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A bee and train question I'm trying to figure...
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19-01-2011 12:43am#1
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Ok guys the penny drops.....dink dink ;-)
I see the train/bee can have different velocitys while traveling together.
As v=d/t both objects will have travelled different distances between the two points of time that the velocity is being measured.
The velocity of an object is defined as the limit of distance/time as t goes to 0 (you will often see this expressed as dx/dt), so you shouldn't be considering two points in time (if you are, then you are considering average velocity, not velocity at some instant).
Take a look at this video.
The car and the wall are in contact from about 0:03 onwards. Notice how, even though the car is in contact with the wall, its velocity is not the same as the wall's. Similarly, the bee will come in contact with the train, but its velocity will not be the same as the train's.0 -
The velocity of an object is defined as the limit of distance/time as t goes to 0 (you will often see this expressed as dx/dt), so you shouldn't be considering two points in time (if you are, then you are considering average velocity, not velocity at some instant).
Take a look at this video.
The car and the wall are in contact from about 0:03 onwards. Notice how, even though the car is in contact with the wall, its velocity is not the same as the wall's. Similarly, the bee will come in contact with the train, but its velocity will not be the same as the train's.
Point taken on the average and instant velocity.
As for the video, the car (as a unit) has not traveled any distance for 0:03 onwards. So surely its velocity is zero
as v = d/t and the d=00 -
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Momentum is a vector quantity consisting of the mass of an object and
it's velocity, written as p = mv. The bolded letters represent the vectors.
A vector is a convenient way to express that the quantity we're dealing
with (i.e. position, velocity, acceleration) depends on both the magnitude
and the direction. So with momentum both the magnitude, i.e. the
numerical value, and the direction matter. In the picture below I have
expressed the momentum of the bee in red and the train in blue, notice
how the blue arrow is much bigger (greater magnitude). This is because
it has more mass and has a higher velocity (in this example).
Box 1 has them approaching each other. Box 2 has them just about to
collide. Box 3 is a moment after collision, notice how both arrows have
gotten smaller because of the collision (I've exaggerated the trains). Box 4
illustrates the moment when the bee has no velocity, at this point the
bee has no momentum. However, that does not mean that the system
has no momentum or that the train has no momentum. Lets do some
math:
[latex] \overline{p}_{(bee)}} \ = \ m_{(bee)} \overline{v}_{(bee)} \ = \ mv_b[/latex]
[latex] \overline{p}_{(train)} \ = \ m_{(train)} \overline{v}_{(train)}\ = mv_t[/latex]
By the conservation of momentum we know that
[latex] m_bv_b_1 \ + \ m_tv_t_1 \ = \ m_bv_b_2 \ + \ m_tv_t_2[/latex]
Since this is an inelastic collision it becomes:
[latex] m_bv_b_1 \ + \ m_tv_t_1 \ = \ (m_b \ + \ m_t)v_t_2[/latex]
At the point at which the bee has zero velocity it has zero momentum,
& the above equation just becomes:
[latex] m_bv_b_1 \ + \ m_tv_t_1 \ = \ (m_b \ + \ m_t)v_t_2[/latex]
[latex] 0 \ + \ m_tv_t_1 \ = \ (m_b \ + \ m_t)v_t_2[/latex]
[latex] m_tv_t_1 \ = \ (m_b \ + \ m_t)v_t[/latex]
So at this particular point we see that the train does not have zero
velocity it has velocity:
[latex] v_t_1 \ = \ \frac{(m_b \ + \ m_t)v_t}{(m_t)}[/latex]
This really is just an instantaneous moment, in the fifth box above you
see the fly is beginning to move with the train but has not reached the
full velocity (and therefore momentum) just yet. I must stress that this
is a miniscule amount of time. The sixth box has the fly and train moving
together now, notice how the fly has a tiny arrow tacked onto the much
bigger arrow, it has the velocity of the train but it's tiny mass forces it to
have a much smaller arrow. The last box considers both the train & the
bee as a whole system, in purple. The thing about vectors is that you can
add them and they will form a whole, notice how I've added the red and
blue arrows and they equal the magnitude and direction of the purple
arrow. They are equal.
Ultimately all of this has to do with forces, Δp = ∫F(t)dt. We see that
momentum is just an algebraically simpler way of understanding what
forces are doing over time, momentum expresses the sum of all the
infinitesimal forces occurring in a system.
Also, I must stress that the mass of the bee compared to the mass of
the train is ridiculous, in the equation
[latex] v_t_1 \ = \ \frac{(m_b \ + \ m_t)v_t}{(m_t)}[/latex]
if we algebraically simplify it:
[latex] v_t_1 \ = \ \frac{(m_b}{(m_t)}v_t + \ \frac{(m_t)}{(m_t)}v_t[/latex]
[latex] v_t_1 \ = \ \frac{(m_b)}{(m_t)}v_t + \ v_t[/latex]
the difference is ridiculous, m_b is like 0.8Kg & the train is like 10,000Kg
so we have [latex] \frac{0.8Kg}{10,000Kg}v_t \ = \ 0.00008v_t[/latex]
and this implies
[latex] v_t_1 \ = \ \frac{(m_b)}{(m_t)}v_t + \ v_t[/latex]
[latex] v_t_1 \ = \ (0.00008)v_t + \ v_t[/latex]
which wouldn't even register, (yes I chose a fat bee).However, what I can't grasp is this, If the bee has a velocity of zero for an instant...surely the train must have a velocity of zero for an instant...
With the bee stuck to the window screen of the train...with no loss of its mass. Surely the train must also be stopped for an instant?
I hope the above answers your question, there is absolutely no way that
the train stops and the equation above clearly shows that.But the Bee must at some stage still have a velocity of zero? As it changes direction
And at the time it has a velocity of zero it will be stuck to the train,so the train will also have a velocity of zero?
This is what I am trying to get to grips with...as how can the bee/train not have a velocity of zero at some stage?
Just to be clear, by the conservation of momentum we see that the
train has a velocity of
[latex] v_t_1 \ = \ \frac{(m_b \ + \ m_t)v_t}{(m_t)}[/latex]
at this moment. The conservation of momentum is just a way to
understand how forces act.v= d / t
Am I correct in saying the concept of velocity gets tricky when dealing with 'instants' in time. Lets say the instant of a collision?
That to measure the velocity, two points are required to measure the distance, and two points in time to get a time measurement. So that it is impossible to measure the velocity of something at the 'instant' of collision?
You really would need to study a bit of calculus to understand how this
works. Yes velocity is an average of the distance between two points
divided by time but through calculus you take an infinite number of
averages over an infinite number of intervals and sum them up
The conservation of momentum does not tell us the bee connot stop the train for an instant. As 'velocity' is essentially an average of velocity between too points. It does not in anyway indicate the train did not stop for a moment in time.
You can't use regular velocity in this situation. Once forces come into play,
which they do in a collision, you are dealing with accelerations. Any form
of acceleration means that velocity cannot be calculated in the regular
way of v = d/t it requires calculus as it's non-linear. Momentum hides the
fact that we're really just dealing with forces and since F = ma and
because a = dv/dt (the rate of change of velocity with respect to time)
we simply cannot think in terms of v = d/t, we are using the idea of
averages over miniscule distances over miniscule time intervals and summing up all the values.From what reduced speed does the train accelerate from and why can it not be zero?
It passes the zero mark as it goes from +10kmh to -100kmh, you have
to understand how direction comes into play, the -100kmh indicates
that the bee is going in the negative direction to that which it was
originally flying in. Look at my picture, the train is always moving from right
to left & we say it has a velocity of -100kmh with regard to our coordinate
system.0 -
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Well there are two routes you could take, you could study non-standard
calculus which as I understand it is calculus using infinitesimals & this was
only put on rigorous ground around the 1960's. A great introduction to this
would be the free book Calculus Made Easy by Silvanus Thompson. You
could then check out the non-standard analysis books that are around,
This one is written by that guy Bell you quoted (I think). I've also come
across other free books that more explicitly use infinitesimals so maybe
they would be a better choice than Thompson.
However, I think that non-standard analysis is mostly studied by students
who already know the more standard calculus & analysis of limits, this is
the kind you'll find in Stewart Calculus, Thomas Calculus, Spivak Calculus
etc... and in the khanacademy videos. This is the kind I think everyone
on here is using & I'd really suggest this because you have so many
choices of books to begin with & an almost unlimited choice of books
that use these ideas in more advanced courses.
That said, I may be getting non-standard analysis wrong because I have
no interest in it yet, the standard kind is hard enough
but the
contents of Bell's book do look good.
Let me stress that calculus/analysis doesn't care about space, it's
irrelevant. It's a mathematical construct that is independent of reality,
it just so happens that it explains nature crazily well. If you start thinking
of distances in the real world at miniscule lengths you're dealing with
quantum mechanics, you'll most likely waste a hell of a lot of time doing
this & I advise teaching yourself partial differential equations, wave
mechanics, special relativity etc... so that you can think about these
things properly* someday. That said, doing this is literally the most difficult
thing I've ever done in my life so don't expect it to be easy :pac:
* Obviously think about it, but math has a way of helping you understand previously
incomprehensible things - no joke!0 -
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