PHYC10001 4 Systems Web

Prof D.N. Jamieson 2011

PHYC10001 Physics 1: AdvancedPart 5 - Systems of Particles

By:

Prof. David N. JamiesonSchool of PhysicsUniversity of Melbourne

Lecture 9 Lecture 10 Lecture 11

2 Prof D.N. Jamieson 2011

At the end of this lecture you will be able to: Calculate the position of the centre of mass of a system of

particles

Compute the position of the centre of mass for a complicated object (water molecule)

Compute the barycentre of the Earth-Moon system

Know the difference between centre of mass motion and internal motion

Describe how a rocket exploits conservation of momentum for propulsion

Use the concept of conservation of momentum to solve problems

8


Centre of mass & Momentum

One particle:v

mvp m dt

dpF

Many particles:iv

imiii m vp

dt

d ii

pF

1v

1m2v 2m

particles all

:momentum totalDefine ipP

particles all :mass Total imM

dtd

dtd i

iPpFF

particles all particles all

:forceNet

avv

mdt

mddtmd )(

p

Unit of p:kg.m/s or N.s


Centre of mass

Define Centre of mass

origin

particles all

iiCM mM rr

1r

2r

Applications seesaw double stars Earth-Moon system

Gravity

kgm 15001 kgm 5.12

1r 2r221100 mrmrrCM then , Put

1

221 m

mrr mrmr 11000 12 , e.g.

Simple

Moon

Binary

Duel

Eject, Ejectj

particles all

imM


pm80

Hml )2/sin(

pm 80

Hml )2/sin(

60

160

60

160

y

x

Centre of Mass: Example

Problem: Find centre of mass of the H2O molecule relative to the O atom

lH

H

O

Data: umH 1umO 16

o106

particles alliiCM mM rr

181116 M

),( CMCMCM yxr

OCM mMy 0OCM mMx 0 Hml )2/cos( Hml )2/cos(16018 CMx upm.120

pml 100

pmpmxCM 7.618120

00 CMy

Solution:

mpm 12101

...2211 xmxmMxCM...2211 ymymMyCM


Centre of Mass: Exercise

Problem: Find centre of mass of Earth-Moon system Data:

Earth mass 1 Moon mass 0.01 Moon orbital radius = 60 x Earth radius

This defines the Barycentre of the Earth-Moon system.

Solution: The barycentre is located 4,671 km from centre of Earth

towards the position of the Moon or 1,707 km below surface.

REarth = 6,350 km


Mercury perihelion advance due to GR: 43 arc/100 years

PSR1913+16 periastron advance due to GR: 4.2o per year


www.exoplanets.org

Kepler planet finder space probe March 2014: 715 new planets!http://kepler.nasa.gov/news/nasakeplernews/index.cfm?FuseAction=ShowNews&NewsID=324


Centre of Mass: Solid bodies

Replace sum with integral, density

particles all

iiCM mM rr

particles allimM

origin

body all

dVrM )( dVM CM )(body all

rrr 1r

1dm

2r

2dmdVdm )( 11 r dzdydxdV ..

)(r

Easy if and body regular )(r



At the end of this lecture you will be able to: Calculate the position of the centre of mass of a

system of particles

Compute the position of the centre of mass for a complicated object (water molecule)

Compute the barycentre of the Earth-Moon system

Know the difference between centre of mass motion and internal motion

Describe how a rocket exploits conservation of momentum for propulsion

Use the concept of conservation of momentum to solve problems

9


Conservation of Momentum

For an isolated system not subject to external forces, momentum is always conserved

Consequence of the symmetry of the laws of physics with position

vp m

particles all


Newtons Third Law: For every action there is an equal and opposite reaction

Unit of momentum: kg.m/s or N.s


Centre of Mass: Motion

Velocity: particles all

iiCM mM rr

dt

md

dtMd iiCM

particles all)(

rr

particles all

hence iiCM mM vv

particles all

ipP

CM

CMi Mdt

Md avF

particles all

Acceleration:

Total Kinetic Energy:

particles all

221

iivmK2

21

CMMv22

21 )(

particles all21

iiCM vmMvK

0 frame M. of C.In CMr0

particles all

iim v0

particles all iim a

Momentum:


Centre of Mass: Kinetic Energy


particles all

221

iivmK2

21

CMMv22

21 )(

particles all21

iiCM vmMvK

CMv

1v

2v

3v

4v

5v

6v

7v


Centre of Mass: Kinetic Energy


particles all

221

iivmK2

21

CMMv22

21 )(

particles all21

iiCM vmMvK

1v

2v

3v

4v5v 6v

7v

Centre of mass frame


Conservation of Momentum: Example

t t + dt)(tP

)( dtt P=

generating over 37 million horsepowerhttp://www.nasa.gov/returntoflight/system/system_SSME.html 28 GW


Conservation of Momentum: Example

Problem: Determine the relationship between the initial mass, mi , the final mass, mf , the exhaust velocity, ve , and the final velocity, vf , of a chemical rocket.

Solution:

Initial Fuel iv

Finalfv

im

Time t vm

Time t+dtdvv dmmudm

Notice the fuelbeing accelerated

to throw awaylater



Solution:

)( vdv evRocket Frame

Time t+dtvdv dmmudm

vm

dmm

vdvf

i

f

i

m

me

v

v 1

fi

f

i

mme

vv mvv )ln(

)ln()ln( ifeif mmvvv

fieif mmvvv /lnThrust dtdmvdtdvm eudmdvvdmmmv .))((

dmvmdv eNsmkg

skgsm 2/../

)()( dttt PP

)(tP

)( dtt P

evdvvu )( Hence


fieif mmvvv /lnConventional Rockets Problems with carrying fuel

To go fast (large vf ) you must: Maximise ( mi/mf ) Maximise ve

For Earth escape you need vf = 11 km/s But best chemical fuels (LO2/LH2) have ve = 3 km/s Hence mi/mf = 55 So mf (payload) can only be 2 % of initial rocket mass

Ouch! Reduce rocket mass by throwing away parts during boost

stage (staging) Better fuels?


Exhaust VelocitiesFuel ve (m/s)Compressed CO2 400

Steam 850

Homebrew solid fuel 1,300

70% H2O2/Kerosene 2,100

Expensive solid fuel 2,400

98% H2O2/Kerosene 2,500

LOX/Kerosene 3,300

LOX/Methane 3,800

LOX/LH2 4,500

O3/LH2 5,200

O1/LH2 7,400

NERVA (5000 MW nuclear) 10,000

Fusion plasma 300,000


High temperature exhaust (large ve) - Nuclear Engines

NERVA nuclear rocket prototype

Built in USA in 1960s

Suffered from failure of the nuclear core

Key idea: Use nuclear heated

exhaust to raise ve


Nuclear Pulse Jet

Momentum from nuclear explosions

Concept tested with chemical explosives


Nuclear Pulse Jet: Interplanetary Society Proposal


Ultra fast exhaust - Light (ve = 3x108 m/s)

Use light as the exhaust

Better still, leave engine behind on Earth

Photon momentum: P = hc/Before

After

Mirror

Photon momentum: P

Mirrormomentum:2P

h = Plancks constant = light wavelengthc = speed of light

P

P P

2


Ultra fast exhaust - Light

Starwisp interstellar probe

1000 km2 mylar with 20 nm Si coating


Ramscoop


Ramscoop Maths


Systems of particles - Summary A system of particles moves through space as if it

was a single particle concentrated at the centre of mass.

The location of the centre of mass is given by:

The Kinetic energy of a system of particles is equal to the centre of mass kinetic energy PLUS the kinetic energy of the internal motion.

An isolated system of particles cannot change its momentum.

Chemical rockets are pathetically weak and the C of M (of any rocket + fuel) stays on launch-pad!

On to collisions

particles all

iiCM mM rr


Generic Collision: Define Impulse

Two bodies approach, interact (time t), then depart Examples of interaction methods:

Mechanical Electrostatic Gravitational

iv1iv2

12F

21F

fv1fv2

dtdf

i

f

i

t

t

p

p 12Fp

dtdt PF )(

dttd )(FP

if ttttF where : average or Constant FJ

dtf

i

t

tif 12FppJ Impulse


Example of elastic collision: Gravitational Slingshot

Can use gravity to bounce off a planet

Pick up twice the orbital speed for a head-on collision

NB: Essential that trajectory does not intersect planet!

Where does thenew momentum come from?

oif vvv 2Exercise: Show that atthe maximum bounce:

Hint: Transform to C of M frame


Example of elastic collision: Supernova

Silicon Burning: O runs out, Si core contracts & heats up until: Tcore ~ 3.5 Billion K density ~ 108 g/cc Ignite Silicon burning: Si melts into a sea of He, p, & n Fuses with rest into Nickel (Ni) & Iron (Fe) Builds a heavy Ni/Fe core. Lasts for about 1 day...

Cross section of a star near the end of its life


Example continued... When burning stops, gravity wins and star collapses

But it compresses inner layers too much They bounce Outer layers are blown off.

Exercise: Repeat analysis and find v2f for m not small compared to M

212

1iMvMgh

Impact

h

ghv i 21

Mm

fv1 fv1

ghv f 222 Small bounce

ghv i 222

Big frame

iv2 fv2

ghv f 21

Big bounce

fv1

ii vv 12

ghv f 232 Small bounce

fff vvv 212

fv2Exercise: Find final speed of small mass

for M>>m

ghvv ii 212 )2()2(122 ghghvvv fii

12


At the end of this lecture you will be able to:

Explain some examples Gravitational slingshot Colinear collisions: Supernova (simplified); Bird/Plane Collisions that did not go as planned Airbags reduce force!

Elastic and Inelastic collisions Collisions in 2D

Exotic collisions Know how a breakdown in conservation of energy was observed in

beta-decay..and how Enrico Fermi saved it Positron annihilation

Summary of collisions

(On to rotations!)

10


iv11m

V21 mm

Example of an inelastic collision Problem: Bird collides with plane, find impulse if bird is 0.5 m

longiv2

2m

Vmmvmvm ii )( 212211

21

2211

mmvmvmV

ii

ivVmm 121 then If

Bird mass 0.5 kg

Aeroplane mass 350,000 kg

Bird speed 1 m/s

Aeroplane speed 1,000 km/hr = 300 m/s

Con

serv

atio

n of

mom

entu

m


Example of an inelastic collision Problem: Bird collides with plane, find impulse if bird is 0.5 m

long

Bird mass 0.5 kg

Aeroplane mass 350,000 kg

Bird speed 1 m/s

Aeroplane speed 1000 km/hr = 300 m/s

m 5.0

Start of collision

End of collision

ii vv 21

svv

t ii 00167.013005.05.0

21

1t

2t

12 ttt

)(0 212ii

if vvmpp J

smkg /.5.150)1300(5.0

tif FppJ

t F

NF 120,9000167.0

5.150 !9 Tonnes


Two spectacular collisions

Train on flask Flask remained intact after collision, suggesting complete success BUT F was reduced because

soft nose of locomotive increased t free recoil of cask increased t

Plane on obstacles Anti-misting fuel exploded on impact, suggesting complete failure BUT F was increased because

operator lost control of plane causing more sudden impact than planned, decreasing t engines impacted with obstacle

Nuclear fuel flask

1500 spectatorsBoeing 720 with dummiesObstacles

t /JF



For an isolated system not subject to external forces, momentum is always conserved

Consequence of the symmetry of the laws of physics with position

vp m

particles all


Newtons Third Law: For every action there is an equal and opposite reaction

Unit of momentum: kg.m/s or N.s


Example : Air Bags Save lives The motion of the passengers must be stopped in a distance of

30 cms to avoid hitting the dashboard. This is h. Consider the following simple analysis as before:

Assume car is travelling at a speed v. the deceleration is a = v2/2h F = ma = mv2/2h Assume v = 70 km/h = 19.4 m/s; h = 0.3m; m=70kg Then F = 44 x 103 N Ouch!

)(2 02

02 xxavv (see lecture 2)


Air Bags cont... When applied to the whole body, (Area, A = 0.1 m2),

P = F/A= 4.4 x 105 N/m2 This is just about the threshold of injury.

What happens when you increase the velocity?

F is proportional to v2, so at 100 km/h passengers may be injured.

Timeline: 0 ms: Impact starts 15 to 30 ms after impact: impact detected and decision to

deploy airbag 60 to 90 ms after impact: pyrotechnics fire and inflate bag ?? ms after impact: body decelerates to zero

Exercise: Calculate the duration of the collision between the

passenger and the inflated bag.


Elastic Collisions

Conserve mechanical energy Potential (gravitational, elastic) plus Kinetic

Exercise show if then:

iv1

02 iv

12F

21F

fv1fv2

if vmmmmv 1

21

211

if vmm

mv 121

12

2

if vv 12 21 mm

01 fv

1:

if vv 12 2if vv 11 21 mm 2:

02 fvif vv 11 21 mm 3:

Special Cases02 iv


Inelastic Collision: Example Problem: A car, mass 1500 kg, speed 30 m/s, collides with a wall and

recoils with speed 3.33 m/s. If the collision took place over a time of 0.2 s, find the average impulse and force on the car during the collision.

Solution:

Initial iv

Final fv

smkgpi /.000,45301500 smkgp f /.000,533.31500

smkgppJ if /.000,50 2/.000,2502.0/000,50 smkg

tJF )000,250( N

Is momentum conserved?

Is kinetic energy conserved?

Mechanical energy converted into heat, sound, light, bent metal..


At the end of this lecture you will be able to:

Wrap-up Systems

Inelastic Collision in 2D

Exotic collisions Positron annihilation Know how a breakdown in conservation of energy was observed in beta-decay..and how

Enrico Fermi saved it Discovery of Higgs Boson

On to Rotations (text: Chapter 12)

Define the angular equivalent to kinematic variables: displacement angle velocity angular velocity acceleration angular acceleration mass moment of inertia

Define the angular equivalent to dynamic variables: kinetic energy kinetic energy force torque

Analyse the slow-down of the Crab nebula pulsar Find moment of inertia from slow-down rate and energy-loss rate

11


Reverse Engineering the Nucleus: 1909/1911

Geiger and Marsden, Proc Roy Soc 1909

Rutherford, Phil Mag 1911


Nucleus (+)

Fast ion beam (+)

Nuclear Physics: Example 1

Ion beam scattering Small angle scattering of ions in electron clouds (drag) Large angle scattering from nucleus

See: www.ph.unimelb.edu.au/~dnj/marcshop


Nuclear Physics: Example 2

Distance of closest approach

min

221

min rqqkmvr beamnucleare :by given at approach Closest

221 mvEmech

minrqqkE beamnuclearemech



Applied Nuclear Physics: Example 3

Solids have crystal structure Therefore the degree of scattering depends on angle Phenomenon of channeling discovered in 1960s.

0 degrees 11 degrees

(random)45 degrees



Applied Nuclear Physics: Example 4

Can use the scattering phenomena to analyse thin films of high-tech materials

Use the energy spectrum of backscattered particles



Collisions in 2 Dimensions

In two (or more) dimensions, just resolve problem in to two (or more) components and solve independently


=

Worked Example: Car & Truck collision

Direction of wreckage

Not an elasticcollision so

kinetic energy isnot conserved

Before After

m1= 830 kg

v1 = 62 kph1p

1p

2p

fp

cosfpsinfp

m2 = 550 kg

v2 = 78 kph2p

Next

ip

Problem: What is vf ?


Cons. Momentum: p1 + p2 = pfX component p1 = pf cos

m1v1 = (m1+ m2) vf cos.(1)Y component p2 = pf sin

m2v2 = (m1+ m2) vf sin.(2)Divide equ (2) by (1)

)(tan

11

22

vmvm

____________________ m1v1 = (m1+ m2) vf cos

Gives = 39.80

p1

p2pf

pfx= pf cos

p fy=

pfsin

=

m1v1

m2v2

Before After


Cons. Momentum: p1 + p2 = pfX component p1 = pf cos

m1v1 = (m1+ m2) vf cos.(1)Y component p2 = pf sin

m2v2 = (m1+ m2) vf sin.(2)

p1

p2pf

pfx= pf cos

p fy=

pfsin

=

m1v1

)(tan

11

22

vmvm

= 39.80

m2v2

Use equ 2 to find vf

sin)( 2122

mmvmv f Gives vf = 48.6 km/hr

Before After


After

Example of an inelastic collisionMatter-Antimatter annihilation

Positron Emission Tomography (PET) makes use of radioactive isotopes that decay by the emission of a positron

When this annihilates with an electron, two gamma rays are produced

e+ e2cmE e

2cmE e

eBC 1111

hcp

hcp

0ip 0fp

Before


PET in practice


Example: Breakdown in conservation of energy?

Energy spectrum of emitted particles:

eBC 1111

Num

ber

eBTotal KKK 11

Expected result

Violation of conservation of energy!!

eB KK 11

eBC 1111True result: Discovery of the neutrino

No!

C11 B11 e AfterBefore


In the C. of M. frame conservation of momentum means:

This is the equation of a straight line through the origin with a negative slope

For an elastic collision kinetic energy is conserved. This means:

This is the equation of an ellipse

These are satisfied simultaneously in an elastic collision.

21212211 )/(0 vmmvvmvm

Centre of mass frame: 2 body collision

)/2()( 12212

21

2222

12112

1 mKvmmvKvmvm


Graphically:

v2Collisions insideobey the law of

conservation ofenergy

Law of conservation ofmomentum, solutions mustlie along this line

Solutions on the edge of the rim correspond to elastic collisions

Viable solutionfor elastic collisions

Totally inelasticcollision is at theorigin.

v1

2121 )/( vmmv

)/2()( 12212

21 mKvmmv


Inventing a new particle to save conservation of energy

Postulated first by Wolfgang Pauli in 1930 to explain how beta decay could conserve energy, momentum, and angular momentum (spin).

Named it "neutron" in keeping with convention employed for naming both the proton and the electron, which in 1930 were known to be respective products for alpha and beta decay.

James Chadwick discovered a much more massive nuclear particle in 1932 and also named it a neutron, leaving two kinds of particles with the same name!

Enrico Fermi, who developed the theory of beta decay, coined the term neutrino (the Italian equivalent of "little neutral one") in 1933 as a way to resolve the confusion.

http://www.fysikbasen.dk/English.php?page=Vis&id=79

Picture of Wolfgang Pauli and Niels Bohr studying a Tippe Top. The picture is taken at the opening of the new institute of physics at the University of Lund on May 31 1951. Credit: Photograph by Erik Gustafson, courtesy AIP Emilio Segre Visual Archives, Margrethe

Bohr Collection (www.aip.org/history/esva)

Pauli Bohr


The ATLAS 4-muon collision event

http://www.atlas.ch/multimedia/4-muon-event.html


Kinetic Energy Momentum relation

Momentum: p = mvKinetic energy = mv2 = p2/2m (a parabola)

E= p2/2m

px

hcp

Photons

phc

Matter


Collisions - Summary

Elastic collision Momentum conserved Energy conserved Kinetic energy conserved

Inelastic collision Momentum conserved Energy conserved Kinetic energy NOT conserved (into heat, sound, springs)

Collisions in 2D (or more) Resolve into vector components and treat each dimension

independently

Rotations

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