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Chapter 7
Linear Momentum and Impulse
Notes
© 2014 Pearson Education, Inc.
Momentum & Force
Momentum is a vector symbolized by the symbol p, and
is defined as
The rate of change of momentum (known as impulse) is
equal to the net force:
Newton’s 2nd Law: The rate of change of momentum of
an object is equal to the net force applied to it
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Law of Conservation of Momentum
The total momentum of an isolated system of objects remains constant.
An isolated system has no EXTERNAL forces acting on it.
Example: Tennis ball and racket
Isolated system: only in the brief instants before and after
the collision
External forces (gravity, air resistance) act on this system
over a longer time scale
Other Examples
Momentum conservation works for a rocket as long as we consider the
rocket and its fuel to be one system, and account for the mass loss of the
rocket. Momentum of fuel out the back balances the forward momentum of
the rocket.
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Other Examples
Would momentum be conserved or not?
If frictionless track – yes.
In real world – no.
If air, track, earth, etc are all part of the system – yes.
Other Examples
Would momentum be conserved or not?
Depends on the definition of the system:
If system = rock and person then no because gravity
accelerates it (external force)
If system = rock, earth, and person then yes because no
external forces and the earth/person accelerates up and
canceling the downward acceleration of the rock
Solving Conservation of Momentum
Collisions and Forces
During a collision, the system’s internal forces are usually
much greater than external forces, so we can ignore
external forces for the brief interval of the collision
Objects are deformed due to the large forces involved
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When the collision occurs, the force jumps from zero to a very large force in
a short time
Though the force is not constant, we can approximate the total force by
using average force over the entire time interval
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Impulse
During a collision, the force is equal to the change in
momentum (impulse) divided by time
Impulse-Momentum Theorem:
Impulse is the
green shaded
area
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Impulse in the Real World
What is the point of these safety items in light of impulse?
Change in momentum is constant (from whatever mv0 to
zero), so increase time it takes, results in decreased force
The impulse tells us that we can get the same change in momentum with a large
force acting for a short time, or a small force acting for a longer time.
This is why you should bend your knees when you land; why you wear boxing
gloves; why landing on a pillow hurts less than landing on concrete, why a
karate chop has to be FAST, and why a slap stings
Practice Problems
7-1. For a top player, the tennis ball must leave the racket on a serve with a
speed of 55m/s (about 120mph). If the ball has a mass of 0.060kg and is in
contact with the racket for about 4ms (4 x 10-3 s), a) estimate the average
force on the ball. b) Would this force be large enough to lift a 60kg person?
C) compare this to the force of gravity on the ball to justify why it can be
ignored at the moment of the collision.
Practice Problems
7-2. Water leaves the hose at a rate of 1.5kg/s with a speed of 20m/s and is
aimed at the side of a car, which stops it. (Ignore splashing back) What is
the force exerted by the water on the car?
Practice Problems
7-3. A 10,000kg railroad car, A, traveling at a speed of 24.0m/s strikes an
identical car, B, at rest. If the cars lock together as a result of the collision,
what is their common speed afterward?
Practice Problems
7-4. A) an empty sled is sliding on frictionless ice when Susan drops
vertically from a tree down onto the sled. When she lands, does the sled
speed up, slow down, or keep the same speed? B)Later, Susan falls
sideways off the sled. When she drops off, does the sled speed up, slow
down, or keep the same speed?
Practice Problems
7-5. Calculate the recoil velocity of a 5.0kg rifle that shoots a 0.020kg
bullet at a speed of 620m/s.
Practice Problems
7-6. Estimate the impulse and the average force delivered by a karate blow
that breaks a board. Assume the hand is 1kg and moves at roughly 10m/s
when it hits the board.
Conservation of Energy
in Collisions Notes
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Momentum is conserved in all
collisions.
Collisions in which kinetic energy
and internal energy are conserved
are called elastic collisions, and
those in which KE and internal E
are not conserved are called
inelastic.
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Elastic Collisions
• When two objects collide, return to their original shapes
and move separately after the collision
• Momentum is still conserved
• Kinetic energy is conserved in elastic
collisions, internal E of objects constant
• Elastic collisions don’t truly exist in
the real world
– i.e. if there is sound, then KE
has been transformed
Elastic Collisions Equations
mAvA,0 + mBvB,0 = mAvA + mBvB
KA,0 + KB,0 = KA+ KB
In a perfectly elastic collision between
balls of equal masses, the balls
exchange velocities
,0
,0
Inelastic Collisions
• A collision in which two objects stick together and move
together after colliding, typically generate thermal energy
which changes internal energy of the objects
• Momentum is still conserved
• Kinetic energy is not conserved in inelastic collisions
– Some energy is transformed into sound, heat, potential, etc.
– Kinetic energy could also be gained such as in an explosion
(inelastic collision in reverse) when potential nuclear or
chemical energy is transformed
Perfectly Inelastic Collision Formula
mAvA,0 + mBvB,0 = (mA+ mB)v
Collisions in 2 or 3 Dimensions
Conservation of energy and momentum can also be used
to analyze collisions in 2D or 3D, but angles are
necessary in addition to masses and initial velocities
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Center of Mass
Until now, we have assumed all objects behave like a particle that undergoes only
translational motion (as in figure a)
However, real objects also rotate and move in other ways that are not just
translational – this is called general motion
But there is one point in the object that despite general motion, that one point only
has translational motion like a particle (as in figure b)
Center of Mass (CM) – the point in an object that
behaves like a particle
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The center of mass
continues to
move according to
the net force
The general motion of an object or system is the sum of the translational
motion of the CM, plus rotational, vibrational, or other forms of motion
about the CM.
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The velocity of the CM of a system cannot be changed by an interaction in
the system. (Velocity of CM in these situations is always zero)
Which is a correct final
location?
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When two objects collide, the velocity of their center of mass will not
change.
If m1=m2 and v1=v2, then momentum and
velocity of CM are zero before and after the
collision
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The center of mass can be found experimentally by suspending an object
from different points. The CM need not be within the actual object (i.e. a
doughnut’s CM is in the center of the hole)
High jumpers have developed a technique where their CM
actually passes under the bar as they go over it. This allows
them to clear higher bars.
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Practice Problems
7-7 (modified to a concept question). A) Ball A and B have equal 2.0kg
masses. Ball A has an initial velocity of 5.0m/s and Ball B has an initial
velocity of -10.0m/s. What are the final velocities of the two balls after they
collide. B) Now Ball B is initially at rest. What is the velocity of each ball
after they collide?
Practice Problems
7-10. A 10,000kg railroad car, A, traveling at a speed of 24.0m/s collides
with an identical car, B, which is at rest. The cars lock together and are
moving at 12.0m/s after the collision. How much energy was transformed
into thermal and other types of energy during this collision?
Practice Problems
7-12 (modified). Three guys of roughly the same mass are on a raft. A)
Approximately where is the CM of the system? B) What would happen if
the middle guy moved backward to 2.0m?
Practice Problems
7-14 (modified). A rocket is shot into the air as shown. At the moment the
rocket reaches its highest point, a horizontal distance 10.0m from its starting
point, a prearranged explosion separates the equal masses. Part I is stopped
in midair by the explosion and falls vertically to Earth. Where does part II
land compared to the starting point?