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Transcript
Chapter 8
Momentum, Impulse,
and Collisions
Goals for Chapter 8
• To determine the momentum of a particle
• To add time and study the relationship of impulse
and momentum
• To see when momentum is conserved and examine
the implications of conservation
• To use momentum as a tool to explore a variety of
collisions
• To understand the center of mass
• To study rocket propulsion
Introduction
• If you watch a football
game, you’ll see collisions,
tackles, many men colliding
at once, maybe just two in
an open area. Are these
situations different?
• Newton told us the forces
result in acceleration of a
mass. We will now study
two new points of view—
momentum and impulse.
8.1 How does momentum relate to mass and velocity?
• Understanding
momentum begins
with the simple
relationship that
momentum is equal
to mass multiplied by
velocity.
p x  mv x
p y  mv y
p z  mv z
The units of the magnitude of
momentum are units of mass times
speed: kg∙m/s. The plural of momentum
is “momenta.”
Don’t forget that momentum is a vector
Newton’s 2nd Law in terms of Momentum
 dp
F 
dt
• The net force (vector sum of all forces) acting on a
particle equals the time rate of change of momentum
of the particle.
• According to the equation, a rapid change momentum
requires a large net force, while a gradual change in
momentum requires less net force.
Impulse
• The impulse, denoted by J, is defined to be the product of
the net force F, acting on an object for a time period ∆t:
 
J  F  t
(when force is constant)
 t2 
J   Fdt
(General Definition of Impulse)
t1
• Impulse is a vector quantity; its direction is the same as the
net force ∑F. The SI unit of impulse is the newton-second
(N∙s). Because 1 N = 1 kg∙m/s2, an alternative set of unit
for impulse is kg∙m/s, the same as the unit of momentum.
The Impulse-Momentum Theorem
 

J  p 2  p1
The change in momentum of a particle during a time
interval equals to the impulse of the net force that
J
acts on the particle during that interval.
• the impulse-momentum theorem is valid even
when the net force varies with time.
The area under in F-t graph equals to J = ∆p
J and p are both vector quantities
Compare momentum and kinetic energy
• The impulse–
momentum
relationship depends
on time of an impact
while the work-energy
theorem focuses on the
distance of force
application.
Example 8.1 Momentum versus kinetic energy
The iceboats have masses m and 2m, and the wind exerts the same
constant horizontal force F on each ice boat. The two iceboats start
from rest and cross the finish line a distance s away.
1. Which iceboat crosses the finish line with greater kinetic energy?
2. Which iceboat crosses the finish line with greater momentum?
Example 8.2 A ball hits a wall
Suppose you throw a ball with a mass of 0.40 kg against a brick wall.
It hits the wall moving horizontally to the left at 30 m/s and rebounds
horizontally to the right at 20 m/s.
a. Find the impulse of the net force on the ball during its collision
with the wall.
b. If the ball is in contact with the wall for 0.010 s, find the average
horizontal force that he wall exerts on the ball during the impact.
Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley
Example 8.3 Kicking a soccer ball
• A soccer ball has a mass of 0.40 kg. initially, it is moving to the
left at 20 m/s, but then it is kicked and given a velocity at 45o
upward and to the right, with a magnitude of 30 m/s. Find the
impulse of the net force and the average net force, assuming a
collision time ∆t = 0.010 s.
Test Your Understanding 8.1
Rank the following situations according to the magnitude of the
impulse of the net force, from largest value to smallest value. In each
situation a 1000-kg automobile is moving along a straight east-west
roads
i. The automobile is initially moving east at 25 m/s and comes to a
stop in 10 s.
ii. The automobile is initially moving east at 25 m/s and comes to a
stop in 5 s.
iii. The automobile is at rest, and a 2000 N net force toward is
applied to it for 10 s.
iv. The automobile is initially moving east at 25 m/s, and a 2000 N
net force toward is applied to it for 10 s
v. The automobile is initially moving east at 25 m/s. Over a 30 s
period, the automobile reverses direction and ends up moving
west at 25 m/s.
8.2 Conservation of momentum
• If the vector sum of the external forces on a system is
zero, the total momentum of the system is constant. It
is a direct consequence of Newton’s 3rd law.
CAUTION conservation of momentum means
conservation of its components
If the vector sum of the
external forces on the system
is zero, then Px, Py, and Pz are
all constant.
Conservation of momentum vs.
conservation of mechanical energy
• Conservation of momentum is more
general than the principle of conservation
of mechanical energy.
• Mechanical energy is conserved only
when the internal forces are
conservative but conservation of
momentum is valid even when the internal
forces are not conservative.
Example 8.4 Recoil of a rifle
• A marksman holds a rifle of mass mR = 3.00 kg loosely in his
hands, so as to let it recoil freely when fired. He fires a bullet of
mass mB = 5.00 g horizontally with a velocity relative the ground of
vRx = 300 m/s. what is the recoil velocity vRx of the rifle? What are
the final momentum and kinetic energy of the bullet? Of the rifle?
Example 8.5 Collision along a straight line
Two gliders move toward
each other on a frictionless
linear air track. After they
collide, glider B moves away
with a final velocity of +2.0
m/s. what is the final velocity
of glider A? How do the
changes in momentum and in
velocity compare for the two
gliders?
Example 8.6 Collision in a horizontal plane
Two battling robots sliding
on a frictionless surface.
Robot A, with mass 20 kg,
initially moves at 2.0 m/s
parallel to the x-axis. It
collides with robot B, which
has mass 12 kg and is
initially at rest. After the
collision, robot A is moving
at 1.0 m/s in a direction that
makes an angle α = 30o with
its initial direction. What is
the final velocity of robot B?
Check your understanding
•
A spring-loaded toy sits at rest on a horizontal
frictionless surface. When the spring releases, the toy
breaks into three equal-mass pieces, A,B, and C,
which slide along the surface, piece A moves off the
negative x-direction, while piece B moves off in the
negative y–direction.
1. What are the signs of the velocity components of piece
C?
2. Which of the three pieces is moving the fastest?
8.3 Momentum Conservation and Collisions
• In physics, we broaden the meaning the term collision to
include any strong interaction between bodies that lasts
a relatively short time.
If the forces between the bodies are much larger than
external forces, as is the case in most collisions, we can
neglect the external forces and treat the bodies as an
isolated system, the momentum is conserved during a
collision.
Elastic Collision
• If the forces between the bodies are conservative during a
collision, then the total kinetic energy of the system is the
same after the collision as before. Such a collision is called an
elastic collision.
Inelastic Collision
• A collision in which the total kinetic energy
after the collision is different (most time is
less) than before the collision is called an
inelastic collision.
Completely inelastic collision
• In a completely inelastic collision, the two bodies stick
together after the collision, they are the same final
velocity v2:
• In a completely inelastic collision, the momentum is
conserved but the kinetic energy is lost – kinetic energy
after the collision is less than the kinetic energy before the
collision.
Caution: an inelastic collision doesn’t have to be completely
inelastic.
Only in elastic collisions, the total kinetic energy before
equals the total kinetic energy after.
Example 8.7 A completely inelastic collision
Suppose we repeat the collision describe in Example 8.5, but this
time equip the gliders so that they stick together instead of bouncing
apart after they collide. Find the common final x-velocity v2x, and
compare the initial and final kinetic energies.
Example 8.8 The ballistic pendulum
A bullet, with mass mB, is fired
into a block of wood with
mass mW, suspended like a
pendulum, and makes a
completely inelastic collision
with it. After the impact of the
bullet, the block swings up to a
maximum height y. Given the
values of y, mB, and mW, what
is the initial speed v1 of the
bullet?
Example 8.9 An automobile collision
A 1000 kg compact car is traveling north at 15 m/s when it
collides with a 2000 kg truck traveling east at 10 m/s. All
occupants are wearing seat belts and there are no injuries,
but the two vehicles are thoroughly tangled and move away
from the impact point as one mass. The insurance adjustor
has asked you to find the velocity of the wreckage just after
impact. What do you tell her?
Test Your Understanding 8.3
• For each situation, state whether the collision
is elastic or inelastic. If it is inelastic, state
whether its completely inelastic.
1. You drop a ball from your hand. It collides
with the floor and bounces back up so that is
just reaches your hand.
2. You drop a different ball from your hand and
let it collide with the ground. This ball
bounces back up the half the height from
which it was dropped.
3. You drop a ball of clay from your hand. When
it collides with the ground, it stops.
8.4 Elastic Collisions
An elastic collision in an isolated system is one in which
kinetic energy as well as momentum is conserved. Elastic
collisions occur when the forces between the colliding
bodies are conservative.
1
1
1
1
2
2
2
2
m A v A1x  m B v B1x  m A v A 2 x  m B v B 2 x
2
2
2
2
m A v A1x m B v B1x  m A v A2 x  mB v B 2 x
Where vA1x and vB1x are x-velocities before the collision, and vA2x
and vB2x are x-velocities after the collision.
Elastic collisions – One body initially at rest
1
1
1
2
2
2
m A v A1x  m A v A2 x  m B v B 2 x
2
2
2
m A v A1x  m A v A2 x  mB v B 2 x
v A2 x 
m A m B
v A1x
m A  mB
vB2 x 
2m A
v A1x
m A  mB
If mA = mB, then vA2x = 0 and vB2x
= vA1x. That is, the body that was
moving stops dead; it gives all its
momentum and kinetic energy to
the body that was at rest.
Elastic collisions and relative velocity
In a straight-line elastic collision of two bodies, the
relative velocities before and after the collision have the
same magnitude but opposite sign.
v B 2 x v A2 x  (v B1x  v A1x )
Example 8.10 An elastic straight-line collision
We repeat the air-track experiment from Example 8.5, but now
we add ideal spring bumpers to the gliders so that the collision
is elastic. What are the velocities of A and B after the collision?
Example 8.11 Moderator in a nuclear reactor
The fission of uranium nuclei in a nuclear reactor produces
high-speed neutrons. Before a neutron can trigger additional
fissions, it has to be slowed down by collisions with nuclei in
the moderator of the reactor. Suppose a neutron (mass 1.0 u)
traveling at 2.6 x 107 m/s undergoes a head-on elastic collision
with a carbon nucleus (mass 12 u) initially at rest. The external
forces during the collision are negligible. What are the
velocities after the collision ( 1u = 1.66 x 10-27 kg)
Example 8.12 A two-dimensional elastic collision
The figure shows an elastic collision of two pucks on a frictionless
air-hockey table. Find the final speed vB2 of puck B and the angles α
and β in the figure.
Test Your Understanding 8.4
• Most present-day nuclear reactors use water as a
moderator. Are water molecules (mass mw = 18.0 u) a
better or worse moderator than carbon atoms? (one
advantage of water is that it also acts as a coolant for the
reactor’s radioactive core.)
8.5 Center of Mass
• Suppose we have several particles with masses m1, m2, and so
on. Let the coordinates of m1 be (x1,y1), those of m2 be (x2,y2),
and so on, we define the center of mass of the system as the
point that has coordinates (xcm, ycm) given by
xcm
m1 x1  m2 x2  m3 x3

m1  m2  m3
mx

 

m
i i
i
i
i
ycm
m y  m2 y2  m3 y3
 1 1
m1  m2  m3
my

 

m
i
i
i
i
i
In statistical language, the center of mass is a mass-weighted
average position of the particles.
Example 8.13 Center of mass of a water molecule
The figure shows a simple model of the structure of a water
molecule. The separation between atoms is d = 9.57 x 10-11 m.
Each hydrogen atom has mass 1.0 u, and the oxygen atom has
mass 16.0 u. Find the position of the center of mass.
Many centers of mass
Motion of center of mass
• When the net external force acting
on the wrench is zero, the center of
mass moves in a straight line with
constant velocity. Total momentum
is conserved.


dp
dv
m
0
dt
dt
• If the net external force on a
system of particles is not zero,




dv dp
 Fext  Macm  M dt  dt
Example 8.14 A tug-of-war on the ice
James and Ramon are standing 20.0 m apart on the slippery
surface of a frozen pond. Ramon has mass 60.0 kg and James
has mass 90.0 kg. Midway between the two man a mug of
their favorite beverage sits on the ice. They pull on the ends of
a light rope that is stretched between them. When James has
moved 6.0 m toward the mug, how far and in what direction
has Ramon moved?
• A shell explodes into two fragments in flight. If air
resistance is ignored, the center of mass continues on
the same trajectory as the shell’s path before exploding.
• The same effect occurs
with exploding fireworks.
Test Your Understanding 8.5
• Will the center of mass in the figure continue on the same
parabolic trajectory even after one of the fragments hits the
ground? Why or why not?
No. If the gravity is the only force acting on the system of two
fragments, the center of mass will follow the parabolic trajectory
of a freely falling object. Once a fragment lands, the ground
exerts a normal force on that fragment.