Question

A jet flies at a speed of 275 m/s. The pilot fires a missile forward off a mounting at a

speed of 700 m/s relative to the ground. The respective masses of the jet and the

missile are 5 000 kg and 50 kg. Calculate the speed of the jet immediately after the

missile had been fired.

Answer 1

by impulse conservation law:

$v(m_1+m_2)=m_1v_1+m_2v_2$

where v is jet speed before the missile had been fired,

v₁ is missile speed,

v₂ is jet speed after the missile had been fired,

m₁ and m₂ are masses of missile and jet.

$v_2=\frac{v(m_1+m_2)-m_1v_1}{m_2}=\frac{275(5000+50)-700\cdot 50}{5000}=270.75$ m/s