Transcript Ch3 - Momentum and Conservation of Momentum

Vern J. Ostdiek
Donald J. Bord
Chapter 3
Energy and Conservation
Laws
Conservation laws
 The most fundamental ideas we have in
physics are conservation laws.

Statements telling us that some quantity
does not change
 Conservation of mass states:
 The total mass of an isolated system is
constant.
 To apply these, we must define a “system.”
Conservation laws, cont’d
 A system is just a collection of objects we
decide to treat at one time.


The tanker and fighter can represent a
system.
The fuel leaving
the tanker goes
into the fighter:
mass is conserved
Linear momentum
 Linear momentum is defined as the product
of an object’s mass and its velocity.
linear momentum  mass  velocity
p  mv

We typically just say momentum.
Linear momentum, cont’d
 Momentum is a measure of an object’s state
of motion.

Consider an object whose momentum is
1 kg·m/s


This could be a 0.005 kg bullet traveling at 200
m/s.
This could be a 0.06 kg tennis ball traveling at
16.7 m/s.
Linear momentum, cont’d
Momentum (continued)
 high mass or high velocity  high momentum
 high mass and high velocity  higher momentum
 low mass or low velocity  low momentum
 low mass and low velocity  lower momentum
Linear momentum, cont’d
 Newton’s 2nd law is closely related to
momentum.

The net external force acting on an object
equals the rate of change of linear momentum:
change in momentum
force 
change in time
p
F
t
Linear momentum, cont’d
 How is this related to F = ma?
 
p  mv
v
F

m
 ma
t
t
t
Example
Example 3.1
Let’s estimate the average
force on a tennis ball as it
is served. The ball’s mass
is 0.06 kg and it leaves
the racquet with a speed
of 40 m/s. High-speed
photography indicates that
the contact time is about
5 milliseconds.
Example
Example 3.1
ANSWER:
The problem gives us:
m  0.06 kg
vi  0 m/s
v f  40 m/s
The force is:
F
t  0.005 s
   0.06 kg 40 m/s
 mv
t
0.005 s
 480 N  108 lb
Linear momentum, cont’d
 This tells why we must exert a force to stop
an object or get it to move.


To stop a moving object, we have to bring its
momentum to zero.
To start moving an object, we have to impart
some momentum to it.
Momentum
When the speed of an object is doubled, its
momentum:
A. remains unchanged in accord with the
conservation of momentum.
B. doubles.
C. quadruples.
D. decreases.
Impulse
 The change in momentum of an object is
equal to the impulse applied to it (force
multiplied by the time interval during which
the force is applied).
 Impulse =

p  F t
The change of momentum, or the Force
multiplied by time, is called “Impulse”.
Impulse

Impulse tells us that we can change the
momentum using various forces and time
intervals.

You can get the same impulse by using a
large force for a short time, or using a small
force for a long time.
Impulse
Impulse
•
•
product of force and contact time
impulse = force  time = Ft
great force for long time  large impulse
same force for short time  smaller impulse
Impulse
When the force that produces an impulse acts
for twice as much time, the impulse is
doubled as well.
Example:
golfer follows through while
hitting the golf ball
•
Impulse
When a car is out of control, it is better to hit a
haystack than a concrete wall. Common
sense, but with a physics reason:
Same impulse occurs
either way, but extension
of hitting time reduces
hitting force.
Conservation of momentum
 The Law of Conservation of Momentum
states:
The total momentum of an isolated system is
constant (no external forces).
A system will have the same momentum both
before and after any interaction occurs. When
the momentum does not change, we say it is
conserved.
Conservation of linear
momentum, cont’d
 This law helps us deal with collisions.
 If the system’s momentum can not change,
the momentum before the collision must
equal that after the collision.
Conservation of linear
momentum, cont’d
 We can write this as:
pbefore  pafter
 To study a collision:
 Add the momenta of the objects before the
collision.
 Add the momenta after the collision.
 The two sums must be equal.
Example
Example 3.2
A 1,000 kg car (car 1) runs into the rear of a stopped
car (car 2) that has a mass of 1,500 kg. Immediately
after the collision, the cars are hooked together and
have a speed of 4 m/s. What was the speed of car 1
just before the collision?
Example
Example 3.2
ANSWER:
The problem gives us:
m1  1,000 kg
m2  1,500 kg
v f  4 m/s
The momentum before:
pbefore  m1v1  1,000 kg v1


The momentum after:




pafter  m1  m2 v2  2,500 kg 4 m/s

Example
Example 3.2
ANSWER:
Conserving momentum
1,000 kg v  2,500 kg 4 m/s
2,500 kg
v 
4 m/s 

1,000 kg
1
1
 10 m/s
Example
Example 3.2
DISCUSSION:
Both cars together have more mass than just
car 1.
Since both move away at 4 m/s, the lighter car
1 must have a greater speed before the
collision.
Conservation of linear
momentum, cont’d
 How do rockets work?

The exhaust exits the rocket
at high speed.


We need high speed because
the gas has little mass.
The rocket moves in the
opposite direction.

Not as fast as the
gas because it has more mass