please explain the steps and the answer SK units (s that must be received no hor
ID: 1884680 • Letter: P
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please explain the steps and the answer
SK units (s that must be received no hor a ckproof watch sent. Which message do you respond to first? bout 50ll R2M.10 (Seeing is not the same as observing!) Imagine that a space station bullet-train running at a very high speed passes two track- ation defines side signs (A and B), as shown in the aerial view below. his event A) ection away measured in ising rain g speed e (call this ct a space- s problem: Sign B 80 ns 45 ns Sign A 60 ns oning on in SR a shock- 0 al to the ed. This Let event A be the passing of the front end of the train by where sign A, and let event B be the passing of the rear end of the train by sign B. An observer is located at the cross marked electro- by an O in the diagram. h later ent D (a) This observer sees (that is, receives light with her eyes) event A to occur at time t- 0 and sees event B to occur at time t 25 ns. When does she observe these events to occur? That is, what would a clock present at sign A read at event A, and what would a clock present at sign B read at event B if these clocks were correctly synchro- nowl- ation, pace- sure nized with the clock at O? (b) In what way is the diagram misleading about the implied time relationship between events A and B? es a after that the ame way (Hint: Remember that the clocks at signs A and B must be synchronized with the clock at O in such a way that they would read the speed of a light signal traveling between them and O to be 1 s/s. Given this, the distance between O and A, and the time that light from event A reached O, can you infer when event A must have happened?)Explanation / Answer
(a) The important thing to note in this problem is that the clocks at A and B are correctly synchronised with the clock at O, which means that we do not have to worry about the time dialation effects of special relativity while doing any kinematical calculations, because it has already been taken care of while synchronising the clocks. So, the distance between A and O is, from Pythagoras' Theorem, 75 ns, and the distance between B and O is 100 ns. We know, time = distance / speed, here, speed of light is given as 1 s/s. So, the observer at O sees the event A to occur at t = 75 ns and the event B to occur at t = 100 ns, given, speed of light is 1 s/s.
(b) The clock at A is correctly synchronised with the clock at O as well as the clock at B is correctly synchronised with the clock at O, so, the clocks at A and B are correctly synchronised with each other too. But, this above diagram is misleading in the sense that looking at this diagram, one may think that there could be possible time dialation between the events at A and B, giving wrong results to the observer at O.
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