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Transcript of Chapter 7 Linear Momentum. MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/2010 2 Linear...
Chapter 7
Linear Momentum
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
2
Linear Momentum
• Definition of Momentum
• Impulse
• Conservation of Momentum
• Center of Mass
• Motion of the Center of Mass
• Collisions (1d, 2d; elastic, inelastic)
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
3
Momentum From Newton’s Laws
Consider two interacting bodies with m2 > m1:
If we know the net force on each body then tm
t netFav
The velocity change for each mass will be different if the masses are different.
F21 F12m1 m2
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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tm 2111 Fv
Rewrite the previous result for each body as:
ttm 211222 FFv
Combine the two results:
ifif mm
mm
222111
2211
vvvv
vv
Momentum
From the 3rd Law
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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The quantity (mv) is called momentum (p).
p = mv and is a vector.
The unit of momentum is kg m/s; there is no derived unit for momentum.
Define Momentum
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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From slide (4):
21
2211
pp
vv
mm
The change in momentum of the two bodies is “equal and opposite”. Total momentum is conserved during the interaction; the momentum lost by one body is gained by the other.
Δp1 +Δp2 = 0
Momentum
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Impulse
Definition of impulse:
t Fp
One can also define an average impulse when the force is variable.
We use impulses to change the momentum of the system.
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Example (text conceptual question 7.2): A force of 30 N is applied for 5 sec to each of two bodies of different masses.
30 N
m1 or m2
Take m1 < m2
(a) Which mass has the greater momentum change?
t FpSince the same force is applied to each mass for the same interval, p is the same for both masses.
Impulse Example
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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(b) Which mass has the greatest velocity change?
Example continued:
m
pv
Since both masses have the same p, the smaller mass (mass 1) will have the larger change in velocity.
(c) Which mass has the greatest acceleration?
t
v
aSince av the mass with the greater velocity change will have the greatest acceleration (mass 1).
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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An object of mass 3.0 kg is allowed to fall from rest under the force of gravity for 3.4 seconds. Ignore air resistance.
What is the change in momentum?
Want p = mv.
(downward) m/s kg 100
m/sec 3.33
vp
v
av
m
tg
t
Change in Momentum Example
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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What average force is necessary to bring a 50.0-kg sled from rest to 3.0 m/s in a period of 20.0 seconds? Assume frictionless ice.
N 5.7
s 0.20
m/s 0.3kg 0.50av
av
av
F
t
m
t
t
vpF
Fp
The force will be in the direction of motion.
Change in Momentum Example
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Impulse
iV
fV
Since we are applying the change of momentum over a time interval consisting of the instant before the impact to the instant after the impact, we ignore the effect of gravity over that time period
f i f i AvgΔP = P - P = M V - V = F Δt
i f
f i
V = 20 m s ; V = -15 m s ; M = 3kg
ΔP = M V - V = 3(-15 - 20)
ΔP = 3 -35 = -105 kg m s
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Impulse and Momentum
The force in impulse problems changes rapidly so we usually deal with the average force
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Work and Impulse Comparison
Work changes ET
Work = 0 ET is conserved
Work = Force x Distance
Impulse changes
Impulse = 0 is conserved
Impulse = Force x Time
P
P
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Conservation of Momentum
v1i
v2i
m1 m2
m1>m2
m1 m2
A short time later the masses collide.
What happens?
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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During the interaction:
N1
w1
F21x
y
1121
11 0
amFF
wNF
x
y
2212
22 0
amFF
wNF
x
y
There is no net external force on either mass.
N2
w2
F12
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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The forces F12 and F21 are internal forces. This means that:
ffii
ifif
2121
2211
21
pppp
pppp
pp
In other words, pi = pf. That is, momentum is conserved. This statement is valid during the interaction (collision) only.
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Momentum Examples
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Generic Zero Initial Momentum Problem
These problems occur whenever all the objects in the system are initially at rest and then separate.
P Ai
p = p + p = 0 P Ai
p' = p' + p' = 0
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Changing the Mass While Conserving the Momentum
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Changing the Mass While Conserving the Momentum
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Changing the Mass While Conserving the Momentum
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Astronaut TossingHow Long Will the Game Last?
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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How Long Will the Game Last?
0 00
V 3VV
2 2
0-V 0V
0V
2
0-V
0 00
V VV
2 2
0 00
V 3VV
2 2
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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The Ballistic Pendulum
Conservation of Momentum
Conservation of Energy
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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The High Velocity Ballistic Pendulum
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Relative Speed Relationship
• The relative speed relationship applies to elastic collisions.
• In an elastic collision the total kinetic energy is also conserved.
• If a second equation is needed to solve a momentum problem, the relative speed relationship is easier to deal with than the KE equation.
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Relative Speed Relationship
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Relative Speed in Problem Solving
Need a second equation when solving for two variables in an elastic collision problem?
The relative speed relationship is preferred over the conservation ofkinetic energy relationship
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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A rifle has a mass of 4.5 kg and it fires a bullet of 10.0 grams at a muzzle speed of 820 m/s.
What is the recoil speed of the rifle as the bullet leaves the barrel?
As long as the rifle is horizontal, there will be no net external force acting on the rifle-bullet system and momentum will be conserved.
m/s 1.82m/s 820kg 4.5
kg 01.0
0
br
br
rrbb
fi
vm
mv
vmvm
pp
Rifle Example
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Center of Mass
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Center of Mass
The center of mass (CM) is the point representing the mean (average) position of the matter in a body.
This point need not be located within the body.
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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The center of mass (of a two body system) is found from:
21
2211
mm
xmxmxcm
This is a “weighted” average of the positions of the particles that compose a body. (A larger mass is more important.)
Center of Mass
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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In 3-dimensions write
ii
iii
cm m
m rr where
iimmtotal
The components of rcm are:
ii
iii
cm m
xmx
ii
iii
cm m
ymy
ii
iii
cm m
zmz
Center of Mass
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Particle A is at the origin and has a mass of 30.0 grams. Particle B has a mass of 10.0 grams. Where must particle B be located so that the center of mass (marked with a red x) is located at the point (2.0 cm, 5.0 cm)?
x
y
A
x
ba
bb
ba
bbaacm mm
xm
mm
xmxmx
ba
bb
ba
bbaacm mm
ym
mm
ymymy
Center of Mass Example
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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cm 8
cm 2g30g 10
g 10
b
bcm
x
xx
cm 20
cm 5g 10g 30
g 10
b
bcm
y
yy
Center of Mass Example
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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The positions of three particles are (4.0 m, 0.0 m), (2.0 m, 4.0 m), and (1.0 m, 2.0 m). The masses are 4.0 kg, 6.0 kg, and 3.0 kg respectively.
What is the location of the center of mass?
x
y
3
2
1
Find the Center of Mass
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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m 92.1
kg 364
m 1kg 3m 2kg 6m 4kg 4321
332211cm
mmm
xmxmxmx
m 38.1
kg 364
m 2kg 3m 4kg 6m 0kg 4321
332211cm
mmm
ymymymy
Example continued:
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Motion of the Center of Mass
For an extended body, it can be shown that p = mvcm.
From this it follows that Fext = macm.
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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For the center of mass:
ii
iii
m
mt
rvr cmcm
Solving for the velocity of the center of mass:
ii
iii
ii
i
ii
m
m
mt
m vr
vcm
Or in component form:
ii
ixii
m
vm ,
xcm,v
ii
iyii
m
vm ,
ycm,v
Center of Mass Velocity
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Center of Mass Reference System
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Center of Mass Treatment
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Center of Mass Treatment
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Center of Mass Treatment
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
45
Center of Mass Treatment
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Center of Mass Reference System
r
1r
2r
R
1m
2m
1 1 2 2
1 2
2 1
m r + m rR =
m + m
r = r - r
21
1 2
12
1 2
mr = R - r
m + m
mr = R + r
m + m
Center of Mass System
Laboratory System
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Collisions in One Dimension
When there are no external forces present, the momentum of a system will remain unchanged. (pi = pf)
If the kinetic energy before and after an interaction is the same, the “collision” is said to be perfectly elastic. If the kinetic energy changes, the collision is inelastic.
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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If, after a collision, the bodies remain stuck together, the loss of kinetic energy is a maximum, but not necessarily a 100% loss of kinetic energy. This type of collision is called perfectly inelastic.
Collisions in One Dimension
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Elastic Collision - Unequal Masses
In the later slides, the initially moving mass, mn , will called be m1 while the initially stationary mass, mc , will be m2.
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Elastic Velocity Ratios
-1.50
-1.00
-0.50
0.00
0.50
1.00
1.50
2.00
0.0 5.0 10.0 15.0 20.0 25.0
Mass Ratio
Ve
loc
ity
Ra
tio
U1
U2
Elastic Collision - KE Transfer
2
1 1
20
1
m- 1
u m= -
mv 1+m
2
20
1
u 2=
mv 1+m
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Elastic Collision - Fractional KE “Loss”
For equal masses all of the KE is transferred from m1 to m2
The curve is like an energy transfer function from m1 to m2. For small m2 the mass m1 just rolls on through. For large m2 the smaller m1 just bounces off without transferring much energy. Efficient energy transfer requires equal masses.
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Fractional KE Loss
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
0 5 10 15 20 25
Mass Ratio
Fra
cti
on
al K
E L
os
s
Fractional KE Loss
U2 Velocity Ratio
Elastic Collision - KE Transfer
For m2 > m1 the velocity of the 2nd mass gets a smaller velocity and even the larger mass can’t overcome the shrinking v2 factor in KE2
2
12
2
1
m4
mf =
m1+
m
2
20
1
u 2=
mv 1+m
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Inelastic Collision - Equal Masses
v1 v2 = 0
u
m1 m2
m1 + m2
11 1 1 2 1
1 2
2 21 1i 1 1 f 1 22 2
2
211f 1 2 12
1 2
21 11f 1 1 i2
1 2 1 2
f 1
i 1 2
mm v = m + m u; u = v
m + m
KE = m v ; KE = m + m u
mKE = m + m v
m + m
m mKE = m v = KE
m + m m + m
KE m=
KE m + m
Before
After
Momentum
Kinetic Energy
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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In a railroad freight yard, an empty freight car of mass m rolls along a straight level track at 1.0 m/s and collides with an initially stationary, fully loaded, boxcar of mass 4.0m. The two cars couple together upon collision.
(a) What is the speed of the two cars after the collision?
m/s 2.0
0
121
1
212111
2121
vmm
mv
vmmvmvmvm
pppp
pp
ffii
fi
Inelastic Collisions in One Dimension
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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(b) Suppose instead that both cars are at rest after the collision. With what speed was the loaded boxcar moving before the collision if the empty one had v1i = 1.0 m/s.
m/s 25.0
00
12
12
2211
2121
ii
ii
ffii
fi
vm
mv
vmvm
pppp
pp
Inelastic Collisions in One Dimension
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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A projectile of 1.0 kg mass approaches a stationary body of 5.0 kg mass at 10.0 m/s and, after colliding, rebounds in the reverse direction along the same line with a speed of 5.0 m/s.
What is the speed of the 5.0 kg mass after the collision?
m/s0.3m/s 5.0m/s 10kg 0.5
kg 0.1
0
112
12
221111
2121
fif
ffi
ffii
fi
vvm
mv
vmvmvm
pppp
pp
One Dimension Projectile
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Body A of mass M has an original velocity of 6.0 m/s in the +x-direction toward a stationary body (B) of the same mass. After the collision, body A has vx = +1.0 m/s and vy = +2.0 m/s.
What is the magnitude of body B’s velocity after the collision?
A
B
vAi
Initial
B
AFinal
Collision in Two Dimensions
Angle is 90o
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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fxfxix
fxfxixix
fxix
vmvmvm
pppp
pp
221111
2121
0
fyfy
fyfyiyiy
fyiy
vmvm
pppp
pp
2211
2121
00
x momentum:
Solve for v2fx: Solve for v2fy:
m/s 00.5
11
2
11112
fxix
fxixfx
vv
m
vmvmv
y momentum:
m/s 00.2
12
112
fyfy
fy vm
vmv
m/s 40.522
22
2 fxfyf vvvThe mag. of v2 is
Collision in Two Dimensions
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Collisions in Two and Three Dimensions
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Totally Inelastic Collision
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Totally Inelastic Collision
If the objects stick together after the collision, then the collision is INELASTIC.
Just because the objects DON’T stick together after the collision, doesn’t mean the collision is ELASTIC.
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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General Glancing CollisionInitially only one object is moving
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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General Glancing CollisionInitially only one object is moving
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Another 3-Vector Problem
Momentum problems where one of the objects is initially at rest yield 3-vector problems that can be solved using triangles, instead of simultaneous equations.
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
65
Combining Conservation of Energy, Momentum, and KE Transfer in Equal Mass Inelastic Collisions
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
66
Δh1Δh2
211 2
1
mg h = mv
v = 2g h 12
mv = (m + m)u
u = v
2122
22
(2m)u = 2mgΔh
u = 2gΔh
2 21 1
2
u v 2gΔh ΔhΔh = = = =
2g 8g 8g 4
Conservation of Energy Conservation of EnergyConservation of Momentum
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Summary
• Definition of Momentum
• Impulse
• Center of Mass
• Conservation of Momentum
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Extra Slides
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
69
Conservation of MomentumTwo objects of identical mass have a collision. Initially object 1 is traveling to the right with velocity v1 = v0. Initially object 2 is at rest v2 = 0.
After the collision: Case 1: v1’ = 0 , v2’ = v0 (Elastic)
Case 2: v1’ = ½ v0 , v2’ = ½ v0 (Inelastic)In both cases momentum is conserved.
Case 1 Case 2
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Conservation of Kinetic Energy?
KE = Kinetic Energy = ½mvo
2
Case 1 Case 2
KE After = KE Before KE After = ½ KE Before
(Conserved) (Not Conserved)
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Where Does the KE Go?
Case 1 Case 2
KE After = KE Before KE After = ½ KE Before
(Conserved) (Not Conserved)
In Case 2 each object shares the Total KE equally.
Therefore each object has KE = 25% of the original KE Before
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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50% of the KE is Missing
Each rectangle represents KE = ¼ KE
Before
MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010
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Turn Case 1 into Case 2Each rectangle represents KE = ¼ KEBefore
Take away 50% of KE. Now the total system KE is correct
But object 2 has all the KE and object 1 has none
Use one half of the 50% taken to speed up object 1. Now it has 25% of the initial KE Use the other half of the 50% taken to slow down object 2. Now it has only 25% of the the initial KE
Now they share the KE equally and we see where the missing 50% was spent.