Post on 14-Jan-2016
Chapter 8Chapter 8
Impulse and MomentumImpulse and Momentum
THE LINEAR MOMENTUMTHE LINEAR MOMENTUM
Momentum = mass times velocity “Think of it as inertia in motion”
vmp
Units - kg m/s or sl ft/s
AN IMPULSEAN IMPULSE
Collisions involve forces (there is a v).
Impulse = force times time.
tFI
Units - N s or lb s
AN IMPULSE CAUSES A CHANGE AN IMPULSE CAUSES A CHANGE MOMENTUMMOMENTUM
Impulse = change in momentum
amF
vmtF
t
vmF
vmtF
pI
)( if vmvmtF
)( if vvmtF
)( if pptF
Case 1Case 1Increasing MomentumIncreasing Momentum
Follow through
Examples:Long Cannons
Driving a golf ballCan you think of others?
pItF
Case 2Decreasing Momentum over a Long Time
Examples:
Rolling with the Punch
Bungee Jumping
Can you think of others?
Ip
tF
tF
Warning – May be dangerous
Case 3Case 3Decreasing Momentum over a Short TimeDecreasing Momentum over a Short Time
Examples:
Boxing (leaning into punch)
Head-on collisions
Can you think of others?
tFIp
BOUNCINGBOUNCING
There is a greater impulse with bouncing.
Example:Pelton Wheel
Pelton Wheel Water Sprinkler
Consider a hard ball and a clay ball that have +10 units of momentum each just before hitting a wall.
The clay ball sticks to the wall and the hard ball bounces off with -5 units of momentum.
Which delivered the most “punch” to the wall?
Initial momentum of the clay ball is 10.Final momentum of clay ball is 0.The change is 0 - 10 = - 10.It received - 10 impulse so itapplied + 10 to the wall.
Initial momentum of the hard ball is 10.Final momentum of hard ball is - 5.The change is - 5 - 10 = - 15.It received - 15 impulse so itapplied + 15 to the wall.
Example:Rifle and bullet
Demo - Rocket balloon Demo - Clackers
Video - Cannon ShootVideo – Scooter Propulsion
CONSERVATION OF LINEAR CONSERVATION OF LINEAR MOMENTUMMOMENTUM
IN COLISIONS AND EXPLOSIONSIN COLISIONS AND EXPLOSIONS
Consider two objects, 1 and 2, and assume that no Consider two objects, 1 and 2, and assume that no external forces are acting on the system composed external forces are acting on the system composed of these two particles.of these two particles.
11111 umvmtF
Impulse applied to object 1
22222 umvmtF
222211110 umvmumvm
Impulse applied to object 2
Total impulseappliedto system
22112211 vmvmumum
or
Apply Newton’s Third Law21
FF
tFtFor 21
In one dimension in component form,In one dimension in component form,
xxxx vmvmumum 22112211
Internal forces cannot cause a change in momentum of the system.
For conservation of momentum, the external forces must be zero.
IN COLLISIONS AND IN COLLISIONS AND EXPLOSIONSEXPLOSIONS
Collisions involve forces internal to colliding bodies.
Inelastic collisions - conserve momentum Totally inelastic collisions - conserve
momentum and objects stick together
A PERFECTLY ELASTIC A PERFECTLY ELASTIC COLLISIONCOLLISION
2222
12112
12222
12112
1 vmvmumum
Perfectly elastic collisions - conserve energy and momentum
Demos Demos
Demo - Momentum ballsDemo - Momentum balls
Demo - Small ball/large ball dropDemo - Small ball/large ball drop
Demo - Funny Balls Demo - Funny Balls
Head-On Totally Inelastic Head-On Totally Inelastic Collision ExampleCollision Example
Let the mass of the truck be 20 times the mass of the car.
Using conservation of momentum, we get
mphvtruck 60 mphvcar 60
vmmphmmphm )21()60()60(20
vmph 21)60(19
)60(21
19mphv
mphv 3.54
Remember that the car and the truck exert equal but oppositely directed forces upon each other.
What about the drivers? The truck driver undergoes the same
acceleration as the truck, that is
t
mph
t
mph
7.5)603.54(
The car driver undergoes the same acceleration as the car, that is
t
mph
t
mphmph
3.114)60(3.54
The ratio of the magnitudes of these two accelerations is
207.5
3.114
Remember to use Newton’s Second Law to Remember to use Newton’s Second Law to see the forces involved.see the forces involved.
For the truck driver his mass times his acceleration gives
F
am
ma
F For the car driver his mass times his greater acceleration gives
Your danger is of the order of twenty times
greater than that of the truck driver.
TRUCKS. Don’t mess with T
COEFFICIENT OF COEFFICIENT OF RESTITUTIONRESTITUTION
For any collision between two bodies moving along a single straight line, the coefficient of restitution e is defined as
xx
xx
uu
vve
21
12
u’s are velocities before impact.v’s are velocities after impact.For perfectly elastic collisions e = 1. For inelastic collisions e < 1.For totally inelastic collisions e = 0.
xx
xx
uu
vve
21
12
Collision between two objects of the same mass. One mass is at rest.
Collision between two objects. One not at rest initially has twice the mass.
Collision between two objects. One at rest initially has twice the mass.
Simple Examples of Head-On Collisions
(Energy and Momentum are Both Conserved)
Collision between two objects of the same mass. One mass is at rest.
Example of Non-Head-On Collisions
(Energy and Momentum are Both Conserved)
If you vector add the total momentum after collision,you get the total momentum before collision.
THE CENTER OF MASSTHE CENTER OF MASS
The center of mass of an object of mass m is the single point that moves in the same way as a point mass would move when subjected to the same external forces that act on the object.
m
Facm
The coordinates of the center of mass areThe coordinates of the center of mass are
i
iicm m
mzz
i
iicm m
mxx
i
iicm m
myy
.
CENTER OF MASS ANDCENTER OF MASS ANDCENTER OF GRAVITYCENTER OF GRAVITY
Center of mass - average position of mass
Earth
.
Center of gravity - average position of weight
Path of center of mass of a rotating object will
be a straight line if no external forces act on
the object.
Locating the Center of GravityLocating the Center of Gravity
Demo - Meterstick
Demo - Map of Texas
Demo - Balancing eagle
Demo - Curious George
Center can be outside of the object.
Examples: high jump and pole vaulting