Transcript Document

Momentum and
Momentum Conservation





Momentum
Impulse
Conservation
of Momentum
Collision in 1-D
Collision in 2-D
March 24, 2009
Linear Momentum
A new fundamental quantity, like force, energy
 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:


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p  mv
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The terms momentum and linear momentum will be
used interchangeably in the text
Momentum depend on an object’s mass and velocity
March 24, 2009
Linear Momentum, cont
Linear momentum is a vector quantity p  mv
 Its direction is the same as the direction of
the velocity
 The dimensions of momentum are ML/T
 The SI units of momentum are kg · m / s
 Momentum can be expressed in component
form:
px = mvx py = mvy pz = mvz

March 24, 2009
Newton’s Law and Momentum

Newton’s Second Law can be used to relate the
momentum of an object to the resultant force
acting on it

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v  (mv )
Fnet  ma  m

t
t

The change in an object’s momentum divided by
the elapsed time equals the constant net force
acting on the object


p change in momentum

 Fnet
t
time interval
March 24, 2009
Impulse

When a single, constant force acts on the
object, there is an impulse delivered to the
 
object



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I  Ft
I is defined as the impulse
The equality is true even if the force is not constant
Vector quantity, the direction is the same as the
direction of the force


p change in momentum

 Fnet
t
time interval
March 24, 2009
Impulse-Momentum Theorem

The theorem states
that the impulse
acting on a system is
equal to the change
in momentum of the
system

 
p  Fnet t  I



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I  p  mv f  mvi
March 24, 2009
Calculating the Change of Momentum
p  pafter  pbefore
 mvafter  mvbefore
 m(vafter  vbefore )
For the teddy bear
p  m0  (v)  mv
For the bouncing ball
p  mv  (v)  2mv
March 24, 2009
How Good Are the Bumpers?
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

March 24, 2009
How Good Are the Bumpers?
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

pi  mvi  (1.5 103 kg)(15m / s)  2.25104 kg  m / s
p f  mvf  (1.5103 kg)(2.6m / s)  0.39104 kg  m / s
I  p f  pi  m vf  m vi
 (0.39104 kg  m / s)  (2.25104 kg  m / s)
 2.64104 kg  m / s
p I
2.64104 kg  m / s
Fav 


 1.76105 N
t t
0.15s
March 24, 2009
Conservation of Momentum

In an isolated and closed system,
the total momentum of the
system remains constant in time.
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Isolated system: no external forces
Closed system: no mass enters or
leaves
The linear momentum of each
colliding body may change
The total momentum P of the
system cannot change.
March 24, 2009
Conservation of Momentum

Start from impulse-momentum
theorem
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
F21t  m1v1 f  m1v1i



F12 t  m2v2 f  m2v2i

Since


F21t  F12t



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 Then m1v1 f  m1v1i  (m2v2 f  m2v2i )

So
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m1v1i  m2v2i  m1v1 f  m2v2 f
March 24, 2009
Conservation of Momentum
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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

 

Fnet t  p  p f  pi


When Fnet  0 then p  0
For an isolated system

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p f  pi
Specifically, the total momentum before the collision will
equal the total momentum after the collision
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m1v1i  m2v2i  m1v1 f  m2v2 f
March 24, 2009
The Archer
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?

pi  p f
m1v1i  m2v2i  m1v1 f  m2v2 f
m1  60.0kg, m2  0.5kg, v1i  v2i  0, v2 f  50m / s, v1 f  ?
0  m1v1 f  m2v2 f
m2
0.5kg
v1 f  
v2 f  
(50.0m / s)  0.417m / s
m1
60.0kg
March 24, 2009
Types of Collisions
Momentum is conserved in any collision
 Inelastic collisions: rubber ball and hard ball
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
Elastic collisions: billiard ball


Kinetic energy is not conserved
Perfectly inelastic collisions occur when the objects
stick together
both momentum and kinetic energy are conserved
Actual collisions

Most collisions fall between elastic and perfectly
inelastic collisions
March 24, 2009
Simple Examples of Head-On Collisions
(Energy and Momentum are Both Conserved)
Collision between two objects of the same mass. One mass is at rest.
Collision between two objects. One at rest initially has twice the mass.
Collision between two objects. One not at rest initially has twice the mass.


p  mv
Example of Non-Head-On Collisions
(Energy and Momentum are Both Conserved)
Collision between two objects of the same mass. One mass is at rest.
If you vector add the total momentum after collision,
you get the total momentum before collision.


p  mv
Collisions Summary
In an elastic collision, both momentum and kinetic
energy are conserved
 In an inelastic collision, momentum is conserved but
kinetic energy is not. Moreover, the objects do not stick
together
 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
 Elastic and perfectly inelastic collisions are limiting cases,
most actual collisions fall in between these two types
 Momentum is conserved in all collisions

March 24, 2009
More about Perfectly Inelastic
Collisions
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When two objects stick together
after the collision, they have
undergone a perfectly inelastic
collision
Conservation of momentum
m1v1i  m2 v2i  (m1  m2 )v f
m1v1i  m2 v2i
vf 
m1  m2

Kinetic energy is NOT conserved
March 24, 2009
An SUV Versus a Compact

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.
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.
(a)
March 24, 2009
An SUV Versus a Compact
(a)
Find the speed of the entangled m  1.80103 kg, v  15m / s
1
1i
cars after the collision.
2
m2  9.0010 kg, v2i  15m / s
pi  p f
m1v1i  m2v2i  (m1  m2 )v f
m1v1i  m2 v2i
vf 
m1  m2
v f  5.00m / s
March 24, 2009
An SUV Versus a Compact
(b)
Find the change in the velocity
of each car.
v f  5.00m / s
m1  1.80103 kg, v1i  15m / s
m2  9.00102 kg, v2i  15m / s
v1  v f  v1i  10.0m / s
v2  v f  v2i  20.0m / s
m1v1  m1 (v f  v1i )  1.8 104 kg  m / s
m2v2  m2 (v f  v2i )  1.8104 kg  m / s
m1v1  m2v2  0
March 24, 2009
An SUV Versus a Compact
(c)
Find the change in the kinetic
3
m

1
.
80

10
kg, v1i  15m / s
energy of the system consisting 1
m2  9.00102 kg, v2i  15m / s
of both cars.
v f  5.00m / s
1
1
2
KEi  m1v1i  m2 v22i  3.04 10 5 J
2
2
1
1
2
KE f  m1v1 f  m2 v22 f  3.38 10 4 J
2
2
KE  KE f  KEi  2.70105 J
March 24, 2009
More About Elastic Collisions

Both momentum and kinetic energy
are conserved
m1v1i  m2 v2i  m1v1 f  m2 v2 f
1
1
1
1
2
2
2
m1v1i  m2 v2i  m1v1 f  m2 v22 f
2
2
2
2


Typically have two unknowns
Momentum is a vector quantity



Direction is important
Be sure to have the correct signs
Solve the equations simultaneously
March 24, 2009
Elastic Collisions

A simpler equation can be used in place of the KE
equation
1
1
1
1
2
2
2
m1v1i  m2 v2i  m1v1 f  m2 v22 f
2
2
2
2
m1 (v12i  v12f )  m2 (v22 f  v22i )
v  v  ( v  v )
m1 (v11i i v1 f )( v21ii  v1 f )  m21(fv2 f  v22i )(f v2 f  v2i )
m1v1i  m2 v2i  m1v1 f  m2 v2 f
m1 (v1i  v1 f )  m2 (v2 f  v2i )
v1i  v1 f  v2 f  v2i
m1v1i  m2 v2i  m1v1 f  m2 v2 f
March 24, 2009
Summary of Types of Collisions

In an elastic collision, both momentum and kinetic
energy are conserved
v1i  v1 f  v2 f  v2i

m1v1i  m2 v2i  m1v1 f  m2 v2 f
In an inelastic collision, momentum is conserved but
kinetic energy is not
m1v1i  m2 v2i  m1v1 f  m2 v2 f

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
m1v1i  m2 v2i  (m1  m2 )v f
March 24, 2009
Problem Solving for 1D Collisions, 1

Coordinates: Set up a
coordinate axis and define
the velocities with respect
to this axis


It is convenient to make
your axis coincide with one
of the initial velocities
Diagram: In your sketch,
draw all the velocity
vectors and label the
velocities and the masses
March 24, 2009
Problem Solving for 1D Collisions, 2

Conservation of
Momentum: Write a
general expression for the
total momentum of the
system before and after
the collision


Equate the two total
momentum expressions
Fill in the known values
m1v1i  m2 v2i  m1v1 f  m2 v2 f
March 24, 2009
Problem Solving for 1D Collisions, 3

Conservation of Energy:
If the collision is elastic,
write a second equation
for conservation of KE, or
the alternative equation

This only applies to perfectly
elastic collisions
v1i  v1 f  v2 f  v2i

Solve: the resulting
equations simultaneously
March 24, 2009
One-Dimension vs Two-Dimension
March 24, 2009
Two-Dimensional Collisions

For a general collision of two objects in twodimensional space, the conservation of momentum
principle implies that the total momentum of the
system in each direction is conserved
m1v1ix  m2 v2ix  m1v1 fx  m2 v2 fx
m1v1iy  m2 v2iy  m1v1 fy  m2 v2 fy
March 24, 2009
Two-Dimensional Collisions
The momentum is conserved in all directions
m1v1ix  m2 v2ix  m1v1 fx  m2 v2 fx
 Use subscripts for





Identifying the object
m1v1iy  m2 v2iy  m1v1 fy  m2 v2 fy
Indicating initial or final values
The velocity components
If the collision is elastic, use conservation of
kinetic energy as a second equation

Remember, the simpler equation can only be used
for one-dimensional situations
v1i  v1 f  v2 f  v2i
March 24, 2009
Glancing Collisions



The “after” velocities have x and y components
Momentum is conserved in the x direction and in the
y direction
Apply conservation of momentum separately to each
direction
mv m v mv m v
1 1ix
2 2 ix
1 1 fx
2 2 fx
m1v1iy  m2 v2iy  m1v1 fy  m2 v2 fy
March 24, 2009
2-D Collision, example
Particle 1 is moving at
velocity v1i and
particle 2 is at rest
 In the x-direction, the
initial momentum is

m1v1i

In the y-direction, the
initial momentum is 0
March 24, 2009
2-D Collision, example cont


After the collision, the
momentum in the x-direction is
m1v1f cos q  m2v2f cos f
After the collision, the
momentum in the y-direction is
m1v1f sin q  m2v2f sin f
m1v1i  0  m1v1 f cos q  m2 v2 f cos f
0  0  m1v1 f sin q  m2 v2 f sin f

If the collision is elastic, apply
the kinetic energy equation
1
1
1
m1v12i  m1v12f  m2 v22 f
2
2
2
March 24, 2009
Problem Solving for TwoDimensional Collisions
 Coordinates:
Set up coordinate axes and
define your velocities with respect to these
axes

It is convenient to choose the x- or y- axis to
coincide with one of the initial velocities
 Draw:
In your sketch, draw and label all
the velocities and masses
March 24, 2009
Problem Solving for TwoDimensional Collisions, 2
Conservation of Momentum: Write
expressions for the x and y components of the
momentum of each object before and after the
collision
 Write expressions for the total momentum
before and after the collision in the x-direction
and in the y-direction

m1v1ix  m2 v2ix  m1v1 fx  m2 v2 fx
m1v1iy  m2 v2iy  m1v1 fy  m2 v2 fy
March 24, 2009
Problem Solving for TwoDimensional Collisions, 3
 Conservation
of Energy: If the collision
is elastic, write an expression for the total
energy before and after the collision



Equate the two expressions
Fill in the known values
Solve the quadratic equations

Can’t be simplified
March 24, 2009
Problem Solving for TwoDimensional Collisions, 4
 Solve



for the unknown quantities
Solve the equations simultaneously
There will be two equations for inelastic
collisions
There will be three equations for elastic
collisions
 Check
to see if your answers are
consistent with the mental and pictorial
representations. Check to be sure your
results are realistic
March 24, 2009
Collision at an Intersection
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.

mc  1.5 103 kg, mv  2.5 103 kg
vcix  25m / s, vviy  20m / s, v f  ?q  ?
March 24, 2009
Collision at an Intersection
mc  1.5 103 kg, mv  2.5 103 kg
vcix  25m / s, vviy  20m / s, v f  ?q  ?
p
p
xi
 mcvcix  mvvvix  mcvcix  3.75104 kg  m / s
xf
 mcvcfx  mvvvfx  (mc  mv )v f cosq
3.75104 kg  m / s  (4.00103 kg)v f cosq
p
p
yi
yf
 mcvciy  mvvviy  mvvviy  5.00104 kg  m / s
 mcvcfy  mvvvfy  (mc  mv )v f sin q
5.00104 kg  m / s  (4.00103 kg)v f sin q
March 24, 2009
Collision at an Intersection
mc  1.5 103 kg, mv  2.5 103 kg
vcix  25m / s, vviy  20m / s, v f  ?q  ?
5.00104 kg  m / s  (4.00103 kg)v f sin q
3.75104 kg  m / s  (4.00103 kg)v f cosq
5.00104 kg  m / s
tanq 
 1.33
4
3.7510 kg  m / s
q  tan1 (1.33)  53.1
5.00104 kg  m / s
vf 
 15.6m / s
3

(4.0010 kg) sin 53.1
March 24, 2009