10/27/09
1
Chapter 8
Conservation of Linear Momentum
Physics 201
October 22, 2009
Conservation of Linear Momentum
•! Definition of linear momentum,
! p
! p = m
! v
Linear momentum is a vector.
Units of linear momentum are kg-m/s.
Can write Newton’s second law in terms of momentum:
d! p
dt=
d(m! v )
dt= m
d! v
dt= m! a
!d! p
dt=! F net
10/27/09
2
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:
From Newton’s second law, we obtain
Psys
! "!!= m
i
"vi
i
! ="pi
i
!
!Fext
i
! =!Fnetext =
d!Psys
dti
!
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 then
Momentum is always conserved (even if forces are nonconservative).
!Psys = mi
!vi
i
! = M!vCM = const
" !""""
!Fext
i
! = 0
10/27/09
3
Collisions
“before” m1 m2
“after” m1 m2
momentum before collision = momentum after collision
Always -
But only if
! F
external= 0
Explosion - I
“before” M
“after” m1 m2
v1 v2
Example: m1 = M/3 m2 = 2M/3
After explosion, which block has larger momentum? (left, right, same)
10/27/09
4
“before” M
“after” m1 m2
v1 v2
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
“after” m1 m2
v1 v2
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?
Explosion - I
10/27/09
5
“before” M
“after” m1 m2
v1 v2
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
“after” m1 m2
v1 v2
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?
Explosion - I
10/27/09
6
Explosion - I
“before” M
“after” m1 m2
v1 v2
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
“after” m1 m2
v1 v2
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!
10/27/09
7
This is like a microscopic explosion ….
Momentum and Impulse
!Fave!t " I definition of impulse
!F = m
!a = m
d!v
dt=d!p
dt# !!p =!F!t
!p ! m
!v
!! For single object….
"! If F = 0, then momentum conserved (p = 0)
!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!
•!For “system” of objects …
10/27/09
8
10/27/09
9
10/27/09
10
Momentum and Impulse
!Fave!t " I definition of impulse
!F = m
!a = m
d!v
dt=d!p
dt# !!p =!F!t
!p ! m
!v
!! For single object….
"! If F = 0, then momentum conserved (p = 0)
!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!
•!For “system” of objects …
10/27/09
11
!Fave!t " I definition of impulse
!F = m
!a = m
d!v
dt=d!p
dt# !!p =!F!t
Let’s estimate the average force during the collision
Club speed: 50 m/s
Assume that impulse is given after 5 cm
--> whiteboard
!Fave =
I
!t=1
!t
!Fdt
ti
t f
"
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
10/27/09
12
Elastic Collision in 1-Dimension
m1v1i + m2
v2i = m1
v1 f + m2
v2 f
1
2m1v1i
2+1
2m2v2i
2=1
2m1v1 f
2+1
2m2v2 f
2
Energy conserved (for elastic
collision only)
Linear momentum is conserved
Initial Final
Elastic Collision Conservation of Momentum
m1v1i + m2v2i = m1v1 f + m2v2 f
m1(v1i ! v1 f ) = m2 (v2 f ! v2i )
Conservation of Kinetic Energy
1
2m1v1i
2+
1
2m2v2i
2=
1
2m1v1 f
2+
1
2m2v2 f
2
m1(v1i
2! v1 f
2 ) = m2 (v2 f
2! v2i
2 )
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.
10/27/09
13
Is this an elastic collision?
v1i ! v2i = !(v
1 f ! v2 f )For elastic collision only:
10/27/09
14
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?
v1i ! v2i = !(v
1 f ! v2 f )Yes, the relative speeds
are approximately the same
before and after collision
For elastic collision only:
10/27/09
15
v1 f =
m1! m
2
m1+ m
2
v1i
v2 f =
2m1
m1+ m
2
v1i
Result:
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:
•! If pAi=0:
!PAi +
!PBi =
!PAf +
!PBf =
!PCM
!PBi =
!PCM
mBv!
Bi = mA + mB( )v!
f
10/27/09
16
!! Two stage process:
1. m collides with M, inelastically. Both M and m then move together with a velocity V
f
(before having risen significantly).
2. Both (m1 + m2) rise a height h, conserving energy E. (no non-conservative forces acting after collision)
Ballistic
Pendulum
What is the initial
velocity vli of the
projectile? Known quantities:
m1, m2, h
•! Stage 1: Momentum is conserved
Energy is not conserved
in x-direction:
!! Stage 2 (after the collision): Energy is conserved
Substituting for V gives:
Ballistic
Pendulum
m1v1i = m
1+ m
2( )Vf
K +U conserved :
1
2m
1+ m
2( )Vf
2= m
1+ m
2( )gh!Vf = 2gh
v1i = 1+
m2
m1
!
"#$
%&2gh
10/27/09
17
Fraction of kinetic energy lost is …………..
If m2<<m1: not much energy is lost
Of m2>>m1:almost all energy is lost
Ballistic
Pendulum
Wthermal =U ! Ki
= m1+ m
2( )gh !1
2m1v1i
2
= m1+ m
2( )gh !1
2m1
m1+ m
2
m1
"
#$%
&'
2
2gh
= m1+ m
2( )ghm1
m2
•! How much energy is dissipated?
Wthermal
Ki
=m1
m1+ m
2
Coefficient of restitution e=1/2
Inelastic collision
10/27/09
18
Coefficient of restitution
v1i ! v2i = !(v
1 f ! v2 f )
Perfectly elastic collision:
The coefficient of restitution is a
measure of the “inelasticity:
Elastic collision: e=1
Perfectly inelastic collision: e=0
e =v1 f ! v2 f
v1i ! v2i
Collisions or Explosions in Two Dimensions
y
x
before after
•! Ptotal,x and Ptotal,y independently conserved
*!Ptotal,x,before = Ptotal,x,after
*!Ptotal,y,before = Ptotal,y,after
Ptotal ,before
! "!!!!!!!= Ptotal ,after
! "!!!!!!
10/27/09
19
Explosions
“before” M
A
Which of these is possible?
A
B
both
B
“after”
Explosions
“before” M
A
Which of these is possible?
A (p appears conserved)
B (p not conserved in y direction)
both
B
“after”
10/27/09
20
Explosions
“before” M
A B
Which of these is possible?
A
B
neither
“after”
Explosions
“before” M
A B
Which of these is possible?
A (p not conserved in y direction)
B
neither
“after”
10/27/09
21
(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:
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)
v1 f
! "!,v2 f
! "!!
10/27/09
22
©2008 by W.H. Freeman and Company
10/27/09
23
•! Assuming –! Collision is elastic (KE is conserved)
–! No spin is imparted –! Balls have the same mass
–! One ball starts out at rest
Shooting Pool...
Shooting Pool •! Elastic collision means conservation of kinetic energy
if m1 = m2:
•! Conservation of momentum:
if
1
2m1v1i
2=p1i
2
2m1
=p1 f
2
2m1
+p2 f
2
2m2
!P1i +!P2i =!P1 f +!P2 f
!P2i= 0
!P1i =!P1 f +!P2 f
p1i
2= p
1 f
2+ p
2 f
2
P1 and p2
Form a right
angle!
10/27/09
24
•! Tip: If you shoot a ball spotted on the
“dot”, you have a good chance of
scratching !
Shooting Pool...
10/27/09
25
v
!
cm =m1v1
"!+ m
2v2
"!"
m1+ m
2
Subtract vcm from
all velocities
Collisions in the CM frame
The transformation to the cm frame is not
necessary, but it is often convenient to switch to
the CM frame
10/27/09
26
Rocket equation
•! The mass is changing
•! Thrust is generated by impulse of exhaust of mass
with velocity v: vdm
10/27/09
27
The Saturn V
The Saturn V rocket:
•! 111 m tall
•! 10m diameter
•! 3000 tons at start
•! Thrust: 34 MN
Rocket equation
•! Mass change
•! Thrust
•! Weight: F = M g
•! Rocket equation:
•! Integration yields:
dM
dt= R = const
M (t) = M0! Rt
dM
M
dM
dt!u
ex=d(M !u
ex)
dt=dP
dt= "F
thrust
Mdv
dt= R !uex " Mg
dv
dt=
R !uex
M0" Rt
" Mg
v = uex! ln
M0
M0" Rt
#
$%&
'(" gt
10/27/09
28
Variable mass
•! Newton’s second law for continuously variable
mass:
•! Where
is the velocity of impacting material relative to
object with mass M at a given time.
!Fnet ,ext
+dM
dt
!vrel= M
dv
!
dt
v
!
rel = u
!! v
!
10/27/09
29