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PES 1110 Fall 2013, Spendier
Lecture 22/Page 1
Today:
- Linear Momentum and Impulse (9.4-9.6)
- go over HW 5, problem 2
So far we have talked about the motion of single objects. Last lecture we have talked
about the center of mass of s system of particles and how one can still apply all the
equations we have learned so far to a system of particles. We just need to calculate the
system’s center of mass. Next we will talk about collisions of objects.
Before look at collisions we need to define linear momentum and impulse.
Thought experiment:
I throw a ping pong ball at you at 10 m/s.
Then I throw a car at you at 10 m/s.
Which is harder to stop? The car is harder to stop.
What is the physical difference in the two situations?
mcar >> mball (m = mass)
I throw a tennis ball at you at 5 m/s.
Then I throw an identical ball at you at 50 m/s.
Which ball is harder to stop? The faster ball.
What is the physical difference between the situations? The speed is different.
The idea of getting an object to stop is similar to the idea of the work required to stop a
moving object. Both depend on mass an speed.

Linear Momentum: p
(later we will talk about “angular” momentum for rotational motion)
Momentum p can be thought of as indicating how difficult an object is to stop
When p increases the mass increases
When p increases the speed increases
We will show that the equation for momentum is:


p  mv
Derivation:
At this point we can all rewrite Newton's second law.


 F  ma
net
write acceleration in terms of velocity (acceleration is the time derivative of velocity)


dv
F

m

dt
net
To go back where Newton started we move the term for mass m inside the derivative
 d  mv 
F

dt
net
PES 1110 Fall 2013, Spendier
Lecture 22/Page 2
Force is the rate at which this quantity mv = p changes. Here p stands for momentum.
 dp
F

dt
net
All the forces added together is the rate at which momentum changes with time. This is a
second way we can write Newton's second law.


Momentum: p  mv units [kg m/s] (no fancy name for units of momentum)
Momentum measures how “hard" it is to change the velocity of an object.
This makes sense, it is harder for heavier objects (big mass) it is harder to change the
velocity so larger momentum.
Example 1:
What is the momentum of a 2 ton truck travelling at 80 km/h west?
Example 2:
Imagine a 1.5 kg ball hitting a wall with a horizontal velocity is 21 m/s and bouncing
back with a horizontal velocity of 18 m/s. What is the change in momentum of the ball?
We know from Newton’s second law,
 dp
F

dt
net
That some net force has acted on the ball to change the momentum. This force has come
form the wall.
PES 1110 Fall 2013, Spendier
Lecture 22/Page 3
Linear Momentum of a System of particles (9.5)
Consider a system of n particles, each with its own mass, velocity, and linear momentum.
The particle may interact with each other, and external forces may act on them,
total linear momentum of system:
  




P  p1  p2  p3  ...  m1v1  m2 v2  m3 v3  ...


P  M vcom
(linear momentum for a system of particles)
If we take the derivative



dvcom
dP
M
 M acom   F
dt
dt
net

The net external force  F of a system of particles changes the linear momentum
net


P  M vcom of the system.
The linear momentum can be changed only by a net external force.

Impulse: J

The momentum p of any particle-like object cannot change unless a net external force
changes it. Push a body to change its momentum or let it collide with a target object to
change its momentum. In a collision, the external force on the body is brief, has large
magnitude, and suddenly changes the body's momentum.
Single collision:
projectile = ball
target = bat
Collision is brief and the ball is experiencing a force that is great enough to reverse the
ball's motion.
Impulse-Momentum Theorem (Average Force Version):
 dp
F

dt
net

Think about the average net force acting on an object, we can call this Favg . An average
force is always a constant. So for the average force we can write (as we did for average
velocity and average acceleration)


p
Favg 
t
PES 1110 Fall 2013, Spendier
Lecture 22/Page 4
Hence after re-arranging:

 
Favg t  p  J units [N s = kg m/s] (momentum and impulse have same units)
The impulse is equal to the change of an object's momentum.


J  p
or the impulse is the average net force multiplied by the elapsed time, ∆t. How long the
net force was applied.
 
J  Favg t
Impulse: area under force curve
When we do the calculus version of the Impulse -Momentum Theory we get:
 tf 
J   F (t ) dt (book page 212)
ti
So if you plot force versus time, the impulse is the area underneath the curve.
Example 3:
A 5.0 kg block is sliding on a frictionless, horizontal surface going 6.0 m/s to the right
when it hits a wall and stops. What impulse is imparted to the block? (Impulses are
always "imparted")
You will see a lot of "before" and "after" pictures in chapter 9. In this example, the
picture before it hits the wall and the picture after it hits the wall. After it stops.
PES 1110 Fall 2013, Spendier
Lecture 22/Page 5
Example 4:
What is the impulse on a ball from a 0.25 kg baseball bat if the ball arrives with
horizontal velocity of 9.85 m/s and leaves with velocity of 11.2 m/s at an angle of 10
degrees down from the horizontal?
PES 1110 Fall 2013, Spendier
Lecture 22/Page 6
Example 5: baseball question continued
If the baseball spends 0.022 is in contact with the bat, what is the average force that the
bat imparts on the ball?
Example 6:
A pool stick applies an average force of 650N to a 0.15 kg pool ball in a time of 3.8 ms.
If the ball is initially at rest, what speed does it move with? (Ignore friction)