Chapter 9 – Center of mass and linear momentumI. The center of mass- System of particles / - Solid bodyII. Newton’s Second law for a system of particlesIII. Linear Momentum - System of particles / - ConservationIV. Collision and impulse- Single collision / - Series of collisionsV. Momentum and kinetic energy in collisionsVI. Inelastic collisions in 1D-Completely inelastic collision/ Velocity of COMVII. Elastic collisions in 1DVIII. Collisions in 2DIX. Systems with varying massX. External forces and internal energy changesI. Center of massThe center of mass of a body or a system of bodies is apoint that moves as though all the mass wereconcentrated there and all external forces were appliedthere.- System of particles:Mxmxmmmxmxmxcom2211212211General:- The center of mass lies somewhere between the two particles.- Choice of the reference origin is arbitrary  Shift of the coordinate systembut center of mass is still at the same relative distance from each particle.M = total mass of the systemI. Center of mass- System of particles:Origin of reference system coincides with m1dmmmxcom2123D:niiicomniiicomniiicomzmMzymMyxmMx111111niiicomrmMr11- Solid bodies: Continuous distribution of matter.Particles = dm (differential masselements).3D:M = mass of the objectAssumption:dVdmVMThe center of mass of an object with a point, line or plane of symmetry lieson that point, line or plane. dVzVzdVyVydVxVxcomcomcom111Uniform objects  uniform densityThe center of mass of an object does not need to lie within the object.Examples: doughnut, horseshoe dmzMzdmyMydmxMxcomcomcom111Volume densityLinear density: λ = M / L  dm = λ dx Surface density: σ = M / A  dm = σ dAProblem solving tactics:(1) Use object’s symmetry.(2) If possible, divide object in several parts. Treat each of these parts as aparticle located at its own center of mass.(3) Chose your axes wisely. Use one particle of the system as origin of yourreference system or let the symmetry lines be your axis.II. Newton’s second law for a system of particlesIt moves as a particle whose mass is equalto the total mass of the system.Motion of the center of mass:comnetaMF-Fnetis the net of all external forces that act on the system. Internal forces(from one part of the system to another are not included).- Closed system: no mass enters or leaves the system during movement.(M=total mass of system).-acomis the acceleration of the system’s center of mass.zcomznetycomynetxcomxnetMaFMaFMaF,,,,,,Proof:(*)............32133221122332211332211nnncomcomnncomcomnncomFFFFamamamamaMdtvdMdtrdMvmvmvmvmvMdtrdMrmrmrmrmrM(*) includes forces that the particles of the system exert on each other(internal forces) and forces exerted on the particles from outside thesystem (external).Newton’s third law  internal forces from third-law force pairs cancelout in the sum (*)  Only external forces.III. Linear momentum- Vector magnitude.vmpLinear momentum of a particle:- System of particles:The time rate of c hange of the momentum of a particle is equal to the netforce acting on the particle and it is in the direction of that force.amdtvmddtpdFnet)(Equivalent to Newton’s second law.nnnvmvmvmvmppppP .......332211321The total linear moment P is the vector sum of the individual particle’s linearmomenta.comvMPThe linear momentum of a system of particles is equal to the product of thetotal mass M of the system and the velocity of the center of mass.dtPdFaMdtvdMdtPdnetcomcomNet external force acting on the system.- Conservation:If no external force acts on a closed, isolated system of particles, the totallinear momentum P of the system cannot change.ifnetPPdtPdFsystemisolatedClosedcteP0),(If the component of the net external force on a closed system is zeroalong an axiscomponent of the linear momentum along that axiscannot change.The momentum is constant if no external forces act on a closed particlesystem. Internal forces can change the linear momentum of portions ofthe system, but they cannot change the total linear momentum of theentire system.Isolated: the net external force acting on the system is zero. If it is notisolated, each component of the linear momentum is conservedseparately if the corresponding component of the net externalforce is zero.Closed: no matter passes through the systemboundary in any direction.IV. Collision and impulseCollision: isolated event in which two or more bodies exert relatively strongforces on each other for a relatively short time.- Single collisionpppdttFJdttFpddttFpddtpdFifttttppfififi)()()(- Measures the strength and duration of the collision force- Vector magnitude.Third law force pairFR=-FL JR=-JLImpulse:- Impulse-linear momentum theoremThe change in the linear momentum of a body in a collision isequal to the impulse that acts on that body.JpppifUnits: kg m/szzizfzyyiyfyxxixfxJpppJpppJppptFJavgFavgsuch that:Area under F(t)-∆t curve = Area under Favg- ∆t- Series of collisionsTarget fixed in place n-projectiles n ∆p = Total change in linearmomentum (projectiles)pnJJsprojectileettargImpulse on the target:J and ∆p have opposite directions,pf<pi ∆pleft J to the right.vmtnptntJFavgn/∆t  Rate at which the projectilescollide with the target.a) Projectiles stop upon impact: ∆v=vf-vi=0-v=-vb) Projectiles bounce: ∆v=vf-vi=-v-v=-2vvtmFtinnmmavg∆m/∆t  Rate at which masscollides with the target.V. Momentum and kinetic energy in collisionsAssumptions: Closed systems (no mass enters or leaves them)Isolated systems (no external forces act on the bodieswithin the system)- Elastic collision:If the total kinetic energy of the system of two collidingbodies is unchanged (conserved) by the collision.- Inelastic collision:The kinetic energy of the system is not conserved some goes into thermal energy, sound, etc.Example: Superball into hard floor.-