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Momentum, Impulse & Collisions
Linear Momentum
Conservation of Linear Momentum
Impulse
Collisions : Elastic & Inelastic Collisions
Collisions in 2-Dimensions
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Linear Momentum
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Conservation of Linear Momentum
The principle of conservation of momentum states thatwhen no external forces act on a system consisting of
two objects that collide with each other, the total
momentum of the system before the collision is equal to
the total momentum of the system afterthe collision.
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Law of Conservation of Linear Momentum
The general statement of the law of conservation of
linear momentum states that
The total momentum of an isolated system of bodies
remains constant
and is given by
pinitial = pfinal
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Newtons Second Law of Motion
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A 10,000 kg railroad car traveling at a speed of 24
m s-1 strikes an identical car at rest. If the cars lock
together as a result of the collision, what is their
common speed afterward ?
EXAMPLE
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SOLUTION
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Impulse
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Impulse
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A 0.375 kg rubber ball traveling horizontally to the right at10 m s-1 hits a wall and bounces back at 6 m s-1 to the left.
What is the total impulse exerted by the wall?
EXAMPLE
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Impulse, I = p = Ft
= mv - mu
= m (v u )
= (0.375 kg)(-6 -10 m s-1)
= (0.375 kg)(-16 m s-1)
I = -6.0 N s
SOLUTION
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A baseball, of mass 0.2 kg, is pitched at 40 m s-1 and is hit
straight back at the pitcher at 90 m s-1. Assume the positive x-
axis points toward the pitcher
(a) Find the impulse exerted by the bat on the ball
(b) If the ball is in contact with the bat for 3.5 ms, find
the average force exerted on the ball(c) How would the result of part (b) change if the
contact time were one-third as long
EXAMPLE
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SOLUTION
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Collisions
Inelastic Collisions
Elastic Collisions
Stationary Target
Moving Target
Collisions in 2-Dimensions
(Glancing Collisions)
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Inelastic Collisions
Collisions in which kinetic energy is not conserved.
Initial kinetic energy is transformed into other types of
energy (thermal, potential etc.)Total final kinetic energy is less than the total initial
kinetic energy.
If two objects stick togetheras a result of the collision,
the collisions is inelastic. Even though the kinetic energy is not conserved, the
total energy is conserved.
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1-Dimensional Inelastic Collisions
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Before Collision
m1 m2 at rest
+x
V
v
After Collision
m1 + m2
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1-Dimensional Inelastic Collisions
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A ballistic pendulum is a device that was used to measure thespeeds of bullets before electronic timing devices weredeveloped.
The device consists of a large block of wood of mass M= 5.4kg, hanging from two long cords. A bullet of mass m = 9.5 g isfired into the block, coming quickly to rest. The block + bulletswing upward, their center of mass rising a vertical distance h =6.3 cm before the pendulum comes momentarily to rest.
(a) What was the speed v of the bullet just prior to thecollisions.
(b) What is the initial kinetic energy? How much
energy remains as mechanical energy?
EXAMPLE
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Elastic Collisions
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Elastic Collisions
then[1]&[2]becomes
m1(v1iv1f)=m2v2f[3]
m1(v1iv1f)(v1i+v1f)=m2v2f2
[4][4][3]
if v
mm
mmv
1
21
21
1
ifv
mm
mv
1
21
1
2
2
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BeforeCollision
m1 m2atrest
v1i
v2i=0
AfterCollision
m1m2
v1f v2f
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Elastic CollisionsMovingTarget
Conservationoflinearmomentumstatesthat
m1v1i+m
2v2i=m
1v1f+m
2v2f[1]
Conservationofkineticenergystatesthat
m1v1i2+m2v2i
2=m1v1f2+m2v2f
2[2]
Combiningtheequations[1]&[2]willgiveus
and
iif vmm
mv
mm
mmv
2
21
2
1
21
21
1
2
iifv
mm
mmv
mm
mv
2
21
12
1
21
1
2
2
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Collisions in 2-Dimensions
Thevectornatureofmomentumespeciallyin2-
dimensionsisveryimportant.
Commontypeofnonhead-oncollisionisthatamovingobjectstrikesasecondobjectinitiallyatrest
Lawofconservationofmomentumstatesthat
x: p1x+p2x=p1x+p2x m1v1=m1v1cos1+m2v2cos2[1]
y:p1y+p2y=p1y+p2y
0=m1v1sin1+m2v2sin2[2]
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Collisions in 2-Dimensions
Ifthecollisionsiselastic,thenthelawofconservation
ofenergystatesthat
m1v12=m1v1
2+m2v22[3]
Fromequations[1],[2]&[3],wecansolveforv1,v2,
1and2ifm1,m2,v1andv2isgiven.
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Example
Abilliardballmovingwithspeedv1=3.0ms-1inthe+x
directionstrikesanequalmassballinitiallyatrest.Thetwoballsareobservedtomoveoffat45,ball1
abovethex-axisandball2belowthex-axis.Thatis
1=45and2=-45.
Whatarethespeedsofthetwoball?
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SolutionAssumethatthetwoballshavethesamespeed(from
symmetry).Conservationofmomentumgives
x: mv1=mv1cos(45)+mv2cos(-45)
v1=v1cos(45)+v2cos(45)=2v1cos(45)
y: 0=mv1sin(45)+mv2sin(-45)
then
v2=-v1[sin(45)/sin(-45)]=v1
so v1=v2=v1/2cos(45)
=(3.0ms-1)/[2(0.7071)]
= 2 1 m s-1
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