1Week 8: Chapter 9Linear Momentum and CollisionsLinear Momentum The linear momentum of a particle, or an object that can be modeled as a particle, of mass m moving with a velocity is defined to be the product of the mass and velocity: The terms momentum and linear momentum will be used interchangeably in the textvmpvLinear Momentum, cont Linear momentum is a vector quantity Its direction is the same as the direction of the velocity The dimensions of momentum are ML/T The SI units of momentum are kg · m / s Momentum can be expressed in component form: px= m vxpy= m vypz= m vzNewton’ Law and Momentum Newton’s Second Law can be used to relate the momentum of a particle to the resultant force acting on itwith constant massdmddmmdt dt dt   vvpFaConservation of Linear Momentum Whenever two or more particles in an isolated system interact, the total momentum of the system remains constant The momentum of the system is conserved, not necessarily the momentum of an individual particle This also tells us that the total momentum of an isolated system equals its initial momentumConservation of Momentum, 2 Conservation of momentum can be expressed mathematically in various ways In component form, the total momenta in each direction are independently conserved pix= pfxpiy= pfypiz= pfz Conservation of momentum can be applied to systems with any number of particles This law is the mathematical representation of the momentum version of the isolated system modeltotal 1 2p = p + p = constant1i 2i 1f 2fp + p= p + p 2Conservation of Momentum, Archer Example The archer is standing on a frictionless surface (ice) Approaches: Newton’s Second Law –no, no information about F or a Energy approach – no, no information about work or energy Momentum – yesArcher Example, 2 Conceptualize The arrow is fired one way and the archer recoils in the opposite direction Categorize Momentum Let the system be the archer with bow (particle 1) and the arrow (particle 2) There are no external forces in the x-direction, so it is isolated in terms of momentum in the x-direction Analyze Total momentum before releasing the arrow is 0Archer Example, 3 Analyze, cont. The total momentum after releasing the arrow is Finalize The final velocity of the archer is negative Indicates he moves in a direction opposite the arrow Archer has much higher mass than arrow, so velocity is much lower120ffppImpulse and Momentum From Newton’s Second Law,  Solving for gives  Integrating to find the change in momentum over some time interval The integral is called the impulse, , of the force acting on an object over tfitfitdt pp p F IddtpFdpddtpFIImpulse-Momentum Theorem This equation expresses the impulse-momentum theorem: The impulse of the force acting on a particle equals the change in the momentum of the particleThis is equivalent to Newton’s Second LawpIMore About Impulse Impulse is a vector quantity The magnitude of the impulse is equal to the area under the force-time curve The force may vary with time Dimensions of impulse are M L / T Impulse is not a property of the particle, but a measure of the change in momentum of the particle3Impulse, Final The impulse can also be found by using the time averaged forceThis would give the same impulse as the time-varying force doestIFImpulse-Momentum: Crash Test Example Categorize Assume force exerted by wall is large compared with other forces Gravitational and normal forces are perpendicular and so do not effect the horizontal momentum Can apply impulse approximationCollisions – Example 1  Collisions may be the result of direct contact The impulsive forces may vary in time in complicated ways This force is internal to the system Observe the variations in the active figure Momentum is conservedCollisions – Example 2 The collision need not include physical contact between the objects There are still forces between the particles This type of collision can be analyzed in the same way as those that include physical contactTypes of Collisions In an elastic collision, momentum and kinetic energy are conserved Perfectly elastic collisions occur on a microscopic level In macroscopic collisions, only approximately elastic collisions actually occur Generally some energy is lost to deformation, sound, etc. In an inelastic collision, kinetic energy is not conserved, although momentum is still conserved If the objects stick together after the collision, it is a perfectly inelastic collisionCollisions, cont In an inelastic collision, some kinetic energy is lost, but the objects do not stick together Elastic and perfectly inelastic collisions are limiting cases, most actual collisions fall in between these two types  Momentum is conserved in all collisions4Perfectly Inelastic Collisions Since the objects stick together, they share the same velocity after the collision11 2 2 1 2ii fmm mmvv v Clicker QuestionIn a perfectly inelastic one-dimensional collision between two moving objects, what condition alone is necessary so that the final kinetic energy of the system is zero after the collision?A. It is not possibleB. The objects must have momenta with the same magnitude but opposite directions.C. The objects must have the same mass.D. The objects must have the same velocity.E. The objects must have the same speed, with velocity vectors in opposite directions.Elastic Collisions Both momentum and kinetic energy are conserved11 2 211 2 22211 2 22211 2 211221122iiffiiffmmmmmmmmvvvvvvvvElastic Collisions, cont Typically, there are two unknowns to solve for and so you need two equations The kinetic energy equation can be difficult to use With some algebraic manipulation, a different equation can be usedv1i–v2i= v1f+ v2f This equation, along with conservation of momentum, can be used to solve for the two unknowns It can only be used with a one-dimensional, elastic collision between two objectsElastic Collisions, final Example of some special cases m1= m2– the particles exchange velocities When a very heavy particle