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Transcript
```Physics 201
8: Linear Momentum
and Collisions
•Linear Momentum and its Conservation
•Impulse and Momentum
•Collisions
•Elastic and Inelastic Collisions in One Dimension
•Two Dimensional Collisions
•The Center of Mass
•Motion of a System of Particles
•Rocket Propulsion
Linear Momentum and its Conservation
Linear momentum
p = mv
Newtons Second law
dv dp
F tot = ma = m
=
dt
dt
(if the mass is constant 
closed system
)
dp
\ F tot = 0 
=0
dt
p = CONSTANT  Dp = 0
F12
1
2
F21
By Newtons Third Law
F12 =  F21
If we consider the two objects as ONE composite
object, then the total force on the composite object
F tot = F 12  F21 = 0
 DP tot = 0 = Ptot, f  P tot,i  P tot, f = P tot,i
P tot = P1  P 2
P1, f  P 2, f = P 1,i  P 2,i
Impulse and Momentum
dp= Fdt


 dp = Dp= Fdt =I
I is called the Impulse of the force
I=

F dt =area under force versus time gra
Dp I
F= =
Dt Dt
Collisions
Basic Assumption
Forces during the collision dominate all
other forces present, so only they need
be considered
This assumption implies that only contact
forces between the objects are
important , i.e. that the system is closed
This in turn implies that the Total Linear
Momentum of the system is conserved
Dptot = 0
Perfectly Inelastic Collisions
The two particles stick together after the
collision
 v1f = v 2f
Thus using Dptot = 0 in one dimension
m1 v1i  m 2v 2i = m1 v f m 2v f = v f (m1  m 2)
m1 v1i  m 2v 2i
 vf =
(m1  m2 )
Elastic Collisions
both Momentum ANDKinetic Energy are conserved
m 1 v 1i  m 2 v 2i = m1 v1f  m 2v 2f
1
1
1
1
m1 v1i2  m 2 v 22i = m 1v 1f2  m 2 v 22f
2
2
2
2
 1 m v 2  v 2 = 1 m v 2  v 2
1 ( 1i
1f )
2 ( 2f
2i )
2
 2
 m1 (v1i  v 1f )(v1i  v1f ) = m 2 (v 2f  v 2i )(v 2f
From conservation of total momentum(1)
eq.

m v  v1f ) = m 2 (v 2f  v 2i )
 1 ( 1i
\ (v1i  v 1f ) = (v 2f  v 2i )

 v1i  v 2i =  (v1f  v 2f )
(1)
(2 )


 v 2i )





Two Dimensional
Collisions


d=impact parameter
use Dp = 0 for all collisions
and DK = 0 for elastic collisions
where p = mv
1
2
and K = m v
2
can always choose a frame of reference
in which one of the particles is at rest
and the moving particles trajectory to be
the x - axis
The Center of Mass
In a system of two or more particles,
the system
moves as though all the mass were concentrated at
a point.That point is called the center of mass
.
m r 1
r =
=
mr

M
m

m (x i  y j  z k) 1
x i  y j z k =
m (x i  y j  z k)
=

M
m

k k
k =1,n
cm
k k
k =1,n
k
k =1, n
k
k
k
k
k =1,n
cm
cm
cm
k
k
k =1, n
k =1,n
k
k
k
Center of mass of extended objects
1
rcm =
M

1
rdm=
M
object

rr(r)dr; r = (x,y,z);; dr =dxdydz
object
dm
r(r) = ; M=
dV

dm
object
c
1
xcm =
M

1
xr(r)dr; ycm =
M
object

1
yr(r)dr; zcm =
M
object

zr(r)dr
object
Motion of a System of Particles
drcm d 1
v cm =
= 
dt
dt M

 1
1
m k r k =
mkvk =
M
M

k
k
ptot
i.e. v cm =
 ptot = M vcm
M
similarly
d 1
a cm =
= 
dt
dt M
dvcm


k
 1
m k v k =
 M


k
1
mkak =
M
Ftot
i.e. a cm =
 F tot = M acm
M

ptot
pk =
M

F tot
k
k
Fk =
M
The effect of an external
force is to change the
momentum of the entire
system. If the external
force is zero the system
maintains a zero or
constant velocity and the
total momentum of the
system is conserved
Rocket Propulsion
A rocket is a system with mass M moving at a
velocityv . When the gases are expelled with exhaust
velocity v e , then the rocket changes its velocity by
d v . Thus through conservation of momentum
M dv =  vedM
M v i = ( M - d M ) v f  d M v e 
M d v =  v d M if M > > d M 


e


dv =  ve

dM
M
 M 
 D v = v e ln  i 
 M f 
Thrust:
F = M
dv
dM
=  ve
dt
dt
```
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