| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1650.1 | COULD THE ANSWER BE ? | TRUCKS::KERVILL_G | | Wed Aug 12 1992 10:13 | 1 |
| (N-1)+(N-2)+(N-3) .. .. ..
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| 1650.2 | the auto insurance situations here could get *very* interesting :-) | HANNAH::OSMAN | see HANNAH::IGLOO$:[OSMAN]ERIC.VT240 | Wed Aug 12 1992 12:03 | 17 |
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Well, let's look at the first two masses. We'll use uppercase letters
for larger masses and lowercase for smaller. Assume masses flow from
left to right. Suppose we have
A b ...
A is larger than b, so A will still be moving to the right after striking
b, but b will be moving to the right a bit faster than A.
Now, b will hit something, and we want to know whether A can possibly strike
b again. If b hits something sufficiently large, A will certainly strike
b again.
Someone want to take it from here ?
/Eric
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| 1650.3 | You mean I've got to give the working????? | TRUCKS::KERVILL_G | | Thu Aug 13 1992 08:18 | 42 |
| Explanation to *.1
Please note that I shall change frame of reference after impacts.
Object 1 2 3 4 ....... N
Obj 1 move right 2 to N are "stationary" and Frame is "stationary"
Consider initial impacts only! there will be (n-1) then N moves off to
the right never to be seen again.
NOW move the Frame of reference to object N-2. In this Frame object
N-1 is moving to the Left towards N-2. the maximum number of impacts
this time is (n-2) and object 1 shoot of to the Left.
Fix the Frame of reference to object 3 and watch object 2 staert the
whole thing again.
(n-1)+(n-2)+(n-3) ....... (n-n) is a posible answer.
Looking at the momentum and energy equations we can ignore the objects
at the extreme edges when they "fly-off" as the next object to be lost
from that same side must have a lower relative velocity.
What I havn't considered in detail are additional impacts as the moving
wave of object collide. I think *.2 is the key if we consider energy
and momentum for the lumped masses. i.e. initally object 1 and Lumped
mass of objects 2 to N.
I think about it over lunch. Over to you Eric.
Gregg
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| 1650.4 | how many collisions of merely 3 objects are possible ? | HANNAH::OSMAN | see HANNAH::IGLOO$:[OSMAN]ERIC.VT240 | Thu Aug 13 1992 11:52 | 32 |
|
Imagine this configuration:
A-> b A
In other words, two very heavy and equal masses A sandwiching a small mass b.
The lefthand A moves to the right and strikes b.
A is large so it is still moving, but only slightly slower. b is now moving
at about the same rate that A is.
A-> b-> A
Now b strikes the righthand A, but since A is so heavy, it starts to move
but hardly. b reverses direction, at almost no loss of speed.
A-> <-b A->
Slightly later, the lefthand A and b collide. A is real heavy, so it continues
moving to the right, hardly slowed down. b on the other hand, experiences
quite a bounce, since it had hit A in a head-on (both moving towards each
other), so b now moves to the right faster than before, and can overtake the
righthand A.
Can this go on forever, with endless collisions ?
I'm of course using a model here where we don't have any loss of energy due
to friction of rolling, and no energy loss during collisions.
/Eric
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| 1650.5 | | DKAS::KOLKER | Conan the Librarian | Thu Aug 13 1992 20:01 | 14 |
| reply .4
What a minute. If a light b strikes a heavy A, doesn't b stop still?
Under what circumstances do we have elastic colisions?
The reason I am asking, I rember a toy where 4 steel balls are hung in
a row. If you pull the left hand ball back and let it strike the other
3 it stops dead, and the right hand ball acheives nearly the same
velocity as the left hand ball had when it struck the middle two.
This is an instance of the conservation of momentum.
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| 1650.6 | | AUSSIE::GARSON | | Fri Aug 14 1992 00:36 | 72 |
| re .5
> What a minute. If a light b strikes a heavy A, doesn't b stop still?
No. b bounces, having a speed close to what it was before the collision
but going in the opposite direction.
> Under what circumstances do we have elastic colisions?
>
> The reason I am asking, I rember a toy where 4 steel balls are hung in
> a row. If you pull the left hand ball back and let it strike the other
> 3 it stops dead, and the right hand ball acheives nearly the same
> velocity as the left hand ball had when it struck the middle two.
>
> This is an instance of the conservation of momentum.
We will assume here that both kinetic energy and momentum are conserved.
These quantities are of course the total for the system. An elastic
collision is one in which kinetic energy is conserved. (Since you can
hear the colliding balls in the above toy you know that conservation of
kinetic energy is only an approximation.)
For simplicity I will analyse the situation where an incoming object with
mass M moving to the right with speed V[1] collides with another object
with mass m stationary (with respect to the observer - but in any case
if you want a relativistic anaysis you better stop off at the PHYSICS
conference).
After the collision the incoming mass has speed V[2] and the formerly
stationary mass speed v.
To the right will be considered positive.
Before collision:
p = MV[1] E = �MV[1]�
After collision:
p = MV[2] + mv E = �MV[2]� + �mv�
Solving these equations gives
2V[1]
v = -----
m
1 + _
M
and
(M-m)
V[2] = (---) V[1]
(M+m)
Thus,
when M >> m, the heavier object slows only slightly and the lighter
object moves off at just under twice the speed the incoming mass had.
when M == m, the incoming mass stops dead and the outgoing mass leaves
with exactly the same speed as the incoming mass had.
when M << m, the lighter object 'bounces' leaving at just under its
original speed and the heavier object moves off very slowly.
All the Physics you need is to redo this derivation assuming that both
objects are initially moving (not necessarily towards each other). I
seem to recall a useful trick for solving this is to notice that the
centre of mass of the system moves with constant velocity.
That then leaves the original question, the answer for which I have no idea.
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| 1650.8 | | TRACE::GILBERT | Ownership Obligates | Fri Aug 14 1992 14:54 | 20 |
| If the two masses are in motion and collide,
M V + M V = M V' + M V' (i.e., conserve momentum)
� � � � � � � �
� M V� + � M V� = � M V'� + � M V'� (i.e., conserve energy)
� � � � � � � �
Solving for the new velocities,
V' = ((M - M )V + 2 M V )/(M + M ) = V - 2(V - V )M /(M + M )
� � � � � � � � � � � � � �
V' = ((M - M )V + 2 M V )/(M + M ) = V + 2(V - V )M /(M + M )
� � � � � � � � � � � � � �
And by the way,
V'- V' = V - V
� � � �
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| 1650.9 | | AUSSIE::GARSON | | Sun Aug 16 1992 23:54 | 10 |
| re .8
Nice work with the Compose key!
Note the assumption that the masses are sliding or in any case not rolling
(else this has to be accounted for).
My guess for the A b A situation is that b will act as a sort of agent,
transfering momentum from the left A to the right A (and get a helluva
banging in the process).
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