"Parse Tree" <parsetree at hotmail.com> wrote in message
news:Wfp%8.6912$DN4.1074154 at news20.bellglobal.com...
> "Cary Kittrell" <cary at afone.as.arizona.edu> wrote in message
> news:ahkjf2$rvp$1 at oasis.ccit.arizona.edu...> >
> > In article "Parse Tree" <parsetree at hotmail.com> writes:
> > <
> > <"Cary Kittrell" <cary at afone.as.arizona.edu> wrote in message
> > <news:ahkcm8$og6$1 at oasis.ccit.arizona.edu...> > <> In article "Parse Tree" <parsetree at hotmail.com> writes:
> > <> <
> > <> <"Cary Kittrell" <cary at afone.as.arizona.edu> wrote in message
> > <> <news:ahk14h$ib0$1 at oasis.ccit.arizona.edu...> > <> <> In article "John Knight" <johnknight at usa.com> writes:
> > <> <>
> > <> <> <Sqr means square-root of the equation in the parenthesis ().
> > <> <> <So, the resulting velocity would be the same, as the same time is
> > <spent
> > <> <on
> > <> <> <the fall, and the tension would be zero.
> > <> <>
> > <> <> The "resulting velocity would be the same" if both masses were
> > <> <experiencing
> > <> <> the same acceleration the instant of release, but they were not.
> The
> > <> <> bottom mass was experiencing -2mg downards due to gravity and +2mg
> > <upwards
> > <> <> due to the tension in the spring. The upper mass is experiencing
a
> > <> <> now unopposed -mg downwards due to gravity and a -2mg downwards
due
> > <> <> to the same spring tension. You figure it out.
> > <> <
> > <> <But they really are experiencing the same acceleration at the
instant
> of
> > <> <release.
> > <>
> > <> That is right as far is acceleration due to gravity is concerned,
> > <> as is implied in my statement above. But each body is experiencing
> > <> additional forces due to the spring, so they will no be subject
> > <> to the same accelerations. If you mentally switch off gravity,
> > <> the two bodies will move towards one another with an acceleration
> > <> proportional to the 2mg tension in the spring. If you now switch
> > <> gravity back on, the whole system will accelerate downards at
> > <> 1 g, but this acts equally on the whole system, so you're back
> > <> to considering things in the frame of referrence of the system
> > <> itself -- as Jet implied.
> > <
> > <This is not true. If you switch off gravity, then each sphere will
stay
> at
> > <rest. Firstly, you're assuming tension again, and secondly, the
tension
> you
> > <assume exists only because of gravity.
> > <
> >
> > This suddenly grows more interesting. I've been reading "spring"; it
> > in fact says "string".
>> Yes, I noticed that. But it's true even with a real spring.
>> > <Regardless, you can simulate this using two balls and a string. Just
put
> > <them on a table and attach them with some string. Then pull on them
and
> > <release. They don't move together with a force proprotionational to
how
> > <much you pulled them apart.
> > <
> >
> > If you assume an infinitely strong string, then you are correct.
Otherwise
> > they will indeed move, unless you've stretched the string inelastically.
> > However, I'm being picky, and you're on to the intention of the
question,
> > I think.
>> They will move, but they won't move the same amount that they are pulled
on.
> Even a spring won't, unless it is perfectly elastic.
>> > <> <Also, you're assuming the value of the unknown.
> > <> <
> > <>
> > <> Um, beg pardon? Assuming the value of what unknown? If you mean
> > <> the spring tension, I simply said that the /initial/ spring
> > <> tension is 2mg, because the lower mass is being pulled downwards
> > <> by a force of 2mg due to gravity. Since it isn't moving initially,
> > <> there must be an equal and opposite force: 2mg of tension in the
> spring.
> > <
> > <The initial spring tension is unknown. You're assuming that the bottom
> > <sphere is suspended from the top one. It simply says that it's
suspended
> at
> > <rest. Which could simply mean that the system is suspended at rest.
Who
> > <knows? Actually, I find many of these questions to be very imprecise.
> > <
> > <Regardless, the acceleration of the system is g. And the acceleration
of
> > <all of the parts are g. Thus the string's tension should be 0.
> >
> > Assuming an infinitely strong string -- one whose relaxation is zero --
> then
> > you are correct.
>> Yes. There are too many assumptions in these questions though. I can see
> why they're difficult. There was another question about probability which
> didn't even seem to specify if the two values involved were independent or
> not.
>>
You "can see why they're difficult"?
To whom are they difficult? To the 29% of American boys (after correcting
for guesses), it obviously wasn't difficult.
Maybe to the 47.7% of Norwegian boys who got it wrong it was difficult, but
you can't claim that the other half found it to be difficult, can you?
http://christianparty.net/timssh04.htm
It's notable that the international average for girls who got it correct,
after correcting for guesses, was only 1.7%, which is lower than the 3%
standard error, which suggests that most women in the world probably agree
that with you that "they're difficult". But is that a fact? No.
What's truly awesome about this forum is that you've already been given the
answer, and you *still* find it "difficult". If you already know the answer
and still find it "difficult", what does it take to get you to understand
it?
Nothing.
You've done a great job of demonstrating the thesis of this thread, which is
that there's no way to educate the uneducable. It's like teaching a pig to
sing. It frustrates the teacher and irritates the pig.
John Knight
