Topic: Engineering Mechanics
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 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?
- A. 0.01 N/mm
- B. 10 N/mm
- C. 20 N/mm
- D. 100 N/mm
Answer: B. 10 N/mm
Explanation: By conservation of momentum: 100*v_s + 900*v_r = 0, so v_r = -v_s/9. Relative velocity v_s - v_r = v_s + v_s/9 = 10*v_s/9 = 1 m/s. Thus, v_s = 0.9 m/s and v_r = -0.1 m/s. Total Kinetic Energy = 0.5 * 100 * (0.9)^2 + 0.5 * 900 * (-0.1)^2 = 40.5 + 4.5 = 45 J. Energy of 10 springs = 10 * (0.5 * k * x²) = 5 * k * (0.03 m)^2 = 0.0045 * k. Equating energies: 0.0045 * k = 45 gives k = 10,000 N/m = 10 N/mm.