Lecture 9.
Chapter 8: Momentum, Impulse, and Collisions
Momentum and Impulse
3/26/2015
1
Momentum and Impulse

In our discussion of work and energy we re-expressed Newton's Second Law
in an integral form called Work-Energy Theorem (Wtotal = DK) which states
that the total work done on a particle equals the change in the kinetic energy
of the particle.

We will once again re-express N2L in an integral form. However, in this case
the independent variable will be time rather than position as was the case
for the Work-Energy Theorem.

 dv
a
dt




dv d
 F  m dt  dt (mv )
The linear momentum of a particle of mass m moving with velocity v is a
vector quantity defined as the product of particle's mass and velocity:


p  mv
3/26/2015
Definition of momentum
2
Momentum and Impulse

Momentum is a vector quantity: it has magnitude (mv) and direction
(the same as velocity vector)



Momentum of a car driving North at 20 m/s is different from momentum
of the same car driving East at the same speed
Ball thrown by a major-league pitcher has greater magnitude of
momentum then the same ball thrown by a child because the speed is
greater
18-wheeler going 65 mph has greater magnitude of momentum than Geo
car with the same speed because the truck’s mass is greater

Units of momentum (SI): mass × speed, kg·m/s

Momentum in
components
p :terms
p of mv
p  mv


x
x
y
y
pz  mvz
N2L in new form: the net force acting on a particle equals the time
rate of change of momentum of the particle:
 dp
 F  dt
N2L in terms of momentum
Momentum and Impulse
 dp
 F  dt


N2L in terms of momentum shows that

Rapid change in momentum requires a large net force

Gradual change in momentum requires less net force
This principle is used in the design of safety air bags in cars




The driver of fast-moving car has a large momentum (the
product of the driver’s mass and velocity).
If car stops suddenly in a collision, driver’s momentum becomes
zero due to collision with car parts (steering wheel and
windshield)
Air bag causes the driver to lose momentum more gradually than
would abrupt collision with the steering wheel, reducing the force
exerted on the driver (and the possibility for injury)
Same principle applies to the padding used to package fragile
objects for shipping
Momentum and Impulse

Consider a particle acted on by a constant net force F during a
time interval Dt from t1 to t2. The impulse of the net force J is
defined to be the product of the net force and the time interval:



J   F (t2  t1 )  FDt

Impulse is a vector quantity.



For constant net force
Its direction is the same as the net force
Its magnitude is the product of the magnitude of the net
force and the length of time the the net force acts
Units of impulse (SI): force × time, N·s

1 N = 1 kg·m/s2  N·s = kg·m/s (same as momentum)
Impulse - Momentum Theorem

If the net force is constant, then dp/dt is also constant and equals to
the total change in momentum during the time interval t2-t1, divided
  
by this time interval:
dp p2  p1
 F  const  dt  t  t
2
1
  
J  p2  p1


 
 F (t2  t1 )  p2  p1
Impulse – Momentum Theorem
Impulse – Momentum Theorem: The change in momentum of a
particle during a time interval equals the impulse of the net force that
acts on the particle during that interval.
Impulse - Momentum Theorem

Impulse – Momentum Theorem also holds when forces are NOT
constant. To see this, lets integrate both sides of N2L
t2 
t2

  
dp
t  Fdt  t dt dt  t dp  p2  p1
1
1
1
t2
 t2

J    F dt
t1
General definition of impulse
Impulse - Momentum Theorem

We can define an average net force
F
such that even when F is NOT constant,
the impulse J is given by
The force on soccer ball
that in contact with player’s
foot from time t1 to t2
 
J  Fav (t2  t1 )

The impulse can be interpreted as the "area"
under the graph of F(t) versus t. The x-
Fx
between t1 and t2 equals the area under Fx–t
component of impulse of the net force
curve, which also equals the area under
t 2 height (Fav)x
rectangle with
J x    Fx dt  ( Fav ) x (t2  t1 )  p2 x  p1x  mv2 x  mv1x
t1
t2
J y    Fy dt  ( Fav ) y (t2  t1 )  p2 y  p1 y  mv2 y  mv1 y
t1
Momentum and Kinetic Energy
  
J  p2  p1

Fundamental difference between momentum and kinetic energy

Impulse - Momentum Theorem:


Changes in a particle’s momentum are due to impulse, which
depends on the time over which the net force acts.
Work – Energy Theorem:

Wtot  K 2  K1
Kinetic energy changes when work done on the particle. The
total work depends on the distance over which the net force
acts.
Momentum and Kinetic Energy
Compared
Consider a particle that starts from rest at t1 so that v=0.


Its initial momentum is p1=mv1=0

Its initial kinetic energy is K1=0.5mv12=0
Let a constant net force equal to F act on that particle from time t1 until
time t2. During this interval the particle moves a distance S in the direction
of the force. From impulse-momentum theorem:

  
J  p2  p1
 


p2  J  p1  J
 
J  F (t2  t1 )
o
The momentum of a particle equals the impulse that accelerated it from the
rest to its present speed
o
The impulse is the product of the net force that accelerated the particle
and the time required for the acceleration
o
Kinetic energy of the particle at t2 is K2=Wtot=FS, the total work done on
the particle to accelerate it from rest
o
The total work is the product of the net force and the distance required to
accelerate the particle
Momentum and Kinetic Energy.
Example
Net force

F

Kinetic energy of a pitched baseball =
the work the pitcher does on it
o
(force × distance the ball moves during
the throw)

Momentum of the ball = the impulse
the pitcher imparts to it
o
(force × time it took to bring the ball up
to speed)
Displacement in time Dt
Kinetic energy gained
 
by ball   F  S
Momentum gained

by ball   FDt
Conservation of Momentum
Conservation of Momentum

Concept of momentum is particularly important in situations when
you have two or more interacting bodies

Consider idealized system of two particles: two astronauts floating
in the zero-gravity environment

Astronauts touch each other (each particle exerts force on another)

N3L: these forces are equal and opposite

Hence, impulses that act on two particles are equal and opposite
Internal and External Forces

Internal forces: the forces that the particles of the system exert on
each other (for any system)

Forces exerted on any part of the system by some object outside it called
external forces

For the system of two astronauts, the internal forces are FBonA, exerted
by particle B on particle A, and FAonB, exerted by particle A on particle B

There are NO external forces: we have isolated system






dp A
dpB
FAonB   FBonA
FBonA 
FAonB 
dt
dt






dp
dp
d ( p A  pB )
FAonB  FBonA  A  B 
0
dt
dt
dt
 

P  p A  pB
 



P  p A  pB  ...  mAv A  mB vB  ...

Total momentum of two particles is the vector
sum of the momenta of individual particles

If system consist of any
number of particles A, B, …
Conservation of Total Momentum

The time rate of change of the total
momentum is zero. Hence the total
momentum of the system is
constant, even though the individual
momenta of the particles that make
up the system can change.



dP
FAonB  FBonA 
0
dt

If there are external forces acting on the
system, they must be included into
equation above along with internal
forces.

Then the total momentum is not constant
in general. But if vector sum of external
forces is zero, these forces do not
contribute to the sum, and dP/dt=0.
Conservation of Total Momentum
Principle of conservation of momentum:

If the vector sum of the external forces on a system is zero, the total
momentum of the system is constant


This principle does not depend on the detailed nature of internal forces that act
between members of the system: we can apply it even if we know very little about
the internal forces.
The principle acts only in internal frame of reference (because we used N2L to
derive it!)
CAUTION Vector sum !
Conservation of Total Momentum
Problem-Solving Strategy
IDENTIFY the relevant concepts:
Before applying conservation of momentum to a problem,
you must first decide whether momentum is conserved!
This will be true only if the vector sum of the external forces
acting on the system of particles is zero.
If this is not the case, you can’t use conservation of
momentum.
Conservation of Total Momentum
Problem-Solving Strategy
SET UP the problem using the following steps:
1.
Define a coordinate system. Make a sketch showing the coordinate
axes, including the positive direction for each. Make sure you are
using an inertial frame of reference. Most of the problems in this
course deal with two-dimensional situations, in which the vectors
have only x- and y-components; all of the following statements can
be generalized to include z-components when necessary.
2.
Treat each body as a particle. Draw “before” and “after” sketches,
and include vectors on each to represent all known velocities. Label
the vectors with magnitudes, angles, components, or whatever
information is given, and give each unknown magnitude, angle, or
component an algebraic symbol. You may use the subscripts 1 and 2
for velocities before and after interaction, respectively.
3.
As always, identify the target variable(s) from among the unknowns.
Conservation of Total Momentum
Problem-Solving Strategy
EXECUTE the solution as follows:
1.
Write equation in terms of symbols equating the total initial x-component
of momentum (before the interaction) to the total final x-component of
momentum (after), using px=mvx for each particle.
2.
Write another equation for y-components, using py=mvy for each particle.
 Remember that the x- and y-components of velocity or momentum are never
added together in the same equation!
 Even when all the velocities lie along a line (such as the x-axis), components of
velocity along this line can be positive or negative; be careful with signs!
3.
Solve these equations to determine whatever results are required. In
some problems you will have to convert from the x- and y-components of
a velocity to its magnitude and direction, or the reverse.
4.
In some problems, energy considerations give additional relationships
among the various velocities.
Conservation of Total Momentum
Problem-Solving Strategy
EVALUATE your answer:
Does your answer make physical sense?
If your target variable is a certain body’s momentum,
check that the direction of the momentum is reasonable.
Conservation of Total Momentum
What Momentum
V=30 MPH
V=0
cannot do
A collision
Before
PCar,i = (1500 lbs) (0 M PH)
= (681 kg) (0 m/s) = 0
PTruck,i = (3500 lbs) (30 M PH)
VT
Vc
After
PTotal,f = Pcivic,f
+ PBlazer,f
= (1,589 kg) (13.4 m/s)
= 21,292 kg  m/s
PTotal,i  Pcivic,i + PBlazer,i
= 21,292 kg  m/s
How do we determine the velocities?
What Momentum
cannot do
PTotal,f = Pcivic,f + PBlazer,f  CONSTANT
Pcar, f = PTotal - PTruck,f
There are many
possibilities
Conservation of
Momentum can’t
tell them apart
Momentum Conservation in
Explosions
Pi=0
Pf=mgvg+mbvb
gun
Pi=Pf
bullet
vg= -vb
Note: vb and vg are
in opposite directions
mb
mg
Continuous Momentum Transfer
A gun shooting a bullet is a
discrete process
We can generalize this to
continuous processes
Rockets operate by shooting out
a continuos stream of gas
Note: Rocket propulsion was
originally thought to be impossible
Nothing to push against !!!
Rockets
Pi = Mv
PF = PRocket + PFuel 
= ( M-Dm)(v+Dv) + Dm(v-vex ) 
= Mv -Dmv  MDv  Dmv  Dmvex  DmDv 
 Mv  MDv  Dmvex
MDv = Dmvex
Rockets
MΔv = Dmvex
Dm  0
Mdv =vex dm
vex
dv = dM
M
vf
Mf
 dv = -v 
ex
vi
Mi
dM
M
 Mi
v f = vi + vex ln 
M
 f




Collisions
Collisions

Collision is any strong interaction between bodies that lasts a
relatively short time

Car’s collision

Balls colliding on a pool table

Neutrons hitting nuclei in reactor core

Bowling ball striking pins

Meteor impact with Earth

If the forces between the bodies are much larger than any external
forces, we can neglect the external forces entirely and treat the
bodies as isolated system

Since a collision constitutes an isolated system (where the net
external force is zero), the momentum of the system is
conserved (the same before and after the collision)

Collision types: inelastic and elastic collisions
Inelastic Collisions

In any collision in which external forces can be neglected,
momentum is conserved and the total momentum before equals
the total momentum after

Collisions are classified according to how much energy is "lost"
during the collision

Inelastic Collisions - there is a loss of kinetic energy due to the
collision
Automobile
collision is inelastic: the structure of the car absorbs as
much of the energy of collision as possible. This absorbed energy
cannot be recovered, since it goes into a permanent deformation of
the car

Completely Inelastic Collisions - the loss of kinetic energy is the
maximum possible. The objects stick together after the collision.

Elastic collisions - the kinetic energy of the system is conserved
Elastic Collision
Elastic collision:

Glider A and glider B approach
each other on a frictionless
surface

Each glider has a steel spring
bumper on the end to ensure an
elastic collision
Elastic Collision
Elastic 1-D collision of two bodies A and B, B is at rest before collision


Momentum of the system is conserved
Kinetic energy of the system is conserved
mAvA1x  mB (0)  mAvA2 x  mB vB 2 x
1
1
1
1
mAv A21x  mB (0) 2  mAv A2 2 x  mB vB2 2 x
2
2
2
2
mB vB2 2 x  mA (vA21x  v A2 2 x )  mA (vA1x  vA2 x )(vA1x  vA2 x )
mB vB 2 x  mA (v A1x  v A2 x )
mB (vA1x  vA2 x )  mA (vA1x  vA2 x )
vB 2 x
2m A

v A1x
mA  mB
Divide equations
v A2 x 
vB 2 x  v A1x  v A2 x
mA  mB
v A1x
mA  mB
Elastic Collision
v A2 x
mA  mB

v A1x
mA  mB
vB 2 x
2m A

v A1x
mA  mB
Interpretation of results



Suppose body A is a ping-pong ball, and body B is bowling ball
We expect A to bounce of after the collision with almost the same
speed but opposite direction, and speed of B will be much smaller
What if situation is reversed? Bowling ball hits ping-pong ball?
Elastic Collision
v A2 x
mA  mB

v A1x
mA  mB
Interpretation of results


Suppose masses of bodies A and B are equal
Then mA=mB, and VA2x=0, VB2x=VA1x
vB 2 x
2m A

v A1x
mA  mB
Elastic Collision
v A2 x
mA  mB

v A1x
mA  mB
vB 2 x  vA1x  v A2 x
vB 2 x
2m A

v A1x
mA  mB
vB 2 x  v A2 x  v A1x

VB2x-VA2x is relative velocity of B to A after the collision, and it is
negative of the velocity of B relative to A before the collision

In a straight-line elastic collision of two bodies, the relative
velocities before and after collision have the same
magnitude, but opposite sign

General case (when velocity of B is not zero):
vB 2 x  vA2 x  (vB1x  vA1x )
Inelastic Collision
Completely inelastic collision:

Glider A and glider B approach
each other on a frictionless
surface

Each glider has a putty on the
end, so gliders stick together
after collision
Center of Mass
Center of Mass




Atemi (Striking the Body): Strike directed at the attacker for
purposes of unbalancing or distraction.
Atemi is often vital for bypassing or “short-circuiting'' an attacker's
natural responses to aikido techniques.
The first thing most people will do when they feel their body being
manipulated in an unfamiliar way is to retract their limbs and drop
their center of mass down and away from the person performing the
technique.
By judicious application of atemi, it is possible to create a “window of
opportunity'' in the attacker's natural defenses, facilitating the
application of an aikido technique.
From Aikido terms
Center of Mass
Consider several particles with masses m1, m2 and so on.

(x1, y1) are coordinates of m1, (x2, y2) are coordinates of m2 …

Center of mass of the system is the point having the coordinates
(xcm, ycm) given by
i i
1 1
2 2
3 3
i
cm
1
2
3
i
i
x
mx

m x  m x  m x  ...


m  m  m  ...
m
my

m y  m y  m y  ...


m  m  m  ...
m
i
ycm
1 1
2
1
2
2
3 3
i
i
3
i
i
Center of Mass

The center of mass of a system of N particles with masses m1, m2,
m3, etc. and positions, (x1, y1, z1), (x2, y2, z2), (x3, y3, z3), etc. is
defined in terms of the particle's position vectors using
i i
1 1
2 2
3 3
i
cm
1
2
3
i
i




m r  m r  m r  ...
r 

m  m  m  ...



m r
m
Center of mass is a mass-weighted average position of particles
If this expression is differentiated with respect to time then the position
vectors become velocity vectors and we have




m1v1  m2v2  m3v3  ...
vcm 

m1  m2  m3  ...

 mi vi
i
m
i
i
Center of Mass




m1v1  m2v2  m3v3  ...
vcm 

m1  m2  m3  ...

 mi vi
i
m
i
i
M  m1  m2  m3  ...





Mvcm  m1v1  m2v2  m3v3  ...  P
Total mass


Total moment of the system
Total moment is equal to total mass
times velocity of the center of mass
For a system of particles on which the net
external force is zero, so that the total
momentum is constant, the velocity of the
center of mass is also constant
Spinning throwing knife on ice:
center of mass follows straight line
External Forces
and Center-of-Mass
Motion



If we differentiate once again, then the derivatives of the velocity terms
will be accelerations
Further, if we use Netwon's Second Law for each particle then the terms
of the form miai will be the net force on particle i.
When these are summed over all particles any forces that are internal
to the system (i.e. a force on particle i caused by particle j) will be
included in the sum twice and these pairs will cancel due to Newton's
Third Law. The result will be





Macm  m1a1  m2 a2  m3a3  ...   Fext


Body
or collection of particles
It states that the net external force on a system of particles equals
the total mass of the system times the acceleration of the center
of mass (Newton's Second Law for a system of particles)
It can also be expressed as the net external force on a system equals
the time rate of change of the total linear momentum of the system.
Center of Mass ?
Center of Mass ?
The segmental method involves computation of the segmental CMs
The whole body CM is computed based on the segmental CMs
8.39
A 0.15 kg glider is moving to the right on a frictionless
surface with a speed of 0.8 m/s. It has a head-on
collision with a 0.3 kg glider that is moving to the left with
a speed of 2.2 m/s. Find the final velocity (magnitude
and direction) of each glider if the collision is elastic.
Set It up
m1= mass of glider 1=0.15 kg
m2= mass of glider 2=0.3 kg
vi1=initial velocity of glider 1=0.8 m/s
vi2=initial velocity of glider 2=-2.2 m/s
Collision is elastic; therefore KE is conserved.
Cons of E and Mom.
m1vi1  m2 v2i  m1v1 f  m2 v2 f
0.54  m1v1 f  m2 v2 f
1.8  0.5v1 f  v2 f
1
1
1
1
m1v12i  m2v 22i  m1v12f  m2v 22 f
2
2
2
2
1.548  m1v12f  m2 v 22 f
1.548  0.15v12f  0.3*( 1.8  0.5v1 f ) 2
1.548  0.15v12f  0.972  0.075v12f  0.54v1 f
0  0.225v12f  0.54v1 f  0.576
2 Solutions
Using the “-” solution, v1f=-3.2 m/s v2f=-0.2 m/s
Using the “+” solution, v1f=0.8 m/s v2f=-2.2 m/s
Physical intuition tells us that the lighter glider (1)
will not push the heavier, faster moving glider
out of the way so the “-” solution is correct.
8-63
Three identical pucks on a horizontal air table have
repelling magnets. They are held together and then
released simultaneously. Each as the same speed at
any instant. One puck moves due west. What is the
magnitude and direction of the other two pucks?
Draw It!
v
v
v
Conservation of momentum in the
north-south direction says:
v
v*sin q
v
q
f
v*sin f
v
v*sin q = v * sin f
qf
From this,
x direction: 0  v  2* v *cos q
y direction: 0  v *sin q  v *sin q
v  2* v *cos q
0.5  cos q
q  60
0
8.71
A 0.1 kg stone rests on a frictionless horizontal
surface. A bullet of mass 6.0 g traveling
horizontally at 350 m/s strikes the stone and
rebounds horizontally at right angles to its
original direction with a speed of 250 m/s.
a) Compute the magnitude and direction of the
velocity of the stone after it is struck
b) Is the collision perfectly elastic?
Draw “Before” and “After”
m1=6 g
m1=6 g
v1i =350 m/s
m2=100 g
v1f =250 m/s
m2=100 g
q
v2i=0
v2f=?
v2f*sin q
Conservation of Mom. In both x and y
directions
x direction: m1v1i  m2 v2 f cos q
2.1  0.1* v2 f cos q
21  v2 f cos q
y direction: 0  m1v1 f  m2 v2 f sin q
1.5  0.1* v2 f sin q
15  v2 f sin q
15 v2 f sin q

 tan q  q  35.50
21 v2 f cos q
152  212  v 22 f  v2 f  25.8 m/s
Part B) Compare KEbefore to KEafter
KEbefore = ½ *(0.006)*3502= 367 J
KEafter = ½ *(0.006)*2502 + ½ *(.1)*25.82
KEafter =221 J
KEbefore not equal KEafter so not perfectly elastic
8.96
A 20 kg projectile is fired at an angle of 600 above
the horizontal with a speed of 80 m/s. At the
highest point in the trajectory, the projectile
explodes into two fragments with equal mass,
one of which falls vertically with zero initial
speed. Ignore air resistance.
a) How far from the point of firing does the other
fragment strike if the terrain is level?
b) How much energy is released during the
explosion?
From Projectiles
We know we must break initial velocity into
horizontal and vertical components
vh=v*cos(600)=0.5*80m/s=40 m/s
vv=v*sin(600)=0.86*80 m/s =69 m/s
In the vertical direction at highest point:
v(t)=0=-g*t+vv or t=vv/g=69/9.8=7 m/s
y(t)=- ½ *g*t2+vv*t= -0.5*9.8*49+69*7 =242 m
x(t)=vh*t=40*7=280 m
Draw Before and After
px1
px
py1=0
px2=0
py=0
py2=0
If “2” is just dropping with zero
speed, then
Y direction: 0=py1+py2 but py2=0 so
py1=0
Draw Before and After
px1
px
py1=0
px2=0
py=0
py2=0
If “2” is just dropping with zero
speed, then
X direction: px=px1+px2 but px2=0 so
px1=px
0.5*m*vx2=mvh which means that
Vx2=2*vh=80 m/s
Finishing part a)
Now, it takes 2*7 seconds for objects to hit ground.
Mass “2” hits ground at 280 m from starting point.
Mass “1” has now accelerated at this point to 2*vh
X(t)=80*7+280=840 m from launch point
Part b) KEafter-Kbefore= Energy of
explosion
KEafter= ½ *(m/2)*(2*vh)2= mvh2
KEbefore= ½ mvh2
Energy then is ½ *m*vh2=0.5*20*40^2
Energy=16000 J