6_momentum and Impuls

7/31/2019 6_momentum and Impuls

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CHAPTER 6

Impulse & Momentum

Pn. Siti Hajar Abd. Rahman

Faculty of Innovative Design and Technology

Universiti Sultan Zainal Abidin,Terengganu

DTS 1113 DYNAMICS


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Which do you think has moremomentum?


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Momentum

Momentum is a measure of how hardit is to stop or turn a moving object.

Momentum is related to both massand velocity.

Momentum is possessed by all moving

objects.


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Calculating Momentum

For one particlep = mv

For a system of multiple particlesP = pi = mivi Momentum is a vector with the same

direction as the velocity vector. The unit of momentum is

kg m/s or Ns


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Sample Problem

Calculate the momentum of a 65-kgsprinter running east at 10 m/s.


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Sample Problem Calculate the momentum of a system

composed of a 65-kg sprinter running east at10 m/s and a 75-kg sprinter running north at9.5 m/s.


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Solutionp = mv= p1 + p2

p1= (65kg)(10 m/s) = 650 kg m/s (east)p2= (75kg)(9.5 m/s) = 712.5 kg m/s (north)

650 kg m/s

712.5 kg m/s

p = (712.52 + 6502)1/2p = 964 kg m/sat tan-1(712.5/650)

= 47.6onorth of east


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Change in momentum

Like any change, change in momentumis calculated by looking at final and

initial momentums. p = pf pi p: change in momentum

pf: final momentum pi: initial momentum


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Momentum change Mini-lab

Using only a meter stick, find the momentumchange of each ball when it strikes the deskfrom a height of exactly one meter.

Turn in: Data for each ball. Include in your data section

the masses Calculation of momentum of each ball just before

and after it strikes the desk. Momentum change for each ball upon striking thedesk.


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Wording

dilemma

In which case isthe magnitude ofthe momentumchange greatest?

In which case is

the change in themagnitude of themomentumgreatest?


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Impulse


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Impulse (J)

Impulse is the product of an externalforce and time, which results in a

change in momentum of a particle orsystem.

J = F t

J =DP

Units: N s or kg m/s(same as momentum)


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Impulsive Forces Usually high

magnitude, shortduration.

Suppose the ball

hits the bat at 90mph and leaves thebat at 90 mph,what is themagnitude of the

momentum change? What is the

change in themagnitude of themomentum?


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Impulse (J) on a graph

Area under the curve


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Sample Problem

Suppose a 1.5-kg brick is dropped on a

glass table top from a height of 20 cm.a) What is the magnitude and direction of the

impulse necessary to stop the brick?

b) If the table top doesnt shatter, and stopsthe brick in 0.01 s, what is the averageforce it exerts on the brick?

c) What is the average force that the brick

exerts on the table top during this period?


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Solution a)

Find the velocity of the brick when it strikes the tableusing conservation of energy.

mgh = mv2

v = (2gh)1/2 = (2*9.8 m/s2*0.20 m) 1/2 = 2.0 m/sCalculate the bricks momentum when it strikes the

table.

p = mv = (1.5 kg)(2.0 m/s) = 3.0 kg m/s (down)

The impulse necessary to stop the brick is the impulsenecessary to change to momentum to zero.

J = Dp = pf pi = 0 3.0 kg m/s = -3.0 kg m/s

or 3.0 kg m/s (up)


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Solution b) and c)

b) Find the force using the otherequation for impulse.

J = Ft3.0 N s = F (0.01 s)

F = 300 N (upward in the same direction

as impulse)c)According the Newtons 3rd law, the

brick exerts an average force of 300

N downward on the table.


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Sample Problem

This force acts on a 1.2 kg object moving at 120.0 m/s.The direction of the force is aligned with the velocity.

What is the new velocity of the object?

0.20 0.40 0.60 0.80 t(s)

F(N)

1,000

2,000


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Solution

Find the impulse from the area under the curve.

A = base * height = (.1 s)(2500 N) = 125 Ns

J = 125 N s

Since impulse is equal to change in momentum and it isin the same direction as the existing momentum,the momentum increases by 125 kg m/s.

Dp = 125 kg m/s

Dp = pf - pi = mvf - mvimvf = mvi + Dp

= (1.2 kg)(120 m/s) + 125 kg m/s = 269 kg m/s

vf = (269 kg m /s) / (1.2 kg) = 224 m/s


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Law of Conservation of Momentum


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Law of Conservation of Momentum

If the resultant external force on asystem is zero, then the vector sum

of the momentums of the objects willremain constant.

Pbefore = Pafter


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Sample problem

A 75-kg man sits in the back of a 120-kgcanoe that is at rest in a still pond. If theman begins to move forward in the canoe at

0.50 m/s relative to the shore, whathappens to the canoe?


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Solution

The momentum before the man moves is equalto the momentum after the man moves.

pb = pa0 = mmvm + mcvc0 = (75 kg)(0.50 m/s) + (120 kg)v

v = - (75 kg)(0.50 m/s)/(120 kg)

v = -0.31 m/sThe canoe slips backward in the water at -0.31

m/s


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External versus internal forces

External forces:forces coming fromoutside the system of particles whosemomentum is being considered.

External forces change the momentum of thesystem.

Internal forces:forces arising frominteraction of particles within a system. Internal forces cannot change momentum of

the system.


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Explosions

When an object separates suddenly, as inan explosion, all forces are internal.

Momentum is therefore conserved in anexplosion.

There is also an increase in kinetic energyin an explosion. This comes from a potential

energy decrease due to chemicalcombustion.


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Recoil

Guns and cannons recoil when fired.

This means the gun or cannon must movebackward as it propels the projectileforward.

The recoil is the result of action-reaction

force pairs, and is entirely due to internalforces. As the gases from the gunpowder explosion

expand, they push the projectile forwards

and the gun or cannon backwards.


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Sample problem

Suppose a 5.0-kg projectile launchershoots a 209 gram projectile at 350

m/s. What is the recoil velocity ofthe projectile launcher?


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Solution

Momentum conservation is used to calculaterecoil speed.

pb = pa0 = mpvp + mlvl0 = (0.209 kg)(350 m/s) + (5.0 kg)v

v = - (0.209 kg)(350 m/s)/(5.0 kg)

v = - 14.6 m/s


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Sample Problem An exploding object breaks into three fragments. A 2.0 kg

fragment travels north at 200 m/s. A 4.0 kg fragment travels east

at 100 m/s. The third fragment has mass 3.0 kg. What is themagnitude and direction of its velocity?


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SolutionThe momentum before is zero, so the momentum after is zero.

This is a vector addition problem. Each fragment has a momentummagnitude of 400 kg m/s according to the formula p = mv.

400 kg m/s

400 kg m/s

400 kg m/s

(4002 + 4002)1/2

566 kg m/sdue southwest

v = p/m= 566/3= 189 m/sdue SW


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Collisions


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Collisions

When two moving objects make contactwith each other, they undergo a collision.

Conservation of momentum is used toanalyze all collisions.

Newtons Third Law is also useful. It tellsus that the force exerted by body A on

body B in a collision is equal and opposite tothe force exerted on body B by body A.


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Collisions

During a collision,external forces areignored.

The time frame ofthe collision is veryshort.

The forces are

impulsive forces(high force, shortduration).


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Collision Types

Elastic collisions Also called hard collisions

No deformation occurs, no kinetic energy lost

Inelastic collisions Deformation occurs, kinetic energy is lost

Perfectly Inelastic (stick together)

Objects stick together and become one object Deformation occurs, kinetic energy is lost


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Sample Problem An 80-kg roller skating

grandma collidesinelastically with a 40-kg kid. What is theirvelocity after thecollision?


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Sample Problem A train of mass 4mmoving 5 km/hrcouples with a

flatcar of mass m atrest. What is thevelocity of the carsafter they couple?


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Elastic Collision

After the collision, there are still twoobjects, with two separate velocities

Kinetic energy remains constant before and

after the collision. Therefore, two basic equations must hold

for all elastic collisions

pb = pa (momentum conservation) Kb = Ka (kinetic energy conservation)


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Sample Problem A 500-g cart on an air track strikes a 1,000-g cart at rest. What

are the resulting velocities of the two carts? (Assume the

collision is elastic, and the first cart is moving at 2.0 m/s whenthe collision occurs.)


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Solution

before after m1v1 = m1v1 + m2v2 1.0 = 0.50v

1+ v

2

m1 v12 = m1v12 + m2v22

2.0 = 0.50v12 + v22

Solve simultaneously

v1 = -0.67 m/s v2 = 1.33 m/s


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Center of Mass


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DEFINITION OF CENTER OF MASS

The center of mass is a point thatrepresents the average location forthe total mass of a system

The center of mass will remainconstant during a collision


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CENTER OF MASS

21

2211

mm

xmxmxcm


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Velocity of the center of mass

D

21

2211

mm

vmvm

vcm

21

2211

mm

xmxm

xcm

DDD


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Velocity of the center of mass

In an isolated system,the total linearmomentum does notchange.

Therefore, thevelocity of the centerof mass does notchange.

21

2211

mm

vmvm

vcm

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6_momentum and Impuls