GREPhysics.NET: GR8677 #69

GR8677 #69

Problem

GREPhysics.NET Official Solution

Special Relativity $\Rightarrow$ }Length Contraction

Things moving relative to a particular rest frame appear shorter, i.e., contracted. Knowing that $\gamma > 1$ , one can deduce this relation without remembering much about special relativity (other than the hackneyed phrase length contraction) $L_{rest}=\gamma L_{moving}$ . So, one has $5=\gamma 3$ for the car, where it is seen to be moving in the reference frame of the garage. This implies that $\gamma=5/3\approx 1.6=\frac{1}{\sqrt{1-(\beta)^2}}$ . Recall the following useful table to memorize, for relating $\beta=v/c$ and $\gamma$

$\begin{eqnarray} \gamma &\Leftrightarrow& \beta \\ 1.005 &\Leftrightarrow& 0.1\\ 1.033 &\Leftrightarrow& 0.25\\ 1.155 &\Leftrightarrow& 0.5\\ 1.51 &\Leftrightarrow& 0.75\\ 2.29 &\Leftrightarrow& 0.9<br /> \end{eqnarray}$

One has $\gamma \approx 1.6 \Rightarrow \beta > 0.75c$ (but less than $c$ ). Choice (C) works best.

Alternate Solutions

ben
2006-07-19 15:54:53

i think an easier way to do it that gives the exact answer without having to memorize a table is to simply solve for v explicitly in the equation 3/sqrt(1-v^2/c^2)=5. just square both sides and you'll find that v^2/c^2=16/25 at which point the answer is obvious.

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Comments

barefoot0
2006-11-27 12:23:00

But 16/25 is .64 not .8 so you would get answer B instead. But ETS said answer C is correct.

barefoot0
2006-11-27 12:25:13

never mind I forgot to take the square root.

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nitin
2006-11-21 04:19:22

trombone,

I think you need to check yourself out. You're acting like a kid, and your aim is clearly to insult me, which is a shame since I did not address you in anyway. Your attitude is that of a moron, and Lubos Motl has given a good definition of it.

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nitin
2006-11-16 11:25:38

Another nonsensical solution...

Ben is right, and I simply don't understand why you decide to change from the fractional form " $\frac{5}{3}$ " to the decimal form " $\approx1.6$ ", which drives you into a long mess!

trombone
2006-11-18 19:16:45

The only nonsensical thing here is you bitching about the method that was used. Post a better solution if you have one, otherwise stop whining.

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tera
2006-08-13 07:38:59

The comment of ben is quite correct iti isd very simply because the square roots become precise

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ben
2006-07-19 15:54:53

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The comment of ben is quite correct iti isd very simply because the square roots become precise

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