Download AP Momentum

THE IMMACULATE RECEPTION
MOMENTUM
AP Physics C: Mechanics
WHAT IS MOMENTUM?

What is its definition?

How do we calculate it?

When do we use this term?

Why was this word invented?

What do we already know about it?

What do we want to know about it?
WHAT IS MOMENTUM?

What is its definition?
Momentum: “quantity of motion”
-Newton
Momentum: “mass in motion”
Momentum: the product of an object’s
mass and its velocity
Momentum: It is a vector!
Momentum: is sometimes called linear momentum
WHAT IS MOMENTUM?
How do we calculate it?
If object is moving in arbitrary direction:
p  mv
What are its units?
mass length 
time
kgm

s


px  mv x
p y  mv y
pz  mv z
WHAT DO WE KNOW ABOUT MOMENTUM?
WHAT IS MOMENTUM?


Why was this word invented?
When do we use this term?
We are yet to make a distinction between a
rhino moving at 5m/s and a hummingbird
moving at 5m/s.
Thus far, how have we handled forces that
are only briefly applied such as collisions?
(we pretended that doesn’t happen)
Some believed that this quantity is conserved
in our universe.
HOW IS MOMENTUM RELATED TO OTHER
PHYSICS CONCEPTS THAT WE HAVE
ALREADY STUDIED?
dv dp
F  ma  m 
dt dt
The time rate of change of linear momentum of a
particle is equal to the net force acting on the particle.

We will soon see that it has many things in common
with Energy, Newton’s 3rd law, and The Calculus.
PAUSE TO THINK ABOUT CALCULUS CONCEPTS:

Why is a derivative involved?
Momentum may be changing non-uniformly with time


What does this say about the slope of a momentum-time
graph?
The slope of a momentum-time graph is net force!
The area under which graph might be meaningful?
The area under a force-time graph is a change in momentum!

So, how might an integral be involved?
The integral of force with respect to time is a change in
momentum!
PAUSE TO THINK ABOUT CALCULUS CONCEPTS:
The integral of force with respect to time is a change in
momentum!
dp
F
dt
We call the lefthand side of this
equation the
IMPULSE of the
force


Fdt  dp
 Fdt   dp
 Fdt  p
I   Fdt  p
tf
ti
PAUSE TO THINK ABOUT CALCULUS CONCEPTS:
The slope of a momentum-time graph is net force!
The area under a
force-time graph
is a change in
momentum or an
impulse
IMPULSE-MOMENTUM THEOREM:
The impulse of a force F equals the change in
momentum of the particle.
I  Ft  p
This is another way of saying that a net force must
be applied to change an objects state of motion.
Why Because
does thisthe force
look different
might be
from theconstant!
last
equation?

A FEW THINGS ABOUT IMPULSE:
It is a vector in the same direction as the change in
momentum.
It is not a property of an object! It is a measure of
the degree to which a force changes a particles
momentum. We say an impulse is given to a
particle.
What are its units?
From the equation we see that they must be the
same as momentum’s units (kgm/s).
Impulse approximation: assume the force is applied
only for an instant and that it is much greater than
other forces present.
ANOTHER QUESTION PLEASE…
TO STOP A SPEEDING TRAIN:
EXPLAIN THESE VIDEOS IN PHYSICS TERMS.
QUICK CONCEPTUAL QUIZ




Can a hummingbird have more momentum than a
rhino?
Why might an out of control truck hit a haystack or
barrels and pile of sand as opposed to a wall as an
emergency stop?
How is a ninja’s ability to break stacks of wood related
to impulse and momentum?
What good is it to know an object’s momentum?
Question
2: If a boxer is able to
make his impact time 5x longer by
“riding” with the punch, how much
will the impact force be reduced?
Ft  mv
By
5x
Ft mv

t
t
mv
F
t
 When
a dish falls, will the impulse
be less if it lands on a carpet than
if it lands on a hard floor?
 No – the same impulse – the force
exerted on the dish is less because
the time of momentum change
increases.
EXAMPLES

Examples of Increasing Impact Time to decrease
Impact Force:
Bend knees when jumping
Gymnasts and wrestlers use mats
Glass dish falling on carpet rather than
concrete
Acrobat safety net
Other examples???
OBSERVING CHANGES IN MOMENTUM:
CONSIDER TWO PARTICLES THAT CAN INTERACT,
BUT ARE OTHERWISE ISOLATED FORM THEIR
SURROUNDINGS.
What do we know about a collision between these
two particles?
Newton’s law says that they exert equal and
opposite forces on each other regardless of
comparative size (mass).
Is it possible for one particle to be in contact with
the second particle for a longer period of time than
the second on the first?
No, so the impulse imparted on each must be the
same.
THEREFORE…
THE PARTICLES MUST UNDERGO THE SAME
CHANGES IN MOMENTUM!
Let’s look at this mathematically.
dp1
F2on1 
dt
dp2
F1on2 
dt
dp1
dp2

dt
dt
dp1 dp2

0
dt
dt

d
 p1  p2   0
dt
d
p1  p2   0
dt
ptot  p1  p2
dptot
0
dt
What does it mean, conceptually,
 for a time derivative of
momentum to be zero
It means that the total
momentum of the system is
constant over time.
aka Momentum is Conserved!
THE LAW OF CONSERVATION OF MOMENTUM
When two isolated, uncharged particles
interact with each other, their total
momentum remains constant.
OR
The total momentum of an isolated system at
all times equals its initial momentum (before
and after collisions).
p1i  p2i  p1f  p2f
FIND THE REBOUND SPEED OF A 0.5 KG BALL
FALLING STRAIGHT DOWN THAT HITS THE FLOOR
MOVING AT 5M/S, IF THE AVERAGE NORMAL FORCE
EXERTED BY THE FLOOR ON THE BALL WAS
0.02S.
I  Ft  p
F
N
 Fg t  mv  v0 
205N  5N0.02s

v

0.5kg
v  3m/s
5m/s
205N FOR
A mass m is moving east with speed v on a smooth
horizontal surface explodes into two pieces. After
the explosion, one piece of mass 3m/4 continues in
the same direction with speed 4v/3. Find the
magnitude and direction for the velocity of the
other piece.
A) v/3 to the left
 B) The piece is at rest.
 C) v/4 to the left
 D) 3v/4 to the left

 E) v/4 to the right


pbefore  pafter
3m 4v  m 
mv  
   v
 4 3 1 4 2
m
mv  mv  v2
4
HOW GOOD ARE BUMPERS?

A car of mass 1500kg is crash-tested into a wall.
It hits the wall with a velocity of -15m/s and
bounces off with a velocity of 2.6m/s. If the
collision lasts for 0.15s, what is the average force
exerted on the car?
I  mv  v0 
I  1500kg2.6m/s  (15m/s)
I  2.64 104 kgm /s
I  Ft
2.64 10 4 kgm /s
F
0.15s
F  1.76 10 5 N
TYPES OF COLLISIONS
Energy is always conserved but may change
types (mv2/2, mgh, kx2/2 etc). There is only one
type of momentum (mv). We identify collisions
based upon their conservation of kinetic energy.
Inelastic
Elastic
• kinetic energy is
NOT constant
• kinetic energy IS
constant
INELASTIC COLLISIONS
These collisions are considered PERFECT when
the objects collide and combine to move as one
object.
Inelastic
Perfectly Inelastic
•Objects bounce but
may be deformed so
kinetic energy is
transformed.
•Objects stick together
PERFECTLY INELASTIC COLLISIONS:
p1i  p2i  p12f
m1v1  m2 v 2  m1  m2 v f
ELASTIC COLLISIONS (IDEALLY)
m1v1i  m2v2i  m1v1f  m2v2f
1
1
1
1
2
2
2
2
m1v1i  m2v2i  m1v1f  m2v2f
2
2
2
2
FOR ELASTIC COLLISIONS, FIND AN EXPRESSION
FOR RELATIVE SPEED OF THE OBJECTS BEFORE
AND AFTER COLLISION.
From momentum conservation…
m1v1i  m2v2i  m1v1f  m2v2f
m1v1i  m1v1f  m2v2f  m2v2i
m1v1i  v1f   m2 v 2f  v 2i 
FOR ELASTIC COLLISIONS, FIND AN EXPRESSION
FOR FINAL SPEED IN TERMS OF INITIAL SPEEDS
AND MASS.
From kinetic energy conservation…
1
1
1
1
2
2
2
2
m1v1i  m2v2i  m1v1f  m2v2f
2
2
2
2
Divide out ½ and move like mass terms to the same side so mass can be
factored out…

2
m1 v1i  v1f
2
 m v
2
2
2f
 v 2i
Factor difference of squares…
m1v1i  v1f v1i  v1f   m2 v 2f  v 2i v 2f  v 2i 

2

m1v1i  v1f   m2 v 2f  v 2i 
m1v1i  v1f v1i  v1f   m2 v 2f  v 2i v 2f  v 2i 
Combine our two results…
v1i  v1f   v2f  v2i 
v1i  v2i   v1f  v2f 


v1i  v2i   v1f  v2f 
The relative speed of the two objects before an
elastic collision equals the negative of their

relative speed after.
SOLVE FOR FINAL SPEEDS IN TERMS OF
INITIAL SPEEDS AND MASS.
m1v1i  m2v2i  m1v1f  m2v2f
1
1
1
1
2
2
2
2
m1v1i  m2v2i  m1v1f  m2v2f
2
2
2
2
TWO-DIMENSIONAL COLLISIONS
Set coordinate system up with x-direction the
same as one of the initial velocities
 Label vectors in a sketch
 Write expressions for components of momentum
v1f
before and after collision for each object

v1fsinθ
v1i
v1fcosθ
θ
φ
m1v1i  m1v1 f cos   m2v2 f cos 
0  m1v1 f sin   m2 v2 f sin 
-v2fsinφ
v2fcosφ
v2f
THE TYPES OF COLLISIONS ARE TREATED
THE SAME MATHEMATICALLY.
p  p
i
f