A satellite of mass 100 kg is attached centrally to the rocket stage of mass 900 kg using a separation system. In space the satellite is to be separated and ejected using 10 numbers of identical spring thrusters having same stroke of 30 mm. Assume springs are fully transferring the energy and angular body rate of the bodies are negligible after separation. What is the stiffness of the spring required to get the satellite axial relative velocity of 1 m/s.?

Correct option is B

​$$\text{Since the system is initially at rest, and the only force is internal (springs), total momentum is conserved.} \\\text{The relative velocity between satellite and rocket becomes 1 m/s after separation.} \\\\\text{Let:} \\\text{Satellite velocity } = v_s \\\text{Rocket stage velocity } = v_r \\v_s - v_r = 1\, \text{m/s} \\\\\text{From conservation of momentum:} \\m_s v_s + m_r v_r = 0 \Rightarrow v_r = -\frac{m_s}{m_r} v_s \\\\\text{Substitute in relative velocity:} \\v_s - \left( -\frac{m_s}{m_r} v_s \right) = 1 \Rightarrow v_s \left( 1 + \frac{m_s}{m_r} \right) = 1 \Rightarrow v_s = \frac{1}{1 + \frac{100}{900}} = \frac{1}{1.1111} \approx 0.9\, \text{m/s} \\\\\text{Now compute total kinetic energy after separation (this is energy from springs):} \\E_k = \frac{1}{2} m_s v_s^2 + \frac{1}{2} m_r v_r^2 \\\\\text{Use:} \\v_r = -\frac{100}{900} \cdot 0.9 = -0.1\, \text{m/s} \\\\\text{So:} \\E_k = \frac{1}{2} \cdot 100 \cdot (0.9)^2 + \frac{1}{2} \cdot 900 \cdot (0.1)^2 \\E_k = 50 \cdot 0.81 + 450 \cdot 0.01 = 40.5 + 4.5 = \boxed{45\, \text{J}}$$$$\text{Total spring energy:} \\E_s = \sum_{i=1}^{10} \frac{1}{2} kx^2 = 10 \cdot \frac{1}{2} kx^2 = 5kx^2 \\\\\text{Equating:} \\5kx^2 = 45 \Rightarrow k = \frac{45}{5x^2} = \frac{45}{5 \cdot (0.03)^2} = \frac{45}{5 \cdot 0.0009} = \frac{45}{0.0045} = {10,\!000\, \text{N/m}} = 10 \text{N/mm}$$​​​