solution: m 1 v 1i m 1 v 1f m 2 v 2f p conserved m 1 v 1i = - m 1 v 1f + m 2 v 2f e conserved ½ m 1...

Solution:Solution:

m1v1i m1v1f m2v2f

P ConservedP Conserved

m1 v1i = - m1 v1f + m2 v2f

E ConservedE Conserved

½ m1 v1i2 = ½ m1 v1f

2 + ½ m2 v2f2

Solve for Solve for ½ m1 v1f2 = Ef

½ m1 v1i2 = ½ m1 v1f

2 + ½ m2 (m1/m2)2(v1i +v1f ) 2

Ef = Ei [(m2 - m1)/(m2+m1)]2

Physics 1710Physics 1710—C—Chapter 9 Momentum & hapter 9 Momentum & ImpulseImpulse

Rutherford Backscattering Rutherford Backscattering Spectrometry:Spectrometry:

m1v1i m1v1f m2v2f

Ef = Ei [(m2 - m1)/(m2+m1)]2 = k Ei

Silicon mSi = 28 u

Uranium mU = 238 uHelium mHe = 4 u

kSi = 0.56

kU = 0.93



Amplifier

Surface Barrier Detector

Pulse Height Pulse Height AnalysisAnalysis +1



0.1

1

10

100

1000

200 400 600 800 1000 1200 1400 1600

Scattered Ion Energy (keV)

Yie

ld (

cou

nts

/μC

)

238U

40Ca238

56Fe

28Si16O

12C


1’ Lecture 1’ Lecture • Impulse is the time integrated force.Impulse is the time integrated force.

• The motion of a system of point particles The motion of a system of point particles is a combination of motion is a combination of motion ofof the center of the center of mass (CM) and the motion mass (CM) and the motion aboutabout the CM. the CM.

• Force equals the time rate of change in Force equals the time rate of change in momentum.momentum.


Impulse and MomentumImpulse and Momentum

d d pp = = FF dtdt

∆∆pp = ∫ = ∫d d pp = ∫ = ∫FF dt = Impulsedt = Impulse

The impulse on a body equals the change in The impulse on a body equals the change in momentum.momentum.


Impulse and MomentumImpulse and Momentum Consider the following scenarios:Consider the following scenarios:

Which will have the greater initial Which will have the greater initial velocity? Scenario A or B?velocity? Scenario A or B?

AA BB


Impulse and “Follow Impulse and “Follow Through”Through”

DemonstrationDemonstration


∆∆pp = ∫ = ∫FF dtdt

∆∆pp =F=Faveave ∆t∆t

For a given force, the greater the For a given force, the greater the time that the force is applied, the time that the force is applied, the

greater will be the impulse and, thus, greater will be the impulse and, thus, the change in momentum.the change in momentum.


∆∆pp = ∫ = ∫FF dtdt

∆∆pp =F=Faveave ∆t∆t

For a impulse, the greater the time For a impulse, the greater the time that the force is applied, the less will that the force is applied, the less will

be the force.be the force.

F F = d = d pp/dt/dt


Impulse and Seat BeltsImpulse and Seat Belts•Seat Belts ( and air bags Seat Belts ( and air bags and crumple zones) increase the stopping and crumple zones) increase the stopping time ∆t.time ∆t.

•If ∆If ∆pp is the same in two instants the is the same in two instants the impulse will be the same. The case with the impulse will be the same. The case with the longer ∆t will exhibit the smaller average longer ∆t will exhibit the smaller average force.force.

Physics 1710Physics 1710—C—Chapter 9 Momentumhapter 9 Momentum

Newton’s Second Law of MotionNewton’s Second Law of Motion(What Newton actually said:)(What Newton actually said:)

∑∑FF = = d d pp/dt/dt

The net external force is equal to the time The net external force is equal to the time rate of change in the linear momentum.rate of change in the linear momentum.


Stopping ForceStopping Force

∆∆p = mvp = mv

∆∆t = s/vt = s/vaveave = s/(v/2) = s/(v/2)

FFave ave = = ∆∆p/ ∆t = mv p/ ∆t = mv 22/(2s) /(2s)

Speed kills? : v Speed kills? : v 22

What about the sudden stop? :1/sWhat about the sudden stop? :1/s

Physics 1710Physics 1710—C—Chapter 9 Momentumhapter 9 Momentum

The Consider Two BodiesThe Consider Two Bodies

⇐② ①⇒ ⇐② ①⇒

FF1212 = - = - FF2121

dd p p11 /dt /dt = - = - dd p p22 /dt /dt

thenthen

∆∆pp11 = - = - ∆p∆p22

Thus, the momentum given to an ejected mass is Thus, the momentum given to an ejected mass is equal and opposite to the momentum given to equal and opposite to the momentum given to the ejecting mass.the ejecting mass.


Impulse Engine:Impulse Engine:

• FFthrust thrust = dp/dt = - d(m v = dp/dt = - d(m v exhaustexhaust)/dt)/dt

• FFthrust thrust = dp/dt = - v = dp/dt = - v exhaustexhaust dm/dt dm/dt


mv exhaustexhaustmv exhaustexhaust

Center of Mass (CM)Center of Mass (CM)

RRCMCM ≡ ∑ ≡ ∑mmii rr/ M/ M

OrOr

RRCMCM ≡ { ≡ {∫ ∫ rrdm }/ Mdm }/ M


The center of mass (CM) of a system of The center of mass (CM) of a system of particles of combined mass M particles of combined mass M

moves like an equivalent particle of mass M moves like an equivalent particle of mass M

would move under the influence of the would move under the influence of the resultant external force on the system.resultant external force on the system.


Total Linear MomentumTotal Linear Momentum

vvCMCM = ( = (1/M) ∑1/M) ∑ii m mii vvii Thus:Thus:

PPCMCM = = ∑∑ii ppii = total = total pp

aaCMCM = = dd vvCMCM / /dt = (1/M) ∑dt = (1/M) ∑ii m mii d d vvii / /dtdt

aaCMCM = = (1/M) ∑(1/M) ∑ii m mii d d vvii / /dtdtThus:Thus:

FFCMCM= = MM a aCMCM = = ∑∑ii m mii aaii


solution: m 1 v 1i m 1 v 1f m 2 v 2f p conserved m 1 v 1i = - m 1 v 1f + m 2 v 2f e conserved ½ m 1...

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