* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Transcript

Elastic Collisions in One Dimension β’ For an elastic collision, both momentum and mechanical energy are conserved. β’ In one dimension, ππ΄ π£π΄1π₯ + ππ΅ π£π΅1π₯ = ππ΄ π£π΄2π₯ + ππ΅ π£π΅2π₯ 1 2 2ππ΄ π£π΄1π₯ 2 2 2 + 12ππ΅ π£π΅1π₯ = 12ππ΄ π£π΄2π₯ + 12ππ΅ π£π΅2π₯ β’ Given the masses and initial velocities, we can solve for the final velocities. β’ We find that the relative velocities before and after the collision are equal magnitude but opposite sign: π£π΄1π₯ β π£π΅1π₯ = β π£π΄2π₯ β π£π΅2π₯ Elastic Collisions in One Dimension β’ For the special case where one body is initially at rest, the velocities after the collision reduce to: ππ΄ β ππ΅ π£π΄2π₯ = π£π΄1π₯ ππ΄ + ππ΅ π£π΅2π₯ 2ππ΄ = π£π΄1π₯ ππ΄ + ππ΅ Collision Example #2 Center of Mass β’ For a collection of particles, we can define the center of mass as the mass-weighted average position of the particles. π1 π₯1 + π2 π₯2 + π3 π₯3 + β― π ππ π₯π π₯cm = = π1 + π2 + π3 + β― π ππ π1 π¦1 + π2 π¦2 + π3 π¦3 + β― π ππ π¦π π¦cm = = π1 + π2 + π3 + β― π ππ β’ In vector form, π1 π1 + π2 ππ + π3 π3 + β― πcm = = π1 + π2 + π3 + β― β’ The sum of the masses is the total mass π. π= ππ π π ππ ππ π ππ Center of Mass Example Center of Mass of Solid Bodies β’ To find the center of mass of a solid body, we replace the sums with integrals. β’ For uniform bodies, the center of mass is the geometric center. β’ The center of mass will lie on an axis of symmetry, if there is one. β’ The center of mass does not have to be within the body. Motion of the Center of Mass β’ Taking the time derivative of the center of mass position gives the center of mass velocity. π1 π1 + π2 ππ + π3 π3 + β― πcm = π1 + π2 + π3 + β― π ππ ππ πcm = π β’ We then see that the total momentum is the total mass times the velocity of the center of mass. ππcm = π1 π1 + π2 ππ + π3 π3 + β― π· = ππcm External Forces and Motion of the Center of Mass β’ Similarly, taking the time derivative of the center of mass velocity gives the center of mass acceleration. π1 π1 + π2 ππ + π3 π3 + β― π ππ ππ πcm = = π1 + π2 + π3 + β― π ππ β’ The sum of all of the forces, internal and external is equal to the mass times acceleration. β’ By Newtonβs third law, for each internal force, there is an equal but opposite reaction force, so the internal forces cancel out. Ξ£πext = ππcm β’ So, the collection of particles moves as if all of the mass were located at the center of mass, and were acted on by only the external forces. External Forces and Motion of the Center of Mass β’ We can also express this in terms of momentum. ππ· Ξ£πext = ππ‘ β’ Only external forces can change the total momentum. β’ This is a restatement of conservation of momentum. Rocket Propulsion β’ In outer space, there is no atmosphere, so how do space ships maneuver? β’ Rocket propulsion β’ Conservation of momentum β’ Burned fuel is ejected at a high velocity from the rear of the rocket, propelling it forward. β’ The mass of the rocket is therefore decreasing. Rocket Propulsion β’ Consider a rocket in outer space (no gravity or air resistance) with a mass π and velocity of magnitude π£. β’ In time ππ‘, the rocket ejects burned fuel and its mass changes by ππ, a negative quantity. β’ A positive mass of βππ is ejected at a relative velocity of βπ£ex β’ From conservation of momentum, we can find the thrust, or force exerted on the rocket, and its acceleration. β’ The mass of the rocket is therefore decreasing. Rocket Propulsion β’ The thrust is given by: ππ πΉ = βπ£ex ππ‘ β’ From Newtonβs second law, the acceleration in the absence of any other forces, such as gravity, is: π£ex ππ π=β π ππ‘ β’ Integrating, we find that the change in velocity is given by: π0 π£ β π£0 = π£ex ln π Rocket Propulsion Example Chapter 8 Summary Momentum, Impulse, and Collisions β’ Momentum: π = ππ β’ Newtonβs second law: β’ Impulse: π= ππ ππ‘ π‘2 π± = π2 β π1 = π‘ Ξ£π ππ‘ = 1 πave Ξπ‘ β’ Conservation of momentum β’ No external forces: π· = ππ΄ ππ΄ + ππ΅ ππ΅ + β― = constant β’ Collisions β Momentum is conserved β’ Elastic β Kinetic energy is conserved β’ Inelastic β Loss of kinetic energy Chapter 8 Summary Momentum, Impulse, and Collisions β’ Center of mass β’ πcm = π1 π1 +π2 ππ +π3 π3 +β― π1 +π2 +π3 +β― ππ· β’ Ξ£πext = ππcm = ππ‘ β’ Rocket propulsion β’ Thrust: πΉ = βπ£ex β’ π£ β π£0 = π£ex ln π0 π ππ ππ‘ = π ππ ππ π ππ