Question

Two missiles $A$ and $B$ powered by solid rocket motors have identical specific impulse, liftoff mass of $5600 \,\text{kg}$ each, and burn durations of $t_A = 30 \,\text{s}$ and $t_B = 70 \,\text{s}$, respectively. The propellant mass flow rates $\dot{m}_A$ and $\dot{m}_B$ are given as: \[ \dot{m}_A = 120 \,\text{kg/s}, \, 0 \leq t \leq 30, \dot{m}_B = 70 \,\text{kg/s}, \, 0 \leq t \leq 70 \] Neglecting gravity and aerodynamic forces, the relation between final velocities $V_A$ and $V_B$ is:

• A. $V_A = 4.1 V_B$
• B. $V_A = V_B$
• C. $V_A = 0.5 V_B$
• D. $V_A = 0.7 V_B$

Hint:

Always compute the propellant mass and use the Tsiolkovsky equation. The ratio of final to initial mass determines the velocity gain.

Answer 1

The Correct Option is C

Step 1: Rocket equation.
\[ \Delta V = I_{sp} g_0 \ln\left(\frac{m_0}{m_f}\right) \]

Step 2: Masses.
Initial mass: $m_0 = 5600 \,\text{kg}$ for both. \[ m_{p,A} = \dot{m}_A t_A = 120 \times 30 = 3600 \Rightarrow m_{f,A} = 5600 - 3600 = 2000 \] \[ m_{p,B} = \dot{m}_B t_B = 70 \times 70 = 4900 \Rightarrow m_{f,B} = 5600 - 4900 = 700 \]

Step 3: Velocity ratios.
\[ V_A \propto \ln\left(\frac{5600}{2000}\right) = \ln(2.8) \approx 1.03 \] \[ V_B \propto \ln\left(\frac{5600}{700}\right) = \ln(8) \approx 2.08 \] \[ \frac{V_A}{V_B} = \frac{1.03}{2.08} \approx 0.5 \] \[ \boxed{V_A = 0.5 V_B} \]