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Two events A and B occur at places separated by 10^(6) km, B occuring 5s after A . (a) Find the velocity of a frame in which these events occur at the same place . (b)What is the time interval between the events in this frame ? |
Answer» Solution : (a) Suppose the events A and B occur at POINTS X and Y at times t_A and t_B where t_B = t_A +5 s. Consider a small train which is at the pointX when the event A occurs. Suppose, this same train moves towards Y and reaches the point Y when the events A and B occur at the same place in the train frame . This frame moves `10^(6)` km in 5sas seen from the original frame . Thus, the VELOCITY of the train frame is ` V =10^(6) km / 5 s = 2 xx 10 s m s ^(-1) ` (b) As the events A and B occur at the same place in the train frame, the time interval between the events measured in this frame is the proper interval . Thus , this time interval is , `= (5 5) (SQRT 1 - v^(2)/ c^(2) ) = (5s) (sqrt 1 - (2/ 3 )^(2) ` `= 3.7s` |
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