Chapter 8 Momentum

2. Chapter 8 Momentum

4. 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)

5. Analysis of a collision

6. 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.

7. An astronaut rescue

8. Rifle recoil

9. COLLISION

10. 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.

11. Elastic Collisions In an elastic collision, momentum AND kinetic energy are conserved. pf = pi and kf = ki

13. 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.

15. USING BOTH CONSERVATION OF MOMENTUMAND CONSERVATION OF TOTAL ENERGY

16. The ballistic pendulum

17. A 2 Dimensional collision problem

18. Work, Kinetic Energy and Potential Energy Kinetic energy is related to motion: K = (1/2) mv2 Potential energy is stored: Gravitational: U = mgh Spring: U = (1/2)kx2

19. Conservation/Conversion • 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

20. Off center collisions

22. 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 other B) The velocity of each ball after the collision C) Total kinetic energy of both balls after the collision D) Total momentum of both balls after the collision

23. The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field.

24. Center of Mass for Two Objects Xcm = (m1x1 + m2x2)/(m1 + m2) = (m1x1 + m2x2)/M

26. 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.

28. 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 + ………).

29. 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 + ………).

30. 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.

31. 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); then Macm = 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