A space vehicle is travelling at 4300 km/hrelative to Earth when the exhausted rocket motor (mass 4m)is disengaged and sent backward with a speed of 82 km/hrelative to the command module ( mass m).What is the speed of the command module relative to Earth just after the separation?

1. The speed of the space vehicle relative to earth$v_i=4300km/hr$.
2. The mass of the rocket motor,$m_1=4mkg$.
3. The relative speed of the rocket motor just after separation$v_r=82km/hr$.
4. The mass of the command module,$m_2=mkg$.
5. Speed of module relative to earth after separation v.

The command module and the rocket motor initially move together and obey the law of conservation of linear momentum as they separate.

${\mathbf{m}}_{\mathbf{1}}{\mathbf{v}}_{\mathbf{i}\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}{\mathbf{v}}_{\mathbf{i}\mathbf{2}}\mathbf{=}{\mathbf{m}}_{\mathbf{1}}{\mathbf{v}}_{\mathbf{f}\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}{\mathbf{v}}_{\mathbf{f}\mathbf{2}}$

The mass of the space vehicle before separation will be = mass of rocket + mass of the module,

$$\begin{gathered} M=4m+m \\ =5m \end{gathered}$$

By law of conservation of linear momentum, we write

Total momentum ofthe space vehicle before separation = momentum of rocket + momentum of module (after separation)

i.e.

$$\begin{gathered} (m_1+m_2)v_i=m_{1}v+m_2(v-v_t) \\ M{v}_i=m_{1}v+m_{2}v-m_2{v}_r \\ v=v_i+\frac{m_2{v}_r}{(m_1+m_2)} \\ =v_i+\frac{4m{v}_r}{5m} \\ =v_i+\frac{4m{v}_r}{5m} \\ =\left(4300+\frac{4\times 82}{5}\right)\mathrm{km}/\mathrm{h} \\ =4365.6\mathrm{km}/\mathrm{h} \\ =4.4\times {10}^3\mathrm{km}/\mathrm{h} \end{gathered}$$