slides lectures

8/10/2019 Slides Lectures

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Contents

1 Chapter 1:IntroductionMotion in one dimensionMotion in three dimensions

2 Chapter 2: Forces and Newtons laws of motion3

Chapter 3: Work and energy 4 Chapter 4: Potential energy and energy conservation5 Chapter 5: Momentum6 Chapter 6: Circular motion7 Chapter 7: Oscillatory motion8 Chapter 8: Law of gravity 9 Chapter 9: Fluids mechanics

Nguyen Thi Hong Van


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Evaluation

Attendance/Attitude: 5 %

Class exercise(s): 10 %

Mid-term test: 25 %

Final exam: 60 %

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Major sources

Major sources1 Classical Mechanics : an introductory course , Richard Fitzpatrick

(professor of Physics, the University of Texas at Austin)

2 R. Resnick, D. Halliday, and K.S. Krane, Fourth edition, Vol. 1(John Wiley & Sons, New York NY, 1992)

3 G.R. Fowles, Third edition (Holt, Rinehart, & Winston, New YorkNY, 1977).

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Chapter 1: Overview of Classical Mechanics (CM)

CM is the study of the motion of bodies in accordance with the generalprincipals rst enunciated by Issac Newton (1687)

CM has many applications in different areas of science such asastronomy , Chemistry , Geology , and Engineering

Types of motion studied in CM:

Translational motion : a body shifts from one point in space toanother

Rotational motion : an extended body changes orientation, withrespect to other bodies in space, without changing position

Oscillatory motion : continually repeats in time with a xed periodCircular motion : a body executes a circular orbit about anotherxed body.

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Chapter 1: Measurements in physics (1)

Units of measurements

Three fundamental quantities which are subject to measurement used inCM:Interval in space lengthQuantities of inertia , or mass , processed by various bodiesInterval in time

SI system (Systemme International) mks units SI system is a standard units for basis quantities. It was establishedin 1960 by an international committeemks system is for length (meter), mass (kilogram) and time (second)

Length : the meter (m) is redened as the distance traveled by light invacuum during a time of 1/299 792 458 second.

Mass : the kilogram (kg), is dened as the mass of a specic platinumiridium alloy cylinder kept at the International Bureau of Weights andMeasures at Sevres, France.Time : the second (s), is dened as 9 192631 770 times the period of vibration of radiation from the cesium-133 atom.

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Chapter 1: Length, mass and time (1)

Approximated value of some measured lengths (m)Distance from the Earth to most remote known normal galaxies: 9 1025

One light-year: 9.46 1015

Mean orbit radius of the Earth about the Sun: 1.50 1011

Mean distance from the Earth to the Moon: 3.84 108

Distance from the equator to the North Pole: 1.00 107

Mean radius of the Earth: 6.37 106

Size of smallest dust particles: 10 4

Diameter of a hydrogen atom: 10 10

Diameter of an atomic nucleus: 10 14

Diameter of a proton: 10 15

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Length, mass and time (2)

Mass of various bodies (kg)

Visible universe 1052

Sun: 1.99 1030

Earth: 5.98 1024

Moon: 7.36 1022

Horse: 103

Human: 102

Mosquito: 10 5

Hydrogen atom: 1.67 10 27

Value of some time interval (s)

Age of the Universe: 5 1017

Age of the Earth: 1.3 1017

One year: 3.16 107

One day: 8.64 104

Period of audible sound waves: 10 3

Period of typical radio waves: 10 6

Duration of a nuclear collision: 10 22

Time for light to cross a proton: 10 24

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Chapter 1: Units of measurements

Units of derived quantities

Velocity: [v ] = ms 1

Acceleration: [a ] = ms 2

Angular momentum: [ p ] = kgms 1

Standard prexesA set of standard prexes is a modication of the mks units of length, mass

and time which is devised to deal more easily with very small and very largequantities (the motion of molecules and the motion of stars in the Galaxy).

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Chapter 1: Density

Density is a property of substance

Density () is dened as the amount of mass contained in a unit volume:

= mV

Units of density: kg / m 3

Density of various substances (103 kg / m3 )

Gold: 19.3Uranium: 18.7Lead: 11.3Copper: 8.92Iron: 7.86

Aluminum: 2.70

Magnesium: 1.75Water: 1.00Air: 0.12 10 2

Nguyen Thi Hong Van


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Chapter 1: Dimensional analysis

Dimension denotes a physical nature of a quantityDimensions of three fundamental quantities length, mass and time are L,

M, T, respectively.We use brackets [ ] to denote the dimension of a quantity.Dimension of speed: [v ] = L/ T Dimension of volume: [V ] = L3

Acceleration: [a ] = L/ T 2

Dimensions can be treated as algebraic quantities.All laws of physics are dimensionally consistent:

A = B [A] = [B ]

Application of dimensional analysis is to check the forms of simple lawsof physics.Exercise:

Find dimensions of density = m / V , momentum p = mv ; x = at 2 / 2Show that the expression v = at is dimensionally correct.

Nguyen Thi Hong Van

Ch 1 U i d i i (1)


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Chapter 1: Uncertainty and signicant gures (1)

Uncertainties of measurementsWhen physical quantities are measured, the measured values are knownonly to within the limits of the experimental uncertainty.The value of uncertainty depends on:

quality of apparatusthe skill of the experimenterthe number of measurements performed

Signicant guresExample:

5.5 2 signicant gures0.1 1 signicant gures

Multiplicity rule: when multiplying several quantities, the number of signicant gures in the nal answer is the same as the number of signicant gures in the least accurate (having the lowest number of signicant gures) of the quantities being multiplied. The same ruleapplies to division.

Nguyen Thi Hong Van

Ch 1 U i d i i (2)


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Chapter 1: Uncertainty and signicant gures (2)

Signicant guresZero may NOT be signicant gures: those used to position the decimalpoint such as: 0.03 (1 signicant gure) and 0.0075 (2 signicant gures).Zero may be signicant gures: when zeros come after other digits.However there is the possibility of mis-interpretation should usescientic notation to indicate the number of signicant gures:

1500 1.5 103 if there are 2 SFs1500 1.50 103 if there are 3 SFs1500 1.500 103 if there are 4 SFs0.00023 2.30 10 4 if there are 3 SFs.

When numbers are added or subtracted, the number of decimal places in

the result should equal the smallest number of decimal places of any termin the sum:123 + 5.35 = 1281.002 - 0.998 = 0.004

Nguyen Thi Hong Van

Ch t 1 M ti i di i (1)


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Chapter 1: Motion in one dimension (1)

Displacement

VelocityVelocity is the rate of change of displacement with timeAverage velocity:

v = x t

Instantaneous velocity:v = lim t 0

x t

Average speed: the ratio of the total distance it travels to the total time ittakes to travel that distance.Instantaneous speed = |v |

Nguyen Thi Hong Van

Ch t 1 M ti i di i (2)


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AccelerationAcceleration is the rate of the change of velocity with timeAverage value:

a = v t

Instantaneous value:

a = lim t 0 v t

= dv dt

= d 2 x dt 2

Equation of kinematics (for motion with a constant acceleration):

v (t ) = v 0 + a.t x (t ) = x 0 + v 0 .t + a .t 2 / 2v (t )2 v 20 = 2 .a .(x (t ) x 0 )

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Motion with constant velocityProblem: given v = const , nd displacement x with time t ?Answer:

v = dx / dt dx = v .dt x = v .t + C C =? : t = 0 x = x 0 (displacement of the object at t = 0 or initialdisplacement). x (t ) = x 0 + v .t

Problem: Plot the graph of displacement versus time for a body movingwith constant velocity?Conclusion with motion with constant velocity:

The graph of x versus t is a straight line.v = x / t gradient of the graph/straight line.a = d 2 x / dt 2 = 0 v = const : consistent with the denition!x increases v > 0 OR object moving to the right,x decreases v < 0 OR object moving to the left,x = const v = 0 OR object remains at rest.

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Motion with constant accelerationProb.: a = const v =?; x =?

a = dv / dt dv = a .dt v (t ) = v 0 + a .t straight line

dx = v .dt x = x 0 + v 0 .t + a.t 2 / 2 parabola.

Conclusion:S = x x 0 = v 0 .t + a .t 2 / 2 the net distance traveled after t

secondsv = v 0 + a.t v 2 = v 20 + 2 .a .S

Free-fall under gravityGalileo: all bodies in free-fall close to the Earth surface acceleratevertically downward with the same acceleration: g = 9 .81m / s 2

???

t = 2hg Nguyen Thi Hong Van

Chapter 1: Motion in three dimensions (1)


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To deal with 3-dimensional motion, it is necessary to set up a suitablecoordinate system.

Vectors

Cartesian coordinates

Cartesian coordinate system consistsof three mutually perpendicular axes.

Vector and scalar quantities

Scalar quantity is specied by a single value with an appropriate unitand has no direction such as temperature, volume, mass, timeA vector quantity has both magnitude and direction. Examples:displacement,velocity, accelerator, force .

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Some properties of vectorEquality of two vectors:

A =

B |

A| = |

B | &

A and

B point in the same direction.

Vector additionSum of two vectors is the vector down from the tail of the rstvector to the tip of the last vector.Parallelogram rule of addition A + B = B + AAssociative law of addition: A + ( B + C ) = ( A + B ) + C

Negative of a vector: A + ( A) = 0 have the same magnitude butpoint in opposite direction.Subtracting vector: A B = A + ( B )

Components of a vector and unit vectorsComponents of a vector are the projections of that vector alongcoordinate axes: A = ( Ax , Ay , Az ).?? Can the components of a vector ever be greater than the magnitude of the vector?hint: Ax = | A| .sin.cos ; Ay = | A| .sin sin ; Az = | A| .cos

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Components of a vector and unit vectorsUnit vectors

a unit vector is a dimensionless vector having a magnitude of exactly1. It is used to specify a given direction and have no other physicalsignicance:

i x ; j y ; k z .| i | = | j | = | k | = 1

i . j = i . k = j . k = 0 A = Ax . i + Ay . j + Az . k .

Vector magnitude: A = A2x + A2y + A2z

Scalar multiplication: S = . r , is a number/scalar

S = (x , y , z ) = ( x , y , z ): multiplying all the components of the original.Scalar product of two vectors:

A. B = Ax .B x + Ay .B y + Az .B z

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Vector displacement, vector velocity and vector accelerationVector displacement: r (t ) = [ x (t ), y (t ), z (t )]Vector velocity:

v (t ) = lim t 0 r (t + t ) r (t )

t =

d r dt

v x = dx

dt ; v y =

dy

dt ; v z =

dz

dt Vector acceleration:

a(t ) = lim t 0 v (t + t ) v (t )

t =

d v dt

= d 2 r dt 2

ax = dv x

dt = d 2 x dt 2

ay = dv y

dt =

d 2 y dt 2

a z = dv z

dt =

d 2 z dt 2

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Nguyen Thi Hong Van



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Relative velocity v 12 velocity of object (1) with respect to frame (2)

v 23 velocity of frame (2) with respect to frame (3) v 13 velocity of object (1) with respect to frame (3)

v 1 3 = v 12 + v 23

Nguyen Thi Hong Van

Chapter 2: Forces and Newtons laws of motion


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Chapter 2: Forces and Newton s laws of motion

Newtons rst law of motionIf the motion of a given body is not disturbed by external inuences then

that body moves with constant velocity r = r 0 + v .t r 0 , v ; constant vectors v = 0 the body simply remains at rest.

Newtons second law of motionMomentum: p = m. v Second law: d

p dt = f net inuence/force

m = const f = m . a

Newtons third law of motion f ab : force exerted on (a) by (b) f ab : force exerted on (b) by (a)

Newtons third law: f ab = f ba

Nguyen Thi Hong Van

Chapter 2: Hookes law


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Chapter 2: Hooke s law

The force exerted on an object by a coiled spring:

f = k . l

l = l l 0 extension of the spring l actual length; l 0 natural length.

Application of Hookes law: to quantify the magnitude and direction of the force exerted on a given body by means of a spring

F 0 = 1 N = 1 kg .m / s 2 = k . l 0 , l 0 critical extension

F l = . l 0 is the value of the force we want to quantify.

Nguyen Thi Hong Van

Chapter 2: Everyday forces


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p y y

Gravitational force: f g = m. g The normal force/normal reaction and weight:

Weight: the weight W of a body is the magnitude of the downwardforce it exerts on any object which supports it.Normal force: under the inuence of the weight of the object,obeying the Newtons third law, the surface of the supporting objectwill act back on the surface of the object this called normalreaction or normal force

Normal force is a force exerted by one surface on another i adirection perpendicular to the surface of contact.The magnitude of the weight = the normal force .Exercise:

Find weight of an object resting on a table?

Find weight of an object of mass m resting on the oor of anelevator moving with acceleration a?Relation between mass and weight : W = m.g in case of accelerator of the frame = 0 and f g = m g is perpendicular to the surface.Friction: f = .R n , where : coefficient of friction and R n is normalreaction of the surface.

Nguyen Thi Hong Van

Chapter 3: Work and energy


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p gy

EnergyPotential energy; kinetic energy; electrical energy; thermal energy;chemical energy; nuclear energy.

An example of energy conservation: energy conservation during free-fall

v 22 v 21 = 2 a . S ; a = g ; S = h2 h1

v 22 v 21 2 g . S

mv 22

2

mv 212

= mgh 2 + mgh 1

1

2

mv 22 + mgh 2 = 1

2

mv 21 + mgh 1

Total energy E of the mass m is conserved: E = mv 2 / 2 + mghK = mv 2 / 2 kinetic energyU = mgh potential energyE conserved K = U

Nguyen Thi Hong Van

Chapter 3: Work and energy


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p gy

Work

W = f . S = | f | .| S | .cos : assuming that the force doesnt vary withposition.

Work-Energy theorem : ?? prove:

K = W

using v 2 = v 20 + 2 . a . S

Find back the energy conservation during free-fall using work-energy

theorem?

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Chapter 3: Work


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pWork of the force varies with position f = f ( r )

W =N

i =1

f i . r i = limN N

i =1

f i . r i =B

A

f ( r ).d r line-integral

Work performed by a 1-dimensional force: From W =x B

x A f (x )dx ,prove: W = K ?f = m

d 2 x dt 2

W =

x B

x A

md 2 x dt 2

.dx =

t B

t A

md 2 x dt 2

.dx dt

.dt =

t B

t A

d dt

m2

dx dt

2

x (t A) = x A , x (t B ) = x B , v A = dx dt t = t A

, v B = dx dt t = t B

W = 12

m .v 2B 12

mv 2A = K

The object is acted by many forces: W i =B

A f i ( r ).d r W = i W i = i

B

A f i ( r ).d r =B

A f i ( r ) .d r =B

A f ( r ).d r ; f ( r ) =i

f i ( r )

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Chapter 4: Potential energy


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p gy

potential energy is the energy associated with a system of objects(consists of two or more objects that exert forces on one another).

Gravitational Potential EnergyGravitational potential energy is the potential energy of theobject - Earth system.gravitational potential energy is dened as the product of themagnitude of the gravitational force mg acting on an object and theheight y of the object

U g = mgh

The work done on any object by the gravitational force is equal tothe negative of the change in the systems gravitational potential

energy: W g = U i U f = U g

only the difference in the gravitational potential energy at the initialand nal locations that mattersHorizontal motion does not affect the value of W g

Nguyen Thi Hong Van

Chapter 4: Conservative and non-conservative force-eld


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A non-uniform force-eld f ( r )acts upon an object which movesalong curved trajectory (frompoint A to point B) labeled bypath 1 and path 2.

Work W 1 performed by f ( r ) along path 1: W 1 =path 1

A B f .d r Work W 1 performed by f ( r ) along path 2: W 2 =

path 2

A B

f .d r

If W 1 = W 2 f ( r ) is a conservative force-eld : the line-integral doesntdepend on the path taken between the end points; Ex: gravitational force.If W 1 = W 2 f ( r ) is a non-conservative force-eld : the line-integraldepends both on the end points A, B and the path taken between them;Ex: frictional force.

Nguyen Thi Hong Van

Chapter 4: Conservative and non-conservative force-eld


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Consequences of conservative and non-conservative force-eldWork done around a close loop for:

Conservative force-eld:

W = A B f .d r + B A f .d r = W 1 W 2 = 0

Non-conservative force-eld:

W = W 1 W 2 = 0

Nguyen Thi Hong Van

Chapter 4: Conservative force and potential energy


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Consider a body moving in a conservative force-eld f ( r ). Pick arbitrarilysome point O in this eld.Dene:

U (R ) = R

O

f .d r (3.1)

(3.1) U (O ) = 0U (R ) dened by (3.1) is unique in conservative force-eld since inthis eld, the work of the eld depends only on the x points butNOT the path taken between them.(3.1) makes no sense in non-conservative force eld since there willbe an innite number of different values corresponding to the innitenumber of different paths the body coild take between O and R.

Consider the object move between A and BFrom the work-energy theorem:

K =B

A f .d r =B

A f .d r =B

O f .d r A

O f .d r = U E = K (A) + U (A) = K (B ) + U (B ) = K + U = constant U represents some form of potential energy

Nguyen Thi Hong Van

Chapter 4: Some conclusions about potential energy


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Possible to associate a potential energy with any conservativeforce-eld.

Any force-eld for which we can dene a potential energy mustnecessary be conservative.

The concept of potential energy is meaning-less in anon-conservative force-eld.

Potential energy dene to within an arbitrary additive constant sowe can choose an arbitrary point at which U = 0.

Potential energy is not a property of the body BUT a property of the force-eld within which the body moves.The potential energy function U can be dened such that the work

done by a conservative force equals the decrease in the potentialenergy of the system:

W = f

i

F .d S = U i U f

Nguyen Thi Hong Van

Chapter 4: Elastic Potential Energy


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Consider a system consisting of a block plus a springForce: f = kx

Work: W = x B

x A f (x )dx = k x B

x A x .dx = [kx 2B / 2 kx 2A/ 2]Potential energy: U (x B ) U (x A) =

x B

x A f (x ).dx = kx 2B / 2 kx 2A/ 2 U (x ) = kx 2 / 2

f = dU dx

In three-dimensional;

f = U

x ,

U

y ,

U

z

Example in gravitational force-eld: U = mgz f = (0 , 0, mg )Total energy of the mass: E = K + U

Nguyen Thi Hong Van

Chapter 4: Conservation of mechanical energy


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The total mechanical energy of a system remains constant in anyisolated system of objects that interact only through conservativeforces.

K i + U i = K f + U f

If there are more than one conservative force acting on an objectwithin a system (sum over all conservative forces present):

K i + U i = K f + U f

Nguyen Thi Hong Van

Chapter 4: Work done by non-conservative force


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If there are some of the forces acting on objects within the systemare not conservative , the total mechanical energy of the systemdoes not remain constant. The change in mechanical energy is equalthe total work done by non-conservative forces present.

Work done by an applied forceWork-kinetic energy theorem: W app + W g = K Gravitational force is conservative W g = U W app = K + U an applied force can transfer energy into or out of the system.

Work done by kinetic friction

Kinetic friction makes total mechanical energy of the systemdissipated (as heat, e.g):

E = K + U = F f . S

Nguyen Thi Hong Van

Chapter 5: Momentum: Two component systems


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Consider a system of twoobjects

m1 : F 1 ,

f 12m2 : F 2 , f 21

Newtons second law for:m 1 : m1 x 1 = F 1 + f 12m 2 : m2 x 2 = F 2 f 21 m 1 x 1 + m2 x 2 = F 1 + F 2 (1)

Center of massx cm =

m1 x 1 + m2 x 2m 1 + m2

(2)

(1) & (2) (m 1 + m2 )x cm = F 1 + F 2dP / dt = F where P = M x cm , M = m 1 + m2 , F = F 1 + F 2If F 1 = 0 = F 2 x cm = 0

v cm = x cm = m1 x 1 + m2 x 2

m 1 + m2= const m 1 x 1 + m2 x 2 = const

P = P 1 + P 2 = const If there is no net external force acting on the system, total momentum of the system is conserved.

Nguyen Thi Hong Van

Chapter 5: Momentum: Multi-component systems


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Consider a system of N mutually interacting point mass objectswhich move in 3-dimensionsInternal force: f ij = f ji ; i , j = 1 , N , i = j External force: F i

Second law of Newton for object i : mi r i = j = i

j =1 , N

f ij + F i

Summing over all objects:

i =1 , N

m i r i = j = i

i , j =1 , N

f ij +i =1 , N

F i =i =1 , N

F i

M r cm = F ; M =i =1 , N

m i , F =i =1 , N

F i ; r cm =

N

i =1m i r i

N

i =1m i

If F = 0 r cm =N

i =1m i r i /

N

i =1m i = const OR P =

N

i =1m i r i = const

In general case: d P / dt = F Nguyen Thi Hong Van

Chapter 5: Collisions in 1-dimension


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m1 and m2 collide in1-dimension

ignore zero net external forcein the direction of movingMomentum conservation gives

m1 v i 1 + m2 v i 2 = m1 v 1 f + m2 v f 2 (1)

In 1-dimention: m1 v i 1 + m2 v i 2 = m1 v f 1 + m2 v f 2

There are many types of collisionElastic collision: Total kinetic energy of the two colliding objects isconserved

1

2m1 v 2i 1 +

1

2m2 v 2i 2 =

1

2m1 v 2f 1 +

1

2m2 v 2f 2

Inelastic collision: some fraction of the initial kinetic energy of thecolliding objects is usually converted into some other form of energy.Totally inelastic collision: two objects stick together after thecollision v f 1 = v f 2

Nguyen Thi Hong Van

Chapter 5: Collisions in 1-dimension: Elastic collision


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In the laboratory frame:

v cm = m1 v i 1 + m2 v i 2

m 1 + m2=

m1 v f 1 + m2 v f 2m 1 + m2

(3)

In the center of mass frame: v = v v cm

v i 1 = m2

m 1 + m2(v i 2 v i 1 ); v i 2 =

m1m 1 + m2

(v i 2 v i 1 )

v f 1 = m2

m 1 + m2(v f 2 v f 1 ); v f 2 =

m1m 1 + m2

(v f 2 v f 1 ) (4)

p i 1 = p i 2 = (v i 2 v i 1 ); p f 1 = p f 2 = (v f 2 v f 1 ); = m1 m 2m 1 + m2 (5)

= m1 m 2m 1 + m2

reduced mass

p i 1 + p i 2 = p f 1 + p f 2 = 0 trivial !This result is valid also for inelastic collision!

The center of mass kinetic energy conservation equation:(p i 1 )

2

2m 1+

(p i 2 )2

2m 2=

(p f 1 )2

2m 1+

(p f 2 )2

2m 2(6)

(5)&(6) v i 2 v i 1 = (v f 2 v f 1 ) (7)

(1)&(7) v f 1 = m1 m 2

m 1 + m2v i 1 +

2m 2

m 1 + m2v i 2 ; v f 2 =

2m 1

m 1 + m2v i 1

m1 m 2

m 1 + m2v i 2

Nguyen Thi Hong Van

Chapter 5: Totally inelastic collisions


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We have: v f = v f 1 = v f 2 m1 v i 1 + m2 v i 2 = m1 v f 1 + m2 v f 2 = ( m1 + m2 )v f

v f = m1 v i 1 + m2 v i 2m1 + m2

= v cm

two objects remain stationary in the center of mass velocity frame.Suppose that v i 2 = 0

v f = m1

m1 + m2v i 1

The fractional loss in kinetic energy of the system due to thecollision is

f = K i K f K i

= m1 v 2i 1 (m1 + m2 )v

2f

m1 v 2i 1= m2

m1 + m2

If m2 m 1 the loss in energy is small

If m2 m 1 the loss is almost 100 %

Nguyen Thi Hong Van

Chapter 5: Collision in 2-dimensions


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Suppose v i 2 = 0; m1 , m2 , v i 1 knownCollision is NOT head-onUnknown: v f 1 , v f 2 , 1 , 2 m1 v i 1 = m1 v f 1 + m2 v f 2 (1)

(1)/ox:m1 v i 1 = m1 v f 1 cos1 + m2 v f 2 cos2(1)/oy: m1 v f 1 sin1 = m2 v f 2 sin2Elastic collision :m1 v 2i 1 / 2 = m1 v

2f 1 / 2 + m2 v

2f 2 / 2 need

one more condition!

Nguyen Thi Hong Van

Chapter 5: Collision in 2-dimensions


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Totally inelastic colliding :Unknown: f , v f m 1 v i 1 + m2 v i 2 = ( m 1 + m2 ) v f (2)(2)/ox: m1 v i 1 + m2 v i 2 cos i = ( m 1 + m2 )v f cos f (2)/oy: m2 v i 2 sin i = ( m 1 + m2 )v f sinf

Nguyen Thi Hong Van

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