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conantee
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« on: December 31, 2009, 10:44:24 AM »

20. A positive kaon (K ^+ ) has a rest mass of 494 \mbox{ MeV/c^2}, whereas a proton has a rest mass of 938 \mbox{ MeV/c^2}. If a kaon has a total energy that is equal to the proton rest energy, the speed of the kaon is most nearly
(A) 0.25 c
(B) 0.40 c
(C) 0.55 c
(D) 0.70 c
(E) 0.85 c
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« Reply #1 on: December 31, 2009, 11:01:38 AM »

The question itself is relatively easy, but the numerical calculation may be a bit tedious. Let's see what we can do about it.
First, one should that the total energy can be written as E = \frac{E_0}{\sqrt{1-v^2/c^2}} where E_0 is the energy at rest (rest mass energy, in other words). [If you cannot remember this equation, just keep in mind that energy reaches infinity when v approaches c, and \sqrt{1-v^2/c^2} appears everywhere in relativity. Then you can work the formula out.]

From the question, we can write equation for kaon's energy and proton rest energy as
E_{kaon} = \frac{E_{0,kaon}}{\sqrt{1-v^2/c^2}}= E_{0,proton}
We can arrange term and solve for v/c:
\frac{v}{c} = \sqrt{1-(\frac{E_{0,kaon}}{E_{0,proton}})^2} \approx \sqrt{1-(\frac{494}{938})^2}

Now, the term we got is ugly. However, with Taylor's approximation and so on, we can find the answer
\frac{v}{c} \approx \sqrt{1-(\frac{494}{938})^2}  \approx 1-\frac{1}{2}(\frac{494}{938})^2 \approx 1-\frac{1}{2} (\frac{1}{2})^2 \approx 0.875
So, the answer should be (E).

The correct answer is (E). 37 of 100 got the correct answer.
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