Center of Mass and Linear Momentum

1. Center of Mass and Linear Momentum Center of Mass Systems of Particles Solid Objects Linear Momentum Collisions and Impulse Elastic, Inelastic and Perfectly Elastic Collisions Rockets pps by C Gliniewicz

2. The center of mass of a system of particles is the point that moves as though all the mass of the object were concentrated there. The center of mass behaves as a point where all the forces acting on the system seems to be acting. Solid objects contain many atoms which we can assume to be the separate particles making up the object. One can use integrals to find the center of mass. If a force acts on the system of particles, Newton’s second law holds. pps by C Gliniewicz

3. Linear momentum is a vector quantity that is the product of the mass (a scalar) and the velocity (a vector). When Newton first expressed his second law, he did so in terms of the time rate of change of momentum. The total momentum of a system of particles is equal to the product of the total mass and the velocity of the center of mass. Rearranging the terms of the last equation and integrating the result gives The impulse is denoted with the letter J. The last equation is the linear momentum-impulse theorem. Momentum is conserved in collisions between two or more objects. pps by C Gliniewicz

4. An elastic collision is a collision in which both momentum and kinetic energy are conserved. No energy is converted into thermal energy, sound or such. In most collisions, some kinetic energy is converted into heat sound, electrical or some other form of energy. These collisions may appear to be elastic if the amount of energy converted is small, but because some kinetic energy is lost these are inelastic collisions. For elastic and inelastic collisions: In a perfectly inelastic collision, the objects stick together after the collision. When this occurs, the maximum amount of kinetic energy is lost. For perfectly inelastic collisions: No matter what type of collision occurs, momentum is always conserved. The velocity of the center of mass of a system of particles remains the same in any collision unless an external force acts on the system. pps by C Gliniewicz

5. One dimensional elastic collisions are the simplest and a moving object hitting a stationary target is the simplest of all. Recall that both kinetic energy and momentum are conserved. Solving these equations simultaneously, one finds If the masses are equal it is obvious that object one stops and object 2 continues with the same velocity that the first object had. If the target is much larger than the first object, the first object bounces backward at nearly the same speed while the target moves slowly ahead. If the projectile is very large compared to the target, it continues at nearly the same speed while the target moves ahead at nearly twice the initial speed of the projectile. pps by C Gliniewicz

6. One dimensional collisions with a moving target require more algebra to solve. One starts with the same assumptions, but with both objects moving. Solving these two equations simultaneously requires more algebra skill, but one finds the following result. One can solve problems in two or even three dimensions by separating the velocity vector into unit vector notation and solving for the result in each dimension separately. Using the total mass and the center of mass velocity (which does not change) one can also solve a multi-dimensional problem using conservation of momentum. pps by C Gliniewicz

7. There are conditions where the total mass of the system remains constant. A rocket, however, loses mass as it burns its fuel. As the fuel is burned, it is ejected in the opposite direction with some velocity relative to the rocket. One can think of the ejected fuel as an explosive collision where the momentum is conserved. The momentum before the explosion is the mass of the rocket multiplied by it velocity. After the explosion, it is the momentum of the fuel plus the momentum of the rocket with its new mass. One uses dM as the mass of fuel lost (a negative value) and dv as the increase in velocity of the rocket. The left term of the final equation is called the thrust of the engine. Rearranging the terms and integrating gives the velocity. pps by C Gliniewicz