Chapter 9. Impulse and Momentum
Chapter Goal: To introduce
the ideas of impulse and
momentum and to learn a
new problem-solving strategy
based on conservation laws.
Ch. 9 Student Learning Objectives
• To understand interactions from the new
perspective of impulse and momentum.
• To learn what is meant by an isolated
system.
• To apply conservation of momentum in
simple situations.
• To understand the basic ideas of inelastic
collisions, explosions, and recoil.
Impulsive Forces and Newton’s 2nd Law
• An impulsive force can be
defined as a large force
exerted during a small
interval of time.
• Collisions and explosions
are examples of impulsive
forces.
• Until now, we’ve been using
Newton’s Law (and
kinematic equations) for
constant forces and
accelerations.
• Impulsive forces are not
constant.
Impulsive Forces and Newton’s 2nd Law
Fnet-x = max
This law applies to non-constant forces as well
Fx (t) = max(t) so Fx (t) = m dvx /dt
multiplying through by dt:
Fx (t) dt = m dvx. Now take the integral of both
sides:
=
m  dv x
Impulsive Forces and Newton’s 2nd Law
 m  dv x
This can be written as:
= mvf – mvi
which is how Newton actually wrote it in the
first place.
You can solve for the velocity of an object, even
if the acceleration is not constant.
Impulse
The impulse upon a particle is defined as:
Impulse has units of N-s
Newton’s Law tells us that the impulse exerted on an object
is equal to ∆ mv
Impulsive forces are often short! Time is sometimes given
in units of ms or µs.
Impulse Problem (EOC #5)
What value of Fmax gives
an impulse of 6 N-s?
A.
B.
C.
D.
.75 N
1.5 N
750 N
1500 N
Impulse Momentum Theorem
(an alternate version of Newton’s 2nd Law)
Jx = ∆px
where
change in momentum = ∆px = mfvf – mivi
Momentum and change in momentum are vector
quantities.
The cart’s change of momentum is
A. 30 kg m/s.
B. 10 kg m/s.
C.–10 kg m/s.
D.–20 kg m/s.
E. –30 kg m/s.
A 10 g rubber ball and a 10 g clay ball are
thrown at a wall and have equal impact speeds.
The rubber ball bounces, the clay ball sticks.
Which ball exerts a larger impulse on the wall?
A. They exert equal impulses because they
have equal momenta.
B. The clay ball exerts a larger impulse because
it gives all its momentum to the wall.
C. Neither exerts an impulse on the
wall because the wall doesn’t change
velocity.
D. The rubber ball exerts a larger impulse
because it bounces.
Graphical explanation of using average force to
replace a non-constant impulsive force
A bouncing ball (EOC #29)
A 200.0 g ball is dropped
from a height of 2.00 m,
bounces and rebounds to
a height of 1.50 m. The
figure shows the impulse
received from the floor.
What is Fmax to 3 sig
figs?
Hint: don’t forget to
include FG in your
calculations, even though
it is not shown in the
graph!
Normal force of floor
on ball
0
Sketch a before, after (and if necessary during)
representation to show objects before/after interaction.
Coordinate system, symbols for knowns and unknowns.
Time usually isn’t important. Position is necessary only
if needed to find velocity (as in this case).
Answers:
viy = - 6.26 m/s
v2y = 5.42 m/s
Fmax = 9.38 x 102 N
Impulse approximation
The figure shows the force
on the ball due to the
floor. Is it reasonable to
neglect the force on the
ball due to the earth
when calculating Fmax?
Calculate Fmax and
compare answers
Force of floor on ball
Impulse approximation
The figure shows the force
on the ball due to the
floor. Is it reasonable to
neglect the force on the
ball due to the earth when
calculating Fmax?
Without including the
impulse due to gravity,
Fmax = 934 N.
If the ball were smaller, it
would be easier to ignore.
Force of floor on ball
Conservation of Momentum
• The change in momentum
of ball 1 is due to the
impulse of ball 2 on ball 1.
• The change in momentum
of ball 2 is due to the
impulse of ball 1 on ball 2.
• If we added the impulse of
ball 1 on ball 2 to the
impulse of ball 2 on ball 1,
what would we get? Why?
Conservation of Momentum
• If we added the change
in momentum of ball 1
to the change in
momentum of ball 2,
what would we get?
• Why?
Conservation of Momentum
Stated mathematically, the law of conservation of
momentum for an isolated system is
The total momentum after an interaction is equal to
the total momentum before the interaction.
The two particles are both moving to the right. Particle
1 catches up with particle 2 and collides with it. The
particles stick together and continue on with velocity vf.
Which of these statements is true?
A. vf = v2.
B. vf is less than v2.
C. vf is greater than v2, but less than v1.
D. vf = v1.
E. vf is greater than v1.
Collisions Problem
• A 2000 kg truck traveling west at 12 m/s
collides with a 1200 kg car traveling east at 16
m/s. They collide and remain stuck together.
What is their final velocity?
Collisions Problem - Answer
• A 2000 kg truck traveling west at 12 m/s
collides with a 1200 kg car traveling east at 16
m/s. They collide and remain stuck together.
What is their final velocity (include direction)?
1.5 m/s west
An explosion in a rigid pipe shoots out
three pieces. A 6 g piece comes out the
right end. A 4 g piece comes out the left
end with twice the speed of the 6 g
piece. From which end does the third
piece emerge?
A. Right end
B. Left end
Explosion Problem - Radioactivity
• The nucleus of an atom travels east at 1.0 x 106
m/s. The nucleus undergoes radioactive decay
and breaks into two masses the smaller being
1/10th the mass of the original nucleus. The
smaller fragment travels east at 2.0 x 107 m/s.
Determine the velocity of the larger fragment.
Explosion Problem - Radioactivity
• The nucleus of an atom travels east a 1.0 x 106
m/s. The nucleus undergoes radioactive decay
and breaks into two masses the smaller being
1/10th the mass of the original nucleus. The
smaller fragment travels east at 2.0 x 107 m/s.
Determine the velocity of the larger fragment.
• Answer: 1.1 x 106 m/s going west.
Momentum Race
Identical constant net forces continuously
push blocks A and B from the
starting point to the finish. Both
blocks start at rest. Block A has 4
times the mass of Block B. Which
block has the larger change in
momentum?
a. A, since it has a greater m
b. B, since it will have a greater v
c. Equal, the mass and velocity change
will make the quantities equal.
d. Too little information given to
determine
Objects A and C are made of different materials, with
different “springiness,” but they have the same mass and
are initially at rest. When ball B collides with object A, the
ball ends up at rest. When ball B is thrown with the same
speed and collides with object C, the ball rebounds to the
left. Compare the velocities of A and C after the collisions.
Is vA greater than, equal to, or less than vC?
A. vA > vC
B. vA < vC
C. vA = vC
EOC# 47
In a ballistics test, a 25-g bullet traveling
horizontally at 1200 m/s goes through a 30.0-cm
thick, 350 kg, initially stationary target and
emerges with a speed of 900.0 m/s. The target is
free to slide on a frictionless surface. Assuming
that the target does not move until after the bullet
has passed through, calculate:
a. How long the bullet was in the target
b. The average force the bullet exerted on the
target
c. The target’s speed after the bullet emerges.
EOC #47
a. 2.86 x 10-4 s
b. Favg = 2.6 x 104 N
c. v target = .021 m/s (assume impulse
approximation and ignore gravitational force
on the bullet).
EOC #68
A 20.0-kg ball hangs on a 2.0-m wire. The
maximum tension the wire can withstand
without breaking is 400.0 N. A 1.0-kg
projectile, traveling horizontally, hits and
embeds itself in the ball. The ball swings up in
a circle. What is the largest speed this
projectile can have and not break the wire?
EOC #68 - Answer
A 20.0-kg ball hangs on a 2.0-m wire. The
maximum tension the wire can withstand
without breaking is 400.0 N. A 1.0-kg
projectile, traveling horizontally, hits and
embeds itself in the ball. What is the largest
speed this projectile can have and not break the
wire?
Vprojectile = 90.3 m/s