Datasets:
vnpthack
/
vnpt_data

Dataset card Data Studio Files
xet
 Community
1
text
stringlengths
0
6.73k
assumptions, the mass must vary in this way. (W© have to say “a few other
assumptions” because we cannot prove anything unless we have some laws which
--- Trang 312 ---
we assume to be true, if we expect to make meaningful deductions.) To avoid
the need to study the transformation laws of force, we shall analyze a collision,
where we need know nothing about the laws of force, except that we shall assume
the conservation of momentum and energy. Also, we shall assume that the
momentum of a particle which is moving is a vector and is always directed in the
direction oŸ the velocity. However, we shall not assume that the momentum is a
constønt tìmes the velocity, as Newton did, but only that it is some ƒwncfion of
velocity. We thus write the momentum vector as a certain coefficient times the
vector velocity:
Dp~Tn,„0®. (16.8)
We©e put a subscript ø on the coeficient to remind us that it is a function of
velocity, and we shall agree to call this coefficient m„ the “mass.” Of course,
when the velocity is small, it is the same mass that we would measure in the
slow-moving experiments that we are used to. Now we shall try to demonstrate
that the formula for rm„ must be rmo/4/1 — 02/2, by arguing from the principle
of relativity that the laws of physics must be the same in every coordinate system.
1 9/2 6/2
¡ l 9/2 g/2
1 1 (b)
Fig. 16-2. Two views of an elastic collision between equal obJects
moving at the same speed In opposite directions.
Suppose that we have two particles, like two protons, that are absolutely equal,
and they are moving toward each other with exactly equal velocities. Theïir total
mmomentum is zero. Now what can happen? After the collision, their directions
of motion must be exactly opposite to each other, because If they are not exactly
opposite, there will be a nonzero total vector momentum, and momentum would
not have been conserved. Also they must have the same speeds, since they are
exactly similar objects; in fact, they must have the same speed they started with,
since we suppose that the energy is conserved in these collisions. 5o the diagram
of an elastic collision, a reversible collision, will look like Fig. 16-2(a): all the
arrows are the same length, all the speeds are equal. We shall suppose that such
--- Trang 313 ---
collisions can always be arranged, that any angle Ø can occur, and that any speed
could be used in such a collision. Next, we notice that this same collision can be
viewecd diferently by turning the axes, and just for convenience we sÖø/l turn
the axes, so that the horizontal splits i evenly, as in Fig. 16-2(b). It is the same
collision redrawn, only with the axes turned.
„v2 :
œ œ x tị tị x
tu tu '94 œ
w 1⁄⁄v v51
() 1ÍŸ1 (b)
Fig. 16-3. Two more views of the collision, from moving cars.
Now here is the real trick: let us look at this collision from the point of view of
someone riding along ín a car that is moving with a speed equal to the horizontal
component of the velocity of one particle. Then how does the collision look?
Tt looks as though particle 1 is just going straight up, because it has lost its
horizontal component, and it comes straight down again, also because i% does
not have that component. That is, the collision appears as shown in Fig. 16-3(a).
Particle 2, however, was going the other way, and as we ride pastf 1% appears tO
ñy by at some terrifc speed and at a smaller angle, but we can appreciate that
the angles before and after the collision are the sœme. Let us denote by u the
horizontal component of the velocity of particle 2, and by œ the vertical velocity
of particle 1.
Now the question is, what is the vertical velocity œtan œ? If we knew that, we
could get the correct expression for the momentum, using the law of conservation
of momentum in the vertical direction. Clearly, the horizontal component of the
1mmomentum is conserved: ¡% is the same before and after the collision for both
particles, and is zero for particle 1. So we need use the conservation law only
for the upward velocity wutanœ. But we cøn get the upward velocity, simply by
looking at the same collision going the other wayl If we look at the collision of
Eig. 16-3(a) from a car moving to the left with speed ứ, we see the same collision,
except “turned over,” as shown in Eig. 16-3(b). Ñow particle 2 is the one that goes
up and down with speed œ, and particle 1 has picked up the horizontal speed ứ.
--- Trang 314 ---
Of course, now we &noœ what the velocity wœtanœ is: iE is 04/1 — w2/c2 (see
Eq. 16.7). We know that the change in the vertical momentum of the vertically
moving particle is
Ap = 2m„+
(2, because it moves up and back down). The obliquely moving particle has a
certain velocity ø whose components we have found to be w and +04/1 — u2/c2,
and whose mass is m„. The change in øerf2cøl momentum of this particle is
therefore AjÈ' = 2m„+0v/1— u2/c2 because, in accordance with our assumed
law (16.8), the momentum component is always the mass corresponding to the
magnitude of the velocity times the component of the velocity in the direction of
interest. Thus in order for the total momentum to be zero the vertical momenta,
must cancel and the ratio of the mass moving with speed ø and the mass moving
with speed + must therefore be
— = V1 u2/e2. (16.9)
°U
Let us take the limiting case that +0 is inÑnitesimal. lÝ u is very tiny indeed, it
1s clear that ø and œ are practically equal. In this case, m„„ —> nọ and rn„ —> my.
The grand result is
¬ .. (16.10)
“ v1—u?/c2
]t is an interesting exercise now to check whether or not Eq. (16.9) is indeed true
for arbitrary values of +, assuming that Eq. (16.10) is the right formula for the
mass. Note that the velocity ø needed in Eq. (16.9) can be calculated rom the
right-angle triangle:
0U —= uẺ + 00Ẻ(1 — u2/c2).
Tlt will be found to check out automatically, although we used it only in the limit
Of smaill ơi.
Now, let us accept that momentum is conserved and that the mass depends
upon the velocity according to (16.10) and go on to find what else we can conclude.
Let us consider what is commonly called an ?melastic collision. For simplicity, we
shall suppose that two objects of the same kind, moving oppositely with equal
speeds +0, hit each other and stick together, to become some new, statlonary