Rev. 08-Feb-04 GB
Physics I
Class 07
07-1

Extra Help for Physics and Calc
There is a new tutoring service available from the
Advising & Learning Assistance Center called
Cluster Group Tutoring.
Sun. and Mon. 6 PM, Academy Hall 2800
PHYS/CALC Cluster Group http://www.rpi.edu/dept/advising/tutoring.html
07-2

What is a System of Objects?
The universe is too large to include all of it in an experiment. We can only concentrate our attention on a small part. If we do things right, we can select a small group of interacting objects in such a way that the phenomenon we want to study is not significantly influenced by anything else.
How to “do things right” is the tricky part.
A “system of objects” is a subset of the universe that we have selected to study a phenomenon.
07-3

Internal and External Forces
Our system here consists of Objects A and B.
Forces between A and B are internal forces.
Forces on A or B from sources outside the system are external forces.
If we change the definition of the system, could that affect which forces are internal and which are external?
F on A from C F on B from C
External Forces
Object A
F on A from B F on B from A
Internal Forces Object B
07-4

The Momentum of a System
The momentum of a system is the sum of all the individual parts:

P
 i
N 

1
 p i
Newton’s Second Law for each object:

F net , i
 m i
 a i
 d d
 p t i
Newton’s Second Law for the system: d d

P t
 i
N 

1 d d
 p t i
 i
N 

1

F net , i
 all
 system

F
07-5

Cancellation of Internal Forces
Some forces in a system are internal , some are external .

F
 

F int
 

F ext all
 system
The internal forces are all in Newton’s Third Law Pairs within the system, so they sum exactly to zero in the system.

F

0
 

F ext
 

F ext all
 system
07-6

Conservation of Momentum
(in a Nutshell)
Only external forces can change the momentum of a system.
d d

P t
 

F ext
If the external forces cancel and/or can be neglected, then momentum is constant (zero time derivative), or as physicists say, conserved .
d

P
0 d t

07-7

One-Dimensional Example
Two Carts on a Track
Two objects are initially at rest, P = 0.
The objects spring apart; the spring force is internal to the system.
After the spring pushes them apart, because P is conserved: p
1 p
2
P
 after
P before
P
 m
1 v
1 p
1

 p
2
 m
2
 v
2
0
 m
1 v
1
 m
2 v
2
07-8

Class #7
Take-Away Concepts
1.
Systems; internal/external forces in systems.
2.
Momentum defined for a system:

P d d


P t

N   p i

1
3.
Newton’s Second Law for a system:
 i

F ext
4.
Conservation of momentum when d P
 d t
P
 after

P

F ext before

0
07-9

Class #7
Problems of the Day
_____1. You are investigating conservation of momentum in one dimension by observing two carts – A and B – collide on a track, similar to the carts and tracks we use in class. The mass of cart A is 1.000 kg and the mass of cart B is 0.500 kg. Friction may or may not be negligible. During the collision of the carts, the force (magnitude) on A from B is Fab and the force on B from A is Fba. Which statement(s) is/are true?
A) Fab > Fba if friction is negligible.
B) Fab > Fba if friction is significant.
C) Fab = Fba if friction is negligible.
D) Fab = Fba if friction is significant.
E) Fab < Fba if friction is negligible.
F) Fab < Fba if friction is significant.
07-10

Answer to Problem 1 for Class #7
The answers are C and D.
The forces between the carts during collision are a Newton’s 3 rd
Law pair. The magnitudes of the forces will be exactly equal (and directions exactly opposite) no matter what masses the carts have and no matter what other forces are acting on the carts.
07-11

Class #7
Problems of the Day
_____2. A rifle of mass M is initially at rest but free to recoil. It fires a bullet of mass m and velocity +v
(relative to the ground) in the X dir. After firing, the velocity of the rifle (relative to the ground) is:
A) –mv
B) –Mv/m
C) –mv/M
D) –v
E) +mv/M
07-12

Answer to Problem 2 for Class #7
The answer is C.
The system is the rifle and the bullet. The initial momentum of the system is zero (at rest).
Momentum is conserved in this system because all external forces are neglected. (The rifle is free to recoil.)
Let the velocity of the rifle be V. Since the total momentum after firing is still zero, we have M V + m v = 0 or V = –mv/M.
07-13

Activity #7 -
Conservation of Momentum
Objectives of the Activity:
1.
Think about how systems are defined and how that affects the classification of internal/external forces.
2.
Use VideoPoint to study conservation of momentum for a two-object system in one dimension.
07-14