Special Relativity: Kinematics

Problem : Someone on a moving train on the earth measures the speed of a meteor in space to be 5×10⁶ m/s. Someone in outer space measures the speed to be 4×10⁶ m/s. Who is right?

Problem : Two spaceships are hurtling towards one another at a constant speed of 0.8c . When they are still 10 000 kilometers apart, one spaceship radios the other to warn them of the impending collision. How much time does it take for the radio wave to reach the other ship, as observed by someone on the receiving ship (assume that the spaceships move little in the time taken for the signal to travel between them)?

Problem : Consider the situation described in Section 1. If the flashes from the sources are observed to occur simultaneously by an observer standing on the ground (at rest relative to the sources), what is the time difference between the events according to an observer on a train speeding past at 0.15c , if that observer measures the distance between the sources to be 1 kilometer?

Solution for Problem 3 >>

The distance between the sources is 1000 meters so here we have l = 500 m. Then t _r = = = 1.96×10^-6 seconds, and t _l = = = 1.45×10^-6 seconds. Thus the time difference is t _r - t _l = 5.12×10^-7 seconds. Even at the immense speed of 45000 kilometers per second, the time difference is hardly noticeable.

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Problem : What if the scenario described in Section 1 is performed with baseballs (which travel at a constant speed b < c ) instead of light pulses. Will the observers still disagree?

Problem : Consider again the scenario described in Section 1. Now consider changing the setup by placing only a single emitter at the center position (where O _A was), and having two receivers placed where the sources formerly were. The source emits two signals, one in each direction (that is, one towards each receiver). An observer at rest with respect to the source and receivers concludes that the source emitted its two signals simultaneously. What does an observer traveling to the right at velocity v observe?

Solution for Problem 5 >>

This situation is exactly analogous to the one described in Section 1. The only difference is that the light on the left of the center point is now traveling to the left, and the light to the right of the center point is moving to the right. Thus the moving observer concludes that the left- moving light takes a time t _l = to reach the left receiver, and the right-moving light takes a time t _r = to reach the right receiver. Thus the 'left' and 'right' times are swapped.

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