Download Chapter 7: Momentum

Chapter 7
Linear Momentum
Linear Momentum
• Definition of Momentum
• Impulse
• Conservation of Momentum
• Center of Mass
• Motion of the Center of Mass
• Collisions (1d, 2d; elastic, inelastic)
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2
Momentum From Newton’s Laws
Consider two interacting bodies with m2 > m1:
F21
m1
F12
m2
Fnet
t
If we know the net force on each body then v  at 
m
The velocity change for each mass will be
different if the masses are different.
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Momentum
Rewrite the previous result for each body as:
m1v1  F21t
m2v2  F12t  F21t
From the 3rd Law
Combine the two results: m1v1  m2 v 2
m1 v1 f  v1i   m2 v 2 f  v 2i 
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Define Momentum
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.
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Momentum
From slide (4):
m1v1  m2 v 2
p1  p 2
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
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Impulse
Definition of impulse:
p  Ft
We use impulses to change the
momentum of the system.
One can also define an average impulse
when the force is variable.
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Impulse Example
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
Take m1 < m2
m1 or m2
(a) Which mass has the greater momentum change?
p  Ft
MFMcGraw-PHY 1401
Since the same force is applied to
each mass for the same interval, p
is the same for both masses.
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Example continued:
(b) Which mass has the greatest velocity change?
p
v 
m
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?
v
a
t
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Since av the mass with the greater
velocity change will have the greatest
acceleration (mass 1).
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Change in Momentum Example
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 = mv.
v  at
v   gt  33.3 m/sec
p  mv  100 kg m/s (downward)
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Change in Momentum Example
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.
p  Fav t
p mv
Fav 

t
t

50.0 kg 3.0 m/s 
Fav 
 7.5 N
20.0 s
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The force will be
in the direction
of motion.
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Impulse


ΔP = Pf - Pi = M V f - Vi = FAvg Δt
Vi
Vi = 20 m s ; V f = -15 m s ; M = 3kg
ΔP = M V f - Vi  = 3(-15 - 20)
ΔP = 3  -35  = -105 kg m s
Vf
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
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Impulse and Momentum
The force in impulse problems changes rapidly so we usually deal with the
average force
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Work and Impulse Comparison
Work changes ET
Work = 0
ET is conserved
Work = Force x Distance
Impulse changes
P
Impulse = 0
P is conserved
Impulse = Force x Time
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Conservation of Momentum
v1i
m1>m2
v2i
m1
m2
A short time later the
masses collide.
m1
m2
What happens?
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During the interaction:
N1
N2
y
F21
F12
w1
F
F
y
 N1  w1  0
x
  F21  m1a1
x
w2
F
F
y
 N 2  w2  0
x
 F12  m2 a2
There is no net external
force on either mass.
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The forces F12 and F21 are internal forces. This means that:
p1  p 2
p1 f  p1i  p 2 f  p 2i 
p1i  p 2i  p1 f  p 2 f
In other words, pi = pf. That is, momentum is conserved.
This statement is valid during the interaction (collision)
only.
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Momentum Examples
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Generic Zero Initial Momentum Problem
p
i
= pP + p A = 0
 p'
i
= p' P + p' A = 0
These problems occur whenever all the objects in the system are
initially at rest and then separate.
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Changing the Mass While Conserving the
Momentum
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Changing the Mass While Conserving the
Momentum
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Changing the Mass While Conserving the
Momentum
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Astronaut Tossing
How Long Will the Game Last?
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How Long Will the Game Last?
-V0
V0
V0
3V
 V0  0
2
2
V0
2
-V0
V0
V
 V0   0
2
2
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V0
3V
 V0  0
2
2
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The Ballistic Pendulum
Conservation of
Energy
Conservation of Momentum
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The High Velocity Ballistic Pendulum
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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.
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Relative Speed Relationship
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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
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Rifle Example
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.
pi  p f
0  mb vb  mr vr
 0.01 kg 
mb
820 m/s  1.82 m/s
vr  
vb  
mr
 4.5 kg 
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Center of Mass
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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.
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Center of Mass
The center of mass (of a two body system) is found from:
xcm
m1 x1  m2 x2

m1  m2
This is a “weighted” average of the positions of the particles
that compose a body. (A larger mass is more important.)
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Center of Mass
mr

m
i i
In 3-dimensions write
rcm
where mtotal   mi
i
i
i
i
The components of rcm are:
m x

m
xcm
i
i
i
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m y

m
i
i i
ycm
i
i
i
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m z

m
i i
i
zcm
i
i
i
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Center of Mass Example
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)?
y
xcm
ma xa  mb xb
mb xb


ma  mb
ma  mb
ycm
ma ya  mb yb
mb yb


ma  mb
ma  mb
x
A
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x
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Center of Mass Example
xcm 
10 g xb
10 g  30g
xb  8 cm
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 2 cm
ycm 
10 g  yb
30 g  10 g
yb  20 cm
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 5 cm
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Find the Center of Mass
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?
y
2
1
x
3
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Example continued:
xcm
m1 x1  m2 x2  m3 x3

m1  m2  m3

4 kg 4 m   6 kg 2 m   3 kg  1 m 

4  6  3 kg
 1.92 m
ycm
m1 y1  m2 y2  m3 y3

m1  m2  m3

4 kg 0 m   6 kg 4 m   3 kg  2 m 

4  6  3 kg
 1.38 m
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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.
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Center of Mass Velocity
For the center of mass:
rcm  v cm
 m r
t 
m
i
i
i
i
i
Solving for the velocity of the center of mass:
ri
t
 i

 mi
 mi
v cm
m v
m
i
i
i
Or in component form:
i
i
i
m v

m
i i,x
v cm, x
i
i
i
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m v

m
i i, y
v cm, y
i
i
i
40
Center of Mass Reference System
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Center of Mass Treatment
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Center of Mass Treatment
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Center of Mass Treatment
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Center of Mass Treatment
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Center of Mass Reference System
Center of Mass System
m1
m1r1 + m2 r2
R=
m1 + m2
r = r2 - r1
r
r1
R
Laboratory System
m2
r2
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m2
r1 = R r
m1 + m2
m1
r2 = R +
r
m1 + m2
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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.
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Collisions in One Dimension
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.
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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.
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Elastic Collision - KE Transfer
Elastic Velocity Ratios
m2
-1
u1
m1
=m2
v0
1+
m1
2.00
1.50
U1
U2
Velocity Ratio
1.00
0.50
0.00
0.0
5.0
10.0
15.0
20.0
25.0
-0.50
-1.00
u2
2
=
v0 1+ m2
m1
-1.50
Mass Ratio
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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.
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Elastic Collision - KE Transfer
Fractional KE Loss
2.0
u2
2
=
v0 1+ m2
m1
1.8
Fractional KE Loss
1.6
U2 Velocity Ratio
Fractional KE Loss
1.4
1.2
1.0
0.8
0.6
f =
0.4
0.2
0.0
0
5
10
15
20
25
m2
4
m1

m2 
 1+

m
1 

2
Mass Ratio
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
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Inelastic Collision - Equal Masses
v1
v2 = 0
Momentum
m1v1 =  m1 + m2  u; u =
Kinetic Energy
m1
m2
KEi = 21 m1v12 ; KE f =
Before
u
m1 + m2
After
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1
2
m1
v1
m1 + m2
 m1 + m2  u 2
2
 m1  2
1
KE f = 2  m1 + m2  
 v1
 m1 + m2 
m1
m1
2
1
KE f = 2
m1v1 =
KEi
m1 + m2
m1 + m2
KE f
m1
=
KEi
m1 + m2
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Inelastic Collisions in One Dimension
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?
pi  p f
p1i  p2i  p1 f  p2 f
m1v1  0  m1v  m2 v  m1  m2 v
 m1 
v1  0.2 m/s
v  
 m1  m2 
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Inelastic Collisions in One Dimension
(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.
pi  p f
p1i  p2i  p1 f  p2 f
m1v1i  m2 v2i  0  0
 m1 
v2i   v1i  0.25 m/s
 m2 
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One Dimension Projectile
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?
pi  p f
p1i  p2i  p1 f  p2 f
m1v1i  0  m1v1 f  m2 v2 f
v2 f
m1
v1i  v1 f 

m2

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1.0 kg
10 m/s   5.0 m/s   3.0 m/s
5.0 kg
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Collision in Two Dimensions
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?
Final
Initial
A
Angle is 90o
A
vAi
B
B
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Collision in Two Dimensions
y momentum:
x momentum:
pix  p fx
piy  p fy
p1ix  p2ix  p1 fx  p2 fx
p1iy  p2iy  p1 fy  p2 fy
m1v1ix  0  m1v1 fx  m2 v2 fx
Solve for v2fx:
v2 fx 
m1v1ix  m1v1 fx
m2
 v1ix  v1 fx
 5.00 m/s
The mag. of v2 is
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0  0  m1v1 fy  m2 v2 fy
Solve for v2fy:
v2 fy 
 m1v1 fy
m2
 v1 fy
 2.00 m/s
v2 f  v 2 2 fy  v 2 2 fx  5.40 m/s
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Collisions in Two and Three
Dimensions
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Totally Inelastic Collision
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Totally Inelastic Collision
If the objects stick together
after the collision, then the
collision is INELASTIC.
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Just because the objects
DON’T stick together after
the collision, doesn’t mean
the collision is ELASTIC.
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General Glancing Collision
Initially only one object is moving
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General Glancing Collision
Initially only one object is moving
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Another 3-Vector Problem
Momentum problems where one of the objects is initially at rest yield 3vector problems that can be solved using triangles, instead of simultaneous
equations.
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Combining Conservation of Energy,
Momentum, and KE Transfer in
Equal Mass Inelastic Collisions
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Δh1
Δh2
Conservation of Energy
Conservation of Momentum
Conservation of Energy
mg h1 = 21 mv 2
mv = (m + m)u
1
2
v = 2g h1
u = 21 v
u 2 = 2gΔh2
(2m)u 2 = 2mgΔh2
u2
v2
2gΔh1 Δh1
Δh2 =
=
=
=
2g 8g
8g
4
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Summary
• Definition of Momentum
• Impulse
• Center of Mass
• Conservation of Momentum
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Extra Slides
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Conservation of Momentum
Two 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
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Conservation of Kinetic Energy?
Case 1
Case 2
KE After = KE Before
KE After = ½ KE Before
(Conserved)
(Not Conserved)
KE = Kinetic Energy = ½mvo2
MFMcGraw-PHY 1401
Chap07b- Linear Momentum: Revised
6/28/2010
70
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
71
50% of the KE is Missing
Each rectangle represents KE = ¼ KE Before
MFMcGraw-PHY 1401
Chap07b- Linear Momentum: Revised
6/28/2010
72
Turn Case 1 into Case 2
Each 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
Use one half of the 50% taken
to speed up object 1. Now it
object 1 has none
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.
MFMcGraw-PHY 1401
Chap07b- Linear Momentum: Revised
6/28/2010
73