•   Momentum
   Momentum and impulse
   Conservation of momentum http://www.physics.wayne.edu/~apetrov/PHY2130/

Last lecture:
1.
Work and energy:
   elastic potential energy
   momentum
Review Problem: An 8.0 kg crate, initially at rest, slides down a slope as in the picture. The work of the force of friction is 300 J. What is the speed of the crate at the bottom of the slope?

An 8.0 kg crate, initially at rest, slides down a slope as in the picture. The work of the force of friction is 300 J. What is the speed of the crate at the bottom of the slope?
Idea: use energy considerations with nonconservative forces present
W nc
= Δ E = E f
− E i
Thus:
At the top: E
At the bottom: E i f
= mgh + 0 mv
= 0 +
2
2
W nc
= mv 2
2
− mgh or v =
2 m
( mgh + W nc
) = 19.2
m / s

•
From Newton ’ s laws: force must be present to change an object ’ s velocity (speed and/or direction)

Wish to consider effects of collisions and corresponding change in velocity
Golf ball initially at rest, so some of the KE of club transferred to provide motion of golf ball and its change in velocity

Method to describe is to use concept of linear momentum scalar vector

p = m v
•
Vector quantity , the direction of the momentum is the same as the velocity’s
•
Applies to two-dimensional motion as well p x
= mv x and p y
= mv y
Size of momentum: depends upon mass
depends upon velocity

•
In order to change the momentum of an object
(say, golf ball), a force must be applied
•
The time rate of change of momentum of an object is equal to the net force acting on it
•

F net

= ma =
 m ( v f
Δ t

− v i
)
=

Δ p
Δ t
 or : Δ p

= F net
Δ t
•
Gives an alternative statement of Newton ’ s second law
•
(F Δ t) is called the impulse
•
Impulse is a vector quantity , the direction is the same as the direction of the force

•
Usually force is not constant, but time-dependent impulse =
Δ t i
•
•
Δ i
= ( )
If the force is not constant, use the average force applied
The average force can be thought of as the constant force that would give the same impulse to the object in the time interval as the actual time-varying force gives in the interval
If force is constant: impulse = F Δ t

•
The most important factor is the collision time or the time it takes the person to come to a rest
•
This will reduce the chance of dying in a car crash
•
Ways to increase the time
•
Seat belts
•
Air bags

The air bag increases the time of the collision and absorbs some of the energy from the body

9
Example: A car of mass 1500 kg collides with a wall and rebounds as shown. If the collision lasts for 0.150 s, Find (a) the impulse delivered to the car due to the collision and (b) the magnitude and direction of the average force exerted on the car.
–   Assume force exerted by wall is large compared with other forces
–
Gravitational and normal forces are perpendicular and so do not effect the horizontal momentum
–
Can apply impulse approximation
(a) Impulse delivered to the car
= Δ

= f
− p i
= ( 1500 kg × 2 .
60 m / s ) − ( 1500 kg × ( − 15 .
0 m / s ))
= 2 .
64 × 10 4 kg .
m / s
(b) The average force exerted on the car
 av
=
Δ

Δ t
=
2 .
64 × 10
0 .
4 kg
150 s
.
m / s
= + 1 .
76 × 10 5 N

Suppose a ping-pong ball and a bowling ball are rolling toward you. Both have the same momentum, and you exert the same force to stop each. How do the time intervals to stop them compare?
1. It takes less time to stop the ping-pong ball.
2. Both take the same time.
3. It takes more time to stop the ping-pong ball.

Suppose a ping-pong ball and a bowling ball are rolling toward you. Both have the same momentum, and you exert the same force to stop each. How do the time intervals to stop them compare?
1. It takes less time to stop the ping-pong ball.
2. Both take the same time. 
3. It takes more time to stop the ping-pong ball.
Note: Because force equals the time rate of change of momentum, the two balls loose momentum at the same rate. If both balls initially had the same momenta, it takes the same amount of time to stop them.

A 50-g golf ball at rest is hit by “ Big
Bertha ” club with 500-g mass.
After the collision, golf leaves with velocity of 50 m/s. a)   Find impulse imparted to ball b)   Assuming club in contact with ball for 0.5 ms, find average force acting on golf ball

Given: mass: m=50 g
= 0.050 kg velocity: v=50 m/s
Find: impulse=?
F average
=?
1. Use impulse-momentum relation: impulse = Δ p = mv f
− mv i
=
=
( 0 .
050
2 .
50 kg kg )( 50
⋅ m s m s ) − 0

2. Having found impulse, find the average force from the definition of impulse:
Δ p = F ⋅ Δ t , thus F =
=
Δ p
Δ t
=
2 .
50
0 .
5 .
00 × 10 3
5 kg ⋅ m
× 10 − 3
N s s

Note: according to Newton ’ s 3 rd law, that is also a reaction force to club hitting the ball: of club
F ⋅ Δ t m v f
= − F
R
− m v i
=
⋅ Δ t ,
−
(
M V or f m v f
+ M V f
− M V i
= m v i
+ M V i
,
) or
CONSERVATION OF MOMENTUM

•
Definition: an isolated system is the one that has no external forces acting on it
Momentum in an isolated system in which a collision occurs is conserved (regardless of the nature of the forces between the objects)
•
A collision may be the result of physical contact between two objects
•
“ Contact ” may also arise from the electrostatic interactions of the electrons in the surface atoms of the bodies

The principle of conservation of momentum states when no external forces act on a system consisting of two objects that collide with each other, the total momentum of the system before the collision is equal to the total momentum of the system after the collision

•
Mathematically: m
1 v
1 i
+ m
2 v
2 i
= m
1 v
1 f
+ m
2 v
2 f
•
Momentum is conserved for the system of objects
•
The system includes all the objects interacting with each other
•
Assumes only internal forces are acting during the collision
•
Can be generalized to any number of objects

Let ’ s go back to our golf ball and club problem:
Ball : Δ p = 2 .
50 kg ⋅ m s , m = 50 gramm
Club :
Δ v = m v
( f
− v i
)
= v
( f
− v i
)
=
− 2 .
50
− 2 .
50
0 .
5 kg kg kg
⋅ m
⋅ m
50 s s , m
= so s
− 5 m s factor of 10 times smaller

18
v
1i v
2i m
1
>m
2 m
1
A short time later the masses collide. m
2 m
1 m
What happens?
2

During the interaction:
N
1
N
2 y
F
21
F
12 w
1 w
2
F y
F x
= N
1
− w
1
= 0
= − F
21
= m
1 a
1
F y
F x
= N
2
− w
2
= F
12
= 0
= m
2 a
2
There is no net external force on either mass.
19 x


F
21

F
21
Δ p
1
=

1
Δ t
= f
−
−
=

F
12
−

F
12
−
1 i
Δ p
=
2
1 i
+
2 i
=
Δ t p

1
2 f f
+
−
20
2 f
2 i
If net external force acting on a system is zero, then the momentum of the system is conserved
If

F ext
= 0 , i
= f

21
Example: A rifle has a mass of 4.5 kg and it fires a bullet of 10.0 grams at a muzzle speed of 820 m/s. What is the recoil speed of the rifle as the bullet leaves the barrel?
Given: m r
= 4.5 kg m b
= 10.0 g
=0.01 kg v b
= 820 m/s v ir
= v ib
= 0 m/s
Find: v r
= ?
Idea: as long as the rifle is horizontal, there will be no net external force acting on the rifle-bullet system and momentum will be conserved. p i
= p f
0 = m b v b
+ m r v r v r
= − m b m r v b
= −
"
$
#
0.01 kg
4.5 kg
%
' 820 m/s
&
= − 1.82 m/s

Suppose a person jumps on the surface of Earth. The Earth
1. will not move at all
2. will recoil in the opposite direction with tiny velocity
3. might recoil, but there is not enough information
provided to see if that could happened

Suppose a person jumps on the surface of Earth. The Earth
1. will not move at all
2. will recoil in the opposite direction with tiny velocity
3. might recoil, but there is not enough information
provided to see if that could happened

Note: momentum is conserved. Let ’ s estimate Earth ’ s velocity after a jump by a 80-kg person. Suppose that initial speed of the jump is 4 m/s, then:
Person : Δ p = 320 kg ⋅ m s
Earth :
V
Δ p
Earth
=
=
M
Earth
V
Earth
− 320 kg
6 × 10 24
⋅ m kg
= s
− 320 kg
= −
⋅ m s ,
5 .
3 × 10 − 23 so m s tiny negligible velocity, in opposite direction

Do you know why police use skid marks to find velocities of vehicles before collision???
See page 246
24

•
The basic equation for rocket propulsion is:
•
•
•   v f
− v i
= v e ln
⎛
⎜⎜
⎝
M
M i f
⎞
⎟⎟
⎠
M i
is the initial mass of the rocket plus fuel
M f
is the final mass of the rocket plus any remaining fuel
The speed of the rocket is proportional to the exhaust speed

•
The thrust is the force exerted on the rocket by the ejected exhaust gases
•
The instantaneous thrust is given by
Ma = M
Δ v
Δ t
= v e
Δ M
Δ t
•
The thrust increases as the exhaust speed increases and as the burn rate ( Δ M/ Δ t) increases