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Chapter 8
Conservation of Linear Momentum
Linear momentum;
Momentum conservation
Impulse
Total kinetic energy of a system
March 9, 2010

Conservation of Linear Momentum
• Definition of linear momentum,
p
p  mv
Linear momentum is a vector (decompose to x,y,z directions).
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
Momentum  force, as if Kinetic energy  work

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:
uuur
r
r
Psys   mi vi   pi
i
i
From Newton’s second law, we obtain
r
 Fext  Fnetext  
i
i
r
dPsys
dt
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
F
ext
i
 0 then
uuuuur
r
r
Psys   mi vi  MvCM  const
i
Momentum is always conserved when no net “external” force.
(even if “internal” forces are non-conservative).
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)
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
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 speed?
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 speed?
mv is the same for each block, so smaller mass has larger speed.
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 speed?
mv is the same for each block, so smaller mass has larger speed.
Is kinetic energy conserved?
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 speed?
mv is the same for each block, so smaller mass has larger speed.
Is kinetic energy conserved?
NO! K was 0 before, it is greater after the explosion.
(internal non-conservative force does some work.)

Momentum
and
Impulse
Momentum
p  mv

For single object….
Fave t  I
F  ma
 If F = 0, then momentum conserved (p = 0)
definition of impulse
dv dp
m

 p  Ft
dt dt
• 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!
Elastic Collision in 1-Dimension
Initial
Final
Linear momentum is conserved
m1v1i  m2 v2i  m1v1 f  m2 v2 f
Energy conserved (for elastic
collision only)
1
1
1
1
2
2
2
m1v1i  m2 v2i  m1v1 f  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
2
2
2
m1v1i  m2 v2i  m1v1 f  m2 v22 f
2
2
2
2
m1 (v1i2  v12f )  m2 (v22 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.
Is this an elastic collision?
For elastic collision only:
v1i  v2i  (v1 f  v2 f )
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
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: after the impact,
speed of club and
speed of golfball
Problem solving strategy:
- Momentum conservation
- Energy conservation (or
use the derived equation
for relative velocities)
Result:
m1  m2
v1 f 
v1i
m1  m2
2m1
v2 f 
v1i
m1  m2
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
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
Kinetic energy of a system of particles:
Where
in terms of the CM velocity and relative velocity to the CM.