Chapter 8
Momentum, Impulse,
and Collisions
PowerPoint® Lectures for
University Physics, Thirteenth Edition
– Hugh D. Young and Roger A. Freedman
Copyright © 2012 Pearson Education Inc.
Goals for Chapter 8
• To learn the meaning of the momentum of a particle
and how an impulse causes it to change
• To learn how to use the principle of conservation of
momentum
• To learn how to solve problems
involving collisions
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Goals for Chapter 8
• To learn the definition of the
center of mass of a system and
what determines how it moves
• To analyze situations, such as
rocket propulsion, in which the
mass of a moving body changes
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Introduction
• In many situations, such as a bullet hitting a carrot, we
cannot use Newton’s second law to solve problems because
we know very little about the complicated forces involved.
• In this chapter, we shall introduce momentum and impulse,
and the conservation of momentum, to solve such problems.
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Momentum and Newton’s second law
• The momentum of a
particle is the product
of its mass and its
velocity:
• Newton’s second law
can be written in
terms of momentum
as
What is the momentum of a 1000kg car going
25 m/s west?
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Momentum and Newton’s second law
• The momentum of a
particle is the product
of its mass and its
velocity:
• Newton’s second law
can be written in
terms of momentum
as
What is the momentum of a 1000kg car going
25 m/s west?
p = mv = (1000 kg)(25 m/s) = 25,000 kgm/s west
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Impulse and momentum
• Impulse of a force is product
of force & time interval
during which it acts.
• Impulse is a vector!
• On a graph of Fx versus
time, impulse equals area
under curve.
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Impulse and momentum
• Impulse-momentum theorem:
Impulse = Change in momentum J of particle during time
interval equals net force acting on particle during interval
• J = Dp = pfinal – pinitial
(note all are VECTORS!)
• J = Net Force x time = (F) x (Dt)
so…
• Dp = (F) x (Dt)
• F = Dp/(Dt)
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Compare momentum and kinetic energy
• Changes in momentum
depend on time over
which net force acts
But…
• Changes in kinetic energy
depend on the distance
over which net force acts.
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Ice boats again (example 8.1)
• Two iceboats race on a frictionless lake; one
with mass m & one with mass 2m.
• Wind exerts same force on both.
• Both boats start from rest, both travel same
distance to finish line.
• Questions!
• Which crosses finish line with more KE?
• Which crosses with more Momentum?
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Ice boats again (example 8.1)
• Two iceboats race on a frictionless lake; one with mass m and
one with mass 2m.
• The wind exerts the same force on both.
• Both boats start from rest, and both travel the same distance to
the finish line?
• Which crosses the finish line with more KE?
In terms of work done by wind?
W = DKE = Force x distance = same;
So ½ m1v12 = DKE1 = W = DKE2 = ½ m2v22
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Ice boats again (example 8.1)
• Two iceboats race on a frictionless lake; one with mass m and
one with mass 2m.
• The wind exerts the same force on both.
• Both boats start from rest, and both travel the same distance to
the finish line?
• Which crosses the finish line with more KE?
In terms of work done by wind?
W = DKE = Force x distance = same;
So ½ m1v12 = DKE1 = W = DKE2 = ½ m2v22
Since m2> m1, v2 < v1 so more massive boat loses
Copyright © 2012 Pearson Education Inc.
Ice boats again (example 8.1)
• Two iceboats race on a frictionless lake; one with mass m and
one with mass 2m.
• The wind exerts the same force on both.
• Both boats start from rest, and both travel the same distance to
the finish line?
• Which crosses the finish line with more p?
In terms of momentum?
Force on the boats is the same for each, but TIME
that force acts is different. The second boat
accelerates slower, and takes a longer time.
Since t2> t1, p2 > p1 so second boat has more momentum
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Ice boats again (example 8.1)
• Check with equations!
• Force of wind = same; distance = same
• Work done on boats is the same, gain in KE same.
• ½ m1v12 = DKE1 = W = DKE2 = ½ m2v22
• And m1v1 = p1; m2v2 = p2
• ½ m1v12 = ½ (m1v1) v1 = ½ p1 v1 & same for ½ p2 v2
• ½ p1 v1 = ½ p2 v2
• Since V1 > V2, P2 must be greater than P1!
p2 > p1
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√
A ball hits a wall (Example 8.2)
• A 0.40 kg ball moves at 30 m/s to the left, then
rebounds at 20 m/s to the right from a wall.
• What is the impulse of net force during the collision,
and if it is in contact for 0.01 s, what is the average
force acting from the wall on the ball?
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Kicking a soccer ball – Example 8.3
• Soccer ball 0.40 kg moving left at 20 m/s, then kicked up & to
the right at 30 m/s at 45 degrees.
If collision time is 0.01 seconds, what is impulse?
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An isolated system
• The total momentum of a system of particles is the vector sum
of the momenta of the individual particles.
• No external forces act on the isolated system consisting of the
two astronauts shown below, so the total momentum of this
system is conserved.
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Conservation of momentum
• External forces (the normal
force and gravity) act on the
skaters, but their vector sum
is zero. Therefore the total
momentum of the skaters is
conserved.
• Conservation of momentum:
If the vector sum of the
external forces on a system
is zero, the total momentum
of the system is constant.
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Remember that momentum is a vector!
• When applying
conservation of
momentum, remember
that momentum is a
vector quantity!
• Use vector addition to
add momenta in
COMPONENTS!
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Recoil of a rifle – 1D conservation of momentum
• A rifle fires a bullet, causing the rifle to recoil.
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Objects colliding along a straight line (example 8.5)
• Two gliders collide on an frictionless track.
• What are changes in velocity and momenta?
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A two-dimensional collision
• Two robots collide
and go off at
different angles.
• You must break
momenta into x & y
components and deal
with each direction
separately
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Elastic collisions
•In an elastic collision, the total
momentum of the system in a
direction is the same after the
collision as before if no external
forces act in that direction:
•Pix = Pfx
•mavai +mbvbi =mavaf +mbvbf
But wait – there’s more!
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Elastic collisions
•In an elastic collision, the
total kinetic energy of the
system is the same after the
collision as before.
•KEi = KEf
•½ mavai2 + ½ mbvbi2 =
½ mavaf2 + ½ mbvbf2 =
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Elastic collisions
•In an elastic collision,
•Difference in velocities initially =
(-) difference in velocities finally
• (vai - vbi) = - (vaf – vbf)
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Elastic collisions
•In an elastic collision,
•Difference in velocities initially =
(-) difference in velocities finally
• (vai - vbi) = - (vaf – vbf)
•Solve generally for final velocities:
vaf = (ma-mb)/(ma+mb)vai + 2mb/(ma+mb)vbi
vbf = (mb-ma)/(ma+mb)vbi + 2ma/(ma+mb)vai
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Elastic collisions
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Elastic collisions
• Behavior of colliding objects is greatly affected by relative
masses.
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Elastic collisions
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Elastic collisions
• Behavior of colliding objects is greatly affected by
relative masses.
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Elastic collisions
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Elastic collisions
• Behavior of colliding objects is greatly affected by
relative masses.
Stationary Bowling Ball!
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Inelastic collisions
• In any collision in which the
external forces can be
neglected, the total momentum
is conserved.
• A collision in which the bodies
stick together is called a
completely inelastic collision
• In an inelastic collision, the
total kinetic energy after the
collision is less than before the
collision.
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Some inelastic collisions
• Cars are intended to have
inelastic collisions so the car
absorbs as much energy as
possible.
• Example 8.7
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The ballistic pendulum
• Ballistic pendulums are
used to measure bullet
speeds
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A 2-dimenstional automobile collision (Example 8.9)
• Two cars traveling at right angles collide.
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An elastic straight-line collision
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Neutron collisions in a nuclear reactor
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A two-dimensional elastic collision (page 258)
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Center of mass of symmetrical objects
It is easy to find the center of
mass of a homogeneous
symmetric object
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Motion of the center of mass
• The total momentum of a system
is equal to the total mass times
the velocity of the center of mass.
• The center of mass of the wrench
at the right moves as though all
the mass were concentrated there.
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Tug-of-war on the ice
• Start 20 meters away; each 10 m from coffee!
• Pull light rope; when James moved 6 m, how far is Ramon?
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External forces and center-of-mass motion
• When a body or collection of particles is acted upon by
external forces, the center of mass moves as though all the
mass were concentrated there.
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Rocket propulsion
• Conservation of momentum holds for rockets, too!
F = dp/dt = d(mv)/dt = m(dv/dt) + v(dm/dt)
• As a rocket burns fuel, its mass decreases (dm< 0!)
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Rocket propulsion
• Initial Values
m = initial mass of rocket
v = initial velocity of rocket
in x direction
mv
= initial x-momentum of rocket
vexh
= exhaust velocity from rocket motor
(This will be constant!)
(v - vexh)
= relative velocity of exhaust gases
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Rocket propulsion
• Changing values
dm
= decrease in mass of rocket from fuel
dv
= increase in velocity of rocket in x-direction
dt
= time interval over which dm and dt change
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Rocket propulsion
• Final values
m + dm
= final mass of rocket less fuel ejected
v + dv
= increase in velocity of rocket
(m + dm) (v + dv) = final x-momentum of rocket (+x direction)
[- dm] (v - vexh) = final x-momentum of fuel (+x direction)
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Rocket propulsion
• Final values
mv
= [(m + dm) (v + dv)] + [- dm] (v - vexh)
initial momentum
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final momentum of
the lighter rocket
final momentum of
the ejected mass of
gas
Rocket propulsion
• Final values
mv
= [(m + dm) (v + dv)] + [- dm] (v - vexh)
mv
= mv + mdv + vdm + dmdv –dmv +dmvexh
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Rocket propulsion
• Final values
mv
= [(m + dm) (v + dv)] + [- dm] (v - vexh)
mv
= mv + mdv + vdm + dmdv –dmv +dmvexh
mv
= mv + mdv + vdm + dmdv –dmv +dmvexh
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Rocket propulsion
• Final values
mv
= mv + mdv + vdm + dmdv –dmv +dmvexh
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Rocket propulsion
• Final values
mv
= mv + mdv + vdm + dmdv –dmv +dmvexh
0
= mdv + dmdv + dmvexh
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Rocket propulsion
• Final values
mv
= mv + mdv + vdm + dmdv –dmv +dmvexh
0
= mdv + dmdv + dmvexh
Neglect the assumed small term:
m(dv) = -dmdv – (dm)vexh
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Rocket propulsion
• Final values
m(dv)
=
– (dm)vexh
Gain in momentum of original rocket is related to rate
of mass loss and exhaust velocity of gas!
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Rocket propulsion
• Final values
m(dv)
=
– (dm)vexh
Gain in momentum of original rocket is related to rate
of mass loss and exhaust velocity of gas!
Differentiate both sides with respect to time!
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Rocket propulsion
• Final values
m(dv)
=
– (dm)vexh
m(dv/dt)
=
– [(dm)/dt] vexh
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Rocket propulsion
• Final values
m(dv)
=
– (dm)vexh
m(dv/dt)
=
– [(dm)/dt] vexh
ma
= Force (Thrust!)
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= – [(dm)/dt] (vexh)
Rocket propulsion
• Final values
m(dv)
=
– (dm)vexh
m(dv/dt)
=
– [(dm)/dt] vexh
ma
= Force (Thrust!)
rate of
change of
mass
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= – [(dm)/dt] (vexh)
Velocity of
gas
exhausted
Rocket propulsion
• Final values
Thrust
=
– [(dm)/dt] (vexh)
Example:
vexhaust = 1600 m/s
Mass loss rate = 50 grams/second
Thrust =?
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Rocket propulsion
• Final values
Thrust
=
– [(dm)/dt] (vexh)
Example Problem 8.62
vexhaust = 1600 m/s
Mass loss rate = 50 grams/second
Thrust = -1600 m/s (-0.05 kg/1 sec) = +80N
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Rocket propulsion
• Gain in speed?
m(dv)
=
– (dm)vexh
dv
=
– [(dm)/m] vexh
integrate both sides
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Rocket propulsion
• Gain in speed?
m(dv)
=
– (dm)vexh
dv
=
– [(dm)/m] vexh
vf - vi
=
(vexh) ln(m0/m)
(integrate both sides)
m0 = initial mass
gain in velocity
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Velocity of gas
exhausted
m0 > m, so ln > 1
“mass ratio”
Rocket propulsion
• Gain in speed?
vf - vi
=
(vexh) ln(m0/m)
m0 = initial mass
gain in velocity
Velocity of gas
exhausted
m0 > m, so ln > 1
“mass ratio”
Faster exhaust, and greater difference in mass, means
a greater increase in speed!
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Rocket propulsion
• Gain in speed?
vf - vi
=
(vexh) ln(m0/m)
Example:
Rocket ejects gas at relative 2000 m/s. What fraction
of initial mass is not fuel if final speed is 3000 m/s?
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Rocket propulsion
• Gain in speed?
vf - vi
=
(vexh) ln(m0/m)
Example:
Rocket ejects gas at relative 2000 m/s. What fraction
of initial mass is not fuel if final speed is 3000 m/s?
ln(mo/m) = (3000/2000) = 1.5 so m/mo = e-1.5 = .223
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Rocket propulsion
• Gain in speed?
vf - vi
=
(vexh) ln(m0/m)
m0 = initial mass
gain in velocity
Velocity of gas
exhausted
m0 > m, so ln > 1
“mass ratio”
STAGE rockets to throw away mass as they use up
fuel, so that m0/m is even higher!
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