Two passenger trains A and B, each 220 m long, pass a 40 m long railroad platfor

Question

Two passenger trains A and B, each 220 m long, pass a 40 m long railroad platform in Winnepeg. The trains are moving in opposite directions at equal speeds of 0.535c with respect to the ground. Train A is traveling west and all tracks are perfectly straight. The Winnepeg stationmaster, knowing the speed of train A, measures its length as it passes him by measuring the difference between the arrival times of the front of the train and the rear of the train. How long does the train seem to be? 1.86 Times 10^2 m What is the difference in arrival times? 1.16 Times 10^-6 s A passenger on train A measures the length of the Winnepeg platform by a similar method. How long does she take to travel from one end of the platform to the other? 2.11 Times 10^-7 s From the point of view of a passenger on train A, how fast is train B moving? (Give your answer as a fraction of the speed of light, e.g. if you get 0.952.) 8.32 Times 10^-1 How long does it take train B to pass the passenger on train A? According to clocks in the station and on both trains it is exactly 12:00 noon when both trains arrive in Winnepeg. How many seconds later should the stationmaster in Vancouver, 1900 kw, to the west, expect train A to arrive? How long will the trip from Winnipeg to Vancouver take according to a clock on train A?

Explanation / Answer

A. Length contraction: L = Lo(1-(v/c)^2)
Proper length, Lo = 220 m
L = 220*(1-(0.535c/c)^2) = 220*(1-0.535^2) = 220*0.845 = 185.9 m

B. t = 185.9/v = 185.9/0.535c = 185.9/(0.535*3.00×10^8) = 1.16×10^-6 s = 1.16 s
When the rear of the train arrives, the front has traveled 185.9 m in 1.16 s

This can also be calculated with time dilation by finding the proper time.
Time dilation: T = To / (1-(v/c)^2)

Under "normal" conditions the time would be: T = d/v = 220/(0.535*3.00×10^8) = 1.37×10^-6 s
To = T*(1-(v/c)^2) = (1.37×10^-6)*(1-0.535c/c)^2) = (1.37×10^-6)*0.845 = 1.16×10^-6 s

---> Difference in time arrival = 1.16 s

C. Platform is 40 m: t = d/v = 40/0.535c = 2.49×10^-7 s
Passenger on train A measures the time to be:
To = (2.49×10^-7)*(1-0.535c/c)^2) = (2.49×10^-7)*0.845 = 2.10×10^-7 s
This means that the passenger measures the platform to be:
d = vt = 0.535c*(2.10×10^-7) = 33.8 m
With length contraction: L = 40*(1-(0.535c/c)^2) = 40*0.845 = 33.8 m

---> Time to travel the platform = 2.10×10^-7 s (0.210 s)

D. Relativistic velocities: Vab = (Va + Vb) / (1 + VaVb/c^2)
Vab = (0.535c + 0.535c) / (1 + 0.535c*0.535c/c^2) = 1.07c / (1+0.535^2) = 0.832c

For a passenger on Train A, it would appear that Train A is stationary and Train B is
moving at 0.832c. Same thing other way around. A passenger on Train B experiences
that her train is the stationary one and that Train A is moving at 0.832c.

E. Time dilation approach:
Time for train B to pass "normally": T = d/v = 220/0.832c = 8.81×10^-7 s
The passenger measures this time to be:
To = (8.81×10^-7)*(1-0.832c/c)^2) = (8.81×10^-7)*0.555 = 4.89×10^-7 s

Length contraction approach: L = 220*0.555 = 122.1 m
t = d/v = 122.1/0.832c = 4.89×10^-7 s

---> Time for Train B to pass passenger = 4.89×10^-7 s (0.489 s)

F. T = d/v = 1900000/0.535c = 1.18×10^-2 s = 11.8 ms

G. To = (1.18×10^-2)*0.845 = 9.97×10^-3 s = 9.97 ms ~10 ms
In the Train A frame, the 1900 km distance is contracted:
L = 1900*0.845 = 1605.5 km => t = 1605500/0.535c = 10 ms

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