Transcript Momentum!!!

Momentum!!!
Physics
Mr. Padilla
Mr. Padilla, what is momentum?
Momentum is the inertia of motion.
Newton’s 1st Law of Motion
A moving objects desire to remain moving.
Momentum is the mass of an object
multiplied by its velocity.
p = mv
Units: (kg)(m/s) = (kgm)/s
Can we do some examples?
 Ted runs at a speed of
7.5m/s. If Ted has a
mass of 80kg, what is
his momentum?
 What is the momentum
of a golf ball with a
mass of .01kg, rolls
down a hill at a speed
of 5m/s?
 Can you think of a time
when Ted and the golf
ball will have the same
momentum?
 p = mv
p = (80kg)(7.5m/s)
p = 600 kgm/s
 p = mv
p = (.01kg)(5m/s)
p = .05 kgm/s
 When velocity = 0
Sample Problem 6A
A 2250 kg pickup truck has a velocity of 25
m/s to the east. What is the momentum of
the truck?
How can we change momentum?
Momentum can change by changing the
objects mass or its velocity.
It is more common to change its velocity by
undergoing acceleration
Since force causes acceleration, applying
a force will cause a change in momentum.
How long the force is applied will
determine how much the momentum is
changed.
Impulse
Impulse is force applied over time.
Impulse (Δp) = Ft
Impulse is also equal to the change in
momentum.
Ft = Δ(mv)
Δp = mvf - mvi
Units = Ft = Ns
Ns = kgm/s
Impulse –Momentum Theorem
FΔt = Δp
FΔt = mvf - mvi
Examples of impulse.
If you were in an out of control car, would
you rather stop by hitting a brick wall or a
large hay stack?
The hay stack is the smart answer.
Both will apply the same change in
momentum, impulse, but the wall will do it
over a shorter time, and so will exert more
force. Ft is the same for both.
Sample Problem 6B
A 1400 kg car moving westward with a
velocity of 15 m/s collides with a utility pole
and is brought to a rest in 0.30 s. Find the
magnitude of the force exerted on the car
during the collision.
Bouncing
Which would change velocity more,
bringing a fast moving object to a stop or
sending it back in the opposite direction?
For the same reasons, momentum is
changed more if an object bounces.
This means there is also a greater impulse
exerted.
Example 1
 300kg of water rushes  Δ(mv) = Ft
down a hill at 15m/s,
(300kg)(15m/s) = F(5s)
it hits a dam and is
4500Ns = F(5s)
900N = F
brought to a stop.
How much force must
the dam exert to stop
it in 5s?
Example 2
 If, in the last example,
the water was turned
and moved back the
way it came at 10m/s,
what would be the
impulse?
 What would be the
force exerted?
 Impulse = Δ(mv)
(300kg)(15m/s) (300kg)(-10m/s)
4500Ns - -3000Ns
7500Ns
 Impulse = Ft
7500Ns = F(5s)
1500N = F
Sample Problem 6C
A 2250 kg car traveling to the west slows
down uniformly from 20.0 m/s to 5.00 m/s.
How long does it take the car to decelerate
if the force on the car is 8450 N to the
east? How far does the car travel during
the deceleration?
Conservation
If no net force or net impulse acts on a
system, the momentum of that system
cannot change.
When momentum, or any quantity does
not change, we say it is conserved.
Law of Conservation of Momentum
In the absence of an external force, the
momentum of a system remains unchanged.
Momentum in a System
 When a cannon fires a cannon-ball, the cannon
exerts a force on the cannon-ball and the
cannon-ball exerts an equal and opposite force
on the cannon according to Newton’s 3rd Law.
 Before the firing, the system is at rest, so the
momentum is zero.
 After it is fired, the net momentum, or total
momentum is still zero.
 Look at it with calculations…
Total initial momentum = total final
momentum
p1i + p2i = p1f + p2f
(m1v1)i + (m2v2)i = (m1v1)f + (m2v2)f
When is Momentum Conserved?
If a system undergoes changes, where all
forces are internal, the net momentum
before and after the event are the same.
Examples:
Cars colliding
Stars exploding
Sample Problem 6D
A 76 kg boater, initially at rest in a
stationary 45 kg boat, steps out of the boat
and onto the dock. If the boater moves out
of the boat with a velocity of 2.5 m/s to the
right, what is the final velocity of the boat?
Homework, Hooray!!!
Practice Problems
6A-D (odds only)
And Now…
Collisions !!!!!!!!
Collisions
When ever 2 objects collide and there are
no external forces the net momentum of
both objects before the collision equals the
net momentum after the collision.
net momentum
before
 net momentum
 mv   mv
i
f
after
Types of Collisions
 3 kinds
 Elastic – Objects collide without being
permanently deformed and with out
generating heat.
Objects bounce perfectly in elastic collisions
 Perfectly Inelastic – Colliding objects get
tangled and coupled together.
 Inelastic – objects deform during collisions,
but move separately after collision
Perfectly Inelastic
Elastic
 2 objects collide and
 Kinetic energy is
move off as one
conserved in perfectly
mass.
elastic collisions
 (m1v1)i + (m2v2)i =
 pi =pf
and
(m1+m2)vf
 KEi = KEf
 KE is NOT constant in
inelastic collisions
Some KE is converted
to sound and internal
energy
Sample Problem 6E
A 1850 kg luxury sedan stopped at a traffic
light is struck from the rear by a compact
car with a mass of 975 kg. The two cars
become entangled as a result of the
collision. If the compact car was moving at
a velocity of 22.0 m/s to the north before
the collision, what is the velocity of the
entangled mass after the collision?
Most collisions involve some external
force.
Most external forces are negligible
during a collision.
Perfectly elastic collisions are not
common in the everyday world.
Heat is usually generated by any collision.
They only occur regularly on the microscopic
level
Sample Problem 6F
Two clay balls collide head-on in a
perfectly inelastic collision. The first ball
has a mass of 0.500 kg and an initial
velocity of 4.00 m/s to the right. The mass
of the second ball is 0.250 kg, and it has
an initial velocity of 3.00 m/s to the left.
What is the final velocity of the composite
ball of clay after the collision? What is the
decrease in kinetic energy during the
collision?
Elastic Collisions
Remember:
Kinetic energy is conserved in elastic collisions
(m1v1)i + (m2v2)i = (m1v1)f + (m2v2)f
or
( ½ m1v12)i + ( ½ m2v22)i = ( ½ m1v12)f + ( ½ m2v22)f
Sample Problem 6G
A 0.015 kg marble moving to the right at
0.225 m/s makes an elastic head-on
collision with a 0.030 kg shooter marble
moving to the left at 0.180 m/s. After the
collision, the smaller marble moves to the
left at 0.315 m/s. Assume that neither
rotates before or after the collision and
that both marbles are moving on a
frictionless surface. What is the velocity of
the 0.030 kg marble after the collision?