91Ó°ÊÓ

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.

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