Spring, 2008 Ho Jung Paik

Physics for Scientists and Engineers

Chapter 9Impulse and Momentum

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Linear MomentumLinear momentum of an object of mass mmoving with a velocity v is defined to be

“Momentum” and “linear momentum” will be used interchangeablyMomentum is a vector quantitySI units of momentum are kg⋅m/s

Momentum can be expressed in component form: px = mvx , py = mvy , pz = mvz

p = mv

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Newton’s 2nd Law & MomentumNewton’s 2nd law as Newton presented it:

This is a more general form than the one that we used so farNewton called m v the quantity of motion

For constant mass (i.e. dm/dt = 0), Newton’s 2nd law becomes

( )vpF mdtd

dtd

==∑

( ) avvF mdtdmm

dtd

===∑

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Conservation of Linear MomentumWhen two or more particles in an isolated system (i.e. no external forces present) interact, the total momentum of the system remains constant

The momentum of the system is conserved, not the momenta of individual particles

Can be expressed mathematically in various waysptotal = p1 + p2 = constantp1i + p2i= p1f + p2f

Conservation of momentum can be applied to systems with any number of particles

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Archer and ArrowAn archer is standing on a frictionless surface

Let the system be the archer with bow (particle 1) and the arrow (particle 2)

There are no external forces in the x-direction, so it is isolated in the x-direction

Total momentum before releasing the arrow is 0⇒ Total momentum after releasing the arrow is p1f + p2f = 0

The archer will move in the opposite direction of the arrow

Agrees with Newton’s 3rd law

Because the archer is much more massive than the arrow, his acceleration and velocity will be much smaller than those of the arrow

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Kaon DecayThe kaon decays into a positive π and a negative πparticle

Total momentum before decay is zero

Therefore, the total momentum after the decay must equal zero:

p+ + p− = 0, or p+ = −p−

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Impulse and MomentumFrom Newton’s 2nd Law F = dp/dt, dp = Fdt

Integrating to find the change in momentum over some time interval

The integral is called the impulse, I, of the force F acting on an object over Δt

This equation expresses the impulse-momentum theorem

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More About ImpulseImpulse is a vector quantity

The magnitude of the impulse is equal to the area under the force-time curve

Dimensions of impulse are ML/T, with units of N s = kg m s-1

Impulse can also be found by using the time averaged force:

tΔ= FI

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Example 1: Deformed BallA hard rubber ball, released at chest height, falls to the pavement and bounces back to nearly the same height. When it is in contact with the pavement, the lower side of the ball is temporarily flattened, with the maximum dent of 1 cm. Estimate the acceleration of the ball while it is in contact with the pavement and the time duration it is deformed.

To find the contact time, we use the impulse equation:

To find Fave, we need to find a. To find a, first find the velocity of the ball at the instant the bottom hits the ground, assuming the ball falls 1.50 m:

ifave pptFI −=Δ=

m/s 42.5)0m 50.1)(m/s 80.9(20)(2 222

−=

−−−+=−+=

f

ifif

vyyavv

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Example 1, contWe use the same equation to find the acceleration, assumed constant:

This is approximately the average acceleration.The momentum of the ball goes from pi = −m(5.42 m/s) to pf = 0 as it is being deformed. Using the impulse theorem:

Since it bounces back to nearly the same height, the time to deform and un-deform is approximately twice this time.

s 1069.3)m/s 1047.1()m/s 42.5(0 3

23−×≅

×−−

≅−

=−

=Δ

−=Δ=

mm

mapp

Fpp

t

pptFI

if

ave

if

ifave

2

222

m/s 1470

0)m 01.00(2)m/s 42.5()(2

=

=−+−=−+=

a

ayyavv ifif

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CollisionsCollision is an event during which two particles come close to each other and interact by means of forces

The time interval during which the velocity changes from its initial to final values is assumed to be short

The interaction force is assumed to be much greater than any external forces present

This means the impulse approximation can be used

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Example 2: Car Crash In a crash test, a car of mass 1500 kg collides with a wall. The initial and final velocities of the car are vi = −15.0 m/s and vf = 2.60 m/s. The collision lasts 0.150 s.Find the impulse and average force on the car.Assume the force exerted by the wall on the car is greater than any other forces that cause its velocity to change.

pi = (1500 kg)(−15.0 m/s) = −2.25 x 104 kg m/s

Pf = (1500 kg)(2.60 m/s) = 0.39 x 104 kg m/s

I = pf - pi = 2.64 x 104 kg m/s

Fave = Δp/Δt = (2.64 x 104 kg m/s)/0.150 s = 1.50 x 105 N

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Collisions – ExamplesCollisions may be the result of direct contact

Impulsive forces may vary in time in complicated ways

Collision need not include physical contact

There are still forces between the particles

Total momentum is always conserved

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Types of CollisionsIn an elastic collision, momentum and kinetic energy are conserved

Perfectly elastic collisions occur on a microscopic levelMacroscopic collisions are only approximately elastic

In an inelastic collision, kinetic energy is not conserved although momentum is conserved

If the objects stick together after the collision, it is a perfectly inelastic collisionElastic and perfectly inelastic collisions are limiting cases; most actual collisions fall in between these two types

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Conservation of MomentumSince there are no external forces,

In an elastic collision,ffii mmmm 22112211 vvvv +=+

( ) fii mmmm vvv 212211 +=+

fiifext PPPPF ==−=Δ ,0t

In a perfectly inelasticcollision, they share the same velocity after the collision

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Example 3: 2D Perf Inel CollisionThe car and the van have masses of 1500 kg and 2500 kg, respectively. Before the collision, the car was moving with a speed of 25.0 m/s in the x-direction and the van with a speed of 20.0 m/s in the y-direction. The car and the van make a perfectly inelastic collision. What is their velocity after the collision?Momentum conservation:

x: m1v1ix = (m1+m2)vf cosθ, 3.75x104 kg m/s = 4000 vf cosθy: m2v2iy = (m1+m2)vf sinθ, 5.00x104 kg m/s = 4000 vf sinθ

⇒ tanθ = 5.00/3.75, θ = 53.1°, vf = 5.00x104/(4000 sin 53.1°) = 15.6 m/s

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ExplosionExplosion is an event during which particles of the system move apart from each other after a brief interaction

Examples: archer and arrow, recoil of rifle, particle decay

Explosion is the reverse of the collision problem

Total momentum is conserved

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Example 4: RadioactivityA 238U nucleus spontaneously decays into a small fragmentthat is ejected with a speed of 1.50 x 107 m/s and a “daughternucleus” that recoils with a speed of 2.56 x 105 m/s.What are the atomic masses of the ejected fragment and the daughter nucleus?The nucleus was initially at rest so the total momentum remains zero.

Combining these two conservation laws,

u 238 ,0)( 21212211 =+=+=+ mmvmmvmvm iff

u 4 u, 234u 238 ,0)u 238( 212

212111 ==×

−==−+ m

vvv

mvmvmff

fff

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Rocket Launch

( )

:equationrocket thegives gIntegratin

or

have weupward,straight goingrocket aFor . thrust, theis quantity theandlocity exhaust ve theis Here

ˆ

law, 2nd sNewton' From rate.burn fuel theis Here

. :decrease tocontinues whereproblemexplosion an is This

0

thrust

ext

0

gRtM

RugMRu

dtdv

dtdvMRuMg

FuRu

uRdtvdMu

dtdM

dtvdMvM

dtd

dtPdjMgF

RRtMMM

−−

=−==+−

−=+===−=

−=

rrr

rr

rr

rr

r

gtRtM

Muv −⎟⎟⎠

⎞⎜⎜⎝

⎛−

=0

0ln

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Example 5: Saturn VSaturn V, the Apollo launch vehicle, has an initial mass of M0 = 2.85 x 106 kg, a payload mass of Mp = 0.27M0, a fuel burn rate of R = 13.84 x 103 kg/s, and a thrust of Fthrust = 34.0 x 106 N. Find (a) the exhaust speed u, (b) burn time tb, (c) acceleration at liftoff and burnout, and (d) final speed of the rocket.

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Example 5, cont

km/s 75.1m/s 1470)31.1)(m/s 1046.2(

s 150m/s 80.927.0

ln :burnoutat speed Final (d)

m/s 3.34m/s 80.9 :burnoutat on Accelerati

m/s 14.2m/s 80.9 :liftoffat on Accelerati (c)

s 150kg/s 1013.84

kg 1085.2)27.01( :Burn time (b)

km/s 46.2kg/s 1013.84m/s kg 100.34 :speedExhaust (a)

3

2

0

0

22

22

0

3

60

3

26

=−×=

×−⎟⎟⎠

⎞⎜⎜⎝

⎛=

=−=

=−=

=×

×−=

−=

=×

×=

v

MMuv

MRu

dtdv

MRu

dtdv

RMMt

RFu

p

Pb

thrust