Question

To illustrate the use of a multistage rocket consider the following:

Answer 1

Answer:

a

 $v_f = 0.916 v_{ex}$

b

 $v_f = 1.05 v_{ex}$

Explanation:

Considering question a

From the question we are told that

   The mass of the rocket is  $m_o$  

    The mass of the fuel which is in the rocket is  m = 0.6 M  

Generally given that the rocket burns all it fuel in a single stage , the final velocity of the rocket is mathematically represented as

          $v_f = v_{ex} ln [\frac{m_o}{m_k} ]$

Here $m_k$ is the mass  of the rocket without fuel which is mathematically evaluated as

        $m_k = m_o- m$

=>      $m_k = m_o -0.6m_o$

=>    $m_k = [1- 0.6] m_o$

=>      $v_f = v_{ex} ln [\frac{m_o}{[1 - 0.6]m_o} ]$    

=>      $v_f = 0.916 v_{ex}$

Considering question b

From the question we are told that

   The mass of the fuel it burn at the first stage is  $m = 0.3 m_o$

     The mass of the first stage fuel tank $m_1 = 0.1 m_o$

    The mass of the fuel at the second stage is  $m_f = 0.3m_o$

Generally the  velocity of the rocket at the first stage  is mathematically represented as

           $v_i = v_{ex} * ln [ \frac{m_o }{[1- m]m_o } ]$

=>         $v_i = v_{ex} * ln [ \frac{m_o }{[1- 0.3]m_o } ]$

=>         $v_i = v_{ex} * ln [ \frac{1 }{0.7 } ]$  

=>         $v_i =0.357 v_{ex}$

Generally the mass of the rocket after first stage is  

    $m_r = m_o - 0.3m_o -0.1m_o$

=>   $m_r = 0.6m_o$

Generally the final  velocity of the rocket at the second stage is  

     $v_f = v_i + v_{ex} * ln [\frac{m_r}{ m_f } ]$

=>   $v_f = v_i + v_{ex} * ln [\frac{0.6 m_o }{0.3mo } ]$

=>   $v_f = 0.357 v_{ex} + 0.693 v_{ex}$

=>   $v_f = 1.05 v_{ex}$