GREPhysics.NET: GR8677 #20

GR8677 #20

Problem

GREPhysics.NET Official Solution

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).

Alternate Solutions

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Comments

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

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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

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