Answer
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Work Step by Step
This is the famous “pole-barn paradox”.
Consider the boy’s frame of reference. The pole is at rest and has a length of 13.0 m. Furthermore, the barn is contracted in size, appearing even shorter, front to back, than its rest length of 10.0 m. In the boy’s frame of reference, the barn is far shorter than the pole. According to the boy, the two ends of the pole can never fit within the barn at the same moment.
Now consider the frame of reference of the barn. It is possible for the pole to be shorter than the barn. Use equation 26–3a to calculate the speed of the boy such that the pole contracts to 10.0 meters.
$$\mathcal{l}=\mathcal{l}_0\sqrt{1-v^2/c^2}$$
$$10.0m=13.0m\sqrt{1-v^2/c^2}$$
$$v=c\sqrt{1-(10m)^2/(13m)^2}=0.639c$$
So, a person at rest with respect to the barn believes that the pole can fit within the barn. The resolution to the paradox is that though the Earth person believes the 2 ends of the pole are inside the barn at the same time, the boy holding the pole sees the front of the pole exit the barn before the back of the pole enters the barn. Events that are simultaneous in one reference frame are not necessarily so in other reference frames. Both observers are correct and there is no paradox.
