Answer:
The speed of the second stage after separation is 4905 m/s
Explanation:
Hi there!
Due to conservation of momentum, the momentum of the system first stage - second stage must remain constant before and after the separation. The momentum of the system is calculated by adding the momentums of each object:
initial momentum = final momentum
m₁₊₂ · v = m1 · v1 + m2 · v2
Where:
m₁₊₂ = mass of the two stage rocket
v = velocity of the rocket
m1 = mass of stage 1
v1 = velocity of stage 1
m2 = mass of stage 2
v2 = velocity of stage 2
We have the following data:
m1 = 3 · m2
m₁₊₂ = m1 + m2 = 3 · m2 + m2 = 4 · m2
v = 1200 m/s
v1 = -35 m/s (let´s consider the backward direction as negative)
v2 = ?
Then, replacing these data in the equation of momentum of the system:
m₁₊₂ · v = m1 · v1 + m2 · v2
4 m2 · 1200 m/s = 3 m2 · (-35 m/s) + m2 · v2
Let´s solve the equation for v2:
divide both sides of the equation by m2:
4 · 1200 m/s = 3 · (-35 m/s) + v2
4800 m/s = -105 m/s + v2
v2 = 4800 m/s + 105 m/s = 4905 m/s
The speed of the second stage after separation is 4905 m/s
