Chapter 8 Momentum
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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
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