Solution: Electricity: 1 kw-hr = (1 x10 3 J/s)(3.6 x10 3 s) = 3.6 x10 6 J 1 kw-hr = (1 x10 3...

Solution:Solution:Electricity:Electricity:• 1 kw-hr = (1 x101 kw-hr = (1 x1033 J/s)(3.6 x10 J/s)(3.6 x1033 s) = 3.6 s) = 3.6 x10x106 6 JJ

• E/$ = E/$ = 3.6 x103.6 x1066/$0.1 = 3.6 x10/$0.1 = 3.6 x107 7 J/$ J/$

Gasoline: Gasoline: • 1 gal = 1 gal = 1.3 x101.3 x108 8 JJ

• E/$ = E/$ = 1.3 x101.3 x1088/$2 = 6.5 x10/$2 = 6.5 x107 7 J/$ J/$

•Cost (gas/elec) ~ 180% for same efficiency.Cost (gas/elec) ~ 180% for same efficiency.

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

Consider this:Consider this:• How high will the tennis ball bounce How high will the tennis ball bounce relative to its starting height when they relative to its starting height when they are dropped as shown below?are dropped as shown below?


1’ Lecture 1’ Lecture • Linear momentum is the product of the Linear momentum is the product of the velocity and the mass and is, therefore, a velocity and the mass and is, therefore, a vector quantity.vector quantity.

• p p = = m m vv

• Momentum is always conserved.Momentum is always conserved.

• Impulse is the time integrated force.Impulse is the time integrated force.


Linear Momentum (Linear Momentum (Latin: “movement”)Latin: “movement”)

p p ≡ ≡ m m vv

ppxx = m v = m vxx

ppyy = m v = m vyy

ppzz = m v = m vzz


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

First law:First law:

In the absence of any net external force In the absence of any net external force (∑(∑FF = = 0)0)the linear momentum the linear momentum p p of a system is conserved of a system is conserved ((d d pp/dt = 0/dt = 0).).


The Collision of Two BodiesThe Collision of Two Bodies

①⇒ ⇐② ①⇒ ⇐②

FF1212 = - = - FF2121

FF2121 + + FF1212 = 0= 0

dd p p11 /dt /dt ++dd p p22 /dt /dt = 0= 0

d(d( p p11 + + pp22 )/dt )/dt = 0= 0

Thus, the total momentum is conserved in a Thus, the total momentum is conserved in a collision.collision.


Conservation of Linear MomentuConservation of Linear Momentumm

pp1i1i + + pp2i2i = = pp1f1f + + pp2f2f

∑∑ppixix =∑p =∑pfxfx

∑∑ppiyiy =∑p =∑pfyfy

∑ ∑ppiziz = ∑p = ∑pfz fz


Big Ball – Little Ball Big Ball – Little Ball CollisionCollision

Demonstration!!!Demonstration!!!

??? ⃘�??? ⃘� ⃗ ⃗ ⃖⃖ ⃝⃝ ??????


Elastic CollisionElastic CollisionP conserved P conserved pp1i1i + p+ p2i 2i = p= p1f1f + p+ p2f2f

mm11vv1i1i + m + m22vv2i2i = m = m11vv1f1f + m + m22vv2f2f

(Assume) E conserved K(Assume) E conserved K1i 1i + K+ K2i 2i = K= K1f1f + K + K2f2f

½ m½ m11vv1i1i22 +½ m +½ m22vv2i2i

22= ½ m= ½ m11vv1f1f22+ ½ m+ ½ m22vv2f2f

22

Solve equations:Solve equations:

vv1f1f = [(m = [(m11 -m -m22)/ (m)/ (m11 +m +m22)]v)]v1i1i +[(2 m +[(2 m22)/ (m)/ (m11 +m +m22)]v)]v2i2i

vv2f2f = [(m = [(m22 –m –m11)/ (m)/ (m11 +m +m22)]v)]v2i2i +[(2 m +[(2 m11)/ (m)/ (m11 +m +m22)]v)]v1i1i




Elastic CollisionElastic Collision




Let mLet m2 2 >> m>> m1 1 ; v; v1i1i = -v = -v2i2i

vv1f1f = [(0-m = [(0-m22)/ (0+m)/ (0+m22)]v)]v1i1i -[(2 m -[(2 m22)/ (0 +m)/ (0 +m22)](-v)](-v1i1i ) )

vv1f1f = (-v = (-v1i1i -2 v -2 v1i1i ) = - 3 v ) = - 3 v1i1i !! !!

KK1f 1f /K/K1i 1i = v= v1f 1f 22 /v /v1i 1i

2 2 = 9 = h= 9 = h1i1i/h/h2f2f

What about a ball that bounces off What about a ball that bounces off of the earth?of the earth?• What is the change in the momentum of What is the change in the momentum of the earth when a 0.60 kg ball falls from a the earth when a 0.60 kg ball falls from a height of 1.0 meter and bounces up?height of 1.0 meter and bounces up?• What is the change in velocity of the earth? What is the change in velocity of the earth? (mass of earth = 6.0x10(mass of earth = 6.0x102424 kg) kg)


Peer Instruction Peer Instruction TimeTime

•What is the change in the momentum of the What is the change in the momentum of the earth when a 0.60 kg ball falls from a height earth when a 0.60 kg ball falls from a height of 1.0 meter and bounces up?of 1.0 meter and bounces up?• What is the change in velocity of the earth? What is the change in velocity of the earth? (mass of earth = 6.0x10(mass of earth = 6.0x102424 kg) kg)


Σpi = Pi +pi = 0 = M Vi + mvi

Vi = - (m/M)vi

vi = √2gh = √2(9.8 m/s2)(1.0 m h ) = 4.4 m/sVi = - (0.6 kg/ 6.0x106.0x102424 kg kg )vi = - 4.4 x 10 – 25m/s

If ball rebounds to same height v1f = v1i; V1f = V1i; ΔV = 8.8 x 10 – 25m/s (~1proton/36yr)

Elastic CollisionElastic CollisionSpecial Case Special Case mm11 = m = m22

vv1f1f = [(2 m = [(2 m22)/ (m)/ (m11 +m +m22)]v)]v2i2i

vv2f2f = [(2 m = [(2 m11)/ (m)/ (m11 +m +m22)]v)]v1i1i

KK1f1f = [(4 m = [(4 m11 m m22)/ (m)/ (m11 +m +m22) ) 22 ] K ] K2i2i

KK2f2f = [(4 m = [(4 m11 m m22)/ (m)/ (m11 +m +m22) ) 22 ] K ] K1i1i

vv1f1f = v = v2i2i K K1f1f = K = K2i2i

vv2f2f = v = v1i 1i KK2f2f = K = K1i1i


Elastic CollisionElastic CollisionSpecial Case Special Case mm11 = m = m22

Newton’s CradleNewton’s Cradle


Totally Inelastic Collisions (no bounce)Totally Inelastic Collisions (no bounce)

[The objects stick together after collision.][The objects stick together after collision.]

vv1f1f = v = vff

vv2f2f = v = vff

mm11 v v1i1i + m + m22 v v2i2i = (m = (m11 + m + m22 ) v ) vff

vvff = (m = (m11 v v1i1i + m + m22 v v2i2i )/ (m )/ (m11 + m + m22 ) )

KKi i = ½ m= ½ m11vv1i1i22 +½ m +½ m22vv2i2i

22

KKf f = ½ (m= ½ (m1 1 + m+ m22 )v)vff

22


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


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

DemonstrationDemonstration


∆∆pp = ∫ = ∫FF dtdt

∆∆pp =F=Faveave ∆t∆t

For a given force the longer it For a given force the longer it isis

applied the greater will be applied the greater will be the impulse and the change in the impulse and the change in

momentum.momentum.


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.


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 Chapter 9 Linear Momentum and Chapter 9 Linear Momentum and CollisionsCollisions

Center of MassCenter of Mass

RRCMCM = ∑ = ∑mmii rr/ M/ M

OrOr

RRCMCM = = ∫∫r r dm / Mdm / M

Physics 1710Physics 1710 Chapter 9 Linear Momentum and Chapter 9 Linear Momentum and CollisionsCollisions

Summary:Summary:• Linear momentum is the product of Linear momentum is the product of the mass and velocity.the mass and velocity.

•Linear momentum is Linear momentum is conservedconserved..•Impulse is the time integral of the Impulse is the time integral of the momentum.momentum.

•The impulse is equal to the change in The impulse is equal to the change in momentum of a system.momentum of a system.

Solution: Electricity: 1 kw-hr = (1 x10 3 J/s)(3.6 x10 3 s) = 3.6 x10 6 J 1 kw-hr = (1 x10 3...

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Transcript of Solution: Electricity: 1 kw-hr = (1 x10 3 J/s)(3.6 x10 3 s) = 3.6 x10 6 J 1 kw-hr = (1 x10 3...