PHY 107 1st Edition Lecture 19 Outline of Last Lecture I. Center of MassII. Linear Momentum Outline of Current LectureIII. Linear MomentumIV. Collision and Impulse Current LectureLinear Momentum:-´p (kg · m/s) → ´p = m´v- The time rate of change of the linear momentum of a particle is equal to the magnitude of the net force acting on the particle and has the direction of the force´Fnet = d ´pdt- Linear momentum of a particle can be changed ONLY by an external force -´p of a system of particles  i-th particle of mass mi, velocity ´vi, and linear momentum ´pi´p = ´p1 + ´p2 + p + … + ´pn = m1´v1 + m2´v2 + m3´v3 + … +mn´vn = M ´vcom- The linear momentum of a system of particles is equal to the product of the total mass M of the system and the velocity ´vcom of the COMConservation of Linear Momentum: These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.[total linear momentum at some point ti] = [total linear momentum at some later point tf] - Note 1: In systems in which ´Fnet ,ext = 0, we can apply the principle of the conservationof linear momentum even when the internal forces are very large as in the case of colliding objects. - Note 2: We will encounter problems (e.g. inelastic collision) in which the mechanical energy is not conserved but the linear momentum is. Collision and Impulse: - Collision forces – act for a brief time, are large, and are responsible for any change in thelinear momentum of the colliding bodies - d´p = ´F(t)dt → ∫titfd ´p = ∫titf´F ( t )dt-∫titfd ´p = ´pf - ´pi = ∆´p (change in momentum) -∫titf´F (t)dt is known as the impulse, ´J-´J = ∫titf´F (t)dt Momentum and Kinetic Energy in Collisions: -´Fnet ,ext = 0- There are 2 classes of collisions: elastic and inelastic  Elastic – no loss of kinetic energy, Ki = Kf Inelastic – if KE is lost during the collision due to conversion into other forms of energy Kf < Ki - One-dimensional inelastic collisions – linear momentum of colliding objects is conserved- One-dimensional completely inelastic collisions – the two objects stick together and move as a single body: v = m1m1+m2v1- One-dimensional elastic collisions – linear momentum and kinetic energy are conserved  v1f = m1−m2m1+m2v1 i + 2 m2m1+m2v2 i v2f = 2 m1m1+m2v1 i + m2−m1m1+m2v2 iSpecial Case Collisions: - Stationary Target:  Substitute v2i = 0 in the solutions for v1f and v2fv1f = m1−m2m1+m2v1iv2f = 2 m1m1+m2v1i- Equal masses (with a stationary target): Ex. playing pool  m1 = m2 = m → v1f = m1−m2m1+m2v1i = m−mm+mv1i = 0 v2f = 2 m1m1+m2v1i= 2 mm+mv1i = v1i The colliding objects exchange