Dynamics of a System of Particles

7/27/2019 Dynamics of a System of Particles

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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


(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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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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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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Dynamics of a System of Particles