4 impulse and momentum

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IMPULSE AND

MOMENTUM

Standard CompetencyAnalyzes the nature phenomenon and its regularity within the

scope of particle’s Mechanics

Base CompetencyAnalyzes the relation between impulse and momentum concepts

to solve the collision problems

Learning ObjectivesAfter completing this chapter, all students should be able to:

1 Formulates the concept of impulse and momentum and

their relationship and their applications in daily life2 Formulate the conservative law of momentum for a

system without external forces

3 Integrates the conservative law of energy and the

conservative law of momentum for several collisionproblems

References[1] John D Cutnell dan Kenneth W. Johnson (2002). Physics 5th Edwith Compliments. John Wiley and Sons, Inc.

[2] Sunardi dan Etsa Indra Irawan (2007). Fisika Bilingual SMA/MAuntuk SMA/MA Kelas XI. CV Yrama Widya hal. 255-298



IMPULSE

Impulse is a force acting on an object that is not constant but

varies with time. In a short, i m pu l s e = time-varying force.

Definition of impulse: The impulse of a force is the product of the average force and time interval

during which the force acts

SI unit of impulse is newton.second (Ns).

Impulse could be describe graphically as an area under the F-t

curve

MOMENTUM (Linier Momentum)

Newton’s observation on OBJECT MOTION IN NATURE revealed

that all the object has mass. While moving, some are fast andsome are slow

I = F Δt

I

t (s)

F (N)

m, v

m, v



Momentum describes how fast or how slow anobject’s moving is. This is

normally applies to an object

that moves in a straight line.

That’s why it is refer as liniermomentum.

Definition of momentum: The linier momentum of the object is

the product of the object’s mass and itsvelocity

If two momentums are configured an angle θ, then

v

p = m v

p

p2

p1

θ

θ cos2 2121

21 p p p p p ++=

v



Exercises

[1] What is the momentum of a 22 grams bird when flying at

the speed of 8.1 m/s?

[2] A 12500 kilograms of loco train is moving smoothly without

any frictions at 18.0 m/s. What is the loco’s momentum?

[3] A bullet of mass of 21.0 grams was fired and crash out at

speed of 210 m/s. Determine its momentum!

[4] A car (m = 1600 kg) moves to the left with velocity of 10

m/s. Determine the momentum of the car

THE IMPULSE-MOMENTUM CHANGE THEOREM

How impulse and momentum are related, is shown in

mathematical description below.

momentum: p = m v

Impulse: I = F Δ

t Newton II Law: F = m a ; a = Δv /Δt

F = m Δv /Δt F Δt = m Δv

This is very popular equation and is refer as The Impulse-

Momentum Change Theorem.

I = Δ p

The greatest change in velocity will occur when theimpulse is the greates

By increasing the amount of force and the amount of time the force applied, the greatest change in velocity

can be achieved



Example

A man is washing his car using

using nozzle which has waterdebit 1.5 kg/s and sprays the

water at speed of 20 m/s.Determine force that delivered by

water to the car’s body.

SolutionIn every second, water delivered from the nozzle has a

momentum of

p x = m v x = (1,5 kg)(20 m/s) = 30 kg.m/s

If the force of water when strikes the car’s body is constant,

Change In Momentum

Consider the two objects (a teddy bear doll and a ball) which

are dropped in a same height and time. The teddy bear will notbounced contrary to the ball.

akhir awal 0 30 kg.m/s

30 N1,0 s

p p p

F t t

−Δ −

= = = = −Δ Δ



Change in momentum: Δ p = pafter − pbefore

Teddy Bear: Δ p = 0 − (−mv ) = mv (apparently not bounces)

Bouncing Ball: Δ p = mv − (−mv ) = 2mv

MOMENTUM AND NEWTON’S II LAW

Newton’s II Law in term of calculus differential

t

p

t

v mt

v mamF

Δ

Δ=

Δ

Δ=

Δ

Δ==

r

r

r

rr

)(

Applying force could change how fast or how slow the object’s

motion.

This equation is only valid for objects that have constant mass.

Here is a more general form in terms of momentum, also usefulwhen the mass is changing:

According to Newton,

force is the rate of change of momentum

To stop such an object, it isnecessary to apply a force

against its motion for a givenperiod of time



APPLICATION OF IMPULSE IN DAILY LIFE

[1] When a force is limited: App ly a f o rce fo r a l ong t im e

Pushing a car

[2] Minimize the force: i nc rease Δt

Catching a ball

[3] Maximize momentum change:

app ly a f o rce fo r a shor t t im e

karate pinalty kick

Followthrough ona golf or asoftballswing

bungee jumping



THE LAW OF CONSERVATION OF MOMENTUM

The momentum of any closed, isolated system DOES NOT

CHANGE.

When there is a collision between two objects, Newton's Third

Law states that the force on one of the bodies is equal and

opposite to the force on the other body.

p1 + p2 = p’ 1 + p’ 2

m1 v 1 + m2 v 2 = m1 v’ 1 + m2 v’ 2

Total momentum before an event (∑ p) is equal to

total momentum after the event (∑ p’ )



Exercises

[1] Calculate the revolving distance of a rifle (m = 5 kg) while

fires a bullet 50 gr at the speed of 120 m/s.

[2] A log wood (length = 40 cm) that stand perpendicular to

the ground being dropped by a hammer (m = 10 kg) at 50

cm above the top of the wood. If the average forceresitance of the ground is 1000 N, determine the number

needed for hammer in order the wood is buried

[3] Ball A with momentum p moves in straight line and collides

ball B which is moves on the same line. After the collision,

momentum of ball A become −3 p. Determines the changein momentum of ball B.

[4] A car (m = 500 kg) moves with velocity 40 m/s. A truck

moves with velocity 10 m/s heading to the car. Estimatethe mass of the truck if the both have collision and move

together at the velocity of 15.5 m/s.

[5] A rain drop comes straight down

with a velocity of v o = −15 m/s andhits the roof of the carperpendiculary. The mass of the

rain per second that strikes the car

roof is 0.060 kg/s. Assuming that

the rain comes to rest upon striking the car (v t = 0 m/s),

find the average force exerted by the rain on the roof

[6] The compressive force per area necessary

to break the tibia in the lower leg, is about

F/A = 1.6 ×10 N.m2

. The smallest crosssectional area of the tibia, about 3.2 cm, is

slightly above the ankle.

Suppose a person of mass m = 60 kg jumps

to the ground from a height h = 2.0 m and

absorbs the shock of hitting the ground bybending the knees. Assume that there is

constant deceleration during the collision.

During the collision, the person lowers hiscenter of mass by an amount Δd = 1 cm.



a) What is the collision time Δt col?b) Find the average force of the ground on the person

during the collision.

c) What is the average impulse of the ground on the

person?d) Will the person break his ankle? How much would you

need to lower your center of mass so you do not break

your ankle?

[7] A rocket can expel 200 kg of mass per second at a velocity

of 4000 m/s. If the launce of the rocket requires anacceleration of 2.0 m/s, what is the greatest mass the

rocket may be at the time of launch? (assume net mass of

rocket is constant)

[8] A 0.04 kg bullet is fired at a velocity of 1200 m/s from a 6

kg gun. What is the recoild velocity of the gun

[9] A neutron (mass = 1.67 x 10−27 kg (collides with a "resting"

helium atom (mass = 6.64 x 10−27 kg) at a velocity of 2 x 106

m/s. The final velocity of the neutron is 1.8 x 106 m/s. What isthe velocity of the helium atom, assuming the collision is in one

dimension?



COLLISION

Collision is likely to mean some sort of automotive disaster

(crash accident).

In Physics, collision is any strong interaction betweenbodies that lasts a relative short time .

Types of Collisions

Two Laws of Conservation which are involve in collision, i.e,:

- Law of Conservation of Mechanical Energy

- Law of Conservation of Momentum

It end up with equation of restitution coefficient

B A

B A

v v

v v e

−

′−′−=

[1] Elastic Collision:head on collison

[2] Inelastic Collision

[1] Elastic Collision:target at rest



ProofingLaw of conservation of momentum

)1()()( KBBB A A A

BBBB A A A A

BB A ABB A A

B AB A

v v mv v m

v mv mv mv m

v mv mv mv m p p p p

′−=′−

−′=′−

′+′=+′+′=+

Law of conservation of energy

)2())(())((

)()(222

2

21

2

212

212

21

KBBBBB A A A A A

BBB A A A

BB A ABB A A

B AB A

v v v v mv v v v m

v v mv v m

v mv mv mv m

k E k E Ek Ek

′−′+=′−′+

−′=′−

′+′=+

′+′=+

)(

))(())((

)()(

)1(

)2(

B AB A

BB A A

BBBBB A A A A A

BBB A A A

v v v v

v v v v

v v v v mv v v v m

v v mv v m

−−=′−′

′+=′+

′−′+=′−′+

′−=′−=

Prove that1

2

h

he =

1)(

)(=

−

′−′−

B A

B A

v v

v v

h2

h1



Exercises

[1] A ball is dropped from height of 15 m above the ground,

then bounced back as height as 0.9 m. Calculate the coefficient

of restitution between ball and the ground

[2] An object drops from the height of 9 m and bounces in 1 m.

Calculate (a) coefficient of restitution, (b) height after thirdbounce

[3] A 5 grams bullet fired and get in through a wood log (m =495 grams) which is hangin in balistic pendulum. The log and

the bullet are then swing at 5 cm height from its stationer

position. If g = 10 m/s2, determine the velocity of the bullet.

[4] A ball is hit by a force of 100 N and makes it flew 200 m/s.

The stick hit the ball within 0,2 seconds. Determine the mass of

the ball

[5] Two objects of similar masses moves in velocity of 10 m/s

and 20 m/s each. The both are heading and exert a completeelastic collision. Determine the final velocity of each object after

collision

[6] A massive cube (m = 4.9 kg) is on a smooth plane hit by a

bullet

[7] An object with mass m1is

initially moving with a velocity v

= 3.0 m/s and collides elasticallywith another object of equal

mass m2 = m1 that is initially atrest. After the collision m

1

moves with an unknown speedv

f1at an angle θ = 30o with

respect to its initial direction of motion. After the collision, m

2

moves with unknown speed v f2

,

at an unknown angle θf (see sketch). Find the final velocities of eachof the masses and the angle θ

f .

4 impulse and momentum

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