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Honors Physics Chapter 9
Momentum and Its
Conservation
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Honors Physics



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Skip anything angular in this chapter
Lecture
Q&A
Next Generation Science Standards


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HS-PS2-2.
Use mathematical representations to support the
claim that the total momentum of a system of
objects is conserved when there is no net force
on the system.
HS-PS2-3.
Apply scientific and engineering ideas to design,
evaluate, and refine a device that minimizes the
force on a macroscopic object during a collision.
Newton’s Second Law
  mv 
v mv


F  ma  m
t
t
t
 F t    mv 
Impulse:
I = Ft
Momentum:
P = mv
 I  P Impulse-Momentum Theorem


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Nothing but Newton’s Second Law
P is a fundamental quantity that is very useful.
Impulse

I = F t

I = Area under curve of Force-Time Graph
(Even if F is not a constant.)
F
Impulse is a vector



Same direction as F
Unit of Impulse
 I    F   t   N  s
5
t
Momentum
P  mv

Momentum is a vector


Same direction as v
Unit of momentum

m
 P    m   v   kg 
s
Remember unit of Impulse
I   N  s
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m

  kg  2   s 
s 




   P   I 
m 

kg 
s
Example: Pg233pp1
A compact car, mass 725 kg, is moving at 115 km/h toward the east. Sketch
the moving car.
a) Find the magnitude and direction of its momentum. Draw an arrow on
your sketch showing the momentum.
b) A second car, with a mass of 2175 kg, has the same momentum. What is
its velocity?
km 1000m hr
m
m  725kg , v  115


 31.9
s
hr
km 3600s
a)P  ?
m
m
4
P  mv  725kg  31.9  2.31 10 kg  , to the east
s
s
m
b) P  2.31  10 kg  , m  2175kg , v  ?
s
m
2.31  104 kg 
P
m
km
s

 10.6  38.2
P  mv  v 
m
2175kg
s
hr
4
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Practice: Pg233pp4
The driver accelerates a 240.0 kg snowmobile, which results in a force
being exerted that speeds up the snowmobile from 6.00 m/s to 28.0
m/s over a time interval of 60.0 s.
a) What is the snowmobile’s change in momentum?
b) What is the impulse on the snowmobile?
c) What is the magnitude of the average force that is exerted on the
snowmobile?
m  240kg , vi  6.00m / s, v f  28.0m / s, t  60.0s
a )P  ?
P  mv
b) I  ?
m
m
 1440kg 
s
s
m
m
Pf  mv f  240kg  28.0  6720kg 
s
s
m
m
m
P  Pf  Pi  6720kg   1440kg   5280kg 
s
s
s
 Pi  mvi  240kg  6.00
c) F  ?
I  F t  F 
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I 5280kg  m / s

 88.0 N
t
60.0s
I  P  5280kg 
m
s
Practice:
The brakes exerts a 640 N force on a car
weighing 15600 N and moving at 20.0 m/s.
The car finally stops.
a) What is the car’s mass?
b) What is its initial momentum?
c) What is the final momentum?
d) What is the change in the car’s momentum?
e) How long does the braking force act on the
car to bring it to a halt?
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F  640 N ,W  15600 N , vi  20.0m / s, v f  0.
a )m  ?
W  mg  m 
Solution
b) Pi  ?
Pi  mvi  1590kg  20.0
W 15600 N

 1590kg
m
g
9.8 2
s
m
m
 3.18  104 kg 
s
s
c ) Pf  ?
Pf  mv f  0
d ) P  ?
P  Pf  Pi  0  3.18  104 kg 
e ) t  ?
m
m
4
 3.18  10 kg 
s
s
I  F t  P
10
P

 t 
F
3.18  104 kg 
640 N
m
s  49.7 s
Closed, Isolated System


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Closed: no objects in or out
Isolated: net external force on system is zero
Conservation of Momentum

The total momentum of any closed, isolated
system does not change.
 Pi  Pf

 P1i  P2i  P1 f  P2 f

 P1  P2  0
 P   P
2
 1
12
The momentum of a single object can change, but the total
momentum of the closed, isolated system does not change.
Special Case: v1f = v2f = vf


Totally/Completely Inelastic Collision: the two
objects stick together after collision.
Final momentum equals to
Pf  P1 f  P2 f  m1v1 f  m2 v2 f
 m1v f  m2 v f 
 m1  m2  v f
Only if two masses stick together.
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Example: Pg238pp14
A 0.105-kg hockey puck moving at 24 m/s is caught and
held by a 75-kg goalie who is initially at rest. With what
speed does the goalie slide on the ice after the catch?
m1  0.105kg , v1i  24
m
, m2  75kg , v2i  0, v1 f  v2 f  v f  ?
s
Pi  Pf
P1i  P2i  Pf
m1v1i  m2 v2i   m1  m2  v f
m1v1i   m1  m2  v f
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m1v1i
vf 
 m1  m2 
m
0.105kg  24
m
s

 0.034
0.105kg  75kg
s
Practice: Pg238pp16
A 0.0350-kg bullet moving at 475 m/s strikes a 2.5-kg bag of
flour that is on ice, initially at rest. The bullet passes through
the bag, and exits it at 275 m/s. The bag was at rest when it
was hit. How fast is the bag moving when the bullet exits?
m
m
m1  35.0 g  0.0350kg , v1i  475 , m2  2.5kg , v2i  0, v1 f  275 ,
s
s
v2 f  ?
Pi  Pf
P1i  P2i  P1 f  P2 f
 m1v1i  m2 v2i  m1v1 f  m2 v2 f
 m2 v2 f  m1v1i  m1v1 f  m1  v1i  v1 f
 v2 f 
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
m1 v1i  v1 f
m2


m
m

0.0350kg   475  275 
m
s
s


 2.8
2.5kg
s
Practice: Pg238pp18
A 0.50-kg ball that is traveling at 6.0 m/s collides head-on with a
1.00-kg ball moving in the opposite direction at a speed of 12.0 m/s.
The 0.50-kg ball bounces backward at 14 m/s after the collision.
Find the speed of the second ball after the collision.
Let original direction of 0.50 kg ball = “+”
m1  0.50kg , v1i  6.0
m
m
m
, m2  1.00kg , v2i  12.0 , v1 f  14 , v2 f  ?
s
s
s
P1i  P2i  P1 f  P2 f
 m1v1i  m2 v2i  m1v1 f  m2 v2 f
 m2 v2 f  m1v1i  m2 v2i  m1v1 f
 v2 f 
m1v1i  m2 v2i  m1v1 f
m2
0.50kg  6.0

16
m
m
m


 1.00kg   12.0   0.50kg   14 
m
s
s
s


  2.0
1.00kg
s
Internal and External Forces

Internal Forces: forces between objects within a
system.
–
–

External Forces: forces from objects outside a system
–
–
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Both force giver and receiver are objects within the system
Internal forces do not change the total momentum of the
system
Force giver is an object outside the system
External forces change the total momentum of the system.
Conservation of Momentum of a
System

Total momentum of a system is conserved only
when there is no net external forces.

Internal forces do not change the total
momentum of a system.
–
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Action and reaction forces cancel out each other.
Example: Pg240pp20
A thread holds a 1.5-kg cart and a 4.5-kg cart together.
After the thread is burned, a compressed spring pushes
the cars apart, giving the 1.5-kg cart a speed of 27 cm/s
to the left. What is the velocity of the 4.5-kg cart?
Let right be the positive direction. then
cm
m1  1.5kg , v1 f  27
, m2  4.5kg ,
s
v1i  v2i  0, v2 f  ?
1.5 kg
4.5 kg
P1i  P2i  P1 f  P2 f
 m1v1i  m2 v2i  m1v1 f  m2 v2 f
 0  m1v1 f  m2 v2 f  m2 v2 f   m1v1 f
 v2 f 
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 m1v1 f
m2
cm 

1.5kg   27

cm
s 


 9.0
, to the right.
4.5kg
s
Practice: Pg240pp21
Carmen and Judi dock a canoe. 80.0-kg Carmen moves
forward at 4.0 m/s as she leaves the canoe. At what
speed and in what direction do the canoe and Judi move
if their combined mass is 115 kg?
Let Carmen = #1, Judi & canoe = #2, also let forward direction = “+”
Then m1 = 80.0kg, v1f = 4.0 m/s, m2 = 115 kg,v1i = v2i = 0, v2f = ?
P1i  P2i  P1 f  P2 f
m1v1i  m2 v2i  m1v1 f  m2 v2 f
0  m1v1 f  m2 v2 f  m2 v2 f   m1v1 f
m

80
kg

4.0
m1v1 f
s   2.8 m “-” indicates backward direction.
v2 f 

m2
115kg
s
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Relative Velocity
What do we mean when we say
• Velocity of bullet with respect to the gun is 1000 m/s, or
• Velocity of bullet relative to gun is 1000 m/s?
m
 vb  vg  1000 ,
s
or
m
vb  vg  1000
s
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Muzzle velocity
Example
You (of mass 50 kg) can throw a rock (of mass 2.0 kg) with a
maximum relative speed of 10 m/s. If you stand on frictionless
ice and throw the rock, what is the maximum possible speed
the rock can have (with respect to the ground)?
Let you = 1, rock = 2, and the direction you throw the rock = “+”, then
m
m1  50kg , m2  2.0kg , v2  v1  10  vm , v1i  v2i  0, v2 f  ?
s
v2  v1  vm  v1  v2  vm  v1 f  v2 f  vm
P1i  P2i  P1 f  P2 f
m1v1i  m2 v2i  m1v1 f  m2 v2 f
 0  m1v1 f  m2 v2 f
 0  m1  v2 f  vm   m2 v2 f
 m1v2 f  m1vm  m2 v2 f  0   m1  m2  v2 f  m1vm
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 v2 f 
m1vm
50kg  10m / s
m

 9.6
m1  m2 50kg  2.0kg
s