PHY300 Chapter 7 physics 5e

Giambattista Physics
Chapter 7
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©McGraw-Hill Education
Chapter 7: Linear Momentum
7.1 A Conservation Law for a Vector Quantity
7.2 Momentum
7.3 The Impulse-Momentum Theorem
7.4 Conservation of Momentum
7.5 Center of Mass
7.6 Motion of the Center of Mass
7.7 Collisions in One Dimension
7.8 Collisions in Two Dimensions

7.1 A Conservation Law for a Vector Quantity
Conservation laws are powerful tools.
If a quantity is conserved, then no matter how complicated the
situation, we can set the value of the conserved quantity at one
time equal to its value at a later time.
The “before-and-after” aspect of a conservation law enables us
to draw conclusions about the results of a complicated set of
interactions without knowing all of the details.
When a vector is conserved, both the magnitude and the
direction of the vector must be constant.

7.2 Momentum
Linear momentum (or just momentum ) is a vector quantity
having the same direction as the velocity:
m

p v
In any interaction between two objects, momentum can be
transferred from one object to the other.
The momentum changes of the two objects are always equal
and opposite, so the total momentum of the two objects is
unchanged by the interaction.
(By total momentum we mean the vector sum of the individual
momenta of the objects.)

Example 7.1
A car weighing 12 kN is driving due north at 30.0 m/s. After
driving around a sharp curve, the car is moving east at 13.6 m/s.
What is the change in momentum of the car?
Strategy
There are two potential pitfalls:
1. momentum depends not on weight but on mass, and
2. momentum is a vector, so we must take its direction into
consideration as well as its magnitude. To find the change in
momentum, we need to do a vector subtraction.

Example 7.1 Solution 1
4
2
1.2 10 N
1220 kg
9.8 m/s
m
W
g
 


i 30.0 m/s, north

v
i i
4
1220 kg 30.0 m/s north
=3.66 10 kg m/s north
m
  
 
p v
f 13.6 m/s, east

v
f f
4
1220 kg 13.6 m/s east
=1.66 10 kg m/s east
m
  
 
p v

2 2
i f
4 2 4 2
4
(3.66 10 kg m/s) (1.66 10 kg m/s)
4.02 10 kg m/s
p p
  
     
  
p
4
f
4
i
opposite 1.66 10 kg m/s
tan 0.454
adjacent 3.66 10 kg m/s
p
p

 
   
 
1
tan 0.454 24.4
 
  
The change in momentum of the car as 4.0 × 104 kg·m/s directed
24° east of south.

7.3 The Impulse-Momentum Theorem
We found that the change in momentum of an object when a
single force acts on it is equal to the product of the force acting
on the object and the time interval during which the force acts:
t
  
p F
The product Ft is given the name impulse.
1 2
1 2
total impulse
( )
t t
t t

    
  
  
F F
F F F

The Impulse-Momentum Theorem Continued
t
  

p F
Impulse When Forces Are Not Constant:
av t
  

p F

Example 7.3
A car moving at 20.0 m/s (44.7 mi/h) crashes into a tree. Find the
magnitude of the average force acting on a passenger of mass 65
kg in each of the following cases.
(a) The passenger is not wearing a seat belt. He is brought to
rest by a collision with the windshield and dashboard that
lasts 3.0 ms.
(b) The car is equipped with a passenger-side air bag. The force
due to the air bag acts for 30 ms, bringing the passenger to
rest.

Example 7.3 Strategy
From the impulse-momentum theorem,
av t
  

p F
where Fav is the average force acting on the passenger and t is
the time interval during which the force acts.
The change in the passenger’s momentum is the same in the
two cases.
What differs is the time interval during which the change occurs.
It takes a larger force to change the momentum in a shorter
time interval.

Example 7.3 Solution
i i 65 kg 20.0 m/s 1300 kg m/s
m
    
p v
His final momentum is zero, so the magnitude of the momentum
change is
1300 kg m/s
  
p
5
av
4
av
1300 kg m/s
(a) No seat belt: 4.3 10 N
0.0030 s
1300 kg m/s
(b) Air bag: 4.3 10 N
0.030 s
t
t
 
   

 
   

p
F
p
F

Graphical Calculation of Impulse
When a force is changing, how can we find the impulse?
For simplicity we consider components along the x-axis. Recall:
av,
av,
• displacement area under ( ) graph
• change in velocity area under ( ) graph
x x
x x x
x v t v t
v a t a t
    
    
By analogy:
av,
• impulse area under ( ) graph
x x
t t
F
F
  

Graphical Calculation of Impulse Continued 1
The area under the Fx(t) graph for a variable force is the impulse.
1
2
area base height 2 s 4 N 8 N s impulse
      

Graphical Calculation of Impulse Continued 2
The average force for a given time interval is the constant force
that would produce the same impulse.
impulse 8 N s
average force 2 N
time interval 4 s

  

Example 7.4
An experimental robotic car of mass 10.2 kg moving at 1.2 m/s in
the +x-direction crashes into a brick wall and rebounds.
A force sensor on the car’s bumper records the force that the
wall exerts on the car as a function of time.
These data are shown in graphical form on the next slide.
(a) What is the maximum magnitude of the force exerted on the
car?
(b) What is the average force on the car during the collision?
(c) At what speed does the car rebound from the wall?

Example 7.4 Continued
i
max av, f
Given: 10.2 kg; 1.2 m/s; graph of ( )
To find : (a) ; (b) ; (c)
x x
x x
m v F t
F F v
 

(a) From the figure, the maximum force is approximately 750 N
in magnitude.

(b)
av, N
10 boxes 2 N s/box 20 N s
0.07 s
impulse 20 N s
300
0.07 s
x
t
F
t
   
 

  


(c)
f i av,
i
f i
f
20 N s
20 N s
1.2 m/s 0.8 m/s
10.2 kg
x x x x
x x x
x x
x
p mv mv F t
p mv p
v v
m m
v
       
  
  
 
   
The car rebounds at a speed of 0.8 m/s.

A Restatement of Newton’s Second Law
We can use the relationship between impulse and momentum
to find a new way to understand Newton’s second law.
Let’s rewrite the impulse-momentum theorem this way:
av
t




p
F
Let the time interval t get smaller and smaller. The average
force is taken over a smaller and smaller time interval,
approaching the instantaneous force.

A Restatement of Newton’s
Second Law Continued
0
lim
t t
 




p
F
When mass is constant, then it can be factored out:
0 0 0
( )
lim lim lim
t t t
m
m m
t t t
     
  
   
  

p v v
F a

7.4 Conservation of Momentum
In a system composed of more than two objects, interactions
between objects inside the system do not change the total
momentum of the system—they just transfer some momentum
from one part of the system to another.
Only external interactions can change the total momentum of
the system.

Conservation of Momentum Continued
If the net external force acting on a system is zero, then the
momentum of the system is conserved.
ext i f
If 0,
 
F p p

Example 7.5
A squid propels itself by filling an internal cavity with water. Then
the mantle, a powerful muscle, squeezes the cavity and expels
water through a narrow opening (the siphon ) at high speed.
Suppose a squid of mass 182 g (including the water that will be
expelled) is initially at rest. It then expels 54 g of water at an
average speed of 62 cm/s (relative to the surrounding water).
Ignoring drag forces, how fast is the squid moving immediately
after expelling the water?

Consider the squid and the water inside its cavity to be a single
system.
Because we assume drag forces on the system are negligible, the
net external force on the system is zero and the momentum of
the system is conserved.

i f s s w w
s
w w
s
s
w w
s
0
182 g 54 g 128 g
(54 g) (62 cm/s)
26 cm/s
128 g
s
m m
m
m
m v
v
m
m
   
  
 

  
p p v v
v
v

7.5 Center of Mass
We can define a point called the center of mass ( cm ) that
serves as an average location of the system.

Definition of center of mass
Vector form:
i i
CM
m
M

 r
r
Component form:
i i i i i i
CM CM CM
, , ,
x
m x m m
y
M
z
y z
M M
  
  
i
where 1, 2, 3, , and
i N M m
   

Center of Mass Continued
The components of the cm are weighted averages of the
position components of the particles.

Example 7.7
Due to the gravitational interaction between the two stars in a
binary star system, each moves in a circular orbit around their
cm.
One star has a mass of 15.0 × 1030 kg; its center is located at x =
1.0 AU and y = 5.0 AU. The other has a mass of 3.0 × 1030 kg; its
center is at x = 4.0 AU and y = 2.0 AU.
Find the cm of the system composed of the two stars. (AU stands
for astronomical unit. 1 AU = the average distance between the
Earth and the Sun = 1.5 × 108 km.)

Given:
2 2
30
1 1 1
30
2
15.0 10 kg 1.0 AU 5.0 AU
3.0 10 kg 4.0 AU 2.0 AU
m x y
y
m x
   
   
To find: CM CM
;
x y

30 30 30
1 2 15.0 10 kg 3.0 10 kg 18.0 10 kg
M m m
     
 
1 2
CM 1 2
30 30
30 30
15.0 10 kg 3.0 10 kg
1.0 AU 4.0 AU
18.0 10 kg 18.0 10 kg
=1.5 AU
x
m m
x x
M M
 
 
  



1 2
CM 1 2
15.0 3.0
5.0 AU 2.0 AU 4.5 AU
18.0 18.0
m m
y
y y
M M

 
   

7.6 Motion of the Center of Mass
i i i i
CM CM
m m t
t
M M
 
 
 

 
r v
r v
CM i i
M m
 
v v
CM
M

p v
CM, CM,
and
x x y y
p
p Mv Mv


The motion of the cm obeys the following statement of Newton’s
second law:
ext CM
M

F a

Example 7.8
A model rocket is fired from the ground in a parabolic trajectory.
At the top of the trajectory, a horizontal distance of 260 m from
the launch point, an explosion occurs within the rocket, breaking
it into two fragments. One fragment, having one third of the
mass of the rocket, falls straight down to Earth as if it had been
dropped from rest at that point.
At what horizontal distance from the launch point does the other
fragment land? Ignore air resistance. [ Hint: The two fragments
land simultaneously.]

Example 7.8 Strategy 1
Strategy 1
Apply conservation of momentum to the explosion. The
momentum of the rocket just before the explosion is equal to the
total momentum of the two fragments just after the explosion.
Why can momentum conservation be assumed here?
The explosion takes place in a very short time interval. As long as
the time interval considered is sufficiently short, the momentum
change of the system can be ignored.

Example 7.8 Strategy 2
Strategy 2
The explosion is caused by an internal interaction between two
parts of the rocket. The motion of the cm of the system is
unaffected by internal interactions, so it continues in the same
parabolic path. Just before the explosion, the rocket is at the top
of its trajectory, so it has py = 0 (with the y-axis pointing up).
Just after the explosion, one fragment is at rest. Then the other
fragment must have py = 0; otherwise, conservation of
momentum would be violated. Then both fragments have vy = 0
just after the explosion. Ignoring air resistance, they land
simultaneously. At that same instant, the cm also reaches the
ground.

i i
i 1 2
2
i 2
3
3
2 i
2
i 1 2
3
2
0 ( )
260 m 260 m 650 m
x x
x x x
x x
x x
y y y
p Mv
p
v
p
M v
v
p p
x
v
p
p
M

 
 

 
    

The piece with mass 1/3 M falls straight down and lands 260 m
from the launch point. After the explosion, the cm continues to
travel just as the rocket itself would have done if it had not
broken apart.
From the symmetry of the parabola, the cm touches the ground
at a distance of 2 × 260 m = 520 m from the launch point. Since
we know the location of the cm and that of one of the pieces, we
can find where the second piece lands.
1 2
CM 1 2
3 3
1 2
CM 1 2
3 3
CM 1
2
3 3 520 m 260 m
650 m
2 2
M M x M x
x x
x x
x
x
x
 
 
  
  

7.7 Collisions in One Dimension
Analyzing Collisions Using Momentum Conservation
We can often use conservation of momentum to analyze
collisions even when external forces act on the colliding objects.
If the net external force is small compared with the internal
forces the colliding objects exert on each other during the
collision, then the change in the total momentum of the two
objects is small compared with the transfer of momentum from
one object to the other.
Then the total momentum after the collision is approximately
the same as it was before the collision.

Example 7.9
A krypton atom (mass 83.9 u) moving with a velocity of 0.80
km/s to the right and a water molecule (mass 18.0 u) moving
with a velocity of 0.40 km/s to the left collide head-on.
The water molecule has a velocity of 0.60 km/s to the right after
the collision. What is the velocity of the krypton atom after the
collision? (The symbol “u” stands for the atomic mass unit.)

1f 2f 1i 2i

 
p p p p
For simplicity we drop the “x” subscripts from the x-
components, remembering that all quantities refer to x-
components:
1 1f 2 2f 1 1i 2 2i
m v m v m v m v
  
Since m1/m2 = 83.9/18.0 = 4.661, we can substitute
m1 = 4.661 m2:
2 1f 2 2f 2 1i 2 2i
4 1 4.6
.66 61
v v
m m v m v
m 
 

1i 2i 2f
1f
4.661
4.661
4.661 0.80 km/s ( 0.40 km/s) 0.60 km/s
4.661
0.59 km/s
v
v v v
 

   


After the collision, the krypton atom moves to the right with a
speed of 0.59 km/s.

Elastic and Inelastic Collisions
A collision in which the total kinetic energy is the same before
and after is called elastic.
There is no conservation law for kinetic energy by itself. The
elastic collision is just a special kind of collision in which no
kinetic energy is changed into other forms of energy.
When the final kinetic energy is less than the initial kinetic
energy, the collision is said to be inelastic.

Perfectly Inelastic Collisions
When a collision results in two objects sticking together, the
collision is perfectly inelastic.
The decrease of kinetic energy in a perfectly inelastic collision is
as large as possible (consistent with the conservation of
momentum).

Problem-Solving Strategy for Collisions
Involving Two Objects 1
1. Draw before and after diagrams of the collision.
2. Collect and organize information on the masses and
velocities of the two objects before and after the collision.
Express the velocities in component form (with correct
algebraic signs).
3. Set the sum of the momenta of the two before the collision
equal to the sum of the momenta after the collision. Write
one equation for each direction:
1 1i 2 2i 1 1f 2 2f
1 1i 2 2i 1 1f 2 2f
x x x x
y y y y
m v m v
m
m v m v
v m v m v m v
  
  

4. If the collision is known to be perfectly inelastic, set the final
velocities equal:
1f 2f 1f 2f
and
x x y y
v
v v v
 
5. If the collision is known to be perfectly elastic, then either
set the final kinetic energy equal to the initial kinetic energy:
2 2 2 2
1 1 1 1
1 1i 2 2i 1 1f 2 2f
2 2 2 2
m v m v m v m v
  
or, for a one-dimensional collision, set the relative speeds
equal:
2i 1i 2f 1f
( )
x x x x
v v
v v
   

6. Solve for the unknown quantities.

Example 7.10
At a Route 3 highway on-ramp, a car of mass 1.50 × 103 kg is
stopped at a stop sign, waiting for a break in traffic before
merging with the cars on the highway.
A pickup of mass 2.00 × 103 kg comes up from behind and hits
the stopped car. Assuming the collision is elastic, how fast was
the pickup going just before the collision if the car is pushed
straight ahead onto the highway at 20.0 m/s just after the
collision?

Given: m1 = 1.50  103 kg; m2 = 2.00  103 kg; before the
collision, v1i = 0; after the collision, v1f = 20.0 m/s
To find: v2i (speed of the pickup just before the collision)

1 1i
m v 2 2i 1 1f 2 2f
2i 1i
m v m v m
v
v
v
  
 2f 1f
( )
v v
  
2 2i 2 1f 2 2f
2 2i 1 2 1f
2 ( )
m v m v m v
m
m v m v
 
 
1 2
2i 1f
2
1500 kg 2000 kg
20.0 m/s 17.5 m/s
2 4000 kg
v
m m
v
m
 
   

7.8 Collisions in Two Dimensions
Most collisions are not limited to motion in one dimension in the
absence of a track or other device to constrain motion to a single
line.
In a two-dimensional collision, we use the same techniques we
used for one-dimensional collisions, as long as we remember
that momentum is a vector.
To apply conservation of momentum, it is usually easiest to work
with x- and y-components.

Example 7.11
A small puck (mass m1 = 0.10 kg) is sliding to the right with an
initial speed of 8.0 m/s on an air table. An air table has many tiny
holes through which air is blown; the resulting air cushion allows
objects to slide with very little friction. The puck collides with a
larger puck (mass m2 = 0.40 kg), which is initially at rest. After
the collision, the pucks move off at angles 1 = 60.0° above and
2 = 30.0° below the initial direction of motion of the small puck.

Example 7.11 Continued
(a) What are the final speeds of the pucks?
(b) Is this an elastic collision or an inelastic collision?
(c) If inelastic, what fraction of the initial kinetic energy is
converted to other forms of energy in the collision?

Masses: 1 2
0.10 kg; 0.40 kg
m m
 
Before Collision: 1i 1i 2i 2i
8.0 m/s; 0
x y x y
v v v v
   
After Collision: 1f 1f 1 1f 1f 1
2f 2f 2 2f 2f 2
1 2
cos ; sin ;
cos ; sin
60.0 and 30.0 )
(
x y
x y
v v v
v v v
v
v
 
 
 
 
  
   
To find: v1f and v2f; total kinetic energy before and
after the collision

(a) Working with components means that we set the total x -
component of momentum before the collision equal to the
total x -component of momentum after the collision.
1f 2f 1i 2i
1 1f 1 2 2f 2 1 1i
2 1
1 1f 1 2f 1 1i
1f 2f
cos cos 0
4
cos60.0 4 cos30.0
3.46 8.0 m/s
0.500
x x x x
x
x
p
m
p p p
v m v m v
m m
v m v
v
m
v
m v
 
  
  

   
 

(a) continued. We treat the y-components in the same way. The
initial momentum is in the x-direction only. Thus, the total
momentum y -component after the collision must be zero.
1f 2f 1i 2i
1 1f 1 1 2f 2
1f 2f
2f 1f 1f
0
sin ( 4 sin ) 0
sin60.0 4 sin30.0 0
sin60.0
0.433
4sin30.0
y y y y
p p p p
m v m v
v v
v v v
 
   
  
   

 


(a) continued
1f 2f
2f 1f 1f
1f 1f 1f
1f
2f 1f
3.46 8.0 m/s
sin60.0
0.433
4sin30.0
3.46(0.433 ) 2.00 8.0 m/s
4.0 m/s
0.433 1.73 m/s 1.7 m/s
0.500
0.500
v
v v v
v v
v
v v
v
v
 

 

  

  

(b) 2
1
i 1 1i
2
2
1
2 (0.10 kg) (8.0 m/s) 3.2 J
m v
K 
  
2 2
1 1
f 1 1f 2 2f
2 2
2 2
1 1
2 2
(0.10 kg) (4.0 m/s) (0.40 kg) (1.73 m/s)
J 0.60 J 1.40 J
0.80
m v m v
K

 
  
 

(c) J 1.40 J 1.8 J
1.8 J
0.56
.
3.2 J
3 2  


PHY300 Chapter 7 physics 5e

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