Answer
(a) $\frac{m}{m_0} = e^{-150}$
(b) The fraction of the initial mass that is not fuel is 0.223.
Work Step by Step
(a) $v-v_0 = v_{ex}~ln\frac{m_0}{m}$
$\frac{v-0}{v_{ex}} = ln\frac{m_0}{m}$
$e^{\frac{v}{v_{ex}}} = \frac{m_0}{m}$
$e^{\frac{(1.00\times 10^{-3})~(3.0\times 10^8~m/s)}{2000~m/s}} = \frac{m_0}{m}$
$\frac{m_0}{m} = e^{150}$
$\frac{m}{m_0} = e^{-150}$
We can see that most of the initial mass must be fuel.
(b) $v-v_0 = v_{ex}~ln\frac{m_0}{m}$
$\frac{v-0}{v_{ex}} = ln\frac{m_0}{m}$
$e^{\frac{v}{v_{ex}}} = \frac{m_0}{m}$
$e^{\frac{3000~m/s}{2000~m/s}} = \frac{m_0}{m}$
$\frac{m_0}{m} = e^{1.5}$
$\frac{m}{m_0} = e^{-1.5}$
$\frac{m}{m_0} = 0.223$
The fraction of the initial mass that is not fuel is 0.223.
