Answer

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**Hint:**The momentum is defined as the product of mass and the velocity. The linear conservation of momentum says that the total momentum of the system is always constant. The impulse is defined as the rate of change of momentum with respect to time.

**Formula used:**The formula of the momentum is given by,

$ \Rightarrow p = m \times v$

Where mass is m and velocity is v.

Formula of the force is given by,

$ \Rightarrow F = m \times a$

Where mass is mass and acceleration is a.

**Complete step by step solution:**

It is asked on the problem to derive expression for velocity of rocket at any instant, derivation for thrust on the rocket and for the speed of the rocket when whole fuels is burnt.

The momentum of rocket at any instant is,

$ \Rightarrow {p_1} = mu$

The momentum of rocket and burnt fuel at time $t + dt$ will be,

$ \Rightarrow {p_2} = \left( {m - dm} \right)\left( {v + dv} \right) + dm\left( {v + dv - u} \right)$

$ \Rightarrow {p_2} = mv + mdv - udm$

The change in momentum in time $dt$ is,

$ \Rightarrow dp = {p_2} - {p_1}$

Replacing the value of ${p_1}$ and ${p_2}$ we get,

$ \Rightarrow dp = \left( {mv + mdv - udm} \right) - mv$

$ \Rightarrow dp = mdv - udm$

Since, impulse is change of momentum therefore we get,

$ \Rightarrow - mgdt = mdv - udm$

$ \Rightarrow dv = u\dfrac{{dm}}{m} - gdt$

As the fuel is always decreasing therefore,

$ \Rightarrow dv = - u\dfrac{{dm}}{m} - gdt$

Integrating on both sides we get.

$ \Rightarrow \int\limits_{{v_o}}^v {dv} = - u\int\limits_{{m_o}}^m {\dfrac{{dm}}{m}} - g\int\limits_0^t {dt} $

$ \Rightarrow v|_{{v_o}}^v = - u\left( {\ln m} \right)|_{{m_o}}^m - gt|_0^t$

$ \Rightarrow v - {v_o} = - u\left( {\ln m - \ln {m_o}} \right) - gt$

$ \Rightarrow v = {v_o} + u\left( {\ln \dfrac{{{m_o}}}{m}} \right) - gt$

$ \Rightarrow v = {v_o} + u{\log _e}\dfrac{{{m_o}}}{m} - gt$

The velocity at any instant is equal to $v = {v_o} + u{\log _e}\dfrac{{{m_o}}}{m} - gt$.

Let us derive the equation of thrust on the rocket.

Using the equation,

$ \Rightarrow mdv = udm$

Dividing the above relation with respect to $dt$ we get,

$ \Rightarrow m\dfrac{{dv}}{{dt}} = u\dfrac{{dm}}{{dt}}$

The force is a product of mass and acceleration.

$ \Rightarrow m \times a = u\dfrac{{dm}}{{dt}}$

$ \Rightarrow F = \dfrac{{udm}}{{dt}}$

The thrust on the rocket is equal to $F = \dfrac{{udm}}{{dt}}$.

Let us now derive the relation of speed of the rocket when whole fuels are burnt.

The speed when whole fuel is burnt is equal to

$ \Rightarrow v = {v_o} + u{\log _e}\dfrac{{{m_o}}}{m}$

Here the mass of the empty rocket is ${m_e}$ and the velocity of the empty rocket is ${v_e}$.

$ \Rightarrow {v_e} = {v_o} + u{\log _e}\dfrac{{{m_o}}}{{{m_e}}}$.

The velocity of the rocket when all the fuel is burnt is equal to ${v_e} = {v_o} + u{\log _e}\dfrac{{{m_o}}}{{{m_e}}}$.

**Note:**The students are advised to understand and remember the formula of the force and formula of the momentum. The rate of change of momentum with respect to time is defined as the force also can be defined as the product of mass and acceleration.

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