Impulse – Change in Momentum Post-Lab
Impulse  Ns 
Impulse - Change in
Momentum Theorem
 
Impulse  1pp
Units:
Change in Momentum  kgm 
s

kgm
2s
Ns
s

kgm
kgm
s
s
Since the graph is linear
and contains (0,0)

Impulse  p
Slope
Impulse – Change in Momentum Post-Lab



Change
in
Momentum

p
Area under the curve of a 
graph of Force vs. time
Impulse
The average force is simply the
constant force that imparts the
same impulse!

F N 

Favg

Favg t
t
t s 
Impulse – Change in Momentum Post-Lab



Change
in
Momentum

p
Area under the curve of a 
graph of Force vs. time
Impulse

Impulse J   the area under the curve of the

graph of F vs. t
– the product of the average (constant)
force and the time that force is
applied
 
J  Favg t
Units:
Ns
Impulse – Change in Momentum Post-Lab



Change
in
Momentum

p
Area under the curve of a 
graph of Force vs. time
Impulse
For a constant (or average) Force
Units:


Ft  p
Ns
Impulse - Change in
Momentum Theorem
kgm
s
Ns   kgm 2  s  kgm
s

s 
Impulse – Change in Momentum Post-Lab


Ft  p
Impulse - Change in
Momentum Theorem
Notice from the lab that this equation holds in any type of collision.
Elastic Collision – a collision during which kinetic energy is conserved.
In other words, no energy is dissipated during the collision.
Inelastic Collision – a collision during which kinetic energy is not conserved.
Some of the energy is dissipated and is usually stored as
thermal energy or sound.
Derivation
For constant (or average) forces
But


Fnet  ma

 v
a
t


v
F m
t
 p
F
t


Ft  p
Newton’s 2nd Law
For constant forces
Which has the greatest change in momentum, an object that bounces or one that stops?
(Assuming they have the same mass and start with the same velocity)
Data Studio
An object that bounces has a greater change in momentum as a result
of the greater impulse imparted to it.
Also, the object changes direction and therefore undergoes a greater
change in velocity!


p  mv

 
p  mv f  vi 
If it stops:

 
p  mv f  vi 


p  mvi
If it bounces:


v f  vi

 
p  m vi  vi 


p  2mvi
Example 1
A 250. N force is applied to a 30. kg object moving at 25 m to bring it to rest.
s
a. How long is the force applied?


Ft  p



Ft  p f  pi


Ft  mvi

mvi
t   
F
t  

30. kg 25 m
 250. N
t  3.0 s
s

Example 1
A 250. N force is applied to a 30. kg object moving at 25 m to bring it to rest.
s
b. How big of an average force would need to be applied if it lasted 6.0 s?
Notice: twice the time
requires only half the force!
 
 F t

t
F


Ft  p



Ft  p f  pi


Ft  mvi


mvi
F 
t

30. kg 25 m

s
F 
6.0 s

F  125 N
