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《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 mass, linear momentum, and impulse , and discuss Newton’s second law for a system of particle and the conservation of linear momentum.

5.1 Center of mass 1. The center of mass of a

body or a system is the point that moves as though all of the mass were concentrated there and all external forces were applied there. The figure

gives

the

explanation for the concept.

2. Suppose there are N particles in the system and their coordinates are i i i z and y x ,,

is given by a position vector:

k z j y i x r i i i i

++=

Here the index identifies the particle, and

k and j i

,,are unit

vectors pointing in the direction of the x, y, and z axes

respectively. So the position of the center of mass of a system of particles can be expressed as:

∑==

++=n

i i i cm cm cm cm r m M

k z j y i x r 11

Where M is the total mass of the system. We can rewrite the above equation using three scalar equations:

∑∑∑======n

i i

i cm

i i

cm n

i i

m M

z y m M y x

m M x 1

11111

3. The center of mass 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 mass element dm , the sum of above equations becomes integrals. And the coordinates of the center of mass are defined as

???===

zdm

M z ydm M y xdm M x cm cm cm 111

Where M is now the mass of the object. The integrals are to be evaluated over all the mass elements in the object. If the object has an uniform density

(mass per volume), then

dV V

dV dm =

=ρ. Where dV is the volume occupied by a mass element dm , and V is the total volume of the object. So we cam give the position of the center of mass of the object as:

???===

zdV

V z ydV V y xdV V x cm cm cm 1

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:

i i n i ij i a m f F

=+∑=1

Where

i F

is the resultant external force acting on the ith particle,

ij f

is the internal force exerted on the ith particle by

the jth particle, and ∑=n i ij

f 1

' 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

∑∑∑∑=======n

i i i n i i i n i i i n

i i r m dt d r m dt

d dt r d m F 1

21221221)(

Since: ∑∑∑=====i i cm

cm n

i i i n

i i r m M r r M r m M m

1,,

, thus we have:

cm cm n i i a M r dt d M F F

===∑=221

Where

cm a

is the acceleration of the center of mass of the

system. We also can rewrite the above equation using three scalar equivalent equations:

∑∑∑===z

cm z

ext y cm y ext x cm x ext Ma F

Ma F

,,,,,, 3. We, therefor, can come to the conclusion that the center of mass of a body or a system moves as if all of the mass 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 p

defined

v m p =

in which m is the mass of the particle and v

is its velocity.

The direction of

is the same as that of the velocity v

, and

its SI unit is the kilogram-meter per second.

(1). Newton actually expressed 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

express it in the equation:

dt p

d F =∑. It’s equivalent to the

expression of Newton’s second law we learned before.

a m dt v d m v m dt d dt p d F

===∑)(

(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

dt p d F / =

is correct, provided

we define the momentum of a particle not as

v m

but as

)

/(1c v v m p -=

, 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 mass, 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

, which is

defined to be the vector sum of the individual particles’ linear momentum. Thus

cm n i i n v M p p p p p P

==++++=∑=1

321.

The linear momentum of a system of particles is equal to the product of the total mass M of the system and the velocity of the center of mass .

If we take the time derivative on both sides of above equation,

we have

∑===ext cm cm F a M dt

v d M dt P

d . 3. Conservation of linear momentum : suppos

e that the sum o

f the external forces actin

g on a system of particles is zero and that no particles leave or enter the system. Since ∑=0ext F

, we

have

0/=dt P d

. Or

i P P word

other in t

cons P

==tan

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 Mass: A Rocket

1. In the system we have dealt with so far, we have assumed that the total mass of the system remains constant. Sometimes it does not. Most of the mass 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： v M

Time = t ＋Δt ：

u dM v

d v dM M

+-++,

dM v d v ud v Md dM u dM v v dMd v Md dM v v M v u dM v d v dM M v M

==-?--+++=+-+++=0)

)(())(( 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:

T Ru Ma dt

M u dt dM ====-

. 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 -=

Take integral on both sides of the equations, we have

i f M M v v M M

u v v M

dM u

dv f i

f i

ln =--=?

*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

dt t F p d )(

Taking integrals on the both sides of the

equation, we have ??=f i

t t p p dt t F p d )(

We call the right side of above equation the impulse J exerted by the force during the time interval i f t t -.

So we can

rewrite the equation as

p p p p d dt t F J i f p p t t f i

f i

?=-===??)(

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:

???

?+-++=+++-=?+=++=+i i f i i f f f i i f

f i i v m m m m v m m m v v m m m v m m m m v v m v m v m v m v m v m v m v m 2211212112221212121122221122221122112211222

1212121

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

closed, isolated system, the linear momentum of each colliding body can change, but the net linear momentum cannot change, regardless 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.

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