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1、物理双语教学课件Chapter 5 Conservation of Linear Momentum 动量守恒 Chapter 5 Conservation of Linear Momentum In this chapter, we will introduce the concepts of center of ma, linear momentum, and impulse, and discu Newtons second law for a system of particle and the conservation of linear momentum. 5.1 Center of
2、 ma 1.The center of ma of a body or a system is the point that moves as though all of the ma were concentrated there and all external forces were applied there.The figure gives us the explanation for the concept.2.Suppose there are N particles in the system and their coordinates are xi,yi,andzi is g
3、iven by a position vector: rrri,j,andkare unit rrrrri=xii+yij+zik Here the index identifies the particle, and vectors pointing in the direction of the x, y, and z axes 1 respectively.So the position of the center of ma of a system of particles can be expreed as: r1rrrrcm=xcmi+ycmj+zcmk=Mrmrii i=1nWh
4、ere M is the total ma of the system.We can rewrite the above equation using three scalar equations: xcm1=M1=M1=Mmxii=1nii=1nni ycmzcmmy imzi=1ii3.The center of ma for a continuum: If an object contains so many particles that we can best treat it as a continuous distribution of matter.The “particles”
5、 then become differential ma element dm, the sum of above equations becomes integrals.And the coordinates of the center of ma are defined as 1xdmM1ycm=ydm M1zcm=zdmMxcm= Where M is now the ma of the object.The integrals are to be evaluated over all the ma elements in the object.If the object has an
6、uniform density r (ma per volume), then 2 dm=rdV=MdV.Where dV is the volume occupied by a ma Velement dm, and V is the total volume of the object.So we cam give the position of the center of ma of the object as: 1xdVV1ycm=ydV V1zcm=zdVVxcm= 5.2 Newtons Second Law for A System of Particles 1.We also
7、suppose there are n particles in the system.According to Newtons second law we have n dynamical equations for n particle.They are: nrvv Fi+fij=miai i=1Where rFi is the resultant external force acting on the ith is the internal force exerted on the ith particle by rfij ni=1particle, rfij the jth part
8、icle, and is the total internal forces acting on the ith particle by the other n-1 particles.2.If we make sum over two sides of the above n equations, we will get vnnvd2rid2d2v Fi=mi2=2(miri)=2dtdti=1i=1i=1dtnvmrii i=1n Since: mi=M,i=1n1vvvmr=Mr,r=iicmcmMi=13 nvmrii, thus we have: nvvrd2v F=Fi=M2rcm
9、=Macm dti=1Where racm is the acceleration of the center of ma of the system.We also can rewrite the above equation using three scalar equivalent equations: F FFext,xext,yext,z=Macm,x=Macm,y =Macm,z3.We, therefor, can come to the conclusion that the center of ma of a body or a system moves as if all
10、of the ma were concentrated there and all external forces were applied there. 5.3 Conservation of linear momentum 1.The linear momentum of a particle is a vector as rrp=mv rp defined rin which m is the ma of the particle and v is its velocity.The direction of rp ris the same as that of the velocity
11、v, and its SI unit is the kilogram-meter per second. (1).Newton actually expreed his second law of motion in terms of momentum: The rate of change of the momentum of a particle is proportional to the net force acting on the particle and is in the direction of that force.Or we can 4 expre it in the e
12、quation: rrdpF=dt.Its equivalent to the expreion of Newtons second law we learned before. rrrdprddvrF=dt=dt(mv)=mdtma (2).Momentum at very high speeds: For particles moving with speeds are near the speed of light, Newtonian mechanics predicts results that do not agree with experiment.In such cases,
13、we must use Einsteins theory of special relativity.In relativity, the formulation rmv1-(v/c)2rrF=dp/dt is correct, provided rmv we define the momentum of a particle not as rp=but as , in which c is the speed of light.2.The linear momentum of a system of particles: Consider now a system of n particle
14、s, each with its own ma, velocity, and linear momentum.The particles may interact with each other, and external forces may act on them as well.The system as a whole has a total linear momentum rP, which is defined to be the vector sum of the individual particles linear momentum.Thus nrrrrrrrP=p1+p2+
15、p3+L+pn=pi=Mvcm. i=1The linear momentum of a system of particles is equal to the product of the total ma M of the system and the velocity of the center of ma.If we take the time derivative on both sides of above equation, 5 we have rrrdvcmrdP=M=Macm=Fext.dtdt3.Conservation of linear momentum: suppos
16、e that the sum of the external forces acting on a system of particles is zero and that no particles leave or enter the system.Since have rdP/dt=0.Or rFext=0, we inotherwordrrPi=Pf rP=constant This important result is called the law of conservation of linear momentum.It tells us that if no net extern
17、al force acts on a system of particles, the total linear momentum of the system remains constant.If a component of the net external force on a closed system is zero along an axis, then the component of the linear momentum of the system along that axis cannot change. *5.4 Systems with Varying Ma: A R
18、ocket 1.In the system we have dealt with so far, we have aumed that the total ma of the system remains constant.Sometimes it does not.Most of the ma of a rocket on its launching pad is fuel, all of which will eventually be burned and ejected from the nozzle of the rocket engine. 2.Our system consist
19、 of the rocket and the exhaust products released during interval dt.The system is closed and isolated. 6 So the linear momentum of the system must be conserved during dt. Time =： MM+dM,Time = tt： -dM,vv vv+dv vvu+vvvvvvMv=(M+dM)(v+dv)+(-dM)(u+v)vvvvvv=Mv+vdM+Mdv+dMdv-vdM-udM vvvdMvMdv-udv=0dv=uMrSin
20、ce the velocity of rocket v is in the opposite direction rwith the velocity of exhaust product u, we can rewrite the above equation as follow: -dMdvu=M=Ma=Ru=Tdtdt.We replace dM/dt by R.Its the fuel consumption rate.We call the term Ru the thrust of the rocket engine and represent it with T. 3.The v
21、elocity of a rocket as it consumes its fuel can be derived from dv=-udMM Take integral on both sides of the equations, we have vfvidv=MfMi-udMMMvf-vi=ulniMf *5.5 Collision A collision is an isolated event in which two or more bodies (the colliding bodies) exert relatively strong forces on each other
22、 for 7 a relatively short time.1.Impulse and linear momentum: If there is a force acting on a object or a particle, according to Newtons second law, we have rrdp=F(t)dt. Taking integrals on the both sides of the equation, we have prrpfitfrrdp=F(t)dt tiWe call the right side of above equation the imp
23、ulse J exerted by the force during the time interval rewrite the equation as tipitf-ti. So we can rrtfrpfrrrrJ=F(t)dt=rdp=pf-pi=Dp 2.Elastic collisions in one dimension: In an elastic collision, both the kinetic energy and the linear momentum of each colliding body can change, but the total kinetic
24、energy and the net linear momentum do not change.We have equations as follow: m1v1i+m2v2i=m1v1f+m2v2f 1111222m1v12i+m2v2=mv+m2v2i11ff2222m1-m22m2v=v+1fm+m1im+mv2i1212v=2m1v+m2-m1v2f1i2im1+m2m1+m2 3.Inelastic collisions in one dimension: An inelastic collision is one in which the kinetic energy of th
25、e system of colliding bodies is not conserved.If the colliding bodies sticks together, the collision is called a completely inelastic collision.In a 8 closed, isolated system, the linear momentum of each colliding body can change, but the net linear momentum cannot change, regardle of whether the co
26、llision is elastic.4.Collisions in two dimensions: Here we consider a glancing collision (it is not head-on) between a projectile body and a target body at rest. 9 物理双语教学课件Chapter 5 Conservation of Linear Momentum 动量守恒 物理双语教学课件Chapter 7 Rolling Torque, and Angular Momentum 力矩与角动量 物理双语教学课件Chapter 7 R
27、olling Torque, and Angular Momentum 力矩与角动量 物理双语教学课件Chapter 9 Oscillations 振动 物理双语教学课件Chapter 10 Waves 波动 物理双语教学课件Chapter 6 Rotation 定轴转动 物理双语教学课件Chapter 15 Electric Fields 电场 物理双语教学课件Chapter 23 Interference 干涉理论 物理双语教学课件Chapter 21 Induction and Inductance, Maxwellx27s Equations 自感互感 麦克斯韦方程 动量守恒定律的应用的教学过程设计物理教案小编举荐 本文来源：网络收集与整理，如有侵权，请联系作者删除，谢谢！第13页 共13页第 13 页 共 13 页第 13 页 共 13 页第 13 页 共 13 页第 13 页 共 13 页第 13 页 共 13 页第 13 页 共 13 页第 13 页 共 13 页第 13 页 共 13 页第 13 页 共 13 页第 13 页 共 13 页