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Lyrx [107]
3 years ago
5

A train with proper length L has clocks at the front and back. A photon is fired from the front to the back. Working in the trai

n frame, we can easily say that if the photon leaves the front of the train when a clock there reads zero, then it arrives at the back when a clock there reads L/c. Now consider this setup in the ground frame, where the train travels by at speed v. Rederive the above frame-independent result (namely, if the photon leaves the front of the train when a clock there reads zero, then it arrives at the back when a clock there reads L/c) by working only in the ground frame.
Physics
1 answer:
tamaranim1 [39]3 years ago
7 0

Explanation:

In train's rest frame, the speed of photon is c and the proper length of the train is L. The time taken by the photon to cross the train is t=\frac{L}{c}

In ground frame, the speed of the photon is given as follows:

v_{x}=\frac{v_{x}+v}{1+\frac{v_{x} \cdot v}{c^{2}}}

=\frac{c+v}{1+\frac{c v}{c^{2}}} \\=c

The speed of light or photon remains same in every frame of reference.

Now, the speed of train is very less as compared to the speed of photon so that v So that, \frac{v}{c} \ll 1

The length contraction in the ground frame is given as follows:

L^{\prime}=L \sqrt{1-\frac{v^{2}}{c^{2}}}

=L

Time taken by the photon to travel the length of the train in ground frame is .

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final speed of car = 0 m/s

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a) distance between deer and car

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   distance travel after you apply brake

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   d₂ = 38 - 0.5 V

   distance travel after you apply brake

   using equation of motion

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    0 = (V)² - 2 x 10 x (38 - 0.5 V)

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  x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}

  V = \dfrac{-10\pm \sqrt{10^2-4(1)(-760)}}{2(1)}

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rejecting the negative term.

hence, maximum speed of the car could be V = 23.01 m/s

 

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