Fractional increase of energy vs momentum with relativity

[Kaguro]

Homework Statement

If the momentum of the electron moving with a velocity 0.9c is increased by 1% then the increase in its energy is

0.81%
0.9%
1%
0.5%

Relevant Equations

$E^2 = p^2c^2 + m_0^2c^4$

$E=mc^2$

My attempt:
$E^2 = p^2c^2 + m_0^2c^4$
$2E dE = 2pc^2 dp$
$\frac{dE}{E} = \frac{pc^2}{E^2}dp=\frac{p^2c^2}{E^2}$ % (dp/p = 1%)
$=\frac{E^2-m_0^2c^4}{E^2}$ %

$=1-\frac{m_0^2c^4}{E^2}$ %
$=1-\frac{m_0^2c^4}{m^2c^4}$ %
$=1-\frac{1}{\gamma ^2}$ %

$=\frac{v^2}{c^2}$%
=0.81 %Solution given:

p=mv
ln(p) = ln(m) + ln(v)
$\frac{dp}{p}=\frac{dm}{m}$ = 1%

Now doing the same with $E=mc^2$,
$\frac{dE}{E}=\frac{dm}{m}$ = 1%

Is taking velocity constant a good idea?

 

[PeroK]

There is a useful identity here: $$\gamma = \sqrt{(\frac{v\gamma}{c})^2 +1}$$ Then it's just some arithmetic.

 

[Kaguro]

That's a nice one.

But it's producing same result as mine and the answer is coming to be 0.81%So is the solution provided( which says 1%) wrong?

 

[PeroK]

That's a nice one.

But it's producing same result as mine and the answer is coming to be 0.81%So is the solution provided( which says 1%) wrong?

Yes, it's definitely $0.81 \%$.

 

[Kaguro]

Thanks very much!