Momentum
the product of mass
Momentum
and velocity of an object
momemtum = mass x velocity
p = mv
m = mass; usually measured in kg
v = velocity; usually measured in m/s or km/hr
p = momentum; kg x m/s
Momentum and Inertia
• Inertia
is a tendency for an object to keep
doing what it’s doing
• The “momentum” of an object is the
quantification of its Inertia or how diffi cult
it is to stop its inertial movement
Momentum
Momentum is a vector quantity –
so direction IS important!
What affects Momentum?
Ex: snowball going
down a mountain;
As it goes down the
mountain its’ mass
increases and it’s
velocity increases
which means it
momentum is
increasing
More force is needed to quickly stop a
baseball thrown at 95 mph than to quickly stop
a baseball thrown at 45 mph, even though
they both have the same mass.
More force is needed to quickly stop a
train
moving at 45 mph than to quickly stop A
car
moving at 45 mph, even though they
both have
the same speed.
Both mass and velocity are
important
factors when considering the force
needed to change the motion of an
Click icon to add picture
An object’s momentum will change if its
mass and/or velocity (speed and
direction) changes.
According to Newton’s laws,
a net force causes an object to accelerate,
or change its velocity.
A net force, therefore, causes a
change in an object’s momentum.
Conceptual Question #1
Which object has the greatest momentum?
A)A
A large truck moving at 50 km/hr
B)A sports car moving at 30 km/hr
C)The Empire State Building
D)Choices
A and B have the same
momentum
Check Your
Understanding
Determine the momentum of
a Pacific leatherback turtle of
mass 8.6 x 102 kg Remember
the same thing as 860 kg,
swimming at a velocity of 1.3
m/s [forward].
This Photo by Unknown Author is licensed under CC BY
Check Your Understanding
Determine the momentum of a Pacific leatherback
turtle of mass 8.6 x 102 kg, swimming at a velocity
of 1.3 m/s [forward].
Answer: 1.1 x 103 or 1,118 kgm/s [fwd]
Example # 1 (Momentum)
What is the velocity of a 5-kg object whose
momentum is -15 kg m/s?
A) -3 m/s
A
B) -5 m/s
C) -15 m/s
D) 5 m/s
E) -60 m/s
What is Impulse?
• Impulse is a force acting
for a given amount of
time to change an
object's momentum
• Impulse
is change in
momentum
•I = F x ∆t = p= mvf- mvi
• Units: kg x m/s
Applications of Impulse
Most problems involving impulse will involve a force
being in contact with an object for a very short period
of time.
Examples:
Boxing
Car Air Bags
Hitting a Baseball
Think about why boxers wear gloves and how the
principles of impulse take into effect.
Gloves allow for the force being applied to be in contact
for a longer period of time, making the person getting
punched not feel as much of a force. If you increase Δt,
Notice how
these people
land from
very high
places and
explain how
this helps
them to
remain
uninjured.
3:42
how
objects
deform
when
they are
in contact
with
other
objects
for very
short
intervals
of
time.2:52
Conceptual Question #2
Air bags are used in
cars because they:
A) Increase the force
with which a
passenger hits the
dashboard.
B) Increase the
B
duration time of the
passenger’s impact.
C) Decrease the
duration time of the
passenger’s impact.
Example #2 (Impulse)
Charlie is playing T-ball. He
swings at a 0.144 kg ball that
is at rest and hits it with a
force of 182N, which is in
contact with the ball for
0.009s. How fast will the ball
be traveling after it is hit?
182 x .009= (.144x vf)-.144x0
1.638=.144 x vf
Vf=11.38
11.38 m/s
11.38 m/s
The importance of “follow through”
Following Through
• Following through means that you continue
your motion after you’ve made contact
• If you stop your motion at impact, you will
reduce the impact time
• An increase in time of collision results in
an increase in change in momentum
Example #3 (Impulse)
-6500x .0013= (.144 x vf)- (.144
x 43)
A 0.144-kg baseball is moving toward home plate with a speed of
-8.45= (.144 x
43 m/s when it is bunted. The bat exerts an average force of -
vf)-
6,500N on the ball for 0.0013s. The pitcher throws in the positive x
(6.192)
direction. (The Force will act in the –-2.258=
direction) .144 x vf
(A) What is the speed of the ball afterVf= -15.68
15.68
the bunt? m/s
m/s
∆p= mvf- mi
(B) -8.45 Kg
What is the change in momentum? ∆p= (.144 x -15.68)-( .144 x
m/s 43)8.45
-2.25792-6.19= ∆p
(C) What is the impulse?-8.45 Kg m/s
∆p=-8.45 kg x m/s
(D) If the bat was to exert more force on the ball, how would the
speed of the ball compare from your answer with (A)?
The ball would move faster, because with more force
there is more impulse meaning there will be a
greater change in momentum
A) PE= KE
Example #4 (Impulse) Mgh= .5mv2
B) PE= KE .015 x 9.8x 1.44= .5 x .015
A 0.015-kg marble is dropped from rest onto the floor2 1.44m
Mgh= .5mv2 xv
below. If the .015
marble bounces
x 9.8x .64= straight
.5 x .015upward
x to a height of
.21168= .0075 v2
0.64m,
v2 V2= 28.224
(A) With what..09408= .0075the
velocity does v2 marble hit the floor?
V= -5.31 m/s
- 5.31
c)V2I=∆p= mvf- mi
= 12.544 m/s
I=V= (.015
3.54xm/s
3.54)-( .015 x - 3.54
5.31)
(B) With what velocity does the marble come up off the floor?m/s
.0531- (-.07965)= I 0.13 kg m/s
(C) What is I=.13kg
the magnitude
x m/s of the impulse for the marble?D) I= F x t
.13= F x .025
(D) If the marble was in contact with the floor
5.2 Nfor 0.025
seconds, what force did the floor exert on the marble? F= 5.2N
(E) If the marble was in contact with the floor for longer than
0.025
The s, would
marble the marble
would come come upfloor
off the off the floor faster
faster. Since or
it is in contact
slower?
with the force for a longer time, the marble will experience a
larger impulse. A larger impulse means a larger change in
Conceptual Question #3
One car crashes into a concrete barrier. Another car
crashes into a highway barrier filled with water at the
same speed. What is the difference between the two
crashes? Select all that apply.
A) Change in Momentum
B
B) Force on the Passengers
&
C
C) Impact time on the Passengers
D) Final Momentum
They have the same initial velocity, they have the same final velocity because they
both come to a stop. Change in momentum would be the same because they have
the same mass both before and after and same initial velocity and final velocities.
Both have the same final momentum because they both come to a stop.
A) ∆dy= .5 ( -
Example #5 (Impulse) 9.8)t2
-2= -4.9t2
A boy who is 2m tall shoots a 0.3 kg dart out of a blow gun horizontally. The dart
lands 8.3m away.
B) I=∆p= mvf- mvi .41= t 2
I= (.03 x 12.97)-( 0) t= .64s
A) How fast did the
I= dart leave
3.89kg 12.97 m/s
from the blow gun?
x m/s dx=vxt
B) What impulse did the dart experience from the person? 3.89 kg m/s 8.3= vx ( .64)
C) If the dart took 0.07s to exit out of the blow gun while it was beingVshot
x= 12.97
out, m/s
the gun ?
55.57
with what average force did the boy blow into
N C) I= F x t
3.89= F x .07
55.57 N
Hail will bounce up so it has
Conceptual Question #4 a final velocity greater than
zero. Giving a larger
A person stands under an umbrella during a Impulse therefore more
Force
rain shower. A few minutes later the
raindrops turn to hail, though the number
of “drops” hitting the umbrella per time and
their speed remains the same. Is the force
required to hold the umbrella in the hail
(a) the same
B
(b) more than, or
(c) less than the force
required in the rain
I= F x t= mvf- mvi
Conceptual Question #5
You are lying in bed and you
want to shut your bedroom
door. You have a bouncy ball
and a blob of clay (both with
the same mass) sitting next to
you. Which one would be
more effective to throw at
your door to close it?
A
(A) the bouncy ball
(B) the blob of clay
(C) it doesn’t matter -- they
will be equally effective
Example #6 (Momentum)
A I=∆p= mvf- mvi
Two groups of canoeists meet inFthe x t= middle
mvf- mvi of a lake.
After a brief visit, a person in canoe 1 pushes
-46 x 1.2 = = (130 on canoe
x vf)- (0) 2
with a force of 46N to separate the-55.2=130vf
canoes. If the mass of
vf canoe 1 = -.42 m/s
canoe 1 and its occupants is 130-kg, and the mass of
55.2=(250vf)-(0)
canoe 2 and its occupants is 250-kg,
Vf canoe 2 = .22 m/s
(A) Find the velocity each canoe moves after 1.2 s of
pushing -0.42 m/s & 0.22
m/s
(B)-54.6
Findkg the momentum
m/s & of each
55 kg m/s Should canoe
be same but opposite but
due to rounding, slightly different
B) P=mv
130 x -.42=-54.6 kg m/s0,
& 250x .22=55 kg m/s Should be
0 to rounding, slightly different
same but opposite but due
Conservation of Momentum
The total momentum in any closed
system will remain constant. 1:59
Before
Momentum After
Car mass = 1000 kg, Truck = 3000 kg Car mass = 1000 kg, Truck = 3000 kg
Car velocity = 20.0 m/s Car velocity = -40.0 m/s
Truck velocity = -20.0 m/s Truck velocity = 0.0 m/s
Car momentum = 20,000 kg m/s Car momentum = -40,000 kg m/s
Truck momentum = -60,000 kg Truck momentum = 0 kg m/s
m/s
Law of Conservation of
Momentum
When two or more objects collide, the collision does NOT
change the total momentum of the two objects. Whatever
momentum is lost by one object in the collision is gained by
the other. The total momentum of the system is conserved.
Newton’s Cradle
Law of Conservation of
Momentum
𝑚1 ⃗𝑣 𝑖 +𝑚2 ⃗𝑣 𝑖 =𝑚1 𝑣⃗ 𝑓 1 +𝑚2 ⃗𝑣 𝑓 2
1 2
The total momentum of the system before the
collision equals the total momentum of the
system after the collision. Thus,
pisystem = pfsystem
Example:
Kangaroos are good runners that can sustain
speeds of 56 km/hr (15. 5m/s). Suppose a
kangaroo is sitting on a log that is floating in
a lake. When the kangaroo gets scared, she
jumps off the log with a velocity of 15m/s
toward the bank. The log moves with a
velocity of 3.8 m/s away from the bank. If
the mass of the log is 250 kg, what is the mass
of the kangaroo?
Conservation of momentum in
collisions
• Elastic Collisions – Bounce
Momentum is conserved
pi = pf
Kinetic Energy is conserved
(no energy is lost)
KEf = KEi
• Inelastic Collisions – Stick Together
Momentum is conserved
pi = pf
Conservation Conservation
of of
EVENT DESCRIPTION Momentum Kinetic Energy
Objects stick No, loses
Inelastic together Kinetic
Collision pf=pi, Yes Energy
Yes
*Elastic Objects bounce off KEf=KEi
Collision each other pf=pi, Yes
No, gains
Explosio Kinetic
n Objects break apart pf=pi,Yes Energy
Inelastic Collisions
• Often occurs when the two colliding
before
objects stick together during a
v1
collision v2
• Energy is lost to heat and sound after
during the collision V1’
• Since the objects stick together, they
have the same final velocity
Example #7 ( Inelastic)
A 1875-kg car going 23 m/s rear ends a 1025-kg
car going 17 m/s on ice in the same direction. The
two cars stick together (vf).
17-23 m/
(A) Between what speeds do we know the vehicles
will be going?
(B) How fast do the two cars move together
immediately after the collision? 20.88 m/s
Example #8 ( Inelastic)
A 1200-kg car moving at 2.5m/s is struck in a head-on collision by a 2600-kg
truck moving at 6.2 m/s. The vehicles stick together.
(A) Which way do we know the vehicles will beLeft
going?
(B) If the vehicles stick together after the collision, what is their speed-3.45
immediately after colliding? ( remember the velocity will be neg. for
22,610 J the
m/sblue
car as it is going to the left
Less,
(C) What is the KE of the cars after the perfectly inelastic collision?
a
perfec
(D) Is the Total KE less, more, or the same as before the collision?
tly
inelas
tic
collisi
on
loses
Example #9 (Inelastic)
On a touchdown attempt, a 95-kg
running back runs toward the end
zone at 3.75 m/s. A 111-kg
linebacker moving at 4.1 m/s meets
the runner in a head-on collision. If
the two players stick together,
(A) In whichIn the direction
direction do weof know
the
Linebacker
the player will move?
-0.48
(B) What is their speed immediately
m/s
after the collision? 1600.81 J &
23.69 J
Example #10 (Inelastic)
A 0.007-kg bullet fired at 284 m/s embeds
itself into a block of mass 0.95-kg that is
attached to a ballistic pendulum.
(A) What is the velocity of the bullet and
2.08 m/s
block after the inelastic collision?
(B) What KE does the block have after the
2.07 J
inelastic collision?
(C) What is the max height that the
0.22 m
pendulum rises from the starting point?
Think KE=PE
(D) If the speed of the bullet was to go
slower or faster before
1) Theit slower
got embedded
the bullet, the lower the block will
into the block, how would
rise. this affect the
height at which the block would go up?
Back in Mute to protect my student’s
the day innocent ears 1:37
they used
ballistic
pendulum
s to
calculate
how fast
bullets
shot out
of
different
Example #11 (Explosions)
A 1-kg firecracker at rest splits into two pieces. Piece #1
has a mass of 0.25-kg. Piece #2 moves with a velocity of
12 m/s to the right.
0.75 kg
-36 m/s
(A)What is the mass of the second piece?
0, 0 because the velocity
(B)What is the velocity of the first piece?
of both pieces going in
opposite directions
(C)What is the initial and final momentum?
cancel out
Conceptual
Question
#6
4:29
Notice how the
watermelon
explodes, and
explain how
momentum is
conserved.
The watermelon
starts with 0
momentum and if
we were able to
add all the
velocities of the
pieces, we would
see that it is still 0
(just like fireworks).
Example #12 (Explosions)
A 57-kg person is on top of a 2.1-kg skateboard and
holds a 1.7-kg bowling ball. Initially the skateboard and
the person are at rest. The person now throws the
bowling ball 18 m/s East.
(A) Which direction will the skateboarder move?
West
(B) With what velocity does the person recoil?
0.52 m/s
(C) If the bowling ball had more or less mass, how would
the recoil speed change when thrown?
1) The more mass the bowling ball has the
faster the person would recoil, because the
bowling ball would now have more
momentum.
2) The less mass the bowling ball has the slower
the person would recoil, because the bowling
Elastic Collison
Some Principles of Elastic Collisions
• KE is conserved.
• When
objects of the same mass hit, they
switch velocities.
• Whena bigger object hits a smaller object,
the bigger object will tend to go in the same
direction. The smaller object will rebound
with an even faster speed.
Some
clips of
collision
s that
are
almost
fully
elastic
2:21
Example #13 (Elastic)
A particle of mass 4.0 kg, initially moving with a velocity of 2.0 m/s East, collides
elastically with a particle of mass 6.0 kg that is initially moving with a velocity of 4.0
m/s West. If the velocity after the collision for the 4-kg particle is 5.2 m/s West,
(A) What is the velocity of the 6-kg particle after the collision?
(B) What is the initial and final KE of the system?
0.8 m/s
56 J
for
both
A golf ball is
stacked on a
bouncy ball,
which is
stacked on a
basketball (As
shown in the
figure). When
they fall to the
ground, how
will the
objects move?
The bouncy ball will rebound up using some of the
momentum of the basketball much higher than it
would on its own
