Special Relativity GRE question (Where is my method flawed?)

[PsychonautQQ]

Homework Statement

http://grephysics.net/ans/9277/70

I'm not interested in how you get the right answer, I'm really just curious if somebody could show me where my thought process was flawed.

Total Energy = E = 100mc^2
Rest Energy = Mc^2
Total Energy = Kinetic Energy + Rest Energy
100mc^2 = p^2/2m + mc^2
99mc^2 = p^2/2m
198(mc)^2 = p^2
sqrt(198)mc = P
and yet the correct answer is
10^4(mc)

where is my method flawed?

 

[phyzguy]

The formula KE = p^2/(2m) and TE = p^2/(2m) + mc^2 are non-relativistic approximations, and are only valid when pc << mc^2. They come about from taking the full relativistic equation and expanding the square root, keeping only the leading term, as follows:
$$E^2 = p^2 c^2 + m^2 c^4$$
$$E = \sqrt{p^2 c^2 + m^2 c^4} = mc^2 \sqrt{1+\frac{p^2 c^2}{m^2 c^4}}$$
$$E \approx mc^2 (1+\frac{p^2 c^2}{2m^2 c^4}) = mc^2 + \frac{p^2}{2m}$$

 

[PsychonautQQ]

The formula KE = p^2/(2m) and TE = p^2/(2m) + mc^2 are non-relativistic approximations, and are only valid when pc << mc^2. They come about from taking the full relativistic equation and expanding the square root, keeping only the leading term, as follows:

$$mc^2 \sqrt{1+\frac{p^2 c^2}{m^2 c^4}}$$
$$E \approx mc^2 (1+\frac{p^2 c^2}{2m^2 c^4})$$

How does this make any sense at all? ignoring the square root and adding a 2 to the denominator?

 

[Chestermiller]

The formula KE = p^2/(2m) and TE = p^2/(2m) + mc^2 are non-relativistic approximations, and are only valid when pc << mc^2. They come about from taking the full relativistic equation and expanding the square root, keeping only the leading term, as follows:

$$mc^2 \sqrt{1+\frac{p^2 c^2}{m^2 c^4}}$$
$$E \approx mc^2 (1+\frac{p^2 c^2}{2m^2 c^4})$$

How does this make any sense at all? ignoring the square root and adding a 2 to the denominator?

Phyzguy is just showing you where your non-relativistic approximation came from. He used the binomial expansion to approximate the square root, and retained only the first two terms in the expansion. Please notice the "approximate" symbol in his equation.

 

[phyzguy]

The formula KE = p^2/(2m) and TE = p^2/(2m) + mc^2 are non-relativistic approximations, and are only valid when pc << mc^2. They come about from taking the full relativistic equation and expanding the square root, keeping only the leading term, as follows:

$$mc^2 \sqrt{1+\frac{p^2 c^2}{m^2 c^4}}$$
$$E \approx mc^2 (1+\frac{p^2 c^2}{2m^2 c^4})$$

How does this make any sense at all? ignoring the square root and adding a 2 to the denominator?

When I expand the square root of (1+ε) in a Taylor series and keep only the first order term, I get:

$$\sqrt{1+\epsilon} \approx 1+\frac{\epsilon}{2}$$

For example: $$\sqrt{1.01} = 1.004987 \ldots \approx 1.005$$

One more comment. You need to have some of these more common approximations at your fingertips in order to do physics. Some of the most important are:
$$\sqrt{1+\epsilon} \approx 1+\frac{\epsilon}{2}$$
$$\frac{1}{1+\epsilon} \approx 1-\epsilon$$
$$sin(\epsilon) \approx \epsilon$$
$$tan(\epsilon) \approx \epsilon$$
$$cos(\epsilon) \approx 1+\frac{\epsilon^2}{2}$$
$$ln(1+\epsilon) \approx \epsilon$$