GREPhysics.NET: GR8677 #20

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GR8677 #20

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Verbatim question for GR8677 #20

Special Relativity $\Rightarrow$ }Rest Energy

If one remembers the formulae from special relativity, arithmetic would be the hardest part of the problem.

The problem is to solve $\gamma m_k c^2 = m_p c^2$ , where $\gamma=\frac{1}{\sqrt{1-\beta^2}}$ and $\beta=\frac{v}{c}$ .

The rest-mass of both the kaon and the proton are given in the problem. Thus, the equation reduces to $\gamma=\frac{938}{494}$ .

Now, because the range of velocities vary significantly between $1.x$ and $2$ , one can't directly approximate that as 2. Boo. So, long-division by hand yields approximately $1.9=\gamma=\frac{1}{\sqrt{1-\beta^2}}$ .

The author of this site prefers to use fractions instead of decimals. Thus $\gamma^2=1.9^2=\left(1+\frac{9}{10}\right)^2= 1+\frac{81}{100} +\frac{180}{100} =\frac{262}{100}$ . Express $\gamma$ in terms of $\beta$ to get $1-\beta^2=\frac{100}{262}\approx\frac{1}{3} \Rightarrow \beta^2=\frac{2}{3}$ .

$8\times 8$ is about 64, so the velocity has to be greater than 0.8c. The only choice left is (E).

If one has the time, one might want to memorize the following:

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

, or perhaps a more elaborate list of $\gamma-\beta$ correlations.

If one knew that before-hand, then one would immediately arrive at choice (E).

See below for user comments and alternate solutions!

Alternate Solutions

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Comments

ernest21
2019-08-10 03:09:34

Its a good approach to solve problems @ home,interesting. wolfgang krauser

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NEC

fredluis
2019-08-09 04:13:03

Both get me closer to 2 compared to the other given answers, but nothing I can think of can give me an exact answer of two. As for the answer explanation here, I am just fairly uncomfortable with the idea that adding two different units together can get the correct answer. tree removal

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joshuaprice153
2019-08-08 07:33:21

Pretty insightful post. Never thought that it was this simple after all. I had spent a good deal of my time looking for someone to explain this subject clearly and youâ€™re the only one that ever did that. Keep it up. house painting

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NEC

JoshWasHere
2014-08-20 08:12:31

In working this out, I approximated $\frac{m_k}{m_p}$ = $\frac{950}{500}$ =1.9 which makes it easier to work with. Then in working out the algebra, I approximated $\frac{2.61}{3.61}$ = $\frac{2.5}{3.6}$ which made taking the square root extremely simple. $\frac{5}{6}$ is nearly .85 c so without too many assumptions or generous approximations, I got nearly the right answer in a short amount of time.

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NEC

dak213
2014-06-30 17:58:42

I think they forgot to add the 1 in the $1.9^2$ equation. I got $\frac{361}{100}$ then solving for $\beta$ I got exactly 0.85c $\Rightarrow$ E

dak213
2014-06-30 18:00:13

Ah, I see this issue has been addressed. I should have read the comments at the bottom.

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NEC

deniskrasnov
2013-01-19 00:08:42

i have a suspicion that ETS only wants you to remember Pythagorean triple 3:4:5
wich translates into:

beta = 0.8 => gamma = 5/3 = 1.66...
beta = 0.6 => gamma = 5/4 = 1.25

and thus, dealing with similar problem one should first check if one of these is enough to identify the correct answer quickly

the first one is sufficient in this case
gamma = 938/494 > 850/500 > 1.7 > 1.66.. => beta > 0.8

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NEC

oliTUTilo
2012-11-01 16:43:56

If we algebraically solve for v (or $\beta$ ) instead of $\gamma$ , as Yosun does, it seems easier to approximate the answer.

$E = \gamma m \Rightarrow{(1-\beta^2)}^{-\frac{1}{2}} = \frac{E}{m} \Rightarrow \beta = {(1-{(\frac{m}{E})}^2)}^{-\frac{1}{2}}$
$\frac{m}{E}$ is approximately $\frac{1}{2}$ , so $\beta$ is approximately ${(\frac{3}{4})}^{-\frac{1}{2}}$ , which is 1.7 something divided by 2, or approximately 0.85. v is $\beta c$ , so badabingbadaboom.

In general, I'd algebraically solve for the variable in question first, gaining the greatest transparency, but redmomatt's method is even a bit faster here.

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NEC

redmomatt
2011-10-14 08:59:20

Why can't we do this....?

$E = \gamma m_0 c^2$

$1000 = \gamma 500$

$.5^2 = 1 - \frac{v^2}{c^2}$

$sqrt{.75} c = v = .85 c$ $\Rightarrow$ (E)

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NEC

checkyoself
2011-10-05 10:46:45

Approximating $\gamma=2$ is fine. You end up with $\beta=\sqrt{0.75}c$ which you know must be greater than 0.75c so the answer must be E.

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NEC

archard
2010-05-25 18:33:08

Instead of doing the long division by hand, you could just say that gamma is less than 2. Solving the inequality you get that v > $\sqrt{3/4}c$ . E is the only answer that satisfies this.

Crand0r
2010-11-12 08:49:13

Actually, you should get that $v\lt \sqrt{\frac{3}{4}}c$ . All of the answers satisfy this, but you also know that $v \approx \sqrt{\frac{3}{4}}c$ , which leaves only E.

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lamejiaa
2009-08-30 09:09:49

The result is good but (19/10)(19/10) = 361/100

walczyk
2011-04-06 17:37:34

^^this is true. you can check this quickly by $20^2 - 19 - 20 = 19^2$ . After a couple steps you will end up with $1-\beta^2 = \frac{100}{361}$ . Do a little math and you'll get $\beta^2 = \frac{261}{361} \approx \frac{260}{360} = \frac{13}{18} \approx 0.73$ Phew, all these relativistic problems really have you scratching away with your pencil, but the most time consuming part of this problem will be the long division, good luckk.

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NEC

jmason86
2009-08-16 16:22:01

If you DO use $\gamma=2$ then you get roughly $v=0.85$ . Not bad, I say. If the listed velocities were closer together, we might have a problem but they are pretty well spaced. Since it takes about 20 seconds to solve through with $\gamma=2$ , I suggest using this, and if the answer came out too far from those listed, to go back and solve for 1.x. If you did have to do this, at least you already solved for v and you could just plug back in without deriving it again.

wittensdog
2009-09-25 17:26:02

I strongly agree - whenever you see spaced out answers, that's always your clue to approximate.

One thing I always remember is that when gamma is 2, the speed is sqrt(3)/2 , which is something like 0.866 or something like that. A value of 2 for gamma comes up in a lot of places, such as when rest energy and kinetic energy are equal, so it's a good thing to remember. Seeing that the ratio is almost two, that the answers are well spaced, and that there's something in the ball park of 86% of c, you should immediately go for it.

Seriously, NEVER do math if you see spaced out answers. One big example I can think of is the drift velocity problem, where each of the answers were spaced out by like 7 orders of magnitude. There was a factor of 1.6(pi) or something like that, but that's certainly not enough to swing 7 OM! Of course it's not quite as dramatic here, but still, no one wants to do long division on the physics GRE.

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NEC

nz_gre
2007-09-24 18:12:08

why are we neglecting the fact that

$E_{tot}^{2}=\left(pc \right)^{2} + m_{0}c^{2}$

And then finding the momentum $\Rightarrow$ velocity?

Surely, if the total kaon energy is equal to the proton rest mass then

938 $^{2}$ = (pc) $^{2}$ + (494) $^{2}$ and we go from there?

bkardon
2007-10-05 13:31:04

The expression for $E_{tot}$ is in fact equivalent to the expresssion
$E_{tot} = \gamma m_0 c^2$

as follows (here $m$ is relativistic mass, $m_0$ is rest mass)

$E_{tot} = m c^2 = \gamma m_0 c^2$
$E_{tot} = \frac{m_o c^2}{sqrt(1 - (v/c)^2)}$
$E^2_{tot} (1 - (v/c)^2) = m^2_0 c^4$
$E^2_{tot} - E^2_{tot} (v/c)^2 = m^2_0 c^4$
$E^2_{tot} = m^2_0 c^4 + v^2 E^2_{tot} / c^2$
Here I will use $E_{tot} = \gamma m_0 c^2 = p c^2 / v$
$E^2_{tot} = m^2_0 c^4 + p^2 c^2$
QED

FutureDrSteve
2011-10-29 13:44:19

Another consideration with your approach is that you still need to calculate $\gamma\$ because once you get down to
$\(pc)^2 = 7.5 E5$
in order to solve for v, you need to use:
p = $\gamma m v$

Since you need the other approach to solve for $\gamma$ , you might as well stick with that method to save time and effort

FutureDrSteve
2011-10-29 13:56:04

Correction: You don't NEED to solve for $\gamma$ using the other method. You can use the definition of $\gamma$ to solve for v, but using the simpler approach will reduce the risk of making a careless mistake, plus you will save lots of time, which, as far as I'm concerned, is the cardinal rule of the PGRE.

ryanjsfx
2014-10-20 11:32:20

I thought using $E^{2} = E_{0}^{2} + (pc)^{2}$ wasn't so bad. Observe (with c set to 1):rnrn $938^{2} = 494^{2} + (\gamma m v)^2$

$938^{2} = 494^{2} + 494^{2} (\gamma v)^{2}$

$4 = 1 + (\gamma v)^{2}$

$3 = \frac{v^{2}}{1-v^{2}}$

$3 - 3v^{2} = v^{2}$

$3 = 4v^{2}$

$\frac{\sqrt{3}}{2} = v$

$\sqrt{3}$ is about 1.7 and 1.7 divided by 2 is about 0.84

Done!

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NEC

leftynm
2006-10-30 17:05:27

This is much simpler than that.

Once you have gamma*m_k = m_p, plug in gamma = [1-(v/c)^2)]^-1, and solve for v/c. You get v/c = sqrt(1 - (m_k/m_p)^2). If you allow m_k/m_p = 1/2, which it just about is, then you get v/c = sqrt(3)/2. From seeing this so much in trig, I know sqrt(3)/2 = 0.866. Then v/c is about 0.866c. So choice E is very close.

blah22
2008-03-22 13:10:35

How is that not, basically, what she did? Just in more detail.

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radicaltyro
2006-10-21 12:59:15

1.9^2 = 3.61 = 361/100, not 262/100

achca
2012-06-26 03:55:13

absolutely right!

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If we algebraically solve for v (or $\beta$ ) instead of $\gamma$ , as Yosun does, it seems easier to approximate the answer. $E = \gamma m \Rightarrow{(1-\beta^2)}^{-\frac{1}{2}} = \frac{E}{m} \Rightarrow \beta = {(1-{(\frac{m}{E})}^2)}^{-\frac{1}{2}}$ $\frac{m}{E}$ is approximately $\frac{1}{2}$ , so $\beta$ is approximately ${(\frac{3}{4})}^{-\frac{1}{2}}$ , which is 1.7 something divided by 2, or approximately 0.85. v is $\beta c$ , so badabingbadaboom. In general, I'd algebraically solve for the variable in question first, gaining the greatest transparency, but redmomatt's method is even a bit faster here.

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