MOMENTUM and
IMPULSE
• LINEAR MOMENTUM
Ch. 6-1 in textbook.
• Momentum describes an object’s
motion
• Momentum  a vector quantity
defines as the product of an
objects mass and velocity.
• The symbol for momentum is:
OBJECTIVES:
•compare the momentum of
different moving objects
• Compare the momentum of
the same objects moving with
different velocities
•Identify examples of change
in the momentum of an
object
•Describe changes in
momentum in terms of force
and time
Linear Momentum…
• The equationfor momentum is:
= mv
Where is momentum, m is mass,
and v is velocity.
Linear Momentum
• Momentum is a VECTOR QUANTITY.
• It’s direction matches that of its velocity.
• The SI units of momentum are Kg x m/s
Linear Momentum
• Think about coasting down a steep hill on your
bike…
– The faster you move, the more momentum you
have and the more difficult it is to come to a stop.
• At the same speed…the larger the mass of the
object the more momentum that object has.
50 mph
50 mph
Sample Problem
• A 2250 kg truck has a velocity of 25 m/s to the
east. What is the momentum of the truck?
GIVEN
M = 2250 kg
V = 25 m/s
=
FORMULA
SUBSTITUTION
=mxv
= 2250 (25)
ANSWER and
UNITS
56250 =
5.6 x 104 Kg x m/s
to the east.
Practice Problems…
• A 21 kg child is riding a 5.9 kg bike with a
velocity of 4.5 m/s to the northwest.
– A. What is the total momentum of the child and
bike together?
– B. What is the momentum of the child?
– C. What is the momentum of the bike?
A 21 kg child is riding a 5.9 kg bike with a velocity of 4.5 m/s to
the northwest.
A. What is the total momentum of the child and bike together?
B. What is the momentum of the child?
C. What is the momentum of the bike?
GIVEN
M(child) = 21 kg
M (bike) = 5.9 kg
V = 4/5 m/s
FORMULA
= mv
SUB
= (21 +5.9)(4.5)
= (26.9)(4.5)
ANS/UNITS
= 121.05 Kg x m/s
to the northwest
A 21 kg child is riding a 5.9 kg bike with a velocity of 4.5 m/s to
the northwest.
A. What is the total momentum of the child and bike together?
B. What is the momentum of the child?
C. What is the momentum of the bike?
GIVEN
M(child) = 21 kg
M (bike) = 5.9 kg
V = 4/5 m/s
FORMULA
= mv
SUB
= (21)(4.5)
ANS/UNITS
= 94 Kg x m/s to
the northwest
A 21 kg child is riding a 5.9 kg bike with a velocity of 4.5 m/s to
the northwest.
A. What is the total momentum of the child and bike together?
B. What is the momentum of the child?
C. What is the momentum of the bike?
GIVEN
M(child) = 21 kg
M (bike) = 5.9 kg
V = 4/5 m/s
FORMULA
= mv
SUB
= (5.9)(4.5)
ANS/UNITS
= 27 Kg x m/s to
the northwest
Change in Momentum
• A change in momentum takes force and time.
F=
t
Force = Change in momentum
time interval
IMPULSE MOMENTUM THEOREM
• (Page 210 in textbook)
• Impulse  for a constant external force, the product of the
force and the time over which it acts on an object.
• Impulse momentum theorem:
Ft=
or
F  t =   = mv(final) – mv(initial)
IMPULSE MOMENTUM THEOREM
• A 1400 kg car moving westward with a
velocity of 15 m/s collides with a utility pole
and is brought to rest in 0.30 s. Find the
magnitude of the force exerted on the car
during the collision.
A 1400 kg car moving westward with a velocity of 15 m/s
collides with a utility pole and is brought to rest in 0.30 s. Find
the magnitude of the force exerted on the car during the
collision.
GIVEN
FORMULA
SUB
M = 1400 kg
F  t =   = mv(final) – mv(initial)
F = (1400)(0) – (1400)(15)
0.30
Vi = 15 m/s
F = mv(final) – mv(initial)
t
F = 21000 kg x m/s
0.30
Vf = 0
t = .30 s
ANS/UNIT
F = 70000 N to the
east
Figure 6-3 in text…
• Stopping times and distance depend on the impulsemomentum theorem.
• Two trucks with the same mass traveling at the same
speed, but one is loaded down…
• The truck with a load in the bed, MUST undergo a
GREATER change in momentum than the truck without
a load when arriving upon the same stop sign.
Figure 6-3 in text…
• The loaded down truck’s momentum is twice as great,
so its change in momentum is also twice as great. The
applied forces are the same, so the time period must
be twice as great as well.
=Ft
Figure 6-3 in text…
• How do the stopping distances compare?
– The loaded truck’s time period is twice as great while its
acceleration is half as much (F-ma).
– Because x = Vi t + ½ a  t 2
– the loaded truck’s stopping distance is 2 times as great as
the empty trucks.
A 2250 kg car traveling to the west
slows down uniformly from 20.0 m/s to
5.0 m/s. How long does it take the car
to decelerate if the force on the car is
8450 N to the east? How far does the
car travel during deceleration?
A 2250 kg car traveling to the west slows down uniformly from 20.0
m/s to 5.0 m/s. How long does it take the car to decelerate if the
force on the car is 8450 N to the east? How far does the car travel
during deceleration?
GIVEN
FORMULA
SUB
M = 2250 kg
F  t =   = mv(final) – mv(initial)
 t= (2250)(-5) – (2250)(-20)
8450
Vi = -20 m/s
(neg b/c going
west)
t = mv(final) – mv(initial)
F
 t = 33750 kg x m/s
8450 kg x m2/s2
Vf = -5 m/s
F = 8450 N to
the east (+)
ANS/UNIT
 t= 3.99s
A 2250 kg car traveling to the west slows down uniformly from 20.0
m/s to 5.0 m/s. How long does it take the car to decelerate if the
force on the car is 8450 N to the east? How far does the car travel
during deceleration?
GIVEN
FORMULA
M = 2250 kg  x = ½ (vi(final) + v(initial))t
SUB
ANS/UNIT
x= ½ (-20 - 5)
(3.99)
Vi = -20 m/s
(neg b/c
going west)
Vf = -5 m/s
T=4
 X= -50 m or 50 m to
the West
A change in momentum over a longer
time requires less force.
• If the speed of a particle is doubled, then the
momentum increases by a factor of 2.
• The thespeed of a particle is doubled, then
the KE increases by a factor of 4.
Impulse and Momentum
• What is the relationship???
Impulse is equivalent to a change in momentum.
When comparing the momentum of 2 moving
objects…
The less massive object will have less momentum
if the velocities are the same.
Which has the GREATEST momentum?
A: Turtle with a mass of 91 kg moving at 1.4 m/s
B: 1.8 kg Roadrunner moving at a 6.7 m/s
C: Hare with a mass of 2.7 kg moving at 7 m/s
D: 270 kg Tortoise moving at 0.5 m/s
Which has the GREATEST momentum?
A: Turtle with a mass of 91 kg moving at 1.4 m/s
127.4 kg*m/s
B: 1.8 kg Roadrunner moving at a 6.7 m/s
12.06 kg*m/s
C: Hare with a mass of 2.7 kg moving at 7 m/s
18.9 kg*m/s
D: 270 kg Tortoise moving at 0.5 m/s
135 kg*m/s
Review…
• If the speed of a particle is doubled…
– By what factor is its momentum changed?
– What happens to its KE?
Review
• 1- Momentum increases by a factor of 2
• KE increases by a factor of 4
• Why the difference?
HOMEWORK
• Page 209 # 1 and 3
• Page 211 # 1 and 3
CONSERVATION of MOMENTUM
• ρ1i + ρ2i = ρ1f + ρ2f
• m 1 v1i + m 2 v2i = m 1 v1f + m 2 v2f
TOTAL INITIAL MOMENTUM = TOTAL FINAL MOMENTUM
COPY DOWN IN NOTES…
• The total momentum of all objects interacting
with one another remains constant regardless
of the nature of the forces between the
objects.
READING
• Read Page 216-217 on momentum
conservation
• Understand that momentum is conserved
during collisions AND for objects that are
pushing away from each other.
EXAMPLE
• 2 Skaters pushing away from each other (fig 6-7)
Both initially at rest with a momentum of ρ1i = ρ2i =0
When they push away from each other, they move in
opposite directions with EQUAL but OPPOSITE
momentum so that the final momentum is also 0.
ρ1f + ρ2f
SAMPLE
• A 76 kg boater, initially at rest in a stationary
45 kg boat, steeps out of the boat and onto
the dock. If the boater moves out of the boat
with a velocity of 2.5 m/s to the right, what is
the final velocity of the boat?
• m1i v1i + m2i v2i =m1f v1f +m2f v2f
Solution
• m1i v1i + m2i v2i =m1f v1f +m2f v2f
• 190 kg m/s + 45 kg ( v2f) = 0
• 45 kg ( v2f) = -190 kg m/s
• ( v2f) = -190 kg m/s
45 kg
( v2f) = -4.2 m/s
The negative sign indicates that the boat is moving left.
(opposite direction of the boater)
HOMEWORK
• Page 219: 1-2
• Page 221: 1 and 3