9. Linear Momentum

GROUP 9

PERSONAL :

JUWITA CANDRA REANEETA SAFITRIE

RIFKA NURHAQI

Pendidikan Biologi Non Reguler

Fakultas Matematika dan Ilmu Pengetahuan Alam

Universitas Negeri Jakarta

2010

ASSALAMUALAIKUM WR. WB

Linear Momentum

and Collisions

Linear Momentum and Its Conservation Impulse and Momentum Collisions in One Dimension Two-Dimensional Collisions The Center of Mass Motion of a System of Particles Rocket Propulsion

CHAPTER OUTLINE

LINEAR MOMENTUM AND ITS CONSERVATION

the momentum of an isolated system is conserved. The momentum of one particle within an isolated system is not necessarily conserved, because other particles in the system may be interacting with it. Always apply conservation of momentum to an isolated system.

IMPULSE AND MOMENTUM

In many physical situations, we shall use what is called the impulse approximation, in which we assume that one of the forces exerted on a particle acts for a short time but is much greater than any other force present.

COLLISIONS IN ONE DIMENSION

PERFECTLY INELASTIC COLLISIONS

Collision between two objects is said to perfectly inelastic when total kinetic energyobjects before and after the collision remain, so the value of the coefficient of restitution equal to 1 (e= 1).So the perfect springy collision applicable law and the law of

conservation of momentumconservation of kinetic energy, the equation used is :

AND

fii vmmvmvm )( 212211

21

2211

mm

vmvmv iif

ELASTIC COLLISIONS

In some springy collision, kinetic energy conservation law does not apply because therechange of kinetic energy before and after the collision.In the collision lening most only apply the law of conservation of momentum alone, and the coefficientcollision resilient partial restitution has a value between zero and one.

ELASTIC COLLISIONS

Momentum:1 1i 2 2i 1 1f 2 2fm v m v m v m v

Energy:2 2 2 2

1 1i 2 2i 1 1f 2 2f

1 1 1 1m v m v m v m v2 2 2 2

ELASTIC COLLISIONS

1 1i 1f 2 2f 2im v v m v v

2 2 2 21 1i 1f 2 2f 2im v v m v v

1 1i 1f 1i 1f 2 2f 2i 2f 2im v v v v m v v v v

1i 1f 2f 2iv v v v

1i 2i 2f 1fv v v v

ELASTIC COLLISIONS – EQUAL MASS

1 1i 2 2i 1 1f 2 2fm v m v m v m v

1i 2i 1f 2fv v v v

1i 2i 2f 1fv v v v

1i 2fv v 2i 1fv v

ELASTIC COLLISION – MASS AT REST

m1 m2

1 1i 1 1f 2 2fm v m v m v 1 1i 1f 2 2fm v v m v

1i 2i 2f 1fv v v v 1i 2f 1fv v v

12f 1i

1 2

2mv v

m m

1 21f 1i

1 2

m mv v

m m

v1

ELASTIC COLLISION – GENERAL CASE

m1 m2

v1 v2

1 2 12f 1i 2i

1 2 1 2

2m m mv v v

m m m m

1 2 21f 1i 2i

1 2 1 2

m m 2mv v v

m m m m

TWO-DIMENSIONAL COLLISIONS

The game of billiards is a familiar example involving multiple collisions of objects moving on a two-dimensional surface. For such twodimensional collisions, we obtain two component equations for conservation of momentum:

The center of mass of the pair of particles described in Figure 9.17 is located on the x axis and lies somewhere between the particles. Its x coordinate is given by

The Center of MassIn this section we describe the overall motion of a mechanical

system in terms of a special point called the center of mass of the system. The mechanical system can be either a group of

particles, such as a collection of atoms in a container, or an extended object, We shall see that the center of mass of the

system moves as if all the mass of the system were concentrated at that point.

the position vector of the ith particle, defined by

the vector position of the center of mass of an extended object in the form

An experimental technique for determining the center of mass of a wrench. The wrench is hung freely first from point A and then from point C. The intersection of the two lines AB and CD locates the center of mass.

MOTION OF A SYSTEM OF PARTICLES velocity of the center of mass of the system:

we conclude that the total linear momentum of the system equals thetotal mass multiplied by the velocity of the center of mass.

Newton’s second law for a system of particles

The center of mass of a system of particles of combined mass M moves like an equivalent particle of mass M would move under the influence of the net external force on the system.

ROCKET PROPULSION

MENTUM OF THE SYSTEM TO THE TOTAL FINAL MOMENTUM, WE OBTAIN

where M represents the mass of the rocket and its remaining fuel after an amount of fuel having mass has been ejected. Simplifying this expression gives

We divide the equation by M and integrate, taking the initial mass of the rocket plus fuel to be Mi and the final mass of the rocket plus its remaining fuel to be Mf. This gives

This is the basic expression for rocket propulsion. First, it tells us that the increase in rocket speed is proportional to the exhaust speed ve of the ejected gases.

Therefore, the exhaust speed should be very high. Second, the increase in rocket speed is proportional to the natural logarithm of the ratio Mi/Mf. Therefore, this ratio should be as large as possible, which means that the mass of the rocket without its fuel should be as small as possible and the rocket should carry as much fuel as possible

The thrust on the rocket is the force exerted on it by the ejected exhaust gases. We can obtain an expression for the thrust

Thank You for your attention

Wassalamualaikum Wr. Wb