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GR8677 #69
Problem
 GREPhysics.NET Official Solution Alternate Solutions

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 $\gamma$

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

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

Alternate Solutions
 casseverhart132019-08-29 02:53:50 Merely a smiling visitant here to share this problem! tree trimmingReply to this comment ben2006-07-19 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.Reply to this comment
casseverhart13
2019-08-29 02:53:50
RusFortunat
2015-10-21 16:04:12
Just a little typo. At last string you have $\\beta >0.75c$when it should be just$\\beta >0.75$
wittensdog
2009-10-07 16:20:35
I've found a way that works for me for remembering how to get the velocity when the lorentz factor is known. If b is the velocity in units of c, aka, v/c, and g is the lorentz factor, then we have,

g = [ 1 - b^+2] ^ - (1/2)

b = [ 1 - g^-2 ] ^ +(1/2)

after working through the math. The way that I remember it is that if you take one of them, swap b and g, and then swap the exponent signs, you get the other one. So they have some kind of nice antisymmetry, or whatever you want to call it.

Hope this helps someone...
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.
 barefoot02006-11-27 12:25:13 never mind I forgot to take the square root.
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.
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!
 trombone2006-11-18 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.
tera
2006-08-13 07:38:59
The comment of ben is quite correct iti isd very simply because the square roots become precise

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.
 Goddar2009-10-04 23:13:11 Same here, i find it very useful on this type of question to express gamma as a fraction: $\gamma \$ = $\frac{a}{b}$ Then in units of c: v = $\frac{sqrt{a^2- b^2}}{a}$ Saves precious time.
 shak2010-07-31 21:27:32 that is the easiest way! i dont understand why Yosun approached this problem in a very complicated way
 neon372010-11-03 12:07:17 Goddar, thats a neat trick. Saves few precious seconds! It's simplicity makes me say... why didnt I think of that!

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