《物理双语教学课件》Chapter 5 Conservation of Linear Momentum 动量守恒_课件适合双语教学

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Chapter 5 Conservation of Linear Momentum

In this chapter, we will introduce the concepts of center of ma, linear momentum, and impulse, and discu Newton’s second law for a system of particle and the conservation of linear momentum.5.1 Center of 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 given by a position vector:

i,j,andkare unit

rixiiyijzik

Here the index identifies the particle, and vectors pointing in the direction of the x, y, and z axes respectively.So the position of the center of ma of a system of particles can be expreed as:

1rcmxcmiycmjzcmkMmrii

i1nWhere M is the total ma of the system.We can rewrite the above equation using three scalar equations:

xcm1M1M1Mmxii1nii1nni

ycmzcmmy

imzi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” then become differential ma element dm, the sum of above equations becomes integrals.And the coordinates of the center of ma are defined as

1xdmM1ycmydm

M1zcmzdmMxcm

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 uniform density

(ma per volume), then 2 dmdV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:

1xdVV1ycmydV

V1zcmzdVVxcm

5.2 Newton’s Second Law for A System of Particles

1.We also suppose there are n particles in the system.According to Newton’s second law we have n dynamical equations for n particle.They are:

n Fi'fijmiai

i1Where Fi is the resultant external force acting on the ith is the internal force exerted on the ith particle by

'fij

ni1particle, fij

the jth particle, 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

nnd2rid2d2 Fimi22(miri)2dtdti1i1i1dtnmrii

i1n Since: miM,i1n1mrMr,riicmcmMi13

nmrii, thus we have:

nd2 FFiM2rcmMacm

dti1Where acm is the acceleration of the center of ma of the system.We also can rewrite the above equation using three scalar equivalent equations:

F

FFext,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 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

pmv

p

defined

in which m is the ma of the particle and v is its velocity.The direction of

p is the same as that of the velocity 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 equation:

dpFdt.It’s equivalent to the expreion of Newton’s second law we learned before.dpddvF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, we must use Einstein’s theory of special relativity.In relativity, the formulation mv1(v/c)2Fdp/dt

is correct, provided

mv we define the momentum of a particle not as p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 particles, 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

P, which is defined to be the vector sum of the individual particles’ linear momentum.Thus

nPp1p2p3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, we have dvcmdPMMacmFext.dtdt3.Conservation of linear momentum: suppose 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 dP/dt0.Or

Fext0, we

inotherwordPiPf

Pconstant

This important result is called the law of conservation of linear momentum.It tells us that if no net external 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 Rocket 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 of the rocket and the exhaust products released during interval dt.The system is closed and isolated.So the linear momentum of the system must be conserved during dt.Time =ｔ：

MMdM,Time = t＋Δt：

dM,v

vdv uvMv(MdM)(vdv)(dM)(uv)MvvdMMdvdMdvvdMudM

dMMdvudv0dvuMSince the velocity of rocket v is in the opposite direction with 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.It’s the fuel consumption rate.We call the term Ru the thrust of the rocket engine and represent it with T.3.The velocity 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 for a relatively short time.1.Impulse and linear momentum: If there is a force acting on a object or a particle, according to Newton’s second law, we have dpF(t)dt.Taking integrals on the both sides of the equation, we have

ppfitfdpF(t)dt

tiWe call the right side of above equation the impulse J exerted by the force during the time interval rewrite the equation as

tipitfti.So we can

tfpfJF(t)dtdppfpip

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 energy and the net linear momentum do not change.We have equations as follow: m1v1im2v2im1v1fm2v2f

1111222m1v12im2v2mvm2v2i11ff2222m1m22m2vv1fmm1immv2i1212v2m1vm2m1v2f1i2im1m2m1m2

3.Inelastic collisions in one dimension: An inelastic collision is one in which the kinetic energy of the system of colliding bodies is not conserved.If the colliding bodies sticks together, the collision is called a completely inelastic collision.In a closed, isolated system, the linear momentum of each colliding body can change, but the net linear momentum cannot change, regardle of whether the collision 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