Chapter 8 Momentum
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Transcript of 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