Chapter 7
Linear Momentum
MFMcGraw Chap07- Linear Momentum: Revised 3/3/2010
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Linear Momentum
• Definition of Momentum
• Impulse
• Conservation of Momentum
• Center of Mass
• Motion of the Center of Mass
• Collisions (1d, 2d; elastic, inelastic)
MFMcGraw Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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Body A has a mass of 3.0 kg and vx = +14.0 m/s. Body B has a mass of 4.0 kg and has vy = 7.0 m/s.
What is the velocity of the center of mass of the two bodies?
Consider a body made up of many different masses each with a mass mi.
The position of each mass is ri and the displacement of each mass is ri = vit.
Center of Mass Velocity
MFMcGraw Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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Applying the previous formulas to the example,
m/s 6
kg 43
m/s 0kg 4m/s 14kg 3
,,,
ba
xbbxaaxcm mm
vmvmv
m/s 4
kg 43
m/s 7kg 4m/s 0kg 3
,,,
ba
ybbyaaycm mm
vmvmv
Center of Mass Velocity
MFMcGraw Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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Summary
• Definition of Momentum
• Impulse
• Center of Mass
• Conservation of Momentum
MFMcGraw Chap07- Linear Momentum: Revised 3/3/2010
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Extra Slides
MFMcGraw Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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 Chap07- Linear Momentum: Revised 3/3/2010
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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
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50% of the KE is Missing
Each rectangle represents KE = ¼ KE
Before
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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.