Linear Momentum(Momentum)
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Momentum
Impulse
Momentum of a System of Objects
Conservation of Momentum
Inelastic Collisions and Explosions
Newton’s First Law tells us that – objects remain in motion with a
constant velocity unless acted upon by a force.
In our experience:
· When two objects of different masses travel with the same
velocity, the one with more mass is harder to stop.
· When objects of the equal masses travel with different speeds,
the faster one is harder to stop.
· Define a new quantity, momentum (p),
into account
that takes these observations
1. Linear Momentum(Momentum)
 The Linear momentum of a moving object
is its mass multiplied by its velocity.
 That means linear momentum increases
with both mass and velocity.
Momentum
(kg m/sec)
p=mv
Velocity (m/sec)
Mass (kg)
 Linear momentum is a vector quantity
System with more than one object:
 The linear momentum for the system is
vector sum of the linear momentum of
every object:
P   Pi
 In 1 dimension:
P  P1  P 2  ....
 m1V 1  m 2V 2  ...
Example:
1. What is the momentum of a 20 kg object with a
velocity of +5.0 m/s?
P= +100 kg m/s
2. What is the momentum of a 20kg object with a
velocity of −5.0m/s?
3. What is the velocity of a 5.0kg object whose
momentum is −15.0 kg∙m/s?
Comparing momentum
A car is traveling at a velocity of 13.5 m/sec (30 mph)
north on a straight road. The mass of the car is 1,300
kg. A motorcycle passes the car at a speed of 30
m/sec (67 mph). The motorcycle (with rider) has a
mass of 350 kg. Calculate and compare the
momentum of the car and motorcycle.
1.
2.
3.
4.
5.
You are asked for momentum.
You are given masses and velocities.
Use: p = m v
Solve for car: p = (1,300 kg) (13.5 m/s) = 17,550 kg m/s
Solve for cycle: p = (350 kg) (30 m/s) = 10,500 kg m/s
–
The car has more momentum even though it is going much slower.
2. Momentum
Change
& Impulse
Suppose that there is an event that changes an
object's momentum.
· from pi - the initial momentum (just before the
event)
· by Δp - the change in momentum
· to pf - the final momentum (just after the event)
The equation for momentum change is:
i
2. Impulse
Impulse is defined as the product force acting on an
object and the time during which the force acts. The
symbol for impulse is I. So, by definition:
I = F Δt
SI unit for impulse is the Newton ·second.
Example: A 50 N force is applied to a 100 kg boulder
for 3 s. The impulse of this force is I = (50 N) (3 s) =
150 N · s.
Note that we didn’t need to know the mass of the
object in the above example.
3. Impulse - Momentum Theorem
The impulse due to all forces acting on an object (the net force) is
equal to the change in momentum of the object:
I = Fnet Δt =  p
We can prove the theorem formally:
Fnet  t = m a  t = m( v / t) t = m v =  p

Impulse - Momentum Example
1. A 0.50-kg cart (#1) is pulled with a 1.0-N force for 1
second; another 0.50 kg cart (#2) is pulled with a 2.0 Nforce for 0.50 seconds.
a). Which cart (#1 or #2) has the greatest acceleration?
b). Which cart (#1 or #2) has the greatest impulse?
c). Which cart (#1 or #2) has the greatest change in momentum?
a). Cart #2
b). The impulse is the same for each cart
c). The momentum change is the same for each cart.
2. A 1.3 kg ball is coming straight at a 75 kg soccer player at 13 m/s
who kicks it in the exact opposite direction at 22 m/s with an average
force of 1200 N. How long are his foot and the ball in contact?
answer: use Fnet t =  p.
Since the ball changes direction,
 p = m  v = m (vf - v0)
= 1.3 [22 - (-13)] = (1.3 kg) (35 m/s)
= 45.5 kg · m /s.
Thus,
(1200N) t = 45.5 kg m/s
t = 45.5 / 1200 = 0.0379 s, which is just under 40 ms.
During this contact time the ball compresses substantially and then decompresses. This
happens too quickly for us to see, though. This compression occurs in many cases, such
as hitting a baseball or golf ball.
3. Now consider a collision of a 0.2 kg tennis ball with
a wall
1). What is the velocity change?
2). What is the momentum change?
3). What is the Impulse?
Vi= 10m/s
Vf=-5 m/s
∆𝑉 = 𝑉𝑓 − 𝑉𝑖 = −5 − 10 = −15
m=0.2 kg
∆𝑃 = 𝑚∆𝑉 = 0.2𝑘𝑔
Impulse: 𝐼 = ∆𝑃 = −3.0 𝑁 ∙ 𝑠
−15
𝑚
𝑠
𝑚
𝑚
= −3.0 𝑘𝑔 ∙
𝑠
𝑠
4. Jennifer, who has a mass of 50.0 kg, is riding at 35.0 m/s in
her red sports car when she must suddenly slam on the
brakes to avoid hitting a deer crossing the road. She strikes
the air bag, that brings her body to a stop in 0.500 s. What
average force does the seat belt exert on her?
Vi= 35m/s, Vf=0 m/s
m=50.0 kg
𝑚
∆𝑉 = 𝑉𝑓 − 𝑉𝑖 = −35
𝑠
∆𝑃 = 𝑚∆𝑉 = 50𝑘𝑔
−35
Impulse: 𝐼 = ∆𝑃 = −1750 𝑁 ∙ 𝑠
T=0.500sec,
𝑚
𝑚
= −1750 𝑘𝑔 ∙
𝑠
𝑠
Force: F=I/t= -3500 N
5. If Jennifer had not been wearing her seat belt and not had an air
bag, then the windshield would have stopped her head in 0.002 s.
What average force would the windshield have exerted on her?
Homework
• http://phet.colorado.edu/en/simulation/collisi
on-lab
Collisions
• A collision occurs when two or more objects hit
each other.
• During a collision, momentum is transferred
from one object to another.
4. CONSERVATION OF LINEAR MOMENTUM
PRINCIPLE OF CONSERVATION OF LINEAR MOMENTUM
The total linear momentum of an isolated system is
constant (conserved).
An isolated system is one for which the sum of
the average external forces acting on the system is zero.
7.2 The Principle of Conservation of Linear Momentum
Conceptual Example 4 Is the Total Momentum Conserved?
Imagine two balls colliding on a billiard
table that is friction-free. Use the momentum
conservation principle in answering the
following questions. (a) Is the total momentum
of the two-ball system the same before
and after the collision? (b) Answer
part (a) for a system that contains only
one of the two colliding
balls.
For a collision occurring between object 1 and object 2 in an
isolated system,
the total momentum of the two objects before the collision is
equal to the total momentum of the two objects after the
collision.
That is, the momentum lost by object 1 is equal to the
momentum gained by object 2.
Or:
P before collision= p after collision
m1 v1 + m2 v2 = m1 va + m2 vb
7.2 The Principle of Conservation of Linear Momentum
Example 6 Ice Skaters
Starting from rest, two skaters
push off against each other on
ice where friction is negligible.
One is a 54-kg woman and
one is a 88-kg man. The woman
moves away with a speed of
+2.5 m/s. Find the recoil velocity
of the man.
7.2 The Principle of Conservation of Linear Momentum
 
Pf  Po
m1v f 1  m2 v f 2  0
vf 2  
vf 2
m1v f 1
m2

54 kg  2.5 m s 

 1.5 m s
88 kg
7.2 The Principle of Conservation of Linear Momentum
Applying the Principle of Conservation of Linear Momentum
1. Decide which objects are included in the system.
2. Relative to the system, identify the internal and external forces.
3. Verify that the system is isolated.
4. Set the final momentum of the system equal to its initial momentum.
Remember that momentum is a vector.
Conservation of Momentum in 1 Dimension
In a isolated system, whenever two or more objects collide (or
when they exert forces on each other without colliding, such as
gravity) momentum of the system (all objects together) is
conserved.
m1 v1 + m2 v2 = m1 va + m2 vb
before:
p
= m1 v1 + m2 v2
v1
(Choosing right as the + direction, m2 has v2 <0, and
momentum.)
v2
m1
after:
p
m2
= m1 va + m2 vb
(va <0, vb >0, and m1 has
va
- momentum.)
vb
m1
m2
-
Sample Problem 1
35 g
7 kg
700 m/s
v=0
A rifle fires a bullet into a giant slab of butter on a frictionless surface. The bullet penetrates
the butter, but while passing through it, the bullet pushes the butter to the left, and the
butter pushes the bullet just as hard to the right, slowing the bullet down. If the butter skids
off at 4 cm/s after the bullet passes through it, what is the final speed of the bullet?
(The mass of the rifle matters not.)
35 g
v=?
4 cm/s
7 kg
continued on next slide
Sample Problem 1
(cont.)
Let’s choose left to be the + direction & use conservation of momentum, converting all units
to meters and kilograms.
35 g
p before = 7 (0) + (0.035) (700)
7 kg
= 24.5 kg · m /s
v=0
35 g
p after = 7 (0.04) + 0.035 v
4 cm/s
v=?
p before = p after
v
700 m/s
7 kg
24.5 = 0.28 + 0.035 v
= 0.28 + 0.035 v
v = 692 m/s
came out positive. This means we chose the correct direction of the
bullet in the “after” picture.
Sample Problem 2
35 g
7 kg
700 m/s
v=0
Same as the last problem except this time it’s a block of wood rather than butter, and the bullet
does not pass all the way through it. How fast do they move together after impact?
v
7. 035 kg
(0.035) (700) = 7.035 v
v = 3.48 m/s
Note: Once again we’re assuming a frictionless surface, otherwise there would be a frictional
force on the wood in addition to that of the bullet, and the “system” would have to include the
table as well.
Sample Problem 3
A 15-kg medicine ball is
thrown at a velocity of 20
km/hr to a 60-kg person
who is at rest on ice. The
person catches the ball
and subsequently slides
with the ball across the
ice. Determine the
velocity of the person and
the ball after the collision.
60 • v + 15 • v = 300
75 • v = 300
v = 4 km/hr
P:
Before Collision
After Collision
Person
0
(60 kg) • v
Medicine
ball
(15 kg) • (20 km/hr)
= 300 kg • km/hr
(15 kg) • v
300 kg • km/hr
300 kg •
km/hr
Total
Sample Problem 4
A crate of raspberry donut filling collides with a tub of lime Kool Aid on a frictionless surface.
Which way on how fast does the Kool Aid rebound? answer: Let’s draw v to the right in
the after picture.
3 (10) - 6 (15) = -3 (4.5) + 15 v
v = -3.1 m/s
Since v came out negative, we guessed wrong in drawing v to the right, but that’s OK as
long as we interpret our answer correctly. After the collision the lime Kool Aid is moving 3.1
m/s to the left.
before
10 m/s
6 m/s
3 kg
15 kg
after
4.5 m/s
v
3 kg
15 kg
Sample Problem 5
A 3000-kg truck moving with a velocity of 10 m/s hits a 1000-kg
parked car. The impact causes the 1000-kg car to be set in
motion at 15 m/s. Assuming that momentum is conserved during
the collision, determine the velocity of the truck immediately after
the collision.
Before Collision
After Collision
Truck
3000 • 10 = 30 000
3000 • v
Car
0
1000 • 15 = 15 000
Total
30 000 kg•m/s
30 000kg•m/s
3000*v + 15 000 = 30 000
3000*v = 15 000
v = 5.0 m/s
example
• http://www.physicsclassroom.com/mmedia/
momentum/crete.cfm