1 Unit 5 Conservation of Momentum Force and Momentum and Conservation of Momentum 9 1 92 9 3 Collisions 9 4 9 5 9 6 Collisions Center of Mass 9 7 9 8 Center of Mass Rocket Propulsion and Quiz Three Review 9 9 9 10 01 14 19 Physics 253 2 Center of Mass and Translational Motion We can prove that the translational motion of a system can be described by considering the effect of the net force on the center of mass In that sense the system is no different than a single force incident on a single particle 01 14 19 We start with the definition of the CM m r rCM i i M which is alternately written as MrCM mi ri Taking the derivative of both sides drCM dri M mi MvCM mi vi dt dt Taking the second derivative dvCM dvi M mi MaCM mi ai dt dt Using the 2nd Law MaCM Fi Physics 253 3 Since all the internal forces cancel by the 3 rd Law the summation involves only the external forces MaCM Fi MaCM FExternal This is Newton s 2nd Law applied to a system the sum of all the forces acting on a system is equal to the total mass of the system times the acceleration of its CM In other words the CM behaves as if all the mass of a system or object resides at the CM and as if all the external forces act directly upon it This makes treatment of the translation motion of a system quite straightforward 01 14 19 Physics 253 4 01 14 19 Physics 253 5 Example A Two Stage Rocket Problem A rocket shot into space reaches it s highest point and a horizontal distance d from it s origin separates into two equal mass segments Imagine the first part just stopped and dropped immediately to the earth Where does part II land Answer Because the explosion is internal the CM will continue on it s path The explosion happened at the top of the trajectory and if the rocket had maintained it s trajectory the CM would have landed 2d from the origin 01 14 19 Thus the average position of the two equal mass fragments at impact must be 2d This can only occur if the second stage travels a total horizontal distance of 3d Notice how we didn t need to resort to any fancy equations of motion and just applied the 2nd Law for systems Physics 253 6 We return to the definition of the CM MrCM mi ri and it s time derivative MvCM mi vi Clearly this is a relationship describing the momentum of the CM MvCM pi P In words the total linear momentum of a system of particles is equal to the product of the total mass M and the velocity of the CM 01 14 19 Taking the derivative of both sides dv dP M CM MaCM dt dt But recall that MaCM FExternal so dP FExternal dt which is Newton s 2nd Law for a system of objects If we know the force we can determine how the momentum changes We derived this before in our discussion of collisions Physics 253 7 Systems of Variable Mass Up until now we ve limited ourselves to systems of constant mass such that dP d mv dv FExt m ma dt dt dt But what if the mass of the system varies Then derivative will have two terms one proportional to dv dt and the other to dm dt 01 14 19 Physics 253 8 Two Examples where dm dt is Nonzero Rockets where dm dt 0 Conveyer belts where dm dt 0 Let s set up a formalism for such cases 01 14 19 Physics 253 9 At time t mass M and velocity v infinitesimal mass dM and velocity u note dM could be positive or negative At time t dt after coalecsing mass M dm with velocity v dv 01 14 19 Time t Time t dt Physics 253 10 Consider t he full system momentum at time t Mv u dM momentum at time t dt M dM v dv change in momentum dP M dM v dv Mv u dM dP Mv Mdv v dM dMdv Mv u dM dP Mdv v dM dMdv u dM Change of momentum with respect to time dP Mdv v dM dMdv u dM dt dt dP dv dM dMdv dM M v u dt dt dt dt dt Ignoring the term with two infinitesi mals dP dv dM dM dv dM M v u M u v dt dt dt dt dt dt 01 14 19 Physics 253 11 dP dv dM M u v dt dt dt dP But since Newton s 2nd Law can be written as FExternal dt dv dM FExternal M u v dt dt Further the relative velocity between th e two objects is u v and from the vantage point of the larger mass M the velocity of the entering mass dM is vR u v Thus dv dM FExternal M vR dt dt dM FExternal Ma vR dt dM Ma FExternal vR dt Note the simularity with previous results and the additional term now depedent on the change of mass 01 14 19 Physics 253 12 dM Ma FExternal vR dt The left side of the equation is just the usual mass times acceleration term The summation refers to all external forces such as gravity friction etc it does not include the force on M due to dM The term which depends on the change of mass with respect to time corresponds to the force exerted on M due to the addition or ejection of mass It is also the change of momentum due to the addition of mass and often called thrust This equation can be used to describe a system subject to changes of mass 01 14 19 Physics 253 13 Example 1 A Conveyer Belt A hopper drops gravel at a rate of 75 0kg s onto a conveyer belt moving at a constant speed of v 2 20m s Determine The force needed to keep the belt moving The power output of the motor drive 01 14 19 Physics 253 14 dv dM We start with FExternal M u v dt dt Remember M is the mass of the larger object which in this case is the belt dv Since it is moving at constant v elocity 0 dt Since the hopper is at rest the material being added to the system is also essentiall y at rest or u 0 dM dM Thus FExternal M 0 0 v v 2 20m s 75 0kg s 165N dt dt In words this is the force the extra mass exerts on the system the motor woul d need to provide at least that much to keep the conveyer moving The power needed to keep the conveyer moving can be calculated just as we did for that of an engine keeping a car moving P Fv 165N 2 2m s 363W 01 14 19 Physics 253 15 Example 2 A Rocket A …