Question

In the pole-in-the-barn paradox, suppose the runner holds a pole of length 15 m in the runner's frame and is running at speed 0.83c. The back and front doors of the barn are a distance of 10 m apart. The farmer watches and quickly closes and opens both doors simultaneously with his door control device just as the front of the pole reaches the back door location in the farmer's reference frame.

The front door at position x = 0 opens at time t = 0 in the farmer's frame, and at x' = 0 at time t' = 0 in the runner's frame. (Use the exact values you enter to make later calculations.)

(a) What is the length of the pole in the farmer's reference frame? (Notice that the pole momentarily fits in the barn in the farmer's frame solely because of length contraction.)

(b) What is the distance between the front and back doors in the runner's frame in which the barn appears to be moving backward? You will find it much smaller than the pole.

(c) Find the time (in nanoseconds after t' = 0) in the runner's frame when the back door closes. That time is positive or negative according to whether the closing and opening of the back door occurs later or sooner, respectively, than the closing and opening of the front door.

(d) How far in the runner's frame has the pole moved from the time the back door opens to the time the front door closes?

(e) The length of the pole minus the distance between doors in the runner's frame minus the distance the pole has moved in part (d) gives the length of the pole still outside the front door when it slams shut in the runner's frame. If this quantity is negative, the pole clears the front door. Evaluate this quantity.

Answer 1