Transcript Concept Questions

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
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
p  mv
SI units
[kg  m  s ]  [N  s]
• Change in momentum
-1
p  mv
tf
• Impulse
I   Fdt
ti
• SI units
[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 dpr
dv
r
F
 m  ma
dt
dt
tf
tf
tf
r
r
r
dp
r
r r
r
I   F dt  
dt   d p  p  p(tf )  p(ti )
dt
t
t
t
i
When
then
i
i
dp
F
0
dt
then
r
r
p(tf )  p(ti )
p  mv  constant (vector)
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 Nonidentical Carts Kinetic Energy
Consider two carts, of masses m and 2m, at rest on an air
track. If you push one cart for 3 s 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: 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.
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  0.11 s
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 the
velocity is
Vy (t)  I y [ti ,t] / m  (199 N  s)(9.80 m  s-2 )/(911 N)  2.14 m  s-1
System of Particles:
Center of Mass
Position and Velocity of Center of Mass
Total mass for discrete or continuous
body (mass density ρ)
i N
msys   mi 
i 1

 dV
body
Position of center of mass
R cm
1

msys
i N
1
mi ri 

msys
i 1

r dV
body
Velocity of center of mass
1
Vcm 
msys
i N
1
m
v


i i
msys
i 1

body
v dV 
psys
msys
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 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 at t = 0.
System of Particles:
Internal and External Forces,
Center of Mass Motion
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
jN
Fint,i   Fi , j
j 1
i j
• The total internal force is the sum of the total internal force on each
particle
jN
i 1
total
Fint   Fint,i   Fi , j
i 1
j 1
i j
• Newton’s Third Law: internal forces cancel in pairs
Fi , j  Fj ,i
• So the total internal force is zero
Finttotal  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
F
total
F
total
ext
F
total
int
F
total
ext
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
p sys   p i
i 1
Total force changes the momentum of the system
iN
F
total
  Fi
iN
total
i 1
dpsys
dpi


dt
i 1 dt
Total force equals total external force
total
Ftotal  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 system
total
ext
F

dpsys
dt
System of Particles: Newton’s
Second and Third Laws
The total momentum of a system remains
constant unless the system is acted on by
an external force
total
ext
F

dpsys
dt
System of Particles:
Translational Motion of the Center of
Mass
Translational Motion of the Center of
Mass
Momentum of system
psys  msys Vcm
System can be treated as point mass located at center of
mass. External force accelerates center of mass
dpsys
dVcm
F 
 msys
 msys A cm
dt
dt
Impulse changes center of mass momentum
total
ext
tf
total
I   Fext
dt  psys  msys (Vcm (tf )  Vcm (ti ))
ti
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
Concept Question: Pushing a
Baseball Bat Recall Issue with
Phrasing
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
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?
System of Particles:
Conservation of Momentum
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.
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  p sys
t
• If the collision time is approximately zero,
tcol
0
then the change in momentum is approximately zero.
p system  0
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?