The figure shows a ship (attached to reference frame S) passing us (standing in

Question

The figure shows a ship (attached to reference frame S) passing us (standing in reference frame S) with velocityV0.946c. A proton is fired at speed 0.947c relative to the ship from the front of the ship to the rear. The proper length of the ship is 788 m. What is the temporal separation between the time the proton is fired and the time it hits the rear wall of the ship according to (a) a passenger in the ship and (b) us? Suppose that, instead, the proton is fired from the rear to the front. What then is the temporal separation between the time it is fired and the time it hits the front wall according to (e) the passenger and (d) us? Proton (a) Number (b) Number (c) Number (d) Number Units Units Units Units CO

Explanation / Answer

given proper length of the spaceship, Lo = 788 m

v = 0.946 c

speed of proton u = 0.947 c

a. for the observer on the ship

length of ship = L

L = Lo*sqroto(1 - v^2/c^2) = 0.3241Lo

L = 255.4432 m

hence

time seperation of events = L/u = 89.913*10^-8 s

b. for the observer on ground

speed of proton, u' = (v - u)/(1 - vu/c^2) = -0.0096c

hence

time seperation of events = Lo/0.0096c = 2.73535*10^-4 s

c. for the observer on the ship

L = Lo*sqroot(1 - v^2/c^2)

and t = L/u = 89.913*10^-8 s

d. for the observer on ground

speed of proton, u' = (v + u)/(1 + vu/c^2) = 0.99849*c

hence

time seperation of events = Lo/0.99849*c = 263.063*10^-8 s

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