Some time, in the not too distant future, relativistic Purple Line trolleys will

Question

Some time, in the not too distant future, relativistic Purple Line trolleys will be zipping across our campus. Consider two such trolleys as they pass by the Stamp Union. Trolley 1 is traveling at 0.6c to the right ( increasing x ) and Trolley 2 is traveling at 0.8c to the left as measured by someone on the steps of the Stamp Union. Each trolley is 10m long in its rest frame ( back at the Trolley round house ) . Define e event #1 to be the front of the trolleys pass each other, and define event #2 to be the back of the trolleys pass each other. How long is each trolley in the Stamp Union frame? Suppose Event #1 occurs at ( x = 0, t = 0 ) in the Stamp frame. When and where does event #2 occur in the Stamp frame. Now calculate the coordinates of these events in each of the trolley frames. What is the velocity of Trolley 2 in the frame of Trolley 1? and vice versa?

Explanation / Answer

Given length of each trolley L = 10 m A ) length of trolley 1 with respect to stamp union is            L = L o 1 - v 2 / c 2                   = 10 ( 1 - 0.6 2 )                 = 8 m length of trolley 2 with respect to stanp union           L = L o 1 - v 2 / c 2                   = 10 ( 1 - 0.8 2 )                 =  6 m B)           time   t =    L / t for trolley 1                    t = 8 m / 0.6 c                      = 4.44*10^ -8 s       event 2 occur in the stamp frame        is              ( 8 m ,      4.44*10^ -8 s ) C ) time   t =    L / t for trolley 2                    t =  6 m / 0.8 c                      =    2.5*10^ -8 s       event 2 occur in the stamp frame        is              (  6 m ,      2.5*10^ -8 s ) D ) Speed of trolley 1 with respect to stamp union is   u = 0.6 c          speed of trolley 2 with respect to stamp union is   u ' = 0.8 c            speed of trolley 2 with respect t o trolley 1 is   v ?          from additional of               u' = u - v / 1- uv /c 2                      v = u' - u / 1- u u' /c 2                        = 0.8 c - 0.6 c / 1 - 0.8*0.6                      = 0.3846 c                 = 10 ( 1 - 0.8 2 )                 =  6 m B)           time   t =    L / t for trolley 1                    t = 8 m / 0.6 c                      = 4.44*10^ -8 s       event 2 occur in the stamp frame        is              ( 8 m ,      4.44*10^ -8 s ) C ) time   t =    L / t for trolley 2                    t =  6 m / 0.8 c                      =    2.5*10^ -8 s       event 2 occur in the stamp frame        is              (  6 m ,      2.5*10^ -8 s ) D ) Speed of trolley 1 with respect to stamp union is   u = 0.6 c          speed of trolley 2 with respect to stamp union is   u ' = 0.8 c            speed of trolley 2 with respect t o trolley 1 is   v ?          from additional of               u' = u - v / 1- uv /c 2                      v = u' - u / 1- u u' /c 2                        = 0.8 c - 0.6 c / 1 - 0.8*0.6                      = 0.3846 c                  t =  6 m / 0.8 c                      =    2.5*10^ -8 s       event 2 occur in the stamp frame        is              (  6 m ,      2.5*10^ -8 s ) D ) Speed of trolley 1 with respect to stamp union is   u = 0.6 c          speed of trolley 2 with respect to stamp union is   u ' = 0.8 c            speed of trolley 2 with respect t o trolley 1 is   v ?          from additional of               u' = u - v / 1- uv /c 2                      v = u' - u / 1- u u' /c 2                        = 0.8 c - 0.6 c / 1 - 0.8*0.6                      = 0.3846 c

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