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Momentum and Impulse
8.01
W06D2
Associated Reading Assignment:
Young and Freedman: 8.1-8.5
Announcements:
No Math Review Night this Week
Next Reading Assignment W06D3:
Young and Freedman: 8.1-8.5
Today’s Reading Questions
(a) Explain the difference between the concepts of
impulse and work.
(b) Explain why the total force on a system of particles
is only due to the sum of external forces.
(a) Under what conditions does the center of mass of
a system of particles move with constant velocity?
Momentum
and Impulse
Obeys a conservation law
Simplifies complicated
motions
Describes collisions
Basis of rocket propulsion &
space travel
Newton’s Definition: 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
p  mv
• Momentum
SI units
[kg  m  s ]  [N  s]
• Change in momentum
-1
p  mv
tf
• Impulse
I   Fdt
ti
• SI units
[N  s]
Integral Version of Newton’s
Second Law
For an object with fixed mass
r
r
r
dv dp
r
F  ma  m 
dt dt
Then impulse produces a change in momentum
tf
r
r
r
dp
r
r r
r
I   F dt  
dt   d p  p  p(tf )  p(ti )
dt
t
t
t
tf
tf
i
i
i
“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”
Concept Question: Pushing
Identical Carts
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: Pushing
Identical Carts Answer
Answer 1: Both objects have the
same mass, are pushed the same
distance, by the same constant force,
so they have the same acceleration.
Since object A started from rest, an
object B has an initial non-zero
speed, object A needs a larger time
interval to reach the finish than the
corresponding time interval for object
B Therefore the impulse on object A
is larger than the corresponding
impulse on object B. Hence object A
has a larger change in momentum.
Momentum, Kinetic Energy, and
Work: Single Particle
Kinetic energy and momentum for a single particle
are related by
1 2 p2
K  mv 
2
2m
Change in kinetic energy and work
tf
2
2
p
r r
p
1 2 1 2
f
W   F  d r  K  mv f  mvi 
 i
2
2
2m 2m
t
i
Concept Question: Pushing Carts
Consider two carts, of masses m and 2m, at rest on an air
track. If you push one cart for 3 seconds and then the other
for the same length of time, exerting equal force on each,
the kinetic energy of the light cart is
1) larger than
2) equal to
3) smaller than
the kinetic energy of the heavy car.
Concept Question: Pushing Carts
Answer 1. The kinetic energy of an object can be written as
1 2 p2
K  mv 
2
2m
Because the impulse is the same for the two carts, the
change in momentum is the same. Both start from rest so
they both have the same final momentum. Since the mass
of the lighter cart is smaller than the mass of the heavier
cart, the kinetic energy of the light cart is larger than the
kinetic energy of the heavy cart.
Concept Question: Stopping Distances
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.
Concept Question: Stopping Distances
Answer 3. The kinetic energy of an object can be written as K = p2/2m.
Because the ping pong ball and the bowling ball have the same
momentum, the kinetic energy of the less massive ping pong ball is
greater than the kinetic energy of the more massive bowling ball. You
must do work on an object to change its kinetic energy. If you exert a
constant force, then the work done is the product of the force with the
displacement of the point of application of the force. Since the work
done on an object is equal to the change in kinetic energy, the ping
pong ball has a greater change in kinetic energy in order to bring it to a
stop, so the you need a longer distance to stop the ping pong ball.
Demo:
Jumping Off the Floor
with a Non-Constant Force
Demo Jumping: Non-Constant Force
• Plot of total external force vs. time for Andy jumping
off the floor. Weight of Andy is 911 N.
Demo Jumping: Impulse
• Shaded area represents impulse of total force
acting on Andy as he jumps off the floor
F total (t )  N(t )  Fgrav
t f 1.23 s
I[ti , t f ] 

ti  0.11s
F total (t ) dt  199 N  s
Demo Jumping: Height
When Andy leaves the
ground, the impulse is
I y [0.11 s,1.23 s]  199 N  s
So the y-component of his
velocity is
vy, f  I y [ti ,t f ] / m  (199 N  s)(9.80 m  s-2 )/(911 N)  2.14 m  s-1
Andy jumped
(1 / 2)mvy,2 f  mgh  h  vy,2 f / 2g  23.3 cm
System of Particles:
Center of Mass
Position and Velocity of Center of Mass
Mass for collection of discrete bodies (system):
i N
msys   mi
i1
Momentum of system:
i N
r
r
psys   mi v i
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
1
Vcm 
msys
i N
r
psys
i1
msys
r
 mi v i 

r
r
psys  msys Vcm
Table Problem: Center of Mass of
Rod and Particle
A slender uniform rod of length d and
mass m rests along the x-axis on a
frictionless, horizontal table. A particle of
equal mass m is moving along the x-axis at
a speed v0. At t = 0, the particle strikes the
end of the rod and sticks to it. Find a vector
expression for the position of the center of
mass of the system at t = 0.
System of Particles:
Internal and External Forces,
Center of Mass Motion
System of Particles: Newton’s
Second and Third Laws
The momentum of a system remains
constant unless the system is acted on by
an external force in which case the
acceleration of center of mass satisfies
r
Fext 
r
dpsys
dt
 msys
r
dVcm
dt
r
 msys A cm
Force on a System of N Particles is
the External Force
The force on a system of particles is the external
force because the internal force is zero
r r
r
r
F  Fext  Fint  Fext
Internal Force on a System of N
Particles is Zero
• The internal force on the ith particle is sum of the interaction forces
with all the other particles
jN
Fint,i   Fi , j
j 1
i j
• The internal force is the sum of the internal force on each particle
i1 r
r
Fint   Fint,i 
j N
r
 Fi, j
i1
j 1
i j
• Newton’s Third Law: internal forces cancel in pairs
Fi , j  Fj ,i
• So the internal force is zero
r
r
Fint  0
External Force and Momentum
Change
The momentum of a system of N particles is defined as the sum of the
individual momenta of the particles
i N
r
r
r
psys   pi  msys Vcm
i1
Force changes the momentum of the system
r
r
dpsys
r
r
dpi
F   Fi  

dt
i1
i1 dt
i N
i N
Force equals external force
r r
F  Fext
Newton’s Second and Third Laws for a system of particles: The
external force is equal to the change inr momentum of the system
r
Fext 
r
dpsys
dt

d(msys Vcm )
dt
r
 msys Acm
External Forces and Constancy of
Momentum Vector
The external force may be zero in one direction but not
others
The component of the system momentum is constant in
the direction that the external force is zero
The component of system momentum is not constant in
a direction in which external force is not zero
Table Problem: Center of Mass of
Rod and Particle Post- Collision
A slender uniform rod of length d and
mass m rests along the x-axis on a
frictionless, horizontal table. A particle
of equal mass m is moving along the xaxis at a speed v0. At t = 0, the particle
strikes the end of the rod and sticks to
it. Find a vector expression for the
position of the center of mass of the
system for t > 0.
Demo : Center of Mass trajectory B78
http://tsgphysics.mit.edu/front/index.php?page=
demo.php?letnum=B%2078&show=0
Odd-shaped objects with their centers of mass
marked are thrown. The centers of mass travel in a
smooth parabola. The objects consist of: a squash
racket, a 16” diameter disk weighted at one point
on its outer rim, and two balls connected with a
rod. This demonstration is shown with UV light.
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
Table Problem: Exploding Projectile
Center of Mass Motion
An instrument-carrying projectile of mass m1 accidentally explodes at the
top of its trajectory. The horizontal distance between launch point and the
explosion is x0. The projectile breaks into two pieces which fly apart
horizontally. The larger piece, m3, has three times the mass of the smaller
piece, m2. To the surprise of the scientist in charge, the smaller piece
returns to earth at the launching station.
a) How far has the center of mass of the system traveled from the launch
when the pieces hit the ground?
b) How far from the launch point has the larger piece graveled when it first
hits the ground?
Concept Question: Pushing a
Baseball Bat
1
2
3
The greatest instantaneous 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
Concept Question: Pushing a
Baseball Bat
1
2
3
Answer 4. The external force is equal to the total
mass times the instantaneous acceleration of the
center-of-mass. It doesn’t matter where the external
force acts with regards to the center-of-mass
acceleration.
Table Problem: Landing Plane and
Sandbag
A light plane of mass 1000 kg makes an emergency landing on a short
runway. With its engine off, it lands on the runway at a speed of 40 ms-1. A
hook on the plane snags a cable attached to a sandbag of mass 120 kg and
drags the sandbag along. If the coefficient of friction between the sandbag
and the runway is μ = 0.4, and if the plane’s brakes give an additional
retarding force of magnitude 1400 N, how far does the plane go before it
comes to a stop?
Strategy: Momentum of a System
1. Choose system
2. Identify initial and final states
3. Identify any external forces in order to determine whether
any component of the momentum of the system is
constant or not
i) If there is a non-zero total external force:
r
dpsys
r total
Fext 
dt
ii) If the total external force is zero then momentum is
constant
p sys,0  p sys,f
Modeling: Instantaneous
Interactions
• Decide whether or not an interaction is instantaneous.
• External impulse changes the momentum of the
system.
I[t , t  tcol ] 
t tcol

Fext dt  (Fext )ave tcol  psys
t
• If the collision time is approximately zero,
tcol
0
then the change in momentum is approximately zero.
p system  0