动量守恒定律1ConservationoflinearMomentumIf课件.ppt

1、Chapter 3 Momentum 动量动量 3-4 Angular Momentum of a Particle and Conservation of Angular Momentum 质点对定点的角动量质点对定点的角动量 角动量守恒定律角动量守恒定律 3-1 Momentum Impulse Momentum Theorem 动量动量 冲量冲量 动量原理动量原理 3-2 Conservation of Momentum 动量守恒定律动量守恒定律 3-3 Collision 碰撞碰撞明确冲量是力对时间的积累效应，掌握动量原理，注意动量的瞬时性、矢量性和相对性。掌握系统动量守恒定律，包括动

2、量分量守恒的情况，会分析动量守恒条件，包括当内力远大于外力时的情况。会用动量守恒定律、机械能守恒定律（或功能原理）解决碰撞等质点在平面内运动的力学问题。教学基本要求教学基本要求 笛卡尔：笛卡尔：Rene Descartes,Rene Descartes,15961650,15961650,法国哲学家、物理法国哲学家、物理学家、数学家和生理学家，解析学家、数学家和生理学家，解析几何的创始人。他论述了动量守几何的创始人。他论述了动量守恒问题，提出宇宙永远保持着同恒问题，提出宇宙永远保持着同量的运动，对碰撞问题做过深入量的运动，对碰撞问题做过深入研究。（研究。（Note:Note:在牛顿力学以前，在

3、牛顿力学以前，碰撞问题的研究和动量守恒定律碰撞问题的研究和动量守恒定律的发现，为建立作用与反作用定的发现，为建立作用与反作用定理准备了一定的条件）理准备了一定的条件）动量动量 is defined as the product of mass and velocity of the particle.This product is given a special name:momentum.It is also sometimes called linear momentum(线动量线动量）to distinguish（区别区别）it from a similar quantity calle

4、d angular momentum（角动量角动量）,to be discussed later.VmmVVmP矢量矢量 是以机械运动来量度机械运动本身，而动能是是以机械运动来量度机械运动本身，而动能是以机械运动转化为一定量的其它形式的运动的能力来以机械运动转化为一定量的其它形式的运动的能力来量度；量度；,在这里是代表简单的机械运动的在这里是代表简单的机械运动的转移，即持续的机械运动的量度，而动能则是已消灭转移，即持续的机械运动的量度，而动能则是已消灭的机械运动的量度。的机械运动的量度。vmPvmP 动量守恒定律不仅适用于宏观物体，而且还适用动量守恒定律不仅适用于宏观物体，而且还适用于微观物体

5、，是物理学中最重要的定律。于微观物体，是物理学中最重要的定律。3-1 Momentum(动量动量)and Impulse(冲量冲量)Momentum Theorem 动量原理动量原理本章讨论力对时间的积累本章讨论力对时间的积累效应。效应。本节讨论冲量与质本节讨论冲量与质点运动状态的变化，质点点运动状态的变化，质点动量增量之间的关系。动量增量之间的关系。1.Linear Momentum and linear Impulse Momentum Theorem 动量动量 冲量冲量 动量原理动量原理The momentum of a particles is defined asThe moment

6、um is a state quantity as the energy,which describes 运动量运动量of the particle in a certain state.VmPVmPVmPFrom the second law of Newton,we have()dVd mVPFmdtdtddtdtPdFThat iswhich is the initial(最初的）最初的）form(形式）形式）of Newtons Second Law when he found it.121221dVmVmPPtFtt(3-1)PddtF We know ,and take time

7、integration（积分积分）from t1to t2,yieldingtFPdd V V t tThe quantity at the left side is defined as the impulse（冲（冲量）量）of the force from t1 to t2,and labeled as ,that is IEquation(3-1)becomes1221dVmVmtFItt(3-2)21dtttFI冲量定义冲量定义叫质点的动量原理。叫质点的动量原理。12PPPIor:(3-1)or(3-2)is called Momentum Theorem of a particle

8、:The increment in momentum of a particle is equal to the impulse delivered(供给）供给）by the net force.粒子动量的增量粒子动量的增量等于作用于它的等于作用于它的力的冲量。力的冲量。12PPPINote:Eq.(3-2)is a vector equation.The components of(3-2)are xxttxxmVmVtFI12d21yyttyymVmVtFI1221d常用来研究碰撞问题常用来研究碰撞问题2.Average Force 平均力平均力(useful to collisions)

9、collision As shown in the below figure,the collision has finished（完成）（完成）in a very small time interval so that the force between two body is usually very great and varies rapidly（快速）（快速）with time.Collision between two ballworkpiecehammer For these cases,it is very useful to introduce the average for

10、ce as discussed as follows.ttdtttFF In Fig.3-1,an impact(冲击）冲击）force is illustrated（图解说明）（图解说明）roughly（粗略地）（粗略地）.When dealing with such a variable force,it is helpful to introduce an average force ,defined by F(t)t1t2FFig.3-1FThis average force can be thought of as the constant force that would give

11、 the same impulse to the object in the time interval as the actual time-varying force gives over this same interval.平均力：可以认为此恒力在相等时间平均力：可以认为此恒力在相等时间内产生了与变力相同的冲量。内产生了与变力相同的冲量。Using the average force,Momentum Theorem can be rewritten as121221dVmVmttFtFItt)(3-3)(3-3)为用平均力表示的动量原理。为用平均力表示的动量原理。From Eq.(3

12、-3),we haveExample 3-1:Example 3-1:如图，两质量分别为如图，两质量分别为m mA A和和 m mB B木块并排放木块并排放置在光滑的水平面上，一子弹水平地穿过两木块，设子置在光滑的水平面上，一子弹水平地穿过两木块，设子弹穿过两木块所用的时间分别为弹穿过两木块所用的时间分别为t tA A和和 t tB B，木块对子弹，木块对子弹的阻力为恒力的阻力为恒力F F，求子弹穿出后两木块的速度大小。，求子弹穿出后两木块的速度大小。解：（解：（1）设子弹穿过）设子弹穿过A后两物块的速度为后两物块的速度为VA，则：，则：ABAAVmmFt)(（2）设子弹穿过）设子弹穿过B后物

13、块后物块B的的速度为速度为VB，则：，则：ABBBBVmVmFtBAAAmmFtVBBBAABmFtmmFtVmAmBExample 3-2:一质量为一质量为m的物体，以初速度的物体，以初速度V0从地面抛出，从地面抛出，抛射角为抛射角为=30，不计空气阻力，则从抛出到接触地面的，不计空气阻力，则从抛出到接触地面的过程中，物体动量增量的大小为过程中，物体动量增量的大小为 ，方向为，方向为 。解：因为解：因为|VV0P0PP则：则：0mVP|，方向竖直向下。，方向竖直向下。0VV12f21f1F2F1m2ma system：m1 and m21F2Fexternal forces:12f21fin

14、ternal forces:3.Theorem of momentum of object system 系统动量原理系统动量原理 Two-particle system:PPt)fF(ttd PPt)fF(ttdEquation(3-2)to m1&m2,it yields（带来）（带来）12f21f1F2F1m2m02112 ffAdding the two equations,and considering Newtons third lawwe have)PP()PP(t)FF(tt d12f21f1F2F1m2m122111221221d21PPPPPPtFFtt)()()(4-9)

15、where22122PPP PPP12f21f1F2F1m2m系统的动量系统的动量It means that the time rate of the momentum of thesystem depends only on the resultant external force,having nothing to do with the internal forces.系统系统的动量变化率仅由合外力决定，与内力无关的动量变化率仅由合外力决定，与内力无关 1221diittiPPtF Generalizing（推广）（推广）above equation to the system incl

16、uding several particles,we haveEq.(3-4)means that the increment in total momentum of the system is equal to the impulse delivered by the resultant external force.This is called as the momentum theorem of a system.上式表明系统所受合外力的冲量等于系统总动量的上式表明系统所受合外力的冲量等于系统总动量的增量，称为系统动量原理。增量，称为系统动量原理。(3-4)3-2 Conservati

17、on of Momentum 动量守恒定律动量守恒定律 在质点动量原理在质点动量原理的基础上，本节将的基础上，本节将讨论两个或两个以讨论两个或两个以上物体组成的系统上物体组成的系统的动量原理并由此的动量原理并由此导出动量守恒定律。导出动量守恒定律。动量守恒定律动量守恒定律1.Conservation of linear Momentum 0iFIf 12iiiiVmVm（3-5a）orEq.(3-5a)is called the conservation law of momentum,which indicates(表明）表明）that when the combined external

18、forces acting in a system equal zero,the total momentum of an isolated system of objects will remain unchanged.12iiPPso 21taniiiimVmVconst上式称为动量守恒定律，它表明：当系统不受外力上式称为动量守恒定律，它表明：当系统不受外力或合外力为零时，系统总动量在运动中保持不变，或合外力为零时，系统总动量在运动中保持不变，内力的作用仅仅改变总动量在各物体之间的分配。内力的作用仅仅改变总动量在各物体之间的分配。这一结论叫动量守恒定律。这一结论叫动量守恒定律。0ixFxi

19、ixiiVmVm120iyFyiiyiiVmVm12IfIf（3-5b）In the form of component equations:上式表明，即使系统所受合外力不为零，但如果合外上式表明，即使系统所受合外力不为零，但如果合外力在某一方向上的分量为零，则系统在该方向的分量力在某一方向上的分量为零，则系统在该方向的分量也是守恒的。也是守恒的。水平方向水平方向有时合外力或它在某方向上的分量并不为零，但有时合外力或它在某方向上的分量并不为零，但合外力（或它在某方向上的分量）比系统内物体的相互合外力（或它在某方向上的分量）比系统内物体的相互作用力（或内力在该方向上的分量）小得多而可忽略时，作用

20、力（或内力在该方向上的分量）小得多而可忽略时，系统的总动量（或动量在该方向的分量）仍可认为是守系统的总动量（或动量在该方向的分量）仍可认为是守恒的。恒的。地面摩擦力较小地面摩擦力较小When external forces 0,we say that the external force do work on the body whose kinetic energy increases;otherwise,W0,w20,w30,w20;(C)w1=0,w20;(D)w1=0,w20,w30为常数，为某一定点到质点的矢径，该质点在 处被释放，由静止开始运动，求它到达无穷远时的速度大小。rrkF

21、r rr解：设质点达无穷远时的速度大小为V，根据动能原理，有0321210rkrrdrkmVr即：即：02mrkV r romExample 3-10:两个质量分别为m1和m2的物块，由绕过滑轮的细绳连接在一起，如图所示。试求当较重的物块落下一段距离h时，每个物体的速度和加速度。h解：（1）系统：m1，m2和地球（2）受力分析：重力和绳子的张力。张力对m1和m2作的功代数和为零，则系统的机械能量守恒。（3）设下降h后，两物体的速度为本V，并选m1和m2初始位置为势能零点，则 VmghmVmghm（4）设它们的加速度为本，考虑到物体作匀加速运动和ahV22可有可有gmmmma2112h即：即：g

22、hmm)mm(V （为什么地球不出现在公式中？）Example 3-11：如图，质量为m的物体，从高出弹簧上端h处由静止落到竖立放置的轻弹簧上，弹簧的倔强系数为k，求弹簧被压缩的最大距离。解：（1）选系统：地球+物体+弹簧；（2）系统的机械能守恒；221maxmaxkxmgxmgh可解出：出：kmgkhgmmgx284222maxmaxxh（3）设弹簧被压缩的最大距离为 ,选初始弹簧上端位置为重力势能和弹性势能的零点，则maxxExample 3-12：如图，质量为m的物体，从高出弹簧上端h处由静止落到竖立放置的轻弹簧上，弹簧的倔强系数为k，求物体可能获得的最大动能。解：（1）选系统：地球+物

23、体+弹簧；（2）系统的机械能守恒；（3）当弹簧被压缩的距离为x时，物体的速度为V,选初始弹簧上端位置为重力势能和弹性势能的零点，则221kxEmgxmghkhxV整理有：mghmgxkx)x(EEkk 显然，有极大值：kgmmghEk222maxhxV 3-7 Collision 碰撞碰撞1.The types of collision:As shown in below figure,collisions can be divided into two kinds:macroscopic and microscopic(宏观的和宏观的和微观的）微观的）.Microscopic 非接触非接触M

24、acroscopic or 接触接触My God!A collision:a relatively strong force(variable)acts on each colliding particle for a relatively short time,and a observable sudden or abrupt(突然的）突然的）change in the motion of the colliding particles occurs.两个或两个以上的物体发生相互作用，使它们的运动两个或两个以上的物体发生相互作用，使它们的运动状态在极短的时间内发生了显著的变化，物理学上称这状

25、态在极短的时间内发生了显著的变化，物理学上称这种相互作用为碰撞。碰撞的物体可以直接接触，也可以种相互作用为碰撞。碰撞的物体可以直接接触，也可以不直接接触。不直接接触。No matter what is the nature of the objects that collide,the common rule of collisions is that the momentum of the system is always conserved.碰撞的共同规律：碰撞的共同规律：在碰撞过程中，碰撞物体间的相互作用力在碰撞过程中，碰撞物体间的相互作用力外力，外力，所以外力可以忽略不计，碰撞物体组成

26、的系统动量守所以外力可以忽略不计，碰撞物体组成的系统动量守恒。恒。tconstan动量The category（分类）（分类）of collision（从能量的角度）（从能量的角度）:(1)Completely elastic collision(briefly elastic弹性弹性):The total kinetic energy of two colliding particles is conserved.(2)Inelastic（非弹性）（非弹性）collision:The total kinetic energy of two colliding particles is not

27、 conserved.If two colliding particles stick（粘连）（粘连）together after collision,this type of collision is termed as completely inelastic or briefly inelastic collision.碰撞的分类：碰撞的分类：动能守恒的碰撞称为弹性碰撞。动能守恒的碰撞称为弹性碰撞。动能不守恒的碰撞称为非弹性碰撞，如果两物动能不守恒的碰撞称为非弹性碰撞，如果两物体碰撞后合二为一，以共同的速度运动，则称体碰撞后合二为一，以共同的速度运动，则称为完全非弹性碰撞。为完全非弹性碰

28、撞。2.Elastic collision in one dimension 一维弹性碰撞一维弹性碰撞自学！自学！3.Elastic collision in two dimension 二维弹性碰撞二维弹性碰撞 As shown in figure,if two bodies(balls球球)do not move along one same straight line after collision,this collision without lost of energy is elastic collision in two dimension.11Vm1021Vm212V22Vxy

29、 1 2There are following specific properties:(1)Their centers are not on the same line before collision;(2)The ball m2,called as target（靶）（靶）,is at rest usually before collision.(3)The initial direction of motion of body m1,called as projectile(抛射的）抛射的）particle,is chosen as the x-axis.11Vm1021Vm212V2

30、2Vxy 1 2 If the collision is elastic,the kinetic energy is conserved so that22221121111coscosVmVmVmx-axisy-axis222211210sinsinVmVm（4-17）222221212111212121VmVmVm（4-18）11Vm1021Vm212V22Vxy 1 211Vm1021Vm212V22Vxy 1 2Three equationsSeven variables:,V,V,V,m,mKnown quantities(often):or,V,m,mExamples 3-7:如图

31、所示，质量为如图所示，质量为 mA的小球沿光滑的弧的小球沿光滑的弧形轨道下滑，与放在轨道水平面端点形轨道下滑，与放在轨道水平面端点P处的静止的小球处的静止的小球B发生弹性碰撞，发生弹性碰撞，B的质量为的质量为mB,A、B两球碰后同时落两球碰后同时落在水平地面上。如果在水平地面上。如果A、B两球的落地点距两球的落地点距P点正下方点正下方O点的距离之比点的距离之比LA/LB=2/5，求它们的质量比，求它们的质量比mA/mB.AABBOPLALB解：（解：（1）全过程可分为：）全过程可分为：A下降、下降、A与与B碰撞和碰撞和A、B下落。下落。（2）设）设A与与B碰撞前的碰撞前的速度为速度为VA0，碰

32、后它们的速碰后它们的速度分别为度分别为VA和和VB，则，则BBAAAAVmVmVm02220212121BBAAAAVmVmVm可解出：可解出：0ABABAAVmmmmVBAAABmmVmV02（3）因两球下落时间相同，即）因两球下落时间相同，即BBAAVLVL，因此有：，因此有：52BABALLVV5BAmmAABBOPLALB3-8 The conservation of Energy能量守恒定律能量守恒定律摩擦无处不在摩擦无处不在 The frictional force is called as a noncoservative force or a dissipative force

33、 which exists everywhere and its work depends on the path.If there exist dissipative forces(internal)such as the internally frictional force,it is sure that the mechanical energy of the system decreases.From equation(3-29)we haveEEEWif非保守（3-30）intintintnoinEEEEEW （3-31）Question:disappears?非非保保守守W No

34、,it is transformed into other energy such as heat energy which leads to the increase of temperature of system so that their internal energy of system has an increment .intETherefore,the following conclusion can be made:2211内机内机EEEEThat is:对一个孤立系统，各种形式的能量可以相互转化，但无论如何转化，能量既不能产生，也不能消灭,保持守恒。We can expre

35、ss this generalized（一般）conservation of energy in words as follows Energy may be transformed from one kind to another in an isolated（孤立）system;but it cannot be created（产生）or destroyed（消失）;the total energy of the system always remains constant.If we toss a small ball into the air,it will follow a para

36、bolic path.What about a grenade?3-9.1 The Center of Mass The center of mass of a system of particles is the point that moves as though(1)all the systems mass were concentrated there and(2)all external forces were applied there.Every part of the grenade moves in a different way except one special poi

37、nt,the center of mass of the grenade,which still moves in a parabolic path.(1)if ,the center of mass lies at the position of The position of the center of mass(com)of this two-particle system can be defined as:dcomx(Fig.5-2a)212commxdmm20m 1m0comx(2)If ,the center of mass lies at the position of 10m

38、 comxd2m(3)If ,the center of mass should be halfway between them.12mm1 2comxdyx1m2mSystems of ParticlesDiscuss For a more generalized situation,in which the coordinate system has been shifted leftward.Note:in spite of the shift of the coordinate system,the center of mass is still the same distance f

39、rom each particle.112212comx mx mxmmyxcomx1xd2x1m2mo112211nnncomiiix mx mx mxm xMM12()MmmFor a system of n particles1122comx mx mxMIf particles are distributed in three dimensions,the Center of mass must be identified by three coordinates:Define the center of mass with the language of vectors.The po

40、sition of a particle is given by a position vector:iiiirxiy jz kThe position of the center of mass of a system of particles is given by a position vector:comcomcomcomrxiyjzk11ncomiiixm xM11ncomiiiym yM11ncomiiizm zM11ncomiiixm xM11ncomiiiym yM11ncomiiizm zMSolid Bodies The three scalar equation can

41、be replaced by a single vector equation11ncomi iirmrM An ordinary solid object contains many particles,can be treated as a continuous distribution of matter.mdmsumintegral(differential mass elements)the coordinates of the center of mass will be 1comxxdmM1comyydmM1comzzdmM(3-26c)Note:the center of ma

42、ss of an object need not lie within the object.We can bypass one or more of these integrals if an object has a point,a line,or a plane of symmetry.The center of mass of such an object then lies at that point,or that line,or in that plane.We consider only uniform objects(uniform density )dmMdVVSubsti

43、tute()dmM V dVinto Eq.3-26c gives1comxxdVV1comyydVV1comzzdVV The center of mass moves like a particle whose mass is equal to the total mass of the system.Then,the motion of it will be governed by Eq.(3-29)3-9.2 The law of motion of center of mass(2)is the total mass of the system.MIf we roll a cue b

44、all at a second billiard ball that is at rest,How do they move after impact?cue ballthe second billiard ball In fact,is the center of mass of the two-ball comFMa(3-29)(Newtons Second Law)Note:(1)is the net force of all external forces acted on the F Eq.5-14 is equivalent to three equations along the

45、 three coordinate axes.(3)is the acceleration of the center of mass of the a,net xcom xFMa,net ycom yFMa,net zcom zFMaGo back and examine the behavior of the billiard balls.no net external force acts on the systemThus,the center of mass must still move forward after the collision with the same speed

46、 and in the same comFMaThe cue ball has begun to roll0netF0comaTwo balls collideinternal forces dont contribute to the net force0netF0comaFor a system of n particles,Differentiating Eq.2 with respect to time leads toDifferentiating Eq.1 with respect to time gives1 12 23 3comn nMrm rm rm rm r1 12233comnnMam am am am a Newtons Second Law applies not only to a system of particles but also to a solid body.Proof of Equation 3-29(1)1 1223 3comnnMvmvm vm vm v(2)(3)We can rewrite Eq.3 as123comnnetMaFFFFF(4)11ncomi iirmrMFrom