Physics I
Class 06
Impulse and
Momentum
Rev. 01-Feb-04 GB
06-1
Momentum of an Object
Definitions
We define momentum
for an object to be:

p  mv
Momentum is a vector. It is in the same direction as velocity.
SI units for momentum: kg m/s.
Later in the semester, we will talk about angular momentum.
Plain “momentum” is also known as linear or translational momentum.
06-2
Change of Momentum
Change of momentum is the difference between the final value
and the initial value.
Beware: We are subtracting vectors!
 



 p  p final  p initial  m v final  v initial 
final
initial
+3

+3
= 0 kg m/s
+3

-3
=
+6 kg m/s
06-3
Connecting Net Force and the
Derivative of Momentum
Newton’s Second
 Law:

 F  Fnet  m a
The time derivative of momentum:


dp d
d
v


 ( m v)  m
 ma
dt dt
dt
We can write Newton’s Second
Law in a new form:

 
dp
 F  Fnet  d t
06-4
The Fundamental Theorem of
Calculus
Recall from Class #1:
Math Fact: Because velocity is the derivative
of displacement, displacement is the area
(integral) under the graph of v versus t.
The Fundamental Theorem of Calculus:
If f is the derivative of g, then
v
v0
t0
t
b
 f (t ) dt  g(b)  g(a )   g
a
06-5
Impulse and the ImpulseMomentum Theorem
Impulse is defined to be the time integral of force. SI units = N s.
Like force, it is a vector. (Net or total force is implied.)


J   F dt
Using the Fundamental Theorem of Calculus:



J   F dt   p
In Physics, this is known as the Impulse-Momentum Theorem.
06-6
Example Problem Using
Impulse-Momentum
An object of mass 0.5 kg is subjected to a
F (N)
force in the +X direction that varies as shown
in the graph from 0 to 7 seconds. Its initial X 5
velocity is zero. What is its final X velocity?
The points on the graph are (0,0), (1,4), (2,2),
(3,5), (4,1), (5,3), (6,3), and (7,0).
0
0
7s t
06-7
Example Problem Using
Impulse-Momentum
An object of mass 0.5 kg is subjected to a
force in the +X direction that varies as shown F (N)
in the graph from 0 to 7 seconds. Its initial X 5
velocity is zero. What is its final X velocity?
Doing this with F=ma would be hard. Doing
it with impulse-momentum is much easier.
0
J = area = 2+3+3.5+3+2+3+1.5 = 18 N s.
0
7s t
18 N s  J   p  p final  p initial
p final  p initial  p final  0  p final  18 kg m / s
v final  p final  m  36 m / s
06-8
Good-bye “Average Force”
Favg
Average force can be defined mathematically,
but it is a concept too easily misused.
We will avoid Average Force in Physics 1.
Never use “Favg = ma” !
Use the Impulse-Momentum Theorem:



J   F dt   p
06-9
Class #6
Take-Away Concepts
1.
Momentum

 defined for an object:
2.
A new way to write Newton’s Second Law:
3.
Impulse defined:
4.
Impulse-Momentum
Theorem:


p  mv

 
dp
 F  Fnet  d t


J   F dt

J   F dt   p
06-10
Class #6
Problems of the Day
___1. For this question, take right as the positive direction. A
0.375 kg rubber ball traveling horizontally to the right at
10 m/s hits a wall and bounces back at 6 m/s to the left.
The total impulse exerted by the wall on the ball is
A)
B)
C)
D)
–6.0 N s.
+6.0 N s.
–1.5 N s.
+1.5 N s.
06-11
Answer to Problem 1 for Class #6
The answer is A.
J = p = p-final – p-initial = m (v-final – v-initial)
v-final = –6 m/s. v-initial = +10 m/s.
J = 0.375 * (–6 – 10) = 0.375 * (–16) = –6.0 N s.
06-12
Class #6
Problems of the Day
___2. Egg Toss Contest: You and a friend are tossing a raw egg
back and forth. For a given speed of toss, you want to catch
the egg without breaking it. To do this, you need to keep the
force on the egg shell low. How can you do that?
A) You can’t. The egg experiences the same change in momentum
no matter how you catch it, therefore it experiences the same
force from your hand.
B) Catch the egg with as stiff an arm as possible to bring it to rest
quickly. That will decrease the time during which the force acts,
thus decreasing the force on the egg.
C) Catch the egg with a lot of arm movement to stretch out the time
of the catch, increasing the time during which the force acts.
06-13
Answer to Problem 2 for Class #6
The answer is C.
The impulse on the egg will be the same no matter how you catch it.
However, impulse is the integral (area under the curve) of force
versus time. If you decrease the time over which the force acts, the
force has to be greater to give the same area under the curve. So
you want to catch the egg (bring it to rest) over a long time period in
order to keep the force low.
This is the same principle behind the automobile air bag. It brings a
person to rest in a longer time than slamming into the steering wheel
or dashboard.
06-14
Class #6
Problems of the Day
3. The Physics Department decides that we need a more spectacular demonstration of
projectile motion for class. We contact a supplier of physics education equipment and
they send us the specifications for a long-range classroom projectile launcher, a more
powerful version of the launcher we saw in class and in the VideoPoint activity. The
mass of the ball fired by the new launcher is 10 g. The graph of net force versus time
for launching the ball is shown below:
force (N)
2
0
0
0.01
0.1
time (s)
Calculate the speed of the ball as it leaves the launcher.
06-15
Answer to Problem 3 for Class #6
You could solve this problem using F = m a and calculus, but you cannot
use the equations of motion for constant acceleration, because acceleration
is not constant in this problem.
Instead, use J = p. In this case, p-initial = 0, so we have J = m v.
J is the area under the curve. Using the formula for a triangle, J = 0.1 N s.
v = J/m = 0.1/0.01 = 10 m/s. (mass in kg)
06-16
Activity #6 Impulse and Momentum
Objectives of the Activity:
1.
2.
3.
Learn how to calibrate sensors to take accurate data.
Take data in a real experiment to study the relationship
between impulse and momentum change.
Learn how to use data analysis features of LoggerPro.
06-17
Class #6 Optional Material
Special Theory of Relativity
Albert Einstein (1879–1955)
In Lecture 03, we mentioned that Albert Einstein
showed that Newton’s Second Law had to be
modified in order to account for the observed
behavior of electromagnetic waves (light) and the
interaction of electromagnetic fields with matter.
Einstein’s Two Postulates of Special Relativity:
1. The laws of physics are the same in all inertial frames.
2. The speed of light, c, is constant in all inertial frames.
06-18
What is an Inertial Frame?
An inertial frame (of reference) is a real or imaginary set of devices
for measuring position and time that are in motion together
according to Newton’s First Law; in other words, these devices are
not accelerating (or rotating).
Neglecting gravity (we’ll talk about that later in the semester), the
track and motion detector that we use in our activities, along with the
clock in your PC when you run LoggerPro, comprise an inertial
reference frame with a special name: the laboratory reference
frame (because this is the frame we use to make measurements).
If we were to set the same equipment up in the Ferris Wheel we
studied earlier, that would not be an inertial reference frame.
Instead, we call that an accelerated frame.
06-19
Where Did Einstein’s
Postulates Come From?
1. The laws of physics are the same in all inertial frames.
This idea goes back to Galileo. Imagine an experiment like our cart track
and hanging weight being performed in an airliner moving uniformly in one
direction at a constant speed and altitude. (Assume no turbulence and the
altitude is low enough so that the force of gravity is about the same as on
the ground.) Our measurements that we take with LoggerPro should be the
same as what we did in class if we set things up carefully.
2. The speed of light, c, is constant in all inertial frames.
Nobody wanted to believe this prior to the Michelson-Morley experiment in
1887. In one of the most famous null results in the history of science, they
showed by careful measurement that the speed of light is independent of the
relative motion of the source and detector. This counter-intuitive result was
not adequately explained until Einstein’s Special Theory of Relativity. Do a
web search and you will see that many people still refuse to believe it!
06-20
Einstein’s Correction to
Newton’s Second Law
Starting from the two postulates, Einstein showed that Newton’s
Second Law would be correct for all velocities if the definition of
momentum was modified:

 
dp
F

F

 net d t

p
m0 
v
2
v
1 2
c
Note that the square root term in the denominator allows the
magnitude of the momentum to become arbitrarily large as the
speed approaches but never quite reaches the speed of light. At
normal speeds much smaller than c, the correction is negligible.
06-21
http://www.amnh.org/exhibitions/einstein/?src=h_h
06-22