Question

A two-stage rocket moves in space at a constant velocity of +4300 m/s. The two stages are then separated by a small explosive charge placed between them. Immediately after the explosion the velocity of the 1160-kg upper stage is +5940 m/s. What is the velocity (magnitude and direction) of the 2100-kg lower stage immediately after the explosion?

Answer 1

Answer:

 $v_{1f}$ = +3,394 103 m / s

Explanation:

We will solve this problem with the concept of the moment. Let's start by defining the system that is formed by the complete rocket before and after the explosions, bone with the two stages, for this system the moment is conserved.

The data they give is the mass of the first stage m1 = 2100 kg, the mass of the second stage m2 = 1160 kg and its final velocity v2f = +5940 m / s and the speed of the rocket before the explosion vo = +4300 m / s

The moment before the explosion

      p₀ = (m₁ + m₂) v₀

After the explosion

      pf = m₁ $v_{1f}$ + m₂ $v_{2f}$

     p₀ = [texpv_{f}[/tex]

     (m₁ + m₂) v₀ = m₁ $v_{1f}$ + m₂ $v_{2f}$

Let's calculate the final speed (v1f) of the first stage

     $v_{1f}$ = ((m₁ + m₂) v₀ - m₂ $v_{2f}$) / m₁

     

     $v_{1f}$ = ((2100 +1160) 4300 - 1160 5940) / 2100

     $v_{1f}$ = (14,018 10 6 - 6,890 106) / 2100

     $v_{1f}$ = 7,128 106/2100

     $v_{1f}$ = +3,394 103 m / s

come the same direction of the final stage, but more slowly