Answered

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Patrick the pole vaulter is running with a 20m long pole toward a 10m long garage with doors at each end. Patrick is running so fast that to an observer O in the rest frame of the garage, the pole appears only 10m long (i.e. the pole would just fit inside the garage).

a. In the frame of observer O, how fast is Patrick running?
b. The observer O closes very quickly the doors on both ends of the garage just as Patrick and the pole are inside, and then immediately opens them again so that Patrick can run out on the other end. However, according to Patrick, the garage is moving towards him and is only 5m long. How can Patrick and his 20m pole possibly make it through without hitting a door?

Sagot :

lublana

Answer with Explanation:

We are given that

Length of pole(l0)=20 m

Length of garage(l)=10 m

a. We have to find the speed by which the Patrick is running.

We know that

[tex]l=l_0\sqrt{1-\frac{v^2}{c^2}}[/tex]

Substitute the values

[tex]10=20\sqrt{1-\frac{v^2}{c^2}}[/tex]

[tex]\frac{10}{20}=\sqrt{1-\frac{v^2}{c^2}}[/tex]

[tex]\frac{1}{2}=\sqrt{1-\frac{v^2}{c^2}}[/tex]

[tex]\frac{1}{4}=1-\frac{v^2}{c^2}[/tex]

[tex]\frac{v^2}{c^2}=1-\frac{1}{4}=\frac{3}{4}[/tex]

[tex]v^2=0.75c^2[/tex]

[tex]v=0.87c[/tex]

b.

l0=10 m

[tex]l=10\sqrt{1-\frac{3}{4}}[/tex]

[tex]l=5m[/tex]

Hence, the length of garage appear to him=5 m long.

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