Chapter 8Momentum

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

Rifle recoil

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.

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

The ballistic pendulum

•

A 2 Dimensional collision problem

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

• Work-Energy Theorem

Wtotal = Kf – Ki

• Conservatives force

Kf + Uf = Ki + Ui

• Non-conservative forces

Kf + Uf = Ki + Ui + Wother

• Onservation of momentum

pf = pi

Conservation/Conversion

Off center collisions

A ball of mass 0.240 kg moving with speed 12.2  m/s collides head-on with an identical stationary ball. (Notice that we do not know the type of collision.)

Which of the following quantities can be calculated from this information alone?

A) The force each ball exerts on the otherB) The velocity of each ball after the collisionC) Total kinetic energy of both balls after the collisionD) Total momentum of both balls after the collision

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