IMPULS & MOMENTUM

M. Djamil

Momentum From Newton’s laws: force must be present to change an

object’s velocity (speed and/or direction) Wish to consider effects of collisions and corresponding

change in velocity

Method to describe is to use concept of linear momentum

scalar vector

Linear momentum = product of mass velocity

Golf ball initially at rest, so some of the KE of club transferred to provide motion of golf ball and its change in velocity

GAYA IMPULS.• Suatu gaya yang bekerja dalam waktu singkat disebut gaya

impuls; misal. Bola yang menumbuk tembok, tumbukan dua benda, dan bentuk tumbukan yag lain.

• Secara matematis, dari Newton II dapat dirumuskan sebagai berikut :

F = m a F = m dv/dt F dt = m dV

IMPULS - MOMENTUM

F

t

tIImpuls Momentum

Cek dimensi & satuannya !

Impulse

In order to change the momentum of an object (say, golf ball), a force must be applied

The time rate of change of momentum of an object is equal to the net force acting on it

Gives an alternative statement of Newton’s second law

(F Δt) is defined as the impulse Impulse is a vector quantity, the direction is the

same as the direction of the force

tFporamt

vvm

t

pF net

ifnet

:)(

Momentum

Vector quantity, the direction of the momentum is the same as the velocity’s

Applies to two-dimensional motion as well

yyxx mvpandmvp

vmp

Size of momentum: depends upon mass depends upon velocity

Example: Impulse Applied to Auto Collisions

The most important factor is the collision time or the time it takes the person to come to a rest This will reduce the chance of dying in a car crash

Ways to increase the time Seat belts Air bags

The air bag increases the time of the collision and absorbs some of the energy from the body

ConcepTest

Suppose a ping-pong ball and a bowling ball are rolling toward you. Both have the same momentum, and you exert the same force to stop each. How do the time intervals to stop them compare?

1. It takes less time to stop the ping-pong ball.2. Both take the same time.3. It takes more time to stop the ping-pong ball.

Answer

Suppose a ping-pong ball and a bowling ball are rolling toward you. Both have the same momentum, and you exert the same force to stop each. How do the time intervals to stop them compare?

1. It takes less time to stop the ping-pong ball.2. Both take the same time.3. It takes more time to stop the ping-pong ball.

Note: Because force equals the time rate of change of momentum, the two balls loose momentum at the same rate. If both balls initially had the same momenta, it takes the same amount of time to stop them.

Problem: Teeing Off

A 50-g golf ball at rest is hit by “Big Bertha” club with 500-g mass. After the collision, golf leaves with velocity of 50 m/s.

a) Find impulse imparted to ballb) Assuming club in contact with

ball for 0.5 ms, find average force acting on golf ball

Problem: teeing off

Given:

mass: m=50 g = 0.050 kgvelocity: v=50 m/s

Find:

impulse=?Faverage=?

1. Use impulse-momentum relation:

2. Having found impulse, find the average force from the definition of impulse:

smkg

smkg

mvmvpimpulse if

50.2

050050.0

N

s

smkg

t

pFthustFp

3

3

1000.5

105.0

50.2,

Note: according to Newton’s 3rd law, that is also a reaction force to club hitting the ball:

iiff

ifif

R

VMvmVMvm

orVMVMvmvm

ortFtF

,

, of club

CONSERVATION OF MOMENTUM

Conservation of Momentum

Definition: an isolated system is the one that has no external forces acting on it

A collision may be the result of physical contact between two objects

“Contact” may also arise from the electrostatic interactions of the electrons in the surface atoms of the bodies

Momentum in an isolated system in which a collision occurs is conserved (regardless of the nature of the forces between the objects)

Conservation of Momentum

The principle of conservation of momentum states when 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 after the collision

v1

v2v’2

v’1

m1m2

F’F

m1 v’1 + m2 v’2 = m1 v1 + m2 v2

Hk Kek. Momentum

Concept TestSuppose a person jumps on the surface of Earth. The Earth

1. will not move at all2. will recoil in the opposite direction with tiny velocity3. might recoil, but there is not enough information provided to see if that could happened

Please fill your answer as question 23 of General Purpose Answer Sheet

Concept Test

Suppose a person jumps on the surface of Earth. The Earth

1. will not move at all2. will recoil in the opposite direction with tiny velocity3. might recoil, but there is not enough information provided to see if that could happened

Note: momentum is conserved. Let’s estimate Earth’s velocity after a jump by a 80-kg person. Suppose that initial speed of the jump is 4 m/s, then:

smkg

smkgV

smkgVMp

smkgp

Earth

EarthEarth

2324

103.5106

320

so,320:Earth

320:Person

tiny negligible velocity, in opposite direction

Types of Collisions

Momentum is conserved in any collision

what about kinetic energy?

Inelastic collisionsKinetic energy is not conserved

○ Some of the kinetic energy is converted into other types of energy such as heat, sound, work to permanently deform an object

Perfectly inelastic collisions occur when the objects stick together○ Not all of the KE is necessarily lost

energylost fi KEKE

Perfectly Inelastic Collisions:

When two objects stick together after the collision, they have undergone a perfectly inelastic collision

Suppose, for example, v2i=0. Conservation of momentum becomes

fii vmmvmvm )( 212211

.20105.2

105

,)2500(0)50)(1000(

:1500,1000ifE.g.,

3

4

21

smkg

smkgv

vkgsmkg

kgmkgm

f

f

fi vmmvm )(0 2111

Perfectly Inelastic Collisions:

What amount of KE lost during collision?

Jsmkg

vmvmKE iibefore

62

222

211

1025.1)50)(1000(2

12

1

2

1

Jsmkg

vmmKE fafter

62

221

1050.0)20)(2500(2

1

)(2

1

JKElost61075.0

lost in heat/”gluing”/sound/…

More Types of Collisions Elastic collisions

both momentum and kinetic energy are conserved

Actual collisionsMost collisions fall between elastic and

perfectly inelastic collisions

More About Elastic Collisions

Both momentum and kinetic energy are conserved

Typically have two unknowns

Solve the equations simultaneously

222

211

222

211

22112211

2

1

2

1

2

1

2

1ffii

ffii

vmvmvmvm

vmvmvmvm

Elastic Collisions:• Using previous example

(but elastic collision is assumed)

smkg

smkgsmkg

vmvmP iibefore

4

2111

100.2

)20)(1500()50)(1000(

smv

smv

f

f

1.31

7.26

2

1

For perfectly elastic collision:

J

JJ

vmvmKE iibefore

6

56

222

211

1055.1

1031025.1

2

1

2

1

222

211

6

22114

2

1

2

11055.1

100.2

ff

ff

vmvmJ

vmvmsmkg

Problem Solving for One Dimensional Collisions

Set up a coordinate axis and define the velocities with respect to this axis It is convenient to make your axis coincide

with one of the initial velocities

In your sketch, draw all the velocity vectors with labels including all the given information

Sketches for Collision Problems

• Draw “before” and “after” sketches

• Label each object – include the direction of

velocity– keep track of subscripts

Sketches for Perfectly Inelastic Collisions

• The objects stick together

• Include all the velocity directions

• The “after” collision combines the masses

Problem Solving for One-Dimensional Collisions.

Write the expressions for the momentum of each object before and after the collision Remember to include the appropriate signs

Write an expression for the total momentum before and after the collision Remember the momentum of the system is what is

conserved

If the collision is inelastic, solve the momentum equation for the unknown Remember, KE is not conserved

If the collision is elastic, you can use the KE equation to solve for two unknown

Glancing Collisions

For a general collision of two objects in three-dimensional space, the conservation of momentum principle implies that the total momentum of the system in each direction is conserved

Use subscripts for identifying the object, initial and final, and components

fyfyiyiy

fxfxixix

vmvmvmvm

andvmvmvmvm

22112211

22112211

Glancing Collisions

• The “after” velocities have x and y components

• Momentum is conserved in the x direction and in the y direction

• Apply separately to each direction

Problem Solving for Two-Dimensional Collisions

Set up coordinate axes and define your velocities with respect to these axes It is convenient to choose the x axis to coincide with

one of the initial velocities

In your sketch, draw and label all the velocities and include all the given information

Write expressions for the x and y components of the momentum of each object before and after the collision

Write expressions for the total momentum before and after the collision in the x-direction Repeat for the y-direction

Problem Solving for Two-Dimensional Collisions, final

Solve for the unknown quantities If the collision is inelastic, additional

information is probably required If the collision is perfectly inelastic, the final

velocities of the two objects is the same If the collision is elastic, use the KE equations

to help solve for the unknowns

Tugas Group :

Group : Holliday: Giancoli Ket.

1 & 6 9.38 7.04

2 & 7 9.35 7.17

3 & 8 9.33 7.20

4 & 9 9.32 7.27

5 & 10 9.30 7.29

Holliday hal : 232 - 233Giancolli hal : 188 - 189

See You Next Week

Prepare :Torque

Rocket Propulsion The operation of a rocket depends on the

law of conservation of momentum as applied to a system, where the system is the rocket plus its ejected fuel This is different than propulsion on the earth

where two objects exert forces on each other road on car train on track

Rocket Propulsion, cont.

The rocket is accelerated as a result of the thrust of the exhaust gases

This represents the inverse of an inelastic collision Momentum is conserved Kinetic Energy is increased (at the expense of

the stored energy of the rocket fuel)

Rocket Propulsion

The initial mass of the rocket is M + Δm M is the mass of the rocket m is the mass of the fuel

The initial velocity of the rocket is v

Rocket Propulsion

The rocket’s mass is M The mass of the fuel, Δm, has been ejected The rocket’s speed has increased to v + Δv

v

- 12

'1

'2

vv

ve

Koefisien Tumbukan / Restitusi :

1 e 0 Harga e

e = 1 elastis sempurna, sedang e = 0 tak elastis !

Contoh : 1. Sebuah bola dijatuhkan dari ketinggian h1 menumbuk lantai dan mantul naik setinggi h2.

Hitunglah koef. restitusinya.Penyelesaian :Kecepatan bola saat akan menumbuk lantai dapat dihitung dariteori kinematika, diperoleh :

v1 = (2 gh1)½

Kecepatan setelah menumbuk lantai dan mengakibatkan bola naik setinggi h2 adalah

v’1 = (2 gh2) ½. v1

v2

h1

h2

v

- 12

'1

'2

vv

ve

Karena v’2 = v2 = 0 - 1

'1

v

ve

1

2

1

2

h g 2

h g 2 -

h

he

2. Dua buah bola masing-masing bermassa m1 = 2 kg, m2 = 3 kg bergerak dalam arah berlawanan dengan kecepatan v1 = 10 m/s dan v2 = 6 m/s. Jika koef. restitusi e = 0.4. Hitunglah kecepatan masing-masing benda setelah tumbukan dan berapa energi sistem yang hilang ?.

m1

m1 m2

m2

v1 v2

v’2v’1

Hukum kekekalan momentum : m1 v1 + m2 v2 = m1 v’1 + m2 v’2

2.10 + 3.(-6) = 2 . v’1 + 3 v’2

2 [kg m/det] = 2 v’1 + 3 v’2 . . . . . . . .(1)

1,6 [ m/det] = v’1 – v’2 . . . . . . . . . . . (2)

Dari kedua pers. maka v’1 dan v’2 dapat diperoleh.Untuk menghitung energi yang hilang berarti dihitung energi kinetik sebelum dan setelah tumbukan.

v

- 12

'1

'2

vv

ve

106

v - 4,0

'1

'2

v