10/27/09
Chapter 8
Conservation of Linear Momentum
Physics 201
October 22, 2009
Conservation of Linear Momentum
•! Definition of linear momentum,
!
!
p = mv
!
p
Linear momentum is a vector.
Units of linear momentum are kg-m/s.
Can write Newton’s second law in terms of momentum:
!
!
!
dp d(mv)
dv
!
=
= m = ma
dt
dt
dt
!
dp !
! = Fnet
dt
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Momentum of a system of particles
•! The total momentum
of a system of
particles is the vector sum of the momenta of
the individual particles:
!!!"
"
"
Psys = ! mi vi = ! pi
i
i
From Newton’s second law, we obtain
!
!
!
dPsys
! Fext = Fnetext = ! dt
i
i
Conservation of Momentum
•! Law of conservation of momentum:
–! If the sum of the external forces on a system is
zero, the total momentum of the system does not
change.
If
!
! Fext = 0 then
i
"""""!
!
!
!
Psys = ! mi vi = MvCM = const
i
Momentum is always conserved (even if forces are nonconservative).
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Collisions
m1
m2
m1
m2
“before”
“after”
momentum before collision = momentum after collision
Always But only if
!
Fexternal = 0
Explosion - I
“before”
M
v1
m1
m2
v2
“after”
Example: m1 = M/3 m2 = 2M/3
After explosion, which block has larger momentum? (left, right, same)
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Explosion - I
“before”
M
v1
m2
m1
v2
“after”
Example: m1 = M/3 m2 = 2M/3
After explosion, which block has larger momentum? (left, right, same)
Each has the same momentum
Explosion - I
“before”
M
v1
m1
m2
v2
“after”
Example: m1 = M/3 m2 = 2M/3
After explosion, which block has larger momentum? (left, right, same)
Each has the same momentum
Which block has larger velocity?
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Explosion - I
“before”
M
v1
m2
m1
v2
“after”
Example: m1 = M/3 m2 = 2M/3
After explosion, which block has larger momentum? (left, right, same)
Each has the same momentum
Which block has larger velocity?
mv is the same for each block, so smaller mass has larger velocity
Explosion - I
“before”
M
v1
m1
m2
v2
“after”
Example: m1 = M/3 m2 = 2M/3
After explosion, which block has larger momentum? (left, right, same)
Each has the same momentum
Which block has larger velocity?
mv is the same for each block, so smaller mass has larger velocity
Is kinetic energy conserved?
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Explosion - I
“before”
M
v1
m2
m1
v2
“after”
Example: m1 = M/3 m2 = 2M/3
After explosion, which block has larger momentum? (left, right, same)
Each has the same momentum
Which block has larger velocity?
mv is the same for each block, so smaller mass has larger velocity
Is kinetic energy conserved?
NO! K was 0 before, it is greater after the explosion.
Explosion - I
“before”
M
v1
m1
m2
v2
“after”
Example: m1 = M/3 m2 = 2M/3
After explosion, which block has larger momentum? (left, right, same)
Each has the same momentum
Which block has larger velocity?
mv is the same for each block, so smaller mass has larger velocity
Is kinetic energy conserved? (green=yes, red=no)
NO!
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This is like a microscopic explosion ….
Momentum and Impulse
!
!
p ! mv
!!
For single object….
!
Fave !t " I
!
!
F = ma
definition of impulse
!
!
dv dp
! !
=m
=
# !p = F!t
dt dt
"! If F = 0, then momentum conserved (p = 0)
•!For “system”
of objects …
!
!
psys = ! pi
i
Internal forces " forces between objects in system
External forces " any other forces
!
!
#psys = Fext #t
!
!
Thus, if Fext = 0, then #psys = 0
i.e. total momentum is conserved!
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Momentum and Impulse
!
!
p ! mv
!!
For single object….
!
Fave !t " I
!
!
F = ma
definition of impulse
!
!
dv dp
! !
=m
=
# !p = F!t
dt dt
"! If F = 0, then momentum conserved (p = 0)
•!For “system”
of objects …
!
!
psys = ! pi
i
Internal forces " forces between objects in system
External forces " any other forces
!
!
#psys = Fext #t
!
!
Thus, if Fext = 0, then #psys = 0
i.e. total momentum is conserved!
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!
Fave !t " I
!
!
F = ma
definition of impulse
!
!
dv dp
! !
=m
=
# !p = F!t
dt dt
Let’s estimate the average force during the collision
Club speed: 50 m/s
tf
!
I
1 !
Assume that impulse is given after 5 cm
Fave =
=
F dt
--> whiteboard
!t !t t
"
i
Some Terminology
•! Elastic Collisions:
collisions that conserve kinetic energy
•! Inelastic Collisions:
collisions that do not conserve kinetic energy
*! Completely
Inelastic Collisons:
objects stick together
n.b. ALL CONSERVE MOMENTUM!!
If external forces = 0
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Elastic Collision in 1-Dimension
Initial
Final
m1v1i + m2 v2i = m1v1 f + m2 v2 f
Energy conserved (for elastic
collision only)
Linear momentum is conserved
1
1
1
1
m1v1i2 + m2 v2i2 = m1v12f + m2 v22 f
2
2
2
2
Elastic Collision
Conservation of Momentum
m1v1i + m2 v2i = m1v1 f + m2 v2 f
m1 (v1i ! v1 f ) = m2 (v2 f ! v2i )
Conservation of Kinetic Energy
1
1
1
1
m1v1i2 + m2 v2i2 = m1v12f + m2 v22 f
2
2
2
2
2
2
2
m1 (v1i ! v1 f ) = m2 (v2 f ! v2i2 )
m1 (v1i ! v1 f )(v1i + v1 f ) = m2 (v2 f ! v2i )(v2 f + v2i )
Combining the above two equations
v1i + v1 f = v2i + v2 f
v1i ! v2i = !(v1 f ! v2 f )
Magnitude of relative velocity is conserved.
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Is this an elastic collision?
For elastic collision only:
v1i ! v2i = !(v1 f ! v2 f )
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What is the speed of the golf ball, in
case of an elastic collision
Club speed: 50 m/s
Mass of clubhead: 0.5kg
Mass of golfball: 0.05kg
Two unknowns:
speed of club and
speed of golfball after impact
Problem solving strategy:
-! Momentum conservation
-! Energy conservation (or
use the derived equation
for relative velocities)
--> whiteboard
Is this an elastic collision?
For elastic collision only:
v1i ! v2i = !(v1 f ! v2 f )
Yes, the relative speeds
are approximately the same
before and after collision
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Result:
m1 ! m2
v1i
m1 + m2
2m1
=
v1i
m1 + m2
v1 f =
v2 f
Special cases:
1)! Golf shot: m1>>m2
Club speed almost unchanged
Ball speed almost 2 x club speed
2) Neutron scatters on heavy nucleus: m1<<m2
neutron scatters back with almost same speed
speed of nucleus almost unchanged
Completely inelastic collision
•! Two objects stick together and move with the center
! ! ! !
!
of mass:
PAi + PBi = PAf + PBf = PCM
•! If pAi=0:
!
!
PBi = PCM
!
!
mB v Bi = ( mA + mB ) v f
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Ballistic
Pendulum
What is the initial
velocity vli of the
projectile?
Known quantities:
m1, m2, h
!!
Two stage process:
1. m collides with M, inelastically. Both M and
m then move together with a velocity Vf
(before having risen significantly).
2. Both (m1 + m2) rise a height h, conserving energy E.
(no non-conservative forces acting after collision)
Ballistic
Pendulum
•! Stage 1: Momentum is conserved
Energy is not conserved
in x-direction:
!!
m1v1i = ( m1 + m2 )V f
Stage 2 (after the collision): Energy is conserved
K + U conserved :
1
( m1 + m2 )Vf2 = ( m1 + m2 ) gh ! Vf = 2gh
2
Substituting for V gives:
!
m $
v1i = # 1 + 2 & 2gh
m1 %
"
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Ballistic
Pendulum
•! How much energy is dissipated?
Wthermal = U ! K i
= ( m1 + m2 ) gh !
1
m1v1i2
2
2
1 " m + m2 %
= ( m1 + m2 ) gh ! m1 $ 1
2gh
2 # m1 '&
= ( m1 + m2 ) gh
m1
m2
Fraction of kinetic energy lost is …………..
If m2<<m1: not much energy is lost
Of m2>>m1:almost all energy is lost
Wthermal
m1
=
Ki
m1 + m2
Inelastic collision
Coefficient of restitution e=1/2
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Coefficient of restitution
Perfectly elastic collision:
v1i ! v2i = !(v1 f ! v2 f )
The coefficient of restitution is a
measure of the “inelasticity:
e=
v1 f ! v2 f
v1i ! v2i
Elastic collision: e=1
Perfectly inelastic collision: e=0
Collisions or Explosions in Two Dimensions
y
x
before
after
!!!!!!!!" !!!!!!!"
Ptotal,before = Ptotal,after
•! Ptotal,x and Ptotal,y independently conserved
*!Ptotal,x,before = Ptotal,x,after
*!Ptotal,y,before = Ptotal,y,after
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Explosions
M
“before”
“after”
A
B
Which of these is possible?
A
B
both
Explosions
M
“before”
“after”
A
B
Which of these is possible?
A
B
(p appears conserved)
(p not conserved in y direction)
both
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Explosions
M
“before”
“after”
B
A
Which of these is possible?
A
B
neither
Explosions
M
“before”
“after”
B
A
Which of these is possible?
A
(p not conserved in y direction)
B
neither
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(Inelastic) Car – truck collision
•! Knowns: m1, m2, v1, v2,
Mcm=m1+m2
•! Unknowns: final velocity vector
(in x and y)
•! Equation(s): Momentum
conservation (in x and y)
•! Strategy: write out conservation
of momentum equation.
( 2 unknowns, 2 equations, ! piece of
cake)
Elastic collision in 2 dimensions
•! Assume we know all initial conditions, mass and
momentum.
!!" !!!"
•! 4 Unknown quantities:
v1 f , v2 f
Equations:
Momentum conservation: 2 (x and y)
Energy conservation: 1
Need one more piece of information to solve the
problem: often a measurement.
What is unspecified above is the impact parameter
(and the precise nature of the interaction)
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©2008 by W.H.
Freeman and
Company
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Shooting Pool...
•! Assuming
–!
–!
–!
–!
Collision is elastic (KE is conserved)
No spin is imparted
Balls have the same mass
One ball starts out at rest
Shooting Pool
•! Elastic collision means conservation of kinetic energy
p12f
p22 f
1
p1i2
2
m1v1i =
=
+
2
2m1 2m1 2m2
if m1 = m2:
p1i2 = p12f + p22 f
!
!
!
!
•! Conservation of momentum: P + P = P + P
1i
2i
1f
2f
if !
!
!
!
P2i = 0
P =P +P
1i
1f
2f
P1 and p2
Form a right
angle!
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Shooting Pool...
•! Tip: If you shoot a ball spotted on the
“dot”, you have a good chance of
scratching !
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Collisions in the CM frame
"!
"!
"
!
m1 v1 + m2 v2
v cm =
m1 + m2
Subtract vcm from
all velocities
The transformation to the cm frame is not
necessary, but it is often convenient to switch to
the CM frame
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Rocket equation
•! The mass is changing
•! Thrust is generated by impulse of exhaust of mass
with velocity v: vdm
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The Saturn V
The Saturn V rocket:
•! 111 m tall
•! 10m diameter
•! 3000 tons at start
•! Thrust: 34 MN
Rocket equation
•! Mass change
•! Thrust
M
dM
dM
= R = const
dt
M (t) = M 0 ! Rt
dM
d(M ! uex ) dP
! uex =
=
= "Fthrust
dt
dt
dt
•! Weight:
F=Mg
dv
•! Rocket equation:
M
dt
= R ! uex " Mg
dv
R ! uex
=
" Mg
dt M 0 " Rt
•! Integration yields: v = uex ! ln #
%
M0 &
" gt
$ M 0 " Rt ('
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Variable mass
•! Newton’s second law for continuously variable
mass:
!
!
dM !
dv
Fnet ,ext +
vrel = M
dt
dt
!
! !
v rel = u ! v
•! Where
is the velocity of impacting material relative to
object with mass M at a given time.
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