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THE IMMACULATE RECEPTION
MOMENTUMAP 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: the product of an object’s mass and its velocity
Momentum: “mass in motion”
Momentum: “quantity of motion”-
Newton
Momentum: It is a vector!
Momentum: is sometimes called linear momentum
WHAT IS MOMENTUM?
How do we calculate it?
p m
v
What are its units?
mass length time
kgm
s
px mvx
If object is moving in arbitrary direction:
py mvy
pz mvz
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?
F m
a
mdv
dt
dp
dt
We will soon see that it has many things in common with Energy, Newton’s 3rd law, and The
Calculus.
The time rate of change of linear momentum of a particle is equal to the net force acting on the
particle.
PAUSE TO THINK ABOUT CALCULUS CONCEPTS: Why is a derivative involved?
What does this say about the slope of a momentum-time graph?
The area under which graph might be meaningful?
So, how might an integral be involved?
Momentum may be changing non-uniformly with time
The slope of a momentum-time graph is net force!
The area under a force-time graph is a change in momentum!
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!
F
dp
dt
F dt d
p
F dt d
p
F dt
p
I
F dt
t i
t f p
We call the left-hand side of this
equation the IMPULSE of the
force
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:
I
F t
p
The impulse of a force F equals the change in momentum of the particle.
This is another way of saying that a net force must be applied to change an objects state of
motion.Why does this look different from the last
equation?
Because the force might be
constant!
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?
By 5x
Ft mv
Ft
t
mv
t
F mv
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.
F2on1 dp1
dt
F1on 2 dp2
dt
1 2dp dp
dt dt
1 2 0dp dp
dt dt 1 2 0
dp p
dt
d
dtp1 p2 0
ptot p1 p2
dptot
dt0
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 205N FOR 0.02S.
I Ft p
FN Fg t m v v0
v 205N 5N 0.02s
0.5kg 5m/s
v 3m/s
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
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.
pbefore pafter
mv 3m
4
4v
3
1
m
4v
2
mv mv m
4v2
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 Ft
F 2.64 104 kgm/s
0.15s
F 1.76 105N
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•kinetic
energy is NOT constant
Elastic•kinetic
energy IS constant
INELASTIC COLLISIONS
These collisions are considered PERFECT when the objects collide and combine to
move as one object. Inelastic•Objects bounce but may be deformed so kinetic energy is transformed.
Perfectly Inelastic•Objects stick together
PERFECTLY INELASTIC COLLISIONS:
p1i p2i p12f
m1v1 m2v2 m1 m2 vf
ELASTIC COLLISIONS (IDEALLY)
m1v1i m2v2i m1v1f m2v2f
1
2m1v1i
2 1
2m2v2i
2 1
2m1v1f
2 1
2m2v2f
2
FOR ELASTIC COLLISIONS, FIND AN EXPRESSION FOR RELATIVE SPEED OF THE OBJECTS BEFORE AND AFTER COLLISION.
m1v1i m2v2i m1v1f m2v2f
From momentum conservation…
m1v1i m1v1f m2v2f m2v2i
m1 v1i v1f m2 v2f v2i
FOR ELASTIC COLLISIONS, FIND AN EXPRESSION FOR FINAL SPEED IN TERMS OF INITIAL SPEEDS AND MASS.
From kinetic energy conservation…
1
2m1v1i
2 1
2m2v2i
2 1
2m1v1f
2 1
2m2v2f
2
m1 v1i2 v1f
2 m2 v2f2 v2i
2 Divide out ½ and move like mass terms to the same side so mass can be factored out…
m1 v1i v1f v1i v1f m2 v2f v2i v2f v2i Factor difference of squares…
m1 v1i v1f v1i v1f m2 v2f v2i v2f v2i
m1 v1i v1f m2 v2f v2i
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.
1
2m1v1i
2 1
2m2v2i
2 1
2m1v1f
2 1
2m2v2f
2
m1v1i m2v2i m1v1f m2v2f
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 before and after collision for each object
v1i
v1f
v2f
v2fcosφ
v1fsinθ
-v2fsinφ
φ
v1fcosθθ
1 1 1 1 2 2cos cosi f fm v m v m v
1 1 2 20 sin sinf fm v m v
THE TYPES OF COLLISIONS ARE TREATED THE SAME MATHEMATICALLY.
pi pf