L9 Momentum PHYS101 - UNIVERSE OF ALI...
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Physics 101 Lecture 9 Linear Momentum and Collisions Dr. Ali ÖVGÜN EMU Physics Department www.aovgun.com
Transcript of L9 Momentum PHYS101 - UNIVERSE OF ALI...
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
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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
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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
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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 = − − =
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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
⋅×=⋅×−−⋅×=
−=−=
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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=
!!!!!
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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.
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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!!!! +=+
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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!!!! +=+
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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 −=−=−=
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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
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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
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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
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More about Perfectly Inelastic Collisions
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 ++=
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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.
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(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
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(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
More About Elastic Collisions
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
+=+
+=+
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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 +=+
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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×=⋅×
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