Question

A two-stage rocket is traveling at 1200 mis with respect to the earth when the first stage runs out of fuel. Explosive bolts release the first stage and push it backward with a speed of 35 mis relative to the second stage. The first stage is three times as massive as the second stage. What is the speed of the second stage after the separation?

Answer 1

Step-by-step explanation:

If the second stage has mass M, the total mass of the first and second stages is 4M. Since the stages are separating at 35 m/s, the lighter second stage must have received most of that velocity change: specifically, 3/4 of it. So Stage Two has added (35 * 0.75) m/s to its velocity. The overall velocity of the second stage is (1200+ (35*0.75) m/s =1226.25 m/s.

Answer 2

The speed of the second stage after separation is 4905 m/s

Step-by-step 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