How is time dilation consistent between both observer?

Question

There is a special variant of time dilation paradox bugs me.

Imagine there are 2 small spaceships moving toward each other in constant speed. Each ship has an observer on-board.
According to both observers, if nothing else happen, the 2 ships are going to collide in exactly 10 seconds.
The relative speed between the 2 ships are high. Each observer may observe a time dilation effect to slow time in the other ship by 50% than his own ship. (If I understand time dilation correctly, this is what happens)
One spaceship has a special timed-bomb on it ticking.
According to the observer on-board this "bomb ship", the bomb ticks at "normal speed" and will implode exactly 9 seconds later. That will prevent the collision from happening.
However, according the the other ship, the bomb's clock ticking 50% slower. Hence the bomb will go off 18 seconds later. That means the ships will collide before the bomb implode.

So what will happen 10 seconds later? Will the 2 ships collide? Or will the bomb ship implode before that?

Update: To keep things closer to the point I puzzled, I would add some initial factor here:

Both ship is approaching to each another on an orbit. They starts the same coordination.
Both ship has a clock synchronized with each another before start. The clocks were counting down to the collision time (0s).
According to these settings, the bomb will go off when the count down clocks on its ship says "1s".

Let's say that the "safe ship" is (A) and "bomb ship" is (B).

As @alfred-centauri stated, their clock will agree each other when they meet (if the bomb wasn't go off) on count down "0s". For time-dilation to stand, both observer should observe an earlier start time so they can match when meet.

i.e. when clock on (A) says "20s", the clock (B) observed on (A) should be "10s" (plus some time for the information to reach (A)).

From (A) point of view, (B) clock must have ran faster before that so it can run slower now. Isn't that strange? Or what did I got wrong? What would these ship observes in their final journey?

The bomb ship will implode before collision, 18 seconds on the non-bomb ship is equal to 9 seconds on the bomb ship — Courage, Apr 19 '16 at 10:40
Let $X$ be the event "Observer 1 says 10 seconds to collision" and let $Y$ be the event "Observer 2 says 10 seconds to collision". According to whom are you assuming these events to be simultaneous? — WillO, Apr 19 '16 at 12:36
I feel this is a logical explanation, but I still have some doubts that need clarify. — Koala Yeung, Apr 19 '16 at 16:05
Does this answer your question? Why isn't the symmetric twin paradox a paradox? — tryst with freedom, Nov 23 '22 at 07:54

score 1 · Answer 1 · answered Apr 19 '16 at 12:53

According to both observers, if nothing else happen, the 2 ships are going to collide in exactly 10 seconds.

For concreteness, stipulate that both spaceship's clocks will read 0 at the instant of the collision (assuming the bomb does not explode first).

Now consider the following question regarding the spaceship without a bomb: when the spaceship's clock reads t = -10s, what is the time observed on the other spaceship's clock?

Clearly, if the other clock is observed to run slow by 50%, the answer is $t' = -5s$. Thus, the timer on the bomb is observed to have $4.5s$ remaining and so, according to both observers, the bomb will explode before the collision.

Let see if I get your right. That means when either of the ships reads "20s" in their count down, they should observe a "10s" on the other ship. So from the "other ship" point of view, when the local time is "20s", the "bomb ship" starts the bomb "10s" count down. — Koala Yeung, Apr 19 '16 at 15:58

How is time dilation consistent between both observer?

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