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Physics 101 Lecture 9 Linear Momentum and Collisions Dr. Ali ÖVGÜN EMU Physics Department www.aovgun.com

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### Transcript of L9 Momentum PHYS101 - UNIVERSE OF ALI OVGUN€¦ · February 13, 2017 Linear Momentum and... Physics 101Lecture 9

Linear Momentum and Collisions Dr. Ali ÖVGÜN

EMU Physics Department

www.aovgun.com February 13, 2017

Linear Momentum and Collisions

q Conservation of Energy

q Momentum q Impulseq Conservation

of Momentumq 1-D Collisionsq 2-D Collisionsq The Center of Mass February 13, 2017

Conservation of Energyq Δ E = Δ K + Δ U = 0 if conservative forces are the only

forces that do work on the system. q The total amount of energy in the system is constant.

q Δ E = Δ K + Δ U = -fkd if friction forces are doing work on the system.

q The total amount of energy in the system is still constant, but the change in mechanical energy goes into “internal energy” or heat.

2222

21

21

21

21

iiifff kxmgymvkxmgymv ++=++

⎟⎠⎞⎜

⎝⎛ ++−⎟

⎠⎞⎜

⎝⎛ ++=− 2222

21

21

21

21

iiifffk kxmgymvkxmgymvdf February 13, 2017

Linear Momentumq This is a new fundamental quantity, like force, energy.

It is a vector quantity (points in same direction as velocity).

q The linear momentum p of an object of mass m moving with a velocity v is defined to be the product of the mass and velocity:

q The terms momentum and linear momentum will be used interchangeably in the text

q Momentum depend on an object’s mass and velocity

vmp !! = February 13, 2017

Linear Momentumq Linear momentum is a vector quantity

n Its direction is the same as the direction of the velocity

q The dimensions of momentum are ML/Tq The SI units of momentum are kg m / sq Momentum can be expressed in component

form:px = mvx py = mvy pz = mvz

m=p vr r February 13, 2017

Newton’s Law and Momentum

q Newton’s Second Law can be used to relate the momentum of an object to the resultant force acting on it

q The change in an object’s momentum divided by the elapsed time equals the constant net force acting on the object

tvm

tvmamFnet Δ

Δ=ΔΔ== )( !!

!!

netFtp !!

==ΔΔ

interval timemomentumin change February 13, 2017

Impulseq When a single, constant force acts on the

object, there is an impulse delivered to the objectn

n is defined as the impulsen The equality is true even if the force is not constantn Vector quantity, the direction is the same as the

direction of the force

Ir

tFI Δ=!!

netFtp !!

==ΔΔ

interval timemomentumin change February 13, 2017

Impulse-Momentum Theorem

q The theorem states that the impulse acting on a system is equal to the change in momentum of the system

if vmvmpI !!!!−=Δ=

ItFp net

!!! =Δ=Δ February 13, 2017

Calculating the Change of Momentum

[ ]0 ( )p m v mvΔ = − − =

( )

after before

after before

after before

p p pmv mvm v v

Δ = −

= −

= −

r r r

For the teddy bear

For the bouncing ball

[ ]( ) 2p m v v mvΔ = − − = February 13, 2017

Ex1: How Good Are the Bumpers?q In a crash test, a car of mass 1.5´103 kg collides with a wall and rebounds as in figure. The initial and final velocities of the car are vi=-15 m/s and vf = 2.6 m/s, respectively. If the collision lasts for 0.15 s, find (a) the impulse delivered to the car due to the collision (b) the size and direction of the average force exerted on the car February 13, 2017

How Good Are the Bumpers?q In a crash test, a car of mass 1.5´103 kg collides with a wall and rebounds as in figure. The initial and final velocities of the car are vi=-15 m/s and vf = 2.6 m/s, respectively. If the collision lasts for 0.15 s, find (a) the impulse delivered to the car due to the collision (b) the size and direction of the average force exerted on the car

smkgsmkgmvp ii /1025.2)/15)(105.1( 43 ⋅×−=−×==

Ns

smkgtI

tpFav

54

1076.115.0

/1064.2 ×=⋅×=Δ

=ΔΔ=

smkgsmkgmvp ff /1039.0)/6.2)(105.1( 43 ⋅×+=+×==

smkgsmkgsmkg

mvmvppI ifif

/1064.2)/1025.2()/1039.0(

4

44

⋅×=⋅×−−⋅×=

−=−= February 13, 2017

Ex2: Impulse-Momentum Theorem

q A child bounces a 100 g superball on the sidewalk. The velocity of the superball changes from 10 m/s downward to 10 m/s upward. If the contact time with the sidewalk is 0.1s, what is the magnitude of the impulse imparted to the superball?

(A) 0(B) 2 kg-m/s(C) 20 kg-m/s(D) 200 kg-m/s(E) 2000 kg-m/s

if vmvmpI !!!!−=Δ= February 13, 2017

Ex3: Impulse-Momentum Theorem 2

q A child bounces a 100 g superball on the sidewalk. The velocity of the superball changes from 10 m/s downward to 10 m/s upward. If the contact time with the sidewalk is 0.1s, what is the magnitude of the force between the sidewalk and the superball?(A) 0(B) 2 N(C) 20 N(D) 200 N(E) 2000 N

tvmvm

tp

tIF if

Δ−

=ΔΔ=

Δ=

!!!!! February 13, 2017

Conservation of Momentumq In an isolated and closed system,

the total momentum of the system remains constant in time.n Isolated system: no external forcesn Closed system: no mass enters or

leavesn The linear momentum of each

colliding body may changen The total momentum P of the

system cannot change. February 13, 2017

Conservation of Momentumq Start from impulse-momentum

theorem

q Since

q Then

q So

if vmvmtF 222212!!!

−=Δ

if vmvmtF 111121!!!

−=Δ

tFtF Δ−=Δ 1221

!!

)( 22221111 ifif vmvmvmvm !!!! −−=−

ffii vmvmvmvm 22112211!!!! +=+ February 13, 2017

Conservation of Momentumq When no external forces act on a system consisting of

two objects that collide with each other, the total momentum of the system remains constant in time

q When thenq For an isolated system

q Specifically, the total momentum before the collision will equal the total momentum after the collision

ifnet ppptF !!!!−=Δ=Δ

0=Δp!0=netF!

if pp !! =

ffii vmvmvmvm 22112211!!!! +=+ February 13, 2017

Ex4: The Archerq An archer stands at rest on frictionless ice and fires a 0.5-kg arrow horizontally at 50.0 m/s. The combined mass of the archer and bow is 60.0 kg. With what velocity does the archer move across the ice after firing the arrow?

ffii vmvmvmvm 22112211 +=+fi pp =

?,/50,0,5.0,0.60 122121 ====== ffii vsmvvvkgmkgm

ff vmvm 22110 +=

smsmkgkgv

mmv ff /417.0)/0.50(

0.605.0

21

21 −=−=−= February 13, 2017

Ex5: Conservation of Momentum

q A 100 kg man and 50 kg woman on ice skates stand facing each other. If the woman pushes the man backwards so that his final speed is 1 m/s, at what speed does she recoil?(A) 0(B) 0.5 m/s(C) 1 m/s(D) 1.414 m/s(E) 2 m/s February 13, 2017

Types of Collisionsq Momentum is conserved in any collisionq Inelastic collisions: rubber ball and hard ball

n Kinetic energy is not conservedn Perfectly inelastic collisions occur when the objects

stick togetherq Elastic collisions: billiard ball

n both momentum and kinetic energy are conserved February 13, 2017

Collisions Summaryq In an elastic collision, both momentum and kinetic

energy are conservedq In a non-perfect inelastic collision, momentum is

conserved but kinetic energy is not. Moreover, the objects do not stick together

q In a perfectly inelastic collision, momentum is conserved, kinetic energy is not, and the two objects stick together after the collision, so their final velocities are the same

q Elastic and perfectly inelastic collisions are limiting cases, most actual collisions fall in between these two types

q Momentum is conserved in all collisions February 13, 2017

q When two objects stick together after the collision, they have undergone a perfectly inelastic collision

q Conservation of momentum

q Kinetic energy is NOT conserved

fii vmmvmvm )( 212211 +=+

21

2211

mmvmvmv ii

f ++= February 13, 2017

Ex6: An SUV Versus a Compactq An SUV with mass 1.80´103 kg is travelling eastbound

at +15.0 m/s, while a compact car with mass 9.00´102

kg is travelling westbound at -15.0 m/s. The cars collide head-on, becoming entangled.

(a) Find the speed of the entangled cars after the collision.

(b) Find the change in the velocity of each car.

(c) Find the change in the kinetic energy of the system consisting of both cars. February 13, 2017

(a) Find the speed of the entangled cars after the collision.

fii vmmvmvm )( 212211 +=+

fi pp =smvkgmsmvkgm

i

i

/15,1000.9

/15,1080.1

22

2

13

1

−=×=

+=×=

21

2211

mmvmvmv ii

f ++=

smv f /00.5+=

An SUV Versus a Compact February 13, 2017

(b) Find the change in the velocity of each car.

smvvv if /0.1011 −=−=Δ

smvkgmsmvkgm

i

i

/15,1000.9

/15,1080.1

22

2

13

1

−=×=

+=×=

smv f /00.5+=

An SUV Versus a Compact

smvvv if /0.2022 +=−=Δ

smkgvvmvm if /108.1)( 41111 ⋅×−=−=Δ

02211 =Δ+Δ vmvm

smkgvvmvm if /108.1)( 42222 ⋅×+=−=Δ February 13, 2017

(c) Find the change in the kinetic energy of the system consisting of both cars.

JvmvmKE iii52

22211 1004.321

21 ×=+=

smvkgmsmvkgm

i

i

/15,1000.9

/15,1080.1

22

2

13

1

−=×=

+=×=

smv f /00.5+=

An SUV Versus a Compact

JKEKEKE if51070.2 ×−=−=Δ

JvmvmKE fff42

22211 1038.3

21

21 ×=+= February 13, 2017

q Both momentum and kinetic energy are conserved

q Typically have two unknownsq Momentum is a vector quantity

n Direction is importantn Be sure to have the correct signs

q Solve the equations simultaneously

222

211

222

211

22112211

21

21

21

21

ffii

ffii

vmvmvmvm

vmvmvmvm

+=+

+=+ February 13, 2017

Elastic Collisionsq A simpler equation can be used in place of the KE

equation

iffi vvvv 2211 +=+

)vv(vv f2f1i2i1 −−=−

222

211

222

211 2

121

21

21

ffii vmvmvmvm +=+

))(())(( 2222211111 ififfifi vvvvmvvvvm +−=+−

)()( 222111 iffi vvmvvm −=−

)()( 22

222

21

211 iffi vvmvvm −=−

ffii vmvmvmvm 22112211 +=+

ffii vmvmvmvm 22112211 +=+ February 13, 2017

Summary of Types of Collisions

q In an elastic collision, both momentum and kinetic energy are conserved

q In an inelastic collision, momentum is conserved but kinetic energy is not

q In a perfectly inelastic collision, momentum is conserved, kinetic energy is not, and the two objects stick together after the collision, so their final velocities are the same

iffi vvvv 2211 +=+ ffii vmvmvmvm 22112211 +=+

ffii vmvmvmvm 22112211 +=+

fii vmmvmvm )( 212211 +=+ February 13, 2017

Ex7: Conservation of Momentum

q An object of mass m moves to the right with a speed v. It collides head-on with an object of mass 3m moving with speed v/3 in the opposite direction. If the two objects stick together, what is the speed of the combined object, of mass 4m, after the collision?(A) 0(B) v/2(C) v(D) 2v(E) 4v February 13, 2017

Problem Solving for 1DCollisions, 1

q Coordinates: Set up a coordinate axis and define the velocities with respect to this axisn It is convenient to make

your axis coincide with one of the initial velocities

q Diagram: In your sketch, draw all the velocity vectors and label the velocities and the masses February 13, 2017

Problem Solving for 1DCollisions, 2

q Conservation of Momentum: Write a general expression for the total momentum of the system before and afterthe collisionn Equate the two total

momentum expressionsn Fill in the known values

ffii vmvmvmvm 22112211 +=+ February 13, 2017

Problem Solving for 1DCollisions, 3

q Conservation of Energy:If the collision is elastic, write a second equation for conservation of KE, or the alternative equationn This only applies to perfectly

elastic collisions

q Solve: the resulting equations simultaneously

iffi vvvv 2211 +=+ February 13, 2017

One-Dimension vs Two-Dimension February 13, 2017

Two-Dimensional Collisionsq For a general collision of two objects in two-

dimensional space, the conservation of momentum principle implies that the total momentum of the system in each direction is conserved

fyfyiyiy

fxfxixix

vmvmvmvmvmvmvmvm

22112211

22112211

+=+

+=+ February 13, 2017

Two-Dimensional Collisionsq The momentum is conserved in all directionsq Use subscripts for

n Identifying the objectn Indicating initial or final valuesn The velocity components

q If the collision is elastic, use conservation of kinetic energy as a second equationn Remember, the simpler equation can only be used

for one-dimensional situations

fyfyiyiy

fxfxixix

vmvmvmvmvmvmvmvm

22112211

22112211

+=+

+=+

iffi vvvv 2211 +=+ February 13, 2017

Glancing Collisions

q The “after” velocities have x and y componentsq Momentum is conserved in the x direction and in the

y directionq Apply conservation of momentum separately to each

direction

fyfyiyiy

fxfxixix

vmvmvmvmvmvmvmvm

22112211

22112211

+=+

+=+ February 13, 2017

2-D Collision, exampleq Particle 1 is moving at

velocity and particle 2 is at rest

q In the x-direction, the initial momentum is m1v1i

q In the y-direction, the initial momentum is 0

1ivr February 13, 2017

2-D Collision, example contq After the collision, the

momentum in the x-direction is m1v1f cos θ+ m2v2f cos φ

q After the collision, the momentum in the y-direction is m1v1f sin θ+ m2v2f sin φ

q If the collision is elastic, apply the kinetic energy equation

φθφθ

sinsin00

coscos0

2211

221111

ff

ffi

vmvmvmvmvm

−=+

+=+

222

211

211 2

121

21

ffi vmvmvm += February 13, 2017

Ex8: Collision at an Intersection

q A car with mass 1.5×103 kg traveling east at a speed of 25 m/s collides at an intersection with a 2.5×103 kg van traveling north at a speed of 20 m/s. Find the magnitude and direction of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision and assuming that friction between the vehicles and the road can be neglected.

??,/20,/25105.2,105.1 33

====×=×=

θfviycix

vc

vsmvsmvkgmkgm February 13, 2017

Collision at an Intersection

??,m/s20,m/s25kg105.2,kg105.1 33

====×=×=

θfviycix

vc

vvvmm

∑ ⋅×==+= m/skg1075.3 4cixcvixvcixcxi vmvmvmp

∑ +=+= θcos)( fvcvfxvcfxcxf vmmvmvmp

θcos)kg1000.4(m/skg1075.3 34fv×=⋅×

∑ ⋅×==+= m/skg1000.5 4viyvviyvciycyi vmvmvmp

∑ +=+= θsin)( fvcvfyvcfycyf vmmvmvmp

θsin)kg1000.4(m/skg1000.5 34fv×=⋅× February 13, 2017

Collision at an Intersection

??,/20,/25105.2,105.1 33

====×=×=

θfviycix

vc

vsmvsmvkgmkgm

33.1/1075.3/1000.5tan 4

4

=⋅×⋅×=

smkgsmkgθ

!1.53)33.1(tan 1 == −θ

m/s6.151.53sin)kg1000.4(

m/skg1000.53

4

⋅×= !fv

θcos)kg1000.4(m/skg1075.3 34fv×=⋅×

θsin)kg1000.4(m/skg1000.5 34fv×=⋅× February 13, 2017

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