chapter 8 momentum

Chapter 8Momentum

Goals for Chapter 8 • To study momentum.• To understand conservation of momentum.• To study momentum changes during

collisions.• To add time and study impulse.• To understand center of mass and how

forces act on the c.o.m.• To apply momentum to rocket propulsion.

Definition of Total Momentum

The total momentum P of any number particles is equal to the vector sum of the momenta of the individual particles:

P = PA + PB + PC + ……. ( total momentum of a system of particles)

Analysis of a collision •

Conservation of Momentum

The total momentum of a system is constant whenever the vector sum of the external forces on the system is zero. In particular, the total momentum of an isolated system is constant.

An astronaut rescue

In collisions, we assume that external forces either sum to zero, or are small enough to

be ignored. Hence, momentum is conserved in

all collisions.

Pf = Pi

Elastic Collisions

In an elastic collision, momentum AND kinetic energy are

conserved.

pf = pi

and kf = ki

A 1 Dimensional collision problem

A 2 Dimensional collision problem

Elastic Collisions

In an elastic collision, momentum AND kinetic energy are

conserved.

pf = pi

and kf = ki

Inelastic Collisions

In an inelastic collision, the momentum of a system is conserved,pf = pi

but its kinetic energy is not,Kf ≠ Ki

Completely Inelastic Collisions

When objects stick together after colliding, the collision is completely inelastic.In completely inelastic collisions, the maximum amount of kinetic energy is lost.

USING BOTH CONSERVATION OF MOMENTUM

ANDCONSERVATION OF TOTAL ENERGY

Work, Kinetic Energy and Potential Energy

Kinetic energy is related to motion:

K = (1/2) mv2

Potential energy is stored:

Gravitational:U = mghSpring:

U = (1/2)kx2

Conservation of total energyKf + Uf = Ki + Ui + Wother

The ballistic pendulum

•

Elastic CollisionsIn an elastic collision,

momentum AND kinetic energy are conserved.

pf = pi

and kf = kiFor head-on collision of 2 objects:

vf,A = [(mB – mA)/(mA+ mB)]vi,A

vf,B = [(2mA)/(mA+ mB)]vi,A

Relative velocity: vf,B – vf,A = -(vi,B – vi,A)

Impulse for constant force

Impulse, duration of the impact

Impulse, duration of the impact

The center of mass of a system of

masses is the point where the system

can be balanced in a uniform gravitational

field.

Center of Mass for Two ObjectsXcm = (m1x1 + m2x2)/(m1 + m2) = (m1x1 + m2x2)/M

Locating the Center of Mass

In an object of continuous,

uniform mass distribution, the center of

mass is located at the

geometric center of the

object. In some cases, this means

that the center of mass is not located within

the object.

Suppose we have several particles A, B, etc., with masses mA, mB, …. Let the coordinates of A be (xA, yA), let those of B be (xB, yB), and so on. We define the center of mass of the system as the point having coordinates (xcm,ycm) given by

xcm = (mAxA + mBxB + ……….)/(mA + mB + ………),

Ycm = (mAyA + mByB +……….)/(mA + mB + ………).

The velocity vcm of the center of mass of a collection of particles is the mass-weighed average of the velocities of the individual particles:

vcm = (mAvA + mBvB + ……….)/(mA + mB + ………).

In terms of components,

vcm,x = (mAvA,x + mBvB,x + ……….)/(mA + mB + ………),

vcm,y = (mAvA,y + mBvB,y + ……….)/(mA + mB + ………).

For a system of particles, the momentum P is the total mass M = mA + mB +…… times the velocity vcm of the center of mass:

Mvcm = mAvA + mBvB + ……… = P

It follows that, for an isolated system, in which the total momentum is constant the velocity of the center of mass is also constant.

Acceleration of the Center of Mass:Let acm be the acceleration of the cener of mass (the rate of change of vcm with respect to time); thenMacm = mAaA + mBaB + ………

The right side of this equation is equal to the vector sum ΣF of all the forces acting on all the particles. We may classify each force as internal or external. The sum of forces on all the particles is then

ΣF = ΣFext + ΣFint = Macm