Transcript Document

Momentum
The (linear) momentum of a particle is
defined as the product of the particle's
mass and velocity


p  mv
m
v
p
Momentum is a quantity describing the
motion of the particle.
The relation between kinetic energy and momentum



mv 2
mv 2
p2
p2

K


2m
2
2m
2m
Newton's second law II
(modern version)
In an inertial reference frame, the rate of change in the
momentum of a particle is equal to the net force exerted on
the particle


dp
Fnet 
dt
classical (non-relativistic) version:





dp dmv 
dv
 ma
Fnet 
m

dt
dt
dt
Impulse
If an object interacts with a particle over a certain time
interval (t1, t2), the integral
 t2 
I   Ft dt
is called the impulse.
t1
The average force over a certain time interval is directly
related to the impulse of this force over this time interval
t2 
Fdt

 t1

I
 t  Fav  t
t
Impulse - Momentum theorem
In an inertial reference frame, the change in the
momentum of a particle is equal to the net impulse
(impulse of the net force)
 t2 
p   Fnet t dt
t1

p2


 t2 

p  pt 2   pt1    dp   Fnet t dt

p1
t1
Puzzle. The length of a rifle is l =1m. When a shot is fired, a
10 g projectile is accelerated to a speed of v = 500 m/s in
time t = 5 ms. What is the average force on the projectile?
Why are the results
different?
Solution 1
mv 2
I
 0  K  Fav  l
2
from which
Solution 2
 
II
mv  0  I  Fav
 t
work-energyfrom which
mv 2
I
Fav =
 1250theorem
Nimpulse-momentum
mv
II
2l
Fav

 1000N
theorem
t
total value of a physical quantity
(for a system of particles)
If a certain physical quantity is defined for particles,
we define the total of this physical quantity for a
system of particles as the sum of the quantity over all
the particles constituting the system.
m1
m2
(total) mass of a system M   m i
p1
i
p2


(total) momentum of a system P   pi
m3
i
(total) kinetic of a system
K   Ki
i
p3
Conservation of Momentum
If a system of particles is isolated, the total momentum
of the system is constant

Pt   const.
because

dP

dt




d 
dpi
  Fnet,i    Fij  0
 pi  
i j i
dt i
i dt
i
Collisions
•
•
•
inelastic
elastic
(maximum loss in
total kinetic energy)
(no loss in total
kinetic energy)
Mechanical interactions between bodies over a finite time
are called collisions.
Collisions do not affect the total momentum of the system.
In case an external force is applied but the collision takes
voyagerof the
place in a time period negligible for the effects
external force, the external force can be ignored.
If either before or after a collision the
particles have equal velocities, the
collision is perfectly inelastic.



m1 v1i  m 2 v 2i  m1  m1 v f
If the total kinetic energy does not
change, the collision is elastic.




m1v1i  m2 v 2i  m1v1f  m2 v 2f
m1v12i m 2 v 22i m1v12f m 2 v 22f



2
2
2
2
collisions
Puzzle. What is the angle between the directions of motion of
momentum
billiard balls after the collision ?
conservation



(1) v1i  v1f  v2f
2
2
2



v1fv2 mv 2f
(2) mv1v2i1iv12m
f
2f
2
2
2
by substitution
90°
j2
j1
v2f
v1f


 


v12f  v22f  2v1f  v2f  v12f  v22f kinetic energy
conservation
therefore
 
v1f  v2f  0
v1i
Center of Mass
z
dm
y
r
x
Center of mass is a (abstract) point
of position related to the
distribution of the particles
according to the following formula
1 

rcm 
 r  dm
MM
where M is the (total) mass of the system.
For discrete systems
1


rcm   mi ri
M i
Example. Three identical particles
z
[0,0,1]
[0,1,0]
y
[1,0,0]
x
rcm
1 1 1 
1

 ( m1,0,0  m0,1,0  m[0,0,1] )   , , 
3 3 3
3m
Example. A thin uniform rod
y
dx
x
x
L
should be the
function.
z
L M
L M
1 L
1 L
1 L M


r
dm
r

Adx


x
dx
,
0
dx
,
0
dx

rcm 







M0
M0
M 0 L

0 L
0 L
1 What
M L2if the  L

,
0
,
0
 rod

,
0
,
0




M  L were
2 not  2
uniform?
useful theorems
x cm
1

 xdm  0
M object
dm
dm
r’
r
x
The center of mass of a homogenous object
must lie on the axis of symmetry.
The position of the center of mass of two
objects is related to the centers of mass of
each object.
 1
1


1 





rcm 
r

dm


 r  dm   r  dm   M1rcm,1  M 2rcm, 2 

M system
M  M1
M2
 M
Newton's second law III
(for a system of particles)




dpi
dP d 
 

    Fi, j  Fi,ext( net)    Fi,ext( net)
  pi  
dt dt i
i dt
i  j
 i
dP
dt
P
Fext
In an inertial reference frame, the
rate of change in the (total)
momentum of a system of particles
is proportional to the total net
external force exerted on the
system.


dP
F

 ext dt
Newton's second law IV
(for a system of particles)


d2  1

d 2ri
d 2rcm

Macm  M
 M 2   miri    mi 2 
2
dt  M i
dt
dt
 i


 


  miai     Fi, j  Fi,ext( net)    Fi,ext( net)
i  j
i
 i
acm
boomerang
Fext
In an inertial reference frame, the
acceleration of the center of mass of a
system of particles is proportional to
the total net external force exerted on
the system.


Macm   Fi,ext( net)
i
Total Momentum and the Center of Mass



d

drcm
d



dri
Mv


M
r

m
v

M

  miri 
P   i i  mi
cm
cm
dt
dt
dt dt i
i
i
The (total) momentum of a system of particles is related to
the velocity of the center of mass of the system.


P  Mvcm
the Moon
work-energy theorem III
(for translational motion)
dWext

2

 Mv cm




dvcm  drcm
  dK T
 Mv cm  dv cm  d
 Fext  drcm  Macm  drcm  M
dt
 2 
T

drcm
The translational kinetic energy of
a system of particles is defined as
1
KT  Mv 2cm
2
W
For a system of particles, the work done on the center
of mass by the net external force is equal to the change
in the translational kinetic energy of the system.
dWext  dKT or Wext  KT
work-energy theorem IV
(the total kinetic energy)
The (total) work done by all the forces
(external and internal) on the particles of
the system (to which they are applied
only) is equal to the change in the total
kinetic energy of the system
dW  dKtot or W  Ktot
T

dr
W
energy
total gravitational potential energy
(near the surface)
U
hcm
U0 = 0
U   U k   mk gh kf  h kf  
k
k
 g  mk h kf   mk h ki  
k

k
 gMh cm,f  Mh cm,i   Mgh cm
Mghcm