10/01/2013phy 113 c fall 2013 -- lecture 101 phy 113 c general physics i 11 am – 12:15 pm mwf olin...

PHY 113 C Fall 2013 -- Lecture 10 110/01/2013

PHY 113 C General Physics I11 AM – 12:15 PM MWF Olin 101

Plan for Lecture 10

Chapter 9 -- Linear momentum

1. Impulse and momentum

2. Conservation of linear momentum

3. Examples – collision analysis

4. Notion of center of mass


Summary of physics “laws”

fiveconservatinoniiff

fiveconservatinon

fiveconservati

fitotal

iffi

veconservati

if

f

i

fitotal

WUmvUmv

WWW

UUW

mvmvdW

dt

dmm

rr

rr

rF

vaF

22

22

2

1

2

1

:forces veconservatiFor

2

1

2

1

:oremenergy the kinetic-Work

:law second sNewton'


Another way to look at Newton’s second law:

sNm/skg :momentumlinear of Units

:momentum"linear " Define

pv

vvaF

mdt

md

dt

dmm

iclicker question:Why would you want to define linear momentum?

A. To impress your friends.B. To exercise your brain.C. It might be helpful.D. To distinguish it from angular momentum.


if

f

i

f

i

ddt

ddt

dt

d

dt

md

dt

dmm

pppF

pF

pvvaF

:calculus little a Using

Relationship between Newton’s second law and linear momentum:

if

f

i

f

i

dt

dt

ppFI

FI

:impulse"" Define

(if m is constant)


smsmm

I

dtI

m

dt

f

if

f

i

f

i

/20/50

1000

velocity?final its isWhat

rest.at initially kg, 50 mass of

object an toimparted is which Ns 1000

of value thehas figure in thisshown

impulse theSuppose :Example

v

ppF

FI


Suppose that a tennis ball with mass m=0.057 kg approaches a tennis racket at a speed of 45m/s. What is the impulse the racket must exert on the ball to return the ball in the opposite direction at the same speed. Assume that the motion is completely horizontal.

Ns

Ns

t

I

NsmvI if

1.173.0

13.5F

s? 0.3for lastedn interactio theif

racket by the exerted force average theisWhat

13.5)45)(057.0(22

pp


Example: A 1500 kg car collides with a wall, with vi= -15m/s and vf=2.6m/s. What is the impulse exerted on the car?

Ns

Ns

t

I

sm

vvmI ifif

51076.115.0

26400F

car? on the exerted force average

theis what s, 0.15 is timecollision theIf

Ns 26400

/156.21500kg

pp

JvvmmvmvΔE ifif52222 106368.1

2

1

2

1

2

1

collision todue lossEnergy


Example of graphical representation of F(t)

NssNsFdttFI 5.130015.0180002

10015.0

2

1)(

:impulse of Estimate

max

0015.0

001.0


Physics of composite systems

ii

i

ii

i

ii

iii

ii dt

d

dt

md

dt

dmm p

vvaF


ii

ii

ii

ii

ii

dt

d

final initial

(constant)

0

: then,0 if that Note

pp

p

p

F


Example – completely inelastic collision; balls moving on a frictionless surface

i

iiv

iviv

vvv

vvv

pp

ˆ 0.125

/5.03.0

ˆ15.0ˆ23.0

ˆ/1 ,ˆ/2

5.0 ,3.0For

21

21

21

2211

212211

final initial

m/s

sm

smsm

kgmkgm

mm

mm

mmmm

f

ii

iif

fii

ii

ii


Energy loss in this example:

840

15.02

123.0

2

1125.05.03.0

2

1

ˆ 0.125

ˆ/1 ,ˆ/2

5.0 ,3.0For

2

1

2

1

2

1

222

21

21

2

22

2

11

2

21

J .

E

m/s

smsm

kgmkgm

mmmmE

f

ii

iif

iv

iviv

vvv


Example – completely elastic collision; balls moving on a frictionless surface

iif

iif

ffii

ffii

iii

iii

ii

ii

vmm

mmv

mm

mv

vmm

mv

mm

mmv

vmvmvmvm

mmmm

vmvm

221

121

21

12

221

21

21

211

222

211

222

211

22112211

2final

2initial

final initial

2

2

:algebra someafter motion, d-1For 2

1

2

1

2

1

2

1

2

1

2

1

vvvv

pp


Completely elastic collision; numerical example:

/25.1/15.03.0

3.05.0

/25.03.0

3.02

/75.1/15.03.0

5.02

/25.03.0

5.03.0

ˆ/1 ,ˆ/2

5.0 ,3.0For

2

2

1

1

21

21

221

121

21

12

221

21

21

211

smsm

smv

smsm

smv

smsm

kgmkgm

vmm

mmv

mm

mv

vmm

mv

mm

mmv

f

f

ii

iif

iif

iviv


iclicker exercise:We have assumed that there is no net force acting on the system. What happens if there are interaction forces between the particles?

A. Analysis still appliesB. Analysis must be modified


Example from homework:

?

final initial

Tf

TfTCfCTiTCiC

ii

ii

mmmm

v

vvvv

pp

Ti

T

CfCiCTf m

mv

vvv


Example from homework: -- continued

2222

2

1

2

1

2

1

2

1

:energy mechanicalin Change

TiTCiCTfTCfCif vmvmvmvmKK

iclicker questionDo you expect Kf-Ki to be

A. >0B. <0


Another example:

i

ii

i final initial pp

before

vf

after


Examples of two-dimensional collision; balls moving on a frictionless surface

-- equations more 2 Need

,,, :Unknowns

,, :Knowns

sinsin0

coscos

21

121

2211

221111

final initial

ff

i

ff

ffi

ii

ii

vv

vmm

vmvm

vmvmvm

pp


Examples of two-dimensional collision; balls moving on a frictionless surface

smsmsm

v

smsmsm

vvv

smsm

vv

vmvm

vmvmvm

smv

smvkgmm

oof

o

fif

o

ff

ff

ffi

f

i

/11.188.17cos

/060.1

88.17sin

/342.0

88.71 060.1

342.0tan

/060.120cos/1/2

coscos

/342.020sin/1

sinsin

sinsin0

coscos

20 ,/1

,/2 ,06.0 :Suppose

1

o

211

21

2211

221111

o2

121


Example: two-dimensional totally inelastic collision

ji

jiv

vvv

vvv

ˆ/5.12ˆ/375.9

ˆ/204000

2500ˆ/254000

1500

221

21

21

1

212211

smsm

smsm

mm

m

mm

m

mmmm

f

iif

fii

m1=1500kg

m2=2500kg

JΔE

mmmmΔE iif

5

22

22

2

22

2

11

2

21

108.4

2025002

1251500

2

1

5.12375.940002

1

2

1

2

1

2

1

:case in thisenergy kinetic of Loss

vvv


iclicker exercise:Can this analysis be used to analyze a real collision?

A. Of course! The laws of physics must be obeyed.B. Of course NOT! In physics class we only deal

with idealized situations which never happen.


Another example of 2-dimensional elastic collision:

m

vi

vf

m0

2

1

2

1

ˆsin2

-- elastic iscollision thecase, In this

22

if

fi

mmE

mv

v

vv

iI

vv


Energy analysis of a simple nuclear reaction :

HeRnRa 42

22286

22688

Q=4.87 MeV


Energy analysis of a simple reaction : HeRnRa 42

22286

22688

Q=4.87 MeV

MeV8.498.04/1222/1

4/1

2

4

1

222

1

2222

222

QQm

pE

m

p

m

p

m

pQ

He

HeHe

uHe

He

Rn

Rn

HeRn ppp


Elastic collision in two dimensions

,,, :Unknowns

,, :Knowns2

1

2

1

2

1

2

1

sinsin0

coscos

21

121

222

211

222

211

2211

221111

final initial

final initial

ff

i

ffii

ff

ffi

ii

ii

ii

ii

vv

vmm

vmvmvmvm

vmvm

vmvmvm

KK

pp


Elastic collision in two dimensions -- example of elastic proton-proton scattering

,, :Unknowns

37 ;

;/105.3 :Suppose

2

1

2

1

2

1

2

1

sinsin0

coscos

21

21

51

222

211

222

211

2211

221111

final initial

final initial

ff

o

i

ffii

ff

ffi

ii

ii

ii

ii

vv

mmm

smv

vmvmvmvm

vmvm

vmvmvm

KK

pp


Elastic collision in two dimensions -- example of elastic proton-proton scattering

o

f

f

oi

ffii

ff

ffi

smv

smv

smv

vvvv

vv

vvv

53

/1011.2

/1080.2

:algebra someAfter

37 ;/105.3 :Given

sinsin0

coscos

52

51

51

22

21

22

21

21

211


The notion of the center of mass and the physics of composite systems

2

2

2

2

2

2

:Define

or


dt

dM

mMM

m

dt

md

dt

dmm

dt

d

dt

md

dt

dmm

CMtotal

ii

ii

iii

CM

i

ii

i

ii

iii

ii

ii

i

ii

i

ii

iii

ii

rFF

rr

rraF

pvv

aF

PHY 113 C Fall 2013 -- Lecture 10 3010/01/2013dt

dM

dt

d

CM

ii

ii

ii

ii

ii

rpp

p

p

FF

final initial

total

(constant)

0

: then,0 If

2

2

:Define

:mass ofcenter for relation General

dt

dM

mMM

m

CMtotal

ii

ii

iii

CM

rFF

rr


Finding the center of mass

ii

iii

CM mMM

m

rr

ji

jiir

jiir

ˆ00.1ˆ750

4

ˆ22ˆ21ˆ)1)(1(

ˆˆˆ

2 ;1 :example In this

321

332211

321

mm.

mmm

mmm

ymxmxm

kgmkgmm

CM

CM


Example from webassign :

x

ym1

m2

m3

m4


Finding the center of massFor a solid object composed of constant density material, the center of mass is located at the center of the object.

10/01/2013phy 113 c fall 2013 -- lecture 101 phy 113 c general physics i 11 am – 12:15 pm mwf olin...

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