Momentum Notes Powerpoint

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Momentum and Impulse
Momentum
• Momentum can be defined as "mass in
motion." All objects have mass; so if an
object is moving, then it has momentum
• Momentum depends upon the variables
mass and velocity
• Momentum = (mass) (velocity)
• p = (m)(v)
• where m = mass and v=velocity
Momentum
• Momentum = (mass) (velocity)
• p = mv
p
m
v
Momentum is a vector quantity
• To fully describe the momentum of a 5-kg
bowling ball moving westward at 2 m/s,
you must include information about both
the magnitude and the direction of the
bowling ball
• p = (m)(v)
• p = (5 kg)(2 m/s west)
Givens:
m = 5kg
• p = 10 kgm / s west
v = 2 m/s west
Elastic and Inelastic Collisions
• When a Ball hits the ground and sticks, the
collision would be totally inelastic
• When a Ball hits the ground and bounces
to the same height, the collision is elastic
• All other collisions are partially elastic
collision
Check Your Understanding
• Determine the momentum of a ...
• 60-kg halfback moving eastward at 9 m/s.
– p = mv = 60 kg ( 9 m/s )
– 540 kgm /s east
Given: m = 60Kg
v= 9 m/s
• 1000-kg car moving northward at 20 m/s.
– p = mv = 1000 kg ( 20 m/s )
Find :
– 20,000 kgm /s north
Given: m = 1000Kg
v= 20 m/s
momentum (p)
Momentum and Impulse
Connection
• To stop an object, it is necessary to apply a force
against its motion for a given period of time
J = F (t) = m D v
J
F
t
Long Time Period:
When momentum is changed over a
long time period, less force is needed:
Short Time Period:
When momentum is changed over a
short time period, a larger force is
needed. This can produce some
drastic results.
Notice how the normally rigid golf ball is
temporarily deformed from the large force
applied over the short time interval.
Bungee jumping
with a non stretch
rope would NOT
be a good idea.
The bungee cord
spreads the change
in momentum over a
longer time so that
the force on you is
less.
When breaking
blocks or
boards, the swift
strike takes
place over a
short period of
time.
This increases
the force, thus
breaking the
Check Your Understanding
• If the halfback experienced a force of 800
N for 0.9 seconds to the north, determine
the impulse
Given: F = 800 N
• J=F(t)=mDv
t = 0.9 s
Find :
• 800N ( 0.9s ) = 720 N*s
Impulse (J)
• the impulse was 720 N*s or
• a momentum change of 720 kg*m/s
Impulse Question #2
• A 0.10 Kg model rocket’s engine is
designed to deliver an impulse of 6.0 N*s.
If the rocket engine burns for 0.75 s, what
is the average force does the engine
produce?
Given: F = 800 N
• J=F(t)=mDv
t = 0.9 s
• 6.0 N*s = F (0.75s)
Find :
• 6.0 N*s/ 0.75s = F
Average
• 8.0 N = F
Force
Impulse Question # 3
• A Bullet traveling at 500 m/s is brought to
rest by an impulse of 50 N*s. What is the
mass of the bullet?
Given: v = 500 m/s
• J=F(t)=mDv
J = 50 N*s
• 50 N*s = m ( 500 m/s – 0 m/s )
Find :
• 50 kg-m/s 2 *s / 500 m/s = m
m=?
• .1 kg = m
Summary
• the impulse experienced by an
object is the (force) x (time)
• the momentum change of an object
is the (mass) x (velocity change)
• the impulse equals the
momentum change
Conservation of
Momentum!
Conservation of Momentum:
In all collisions or interactions, momentum of a
system is always conserved.
You cannot gain or lose any momentum - what
you started with (total) is what you will end with!
You may have previously learned about
conservation of mass or energy from chemistry
class...
Since momentum is a vector quantity, direction
must be taken into account to see that
momentum truly is conserved.
Conservation of
Momentum Problems:
When solving problems involving the
conservation of momentum, the most important
thing to consider is:
Total momentum
before collision
=
Total momentum
after collision
Sample Problem:
A 300 kg cannon fires a 10 kg projectile at 200
m/s. How fast does the cannon recoil
backwards?
BOOM
Solution:
The momentum of the projectile must equal the
momentum of the cannon.
They must be equal since they must cancel
each other out.
p before = p after
BOOM
Solution:
p before = p after
Givens:
m (cannon) = 300 kg
m (cannonball) = 10 kg
v (cannonball) = 200 m/s
v (cannon) = ?
p before = p after
0 = mcannonvcannon + mprojvproj
0= (300 kg) (vcannon) + (10kg)
(200m/s)
-(2000kgm/s)
vcannon = ---------------300 kg
vcannon = -6.67 m/s
Q: Why does the cannon move so
much slower compared to the
projectile?
A: It is much more massive, more
inertia.
Q: What does the negative sign
indicate?
A:The cannon moves in the opposite
direction compared to the projectile.
Another Problem:
A 5kg fish swims toward and swallows a 1kg
fish at rest (it wasn’t paying attention). The big
fish initially swims at 1m/s. How fast will it be
swimming after having lunch?
Solution:
p before = p after
Givens:
m (big fish) = 5 kg
m (small fish) = 1 kg
v (big fish before) = 1 m/s
v (little fish before) = 0
v (total after) = ?
m1v1 + m2v2 = m1v1’ + m2v2’
p before = p after
(5kg) (1m/s) + (1 kg) (0m/s) = (6kg)
(v)
6 kg represents the combined mass of the fish
5kgm/s = 6kg (v)
v = 0.83 m/s
More Complicated
Momentum
Conservation
Collisions do not always take place in a nice
neat line:
Often, collisions take place in 2 or 3 dimensions:
Momentum is always conserved.
Although the mathematics needed to show this
may be complicated, the general idea can
easily be conveyed.
Another Example:
One ball collides into another. By using
momentum vector components, you can predict
the result:
After impact:
Before impact:
Total P before
Y components cancel out
X components add up to
previous P

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