[Fruityloop (OP)]

Imagine the following situation:
We have two ships that are 5 light-years in length which are moving past each other at 0.8c.
Because of length contraction each ship will view the other ship as 3 light-years in length.
5*(1-o.8^2)^0.5 = 3
We will call the ships A and B.  We are only interested in two events.  The first event is
the front of ship B passing the tail of ship A.  The second event is the tail of ship B passing
the front of ship A.
Let us see what happens and take a look shall we?
I used F for front and T for tail.
First we will look from the reference frame where A is at rest.

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                             T----------F  (ship A)    (5 light-years in length)
 <---(0.8c)                      F------T  (ship  B)   (3 light-years in length)
later on.....
                             T----------F 
 <---(0.8c)                  F------T
So we see that the tail of ship B passes the front of ship A,
and then later on the front of ship B passes the tail of ship A.
Now let's look from the reference frame where B is at rest.

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                             T------F    ----->(0.8c)     (ship A)   (3 light-years in length)
                             F----------T                 (ship B)   (5 light-years in length)
later on...
                                 T------F    ------>(0.8c)
                             F----------T
So we see that the front of ship B passes the tail of ship A
and then later on the tail of ship B passes the front of ship A.
Now if you look carefully you will notice that the events
are in reverse order between reference frames!
Now imagine that there is some kind of structure extending up
from the tail of ship B and there is some kind of structure extending down
from the front of ship A.  The structure could be a pole, a wall, etc., whatever.
Now the structures are going to collide when the tail of ship B passes the front of
ship A.  Now the collision of the structures will be simultaneous between both
frames of reference since ship B hitting ship A and ship A hitting
ship B occur at the same time and place.  Now at that moment where is the
front of ship B?  The front of ship B has both already passed the tail of ship A
and has not yet reached the tail of ship A as one can easily see by looking at
the diagram above.  So if you believe that relativity is true, then you must
believe that the front of ship B can be in two places at the same time!
Maybe it's the tail of ship A that's in two places at the same time or
some combination of the two!

[Colin2B]

So if you believe that relativity is true, then you must
believe that the front of ship B can be in two places at the same time!

Incorrect.

You have obviously had some exposure to relativity, but clearly have not fully understood it. This may be down to poor teaching or an unwillingness to learn. I'll give you the benefit of the doubt, but suggest that you go back to your teacher and explain that you haven't understood, because this would be a good example for class discussion.

It's a good example because it takes us back to the first principles of relativity. In your first or second lesson the teacher will have introduced you to the principle of simultaneity. The usual example, which I'm sure you can find on the net, is a train passing a platform.
On the front and back walls of the carriage are 2 clocks which the observer in the carriage wants to synchronise. This is done by setting off a flash bulb at the exact centre of the carriage and when the light reaches the clocks they start counting. So each clock is now showing the same time as the other, they are synchronised.
However, to the observer on the platform the light has the same speed as measured by the observer on the train, so they see the back wall moving towards the flash and the front wall moving away and so the light hits the clocks at different times. The observer on the platform says that the clocks do not show the same time, they are not synchronised.
(Note that this is very different from Galilean relativity where if 2 balls are thrown at the front & back walls, both observers will agree that they arrive at the same time. I'll leave you to work out why.)

This difference in time on the clocks is also why we see length contraction. If we measure the length of the train by using the speed of the train and timing when the front and back pass a point on the platform, then if we disagree on the time shown by the clocks we will disagree on the length of the train. For some reason many teachers teach time dilation and length contraction as 2 separate phenomena, I don't understand why because they are 2 aspects of the same thing.

If you understand the above you will understand almost all of special relativity (and quite o lot of GR as well),  if you don't understand it discuss it with teacher and go over it until you do. This is important because most of the situations in relativity which puzzle people boil down to not understanding these basic principles.

Now you are in a good position to go back and solve your 'puzzle'. I won't do the arithmetic for you, but I'll give you another hint, synchronise the clocks when the noses of the 2 ships pass each other, then work out the times at the front and back of each ship for each observer for each event, you'll see that the problem disappears and your concept of 'at the same time' is faulty.

I'm assuming that you that you really do want to learn, so come back when you've done it and show us the results, otherwise don't bother posting any more silly 'puzzles' because they are all based on the same misunderstandings.

[Fruityloop (OP)]

Thank you for the response.  However, I must politely disagree.
Events, such as collisions, which are not separated by space or time
are simultaneous between frames of reference.  But I would like to
point out another paradox.  Let's assume that the ships are passing
over a large star or planet so that there is a gravitational field.  There
is a passenger at the front of ship A holding a red ball and there is a
passenger at the tail of ship A holding a blue ball.  There is also a hole
in the middle of ship A which is deep enough for the balls to fall into.
There is a window for the passengers to see what's happening outside.
When the tail of ship A passes the front of ship B, the passenger at the
tail starts rolling the blue ball towards the middle of the ship.  When the
front of ship A passes the tail of ship B, the passenger at the front of the ship
starts rolling the red ball towards the middle of the ship.  The two balls
are rolled at the same speed.  Now I think you can see what's going to happen.
According to the passengers aboard ship A, the red ball will wind up
of the bottom of the hole, beneath the blue ball.  According to the passengers
aboard ship B, the blue ball will wind up on the bottom of the hole, beneath
the red ball.  In order to resolve this paradox,  I guess we can assume that for
the passengers aboard ship B the blue ball is rolling more slowly than the
red ball, which enables the red ball to wind up on the bottom.  However, this is just
an ad hoc solution which doesn't seem to have any scientific basis for assuming
such a thing.

[Colin2B]

Why are you avoiding looking at the first question by posting another.
I take it that you have not bothered to do the calculations I suggested. If you had you would see where you are going wrong. I can only assume that you are not interested in the answer, but are just trolling.

[Fruityloop (OP)]

For the reference frame of ship B, there is only one moment in time when the tail of ship B reaches, and is lined up with, the front of ship A.   For the reference frame of Ship A there is also only one moment when the front of ship A reaches, and is lined up with, the tail of ship B.  Therefore, this moment must be the same moment for both ship A and ship B and must be simultaneous for both ships.  Now at that moment, if we ask an observer who is at the front of ship B, "Where is the tail of ship A?", he will say "The tail of ship A has already passed me." At that moment, if we ask an observer who is at the tail of ship A, "Where is the front of ship B?".  He will say, "I have not yet reached him."  So the tail of ship A has both already passed the front of ship B and has not yet passed the front of ship B.  Obviously, this is being in two places at the same time.  Can someone explain how this isn't being in two places at once?
Or if we asked both observers to look at each other, how would they do it?
I am going to give you two rulers, make ruler #1 longer than ruler #2 and make ruler #1 shorter than ruler #2.   A five-year old might realize this is impossible until he learns relativity!