lesson 9: impulse, momentum, center of mass, collisions ...lesson 9: impulse, momentum, center of...
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Lesson 9: Impulse, Momentum, Center of Mass, Collisions (Sections 7.1-7.7)
Lesson 9, page 1
Chapter 7 Linear Momentum
From page 234
In Chapter 4 we learned how to determine the acceleration of an object by finding the net
force acting on it and applying Newton’s second law of motion. If the force happens to
be constant, then the resulting constant acceleration enables us to calculate changes in
velocity and position. Calculating velocity and position changes when the forces are not
constant is much more difficult. In many cases, the forces cannot even be easily
determined. Conservation of energy is one tool that enables us to draw conclusions about
motion without knowing all the details of the forces acting….
In this chapter we develop another conservation law. Conservation laws are
powerful tools. If a quantity is conserved, then no matter how complicated the situation,
we can set the value of the conserved quantity at one time equal to its value at a later
time….
The new conserved quantity, momentum, is a vector quantity, in contrast to
energy, which is a scalar. When momentum is conserved, both the magnitude and
direction of the momentum must be constant. Equivalently, the x- and y-components of
momentum are constant. When we find the total momentum of more than one object, we
must add the momentum vectors according to the procedure by which vectors are always
added.
Definition of linear momentum
An object with mass m and moving with velocity v
has linear momentum p
vp
m
Impulse-Momentum Theorem
Starting with Newton’s second law
aF
m
and using the definition of acceleration
t
va
we have
tm
vF
If the mass is constant, we can pull it inside the operator
t
m
)( vF
Lesson 9: Impulse, Momentum, Center of Mass, Collisions (Sections 7.1-7.7)
Lesson 9, page 2
t
pF
ttotal FJp
This states that the change in linear momentum is caused by the impulse. The quantity
tFJ
is called the impulse. For situations where the force is not constant, we use the average force,
tavFJ
Restatement of Newton’s second law
There is a more general form of Newton’s second law
tt
pF
0lim
The net force is the rate of change of momentum.
Many Particles
If many particles m1, m2, etc. are moving with velocities ,, 21 vv
etc., the total linear momentum
of the system is the vector sum of the individual momenta,
i
ii
i
i m vpp
Linear momentum is a vector quantity. We have to use our familiar rules for vector addition
when dealing
with
momentum.
The
applications of
the impulse-
momentum
theorem are
unlimited. In
an automobile
we have
Lesson 9: Impulse, Momentum, Center of Mass, Collisions (Sections 7.1-7.7)
Lesson 9, page 3
crumple zones, air bags, and bumpers. What do seat
belts do?
Here is another example:
The idea is to lengthen the time t during which the
force acts, so that the force is diminished while
changing the momentum a defined amount.
Problem 17. An automobile traveling at a speed of
30.0 m/s applies its brakes and comes to a stop in 5.0
s. If the automobile has a mass of 1.0×103 kg, what
is the average horizontal force exerted on it during
braking. Assume the road is level.
Solution: A FBD for the situation would be
We would use the impulse-momentum theorem
t Fp
To use a vector equation we need to take components
tFpxx
The change in momentum is
m/skg3000)m/s0.3)(kg1000(0 ixfxixfxx mvmvppp
N600m/skg600s5
m/skg3000 2
,
,
t
pF
tF
tFp
xxav
xav
xx
The average braking force acts to the left and is 600 N. Additionally, the work done on the car to
stop it is
mg
Lesson 9: Impulse, Momentum, Center of Mass, Collisions (Sections 7.1-7.7)
Lesson 9, page 4
J1050.4)m/s30)(kg1000(00)( 52
2
1 ifnc KKUKW
and the corresponding power is
W000,90s0.5
J1050.4 5
t
WP
Conservation of Linear Momentum
Consider the collision between two pucks. When they collide, they exert forces on each other,
12F
and 21F
By Newton’s third law,
2112 FF
The total force acting on both pucks is then
02112 FFF
(Recall that since ay = 0 for both pucks, the y-component of the net force is zero.) By the
impulse-momentum theorem
0
tFp
Which leads to
fi pp
Lesson 9: Impulse, Momentum, Center of Mass, Collisions (Sections 7.1-7.7)
Lesson 9, page 5
In a system composed of more than two objects, interactions between objects inside the system
do not change the total momentum of the system – they just transfer some momentum from one
part of the system to another. Only external interactions can change the total momentum of the
system.
The total momentum of a system is the vector sum of the momenta of each object in the
system
External interactions can change the total momentum of a system.
Internal interactions do not change the total momentum of a system.
The Law of Conservation of Linear Momentum
If the net external force acting on a system is zero, then the momentum of the system is
conserved.
fiext ppF
,0If
Linear momentum is always conserved for an isolated system. Of course, we deal with
components
fyiyfxixpppp and
I like to say that momentum is conserved in collisions and explosions.
Note: The total momentum of the system is conserved. The momentum of an individual particle
can change.
Problem 23. A rifle has a mass of 4.5 kg and it fires a bullet of mass 10.0 g at a muzzle speed of
820 m/s. What is the recoil speed of the rifle as the bullet leaves the gun barrel?
Solution: Draw a sketch of the initial and final situations.
Since only internal forces act, linear momentum is conserved.
fi pp
The motion is in the x-direction,
fxix pp
Initially, everything is at rest and pix = 0. The final momentum is the vector sum of the momenta
of the bullet and the rifle. From the diagram,
Initial Final
Lesson 9: Impulse, Momentum, Center of Mass, Collisions (Sections 7.1-7.7)
Lesson 9, page 6
rrbbfx vmvmp
Using conservation of momentum
m/s82.1kg5.4
)m/s820)(kg010.0(
0
r
bbr
rrbb
fxix
m
vmv
vmvm
pp
The heavier the rifle, the smaller the recoil speed.
Center of Mass
From page 245:
We have seen that the momentum of an isolated system is conserved even though parts of
the system may interact with other parts; internal interactions transfer momentum
between parts of the system but do not change the total momentum of the system. We
can define a point called the center of mass (CM) that serves as an average location of the
system….
What if a system is not isolated, but has external interactions? Again imagine all
of the mass of the system concentrated into a single point particle located at the CM. The
motion of this fictitious point particle is determined by Newton’s second law, where the
net force is the sum of all the external forces acting on any part of the system. In the case
of a complex system composed of many
parts interacting with each other, the
motion of the CM is considerably
simpler than the motion of an arbitrary
particle of the system.
For the two particles pictured to the right, the
CM is
M
xmxm
mm
xmxmxCM
2211
21
2211
Notice that the CM is closer to the more
massive object.
For many particles the definition is generalized
to
M
m ii
CM
rr
Lesson 9: Impulse, Momentum, Center of Mass, Collisions (Sections 7.1-7.7)
Lesson 9, page 7
The more useful component form is
M
xmx
ii
CM
M
ymy
ii
CM
M
zmz
ii
CM
Problem 38. Find the x-coordinate of the CM of the composite object shown in the figure. The
sphere, cylinder, and rectangular solid all have uniform composition. Their masses and
dimensions are: sphere: 200 g, diameter = 10 cm; cylinder: 450 g, length = 17 cm, radius = 5.0
cm; rectangular solid: 325 g, length in x-direction = 16 cm, height = 10 cm, depth = 12 cm.
Solution: Because the different pieces have uniform composition, the center of mass of each
piece is located at the geometrical center of that piece. (See the note at the top of page 240.) For
the sphere, ms = 200 g, xs = 5 cm. For the cylinder, mc = 450 g, xc = 10 cm + 8.5 cm = 18.5 cm.
For the rectangular solid, mr = 325 g, xr = 10 cm + 17 cm + 8 cm = 35 cm. The center of mass is
cm2.21
)g325()g450()g200(
)cm35)(g325()cm5.18)(g450()cm5)(g200(
rcs
rrccss
ii
CM
mmm
xmxmxm
M
xmx
Motion of the Center of Mass
How is the velocity of the CM related to the velocities of the various parts of the system of
particles?
During a short time interval t, each particle changes position by ir
. The CM changes
M
m ii
CM
rr
Divide each side by the time interval t
M
tm
t
ii
CM
r
r
In general, the velocity is defined as
Lesson 9: Impulse, Momentum, Center of Mass, Collisions (Sections 7.1-7.7)
Lesson 9, page 8
tt
rv
0lim
Using the definition of velocity
iiCM
ii
CM
mM
M
m
vv
vv
The right hand side is the total linear momentum of the system of particles. We have the very
useful relation
CMtotal Mvp
For a complicated system consisting of many particles moving in different directions, the total
linear momentum of the system can be found from the total mass of the system and the velocity
of the center of mass.
(Page 247) … we have showed that, for an isolated system, the total linear momentum of
the system is conserved. In such a system, the equation above implies that the CM must
move with constant velocity regardless of the motions of the individual particles. On the
other hand, what if the system is not isolated? If a net force acts on a system, the CM
does not move with constant velocity. Instead, it moves as if all the mass were
concentrated into a fictitious point particle with all the external forces acting on that
point. The motion of the CM obeys the following statement of Newton’s second law:
CMext MaF
A complicated system is reduced to treating the system as a point particle located at its center of
mass reacting only to external forces!
Collisions in One Dimension
In general, conservation of momentum is not enough to predict what will happen after a
collision. Here is a collision between two objects with the same mass. Many other outcomes are
possible besides the two given,
Lesson 9: Impulse, Momentum, Center of Mass, Collisions (Sections 7.1-7.7)
Lesson 9, page 9
Some vocabulary
A collision in which the total kinetic energy is the same before and after is called elastic.
See (a) above.
When the final kinetic energy is less than the initial kinetic energy, the collision is said to
be inelastic. Collisions between two macroscopic objects (for example, billiard balls) are
generally inelastic to some degree, but sometimes the change in kinetic energy is so small
that we treat the collision as elastic.
When a collision results in two objects sticking together, the collision is perfectly
inelastic. See (b) above. The decrease of kinetic energy is a perfectly inelastic collision
is as large as possible (consistent with the conservation of momentum).
Problem-Solving Strategy for Collisions Involving Two Objects (pages 251-2)
1. Draw before and after diagrams of the collision.
2. Collect and organize information on the masses and velocities of the two objects before
and after the collision. Express the velocities in component form (with correct algebraic
signs).
3. Set the sum of the momenta of the two before and after the collision equal to the sum of
the momenta after the collision. Write one equation for each direction:
fxfxixix vmvmvmvm 22112211
fyfyiyiy vmvmvmvm 22112211
4. If the collision is known to be perfectly inelastic, set the final velocities equal:
fyfyfxfx vvvv 2121 and
5. If the collision is known to be perfectly elastic, then set the final kinetic energy equal to
the initial kinetic energy:
2
22212
11212
22212
1121
ffii vmvmvmvm
or, for a one-dimensional collision, set the relative speeds equal:
)( 1212 fxfxixix vvvv
Lesson 9: Impulse, Momentum, Center of Mass, Collisions (Sections 7.1-7.7)
Lesson 9, page 10
6. Solve for the unknown quantities.
Example: A one-dimensional elastic collision.
Conservation of momentum gives
ffii vmvmvmvm 22112211
Elastic collision means that kinetic energy is conserved
2
222
12
112
12
222
12
112
1ffii vmvmvmvm
Cancelling the ½ and grouping like masses on the same side of the equation,
)()(2
2
2
22
2
1
2
11 iffi vvmvvm
Doing the same with the linear momentum equation
)()( 222111 iffi vvmvvm
Divide the two equations
iffi
if
if
fi
fi
vvvv
vvm
vvm
vvm
vvm
2211
222
2
2
2
22
111
2
1
2
11
)(
)(
)(
)(
This can be rewritten as
ffii vvvv 1221
The equation states that m1 approaches m2 with the same speed as m2 moves away from m1 after
the collision. Important result. Solve the above for v2f
iiff vvvv 2112
and substitute into the momentum equation
v1i v2i m1 m2
v1f v2f m1 m2
initial final
Lesson 9: Impulse, Momentum, Center of Mass, Collisions (Sections 7.1-7.7)
Lesson 9, page 11
iif
fii
iiff
iiff
ffii
vmm
mv
mm
mmv
vmmvmvmm
vmvmvmvm
vvvmvm
vmvmvmvm
2
21
21
21
211
12122121
22121211
211211
22112211
2
)(2)(
)(
Using very similar reasoning, or swapping the 1 and 2 subscripts, we can find an expression for
the final speed of m2
iif vmm
mmv
mm
mv 2
21
121
21
12
2
Be careful when you use these equations. A mass moving to the left would have a negative
speed.
If m2 is initially at rest, v2i = 0 and we have
if
if
vmm
mv
vmm
mmv
1
21
12
1
21
211
2
After the collision, m2 moves to the right.
If m1 > m2, m1 continues to move to the right.
If m1 = m2, m1 stops and m2 to move to the right with speed v1i.
If m1 < m2, m1 bounces back to the left.
Example: The masses are m1 = 2 kg and m2 = 3 kg and they collide elastically in one dimension.
Before the collision, m1 is moving at 4 m/s to the right and m2 is moving at 1 m/s to the left.
What are the velocities of m1 and m2 after the collision?
Solution: Use the equations derived above,
m/s2
)m/s1(kg3kg2
kg)3(2)m/s4(
kg3kg2
kg3kg2
22
21
21
21
211
iif v
mm
mv
mm
mmv
Lesson 9: Impulse, Momentum, Center of Mass, Collisions (Sections 7.1-7.7)
Lesson 9, page 12
m/s3
)m/s1(kg3kg2
kg2kg3)m/s4(
kg3kg2
)kg2(2
22
21
121
21
12
iif v
mm
mmv
mm
mv
To review relative velocity, the velocity of 2 relative to 1 is
1221 GG vvv
Taking components,
m/s5m/s)4(m/s11221 xGGxx vvv
As seen from m1, m2 is moving to the left at 5 m/s. After the collision,
m/s5m/s)2(m/s31221 xGGxx vvv
The masses approach at 5 m/s and separate at 5 m/s.
1 m/s 4 m/s m1 m2
3 m/s 2 m/s m1 m2