block of woodwood

– Momentum equation will have two Momentum equation will have two unknownsunknowns

– Use conservation of energy from Use conservation of energy from the pendulum to find the velocity the pendulum to find the velocity just after the collisionjust after the collision

– Then you can find the speed of Then you can find the speed of the bulletthe bullet

Ballistic PendulumBallistic Pendulum• A multi-flash photograph of A multi-flash photograph of

a ballistic penduluma ballistic pendulum

• AnalyzeAnalyze

– Solve resulting system of Solve resulting system of equationsequations

• FinalizeFinalize

– Note different systems Note different systems involvedinvolved

– Some energy was Some energy was transferred during the transferred during the perfectly inelastic perfectly inelastic collisioncollision

V. Momentum and kinetic energy in collisionsV. Momentum and kinetic energy in collisions

Assumptions:Assumptions: Closed systemsClosed systems (no mass enters or leaves (no mass enters or leaves them) them)

Isolated systemsIsolated systems (no external forces act on (no external forces act on the bodies within the system) the bodies within the system)

Elastic collision:Elastic collision: If the total kinetic energy of the system of If the total kinetic energy of the system of two colliding bodies is unchanged two colliding bodies is unchanged (conserved) by the collision.(conserved) by the collision.

Inelastic collision:Inelastic collision: The kinetic energy of the system is not The kinetic energy of the system is not conserved conserved some goes into thermal some goes into thermal energy, sound, etc.energy, sound, etc.

Example:Example: Ball into hard floor.Ball into hard floor.

Completely inelastic collision:Completely inelastic collision: After the collision the bodies lose After the collision the bodies lose energy and stick together.energy and stick together.

Example:Example: Ball of wet putty into floorBall of wet putty into floor

VII. Elastic collisions in 1DVII. Elastic collisions in 1D)()( collisionafterenergykineticTotalcollisionbeforeenergykineticTotal

ffi vmvmvm 221111 Closed, isolated Closed, isolated system system

In an elastic collision, the In an elastic collision, the kinetic energy of each kinetic energy of each colliding colliding body may change, body may change, but the total kinetic energy of but the total kinetic energy of the system does not change.the system does not change.

Stationary target:Stationary target:

Linear momentumLinear momentum

222

211

211 2

1

2

1

2

1ffi vmvmvm Kinetic energyKinetic energy

)2())(()(

)1()(

11111222

21

211

22111

fififfi

ffi

vvvvmvmvvm

vmvvm

ifif

ifiiff

fif

fif

vmm

mvv

mm

mmv

vvvm

mvvvin

vvm

mvFrom

vvvDividing

121

121

21

211

1112

1121

112

12

112

2

)()3()1(

)()1(

)3()1/()2(

v2f >0 alwaysv1f >0 if m1>m2 forward mov.v1f <0 if m1<m2 rebounds

Stationary target:Stationary target:

Equal masses:Equal masses: mm11=m=m2 2 v v1f1f=0 and v=0 and v2f 2f = v= v1i1i In head-on In head-on

collisions bodies of equal masses simply exchange collisions bodies of equal masses simply exchange velocities.velocities.

Massive target:Massive target: mm22>>m>>m11 vv1f 1f ≈ -v≈ -v1i1i and and vv2f 2f

≈ (2m≈ (2m11/m/m22)v)v1i1i Body 1 bounces back with Body 1 bounces back with

approximately same speed. Body 2 moves forward approximately same speed. Body 2 moves forward at low speed.at low speed.

Massive projectile:Massive projectile: mm11>>m>>m22 vv1f 1f ≈ v≈ v1i1i and and

vv2f 2f ≈ 2v≈ 2v1i1i Body 1 keeps on going scarcely lowed by Body 1 keeps on going scarcely lowed by

the collision. Body 2 charges ahead at twice the the collision. Body 2 charges ahead at twice the initial speed of the projectile.initial speed of the projectile.

VII. Elastic collisions in 1DVII. Elastic collisions in 1D

ffii vmvmvmvm 22112211 Closed, isolated system Closed, isolated system

iif

iif

fifififi

fifi

vmm

mmv

mm

mv

vmm

mv

mm

mmvDividing

vvvvmvvvvm

vvmvvm

221

121

21

12

221

21

21

211

2222211111

222111

2

2)1/()2(

)2())(())((

)1()()(

Moving target:Moving target:

Linear momentumLinear momentum

222

211

222

211 2

1

2

1

2

1

2

1ffii vmvmvmvm Kinetic energyKinetic energy

VIII. Collisions in 2DVIII. Collisions in 2D

Closed, isolated system Closed, isolated system

ffii PPPP 2121

Linear momentum conservedLinear momentum conserved

ffii KKKK 2121 Kinetic energy conservedKinetic energy conserved

Elastic collision Elastic collision

Example:Example:

222

211

211 2

1

2

1

2

1ffi vmvmvm

22211111 coscos ffi vmvmvmaxisx

222111 sinsin0 ff vmvmaxisy

If the collision is elastic If the collision is elastic

Two blocks of masses Two blocks of masses MM and and 33MM are placed on a horizontal, are placed on a horizontal, frictionless surface. A light spring is attached to one of them, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between and the blocks are pushed together with the spring between them. A cord initially holding the blocks together is burned; after them. A cord initially holding the blocks together is burned; after this, the block of mass this, the block of mass 33MM moves to the right with a speed of moves to the right with a speed of 2.00 m/s2.00 m/s. (a) What is the speed of the block of mass . (a) What is the speed of the block of mass MM? (b) ? (b) Find the original elastic potential energy in the spring if Find the original elastic potential energy in the spring if MM = = 0.350 kg0.350 kg..

A tennis player receives a shot with the ball (A tennis player receives a shot with the ball (0.0600 kg0.0600 kg) ) traveling horizontally at traveling horizontally at 50.0 m/s50.0 m/s and returns the shot with and returns the shot with the ball traveling horizontally at the ball traveling horizontally at 40.0 m/s40.0 m/s in the opposite in the opposite direction. (a) What is the impulse delivered to the ball by the direction. (a) What is the impulse delivered to the ball by the racquet? (b) What work does the racquet do on the ball?racquet? (b) What work does the racquet do on the ball?

Two blocks are free to slide along the frictionless wooden track Two blocks are free to slide along the frictionless wooden track ABCABC shown in Figure. A block of mass shown in Figure. A block of mass mm11 = 5.00 kg = 5.00 kg is released is released

from from AA.. Protruding from its front end is the north pole of a Protruding from its front end is the north pole of a strong magnet, repelling the north pole of an identical magnet strong magnet, repelling the north pole of an identical magnet embedded in the back end of the block of mass embedded in the back end of the block of mass mm22 = 10.0 = 10.0

kgkg, initially at rest. The two blocks never touch. Calculate the , initially at rest. The two blocks never touch. Calculate the maximum height to which maximum height to which mm11 rises after the elastic collision. rises after the elastic collision.

As shown in Figure, a bullet of mass As shown in Figure, a bullet of mass mm and speed and speed vv passes passes completely through a pendulum bob of mass completely through a pendulum bob of mass MM.. The bullet The bullet emerges with a speed of emerges with a speed of vv/2/2. The pendulum bob is suspended . The pendulum bob is suspended by a stiff rod of lengthby a stiff rod of length ℓℓ and negligible mass. What is the and negligible mass. What is the minimum value of minimum value of vv such that the pendulum bob will barely such that the pendulum bob will barely swing through a complete vertical circle?swing through a complete vertical circle?

A small block of mass A small block of mass mm11 = 0.500 kg = 0.500 kg is released from rest at is released from rest at

the top of a curve-shaped frictionless wedge of mass the top of a curve-shaped frictionless wedge of mass mm2 2 = 3.00 kg= 3.00 kg, which sits on a frictionless horizontal surface as , which sits on a frictionless horizontal surface as

in Figure (a). When the block leaves the wedge, its velocity is in Figure (a). When the block leaves the wedge, its velocity is measured to be measured to be 4.00 m/s4.00 m/s to the right, as in Figure (b). to the right, as in Figure (b). (a) What is the velocity of the wedge after the block reaches (a) What is the velocity of the wedge after the block reaches the horizontal surface? (b) What is the height the horizontal surface? (b) What is the height hh of the wedge?of the wedge?

Two-Dimensional CollisionsTwo-Dimensional Collisions

• The momentum is conserved in all directionsThe momentum is conserved in all directions• Use subscripts forUse subscripts for

– Identifying the objectIdentifying the object– Indicating initial or final valuesIndicating initial or final values– The velocity componentsThe velocity components

• If the collision is elastic, use conservation of If the collision is elastic, use conservation of kinetic energy as a second equationkinetic energy as a second equation– Remember, the simpler equation can only be used for Remember, the simpler equation can only be used for

one-dimensional situationsone-dimensional situations

Two-Dimensional Collision, exampleTwo-Dimensional Collision, example

• Particle 1 is moving at Particle 1 is moving at velocity and velocity and particle 2 is at restparticle 2 is at rest

• In the In the xx-direction, the -direction, the initial momentum is initial momentum is mm11vv11ii

• In the In the yy-direction, the -direction, the initial momentum is initial momentum is 00

1iv

Two-Dimensional Collision, Two-Dimensional Collision, exampleexample

• After the collision, the After the collision, the momentum in the momentum in the xx-direction -direction is is mm11vv11ff cos cos θθ + + mm22vv22ff cos cos φφ

• After the collision, the After the collision, the momentum in the momentum in the yy-direction -direction is is mm11vv11ff sin sin θθ+ + mm22vv22ff sin sin ff

• If the collision is elastic, If the collision is elastic, apply the kinetic energy apply the kinetic energy equationequation

• This is an example of a This is an example of a glancing collisionglancing collision

Problem-Solving Strategies – Two-Problem-Solving Strategies – Two-Dimensional CollisionsDimensional Collisions

• ConceptualizeConceptualize– Imagine the collisionImagine the collision– Predict approximate directions the particles will Predict approximate directions the particles will

move after the collisionmove after the collision– Set up a coordinate system and define your Set up a coordinate system and define your

velocities with respect to that systemvelocities with respect to that system• It is usually convenient to have the It is usually convenient to have the xx-axis coincide -axis coincide

with one of the initial velocitieswith one of the initial velocities

– In your sketch of the coordinate system, draw In your sketch of the coordinate system, draw and label all velocity vectors and include all the and label all velocity vectors and include all the given informationgiven information


• CategorizeCategorize– Is the system isolated?Is the system isolated?– If so, categorize the collision as elastic, inelastic or perfectly If so, categorize the collision as elastic, inelastic or perfectly

inelasticinelastic

• AnalyzeAnalyze– Write expressions for the Write expressions for the xx- and - and yy-components of the -components of the

momentum of each object before and after the collisionmomentum of each object before and after the collision• Remember to include the appropriate signs for the Remember to include the appropriate signs for the

components of the velocity vectorscomponents of the velocity vectors– Write expressions for the total momentum of the system in the Write expressions for the total momentum of the system in the

xx-direction before and after the collision and equate the two. -direction before and after the collision and equate the two. Repeat for the total momentum in the Repeat for the total momentum in the yy-direction.-direction.


– If the collision is inelastic, kinetic energy of the system If the collision is inelastic, kinetic energy of the system is not conserved, and additional information is probably is not conserved, and additional information is probably neededneeded

– If the collision is perfectly inelastic, the final velocities of If the collision is perfectly inelastic, the final velocities of the two objects are equal. Solve the momentum the two objects are equal. Solve the momentum equations for the unknowns.equations for the unknowns.

– If the collision is elastic, the kinetic energy of the If the collision is elastic, the kinetic energy of the system is conservedsystem is conserved

• Equate the total kinetic energy before the collision to the Equate the total kinetic energy before the collision to the total kinetic energy after the collision to obtain more total kinetic energy after the collision to obtain more information on the relationship between the velocitiesinformation on the relationship between the velocities


• FinalizeFinalize– Check to see if your answers are Check to see if your answers are

consistent with the mental and pictorial consistent with the mental and pictorial representationsrepresentations

– Check to be sure your results are realisticCheck to be sure your results are realistic

Two-Dimensional Collision ExampleTwo-Dimensional Collision Example• ConceptualizeConceptualize

– See pictureSee picture– Choose East to be the positive Choose East to be the positive

xx-direction and North to be the -direction and North to be the positive positive yy-direction-direction

• CategorizeCategorize– Ignore frictionIgnore friction– Model the cars as particlesModel the cars as particles– The collision is perfectly The collision is perfectly

inelasticinelastic• The cars stick togetherThe cars stick together

Two dimensional collision

m1 = 1500.0kg

m2 = 2500.0 kg

Find vf .

An unstable atomic nucleus of mass An unstable atomic nucleus of mass 17.0 17.0 10 10–27–27 kg kg initially initially at rest disintegrates into three particles. One of the at rest disintegrates into three particles. One of the particles, of mass particles, of mass 5.00 5.00 10 10–27–27 kg kg, moves along the, moves along the yy--axis axis with a speed of with a speed of 6.00 6.00 10 1066 m/s m/s. Another particle, of mass . Another particle, of mass 8.40 8.40 10 10–27–27 kg kg, moves along the , moves along the xx--axis with a speed of axis with a speed of 4.00 4.00 10 1066 m/s m/s. Find (a) the velocity of the third particle and . Find (a) the velocity of the third particle and (b) the total kinetic energy increase in the process.(b) the total kinetic energy increase in the process.

The Center of MassThe Center of Mass

• There is a special point in a system or There is a special point in a system or object, called the object, called the center of masscenter of mass, that , that moves as if all of the mass of the system moves as if all of the mass of the system is concentrated at that pointis concentrated at that point

• The system will move as if an external The system will move as if an external force were applied to a single particle of force were applied to a single particle of massmass MM located at the center of mass.located at the center of mass.MM is the total mass of the system is the total mass of the system

II. Newton’s second law for a system of particlesII. Newton’s second law for a system of particles

Center of the mass of the system moves as a particle Center of the mass of the system moves as a particle whose mass is equal to the total mass of the system.whose mass is equal to the total mass of the system.

Motion of the center of mass:Motion of the center of mass:

comnet aMF

FFnetnet is the is the net of all external forcesnet of all external forces that act on the that act on the

system. Internal forces (from one part of the system to system. Internal forces (from one part of the system to another are not included).another are not included).

The system is closed: no mass enters or leaves the The system is closed: no mass enters or leaves the system during the movement. (system during the movement. (MM=total system mass).=total system mass).

aacomcom is the acceleration of the system’s center of is the acceleration of the system’s center of

mass.mass.

I. Center of massI. Center of mass

The center of mass of a body or a system of The center of mass of a body or a system of bodies is the point that moves as though all the bodies is the point that moves as though all the mass were concentrated there and all external mass were concentrated there and all external forces were applied there.forces were applied there.


Origin of reference Origin of reference system coincides system coincides withwith mm11

dmm

mxcom

21

2

Two particles of massesTwo particles of masses mm11 and and mm22 separated by a distanceseparated by a distance dd


M

xmxm

mm

xmxmxcom

2211

21

2211

General:General:

The center of mass lies somewhere between the two particles.The center of mass lies somewhere between the two particles.

Choice of the reference Choice of the reference origin is arbitrary origin is arbitrary Shift Shift of the coordinate system of the coordinate system but center of mass is still but center of mass is still at the same distance at the same distance from each particle. from each particle.

MM = total mass of the system= total mass of the system

3D:3D:

n

iiicom

n

iiicom

n

iiicom zm

Mzym

Myxm

Mx

111

111


We can extend this equation to a general situation forWe can extend this equation to a general situation for nn particles particles that strung along that strung along xx-axis. The total mass of the system-axis. The total mass of the system M=mM=m11+m+m22+m+m33+……+m+……+mnn The location of center of the mass:The location of center of the mass:

n

iii

nncom xm

MM

xmxmxmxmx

1

332211 1........

3D3D: : The vector formThe vector form

Position of the Position of the particle:particle:

n

iiicom rm

Mr

1

1 MM = total mass = total mass of the objectof the object


kzjyixr iiiiˆˆˆ

Position COM:Position COM: kzjyixr comcomcomcomˆˆˆ

Center of Mass, Extended ObjectCenter of Mass, Extended Object

• Similar analysis can be Similar analysis can be done for an extended done for an extended objectobject

• Consider the extended Consider the extended object as a system object as a system containing a large containing a large number of particlesnumber of particles

• Since particle Since particle separation is very small, separation is very small, it can be considered to it can be considered to have a constant mass have a constant mass distributiondistribution

Center of Mass, positionCenter of Mass, position

• The center of mass in three dimensions can be The center of mass in three dimensions can be located by its position vector, located by its position vector, – For a system of particles,For a system of particles,

is the position of the is the position of the iithth particle, defined byparticle, defined by

– For an extended object, For an extended object,

CM

1dm

M r r

ˆ ˆ ˆi i i ix y z r i j k

CMr

1CM i i

i

mM

r r

ir

Solid bodies:Solid bodies:

Continuous distribution of matter. Particles = Continuous distribution of matter. Particles = dmdm (differential mass elements).(differential mass elements).

3D:3D: dmzM

zdmyM

ydmxM

x comcomcom111

MM = mass of the object= mass of the object

Assumption:Assumption:V

M

dV

dm

dVzV

zdVyV

ydVxV

x comcomcom111

Uniform objects Uniform objects uniform density uniform density

Motion of a System of ParticlesMotion of a System of Particles

• Assume the total mass, Assume the total mass, MM, of the system remains , of the system remains constantconstant

• We can describe the motion of the system in We can describe the motion of the system in terms of the velocity and acceleration of the terms of the velocity and acceleration of the center of mass of the systemcenter of mass of the system

• We can also describe the momentum of the We can also describe the momentum of the system and Newton’s Second Law for the systemsystem and Newton’s Second Law for the system

Velocity and Momentum of a Velocity and Momentum of a System of ParticlesSystem of Particles

• The velocity of the center of mass of a system of The velocity of the center of mass of a system of particles isparticles is

• The total momentum of the system can be The total momentum of the system can be expressed asexpressed as

• The total linear momentum of the system equals The total linear momentum of the system equals the total mass multiplied by the velocity of the the total mass multiplied by the velocity of the center of masscenter of mass

CMCM

1i i

i

dm

dt M r

v v

CM toti i ii i

M m v v p p

Acceleration of the Center of MassAcceleration of the Center of Mass

• The acceleration of the center of mass can be The acceleration of the center of mass can be found by differentiating the velocity with found by differentiating the velocity with respect to timerespect to time

CMCM

1i i

i

dm

dt M v

a a

Forces In a System of ParticlesForces In a System of Particles

• The acceleration can be related to a forceThe acceleration can be related to a force

• If we sum over all the internal forces, they If we sum over all the internal forces, they cancel in pairs and the net force on the system cancel in pairs and the net force on the system is caused only by the external forcesis caused only by the external forces

CM ii

M a F

Newton’s Second Law for a System of Newton’s Second Law for a System of ParticlesParticles

• Since the only forces are external, the net Since the only forces are external, the net external force equals the total mass of the external force equals the total mass of the system multiplied by the acceleration of the system multiplied by the acceleration of the center of mass:center of mass:

• The center of mass of a system of particles of The center of mass of a system of particles of combined mass combined mass MM moves like an equivalent moves like an equivalent particle of mass particle of mass MM would move under the would move under the influence of the net external force on the systeminfluence of the net external force on the system

ext CMM F a

Impulse and Momentum of a System of Impulse and Momentum of a System of ParticlesParticles

• The impulse imparted to the system by The impulse imparted to the system by external forces isexternal forces is

• The total linear momentum of a system of The total linear momentum of a system of particles is conserved if no net external force particles is conserved if no net external force is acting on the systemis acting on the system

ext CM totdt M d I F v p

0C M to t extM constan t w hen v p F

The center of mass of an object with a point, line or The center of mass of an object with a point, line or plane of symmetry lies on that point, line or plane.plane of symmetry lies on that point, line or plane.

The center of mass of an object does not need to lie The center of mass of an object does not need to lie within the object (Examples: doughnut, horseshoe )within the object (Examples: doughnut, horseshoe )

Finding Center of Mass, Irregularly Finding Center of Mass, Irregularly Shaped ObjectShaped Object

• Suspend the object Suspend the object from one pointfrom one point

• Then suspend from Then suspend from another pointanother point

• The intersection of The intersection of the resulting lines is the resulting lines is the center of massthe center of mass

Center of GravityCenter of Gravity

• Each small mass element of an extended object Each small mass element of an extended object

is acted upon by the gravitational forceis acted upon by the gravitational force

• The net effect of all these forces is equivalent to The net effect of all these forces is equivalent to

the effect of a single force acting through a the effect of a single force acting through a

point called the point called the center of gravitycenter of gravity

– If is constant over the mass distribution, the If is constant over the mass distribution, the

center of gravity coincides with the center of masscenter of gravity coincides with the center of mass

Mg

g

Center of Mass, RodCenter of Mass, Rod

Show that the center of mass Show that the center of mass of a rod of mass of a rod of mass M M and length and length L L lies midway between its lies midway between its ends, assuming the rod has a ends, assuming the rod has a uniform mass per unit length. uniform mass per unit length.

Problem solving tactics:Problem solving tactics:

(1) Use object’s symmetry.(1) Use object’s symmetry.

(2) If possible, divide object in several parts. Treat each of (2) If possible, divide object in several parts. Treat each of

these parts as a particle located at its own center of mass.these parts as a particle located at its own center of mass.

(3)(3) Chose your axes wisely. Use one particle of the system as Chose your axes wisely. Use one particle of the system as

origin of your reference system or let the symmetry lines be origin of your reference system or let the symmetry lines be

your axis.your axis.

A water molecule consists of an oxygen atom with two A water molecule consists of an oxygen atom with two hydrogen atoms bound to it. The angle between the two hydrogen atoms bound to it. The angle between the two bonds is bonds is 106106. If the bonds are . If the bonds are 0.100 nm0.100 nm long, where is long, where is the center of mass of the molecule?the center of mass of the molecule?

A uniform piece of sheet steel is shaped as in Figure. A uniform piece of sheet steel is shaped as in Figure. Compute the Compute the xx and and yy coordinates of the center of coordinates of the center of mass of the piece.mass of the piece.

Rocket PropulsionRocket Propulsion

The operation of a rocket depends upon the law of The operation of a rocket depends upon the law of conservation of linear momentum as applied to a system conservation of linear momentum as applied to a system of particles, where the system is the rocket plus its of particles, where the system is the rocket plus its ejected fuel.ejected fuel.

IV. Systems with varying massIV. Systems with varying mass

The initial mass of the rocket The initial mass of the rocket plus all its fuel is plus all its fuel is MM + + ΔΔmm at at time time ttii and velocity and velocity vv

The initial momentum of the The initial momentum of the system is system is ppii = ( = (MM + + ΔΔmm) v) v


At some time At some time tt + + ΔΔtt, the rocket’s mass has been , the rocket’s mass has been reduced to reduced to MM and an amount of fuel, and an amount of fuel, ΔΔmm has been has been ejected.ejected.

The rocket’s speed has increased by The rocket’s speed has increased by ΔΔvv

Because the gases are given some momentum when Because the gases are given some momentum when they are ejected out of the engine, the rocket receives they are ejected out of the engine, the rocket receives a compensating momentum in the opposite directiona compensating momentum in the opposite direction

Therefore, the rocket is accelerated as a result of the Therefore, the rocket is accelerated as a result of the “push” from the exhaust gases“push” from the exhaust gases

In free space, the center of mass of the system (rocket In free space, the center of mass of the system (rocket plus expelled gases) moves uniformly, independent of plus expelled gases) moves uniformly, independent of the propulsion processthe propulsion process


IV. Systems with varying massIV. Systems with varying mass

)()( dvvdMMudMMv

Example:Example: most of the mass of a rocket on its launching is fuel most of the mass of a rocket on its launching is fuel that gets burned during the travel.that gets burned during the travel.

System: System: rocket + exhaust productsrocket + exhaust productsClosed and isolatedClosed and isolated mass of this mass of this system does not change as the system does not change as the rocket accelerates. rocket accelerates. PP=const =const PPii=P=Pff

Linear momentum of Linear momentum of exhaust products exhaust products released during the released during the intervalinterval dtdt

Linear momentumLinear momentumof rocket at the of rocket at the end ofend of dtdt

AfterAfter dtdt

MdvdMvMvdvdMvdMMdvMvdMvdvdMvdMMv

dvvdMMvdvvdMMv

relrel

rel

))((])[(

RR=Rate at which the rocket losses mass==Rate at which the rocket losses mass= -dM/dt-dM/dt = rate of fuel = rate of fuel consumptionconsumption

MavRdt

dvMv

dt

dMrelrel First rocket First rocket

equationequation

Velocity of rocket relative to frameVelocity of rocket relative to frame = (velocity of rocket relative= (velocity of rocket relative to products)+ (velocity of products relative to frame)to products)+ (velocity of products relative to frame)

relrel vdvvuuvdvv )()(

)()( dvvdMMudMMv

dt

dvMv

dt

dMdMvMdv relrel

f

irelif

ifrel

v

v

M

M

rel

relrel

M

Mvvv

MMvM

dMvdv

vM

dMdv

dt

dvMv

dt

dM

f

i

f

i

ln

lnln

Second rocket equationSecond rocket equation

• The basic equation for rocket propulsion is The basic equation for rocket propulsion is

• The increase in rocket speed is proportional to The increase in rocket speed is proportional to the speed of the escape gases (the speed of the escape gases (vvee))– So, the exhaust speed should be very highSo, the exhaust speed should be very high

• The increase in rocket speed is also proportional The increase in rocket speed is also proportional to the natural log of the ratio to the natural log of the ratio MMi i / / MMff

– So, the ratio should be as high as possible, meaning So, the ratio should be as high as possible, meaning the mass of the rocket should be as small as possible the mass of the rocket should be as small as possible and it should carry as much fuel as possibleand it should carry as much fuel as possible

ln if i e

f

Mv v v

M

ThrustThrust

• The thrust on the rocket is the force exerted The thrust on the rocket is the force exerted on it by the ejected exhaust gaseson it by the ejected exhaust gases

Thrust =Thrust =

• The thrust increases as the exhaust speed The thrust increases as the exhaust speed increasesincreases

• The thrust increases as the rate of change of The thrust increases as the rate of change of mass increasesmass increases– The rate of change of the mass is called the The rate of change of the mass is called the rate rate

of fuel consumptionof fuel consumption or or burn rateburn rate

e

dv dMM v

dt dt

The first stage of a Saturn V space vehicle consumed fuel The first stage of a Saturn V space vehicle consumed fuel and oxidizer at the rate of and oxidizer at the rate of 1.50 1.50 10 1044 kg/s kg/s, with an exhaust , with an exhaust speed of speed of 2.60 2.60 10 1033 m/s m/s. (a) Calculate the thrust produced . (a) Calculate the thrust produced by these engines. (b) Find the acceleration of the vehicle by these engines. (b) Find the acceleration of the vehicle just as it lifted off the launch pad on the Earth if the vehicle’s just as it lifted off the launch pad on the Earth if the vehicle’s initial mass was initial mass was 3.00 3.00 10 1066 kg kg. . Note: Note: You must include the You must include the gravitational force to solve part (b).gravitational force to solve part (b).

Model rocket engines are sized by thrust, thrust duration, and Model rocket engines are sized by thrust, thrust duration, and total impulse, among other characteristics. A size C5 model total impulse, among other characteristics. A size C5 model rocket engine has an average thrust of rocket engine has an average thrust of 5.26 N5.26 N, a fuel mass of , a fuel mass of 12.7 grams12.7 grams, and an initial mass of , and an initial mass of 25.5 grams25.5 grams. The duration . The duration of its burn is of its burn is 1.90 s1.90 s. (a) What is the average exhaust speed of . (a) What is the average exhaust speed of the engine? (b) If this engine is placed in a rocket body of the engine? (b) If this engine is placed in a rocket body of mass mass 53.5 grams53.5 grams, what is the final velocity of the rocket if it is , what is the final velocity of the rocket if it is fired in outer space? Assume the fuel burns at a constant rate.fired in outer space? Assume the fuel burns at a constant rate.

A rocket for use in deep space is to be capable of boosting a A rocket for use in deep space is to be capable of boosting a total load (payload plus rocket frame and engine) of total load (payload plus rocket frame and engine) of 3.00 3.00 metric tonsmetric tons to a speed of to a speed of 10 000 m/s10 000 m/s. (a) It has an engine . (a) It has an engine and fuel designed to produce an exhaust speed of and fuel designed to produce an exhaust speed of 2 000 m/s2 000 m/s. . How much fuel plus oxidizer is required? (b) If a different fuel How much fuel plus oxidizer is required? (b) If a different fuel and engine design could give an exhaust speed of and engine design could give an exhaust speed of 5 000 m/s5 000 m/s, , what amount of fuel and oxidizer would be required for the what amount of fuel and oxidizer would be required for the same task?same task?

A A 60.0-kg60.0-kg person running at an initial speed of person running at an initial speed of 4.00 m/s4.00 m/s jumps onto a jumps onto a 120-kg120-kg cart cart initially at rest. The person slides on the cart’s top surface and finally comes to rest initially at rest. The person slides on the cart’s top surface and finally comes to rest relative to the cart. The coefficient of kinetic friction between the person and the cart is relative to the cart. The coefficient of kinetic friction between the person and the cart is 0.4000.400. Friction between the cart and ground can be neglected. (a) Find the final . Friction between the cart and ground can be neglected. (a) Find the final velocity of the person and cart relative to the ground. (b) Find the friction force acting velocity of the person and cart relative to the ground. (b) Find the friction force acting on the person while he is sliding across the top surface of the cart. (c) How long does on the person while he is sliding across the top surface of the cart. (c) How long does the friction force act on the person? (d) Find the change in momentum of the person the friction force act on the person? (d) Find the change in momentum of the person and the change in momentum of the cart. (e) Determine the displacement of the person and the change in momentum of the cart. (e) Determine the displacement of the person relative to the ground while he is sliding on the cart. (f) Determine the displacement of relative to the ground while he is sliding on the cart. (f) Determine the displacement of the cart relative to the ground while the person is sliding. (g) Find the change in kinetic the cart relative to the ground while the person is sliding. (g) Find the change in kinetic energy of the person. (h) Find the change in kinetic energy of the cart. (What kind of energy of the person. (h) Find the change in kinetic energy of the cart. (What kind of collision is this, and what accounts for the loss of mechanical energy?)collision is this, and what accounts for the loss of mechanical energy?)

A A 2.00-kg2.00-kg block situated on a frictionless incline is block situated on a frictionless incline is connected to a spring of negligible mass having a spring connected to a spring of negligible mass having a spring

constant of constant of 100 N/m100 N/m. The pulley is frictionless.. The pulley is frictionless. The block The block is released from rest with the spring initially unstretched. is released from rest with the spring initially unstretched. (a) How far does it move down the incline before coming (a) How far does it move down the incline before coming to rest? (b) What is its acceleration at its lowest point? Is to rest? (b) What is its acceleration at its lowest point? Is the acceleration constant? the acceleration constant?

A A 2.00-kg2.00-kg block situated on a block situated on a roughrough incline is connected to a spring of incline is connected to a spring of negligible mass having a spring constant of negligible mass having a spring constant of 100 N/m100 N/m. The pulley is . The pulley is frictionless.frictionless. The block is released from rest when the spring is unstretched. The block is released from rest when the spring is unstretched. The block moves The block moves 20.020.0 cmcm down the incline before coming to rest. Find the down the incline before coming to rest. Find the coefficient of kinetic friction between the block and incline.coefficient of kinetic friction between the block and incline.

115.0

)2.0)(37)(sin/8.9)(2(2

)2.0)(/100()2.0)(37)(cos/8.9)(2(

00

sin2

1cos

cos

cos

022

02

2

k

k

fi

k

kk

msmkgmmN

msmkg

Kvv

mgxkxKmgx

mgf

mgn

Example: Tennis RacketExample: Tennis Racket

AA 50 g50 g ball is struck by a racket. If the ball is ball is struck by a racket. If the ball is initially travellinginitially travelling at at 5 m/s5 m/s up and ends up up and ends up travelling attravelling at 5 m/s5 m/s down afterdown after 0.1 s0.1 s of contact what of contact what is the (average) force exerted by the racket on the is the (average) force exerted by the racket on the ball? What is the impulse?ball? What is the impulse?

Example: Bowling ballsExample: Bowling balls

Two particles massesTwo particles masses mm and and 3m3m are moving towards each are moving towards each other at the same speed, but opposite velocity, and collide other at the same speed, but opposite velocity, and collide elasticallyelastically

After collision,After collision, mm moves off at right anglesmoves off at right angles

What areWhat are vvmfmf andand vv3mf3mf? At what angle is? At what angle is 3m3m scattered?scattered?

m 3m

m

vo vo

vfm

x

y

Sulfur dioxideSulfur dioxide (SO(SO22)) consist of two oxygen atoms (each consist of two oxygen atoms (each

of massof mass 16u16u) and single sulfur atom (of mass) and single sulfur atom (of mass 32u32u). The ). The center-to-center distance between the atoms iscenter-to-center distance between the atoms is 0.143nm0.143nm. Angle is given on the picture. Find the. Angle is given on the picture. Find the x-x- andand yy-coordinate of center of the mass of this molecule.-coordinate of center of the mass of this molecule.

linear momentum and collisions chapter 9. i.linear momentum and its conservation ii.impulse and...

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