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Momentum and Impulse
8.01
W07D1 Fall 2006
Momentum
and
Impulse
Obeys a conservation law
Simplifies complicated
motions
Describes collisions
Basis of rocket propulsion
& space travel
Quantity of Motion
DEFINITION II Newton’s Principia
The quantity of motion is the measure of the
same, arising from the velocity and quantity of
matter conjointly.
The motion of the whole is the sum of the
motions of all the parts; and therefore in a body
double in quantity, with equal velocity, the motion
is double; with twice the velocity, it is quadruple.
Momentum and Impulse:
Single Particle
• Momentum
r
r
p  mv
SI units
[kg  m  s-1 ]  [N  s]
• change in momentum
• average force
• impulse
SI units
r
r
p  mv
r
r
Fave  p / t
r r
I  Fave t
[N  s]
tf
r
r
I   Fdt
ti
Impulse Demo
To explain the demo and the example we
need concepts of momentum and impulse
An explanation is easier if we also use the
idea of center of mass
Jumping: Non-Constant Force
• Plot of total external force vs. time for Andy jumping
off the floor. Weight of Andy is 911 N.
Jumping: Impulse
• Shaded area represents impulse of total force
acting on Andy as he jumps off the floor
F
t f 1.23 s
I[ti ,t f ] 

ti  0.11 s
total
r
r
(t)  N(t)  Fgrav
r total
F (t) dt  199 N  s
Newton’s Second Law
The change of motion is proportional to the motive
force impresses, and is made in the direction of
the right line in which that force is impressed,
r
r total dpr
dv
r
F 
m
 ma
dt
dt
tf
r
r total
r r
r
I   F dt  p  p(tf )  p(ti )
ti
When
then
r dpr r
F
0
dt
r
r
p  mv  constant (vector)
Concept Question: Impulse
and Change in Momentum
Identical constant forces push
two identical objects A and B
continuously from a starting
line to a finish line. If A is
initially at rest and B is initially
moving to the right,
1. Object A has the larger
change in momentum.
2. Object B has the larger
change in momentum.
3. Both objects have the same
change in momentum
4. Not enough information is
given to decide.
Concept Question: Same
Momentum, Different Masses
Suppose a ping-pong ball and a bowling ball are
rolling toward you. Both have the same momentum,
and you exert the same force to stop each. How do
the distances needed to stop them compare?
1. It takes a shorter distance to stop the ping-pong ball.
2. Both take the same distance.
3. It takes a longer distance to stop the ping-pong ball.
Internal Force on a System of
N Particles
• The total internal force on the ith particle is sum of the
interaction forces with all the other particles
r
Fint,i 
j N
r
 Fi, j
j 1
i j
• The total internal force is the sum of the total internal
force on each particle
j N
r total i1 r
F   Fint,i 
int
i1
r
 Fi, j
j 1
i j
• Newton’s Third Law: internal forces cancel in pairs
r
r
Fi, j  Fj,i
• So the total internal force is zero
r total r
Fint  0
Total Force on a System of N
Particles is the External Force
The total force on a system of particles is the sum of the
total external and total internal forces. Since the total
internal force is zero
r total r total r total r total
F  Fext  Fint  Fext
External Force and Momentum
Change
The total momentum of a system of N particles is defined as the
sum of the individual momenta of the particles
i N
r
r
psys   pi
i1
Total force changes the momentum of the system
r total
r total
F   Fi
i N
i1
r
r
dpsys
dp i


dt
i1 dt
i N
Total force equals total external force
r total r total
F  Fext
Newton’s Second and Third Laws for a system of particles: The
total external force is equal to the change in momentum of the
r
system
dp
r
total
Fext

sys
dt
Position and Velocity of Center
of Mass
• Total mass
i N
msys   mi
i1
• Position of center of mass
r
1 i N r
R cm 
mi ri

msys i1
• Velocity of center of mass
r
i N
p
r
1
r
sys
Vcm 
m
v


msys i1 i i msys
Translational Motion of the
Center of Mass of Rigid Body
• Momentum of system
r
r
psys  msys Vcm
• System can be treated as point mass located at
center of mass. External force accelerates
center of mass r
r
r
dpsys
r total
dVcm
Fext 
 msys
 msys Acm
dt
dt
• Impulse
changes center of mass momentum
tf
r
r total
r
r
r
I   Fext dt  psys  msys ( Vcm (tf )  Vcm (ti ))
ti
Jumping: Center of Mass Velocity
• Plot of center of mass
velocity vs. time starting
from rest Vcm, y (ti )  0
• Impulse for time interval
[ti, t] equals change in
momentum of system.
t
I y [ti ,t] 

Fytotal (t) dt  Psys,y  m(Vcm, y (t)  Vcm, y (ti ))  mVcm, y (t)
ti  0.11 s
• Center of mass velocity at
time t is then
Vcm, y (t)  I y [ti ,t] / m
Jumping: Center of Mass Velocity
• When Andy leaves the
ground, the impulse is
I y [0.11 s,1.23 s]  199 N  s
• So the center of mass
velocity is
Vcm, y (t)  I y [ti ,t] / m  (199 N  s)(9.80 m  s-2 )/(911 N)  2.14 m  s-1
What was the magnitude of the displacement of
Andy’s center of mass after he left the floor?
2
Ycm  Vcm,y
(t f ) / 2g  (2.14 m  s-1 )2 / (2)(9.80 m  s-2 )  0.23 m
What is the
shape of the
trajectory of
the center of
mass of this
coin?
CM moves as though all external
forces on the system act on the
CM
so the jumper’s cm follows a parabolic trajectory of a
point moving in a uniform gravitational field
Fosbury Flop
Concept Question: Pushing
Baseball Bat
1
2
3
The greatest acceleration of the center of mass
will be produced by pushing with a force F at
1. Position 1
2. Position 2
3. Position 3
4. All the same