• Examples are good, but some of them
need to be put into the ws so students
follow along.
• Impulse slides
– Jumping 20m in/on to water versus concrete
– Firemen catching a person with a blanket
– Roll start a car (the longer you push, the
faster it goes). Do a few trails and let it start
on the last one.
Situation A
Situation B
Momentum
Inertia in Motion
Definition of Momentum
• Momentum is the product of mass and
velocity.
p  mv
Momentum
 kg m s 
Mass
 kg 
Velocity
ms 
Momentum Depends on Velocity
• An object must be moving in order to have
momentum.
• Consider the truck at rest. pTruck  0
• Consider the skate boarder. He has momentum.
pSkateboarder  0
Momentum Depends on Mass
• Larger objects traveling at the same velocity as
smaller objects will have greater momentum.
pTruck  pMan
The Nature of Momentum
• Momentum is a vector quantity.
– It has magnitude (how big).
– It has direction (right(+) or left(-)).
• Direction is denoted by sign (+ or -).
pTruck
Bus 00
Calculating Momentum
• As mentioned before,
momentum is the product
of mass and velocity.
pTank  22500kg m s
p  mv
v  2.5 m s
pTank  9000kg 2.5 m s 
m  9000kg
Practice Problem 1
• Calculate the momentum of the tank if
its velocity is 3.0 m s to the right.
m  9000kg
v  3.0 m s
p  mv
p  9000kg 3.0 m s 
p  27000kg m s
Practice Problem 2
• Calculate the momentum of the tank if its
velocity is 2.0 m s to the left.
m  9000kg
v  2.0 m s
p  mv
p  9000kg  2.0 m s 
p  18000kg m s
Collisions
• The collisions that we will study will involve two objects.
• Each object possesses its own momentum.
• Collisions are broken down into two categories:
– Elastic (Objects Separate)
– Inelastic (Objects Stick Together)
Momentum – Elastic Collisions
• Elastic Collision – a collision in which the colliding bodies
do not stick together.
• The equation used for elastic collisions is as follows.
m1v1  m2v2  m v  m v
'
1 1
'
2 2
Momentum – Inelastic Collisions
• Inelastic Collision – a collision in which the colliding bodies
stick together.
• The equation used for inelastic collisions is as follows.
m1v1  m2v2  MV '
Elastic Collisions
• The total momentum of the system remains
constant.
• Elastic collision, objects move off with
separate velocities.
• This means that they can bounce.
• Or they can be projected from rest (like a cannon).
Momentum
• Identify the number and types of collisions in the
animation below.
Momentum
• Identify the number and types of collisions in the
animation below.
The Law of Conservation of Momentum
In a closed and isolated system, the total momentum of
all the objects present remains constant.
pTotBefore  pTotAfter
pBlue
ptotal  2.0kg m s
 8.0kg m s pRed  6.0kg m s
Elastic Collision Equation
pTotBefore  pTotAfter
m1v1  m2v2  m1v1'  m2v2'
Before
After
Subsripts:
I: Initial (Before)
Initial Momentum
Initial Momentum
Object 1
F: Final (After) Object 2
Final Momentum
Object 1
Final Momentum
Object 2
Sample Problem (Elastic)
What is the initial momentum of the blue ball?
8.0
6.0
6.0
?
8.0
'
m1v1  m2v2  m1v1  m2v2
What
isthe
the
final
momentum
the
red
What
What
isisisthe
final
final
velocity
momentum
of theofof
red
ball?
blue
What
the
initial
momentum
ofthe
the
redball?
ball?
ball?
m m
p
m
8.0
kg
s s
p  mv p ?m
1.0
kg
8.0

8.0
kg


mvs
mv

v

m
mm4.0 m s
m
m
m
v


8.0
kg
p  mv
s 1.0
kgms 
6.0


6.0
kg
p  mv
2.0kg
3.0


6.0
kg
s
2F
s
s
s2.0kg
ss
8.0kg  6.0
kg
v1F = -6.0m/s
v1I = 8.0m/s
1.0kg
1.0kg
 6.0kg
?
v2F = ?
v2I
=
-3.0m/s
2.0kg
Before
After
2.0kg
Practice Problem 3 (Elastic)
• Two air track cars collide and bounce.
• The red car is initially traveling at a velocity of 0.50m/s to the right
while the yellow car is traveling at a velocity of 1.5m/s to the left.
• After the collision the red car is moving 2.0m/s to the left with the
yellow car following, but not touching.
red car?
What is the final momentum of the yellow
car?
What is the initial momentum of the yellow
red car?
car?
What is the final velocity of the yellow car?
Collide
'
m m 
mm
m
p 0.5
m
0.15

1.2


0.6
?
p

mv

0.3
0.8
kg

2.0
1.5
0.15

0.6
1.2
kg
kg

0.45
kg



v


0.5625
s
s
s
s
s
s
p  mv  v 
2
v

2F
?  0.45kg m sm
0.8kg
mred = 0.3kg
0.15 1.2
0.6
0.45
m1v1I  m2v2 I  m1v1F  m2v2 F
0.3kg
Air Track
myellow = 0.8kg
0.8kg
Sample Problem (Elastic)
• This problem will demonstrate the “cannon type” scenario.
0
?'
0
1500
1500
m1v1  m2v2  m1v1  m2v2'
– The cannon and cannonball are initially at rest.
– The ball then is fired at a velocity of 150m/s.
What is the velocity
of the
cannon?
afterafter
firing?
initialmomentum
final
momentum
momentum
of cannon
theofof
cannon
the
theimmediately
cannonball?
cannonball?
immediately
firing?

?300
kgp
1500kg
kgmmss
p  0mv
mv
10
kg
150
1500
00mmmsssm
0

1500
kg
s 0
p  mv  v 
v
m
?  1500kg sm
300kg
v  5.0 m s
Fire!
mcannon = 300kg
mball = 10kg
Practice Problem 4 (Elastic)
• A gun (m = 1.5kg), initially at rest, is fired sending its
bullet (m = 0.1kg) at a velocity of 180 m s.
What is the final
initialvelocity
total
momentum
momentum
of the
ofof
of the
the
gun?
gun
bullet?
gun?
system
and bullet?
before
after
firing?
firing?
pGun  mv  1.5kg  0 m s   0
0
18'
0
'18
m1v1  m2v2  m1v1  m2v2
pBullet  mv   0.1kg  0 m s   0
p  mv   0.1kg 180 m s   18kg m s
0  0  18  x
18  x
p 18kgm s
p  mv  v  
 12 m s
m
1.5kg
Before
After
0 0
Homework
• WS 9a
– #’s 1,2, 5-17
Inelastic Collisions
• The Law of Conservation of Momentum: that the total
momentum of the system remains constant, no matter
what the collision.
• In an inelastic collision, objects stick together.
• This means that they have the same final velocity (vF).
Collide
v2F
vF
S
N
Inelastic Collision Equation
• The inelastic collision equation takes
into account the momentum of each
object before and after the collision.
• Because the objects are combined after
the collision, and they have the same
final velocity (vF).
Inelastic Collision Equation
pTotBefore  pTotAfter
m1v1  m2v2   m1  m2  vF
Before
Initial
Momentum
Initial
Momentum
Object 1
Object 2
After
Final Combined
Momentum
Objects 1 & 2
Sample Problem (Inelastic)
The big rig backs up with an initial velocity of
2.0m/s and collides with the trailer, which is
initially at rest.
0.8
4000 m v 0m v   m 5000
m v
1 1
2 2
I
1
2
F
Whatisisisisthe
thefinal
initial
momentum
of
the
trailer?
What
the
initial
momentum
the
truck?
What
What
the
combined
velocity
mass
of the
ofof
truck
the
truck
and
and
trailer
trailer?
combo?
m  kg
m kg
4000
02000
20.0kg
  kg
pm1 mv
m2ms2000
kg
5000
5000
kg

3000
kg
3000
0
sv F 4000
s
4000kg m s 5000kg v F
v  0.8 m

5000kg
5000kg
F
mTruck = 2000.0kg
mTrailer = 3000.0kg
Before
After
s
Practice Problem 5 (Inelastic)
• Happy Penguin (m = 7.0kg) slides across the ice on a magnet
with a velocity of 3.0m/s.
• Cool Penguin (m = 8.0kg) comes up from behind with a velocity
of 5.0m/s. He, also having a magnet, will collide and stick.
What is the final
initialvelocity
combined
momentum
momentum
mass
of the
of
ofofthe
penguin
Happy
Cool
twoPenguin?
Penguin?
penguins?
pair?
m
p 1.0mv
m15
kg538
..00.0kg 40
21
15..00kg
kg
61

.087v..0F0kg
2
61.0 15.0v F
21.0
40.0
15.0 4.07

m1v1  m2v2   m1  m2  vF
15.0
15.0
vF  4.07 m s
m
m
ss
N
S
S
N
m
m
ss
ICE
Collide
Momentum
 While climbing a cliff, a super model (m =
51.0 kg) slips and falls.
 She falls for 2s before she is rescued by
Super Doctor Physics (m = 63.0 kg, v =
27.85 m/s).
 What was their velocity immediately after
the collision?
Homework
• WS 2
– # 1,2
• WS 3
– #1
Impulse (Dmv)
• When objects collide, their velocities
change. This means that their individual
momentums will also change.
• Impulse simply defines the change in
momentum for an individual object
(denoted by Dmv).
Impulse (Dmv)
• There are two ways to calculate impulse:
– You can subtract initial momentum from final
momentum.
Dmv  p After  pBefore
– Or you can take the product of force and time.
Dmv  Ft
• The units for impulse are in Ns, which is equal to
kgm/s
Sample Problem 1 (Impulse)
• A golf ball, initially at rest, is struck by a
golf club.
3.5 3.5
0
Dp  pAfter  pBefore
– The golf ball (m = 0.1kg) rolls away with a
velocity of 35m/s.
What is the impulse
initialmomentum
final
momentum
on the ball?
ofofthe
theball?
ball?
m  0kg m 
mm
mm3.5
m
D
p  mv
kg
kg
pAfter
pBefore
3.5
mv


0.1
0.1
kg

35
0


0

kg
3.5
kg






s
s
s
s
s
s
s
F I
Swing
Practice Problem 6 (Impulse)
• A soccer ball (m = 0.4kg) approaches a player with a
velocity of 13m/s (left). The player kicks the ball giving it
a new velocity of 20m/s (right).
m
What is the impulse
initialmomentum
final
momentum
felt by the
ofofsoccer
the
thesoccer
soccer
ball?ball?
ball?
m m 
m m
20
8.0
kg
p  mv   0.4D
kgp

5.2
kg


p13

p
s
s s
s
F
I
Dp  8.0kg m s   5.2kg m s 
Kick
Dp  13.2kg m s
pI  5.2kg s
pF  8.0kg m s
Sample Problem 2 (Impulse)
• A baseball is struck by a bat with a force of
3000N for a duration of 0.05s. What is the
change in momentum of the baseball?
F  3000 N
t  0.05 s
Dp  Ft
Hit
Dp  3000N  0.05s 
Dp  150 Ns
or
Dp  150kg m s
Practice Problem 7 (Impulse)
• A tennis ball is struck with a force of 400N
for a time of 0.15s.
What is the impulse felt by the tennis ball?
Dp  Ft
Dp   400N  0.15s   60Ns
Hit
Impulse Worksheet 1-5
I
Summary
• All problems surrounding momentum are
based on the following principles:
–
–
–
–
Momentum depends on both mass and velocity.
Momentum is conserved, regardless of collision.
Collision type determines the equation used.
Impulse is simply change in momentum.
Trains are stupid
cuz they can’t do
Physics!